This course develops the theory of characteristic classes from the curvature of connections on fibre bundles, showing how geometric data on a bundle produce topological invariants of the underlying manifold. The central goal is to explain why curvature is not just a local analytic quantity, but also a generator of cohomology classes that detect global bundle structure. Along the way, the course treats complex and real vector bundles, oriented bundles, and the special role of characteristic numbers as computable global measurements. The chapters build from the basic passage from curvature to cohomology, then construct the Chern-Weil homomorphism as the main machine translating invariant polynomials into closed differential forms. From there, the course studies the standard characteristic classes: Chern classes for complex bundles, the Euler class via Pfaffian curvature, and Pontryagin classes for real bundles. Later chapters move from primary classes to secondary and transgressive phenomena, including Chern-Simons forms and invariants, and then connect these ideas to local index-theoretic forms and explicit geometric computations in representative bundles. The final chapters emphasize applications and interpretation. They examine characteristic classes as obstructions to geometric structures and as numerical invariants extracted from manifolds, with special attention to four-dimensional geometry and instanton examples. The course closes by synthesizing the subject into a coherent picture: characteristic classes are the bridge between local differential geometry and global topology, encoding both hidden obstructions and measurable geometric content. # Introduction This course studies a central bridge in differential geometry: curvature is local and analytic, while characteristic classes are global and topological. The guiding question is how a differential form built from the curvature of a connection can define a cohomology class that does not depend on the connection. The answer is the Chern-Weil construction, which turns invariant polynomials on Lie algebras into characteristic classes of principal and vector bundles. The preceding fibre bundles material supplied principal bundles, associated vector bundles, and connections. This chapter fixes the viewpoint for the course: connections are not only devices for parallel transport, but also a calculus for producing cohomological invariants. Chapters 1 and 2 make the Chern-Weil mechanism precise, Chapters 3 through 6 specialise it to Chern, Euler, Pontryagin, and characteristic-number calculations, and Chapters 7 and 8 study transgression and Chern-Simons forms as secondary invariants measuring choices of connection. ## The Organising Problem A vector bundle or principal bundle can be described locally by transition functions, but characteristic classes should detect global twisting in a way that survives changes of trivialisation. Curvature forms are defined after choosing a connection, so the first issue is to understand why they can encode topological information rather than only auxiliary geometric data. [explanation: Local Curvature And Global Invariants] Let $P \to M$ be a principal $G$-bundle over a smooth manifold $M$, and let $A$ be a connection on $P$ with curvature $F_A$. The curvature is a horizontal, equivariant $\frak g$-valued $2$-form on $P$, or equivalently an adjoint-bundle-valued $2$-form on $M$. If $q$ is a polynomial on $\frak g$ invariant under the adjoint action of $G$, then $q(F_A)$ descends to an ordinary differential form on $M$. The two structural facts behind the course are that $q(F_A)$ is closed and that its de Rham cohomology class is independent of $A$. Closedness is a differential consequence of the Bianchi identity. Independence from the connection is obtained by comparing two connections along a path and writing the derivative of the resulting forms as an exact form. [/explanation] The construction needs invariant polynomials because curvature changes by conjugation under changes of local frame. A polynomial expression that is not invariant would depend on the chosen trivialisation and therefore would not define a differential form on the base. For instance, on a rank two vector bundle the function sending a matrix to its $(1,1)$-entry changes after conjugating the curvature matrix by a change-of-frame matrix, while $\operatorname{tr}(X)$ and $\operatorname{tr}(X^2)$ are unchanged by conjugation. [definition: Invariant Polynomial] Let $G$ be a Lie group with [Lie algebra](/page/Lie%20Algebra) $\frak g$. An invariant polynomial of degree $k$ on $\frak g$ is a [homogeneous polynomial](/page/Homogeneous%20Polynomial) map $q:\frak g \to \bb R$ or $q:\frak g \to \bb C$ such that \begin{align*} q(\operatorname{Ad}_g X) = q(X) \end{align*} for all $g \in G$ and $X \in \frak g$. [/definition] After polarisation, such a polynomial is often treated as a symmetric $k$-linear form on $\frak g$ that is invariant under the adjoint action in every argument. This is the form in which it is evaluated on curvature forms. [example: Trace Polynomials] Let $E \to M$ be a complex vector bundle with a connection whose curvature is represented in one local frame by a matrix-valued $2$-form $F$. If the local frame is changed by an invertible matrix-valued function $g$, then the curvature matrix in the new frame is $F' = g^{-1}Fg$. For an ordinary matrix $X$ and $k \ge 1$, the trace polynomial $q_k(X)=\operatorname{tr}(X^k)$ is invariant under conjugation because \begin{align*} (g^{-1}Xg)^2 = g^{-1}Xgg^{-1}Xg = g^{-1}X^2g. \end{align*} Repeating the same cancellation $gg^{-1}=I$ gives \begin{align*} (g^{-1}Xg)^k = g^{-1}X^kg. \end{align*} Therefore \begin{align*} \operatorname{tr}((g^{-1}Xg)^k)=\operatorname{tr}(g^{-1}X^kg). \end{align*} Using cyclicity of trace, $\operatorname{tr}(AB)=\operatorname{tr}(BA)$, with $A=g^{-1}$ and $B=X^kg$, we get \begin{align*} \operatorname{tr}(g^{-1}X^kg)=\operatorname{tr}(X^kgg^{-1})=\operatorname{tr}(X^k). \end{align*} The same algebra applies to the matrix-valued curvature form, with matrix multiplication combined with wedge product. Since $F$ is a $2$-form, it commutes in sign with the degree-$0$ matrix entries of $g$ and $g^{-1}$, so \begin{align*} (F')^k=(g^{-1}Fg)^k=g^{-1}F^kg. \end{align*} Taking traces gives \begin{align*} \operatorname{tr}((F')^k)=\operatorname{tr}(g^{-1}F^kg)=\operatorname{tr}(F^kgg^{-1})=\operatorname{tr}(F^k). \end{align*} Thus the local forms $\operatorname{tr}(F^k)$ agree on overlaps and patch to a globally defined $2k$-form on $M$. These trace forms are the basic curvature expressions from which Chern character forms are built. [/example] This example explains why matrix operations appear throughout the subject. The topological invariant is not the curvature matrix itself, since that matrix changes with the frame, but a conjugation-invariant expression in that matrix. ## Connections, Curvature, And Characteristic Forms The immediate problem is to turn a connection-dependent curvature form into a closed ordinary form on the base. The course uses the same mechanism repeatedly: choose a connection, apply an invariant polynomial to its curvature, and then prove that the resulting de Rham class is independent of the choice. [definition: Characteristic Form] Let $P \to M$ be a principal $G$-bundle with connection $A$ and curvature $F_A$. Let $q$ be an invariant polynomial of degree $k$ on $\frak g$. The characteristic form associated to $(q,A)$ is the differential form $q(F_A) \in \Omega^{2k}(M)$ obtained by evaluating the polarised form associated to $q$ on $F_A,\dots,F_A$. [/definition] The notation suppresses the descent from $P$ to $M$. In local trivialisations it is computed by applying $q$ to the local curvature matrix or Lie-algebra-valued curvature form. For this construction to enter de Rham cohomology, the form must be closed; the next theorem proves exactly this property from the Bianchi identity and the adjoint-invariance built into $q$. [quotetheorem:9757] [citeproof:9757] The hypotheses in the closedness theorem do real work. The connection hypothesis supplies the covariant [exterior derivative](/theorems/1525) and the Bianchi identity $d_AF_A=0$; a general Lie-algebra-valued $2$-form has no reason to satisfy such an identity. The invariant-polynomial hypothesis is equally necessary: if a non-invariant expression such as a single matrix entry of $F_A$ is used in a local frame, then a change of frame can alter the expression, so the resulting local forms do not patch to a well-defined form on $M$. The theorem also does not say that $q(F_A)$ is exact, nor that its periods are integral; it only puts the characteristic form into de Rham cohomology. The second half of the construction addresses the more serious topological question, because a closed form depending on the arbitrary choice of $A$ would still not be an invariant of the bundle itself. [quotetheorem:9758] [citeproof:9758] This theorem is the reason characteristic classes belong to topology rather than to a chosen metric or connection. The invariant-polynomial hypothesis is again essential: without it, the differentiated expression along $A_t$ contains commutator terms that do not combine into the exterior derivative of a globally defined transgression form. The connection hypothesis is also not cosmetic, since the comparison uses that the space of connections is affine over $\Omega^1(M;\operatorname{ad}P)$, making $A_1-A_0$ a well-defined adjoint-bundle-valued $1$-form and making the path $A_t$ legitimate. A concrete failure mode is obtained by trying to use a non-conjugation-invariant matrix entry of the curvature; even if two local computations differ by an exact form in one trivialisation, a different frame changes the expression itself, so no bundle invariant has been produced. The theorem is still a de Rham statement: without the standard normalising constants and integral characteristic class theory, it does not give an integral lift, a canonical representative, or a secondary invariant independent of auxiliary choices. The same computation also introduces transgression forms, which later become Chern-Simons forms when one records the connection-dependent primitive itself. ## The Main Families Of Classes The next problem is to identify which invariant polynomials produce the characteristic classes most commonly used in geometry and topology. Different structure groups carry different natural invariant polynomials, and these lead to different families of characteristic classes. [definition: Chern-Weil Class] Let $P \to M$ be a principal $G$-bundle with Lie algebra $\frak g$, and let $q:\frak g \to \bb R$ or $q:\frak g \to \bb C$ be an invariant polynomial of degree $k$. The Chern-Weil class associated to $q$ is \begin{align*} \operatorname{cw}_q(P) := [q(F_A)] \in H^{2k}_{\mathrm{dR}}(M), \end{align*} where $A$ is any connection on $P$. [/definition] The previous theorem is what makes this definition well-posed. Specific normalisations of $q$ are chosen so that the resulting de Rham classes agree with integral characteristic classes after applying the natural map from integral cohomology to real or complex de Rham cohomology. [example: First Chern Class Of A Line Bundle] Let $L \to M$ be a Hermitian complex line bundle with a unitary connection. In a local unitary frame, the connection is an $i\bb R$-valued $1$-form $A$, and because the structure group is $U(1)$, the curvature is \begin{align*} F_A=dA+A\wedge A. \end{align*} Here $A\wedge A=0$, since $A$ is a scalar-valued $1$-form and every $1$-form wedges with itself to zero. Thus locally \begin{align*} F_A=dA. \end{align*} Taking one more exterior derivative gives \begin{align*} dF_A=d(dA)=0. \end{align*} Multiplying by the constant $i/(2\pi)$ preserves closedness, so \begin{align*} d\left(\frac{i}{2\pi}F_A\right)=\frac{i}{2\pi}dF_A=0. \end{align*} The factor $i/(2\pi)$ is the standard normalisation for a unitary line bundle: since $F_A$ is $i\bb R$-valued, the form $(i/2\pi)F_A$ is real-valued and represents $c_1(L)$ in de Rham cohomology. If $\Sigma$ is a closed oriented surface and $L\to \Sigma$ has degree $d$, then the de Rham pairing with the fundamental class gives \begin{align*} \int_\Sigma \frac{i}{2\pi}F_A=\langle c_1(L),[\Sigma]\rangle=d. \end{align*} Thus the curvature of a unitary connection records the topological degree after the single normalising factor $i/(2\pi)$. [/example] Line bundles give the most direct calculation because the structure group is abelian. For higher rank bundles the determinant and trace combine the eigenvalue-like information of the curvature into total Chern forms. [example: Vanishing First Chern Form For Special Unitary Bundles] Let $E \to M$ be a rank $n$ complex vector bundle whose structure group is reduced to $SU(n)$, and let $A$ be an $SU(n)$-connection. In any local $SU(n)$-frame, the connection form is $\frak{su}(n)$-valued, so its curvature form $F_A$ is also $\frak{su}(n)$-valued. By definition, \begin{align*} \frak{su}(n)=\{X\in \frak{gl}_n(\bb C): X^*+X=0 \text{ and } \operatorname{tr}(X)=0\}. \end{align*} Thus, for every pair of tangent vectors $v,w\in T_pM$, the matrix $F_A(v,w)$ lies in $\frak{su}(n)$, and hence \begin{align*} \operatorname{tr}(F_A(v,w))=0. \end{align*} Since the $2$-form $\operatorname{tr}(F_A)$ is defined by evaluating $F_A$ on tangent vectors and then taking the matrix trace, this gives \begin{align*} (\operatorname{tr}(F_A))_p(v,w)=\operatorname{tr}(F_A(v,w))=0 \end{align*} for all $p\in M$ and all $v,w\in T_pM$. Therefore \begin{align*} \operatorname{tr}(F_A)=0. \end{align*} With the usual Chern-Weil normalisation, the first Chern form of a unitary connection is proportional to $\operatorname{tr}(F_A)$, namely \begin{align*} c_1(A)=\frac{i}{2\pi}\operatorname{tr}(F_A). \end{align*} Substituting $\operatorname{tr}(F_A)=0$ gives \begin{align*} c_1(A)=\frac{i}{2\pi}\cdot 0=0. \end{align*} Thus an $SU(n)$-connection produces the zero de Rham representative for $c_1(E)$, reflecting that the determinant direction has no curvature trace. [/example] Real oriented bundles lead to Euler classes through the Pfaffian of the curvature of an oriented orthogonal connection. For instance, the Euler form of an oriented rank two real bundle is a normalised curvature $2$-form whose integral over a closed oriented surface gives the Euler number. Real bundles with orthogonal connections lead to Pontryagin classes after complexification or through invariant polynomials on $\frak{so}(n)$; the first Pontryagin form is represented, up to the course's normalisation convention, by a trace expression built from $F_A \wedge F_A$. The course treats these constructions as parallel instances of the same mechanism rather than as unrelated definitions. ## Transgression And Secondary Invariants The independence of connection raises a new question: if two characteristic forms differ by an exact form, can the primitive carry geometric information? Transgression answers this question by keeping track of the exact term produced when connections vary. [definition: Transgression Form] Let $A_0$ and $A_1$ be two connections on a principal $G$-bundle $P \to M$, and let $q$ be an invariant polynomial of degree $k$. A transgression form for $(q,A_0,A_1)$ is a differential form $T_q(A_0,A_1) \in \Omega^{2k-1}(M)$ satisfying \begin{align*} dT_q(A_0,A_1)=q(F_{A_1})-q(F_{A_0}). \end{align*} [/definition] The transgression form depends on choices, but this dependence is controlled. When one of the connections is a reference connection, the resulting secondary form records geometric data invisible to the primary cohomology class. [example: Chern-Simons Form In Degree Three] Let $A$ be a matrix-valued connection $1$-form, and write products using wedge product of forms together with matrix multiplication. Its curvature is \begin{align*} F_A=dA+A\wedge A. \end{align*} The degree-three Chern-Simons form for the trace polynomial is \begin{align*} \operatorname{CS}(A)=\operatorname{tr}\left(A\wedge dA+\frac{2}{3}A\wedge A\wedge A\right). \end{align*} We compute its exterior derivative term by term. Since $d^2A=0$ and $\deg A=1$, \begin{align*} d(A\wedge dA)=dA\wedge dA-A\wedge d^2A=dA\wedge dA. \end{align*} For the cubic term, the graded Leibniz rule gives \begin{align*} d(A\wedge A\wedge A)=dA\wedge A\wedge A-A\wedge dA\wedge A+A\wedge A\wedge dA. \end{align*} Using graded cyclicity of the trace, $\operatorname{tr}(\alpha\wedge\beta)=(-1)^{\deg\alpha\deg\beta}\operatorname{tr}(\beta\wedge\alpha)$ for homogeneous matrix-valued forms, we have \begin{align*} \operatorname{tr}(A\wedge dA\wedge A)=-\operatorname{tr}(dA\wedge A\wedge A). \end{align*} Also, \begin{align*} \operatorname{tr}(A\wedge A\wedge dA)=\operatorname{tr}(dA\wedge A\wedge A). \end{align*} Therefore \begin{align*} d\,\operatorname{CS}(A)=\operatorname{tr}(dA\wedge dA)+2\operatorname{tr}(dA\wedge A\wedge A). \end{align*} Now expand the primary trace form: \begin{align*} \operatorname{tr}(F_A\wedge F_A)=\operatorname{tr}\left((dA+A\wedge A)\wedge(dA+A\wedge A)\right). \end{align*} Multiplying out gives \begin{align*} \operatorname{tr}(F_A\wedge F_A)=\operatorname{tr}(dA\wedge dA)+\operatorname{tr}(dA\wedge A\wedge A)+\operatorname{tr}(A\wedge A\wedge dA)+\operatorname{tr}(A\wedge A\wedge A\wedge A). \end{align*} The third term equals $\operatorname{tr}(dA\wedge A\wedge A)$ by graded cyclicity, while the last term vanishes because cycling the first degree-$1$ factor to the end gives \begin{align*} \operatorname{tr}(A\wedge A\wedge A\wedge A)=-\operatorname{tr}(A\wedge A\wedge A\wedge A). \end{align*} Hence \begin{align*} \operatorname{tr}(F_A\wedge F_A)=\operatorname{tr}(dA\wedge dA)+2\operatorname{tr}(dA\wedge A\wedge A)=d\,\operatorname{CS}(A). \end{align*} Thus, before inserting whatever normalising constants are chosen for the characteristic class, the Chern-Simons $3$-form is a local primitive for the degree-$4$ trace form $\operatorname{tr}(F_A\wedge F_A)$. [/example] These secondary forms are central in geometry and mathematical physics. They appear in gauge theory, three-dimensional topology, index theory, and boundary correction terms. A precise model is the Atiyah-Patodi-Singer index theorem, where changing the connection changes local characteristic forms by an exact transgression term, and the boundary contribution records the corresponding secondary correction rather than only the primary cohomology class. ## Geometric Calculations And Course Roadmap The final problem of the course is computational: after the theory constructs characteristic classes, how are they evaluated on concrete bundles and manifolds? The answer combines local curvature formulae, naturality, splitting principles, and integration over cycles. [explanation: Course Roadmap] The first part develops curvature forms for associated vector bundles, invariant polynomials, and the Chern-Weil homomorphism. The second part specialises the construction to Chern classes of complex bundles, Euler classes of oriented real bundles, and Pontryagin classes of real bundles. The third part studies transgression and Chern-Simons forms, then applies the theory to homogeneous spaces, tangent bundles, and Riemannian curvature. A recurring calculation has the following pattern. First choose a connection adapted to the geometry, such as a unitary connection, an orthogonal connection, or a Levi-Civita connection. Next compute its curvature in a convenient frame. Then apply the relevant invariant polynomial, simplify using symmetry, and integrate the resulting closed form over cycles when numerical characteristic numbers are desired. [/explanation] The course assumes fluency with smooth manifolds, de Rham cohomology, Lie groups and Lie algebras, principal bundles and connections, vector bundles, and basic Riemannian geometry. These prerequisites are used actively: de Rham cohomology supplies the target of the invariants, Lie theory supplies the invariant polynomials, and bundle theory supplies the curvature forms whose algebra drives the construction. [remark: Normalisation Conventions] Characteristic forms often include factors of $2\pi$, $i$, signs, and Pfaffian normalisations. The course will state each convention at the point where the corresponding class is introduced. The guiding principle is that the de Rham representative should match the standard integral characteristic class under the comparison map to real or complex cohomology. [/remark] This introduction leaves the main definitions in a deliberately reusable form. The rest of the course repeatedly returns to the same sequence: curvature, invariant polynomial, closed form, connection-independent class, and geometric calculation. The first chapter now turns this sequence into a theorem-level mechanism. It proves that invariant polynomials applied to curvature produce closed forms, and that the resulting cohomology classes depend only on the underlying bundle rather than on the auxiliary connection. # 1. From Curvature To Cohomology This opening chapter explains how curvature, which is initially a differential-geometric object attached to a connection, produces de Rham cohomology classes. The central mechanism is that an invariant polynomial on a Lie algebra can be evaluated on the curvature form, producing a differential form on the base manifold. The two structural facts behind the construction are the Bianchi identity and invariance of the polynomial: together they force the resulting form to be closed, and a homotopy argument shows that its cohomology class does not depend on the chosen connection. As previewed in the introduction, the course has already assumed familiarity with principal bundles, associated vector bundles, and connections. Here we fix the notation that will be used throughout the characteristic class constructions, then prove the first Chern-Weil results in the concrete language of matrix-valued forms. ## Curvature Forms For Associated Vector Bundles The first problem is to turn the curvature of a connection into an object that can be inserted into algebraic expressions such as traces, determinants, and symmetric polynomials. For a vector bundle, this means treating curvature as an endomorphism-valued two-form and keeping careful track of how matrix multiplication interacts with the exterior product. [definition: Endomorphism-Valued Differential Form] Let $E \to M$ be a smooth vector bundle. An endomorphism-valued $k$-form on $M$ is an element of $\Omega^k(M;\operatorname{End}(E))$, the space of smooth sections of $\Lambda^kT^*M \otimes \operatorname{End}(E)$. [/definition] Such a form assigns to tangent vectors $v_1,\dots,v_k \in T_pM$ an endomorphism of the fibre $E_p$. To multiply curvature terms later, we need a product that remembers both the alternating form degree and the order of endomorphism composition. [definition: Product Of Endomorphism-Valued Forms] Let $E\to M$ be a smooth vector bundle. For each $p,q\geq 0$, the product of endomorphism-valued forms is the map \begin{align*} \wedge:\Omega^p(M;\operatorname{End}(E))\times\Omega^q(M;\operatorname{End}(E)) \to \Omega^{p+q}(M;\operatorname{End}(E)) \end{align*} defined by sending $(A,B)$ to $A\wedge B$, where $A\wedge B$ is obtained by alternating the [tensor product](/page/Tensor%20Product) of the form parts and composing the endomorphism parts in the order $A$ then $B$. [/definition] This product is associative but not graded-commutative in general, because endomorphisms need not commute. In a local frame of $E$, endomorphism-valued forms are represented by matrices whose entries are ordinary differential forms, and the product is matrix multiplication with wedge product in the entries. [example: Matrix Product Of Form-Valued Endomorphisms] Let $E=M\times \mathbb C^2$, and write $A$ and $B$ in the standard frame by their nonzero entries \begin{align*} A_{11}=\alpha,\quad A_{22}=\beta,\quad B_{12}=\gamma,\quad B_{21}=\delta. \end{align*} Here $\alpha,\beta\in\Omega^1(M)$ and $\gamma,\delta\in\Omega^2(M)$. By the matrix product rule for endomorphism-valued forms, the $(1,2)$-entry of $A\wedge B$ is the sum over the intermediate index: \begin{align*} (A\wedge B)_{12}=A_{11}\wedge B_{12}+A_{12}\wedge B_{22}. \end{align*} Substituting the entries gives \begin{align*} (A\wedge B)_{12}=\alpha\wedge\gamma+0\wedge 0=\alpha\wedge\gamma. \end{align*} In the reversed order, the same entry is \begin{align*} (B\wedge A)_{12}=B_{11}\wedge A_{12}+B_{12}\wedge A_{22}. \end{align*} Substituting again gives \begin{align*} (B\wedge A)_{12}=0\wedge 0+\gamma\wedge\beta=\gamma\wedge\beta. \end{align*} Since $\deg(\gamma)=2$ and $\deg(\beta)=1$, the ordinary form sign in $\gamma\wedge\beta=(-1)^{2\cdot 1}\beta\wedge\gamma=\beta\wedge\gamma$ is positive, but the matrix multiplication has still selected $\beta$ rather than $\alpha$. Thus the order of the endomorphism factors matters even when the exterior-form degrees introduce no sign change in this entry. [/example] We now attach this algebra to a connection. The preceding example explains the local multiplication rule; curvature is the geometric object whose powers will be multiplied in exactly this sense. [definition: Curvature Of A Vector Bundle Connection] Let $E \to M$ be a smooth vector bundle with connection $\nabla$. The curvature of $\nabla$ is the endomorphism-valued two-form $F_\nabla \in \Omega^2(M;\operatorname{End}(E))$ defined by \begin{align*} F_\nabla(X,Y)s = \nabla_X\nabla_Ys-\nabla_Y\nabla_Xs-\nabla_{[X,Y]}s \end{align*} for vector fields $X,Y\in\mathfrak X(M)$ and smooth sections $s\in\Gamma(E)$. [/definition] In a local frame, the connection can be written as $\nabla=d+A$, where $A$ is a matrix of one-forms. The curvature then has the familiar gauge-theoretic expression \begin{align*} F_\nabla = dA + A\wedge A. \end{align*} This formula is local, but the object $F_\nabla$ is globally defined as an endomorphism-valued two-form. [example: Curvature Of A Line Bundle Connection] Let $L\to M$ be a complex line bundle and choose a local unitary frame, so a unitary connection is written as $\nabla=d+A$ with $A\in\Omega^1(M;i\mathbb R)$. In this frame the curvature is \begin{align*} F_\nabla=dA+A\wedge A. \end{align*} Because the fibre endomorphism algebra of a line is $\mathbb C$, the product term is scalar multiplication together with the exterior product. For vector fields $X,Y$, \begin{align*} (A\wedge A)(X,Y)=A(X)A(Y)-A(Y)A(X)=0. \end{align*} Thus the local curvature form is \begin{align*} F_\nabla=dA. \end{align*} If the unitary frame is changed by $e'=u e$ with $u:M\to U(1)$, then writing a section as $s=f e=f'e'$ gives $f=u f'$. Therefore \begin{align*} df+Af=u\,df'+(du)f'+Au f'=u\left(df'+(u^{-1}du+A)f'\right). \end{align*} So the new connection one-form is \begin{align*} A'=A+u^{-1}du. \end{align*} Taking exterior derivatives gives \begin{align*} dA'=dA+d(u^{-1}du). \end{align*} Since $d(u^{-1})=-u^{-2}du$ and $d^2u=0$, \begin{align*} d(u^{-1}du)=d(u^{-1})\wedge du+u^{-1}d^2u=-u^{-2}du\wedge du=0. \end{align*} Hence $dA'=dA$, so the local two-forms agree on overlaps and assemble into a global imaginary-valued curvature two-form. [/example] The line bundle computation shows how local connection forms give global curvature. To connect this with principal bundles, we need the corresponding statement for a representation of the structure group, since characteristic classes are often computed after passing to an associated vector bundle. [quotetheorem:6271] [citeproof:6271] This result is the reason characteristic forms can be introduced either on principal bundles using $\mathfrak g$-valued curvature, or on associated vector bundles using matrix-valued curvature. The representation hypothesis is essential: without a fixed representation, a $\mathfrak g$-valued curvature form has no canonical action on the fibres of an arbitrary vector bundle. The theorem does not say that every endomorphism-valued curvature form arises from every principal connection; it says that associated bundles inherit curvature functorially through $d\rho$. This bridge lets the next section ask a precise algebraic question: which functions of matrix-valued curvature remain meaningful after changing frames? ## Invariant Polynomials And The Bianchi Identity The next question is why algebraic expressions in curvature should produce closed differential forms. A generic polynomial in the matrix entries is not geometrically meaningful: for instance, the function $A\mapsto A_{11}$ on $\mathfrak{gl}(2,\mathbb C)$ changes under conjugation, so evaluating it on a curvature matrix depends on the chosen local frame. The answer is not a special feature of trace alone: it comes from applying an invariant polynomial to a curvature form that satisfies a covariant differential identity. [definition: Invariant Polynomial] Let $G$ be a Lie group with Lie algebra $\mathfrak g$. A homogeneous polynomial $P:\mathfrak g\to\mathbb R$ or $P:\mathfrak g\to\mathbb C$ of degree $k$ is invariant if \begin{align*} P(\operatorname{Ad}_g X)=P(X) \end{align*} for all $g\in G$ and all $X\in\mathfrak g$. [/definition] The polynomial is usually used through its polarisation, a symmetric $k$-linear form still denoted $P$. To use invariance in differential-form calculations, we need its infinitesimal version, because connection terms appear through Lie brackets. [quotetheorem:9759] [citeproof:9759] This infinitesimal identity supplies the algebraic cancellation. Invariance cannot be dropped: for $P(A)=A_{11}$ on $\mathfrak{gl}(2,\mathbb C)$, take \begin{align*} Y=E_{12} \quad\text{and}\quad X=E_{21}, \end{align*} where $E_{ij}$ denotes the elementary matrix with a single $1$ in the $(i,j)$ entry. Then $[Y,X]=\operatorname{diag}(1,-1)$, so the infinitesimal identity would demand $P([Y,X])=0$, while $P([Y,X])=1$. Thus the theorem is not a closedness statement by itself, does not apply to arbitrary polynomials in matrix entries, and does not replace the differential input supplied by curvature. Its scope is algebraic: it cancels bracket insertions once they have appeared. The geometric side of the same cancellation requires differentiating endomorphism-valued forms using the connection, rather than using the ordinary exterior derivative alone. In the Chern-Weil closedness proof below, this identity removes the commutator terms introduced when the ordinary exterior derivative is traded for the covariant derivative. [definition: Covariant Exterior Derivative] Let $E\to M$ be a vector bundle with connection $\nabla$. The covariant exterior derivative \begin{align*} d_\nabla:\Omega^k(M;E)\to\Omega^{k+1}(M;E) \end{align*} extends $\nabla$ by the graded Leibniz rule with respect to ordinary differential forms. [/definition] For endomorphism-valued forms, the induced connection on $\operatorname{End}(E)$ gives a covariant exterior derivative also denoted $d_\nabla$. Ordinary closedness is the wrong expectation. In a local frame with connection matrix $A$, if \begin{align*} F=dA+A\wedge A, \end{align*} then the ordinary exterior derivative is typically \begin{align*} dF=dA\wedge A-A\wedge dA, \end{align*} which need not vanish when the matrices in $A$ do not commute. The covariant derivative corrects this defect by adding commutator terms: \begin{align*} d_\nabla B=dB+A\wedge B-(-1)^qB\wedge A, \end{align*} for $B\in\Omega^q(M;\operatorname{End}(E))$. The next result states that curvature is closed for this corrected derivative, which is exactly the geometric identity needed for Chern-Weil closedness. [quotetheorem:1541] [citeproof:1541] The Bianchi identity says that curvature is covariantly closed, not ordinarily closed. Replacing $d_\nabla$ by $d$ fails even for the local connection matrix \begin{align*} A_{12}=x\,dy,\qquad A_{21}=z\,dx,\qquad A_{11}=A_{22}=0 \end{align*} on a coordinate chart in $\mathbb R^3$. Here $F=dA+A\wedge A$ has $(1,1)$-entry $-z\,dx\wedge x\,dy=-xz\,dx\wedge dy$, so \begin{align*} (dF)_{11}=-x\,dx\wedge dy\wedge dz. \end{align*} This nonzero component records the failure of ordinary closedness. The theorem also does not say that every endomorphism-valued two-form is covariantly closed: for the flat connection on a product bundle, $d_\nabla=d$, and a form such as $x\,dy\wedge dz$ has nonzero exterior derivative. The point is more precise: the curvature of a connection is closed for the covariant derivative defined by that same connection. This is the differential half of the Chern-Weil argument. When an invariant polynomial is applied to curvature, Bianchi removes the covariant-derivative contribution, while infinitesimal invariance cancels the bracket terms that separate $d_\nabla$ from $d$; together they turn covariant closedness of curvature into ordinary closedness of the scalar characteristic form. [example: Trace Powers Are Invariant] Let $E\to M$ be a complex vector bundle of rank $r$, and let $P_k(A)=\operatorname{tr}(A^k)$ on $\mathfrak{gl}(r,\mathbb C)$. For $B\in GL(r,\mathbb C)$, we first compute the power of the conjugate. The case $k=1$ is $BAB^{-1}=BA^1B^{-1}$. If $(BAB^{-1})^k=BA^kB^{-1}$, then \begin{align*} (BAB^{-1})^{k+1}=(BA^kB^{-1})(BAB^{-1})=BA^k(B^{-1}B)AB^{-1}=BA^{k+1}B^{-1}. \end{align*} Thus $(BAB^{-1})^k=BA^kB^{-1}$ for every $k\geq 1$. It remains to see that trace is unchanged by conjugation. For any matrix $C$, \begin{align*} \operatorname{tr}(BCB^{-1})=\sum_i\sum_{j,\ell}B_{ij}C_{j\ell}(B^{-1})_{\ell i}. \end{align*} Reordering the finite sum gives \begin{align*} \operatorname{tr}(BCB^{-1})=\sum_{j,\ell}C_{j\ell}\sum_i(B^{-1})_{\ell i}B_{ij}. \end{align*} Since $\sum_i(B^{-1})_{\ell i}B_{ij}=(B^{-1}B)_{\ell j}=\delta_{\ell j}$, this becomes \begin{align*} \operatorname{tr}(BCB^{-1})=\sum_{j,\ell}C_{j\ell}\delta_{\ell j}=\sum_j C_{jj}=\operatorname{tr}(C). \end{align*} Applying this with $C=A^k$ gives \begin{align*} \operatorname{tr}((BAB^{-1})^k)=\operatorname{tr}(BA^kB^{-1})=\operatorname{tr}(A^k). \end{align*} Therefore $P_k$ is invariant under the adjoint action. Evaluating this invariant polynomial on the curvature of a connection gives \begin{align*} \operatorname{tr}(F_\nabla^k)\in\Omega^{2k}(M;\mathbb C), \end{align*} where $F_\nabla^k$ means $k$ repeated products using wedge product in the differential-form entries and matrix multiplication in the endomorphism entries. [/example] The determinant gives another family of invariant polynomials. It is especially important for unitary connections because it packages the Chern forms into a single generating expression. [example: Determinant Polynomial For A Unitary Connection] Let $E\to M$ be a Hermitian complex vector bundle of rank $r$ with unitary connection $\nabla$. For $B\in U(r)$ and $A\in\mathfrak u(r)$, first observe that the formal scalar $t$ commutes with the matrix entries, so \begin{align*} B\left(I+\frac{i}{2\pi}tA\right)B^{-1}=BB^{-1}+\frac{i}{2\pi}tBAB^{-1}=I+\frac{i}{2\pi}tBAB^{-1}. \end{align*} Using multiplicativity of determinant and $\det(B^{-1})=\det(B)^{-1}$ gives \begin{align*} \det\left(I+\frac{i}{2\pi}tBAB^{-1}\right)=\det(B)\det\left(I+\frac{i}{2\pi}tA\right)\det(B^{-1}). \end{align*} Therefore \begin{align*} \det\left(I+\frac{i}{2\pi}tBAB^{-1}\right)=\det\left(I+\frac{i}{2\pi}tA\right). \end{align*} Thus the polynomial $\det(I+\frac{i}{2\pi}tA)$ is invariant under conjugation by $U(r)$. Now choose a local unitary frame for $E$. The curvature $F_\nabla$ is represented by an $\mathfrak u(r)$-valued matrix of two-forms, so applying the determinant polynomial gives \begin{align*} \det\left(I+\frac{i}{2\pi}tF_\nabla\right)=\sum_{\sigma\in S_r}\operatorname{sgn}(\sigma)\prod_{a=1}^r\left(\delta_{a,\sigma(a)}+\frac{i}{2\pi}t(F_\nabla)_{a,\sigma(a)}\right). \end{align*} In this expansion, a term containing exactly $j$ curvature entries contributes a product of $j$ two-forms, hence has differential-form degree $2j$, and it is multiplied by $t^j$. Collecting the terms with the same power of $t$ gives \begin{align*} \det\left(I+\frac{i}{2\pi}tF_\nabla\right)=1+c_1(\nabla)t+\cdots+c_r(\nabla)t^r. \end{align*} The coefficient $c_j(\nabla)$ is therefore a differential form of degree $2j$: \begin{align*} c_j(\nabla)\in\Omega^{2j}(M;\mathbb C). \end{align*} Conjugation-invariance is what makes these locally computed coefficients independent of the chosen unitary frame, so $c_j(\nabla)$ is the $j$th Chern-Weil form of the unitary connection. [/example] ## Closed Forms From Curvature We can now answer the first main question of the chapter: when an invariant polynomial is evaluated on curvature, why does it define a de Rham cohomology class? The local computations above combine into a compact theorem. [definition: Chern-Weil Form] Let $P$ be an invariant homogeneous polynomial of degree $k$ on the Lie algebra $\mathfrak g$ of a Lie group $G$, and let the associated symmetric $k$-linear form be denoted \begin{align*} P:\mathfrak g^k\to K, \end{align*} where $K=\mathbb R$ or $K=\mathbb C$. Let $Q\to M$ be a principal $G$-bundle with connection curvature $\Omega\in\Omega^2(Q;\mathfrak g)$. Applying the polarised form to $\Omega$ gives the $K$-valued $2k$-form $P(\Omega,\dots,\Omega)$ on $Q$. The Chern-Weil form associated to $P$ and the connection is the unique descended form on $M$ whose pullback to $Q$ is $P(\Omega,\dots,\Omega)$. [/definition] The definition uses the standard fact that $P(\Omega,\dots,\Omega)$ is horizontal and $G$-invariant on the principal bundle, hence descends to the base manifold. We write the descended form as $P(\Omega)$ when no confusion can arise. The next theorem proves the decisive analytic property of this descended form: it is closed. [quotetheorem:7039] [citeproof:7039] Closedness is the first step from geometry to topology: it allows $P(\Omega)$ to define a de Rham cohomology class. Both hypotheses are doing work. If $F$ is replaced by an arbitrary endomorphism-valued two-form rather than the curvature of the same connection, the flat-connection example $F=x\,dy\wedge dz$ leaves an uncancelled $d_A F=dx\wedge dy\wedge dz$ term. If invariance is dropped, the polynomial $A\mapsto A_{11}$ on $\mathfrak{gl}(2,\mathbb C)$ changes under conjugation and its commutator terms do not cancel. The theorem does not identify an integral characteristic class or say that every closed form occurs in this way; it produces de Rham representatives from curvature. To show that this class is intrinsic to the bundle, not to the chosen connection, we need compare two connections. [quotetheorem:9760] [citeproof:9760] The differential form inside the exterior derivative is a transgression form. The invariant-polynomial hypothesis is essential: with the non-invariant entry polynomial $A\mapsto A_{11}$, a change of frame can alter the two endpoint forms by non-exact local conjugation terms, so no frame-independent transgression formula is available. The affine-path hypothesis is also part of the statement; choosing a non-smooth path of connection matrices would make $\frac{d}{dt}F_{\nabla^t}$ undefined and the integral formula meaningless. The theorem does not say that the two curvature forms are equal, nor that the transgression form is unique; it says their invariant-polynomial evaluations differ by an exact form. This exact-difference statement is the missing step between closed representatives and characteristic classes. Closedness alone gives a cohomology class only after a connection has been chosen, so it still leaves open whether a different connection would produce a different class. The homotopy formula answers the comparison problem at the level of forms, but the course now needs the invariant itself: a statement that the resulting de Rham class belongs to the bundle rather than to the auxiliary connection. The next theorem extracts precisely that topological conclusion. Later chapters will return to the same transgression form as a Chern-Simons form, where changing the path of connections changes the representative but not the cohomological conclusion needed here. [quotetheorem:1545] [citeproof:1545] This theorem is the conceptual turning point: curvature depends on a connection, but the cohomology class extracted from curvature by an invariant polynomial depends only on the bundle. The existence of two connections is not extra structure in the final invariant; it is the comparison device used to prove independence from that structure. The theorem does not say the representative form is canonical, since different connections usually give different differential forms. The rest of the course identifies these de Rham classes with familiar characteristic classes, and later compares them with integral classes coming from classifying spaces. [example: First Chern Form Of A Hermitian Line Bundle] Let $L\to M$ be a Hermitian line bundle with unitary connection $\nabla$. In a local unitary frame the connection form is $i\mathbb R$-valued, so its curvature is also $i\mathbb R$-valued. Thus there is a real two-form $\omega$ with $F_\nabla=i\omega$, and \begin{align*} c_1(\nabla)=\frac{i}{2\pi}F_\nabla=\frac{i}{2\pi}(i\omega)=-\frac{1}{2\pi}\omega. \end{align*} Hence $c_1(\nabla)$ is a real-valued two-form. Now let $\nabla^0$ and $\nabla^1$ be two unitary connections on $L$. Their difference \begin{align*} a=\nabla^1-\nabla^0 \end{align*} is a globally defined $i\mathbb R$-valued one-form, because the difference of two connections is tensorial. In a local unitary frame write \begin{align*} \nabla^0=d+A_0,\qquad \nabla^1=d+A_1. \end{align*} Then $a=A_1-A_0$ in this frame, and for a line bundle the curvature formula reduces to $F_{\nabla^j}=dA_j$ because the scalar term $A_j\wedge A_j$ vanishes. Therefore \begin{align*} F_{\nabla^1}-F_{\nabla^0}=dA_1-dA_0=d(A_1-A_0)=da. \end{align*} Multiplying by $\frac{i}{2\pi}$ gives \begin{align*} c_1(\nabla^1)-c_1(\nabla^0)=\frac{i}{2\pi}(F_{\nabla^1}-F_{\nabla^0})=\frac{i}{2\pi}da=d\left(\frac{i}{2\pi}a\right). \end{align*} Thus the first Chern forms of two unitary connections differ by an exact two-form, so the de Rham class $[c_1(\nabla)]\in H^2_{\mathrm{dR}}(M;\mathbb R)$ depends only on the Hermitian line bundle $L$, not on the chosen unitary connection. [/example] ## Naturality Of Chern-Weil Forms A final structural question is how Chern-Weil forms behave under smooth maps. Characteristic classes should be functorial: pulling a bundle back to another manifold should pull back its characteristic forms. [quotetheorem:9761] [citeproof:9761] Naturality will be used repeatedly when characteristic classes are compared across bundle maps, restrictions to submanifolds, classifying maps, and associated bundles. The pulled-back connection hypothesis is necessary: an unrelated connection on $f^*E$ need not have curvature equal to $f^*F_\nabla$, although connection-independence says it gives the same cohomology class after applying $P$. The theorem does not assert that pullback detects all characteristic classes; it describes compatibility of this construction with a given smooth map. This compatibility is what later allows universal classes on classifying spaces to control characteristic classes on arbitrary bundles. [example: Restricting Trace Powers To A Submanifold] Let $i:N\hookrightarrow M$ be an embedded submanifold and let $E\to M$ be a complex vector bundle with connection $\nabla$. By *[Naturality Of Chern-Weil Forms](/theorems/9761)*, the pulled-back connection $i^*\nabla$ on $i^*E\to N$ has curvature \begin{align*} F_{i^*\nabla}=i^*F_\nabla. \end{align*} The product on endomorphism-valued forms is matrix multiplication with wedge product in the entries, and [pullback preserves wedge products](/theorems/3914). Hence the powers are compatible with restriction: for $k=1$ this is the displayed identity, and if $F_{i^*\nabla}^k=i^*(F_\nabla^k)$, then \begin{align*} F_{i^*\nabla}^{k+1}=F_{i^*\nabla}^k\wedge F_{i^*\nabla}=i^*(F_\nabla^k)\wedge i^*F_\nabla=i^*(F_\nabla^k\wedge F_\nabla)=i^*(F_\nabla^{k+1}). \end{align*} Thus \begin{align*} F_{i^*\nabla}^k=i^*(F_\nabla^k). \end{align*} Taking trace also commutes with pullback because in a local frame the trace is the sum of diagonal entries: \begin{align*} \operatorname{tr}(i^*(F_\nabla^k))=\sum_a i^*(F_\nabla^k)_{aa}=\sum_a i^*((F_\nabla^k)_{aa})=i^*\left(\sum_a(F_\nabla^k)_{aa}\right)=i^*\operatorname{tr}(F_\nabla^k). \end{align*} Combining the two identities gives \begin{align*} \operatorname{tr}(F_{i^*\nabla}^k)=i^*\operatorname{tr}(F_\nabla^k). \end{align*} So the trace-power characteristic form on the restricted bundle is exactly the restriction of the trace-power characteristic form on $M$, with no additional correction term. [/example] The chapter has established the Chern-Weil mechanism in its first form: invariant polynomials applied to curvature give closed forms, their cohomology classes are independent of the connection, and the construction is natural under pullback. The next chapter packages this mechanism into the Chern-Weil homomorphism and studies the algebra of invariant polynomials systematically. # 2. The Chern-Weil Homomorphism This chapter turns the curvature constructions of Chapter 1 into an actual cohomological machine. The guiding question is: which algebraic expressions in curvature give characteristic classes of principal bundles, and why do they ignore the auxiliary choice of connection? We first organise the allowable expressions as an algebra of invariant polynomials, then use connections on principal bundles to evaluate them as closed differential forms, and finally check that the construction behaves correctly under the standard operations on bundles. ## Invariant Polynomials and Universal Characteristic Forms The starting problem is to identify which polynomial functions on a Lie algebra can be evaluated on curvature without depending on a choice of local trivialisation. Curvature transforms by the adjoint action under a change of gauge, so only polynomials invariant under that action can produce globally defined forms. [definition: Ad-Invariant Polynomial] Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. A homogeneous polynomial $P: \mathfrak{g} \to \mathbb{R}$ of degree $k$ is Ad-invariant if \begin{align*} P(\operatorname{Ad}_g X) = P(X) \end{align*} for all $g \in G$ and $X \in \mathfrak{g}$. [/definition] This definition isolates the gauge-invariant algebraic data, but curvature substitutions are not made by evaluating a single Lie-algebra element. A degree-$k$ polynomial must be polarised so that $k$ Lie-algebra-valued form factors can be inserted separately and then wedged. The obstruction is that this multilinear expression still has to survive a simultaneous change of gauge in every input. [definition: Symmetric Invariant Multilinear Form] Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. A symmetric $k$-[linear map](/page/Linear%20Map) \begin{align*} \widetilde P: \mathfrak{g}^k \to \mathbb{R} \end{align*} is invariant if \begin{align*} \widetilde P(\operatorname{Ad}_g X_1, \dots, \operatorname{Ad}_g X_k) = \widetilde P(X_1, \dots, X_k) \end{align*} for all $g \in G$ and $X_1, \dots, X_k \in \mathfrak{g}$. [/definition] The polynomial and multilinear viewpoints are interchangeable by polarisation, with $P(X)=\widetilde P(X,\dots,X)$. Since curvature is a Lie-algebra-valued differential form rather than a Lie algebra element, we next fix the convention for substituting curvature into an invariant polynomial. [definition: Evaluation of an Invariant Polynomial on Curvature] Let $M$ be a smooth manifold. Let $P$ be represented by a symmetric invariant $k$-linear form $\widetilde P$, and let $\Omega \in \Omega^2(M;\mathfrak{g})$ be a $\mathfrak{g}$-valued $2$-form. Define \begin{align*} P(\Omega) := \widetilde P(\Omega,\dots,\Omega) \in \Omega^{2k}(M), \end{align*} where the form part is multiplied using wedge products and the Lie algebra part is evaluated by $\widetilde P$. [/definition] This notation suppresses the alternating signs coming from differential forms, but no sign ambiguity occurs here because curvature has even degree. The construction is local at this stage; examples from matrix groups show what these forms look like before we package all invariant polynomials into one algebra. [example: Trace Polynomials for Matrix Groups] Let $G \subset GL(n,\mathbb{C})$ be a matrix Lie group with Lie algebra $\mathfrak{g}\subset \mathfrak{gl}(n,\mathbb{C})$. For $k\geq 1$, define $P_k(X)=\operatorname{tr}(X^k)$. If $h\in G$, then \begin{align*} (hXh^{-1})^k=hX(h^{-1}h)X(h^{-1}h)\cdots Xh^{-1}=hX^kh^{-1}. \end{align*} Using the cyclic trace identity $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ with $A=hX^k$ and $B=h^{-1}$ gives \begin{align*} P_k(hXh^{-1})=\operatorname{tr}(hX^kh^{-1})=\operatorname{tr}(h^{-1}hX^k)=\operatorname{tr}(X^k)=P_k(X). \end{align*} Thus $P_k$ is invariant under conjugation, hence under the adjoint action of any matrix subgroup. If $\Omega=(\Omega_{ij})$ is a matrix-valued curvature $2$-form, its $k$-fold product is formed by matrix multiplication with wedge products in the entries. For instance, \begin{align*} (\Omega\wedge\Omega)_{ij}=\sum_{\ell=1}^n \Omega_{i\ell}\wedge\Omega_{\ell j}. \end{align*} Therefore \begin{align*} P_k(\Omega)=\operatorname{tr}(\Omega\wedge\cdots\wedge\Omega) \end{align*} with $k$ curvature factors. When $\Omega$ is the curvature of a connection, this is exactly the kind of invariant curvature expression to which the Chern-Weil closedness argument applies, so the trace powers become the basic closed candidate characteristic forms used later in the construction of Chern characters. [/example] For complex matrix groups, $\operatorname{tr}(X^k)$ is naturally complex-valued. In the real Chern-Weil algebra above, it is understood either after passing to the underlying real and imaginary parts as real-valued invariant polynomials, or after complexifying the target cohomology to $H^*_{\mathrm{dR}}(M;\mathbb{C})$. The later Chern character formulas use the complex-valued convention, while real characteristic classes are obtained from the appropriate real combinations and normalising constants. Trace polynomials show that many useful characteristic forms are generated by a common algebraic principle. To define the Chern-Weil map as a homomorphism rather than as separate formulas, we now collect all invariant polynomials into a graded algebra. [definition: Algebra of Invariant Polynomials] For a Lie group $G$ with Lie algebra $\mathfrak{g}$, define \begin{align*} I^k(G) = \{P \in \operatorname{Sym}^k(\mathfrak{g}^*) : P \text{ is Ad-invariant}\}. \end{align*} Define \begin{align*} I(G) = \bigoplus_{k\ge 0} I^k(G). \end{align*} The multiplication is pointwise multiplication of polynomial functions. [/definition] This graded algebra is independent of any bundle. It is the universal source from which Chern-Weil classes are obtained after choosing a principal bundle and a connection. [remark: Degree Convention] A polynomial in $I^k(G)$ has polynomial degree $k$, but its Chern-Weil form has de Rham degree $2k$. Thus the Chern-Weil map doubles degrees: invariant algebra degree $k$ lands in $H^{2k}_{\mathrm{dR}}(M)$. [/remark] The word universal here means that the same polynomial can be evaluated on the curvature of every principal $G$-bundle, just as Chapter 1 used the same trace and determinant polynomials across local frames. The result is a family of characteristic forms indexed by bundles and connections, rather than a form living on a single fixed manifold. ## The Chern-Weil Map for Principal Bundles We now ask how an invariant polynomial becomes a cohomology class on the base of a principal bundle. The obstruction is that curvature naturally lives on the total space as a horizontal, equivariant $\mathfrak{g}$-valued form; invariant polynomials are exactly what descends it to the base. [definition: Curvature of a Principal Connection] Let $\pi:P \to M$ be a principal $G$-bundle and let $A \in \Omega^1(P;\mathfrak{g})$ be a principal connection form. Its curvature is \begin{align*} F_A = dA + \frac{1}{2}[A \wedge A] \in \Omega^2(P;\mathfrak{g}). \end{align*} [/definition] The curvature form $F_A$ is horizontal and $G$-equivariant. Since differential forms on the base correspond to basic forms on the total space, the next obstruction is descent: a horizontal form alone is not enough, because it may still change under the right $G$-action. For instance, a non-Ad-invariant linear functional on $\mathfrak{g}$ applied to $F_A$ generally changes under a gauge transformation, so it cannot define a form on the base. The next theorem proves that Ad-invariance is exactly the algebraic condition that removes this obstruction and makes $P_0(F_A)$ descend to $M$. [quotetheorem:9762] [citeproof:9762] This theorem explains why the construction is intrinsic on $M$, even though the formula begins with a connection form on $P$. Horizontality removes all vertical dependence, invariance under the principal action removes the remaining ambiguity along the fibres, and the uniqueness of descent means that the base form is determined by its pullback to $P$. If either condition is absent, the expression still lives naturally on the total space rather than on the base: a horizontal but non-invariant form can change along an orbit, while an invariant form with vertical components cannot be tested against tangent vectors of $M$ alone. The local gauge picture shows the necessity of Ad-invariance concretely. On a trivialising [open set](/page/Open%20Set), write the curvature as a $\mathfrak{g}$-valued form $F$. If the trivialisation is changed by a smooth gauge $g:U\to G$, the local curvature becomes $\operatorname{Ad}_{g^{-1}}F$. A polynomial $Q$ that is not Ad-invariant gives local forms $Q(F)$ and $Q(\operatorname{Ad}_{g^{-1}}F)$ on the same overlap, and these need not agree. Thus the local expressions fail the gluing condition for a global form on $M$. The next step is to see why the descended form is closed. [quotetheorem:9763] [citeproof:9763] The Bianchi identity is essential here: a general Lie-algebra-valued $2$-form inserted into an invariant polynomial need not produce a closed scalar form. For example, on a trivial bundle, an arbitrary matrix-valued $2$-form that does not satisfy the curvature Bianchi identity can have trace powers with nonzero exterior derivative. The theorem does not identify the resulting cohomology class as topological; it only proves that a fixed connection gives a closed representative. To obtain a characteristic class, we must next prove that changing the auxiliary connection changes the representative only by an exact form. [quotetheorem:9764] [citeproof:9764] The existence of the path matters because the proof differentiates a family of curvature forms and integrates the derivative. For a concrete model, let $A_t=A_0+t\alpha$ on a line bundle, where $\alpha$ is an imaginary-valued $1$-form on $M$. Then $F_{A_t}=F_{A_0}+t\,d\alpha$, so \begin{align*} \frac{d}{dt}F_{A_t}=d\alpha \end{align*} is exactly the exterior derivative that the transgression integrates. Knowing only the two endpoint curvatures gives no canonical intermediate derivative, hence no canonical transgression form. The theorem does not say that the differential forms themselves are equal; in this line-bundle example, changing the connection by $\alpha$ changes the curvature representative by the exact $2$-form $d\alpha$. It also assumes the two connections are already given, while the later definition of the Chern-Weil map requires working in a setting where such connections are available, for example on principal bundles over paracompact manifolds. With this independence result in hand, the curvature expression can finally be promoted from a connection-dependent representative to a connection-independent cohomology class. [definition: Chern-Weil Homomorphism] Let $\pi:P \to M$ be a principal $G$-bundle equipped with at least one principal connection, for example with $M$ paracompact. The Chern-Weil homomorphism is the graded algebra map \begin{align*} \operatorname{cw}_P:I(G) \to H^{\mathrm{even}}_{\mathrm{dR}}(M) \end{align*} defined by \begin{align*} \operatorname{cw}_P(P_0)=[P_0(F_A)_M] \end{align*} for any principal connection $A$ on $P$. [/definition] The definition is meaningful because the previous theorem proves independence from $A$. The algebra property comes from the fact that products of invariant polynomials evaluate as wedge products of their curvature forms. [quotetheorem:9765] [citeproof:9765] Each ingredient of the theorem is necessary: without descent there is no form on $M$, without closedness there is no de Rham class, and without connection independence the output would depend on auxiliary differential-geometric data. For a concrete descent failure, applying a non-invariant polynomial to curvature in two different local frames gives local formulas that need not glue, so no homomorphism to $H^{\mathrm{even}}_{\mathrm{dR}}(M)$ is obtained. For a closedness failure, a scalar-valued even form $\alpha$ with $d\alpha\neq 0$ cannot define a de Rham cohomology class, even if it is globally defined. For a connection-independence failure, two choices of connection could give closed representatives in different cohomology classes, so the assignment would depend on the chosen geometric input rather than only on the bundle. The theorem does not classify principal bundles, nor does it prove that every real characteristic class arises from an invariant polynomial. It gives a systematic source of characteristic classes, and later chapters identify the trace and determinant polynomials that recover Chern, Pontryagin, Euler, and Chern character classes. [example: Curvature Representative for a U(1)-Bundle] Let $\pi:P\to M$ be a principal $U(1)$-bundle. Identify $\mathfrak{u}(1)=i\mathbb{R}$. Since $U(1)$ is abelian, for every $h\in U(1)$ and $X\in i\mathbb{R}$ one has \begin{align*} \operatorname{Ad}_hX=hXh^{-1}=hh^{-1}X=X, \end{align*} where the middle equality uses commutativity in $U(1)$. Therefore every polynomial $P:\mathfrak{u}(1)\to\mathbb{R}$ is Ad-invariant, because \begin{align*} P(\operatorname{Ad}_hX)=P(X). \end{align*} For a connection $A$, the curvature $F_A$ is an $i\mathbb{R}$-valued basic $2$-form, so it descends to an $i\mathbb{R}$-valued $2$-form on $M$. The standard normalization for the first Chern form is \begin{align*} \frac{1}{2\pi i}F_A. \end{align*} Because $i^2=-1$, multiplying numerator and denominator by $-i$ gives \begin{align*} \frac{1}{i}=\frac{-i}{i(-i)}=\frac{-i}{-i^2}=\frac{-i}{1}=-i. \end{align*} Hence the same representative can be written as \begin{align*} \frac{1}{2\pi i}F_A=-\frac{i}{2\pi}F_A. \end{align*} Under the convention that unitary curvature is $\mathfrak{u}(1)$-valued, this real $2$-form represents the real first Chern class. If the opposite curvature sign convention is used, $F_A$ is replaced by $-F_A$, so the displayed representative changes sign. If the connection is changed, Chern-Weil connection independence says that the two normalized curvature forms differ by an exact $2$-form, so the resulting de Rham cohomology class is independent of the chosen connection. [/example] This example is the lowest-degree instance of Chern-Weil theory. It also explains why curvature is often described as a differential-form representative of a topological class. [example: Vanishing First Chern Form for SU(n)-Bundles] Let $\pi:P\to M$ be a principal $SU(n)$-bundle with connection $A$, and let $E=P\times_{SU(n)}\mathbb{C}^n$ be the associated bundle for the defining representation. The Lie algebra is \begin{align*} \mathfrak{su}(n)=\{X\in \mathfrak{gl}(n,\mathbb{C}) : X^*+X=0 \text{ and } \operatorname{tr}(X)=0\}. \end{align*} Thus, for any point $p\in P$ and tangent vectors $u,v\in T_pP$, the curvature value satisfies $F_A(u,v)\in\mathfrak{su}(n)$, so \begin{align*} \operatorname{tr}(F_A(u,v))=0. \end{align*} Since this holds for every pair $u,v$, the scalar $2$-form obtained by applying the degree-one trace polynomial is identically zero: \begin{align*} \operatorname{tr}(F_A)=0. \end{align*} In the defining representation, the induced vector-bundle curvature is represented by the same matrix-valued curvature form, so the first Chern form is \begin{align*} \frac{1}{2\pi i}\operatorname{tr}(F_A)=\frac{1}{2\pi i}\cdot 0=0. \end{align*} Thus the first Chern form associated to the defining complex vector bundle vanishes; the determinant-one condition forces the curvature matrices to remain trace-free. [/example] The $SU(n)$ example illustrates how a reduction of structure group imposes algebraic constraints on the allowed curvature expressions. We now make this functorial behaviour systematic. ## Functoriality, Reductions, and Associated Bundles The Chern-Weil class should be compatible with maps of manifolds and with changes in the structure group. The main question here is whether the construction depends only on the natural operations on bundles, rather than on choices made during a calculation. [quotetheorem:9766] [citeproof:9766] The pullback hypothesis is essential because it supplies a canonical connection whose curvature is literally the pulled-back curvature; without this relation, two unrelated bundles over $N$ need not have related characteristic forms. For example, pulling back a nontrivial line bundle along a constant map gives a trivial bundle, so naturality predicts the zero class on the point rather than preserving the original class. The theorem does not say that characteristic forms are unchanged under arbitrary maps of total spaces, only under the standard pullback operation on principal bundles. This compatibility is what allows Chern-Weil classes to be integrated over cycles, compared across classifying maps, and transported through the usual functorial constructions of topology. [definition: Reduction of Structure Group] Let $H\subset G$ be a Lie subgroup. A reduction of structure group of a principal $G$-bundle $P\to M$ to $H$ is a principal $H$-bundle $Q\to M$ together with a $G$-bundle isomorphism \begin{align*} Q\times_H G \cong P. \end{align*} [/definition] A reduction says that the transition functions of the bundle can be chosen inside the smaller group $H$. This description is geometric, while Chern-Weil theory is expressed through connections, curvature forms, and invariant polynomials. The issue is therefore not just whether a reduction exists, but whether characteristic forms computed on the reduced bundle agree with those computed after extending the structure group back to $G$. The necessary comparison is functorial at the level of Lie group homomorphisms: extending an $H$-bundle to a $G$-bundle pushes connections and curvature through the induced Lie algebra map, while invariant polynomials restrict in the opposite direction. The next compatibility result is needed to make reductions useful for Chern-Weil theory: it identifies the characteristic classes before and after extension of structure group, so reductions can justify vanishing and simplification statements rather than merely changing notation. [quotetheorem:9767] [citeproof:9767] The homomorphism hypothesis is necessary because curvature can be pushed from $\mathfrak{h}$ to $\mathfrak{g}$ only through a Lie algebra map compatible with brackets; an arbitrary smooth map of Lie algebras would not respect the curvature formula. For example, restricting from $U(n)$ to $SU(n)$ kills the trace polynomial, while a non-compatible linear map would not correspond to any induced principal connection. The theorem does not say that every $H$-invariant polynomial comes from a $G$-invariant one, so reductions may reveal characteristic forms unavailable from the larger group alone. For a reduction $Q\subset P$, it says that characteristic classes of $P$ computed from $G$-invariant polynomials can be computed on $Q$ after restricting the polynomials to $\mathfrak{h}$, giving the mechanism behind many vanishing results. [example: SU(n) Reduction and the First Chern Class] Suppose the $U(n)$-frame bundle of $E$ is obtained from a principal $SU(n)$-bundle $Q\to M$ by extension of structure group. The degree-one polynomial used for the first Chern form is the normalized trace \begin{align*} P_1(X)=\frac{1}{2\pi i}\operatorname{tr}(X) \end{align*} on $\mathfrak{u}(n)$. Its restriction to $\mathfrak{su}(n)$ is zero because \begin{align*} \mathfrak{su}(n)=\{X\in \mathfrak{gl}(n,\mathbb{C}):X^*+X=0\text{ and }\operatorname{tr}(X)=0\}. \end{align*} Thus, for every $X\in\mathfrak{su}(n)$, \begin{align*} P_1(X)=\frac{1}{2\pi i}\operatorname{tr}(X)=\frac{1}{2\pi i}\cdot 0=0. \end{align*} Let $A$ be an $SU(n)$-connection on $Q$. Its curvature satisfies $F_A(u,v)\in\mathfrak{su}(n)$ for every point and every pair of tangent vectors $u,v$, so the descended first Chern form satisfies \begin{align*} P_1(F_A)(u,v)=\frac{1}{2\pi i}\operatorname{tr}(F_A(u,v))=\frac{1}{2\pi i}\cdot 0=0. \end{align*} Since this holds on every pair of tangent vectors, $P_1(F_A)=0$ as a $2$-form. Therefore the Chern-Weil representative of $c_1(E)$ is the zero form, and the image of $c_1(E)$ in real de Rham cohomology is \begin{align*} [c_1(E)]_{\mathbb{R}}=[0]=0. \end{align*} The determinant-one reduction forces the curvature matrices to be trace-free, so the trace polynomial has no degree-one contribution. [/example] This reduction example explains vanishing through a smaller structure group. To compute characteristic forms for ordinary vector bundles, we also need the construction that turns a principal bundle and a representation into a vector bundle. [definition: Associated Vector Bundle] Let $\rho:G\to GL(V)$ be a finite-dimensional representation of a Lie group $G$, and let $P\to M$ be a principal $G$-bundle. The associated vector bundle is \begin{align*} E=P\times_\rho V\to M. \end{align*} [/definition] The connection on $P$ induces a connection on $E$, but Chern-Weil forms can only be compared if their curvatures are related by the representation. The possible obstruction is that a vector-bundle curvature matrix depends on a local frame, while the principal curvature takes values in $\mathfrak g$. The representation must convert the latter into the former in a way compatible with the induced connection. [quotetheorem:9768] [citeproof:9768] The representation hypothesis is necessary: a vector bundle associated to $P$ is built from a genuine linear action of $G$ on $V$, and the induced curvature formula uses the derivative of that action. If one chose unrelated local vector-bundle connection matrices, their curvature traces would not be forced to agree with any principal-bundle Chern-Weil form. The theorem does not say that all vector-bundle connections arise from a fixed principal connection under every representation; it identifies the compatible one induced from $A$. This is the computational reason that Chern-Weil theory for principal bundles and for vector bundles are two versions of the same construction, allowing later chapters to pass between unitary frames, matrix curvature forms, curvature integrals, and intrinsic characteristic classes. [remark: Integral Versus Real Classes] Chern-Weil theory naturally produces real de Rham cohomology classes. Statements identifying these classes with integral characteristic classes require normalising constants, such as powers of $2\pi i$ in Chern class formulas, and the de Rham comparison map from integral cohomology modulo torsion to real cohomology. [/remark] This comparison is also the point where Chern-Weil theory begins to interact with other parts of geometry. In gauge theory, the same curvature polynomials become Yang-Mills densities, instanton numbers, and topological charges constraining spaces of connections. In index theory, characteristic forms are the local ingredients in formulas such as the Atiyah-Singer index theorem, where curvature representatives of Chern characters, Todd classes, and the A-hat class $\hat A$ are integrated to compute analytic indices. These applications use the same mechanism developed here: invariant algebraic expressions in curvature produce closed forms whose cohomology classes are stable under changes of connection. The chapter has established the formal Chern-Weil homomorphism and its basic functorial properties. The next chapter specialises the invariant polynomials for $U(n)$ to construct the total Chern class of a complex vector bundle. # 3. Chern Classes Of Complex Vector Bundles Chapter 2 constructed the Chern-Weil homomorphism: invariant polynomials applied to curvature give closed differential forms whose de Rham cohomology classes do not depend on the chosen connection. This chapter specialises that machine to complex vector bundles and packages the resulting classes into the total Chern class. The main point is that the determinant polynomial records all elementary symmetric functions of curvature, and therefore behaves well under pullback, [direct sum](/page/Direct%20Sum), and the splitting principle. ## The Total Chern Form The first problem is to turn the curvature of a connection on a complex vector bundle into a sequence of even-degree cohomology classes with integral normalisation. The determinant is the correct invariant polynomial because it converts endomorphism-valued curvature into scalar forms and because its coefficients are the elementary symmetric polynomials. Let $E \to M$ be a complex vector bundle of rank $r$ over a smooth manifold $M$, and let $\nabla$ be a connection on $E$. Its curvature is an element $F_\nabla \in \Omega^2(M;\operatorname{End}(E))$. In a local frame it is represented by an $r \times r$ matrix of $2$-forms, and wedge product of forms is combined with matrix multiplication. [definition: Total Chern Form] Let $E \to M$ be a complex vector bundle of rank $r$ with connection $\nabla$ and curvature $F_\nabla$. The total Chern form of $(E,\nabla)$ is \begin{align*} c(E,\nabla;t) = \det\left(I + \frac{t}{2\pi i}F_\nabla\right) = 1 + c_1(E,\nabla)t + \cdots + c_r(E,\nabla)t^r, \end{align*} where $c_k(E,\nabla) \in \Omega^{2k}(M;\mathbb C)$ is the $k$-th Chern form. [/definition] The definition gives differential forms, but characteristic classes must be cohomology classes attached to the bundle rather than to a chosen connection. Two obstructions have to be removed: the determinant coefficients must be closed, and changing the connection must not change their de Rham classes. Otherwise the construction would produce connection-dependent curvature data rather than characteristic classes of $E$. The next obstruction is therefore not how to write the forms, but why they survive passage to cohomology in a way independent of the auxiliary connection. The required result is the Chern-Weil statement for these determinant polynomials: it turns the locally defined curvature expression into closed global forms whose de Rham classes depend only on the bundle. [quotetheorem:9769] [citeproof:9769] This theorem is the first passage from geometry to topology, but only after tensoring cohomology with $\mathbb C$. The connection is needed because curvature is the local differential-geometric object to which invariant polynomials can be applied, while Ad-invariance is what makes the determinant coefficients independent of the chosen frame. Without invariant polynomials, a formula in the entries of the curvature matrix would usually change under a change of local trivialisation and would not define a global form. The conclusion is therefore a de Rham conclusion: it gives closed forms and connection-independent complex cohomology classes, but it cannot detect torsion and it does not identify an integral cohomology class. That gap matters because the algebraic rules later in the chapter are stated in $H^*(M;\mathbb Z)$, not merely in de Rham cohomology. De Rham representatives cannot distinguish two integral classes that differ by torsion, and a closed complex-valued form can define a valid de Rham class without being the image of any class in $H^*(M;\mathbb Z)$; on $S^2$, for instance, a real multiple $\lambda\omega$ of an area form with total integral $1$ is integral only when $\lambda\in\mathbb Z$. The determinant normalisation is therefore not a cosmetic factor: it is what makes curvature forms land on the integral lattice for genuine complex vector bundles. The next theorem supplies the missing bridge. It says that the Chern-Weil de Rham class is the image of a canonical integral Chern class, so later identities may be proved integrally and then checked against curvature forms after applying the coefficient map. This comparison is usually proved by comparing the Chern-Weil representatives with universal Chern classes on classifying spaces, or equivalently by checking the universal bundle and using naturality. [quotetheorem:9770] This theorem is what justifies using the same notation $c_k(E)$ in integral cohomology and in de Rham cohomology. Its hypotheses are doing real work: the class must come from a complex vector bundle and from the determinant Chern-Weil construction with the stated $2\pi i$ normalisation. An arbitrary closed $2k$-form need not have integral periods, so it need not represent a Chern class even if it is closed and natural-looking. There is also a limitation in the opposite direction: the de Rham representative does not remember every integral feature. The coefficient map kills torsion, so a flat complex line bundle with torsion first Chern class can have zero Chern-Weil form in de Rham cohomology while its integral class is nonzero. Thus integrality is a comparison statement, not an assertion that differential forms alone classify complex vector bundles. For the rest of the chapter, formulas over $H^*(M;\mathbb Z)$ are to be understood as identities among these integral Chern classes, with the Chern-Weil forms giving their de Rham representatives after applying the coefficient map. This is the point at which curvature calculations become legitimate input for the algebraic rules that follow: direct sums, pullbacks, duals, and projective-space computations are all carried out integrally, then checked by their Chern-Weil representatives when a connection is present. [example: Line Bundles] Let $L \to M$ be a complex line bundle with connection $\nabla$. In a local frame, $\operatorname{End}(L)$ is identified with $1\times 1$ complex matrices, so the curvature matrix has the form $(F_\nabla)$ for a scalar $2$-form $F_\nabla$. Therefore the determinant in the definition of the total Chern form is the determinant of a $1\times 1$ matrix: \begin{align*} c(L,\nabla;t)=\det\left(1+\frac{t}{2\pi i}F_\nabla\right). \end{align*} For a $1\times 1$ matrix $(a)$, $\det(a)=a$, so \begin{align*} \det\left(1+\frac{t}{2\pi i}F_\nabla\right)=1+\frac{t}{2\pi i}F_\nabla. \end{align*} Comparing coefficients with \begin{align*} c(L,\nabla;t)=1+c_1(L,\nabla)t \end{align*} gives \begin{align*} c_1(L,\nabla)=\frac{F_\nabla}{2\pi i}. \end{align*} There are no powers $t^k$ with $k\ge 2$ in this degree-one determinant polynomial, so $c_k(L,\nabla)=0$ for $k\ge 2$, and hence $c_k(L)=0$ for $k\ge 2$ in cohomology. Thus a line bundle contributes exactly one Chern root, namely its first Chern class, which is why direct sums of line bundles form the base case for reconstructing higher-rank Chern classes under the splitting principle. [/example] For higher rank bundles the determinant expands into trace expressions. The first two coefficients already show the pattern of symmetric polynomial invariants. [example: First Two Chern Forms] Let $E\to M$ have rank $r$, and write the curvature matrix in a local frame as $F=(F_{ij})$, where each $F_{ij}$ is a $2$-form. Put $\alpha=\frac{1}{2\pi i}$. Since the entries of $F$ have even degree, they commute under wedge product, so the ordinary determinant expansion applies to $I+\alpha tF$ with multiplication interpreted as wedge product. Using the permutation formula for the determinant, the coefficient of $t$ can only come from the identity permutation by choosing one diagonal entry $\alpha tF_{ii}$ and the remaining diagonal entries $1$. Hence \begin{align*} c_1(E,\nabla)=\alpha\sum_{i=1}^r F_{ii} \end{align*} and therefore \begin{align*} c_1(E,\nabla)=\frac{1}{2\pi i}\operatorname{tr}(F). \end{align*} For the coefficient of $t^2$, there are two kinds of terms. The identity permutation contributes the product of two chosen diagonal entries: \begin{align*} \alpha^2\sum_{1\le i<j\le r}F_{ii}\wedge F_{jj}. \end{align*} A transposition $(ij)$ has sign $-1$ and contributes the off-diagonal product \begin{align*} -\alpha^2 F_{ij}\wedge F_{ji}. \end{align*} Summing over all transpositions gives \begin{align*} c_2(E,\nabla)=\alpha^2\sum_{1\le i<j\le r}\left(F_{ii}\wedge F_{jj}-F_{ij}\wedge F_{ji}\right). \end{align*} This is the same as the trace formula. First, \begin{align*} (\operatorname{tr}F)^2=\left(\sum_i F_{ii}\right)\wedge\left(\sum_j F_{jj}\right)=\sum_i F_{ii}\wedge F_{ii}+2\sum_{i<j}F_{ii}\wedge F_{jj}. \end{align*} Also, by matrix multiplication with wedge product, \begin{align*} \operatorname{tr}(F\wedge F)=\sum_i(F\wedge F)_{ii}=\sum_{i,k}F_{ik}\wedge F_{ki}. \end{align*} Splitting the last sum into diagonal and off-diagonal pairs gives \begin{align*} \operatorname{tr}(F\wedge F)=\sum_i F_{ii}\wedge F_{ii}+\sum_{i<j}\left(F_{ij}\wedge F_{ji}+F_{ji}\wedge F_{ij}\right). \end{align*} Because $F_{ij}$ and $F_{ji}$ are both $2$-forms, $F_{ji}\wedge F_{ij}=F_{ij}\wedge F_{ji}$. Therefore \begin{align*} (\operatorname{tr}F)^2-\operatorname{tr}(F\wedge F)=2\sum_{i<j}\left(F_{ii}\wedge F_{jj}-F_{ij}\wedge F_{ji}\right). \end{align*} Multiplying by $\frac{1}{2}\alpha^2$ gives \begin{align*} c_2(E,\nabla)=\frac{1}{2}\left(\frac{1}{2\pi i}\right)^2\left((\operatorname{tr}F)^2-\operatorname{tr}(F\wedge F)\right). \end{align*} Under a change of local frame, the curvature matrix is conjugated, and the determinant coefficients are invariant under conjugation. Thus these local expressions define global differential forms, giving the standard low-degree Chern-Weil representatives for $c_1(E)$ and $c_2(E)$. [/example] ## Chern Roots And The Splitting Principle The determinant formula suggests pretending that the curvature matrix has eigenvalues. For a vector bundle, this language is not literally a global diagonalisation of curvature; it is a bookkeeping device justified by the splitting principle. [definition: Chern Roots] Let $E \to M$ be a complex vector bundle of rank $r$. Formal symbols $x_1,\dots,x_r$ of degree $2$ are called Chern roots of $E$ if the total Chern class is written formally as \begin{align*} c(E;t)=\prod_{j=1}^r(1+x_jt). \end{align*} [/definition] The Chern roots are not additional geometric data on $M$. They encode the elementary symmetric polynomial relation \begin{align*} c_k(E)=e_k(x_1,\dots,x_r), \end{align*} where $e_k$ is the $k$-th elementary symmetric polynomial. [example: Rank Two Bundle] For a rank two complex vector bundle $E$, write its formal Chern roots as $x_1,x_2$. By definition, the total Chern class is encoded by the product \begin{align*} c(E;t)=(1+x_1t)(1+x_2t). \end{align*} Expanding the product term by term gives \begin{align*} (1+x_1t)(1+x_2t)=1\cdot 1+1\cdot x_2t+x_1t\cdot 1+x_1t\cdot x_2t. \end{align*} Since the roots have even cohomological degree, their products commute, so this becomes \begin{align*} (1+x_1t)(1+x_2t)=1+(x_1+x_2)t+x_1x_2t^2. \end{align*} Comparing this with the rank two total Chern class \begin{align*} c(E;t)=1+c_1(E)t+c_2(E)t^2 \end{align*} gives \begin{align*} c_1(E)=x_1+x_2. \end{align*} The coefficient of $t^2$ gives \begin{align*} c_2(E)=x_1x_2. \end{align*} Thus, for rank two bundles, the first Chern class is the sum of the two roots and the second Chern class is their product; root calculations may be done in $x_1,x_2$ and then rewritten using $c_1(E)$ and $c_2(E)$. [/example] The reason this formal notation is reliable is that every bundle becomes a sum of line bundles after pulling back to a suitable auxiliary space. The relevant condition is that the pullback in cohomology remains injective, so identities proved upstairs descend downstairs. [quotetheorem:7041] [citeproof:7041] The splitting principle is not a claim that $E$ splits on $M$; it is a method for proving identities after a controlled pullback. The injectivity of $\pi^*$ is essential: if a class vanished after pullback without injectivity, an identity proved upstairs could lose information downstairs. For example, many vector bundles do not split as sums of line bundles on the original base, so treating their roots as actual classes on $M$ would be unjustified. What the theorem permits is more precise: prove a universal identity after pullback to the flag bundle, then descend it because integral cohomology classes on $M$ are faithfully detected upstairs. To use the method systematically, we need to know which expressions in the formal roots descend from the flag bundle and how they are translated into ordinary Chern classes. The key point is symmetry: permuting the line summands changes the labels of the roots, so only symmetric expressions can be intrinsic to the original bundle. The [fundamental theorem of symmetric polynomials](/theorems/5179) gives the translation before evaluation in cohomology, while the splitting principle supplies the descent of the resulting cohomology class. [quotetheorem:5179] [citeproof:5179] This rule is the main computational advantage of Chern roots. It works because Chern classes are elementary symmetric functions of the roots, so every symmetric root expression has an intrinsic meaning downstairs. The uniqueness is formal uniqueness before evaluating the polynomial: after evaluation, relations in $H^*(M;R)$ may identify different polynomial expressions. For example, in $H^*(\mathbb C P^n;\mathbb Z)$ the relation $h^{n+1}=0$ means that two polynomials differing by a multiple of $h^{n+1}$ define the same class. The [symmetry condition](/theorems/1360) is still necessary: an expression such as $x_1$ alone depends on an ordering of the line summands on the flag bundle and usually has no well-defined descendant on $M$. Thus root notation is a calculation device, not extra structure; before a formula is interpreted as a class of $E$, the final expression must be symmetric in the roots. ## Naturality And Pullbacks Characteristic classes should commute with maps of manifolds; otherwise they would not be intrinsic to vector bundles as objects over spaces. The Chern-Weil construction has this property because connections, curvature, determinant polynomials, and de Rham pullback all interact functorially. [quotetheorem:9771] [citeproof:9771] Naturality lets us compute on universal or convenient spaces and pull results back. The hypothesis that $f^*E$ is the pullback bundle is essential: the statement is not about an arbitrary smooth map between total spaces of vector bundles, but about the functorial operation that carries fibres of $E$ over $f(y)$ to fibres over $y\in N$. The integral proof is topological, via classifying maps, while Chern-Weil theory verifies the same formula after de Rham comparison because pullback connections have pullback curvature. This is why tautological examples on [projective space](/page/Projective%20Space) control many later calculations: once a bundle is obtained by pullback from a universal or geometric model, its Chern classes are pulled back from the model as well. [example: Tautological Line Bundle Over Complex Projective Space] Let $\gamma^1\to \mathbb C P^n$ be the tautological line bundle, so the fibre over a point $[\ell]$ is the line $\ell\subset \mathbb C^{n+1}$. Its dual $(\gamma^1)^*$ is the hyperplane line bundle $\mathcal O(1)$. We choose the positive generator $h\in H^2(\mathbb C P^n;\mathbb Z)$ by the convention \begin{align*} h=c_1(\mathcal O(1))=c_1((\gamma^1)^*). \end{align*} For a line bundle $L$, dualising inverts the transition functions, so the transition-function representative of $c_1(L^*)$ is the negative of the representative of $c_1(L)$. Applying this to $L=\gamma^1$ gives \begin{align*} c_1((\gamma^1)^*)=-c_1(\gamma^1). \end{align*} Substituting $c_1((\gamma^1)^*)=h$ into this identity gives \begin{align*} h=-c_1(\gamma^1). \end{align*} Therefore \begin{align*} c_1(\gamma^1)=-h. \end{align*} Thus the tautological line bundle has negative first Chern class, while its dual is the basic positive line bundle on projective space. With this convention, \begin{align*} H^*(\mathbb C P^n;\mathbb Z)\cong \mathbb Z[h]/(h^{n+1}), \end{align*} so $1,h,h^2,\dots,h^n$ generate the cohomology groups through degree $2n$, and $h^{n+1}=0$. [/example] ## Whitney Product Formula And Dual Bundles The determinant formula should also recognise direct sums of bundles. For smooth complex vector bundles, the integral Chern classes are normalized so that the total class satisfies the Whitney sum formula \begin{align*} c(E\oplus F;t)=c(E;t)c(F;t). \end{align*} This is the property that lets computations pass from line bundles to split bundles and then, by the splitting principle, to arbitrary bundles. The quoted result below gives the corresponding analytic statement for a short exact sequence of holomorphic vector bundles in complex de Rham cohomology; it is a model for the same multiplicativity mechanism, not the sole source of the integral topological formula used in the following examples. [quotetheorem:7052] [citeproof:7052] In the rest of this section, "Whitney sum formula" refers to the integral formula for smooth complex vector bundles. It makes Chern classes multiplicative for direct sums rather than additive. The direct-sum hypothesis matters because, after applying the splitting principle, the roots of $E\oplus F$ are the combined list of the roots of $E$ and $F$; at the curvature level it corresponds to block-diagonal curvature and determinants of block matrices. There is no analogous naive formula $c(E\otimes F)=c(E)c(F)$ for tensor products: if $L$ and $M$ are line bundles, then $c_1(L\otimes M)=c_1(L)+c_1(M)$ rather than $c_1(L)c_1(M)$. Tensor products require Chern-root calculations, where the roots of $E\otimes F$ are sums $x_i+y_j$, not the separate roots multiplied term by term. [example: Sum Of Line Bundles] Let $E=L_1\oplus\cdots\oplus L_r$ be a direct sum of complex line bundles, and put $x_j=c_1(L_j)$. For each line bundle $L_j$, the total Chern class is \begin{align*} c(L_j;t)=1+x_jt. \end{align*} Applying the *Whitney Sum Formula* first to $L_1\oplus L_2$, then to $(L_1\oplus L_2)\oplus L_3$, and continuing inductively gives \begin{align*} c(E;t)=c(L_1;t)c(L_2;t)\cdots c(L_r;t). \end{align*} Substituting $c(L_j;t)=1+x_jt$ into each factor gives \begin{align*} c(E;t)=\prod_{j=1}^r(1+x_jt). \end{align*} To read off the coefficient of $t^k$, choose the $x_jt$ term from exactly $k$ of the $r$ factors and choose $1$ from the remaining factors. For a subset $\{j_1<\cdots<j_k\}\subseteq\{1,\dots,r\}$, that choice contributes \begin{align*} (x_{j_1}t)(x_{j_2}t)\cdots(x_{j_k}t)=x_{j_1}x_{j_2}\cdots x_{j_k}t^k. \end{align*} Summing over all $k$-element subsets gives \begin{align*} c_k(E)=\sum_{1\le j_1<\cdots<j_k\le r}x_{j_1}\cdots x_{j_k}=e_k(x_1,\dots,x_r). \end{align*} Thus, for an actual direct sum of line bundles, the formal Chern roots are the genuine first Chern classes $c_1(L_1),\dots,c_1(L_r)$. [/example] The same root language also has to handle dualisation. For a line bundle the dual reverses the first Chern class, but for a higher-rank bundle the Chern classes are elementary symmetric functions of all roots, so the sign change must be translated degree by degree. The question is whether dualising simply negates every class or introduces the parity-dependent sign forced by the degree. [quotetheorem:9772] [citeproof:9772] This sign rule is often the fastest way to handle exact sequences involving duals or cotangent bundles. Its scope is narrow: it uses the fact that dualising a line bundle changes the Chern root $x$ to $-x$, and then the splitting principle propagates that sign through elementary symmetric polynomials. It does not determine Chern classes of arbitrary tensor products or Hom bundles by itself; for those, one must again split into roots and compute with the relevant root transformations. In projective geometry the rule pairs naturally with the tautological bundle and its dual, where the hyperplane class is the positive generator. ## The Euler Sequence And The Tangent Bundle Of Projective Space The last calculation in this chapter illustrates how Chern classes interact with an exact sequence of vector bundles. Complex projective space has a canonical exact sequence relating its tangent bundle to the tautological line bundle, and the Whitney formula turns the sequence into a closed formula for the total Chern class. [quotetheorem:7018] [citeproof:7018] This sequence converts the tangent bundle computation into a direct sum of copies of the hyperplane line bundle. The projective-space geometry is essential: tangent vectors at a line are maps from that line to the quotient by it, which is why the quotient bundle and the tautological line bundle appear. The statement concerns the complex tangent bundle of $\mathbb C P^n$; the underlying real tangent bundle has real characteristic classes related to these Chern classes, but it is not the object being computed here. The rank discrepancy in the Euler sequence is handled by the initial copy of the constant line bundle, whose total Chern class is $1$. [example: Chern Class Of The Tangent Bundle Of Complex Projective Space] Let $h=c_1((\gamma^1)^*)\in H^2(\mathbb C P^n;\mathbb Z)$. The *[Euler Sequence For Complex Projective Space](/theorems/7018)* gives \begin{align*} 0\to \underline{\mathbb C}\to ((\gamma^1)^*)^{\oplus(n+1)}\to T\mathbb C P^n\to 0. \end{align*} As smooth complex vector bundles this short exact sequence splits, so \begin{align*} ((\gamma^1)^*)^{\oplus(n+1)}\cong \underline{\mathbb C}\oplus T\mathbb C P^n. \end{align*} Applying the *Whitney Sum Formula* gives \begin{align*} c(((\gamma^1)^*)^{\oplus(n+1)};t)=c(\underline{\mathbb C};t)c(T\mathbb C P^n;t). \end{align*} The constant line bundle has total Chern class $c(\underline{\mathbb C};t)=1$, hence \begin{align*} c(T\mathbb C P^n;t)=c(((\gamma^1)^*)^{\oplus(n+1)};t). \end{align*} For the line bundle $(\gamma^1)^*$, the definition of $h$ gives \begin{align*} c((\gamma^1)^*;t)=1+ht. \end{align*} Applying the *Whitney Sum Formula* repeatedly to the direct sum of $n+1$ copies gives \begin{align*} c(((\gamma^1)^*)^{\oplus(n+1)};t)=(1+ht)^{n+1}. \end{align*} By the [binomial theorem](/theorems/750), \begin{align*} (1+ht)^{n+1}=\sum_{k=0}^{n+1}\binom{n+1}{k}h^kt^k. \end{align*} In $H^*(\mathbb C P^n;\mathbb Z)\cong \mathbb Z[h]/(h^{n+1})$, the relation $h^{n+1}=0$ removes the $k=n+1$ term, so \begin{align*} c(T\mathbb C P^n;t)=\sum_{k=0}^{n}\binom{n+1}{k}h^kt^k. \end{align*} Comparing this with \begin{align*} c(T\mathbb C P^n;t)=\sum_{k=0}^{n}c_k(T\mathbb C P^n)t^k \end{align*} gives \begin{align*} c_k(T\mathbb C P^n)=\binom{n+1}{k}h^k \end{align*} for $0\le k\le n$. In particular, $\binom{n+1}{1}=n+1$ and $\binom{n+1}{n}=n+1$, so \begin{align*} c_1(T\mathbb C P^n)=(n+1)h \end{align*} and \begin{align*} c_n(T\mathbb C P^n)=(n+1)h^n. \end{align*} Thus the Euler sequence turns the tangent bundle calculation into the binomial expansion of the hyperplane class. [/example] The calculation also explains why the normalisation in the Chern-Weil determinant matters. On $\mathbb C P^n$, the hyperplane class $h$ is integral, and the curvature representative of $c_1((\gamma^1)^*)$ integrates to $1$ over a projective line. # 4. Euler Class And Pfaffian Curvature This part of the course develops characteristic classes by moving between three languages: topology of vector bundles, differential forms from connections, and geometric representatives such as zero sets of sections. Chapters 1 through 3 introduced Chern-Weil theory and explained how invariant polynomials applied to curvature produce de Rham cohomology classes, with Chern classes arising from determinant polynomials. The present chapter applies that same framework to oriented real bundles, where the Pfaffian is the relevant invariant polynomial and the resulting class is the Euler class. The prerequisites are the Thom class, orientations of real vector bundles, basic transversality, and the Chern-Weil homotopy argument; the goal is to connect these tools to Gauss-Bonnet-Chern and Poincare-Hopf. ## Oriented Real Vector Bundles And The Pfaffian The first question is what extra structure on a real bundle is needed before a top-degree characteristic class can have a sign. Complex vector bundles carry a canonical orientation, but real bundles do not. The Euler class begins with a choice of orientation and is therefore sensitive to reversing that choice. [definition: Oriented Real Vector Bundle] Let $E \to M$ be a real vector bundle of rank $r$. An orientation of $E$ is a choice of orientation of each fibre $E_p$ varying locally constantly with respect to oriented local frames. An oriented real vector bundle is a real vector bundle equipped with such a choice. [/definition] For an oriented rank $r$ bundle, local frames may be chosen with transition functions in $GL^+(r,\mathbb R)$. After choosing a bundle metric, the structure group reduces further to $SO(r)$, which is the group seen by metric connections. The curvature matrices for such connections are skew-symmetric, so the next task is to identify the invariant polynomial on skew-symmetric matrices that remembers orientation. [definition: Pfaffian] Let \begin{align*} \operatorname{Skew}_{2m}(\mathbb R)=\{A\in M_{2m}(\mathbb R):A^\top=-A\}. \end{align*} The Pfaffian is the polynomial map $\operatorname{Pf}:\operatorname{Skew}_{2m}(\mathbb R)\to \mathbb R$ defined by \begin{align*} \operatorname{Pf}: A \mapsto \frac{1}{2^m m!}\sum_{\sigma \in S_{2m}} \operatorname{sgn}(\sigma) A_{\sigma(1)\sigma(2)}\cdots A_{\sigma(2m-1)\sigma(2m)}. \end{align*} [/definition] The Pfaffian is the square root of the determinant polynomial on skew-symmetric matrices, with a sign fixed by the chosen orientation. To use it in Chern-Weil theory, we need both the determinant relation and the change-of-frame law, since these show how the expression behaves under orthonormal frame changes. [quotetheorem:9773] [citeproof:9773] The second formula says that the Pfaffian is invariant under $SO(2m)$, and changes sign under orientation reversal. The restriction to skew-symmetric matrices is essential: on arbitrary matrices there is no polynomial square root of $\det A$ with this alternating change-of-frame behaviour. For example, the identity matrix has $\det I=1$, but conjugating a chosen square root by a reflection would have to change its sign even though the matrix remains $I$; this contradicts any attempt to make the same signed expression depend only on an arbitrary matrix. Even dimension is also essential, because odd-dimensional skew-symmetric matrices always have determinant zero, so no nonzero top-degree Pfaffian invariant can be obtained. In dimension $3$, for instance, a skew-symmetric matrix has entries $A_{12}=a$, $A_{13}=b$, $A_{23}=c$, with $A_{ji}=-A_{ij}$ and $A_{ii}=0$. Such a matrix satisfies $\det A=0$ for all $a,b,c\in \mathbb R$; there is no signed top-degree polynomial whose square recovers a nonzero determinant invariant. The distinction between $SO(2m)$ and $O(2m)$ is exactly the orientation issue: an orthogonal change of frame with determinant $-1$ reverses the sign of the Pfaffian, so the expression is not an unoriented invariant. A rank-two illustration is already decisive: replacing an oriented orthonormal frame by $(e_2,e_1)$ changes the curvature entry $F_{12}$ to $-F_{12}$, so the local Pfaffian form changes sign rather than gluing to the same form. The transformation law is stated for arbitrary $B\in GL(2m,\mathbb R)$ because it records how the Pfaffian changes under every [change of basis](/page/Change%20Of%20Basis) of the ambient [vector space](/page/Vector%20Space); the Chern-Weil application then restricts this law to oriented orthonormal changes of frame, where $\det B=1$. Thus it gives an Ad-invariant polynomial on $\mathfrak{so}(2m)$ once an orientation has been fixed, but it does not define a signed Euler form on an unoriented real bundle. This sign change is already visible in the smallest possible rank, where all the notation collapses to a single curvature entry. [example: Pfaffian In Rank Two] Let $A$ be the skew-symmetric $2\times 2$ matrix with diagonal entries $A_{11}=A_{22}=0$, upper-right entry $A_{12}=a$, and lower-left entry $A_{21}=-a$. Here $2m=2$, so $m=1$. The defining formula for the Pfaffian gives \begin{align*} \operatorname{Pf}(A)=\frac{1}{2}\sum_{\sigma\in S_2}\operatorname{sgn}(\sigma)A_{\sigma(1)\sigma(2)}. \end{align*} The two permutations in $S_2$ are the identity, with sign $+1$, and the transposition $(12)$, with sign $-1$. Therefore \begin{align*} \operatorname{Pf}(A)=\frac{1}{2}\left(A_{12}-A_{21}\right). \end{align*} Substituting $A_{12}=a$ and $A_{21}=-a$ gives \begin{align*} \operatorname{Pf}(A)=\frac{1}{2}\left(a-(-a)\right)=a. \end{align*} Thus, for an oriented rank-two bundle with curvature matrix entry $F_{12}$ in a positive oriented orthonormal frame, the Euler form is \begin{align*} e(\nabla)=\operatorname{Pf}\left(\frac{F_\nabla}{2\pi}\right)=\frac{F_{12}}{2\pi}. \end{align*} Changing the orientation changes the sign of the chosen oriented frame and replaces $F_{12}$ by $-F_{12}$, so this rank-two computation is the local model for the tangent bundle of an oriented surface. [/example] ## Euler Form Of A Metric Connection The next problem is to turn the Pfaffian polynomial into a differential form on the base manifold. A metric connection on an oriented Euclidean bundle has curvature valued in $\mathfrak{so}(2m)$, so the Pfaffian can be applied to the curvature matrix in an oriented orthonormal frame. [definition: Euler Form] Let $E \to M$ be an oriented real vector bundle of rank $2m$ with bundle metric and metric connection $\nabla$. Let $F_\nabla \in \Omega^2(M;\mathfrak{so}(E))$ be the curvature. In an oriented local orthonormal frame, write $F_\nabla=(F_{ij})$ as a skew-symmetric matrix of $2$-forms. The Euler form of $\nabla$ is \begin{align*} e(\nabla)=\operatorname{Pf}\left(\frac{F_\nabla}{2\pi}\right) =\frac{1}{(2\pi)^m}\operatorname{Pf}(F_\nabla) \in \Omega^{2m}(M). \end{align*} [/definition] The transformation law for the Pfaffian makes this local expression independent of the oriented orthonormal frame. If the rank of $E$ is odd, the Chern-Weil Euler form is taken to be zero. The remaining Chern-Weil question is whether this differential form is closed and whether its cohomology class depends on the connection. [quotetheorem:9774] [citeproof:9774] This theorem is the Chern-Weil realisation of the Euler class: it gives a closed differential form whose de Rham cohomology class is the real image of the integral class $e(E)\in H^{2m}(M;\mathbb Z)$. It does not, by itself, prove that the integral of this form counts zeros of sections; that interpretation needs the Thom-class and transversality arguments below. The orientation hypothesis is not cosmetic: without it, the Pfaffian changes sign between local orthonormal frames and the local top-degree forms do not glue to a globally signed form. The even-rank hypothesis is also structural, since the Pfaffian is defined only for $2m\times 2m$ skew-symmetric matrices; in odd rank the corresponding Euler class in real cohomology is zero because changing orientation on a stabilised odd-rank frame forces the putative class to equal its negative. The metric-connection hypothesis is equally important: metric compatibility makes the connection form and curvature form take values in $\mathfrak{so}(2m)$ in oriented orthonormal frames, which is the domain on which the Pfaffian is the required invariant polynomial. A general non-metric connection has $\mathfrak{gl}(2m,\mathbb R)$-valued curvature, so applying the same Pfaffian expression to its matrix entries would not be frame-independent; one first replaces it by a connection on the [oriented orthonormal frame bundle](/theorems/6245), or works with the metric part that gives the same Euler class after transgression. The theorem therefore converts a topological class into a curvature representative exactly in the setting where a signed top-degree curvature polynomial exists, which is the bridge to Gauss-Bonnet-Chern after the zero-counting interpretation has been established. [example: Curvature Computation For The Round Two-Sphere] Let $S^2$ carry the round metric of radius $1$ with its usual orientation, and let $\nabla$ be the Levi-Civita connection on $TS^2$. Choose a local oriented orthonormal coframe $(\theta^1,\theta^2)$, so the Riemannian area form is \begin{align*} \Omega_{12}=\theta^1\wedge \theta^2. \end{align*} For an oriented Riemannian surface, the curvature matrix of the Levi-Civita connection is skew-symmetric with \begin{align*} (F_\nabla)_{12}=K\theta^1\wedge\theta^2. \end{align*} Here $K$ is the Gaussian curvature. On the unit round sphere, $K=1$, so \begin{align*} (F_\nabla)_{12}=\theta^1\wedge\theta^2=\Omega_{12}. \end{align*} Skew-symmetry gives the other off-diagonal entry: \begin{align*} (F_\nabla)_{21}=-(F_\nabla)_{12}=-\Omega_{12}. \end{align*} In rank two, the Pfaffian of a skew-symmetric matrix is its upper-right entry, as in the rank-two Pfaffian computation above. Therefore \begin{align*} e(\nabla)=\operatorname{Pf}\left(\frac{F_\nabla}{2\pi}\right)=\left(\frac{F_\nabla}{2\pi}\right)_{12}. \end{align*} Substituting $(F_\nabla)_{12}=\Omega_{12}$ gives \begin{align*} e(\nabla)=\frac{\Omega_{12}}{2\pi}. \end{align*} Integrating over $S^2$, \begin{align*} \int_{S^2}e(\nabla)=\int_{S^2}\frac{\Omega_{12}}{2\pi}. \end{align*} Pulling out the constant factor gives \begin{align*} \int_{S^2}e(\nabla)=\frac{1}{2\pi}\int_{S^2}\Omega_{12}. \end{align*} Since $\Omega_{12}$ is the Riemannian area form and the unit sphere has area $4\pi$, \begin{align*} \int_{S^2}e(\nabla)=\frac{1}{2\pi}\cdot 4\pi. \end{align*} Cancelling $2\pi$ gives \begin{align*} \int_{S^2}e(\nabla)=2. \end{align*} This matches $\chi(S^2)=2$, for instance from the CW decomposition of $S^2$ with one $0$-cell and one $2$-cell. [/example] The sphere computation also records the role of orientation: the sign of the local curvature form depends on the oriented orthonormal frame. This motivates separating the orientation convention from the curvature calculation. [remark: Orientation Dependence] Reversing the orientation of $E$ changes the sign of the Pfaffian and therefore changes $e(E)$ to $-e(E)$. For tangent bundles this matches the fact that integration over an oriented manifold also changes sign when the orientation of the manifold is reversed, so the numerical Gauss-Bonnet-Chern integral is orientation-compatible. [/remark] ## Thom Classes And Zero-Counting The curvature formula identifies a de Rham representative, but the geometric meaning of the Euler class comes from the Thom class. The guiding question is: what cohomology class is detected by the zero set of a section? [definition: Thom Class] Let $E \to M$ be an oriented real vector bundle of rank $r$. A Thom class of $E$ is a class $u_E \in H^r_c(E)$ whose restriction to each fibre $E_p \cong \mathbb R^r$ is the positive generator of $H^r_c(E_p)$ determined by the orientation. [/definition] The Thom class is the universal cohomology class concentrated along the fibres of $E$, but the Euler class must live on the base $M$. This motivates the following definition: pull the Thom class back along the zero section, whose image records where a section is meant to vanish. [definition: Euler Class From The Thom Class] Let $s_0:M\to E$ be the zero section of an oriented rank $r$ real vector bundle. The Euler class of $E$ is \begin{align*} e(E)=s_0^*u_E \in H^r(M). \end{align*} [/definition] This definition explains the obstruction-theoretic role of $e(E)$. If a nowhere-zero section exists, the zero section can be separated from that section inside the total space, forcing the Euler class to vanish after the usual Thom-class argument. [quotetheorem:2284] [citeproof:2284] The converse does not hold in this generality without additional hypotheses on the obstruction problem being asked. The Euler class is the primary obstruction to extending a nonzero section from the $(r-1)$-skeleton to the $r$-skeleton; if the base has higher-dimensional cells or if a boundary value has already been prescribed, later obstruction groups can enter. A useful model is a rank-$2$ oriented bundle with $e(E)=0$: after the Euler obstruction vanishes, asking for two everywhere linearly independent sections is a stronger problem, and the remaining obstruction is encoded by the complementary bundle rather than by $e(E)$ alone. Similarly, a prescribed nonzero section over a subspace can fail to extend across higher cells even when the absolute Euler class vanishes, because the relative obstruction groups remember the chosen boundary data. For a rank-$r$ oriented bundle over an $r$-dimensional CW complex, this primary obstruction is the only obstruction for the existence of one nonzero section, so $e(E)=0$ is sufficient in that restricted setting. Outside that range, the theorem is best read as a necessary condition for a nowhere-zero section rather than a complete existence theorem. It is already strong enough to give many nonexistence results for vector fields on manifolds, where the tangent bundle supplies a geometrically meaningful source of sections. [remark: Preview Of The Obstruction] The obstruction-theoretic meaning of the top Chern class will become precise only after the [Gauss-Bonnet-Chern theorem](/theorems/9776) is stated below. The following discussion is therefore a preview of that later theorem's application, not a local consequence of the material already proved. [/remark] With that warning in place, the Euler class can still be used as a guide to the geometric obstruction one expects. The next example shows the simplest closed manifold where the obstruction is visible numerically: the tangent bundle of an even-dimensional sphere has nonzero Euler number, so it cannot admit a nowhere-zero vector field. [example: Obstruction To Nowhere-Zero Vector Fields] By *Gauss-Bonnet-Chern Theorem* applied to the tangent bundle of the closed oriented manifold $S^{2m}$, \begin{align*} \int_{S^{2m}} e(TS^{2m})=\chi(S^{2m}). \end{align*} The sphere $S^{2m}$ has a CW decomposition with one $0$-cell and one $2m$-cell, so \begin{align*} \chi(S^{2m})=1+(-1)^{2m}=1+1=2. \end{align*} Therefore \begin{align*} \int_{S^{2m}} e(TS^{2m})=2. \end{align*} If $e(TS^{2m})$ were zero in $H^{2m}(S^{2m})$, then its pairing with the fundamental class would be zero, so the displayed integral would be $0$. Since it is $2$, we have \begin{align*} e(TS^{2m})\ne 0. \end{align*} By *[Euler Class Vanishes for Non-Vanishing Sections](/theorems/2284)*, $TS^{2m}$ therefore admits no nowhere-zero section, equivalently $S^{2m}$ admits no nowhere-zero tangent vector field. In contrast, view \begin{align*} S^{2m-1}=\{z=(z_1,\dots,z_m)\in \mathbb C^m:\sum_{j=1}^m |z_j|^2=1\}. \end{align*} Define a vector field by \begin{align*} X_z=iz. \end{align*} The tangent space to the unit sphere at $z$ is \begin{align*} T_zS^{2m-1}=\left\{v\in \mathbb C^m:\operatorname{Re}\sum_{j=1}^m v_j\overline{z_j}=0\right\}. \end{align*} For $v=iz$, \begin{align*} \operatorname{Re}\sum_{j=1}^m (iz_j)\overline{z_j} =\operatorname{Re}\left(i\sum_{j=1}^m |z_j|^2\right) =\operatorname{Re}(i\cdot 1) =0. \end{align*} Thus $X_z\in T_zS^{2m-1}$. Also, \begin{align*} |X_z|^2=\sum_{j=1}^m |iz_j|^2=\sum_{j=1}^m |z_j|^2=1, \end{align*} so $X_z$ never vanishes. This exhibits the parity contrast: even-dimensional spheres are obstructed by their nonzero Euler class, while odd-dimensional spheres have an explicit nowhere-zero tangent vector field. [/example] The strongest zero-counting statement needs transversality. A transverse section has a zero set which is a submanifold of the expected codimension; in the equal-rank case its zeros are isolated and carry signs. The sphere example above only detects the global obstruction to removing all zeros. It says that zeros must occur on $S^{2m}$, but it does not distinguish a single positive zero from several positive and negative zeros whose signed total is the same. To turn the Euler class from a nonexistence test into an enumerative formula, each isolated transverse zero must contribute a local sign that can be added over the zero set. The next definition supplies exactly this missing local contribution. To turn an isolated zero into a number, the local intersection between $s(M)$ and the zero section must be oriented. Transversality makes the derivative of the section identify the tangent space of the base with the fibre at the zero, so the orientation of $M$ and the orientation of $E$ can be compared at the zero. The resulting sign is the unit that will be summed in the Euler-class formula. [definition: Index Of A Transverse Zero] Let $E \to M$ be an oriented real vector bundle of rank $n$ over an oriented $n$-manifold $M$, and let $s:M\to E$ be a smooth section transverse to the zero section. If $p\in M$ satisfies $s(p)=0$, the index $\operatorname{ind}_p(s)$ is $+1$ or $-1$ according to whether the induced map \begin{align*} Ds_p:T_pM\to E_p \end{align*} preserves or reverses orientation. [/definition] This local sign is the differential-topological shadow of the Thom class. The definition supplies the contribution of each isolated zero, and the remaining question is whether the sum of these local signs recovers the global Euler class. The answer is the intersection formula between the section and the zero section. [quotetheorem:9775] [citeproof:9775] The hypotheses are exactly those needed for the displayed formula to be a finite signed count. Closedness supplies the ordinary fundamental class $[M]$ and avoids boundary terms; for manifolds with boundary one must use relative classes or impose boundary conditions on the section. The equality of rank and dimension makes a transverse zero set zero-dimensional, while in lower rank the zero set would usually be a positive-dimensional submanifold Poincare dual to $e(E)$. This is the intersection-theoretic framework behind the formula: $e(E)$ is the cohomology class dual to the intersection of a section with the zero section, and the isolated signed count is the zero-dimensional case of that duality. Transversality prevents degenerate zeros from having ambiguous local contribution; if a zero is nontransverse, it must be perturbed or assigned a local degree rather than read directly from $Ds_p$. ## Tangent Bundles And Classical Theorems The central application is the tangent bundle of an oriented Riemannian manifold. The Euler form of the Levi-Civita connection is made from the Riemann curvature tensor, and its integral is forced to be topological. [quotetheorem:9776] [citeproof:9776] For surfaces this is the classical Gauss-Bonnet theorem. Compactness is used to make the integral and Euler characteristic finite, while closedness is what prevents additional boundary curvature and geodesic-curvature terms from appearing. Orientation is needed to integrate the Euler form as a signed top-degree form; reversing the orientation reverses both the form and the fundamental class, so the numerical equality remains consistent. In higher dimensions the Pfaffian packages all sectional curvature data into the correct top-degree invariant polynomial, but the formula is not a pointwise curvature sign statement: positive and negative local curvature contributions may cancel in the integral. [example: Euler Class Of Even-Dimensional Spheres] Let $\eta\in H^{2m}(S^{2m};\mathbb Z)$ be the positive generator determined by the standard orientation, normalized by \begin{align*} \langle \eta,[S^{2m}]\rangle=1. \end{align*} Since $H^{2m}(S^{2m};\mathbb Z)\cong \mathbb Z$, there is an integer $k$ such that \begin{align*} e(TS^{2m})=k\eta. \end{align*} Pairing both sides with the fundamental class gives \begin{align*} \langle e(TS^{2m}),[S^{2m}]\rangle=\langle k\eta,[S^{2m}]\rangle. \end{align*} By linearity of the Kronecker pairing, \begin{align*} \langle k\eta,[S^{2m}]\rangle=k\langle \eta,[S^{2m}]\rangle=k. \end{align*} By *Gauss-Bonnet-Chern Theorem* applied to the tangent bundle, \begin{align*} \langle e(TS^{2m}),[S^{2m}]\rangle=\int_{S^{2m}}e(TS^{2m})=\chi(S^{2m}). \end{align*} The sphere $S^{2m}$ has a CW decomposition with one $0$-cell and one $2m$-cell, so \begin{align*} \chi(S^{2m})=1+(-1)^{2m}=1+1=2. \end{align*} Hence $k=2$, and therefore \begin{align*} e(TS^{2m})=2\eta. \end{align*} If the orientation of $S^{2m}$ is reversed, the positive generator changes from $\eta$ to $-\eta$, so the same underlying Euler class is written as twice the positive generator for the new orientation with the corresponding sign convention. The round Levi-Civita curvature representative gives the same class in de Rham cohomology, so its integral is also \begin{align*} \int_{S^{2m}}\operatorname{Pf}\left(\frac{F_{\nabla^{LC}}}{2\pi}\right)=2. \end{align*} [/example] The sphere example shows how the tangent-bundle Euler class converts a curvature integral into a vector-field obstruction. The general vector-field theorem is obtained by replacing the round sphere calculation with an arbitrary vector field having isolated nondegenerate zeros. [quotetheorem:3940] [citeproof:3940] The nondegeneracy hypothesis is the vector-field version of transversality: it ensures that each local derivative $DX_p:T_pM\to T_pM$ is an isomorphism and hence has a well-defined sign. If zeros are isolated but degenerate, the theorem can still be recovered by replacing the sign of $DX_p$ with the local degree of the vector field, but that is extra data not contained in the displayed statement. Closedness is again essential; on a compact manifold with boundary, outward-pointing or inward-pointing boundary conditions change the formula, and boundary index terms may appear. This chapter therefore gives three equivalent faces of the same invariant: the Pfaffian curvature form, the Thom-class pullback, and the signed zero-count of transverse sections. The same pattern reappears in index theory, where analytic local data integrate to topological invariants, and in differential topology, where local degrees control global intersection numbers. The equivalence is the prototype for later characteristic-class calculations, where Chern classes and Pontryagin classes will again be represented by curvature polynomials and interpreted through topology. It is also the first obstruction-theoretic model: the Euler class is a primary obstruction, while later characteristic classes package higher or more structured obstructions to simplifying bundles. # 5. Pontryagin Classes And Real Bundles This chapter brings the real theory of characteristic classes into the Chern-Weil framework developed for complex bundles in Chapters 2 and 3 and for Euler classes in Chapter 4. The guiding question is how curvature on a real vector bundle produces cohomology classes when the natural invariant polynomials for orthogonal matrices involve even-degree data. The answer is that Pontryagin classes are the real shadows of Chern classes after complexification, with sign conventions determined by the eigenvalue pairing of skew-symmetric curvature. ## Pontryagin Forms From Real Curvature And Complexification For a complex vector bundle, the determinant polynomial applied to curvature packages the Chern classes. A real vector bundle has curvature valued in a real matrix Lie algebra, and after choosing a metric connection this curvature is skew-symmetric. The first problem is to extract differential forms from this real curvature in a way that is invariant under change of orthonormal frame and stable under changing the connection. Let $E \to M$ be a rank $r$ real vector bundle over a smooth manifold $M$. To use the Chern-Weil machinery already built for complex bundles, we pass from $E$ to the complex vector bundle $E_{\mathbb C}=E\otimes_{\mathbb R}\mathbb C$. [definition: Complexification Of A Real Vector Bundle] Let $E \to M$ be a real vector bundle. Its complexification is the complex vector bundle \begin{align*} E_{\mathbb C}:=E\otimes_{\mathbb R}\mathbb C \to M. \end{align*} If $\nabla$ is a connection on $E$, the induced connection on $E_{\mathbb C}$ is the $\mathbb C$-linear extension of $\nabla$. [/definition] Complexification keeps the same base manifold and doubles the real fibre dimension, but it gives access to Chern forms. If $F_\nabla$ is the curvature of $\nabla$, then the curvature of the induced connection on $E_{\mathbb C}$ is the complex-linear extension of $F_\nabla$. The next step is to choose which Chern components of $E_{\mathbb C}$ should be regarded as the real characteristic forms of $E$. [definition: Pontryagin Form] Let $E \to M$ be a real vector bundle and let $\nabla$ be a connection on $E$. The $k$-th Pontryagin form of $(E,\nabla)$ is the differential form \begin{align*} p_k(\nabla):=(-1)^k c_{2k}(\nabla^{\mathbb C})\in \Omega^{4k}(M), \end{align*} where $c_{2k}(\nabla^{\mathbb C})$ is the $2k$-th Chern-Weil form of the complexified connection on $E_{\mathbb C}$. [/definition] This definition is arranged so that the resulting cohomology class agrees with the standard topological convention for Pontryagin classes. Since $c_{2k}(\nabla^{\mathbb C})$ is a closed $4k$-form, the form $p_k(\nabla)$ is also closed and represents a de Rham cohomology class in $H^{4k}_{\mathrm{dR}}(M)$. The remaining issue is whether this representative depends on the chosen connection. [quotetheorem:9777] [citeproof:9777] The theorem makes Pontryagin classes legitimate characteristic classes of real vector bundles because it removes the arbitrary choice of connection from the final cohomology class. The passage to cohomology is essential: different connections can produce different closed forms, and their difference may be a nonzero exact form even though the represented class is fixed. Thus the theorem does not say that the differential form $p_k(\nabla)$ is canonical; it says that changing the connection changes the representative only by an exact correction. The theorem also does not assert an integral statement: in this chapter $p_k(E)$ is a de Rham class, and questions about integral Pontryagin classes require additional topological input. The degree restriction comes from the same source, since $c_{2k}(E_{\mathbb C})$ has real cohomological degree $4k$. [example: First Pontryagin Form From Curvature] Let $E\to M$ have a metric connection $\nabla$, and write its curvature matrix in a local oriented orthonormal frame as $\Omega=(\Omega_{ij})$. Since the connection is metric, $\Omega$ is skew-symmetric, so $\Omega_{ij}=-\Omega_{ji}$. In particular $\Omega_{ii}=-\Omega_{ii}$, hence $\Omega_{ii}=0$ and \begin{align*} \operatorname{tr}(\Omega)=\sum_i \Omega_{ii}=0. \end{align*} For the complexified connection, the Chern-Weil total form is obtained from \begin{align*} \det\left(I+\frac{i}{2\pi}\Omega\right). \end{align*} Set $A=\frac{i}{2\pi}\Omega$. The degree-two Chern form is the coefficient \begin{align*} c_2(\nabla^{\mathbb C})=\frac{1}{2}\left((\operatorname{tr}A)^2-\operatorname{tr}(A\wedge A)\right). \end{align*} Because $\operatorname{tr}A=\frac{i}{2\pi}\operatorname{tr}(\Omega)=0$, this becomes \begin{align*} c_2(\nabla^{\mathbb C})=-\frac{1}{2}\operatorname{tr}(A\wedge A). \end{align*} Now \begin{align*} A\wedge A=\left(\frac{i}{2\pi}\right)^2\Omega\wedge\Omega=-\frac{1}{4\pi^2}\Omega\wedge\Omega. \end{align*} Therefore \begin{align*} c_2(\nabla^{\mathbb C})=\frac{1}{8\pi^2}\operatorname{tr}(\Omega\wedge\Omega). \end{align*} By the definition of the first Pontryagin form, $p_1(\nabla)=-c_2(\nabla^{\mathbb C})$, so \begin{align*} p_1(\nabla)=-\frac{1}{8\pi^2}\operatorname{tr}(\Omega\wedge\Omega). \end{align*} The linear trace term vanishes because the curvature is skew-symmetric, so the construction produces no degree-two real class here. The first surviving invariant trace expression is quadratic in curvature, which is why $p_1(\nabla)$ is a degree-four form. [/example] The local formula is useful for calculations, but the intrinsic definition through complexification is the safest way to manage signs. The next section turns this definition into algebraic relations among characteristic classes. ## Chern Classes Of Complexified Real Bundles The complexified bundle $E_{\mathbb C}$ is not an arbitrary complex bundle: it carries a conjugation induced by complex conjugation on the factor $\mathbb C$. The main question is what this extra real structure forces on its Chern classes. The key phenomenon is that the Chern roots appear in opposite pairs, so odd Chern classes vanish and even Chern classes encode Pontryagin classes. [definition: Conjugate Complex Vector Bundle] Let $V \to M$ be a complex vector bundle. The conjugate bundle $\overline V \to M$ has the same underlying real vector bundle as $V$. For each $m\in M$, its fibrewise scalar multiplication is the map \begin{align*} \mathbb C\times \overline V_m\to \overline V_m \end{align*} given by \begin{align*} (\lambda,v)\mapsto \lambda\cdot_{\overline V}v:=\overline{\lambda}v, \end{align*} where the product on the right is the original scalar multiplication in $V_m$. [/definition] The conjugate bundle reverses complex scalar multiplication while leaving the underlying real bundle unchanged. Since Chern classes detect complex structure, we need a sign rule comparing $V$ and $\overline V$ before applying conjugation to $E_{\mathbb C}$. [quotetheorem:9778] [citeproof:9778] The sign rule is now ready to be applied to real bundles. The hypothesis that the bundle is conjugated as a complex bundle is essential: an unrelated complex bundle with the same rank need not have Chern classes related by signs. The theorem also does not say that $V$ and $\overline V$ are usually isomorphic; for a line bundle with $c_1(V)\ne 0$, the sign change is precisely the obstruction to such an isomorphism. For a complexification, however, conjugation comes from the real form itself, so the sign rule becomes a compatibility condition on the Chern classes of $E_{\mathbb C}$, and the next theorem identifies the resulting forced vanishing. [quotetheorem:9779] [citeproof:9779] The vanishing theorem explains why the real theory occupies degrees divisible by $4$, and the real-coefficient hypothesis is doing work. Over integral cohomology the same argument gives $2c_{2k+1}(E_{\mathbb C})=0$, so it proves only that the odd Chern classes are two-torsion unless further information is available. Thus the de Rham setting removes torsion phenomena and leaves the even Chern classes as the only visible Chern data of a complexified real bundle. The next result records the exact translation into Pontryagin classes. [quotetheorem:9780] [citeproof:9780] The formula has two complementary uses. It defines real characteristic classes from the already established complex theory, and it lets us compute real classes by passing through Chern roots. The sign $(-1)^k$ is not cosmetic: without it, an oriented two-plane bundle would satisfy $p_1=-e^2$ rather than the Euler-square convention used in topology. The theorem does not reconstruct all Chern data of a real bundle's complexification, since the odd classes have disappeared over de Rham cohomology; it identifies exactly the even part that survives as Pontryagin data. [example: Complexification Of An Oriented Two Plane Bundle] Let $E \to M$ be an oriented real rank-two vector bundle. After choosing a metric, each oriented fibre is a real Euclidean plane with a preferred rotation by $\pi/2$, so $E$ becomes a complex line bundle $L$. The complexification then splits into the two eigenspaces of this fibrewise complex structure: \begin{align*} E_{\mathbb C}\cong L\oplus \overline L. \end{align*} Write $e=e(E)$. With the convention $c_1(L)=e$, the conjugate line bundle has first Chern class $c_1(\overline L)=-e$, so \begin{align*} c(L)=1+e. \end{align*} Also, \begin{align*} c(\overline L)=1-e. \end{align*} Using the Whitney product formula for the direct sum $L\oplus \overline L$, we get \begin{align*} c(E_{\mathbb C})=c(L)c(\overline L). \end{align*} Substituting the two line-bundle Chern classes gives \begin{align*} c(E_{\mathbb C})=(1+e)(1-e). \end{align*} Expanding the product, \begin{align*} (1+e)(1-e)=1-e+e-e^2. \end{align*} The middle terms cancel, hence \begin{align*} c(E_{\mathbb C})=1-e^2. \end{align*} Therefore the degree-four component is \begin{align*} c_2(E_{\mathbb C})=-e^2. \end{align*} By the definition of the first Pontryagin class, \begin{align*} p_1(E)=-c_2(E_{\mathbb C}). \end{align*} Substituting $c_2(E_{\mathbb C})=-e^2$ gives \begin{align*} p_1(E)=-(-e^2)=e^2. \end{align*} Thus, for an oriented real two-plane bundle, the first Pontryagin class is exactly the square of the Euler class. [/example] Rank two already shows the geometric meaning of the sign convention: Pontryagin classes are designed so that the [first Pontryagin class of an oriented real two-plane bundle](/theorems/9783) is the square of its Euler class. Higher-rank calculations follow the same paired-root pattern. ## Whitney Formula And Stable Invariance Characteristic classes become useful because they behave predictably under direct sums. For Chern classes this was the Whitney product formula; for real bundles the question is whether complexification converts direct sums into a product formula for Pontryagin classes without introducing odd correction terms. [quotetheorem:9781] [citeproof:9781] The formula implies that adding a product real bundle does not change Pontryagin classes, but the product hypothesis is essential. Adding a nontrivial summand can change the total Pontryagin class by the factor $p(F)$; for instance, if $F$ is an oriented two-plane bundle with $e(F)^2\ne 0$, then $p_1(E\oplus F)=p_1(E)+e(F)^2$. There is therefore no cancellation statement for arbitrary direct sums. This matters because many bundle identities in geometry are not literal isomorphisms but become isomorphisms after adding product summands; the next theorem turns that observation into a stable invariance principle. [quotetheorem:9782] [citeproof:9782] Stable invariance is especially important for tangent bundles, because many natural tangent bundle identities hold only after adding a product line. The theorem does not say that stably equivalent bundles are isomorphic: a nontrivial real line bundle $L$ satisfies $L\oplus L\cong \varepsilon^2$ in examples such as the tautological line bundle over $\mathbb{RP}^1$, but $L$ itself is not isomorphic to the product line bundle. Nor are all characteristic classes stable in the same way; for an oriented two-plane bundle with $e(E)\ne 0$, adding a product line changes the rank from $2$ to $3$, so the Euler class of the new bundle lives in a different degree and cannot be recovered as the old class. Real projective space gives a standard example where the tangent bundle is accessed through a stable identity rather than through a direct splitting. [example: Pontryagin Classes Of Real Projective Space] Let $\gamma^1\to \mathbb{RP}^n$ be the tautological real line bundle and let $T\mathbb{RP}^n$ be the tangent bundle. The standard stable tangent bundle identity is \begin{align*} T\mathbb{RP}^n\oplus \varepsilon^1\cong (n+1)\gamma^1. \end{align*} A real line bundle has no positive-degree Pontryagin classes: after choosing a metric connection, its curvature matrix is a $1\times 1$ skew-symmetric matrix, hence its only entry $\Omega_{11}$ satisfies $\Omega_{11}=-\Omega_{11}$, so $\Omega_{11}=0$. Therefore every positive-degree Pontryagin form of $\gamma^1$ vanishes, and \begin{align*} p(\gamma^1)=1. \end{align*} The product line bundle also has total Pontryagin class \begin{align*} p(\varepsilon^1)=1. \end{align*} Apply the *Whitney Formula For Pontryagin Classes* to the stable identity. The left-hand side gives \begin{align*} p(T\mathbb{RP}^n\oplus \varepsilon^1)=p(T\mathbb{RP}^n)p(\varepsilon^1). \end{align*} Substituting $p(\varepsilon^1)=1$ gives \begin{align*} p(T\mathbb{RP}^n\oplus \varepsilon^1)=p(T\mathbb{RP}^n). \end{align*} For the right-hand side, $(n+1)\gamma^1$ is the direct sum of $n+1$ copies of $\gamma^1$, so repeated use of the Whitney formula gives \begin{align*} p((n+1)\gamma^1)=p(\gamma^1)^{n+1}. \end{align*} Substituting $p(\gamma^1)=1$ gives \begin{align*} p((n+1)\gamma^1)=1^{n+1}=1. \end{align*} Since isomorphic bundles have the same Pontryagin classes, the stable identity gives \begin{align*} p(T\mathbb{RP}^n)=1. \end{align*} Equivalently, $p_k(T\mathbb{RP}^n)=0$ for every $k>0$ over real coefficients. This is not merely a zero computation: the stable tangent bundle is built from the nontrivial tautological line bundle, but de Rham Pontryagin classes cannot see the torsion information carried by $\gamma^1$. The result also agrees with the cohomological fact that $H^{4k}_{\mathrm{dR}}(\mathbb{RP}^n)=0$ whenever $0<4k<n$. [/example] The projective-space computation illustrates a general limitation: over de Rham cohomology, torsion characteristic classes vanish. Pontryagin classes can still carry substantial information on manifolds with nonzero cohomology in degrees divisible by $4$. ## Orientation Phenomena And Basic Calculations Pontryagin classes exist for all real vector bundles, with no orientation hypothesis. Orientation enters through the Euler class, and the main interaction is that in real rank two the first Pontryagin class is the square of the Euler class. This section records the resulting low-dimensional calculations and a central example from complex projective space. [quotetheorem:9783] [citeproof:9783] This theorem is the bridge between orientation-sensitive and orientation-insensitive classes. Reversing the orientation of $E$ changes $e(E)$ to $-e(E)$, but $e(E)^2$ is unchanged, as required for $p_1(E)$. [example: Tangent Bundle Of The Two Sphere] For the oriented tangent bundle $TS^2\to S^2$, the Euler class is the degree-two class whose integral is the Euler characteristic: \begin{align*} \int_{S^2} e(TS^2)=\chi(S^2)=2. \end{align*} Thus $e(TS^2)$ is nonzero in $H^2_{\mathrm{dR}}(S^2)$. Since $TS^2$ is an oriented real rank-two bundle, *First Pontryagin Class Of An Oriented Two Plane Bundle* gives \begin{align*} p_1(TS^2)=e(TS^2)^2. \end{align*} The square lies in degree four: \begin{align*} e(TS^2)^2\in H^4_{\mathrm{dR}}(S^2). \end{align*} But $S^2$ has real dimension $2$, so every degree-four differential form on $S^2$ is zero, and hence \begin{align*} H^4_{\mathrm{dR}}(S^2)=0. \end{align*} Therefore \begin{align*} e(TS^2)^2=0 \end{align*} and consequently \begin{align*} p_1(TS^2)=0. \end{align*} The Euler class detects the oriented rank-two tangent bundle in top degree, while the first Pontryagin class has no nonzero degree-four cohomology group in which to live. Thus $p_1=e^2$ remains the correct structural identity even though dimensional reasons force this particular square to vanish. [/example] The next calculation is the standard complex projective space example. Although $\mathbb{CP}^n$ is a complex manifold, its Pontryagin classes refer to the underlying real tangent bundle. [quotetheorem:9784] [citeproof:9784] This computation shows how Pontryagin classes detect curvature-like information that survives after forgetting the complex structure. The Euler sequence hypothesis is doing real work: replacing $\mathbb{CP}^n$ by a general almost complex manifold would not give the same closed formula without comparable control of the Chern classes. The result also gives a quick obstruction: if a real vector bundle is stably equivalent to $T\mathbb{CP}^n$, its first Pontryagin class must be $(n+1)h^2$. This is the point at which the theory connects to stable tangent bundle questions and, in dimension four, to curvature and signature formulas. [example: The Case Of The Complex Projective Plane] For $\mathbb{CP}^2$, the formula *[First Pontryagin Class Of Complex Projective Space](/theorems/9784)* gives, with $n=2$, \begin{align*} p_1(T\mathbb{CP}^2)=(2+1)h^2. \end{align*} Since $2+1=3$, this is \begin{align*} p_1(T\mathbb{CP}^2)=3h^2. \end{align*} The class $h=c_1(\mathcal O(1))$ is normalized by \begin{align*} \int_{\mathbb{CP}^2}h^2=1. \end{align*} Using linearity of integration, \begin{align*} \int_{\mathbb{CP}^2}p_1(T\mathbb{CP}^2)=\int_{\mathbb{CP}^2}3h^2. \end{align*} Pulling out the scalar gives \begin{align*} \int_{\mathbb{CP}^2}3h^2=3\int_{\mathbb{CP}^2}h^2. \end{align*} Substituting the normalization, \begin{align*} 3\int_{\mathbb{CP}^2}h^2=3\cdot 1=3. \end{align*} Thus the Pontryagin number is \begin{align*} \int_{\mathbb{CP}^2}p_1(T\mathbb{CP}^2)=3. \end{align*} In dimension four, the Hirzebruch signature formula reads $\sigma(M)=\frac{1}{3}\int_M p_1(TM)$, so for $\mathbb{CP}^2$ it gives \begin{align*} \sigma(\mathbb{CP}^2)=\frac{1}{3}\cdot 3=1. \end{align*} This agrees with the intersection form on $H^2(\mathbb{CP}^2;\mathbb R)$, which is generated by $h$ and satisfies $\int_{\mathbb{CP}^2}h^2=1$. [/example] The chapter therefore completes the basic dictionary among Euler, Chern, and Pontryagin classes. Chern classes govern complex bundles, Euler classes detect orientation in top degree for real bundles, and Pontryagin classes provide stable even-degree characteristic classes for real bundles through complexification. # 6. Characteristic Numbers This chapter turns characteristic classes into numerical invariants by integrating their top-degree parts over compact manifolds. Chapters 3 through 5 constructed Chern, Euler, and Pontryagin classes as cohomology classes represented by curvature forms; here the question is what information remains after pairing those classes with the fundamental class. The resulting integers and rational numbers are characteristic numbers, and they are among the most computable invariants produced by Chern-Weil theory. The guiding theme is that characteristic classes live in cohomology, while closed manifolds provide a canonical way to evaluate top-degree cohomology classes. This evaluation connects bundle geometry to global topology: Chern numbers give computable obstructions to complex tangent-bundle equivalence, Pontryagin numbers detect oriented cobordism classes, and the signature theorem relates curvature-derived classes to the intersection form. These numbers are powerful but deliberately compressed invariants, so the chapter also keeps track of what they fail to distinguish. ## Pairing Characteristic Classes with Fundamental Classes A characteristic class becomes a number only when its degree matches the dimension of the manifold. The basic problem is therefore to formalise the operation that takes a cohomology class in top degree and evaluates it on the orientation class of a closed manifold. [definition: Fundamental Class] Let $M$ be a closed connected oriented smooth $n$-manifold. The fundamental class of $M$ is the homology class $[M] \in H_n(M;\mathbb Z)$ determined by the chosen orientation. [/definition] The fundamental class is the homological object supplied by the manifold itself. To use it with characteristic classes, we need the corresponding cohomology-homology evaluation map, because Chern-Weil theory naturally produces cohomology classes rather than homology classes. [definition: Pairing with the Fundamental Class] Let $M$ be a closed oriented smooth $n$-manifold and let $R$ be a coefficient ring. The pairing with the fundamental class is the evaluation map \begin{align*} \langle -, [M]\rangle : H^n(M;R) \to R, \qquad \alpha \mapsto \langle \alpha,[M]\rangle. \end{align*} [/definition] This evaluation is the restriction to $[M]$ of the cohomology-homology pairing \begin{align*} \langle -,-\rangle : H^n(M;R)\times H_n(M;R) \to R. \end{align*} For real de Rham classes, if $\alpha=[\omega]$ for a closed $n$-form $\omega$, then the pairing is represented by \begin{align*} \langle [\omega],[M]\rangle = \int_M \omega. \end{align*} This formula is the bridge between the Chern-Weil representatives of earlier chapters and the numerical invariants of this chapter. Before naming the general construction, it is useful to see that it already recovers a familiar invariant. [example: Euler Class of a Surface] Let $\Sigma_g$ be a closed oriented surface of genus $g$. The Euler class $e(T\Sigma_g)\in H^2(\Sigma_g;\mathbb Z)$ is a top-degree characteristic class, so it can be paired with the fundamental class $[\Sigma_g]$. By the *Poincare-Hopf theorem*, this pairing equals the Euler characteristic: \begin{align*} \langle e(T\Sigma_g),[\Sigma_g]\rangle=\chi(\Sigma_g). \end{align*} Using the standard CW decomposition of $\Sigma_g$ with one $0$-cell, $2g$ one-cells, and one $2$-cell, the Euler characteristic is \begin{align*} \chi(\Sigma_g)=1-2g+1. \end{align*} Therefore \begin{align*} \chi(\Sigma_g)=2-2g. \end{align*} Combining the two equalities gives \begin{align*} \langle e(T\Sigma_g),[\Sigma_g]\rangle=2-2g. \end{align*} Thus the Euler-class characteristic number of the tangent bundle recovers the familiar Euler characteristic of the surface. [/example] The example also shows why compactness and orientation are part of the basic setup. Compactness makes integration finite and homological evaluation available in top degree, while orientation chooses the sign of the fundamental class. The next definition packages the common pattern: multiply characteristic classes until top degree, then evaluate. [definition: Characteristic Number] Let $M$ be a closed oriented smooth $n$-manifold and let $E\to M$ be a vector bundle with characteristic classes $a_i(E)\in H^{d_i}(M;R)$. A characteristic number of $E$ is a number of the form \begin{align*} \left\langle a_{i_1}(E)a_{i_2}(E)\cdots a_{i_r}(E),[M]\right\rangle, \end{align*} where $d_{i_1}+d_{i_2}+\cdots+d_{i_r}=n$. [/definition] Characteristic numbers are therefore not new characteristic classes; they are top-degree evaluations of products of already constructed classes. The product condition forces the total degree to equal the dimension of the manifold, so lower-degree information contributes only through cup products. The first structural question is whether these numbers depend on the chosen presentation of the manifold or bundle. [quotetheorem:9785] [citeproof:9785] The same proof explains exactly where the hypotheses enter. If $f$ reverses orientation, then $f_*[M]=-[N]$, so a nonzero top-degree characteristic number changes sign rather than being preserved. For example, on $\mathbb{CP}^2$ the number $\langle c_1(T\mathbb{CP}^2)^2,[\mathbb{CP}^2]\rangle=9$ becomes $-9$ after reversing the orientation of the fundamental class. The theorem also does not say that any two manifolds with the same characteristic numbers are diffeomorphic. Characteristic numbers only record evaluations of selected top-degree products; they discard the rest of the [cohomology ring](/theorems/2271) and all lower-degree class data that cannot be paired with $[M]$ directly. Thus the invariant is robust under the stated geometric equivalence, but it is far from a classification theorem. ## Chern Numbers, Pontryagin Numbers, and Euler Characteristic The next problem is to identify the standard characteristic numbers attached to the main classes of the course. Complex manifolds have Chern numbers, oriented real manifolds have Pontryagin numbers, and even-dimensional oriented manifolds have the Euler number of the tangent bundle. [definition: Chern Number] Let $M$ be a compact complex manifold of complex dimension $m$. For a partition $i_1+\cdots+i_r=m$ with $i_j\ge 1$, the associated Chern number is \begin{align*} \left\langle c_{i_1}(TM)c_{i_2}(TM)\cdots c_{i_r}(TM),[M]\right\rangle, \end{align*} where $TM$ is the holomorphic tangent bundle regarded as a complex vector bundle. [/definition] Since $c_i(TM)\in H^{2i}(M;\mathbb Z)$, the condition $i_1+\cdots+i_r=m$ makes the product a class in $H^{2m}(M;\mathbb Z)$. Chern numbers are especially useful because the complex structure supplies a canonical orientation. The natural next question is how much of the complex-geometric structure is remembered by smooth maps. [quotetheorem:9786] [citeproof:9786] The theorem says that Chern numbers are controlled by the complex tangent bundle, not by a coordinate description of the manifold. The extra hypothesis is essential: an arbitrary smooth diffeomorphism between complex manifolds need not identify the two holomorphic tangent bundles as complex vector bundles, so naturality for Chern classes has no input. In particular, the theorem should not be read as saying that Chern numbers are invariants of a bare smooth manifold independent of the chosen stable complex or almost-complex tangent data. A useful warning comes from changing the almost-complex or stable-complex structure on a fixed smooth manifold: the underlying smooth tangent bundle is the same real bundle, but the chosen complex-bundle data can change the Chern classes that enter the evaluation. This distinction is not a technical nuisance; it is the reason the real analogue uses different classes. If the manifold is only oriented and smooth, there is no preferred complex tangent bundle from which Chern classes could be formed. The next construction replaces that missing complex data with Pontryagin classes of the real tangent bundle, giving characteristic numbers that are intrinsic to the oriented smooth tangent bundle. [definition: Pontryagin Number] Let $M$ be a closed oriented smooth manifold of dimension $4k$. For a partition $i_1+\cdots+i_r=k$ with $i_j\ge 1$, the associated Pontryagin number is \begin{align*} \left\langle p_{i_1}(TM)p_{i_2}(TM)\cdots p_{i_r}(TM),[M]\right\rangle. \end{align*} [/definition] Pontryagin numbers capture the part of the tangent-bundle story built from classes in degrees $4,8,12,\dots$. They do not cover the top-degree invariant coming from the Euler class, so we need a separate definition for the Euler-class pairing before comparing it with the Euler characteristic. [definition: Euler Number] Let $M$ be a closed oriented smooth manifold of dimension $n$. The Euler number of $M$ is \begin{align*} \left\langle e(TM),[M]\right\rangle. \end{align*} [/definition] The Euler number is a characteristic number of the tangent bundle, but it is often written separately because it should match an invariant already known from algebraic topology. This motivates the comparison between the Euler class pairing and the alternating sum of Betti numbers. [quotetheorem:9787] [citeproof:9787] The theorem gives a model for the whole subject: curvature produces a closed differential form, cohomology makes the construction independent of choices, and pairing with $[M]$ produces a topological number. The hypotheses are doing real work. Closedness prevents boundary correction terms; for a compact manifold with boundary, Gauss-Bonnet includes an additional boundary contribution, so the integral of the Euler form alone need not equal $\chi(M)$. Orientation is needed to integrate the Euler form as a top-degree form, although the final Euler characteristic is independent of orientation once the Euler class is interpreted with the appropriate local coefficients. The theorem also does not say that the Euler characteristic determines the manifold. For instance, many non-diffeomorphic closed oriented surfaces have different Euler characteristics, but in higher dimensions there are numerous non-diffeomorphic manifolds with the same Euler characteristic. The following computations show how the definitions work in projective examples where the cohomology ring is explicit. [example: Chern Numbers of Complex Projective Space] Let $H\in H^2(\mathbb{CP}^m;\mathbb Z)$ be the hyperplane class, normalised by $\langle H^m,[\mathbb{CP}^m]\rangle=1$. By the *Euler sequence*, \begin{align*} c(T\mathbb{CP}^m)=(1+H)^{m+1}. \end{align*} For $\mathbb{CP}^2$, we expand \begin{align*} (1+H)^3=1+3H+3H^2+H^3. \end{align*} Since $H^3=0$ in $H^*(\mathbb{CP}^2;\mathbb Z)$, the positive-degree Chern classes are \begin{align*} c_1(T\mathbb{CP}^2)=3H,\qquad c_2(T\mathbb{CP}^2)=3H^2. \end{align*} Thus \begin{align*} c_1(T\mathbb{CP}^2)^2=(3H)^2=9H^2. \end{align*} Pairing with the fundamental class gives \begin{align*} \langle c_1^2,[\mathbb{CP}^2]\rangle=\langle 9H^2,[\mathbb{CP}^2]\rangle=9\langle H^2,[\mathbb{CP}^2]\rangle=9. \end{align*} Similarly, \begin{align*} \langle c_2,[\mathbb{CP}^2]\rangle=\langle 3H^2,[\mathbb{CP}^2]\rangle=3\langle H^2,[\mathbb{CP}^2]\rangle=3. \end{align*} For $\mathbb{CP}^3$, we expand \begin{align*} (1+H)^4=1+4H+6H^2+4H^3+H^4. \end{align*} Since $H^4=0$ in $H^*(\mathbb{CP}^3;\mathbb Z)$, this gives \begin{align*} c_1=4H,\qquad c_2=6H^2,\qquad c_3=4H^3. \end{align*} The degree-six Chern monomials are then \begin{align*} c_1^3=(4H)^3=64H^3. \end{align*} Therefore \begin{align*} \langle c_1^3,[\mathbb{CP}^3]\rangle=\langle 64H^3,[\mathbb{CP}^3]\rangle=64\langle H^3,[\mathbb{CP}^3]\rangle=64. \end{align*} For the mixed monomial, \begin{align*} c_1c_2=(4H)(6H^2)=24H^3. \end{align*} Hence \begin{align*} \langle c_1c_2,[\mathbb{CP}^3]\rangle=\langle 24H^3,[\mathbb{CP}^3]\rangle=24\langle H^3,[\mathbb{CP}^3]\rangle=24. \end{align*} Finally, \begin{align*} \langle c_3,[\mathbb{CP}^3]\rangle=\langle 4H^3,[\mathbb{CP}^3]\rangle=4\langle H^3,[\mathbb{CP}^3]\rangle=4. \end{align*} In these examples, every Chern number is the coefficient of the top power of the hyperplane class. [/example] Projective space is the standard testing ground because the cohomology ring has one generator and every characteristic number reduces to extracting the coefficient of $H^m$. Products of projective spaces add a second useful feature: the tangent bundle splits as an external sum. [example: Chern Numbers of $\mathbb{CP}^1\times\mathbb{CP}^1$] Let $a,b\in H^2(\mathbb{CP}^1\times\mathbb{CP}^1;\mathbb Z)$ be the pullbacks of the hyperplane classes from the two factors. Since the square of the hyperplane class on $\mathbb{CP}^1$ vanishes, the pullback classes satisfy $a^2=0$ and $b^2=0$, while the product class is normalized by $\langle ab,[\mathbb{CP}^1\times\mathbb{CP}^1]\rangle=1$. The tangent bundle of a product splits as the direct sum of the pulled-back tangent bundles of the two factors, so multiplicativity of total Chern classes gives \begin{align*} c(T(\mathbb{CP}^1\times\mathbb{CP}^1))=(1+2a)(1+2b). \end{align*} Expanding the product gives \begin{align*} (1+2a)(1+2b)=1+2a+2b+4ab. \end{align*} The degree-two part is therefore \begin{align*} c_1(T(\mathbb{CP}^1\times\mathbb{CP}^1))=2a+2b. \end{align*} The degree-four part is \begin{align*} c_2(T(\mathbb{CP}^1\times\mathbb{CP}^1))=4ab. \end{align*} Now compute the square of the first Chern class: \begin{align*} c_1^2=(2a+2b)^2. \end{align*} Expanding the square gives \begin{align*} (2a+2b)^2=(2a)^2+2(2a)(2b)+(2b)^2. \end{align*} Thus \begin{align*} (2a+2b)^2=4a^2+8ab+4b^2. \end{align*} Using $a^2=0$ and $b^2=0$, this becomes \begin{align*} c_1^2=8ab. \end{align*} Pairing with the fundamental class and using linearity gives \begin{align*} \langle c_1^2,[\mathbb{CP}^1\times\mathbb{CP}^1]\rangle=\langle 8ab,[\mathbb{CP}^1\times\mathbb{CP}^1]\rangle. \end{align*} Hence \begin{align*} \langle c_1^2,[\mathbb{CP}^1\times\mathbb{CP}^1]\rangle=8\langle ab,[\mathbb{CP}^1\times\mathbb{CP}^1]\rangle=8. \end{align*} Similarly, \begin{align*} \langle c_2,[\mathbb{CP}^1\times\mathbb{CP}^1]\rangle=\langle 4ab,[\mathbb{CP}^1\times\mathbb{CP}^1]\rangle. \end{align*} Therefore \begin{align*} \langle c_2,[\mathbb{CP}^1\times\mathbb{CP}^1]\rangle=4\langle ab,[\mathbb{CP}^1\times\mathbb{CP}^1]\rangle=4. \end{align*} Thus the two Chern numbers of $\mathbb{CP}^1\times\mathbb{CP}^1$ are $\langle c_1^2,[\mathbb{CP}^1\times\mathbb{CP}^1]\rangle=8$ and $\langle c_2,[\mathbb{CP}^1\times\mathbb{CP}^1]\rangle=4$. [/example] These two complex surfaces have different Chern numbers, so they cannot be biholomorphic or complex-tangent diffeomorphic. The same computations also foreshadow how characteristic numbers interact with intersection forms in real dimension four. ## Cobordism Invariance and Geometric Meaning The final problem is to understand which characteristic numbers survive deformation through a manifold with boundary. Cobordism asks whether a closed manifold can appear as the boundary of a compact manifold one dimension higher, and characteristic numbers provide systematic obstructions. [definition: Oriented Cobordism] Two closed oriented smooth $n$-manifolds $M_0$ and $M_1$ are oriented cobordant if there exists a compact oriented smooth $(n+1)$-manifold $W$ with \begin{align*} \partial W = M_1 \sqcup (-M_0) \end{align*} as oriented manifolds. [/definition] The sign in the boundary convention is what makes characteristic numbers subtract across the two boundary components. A class that extends over $W$ has zero total pairing with the boundary, which is the algebraic mechanism behind cobordism invariance. This leads to the main cobordism result for Pontryagin numbers. [quotetheorem:9788] [citeproof:9788] This theorem is stronger than diffeomorphism invariance because cobordism is a much coarser relation. The closedness hypothesis is needed because Pontryagin numbers are defined by pairing with a fundamental class in top degree; for manifolds with boundary, [Stokes' theorem](/theorems/1530) produces boundary terms rather than a closed-manifold invariant. Orientation is equally essential: reversing the orientation changes the sign of the fundamental class, so each nonzero Pontryagin number in dimension $4k$ changes sign, while a zero value remains zero. The dimension condition is not cosmetic. Pontryagin classes live in degrees divisible by $4$, so a product can pair with $[M]$ only when the dimension is divisible by $4$. The result also has a sharp limitation: it says Pontryagin numbers descend to oriented cobordism classes, not that they classify smooth manifolds up to diffeomorphism or homotopy equivalence. To see what extra geometry these numbers measure, the next step is to combine them in a special universal polynomial whose value is not merely an abstract cobordism invariant, but the signature of the middle-dimensional intersection form. [quotetheorem:9789] The course uses this theorem as a structural statement rather than proving it in full. Its proof belongs to index theory: the signature is the index of the signature operator, and the Atiyah-Singer index theorem identifies the local index density with the $L$-polynomial in the Pontryagin forms. The restriction to dimension $4k$ is forced by the intersection form: only in middle degree $2k$ on a $4k$-manifold is the cup-product pairing symmetric and therefore equipped with a signature. Orientation fixes the sign of both the fundamental-class pairing and the intersection form. The theorem is not a classification theorem. It extracts one integer from the middle-dimensional intersection form and expresses that integer as a Pontryagin number combination. Distinct intersection forms can have the same signature, and manifolds with the same signature can still differ in their integral intersection form, homotopy type, or smooth structure. In dimension four, the theorem becomes a concrete formula for the first Pontryagin number. [example: Signature and Pontryagin Number in Dimension Four] For a closed oriented smooth $4$-manifold $M$, the degree-four part of the Hirzebruch $L$-class is $L_1=p_1/3$. Applying the *[Hirzebruch Signature Theorem](/theorems/9789)* in dimension $4$ gives \begin{align*} \sigma(M)=\left\langle L_1(TM),[M]\right\rangle. \end{align*} Substituting $L_1(TM)=p_1(TM)/3$ gives \begin{align*} \sigma(M)=\left\langle \frac{1}{3}p_1(TM),[M]\right\rangle. \end{align*} By linearity of the fundamental-class pairing, \begin{align*} \left\langle \frac{1}{3}p_1(TM),[M]\right\rangle=\frac{1}{3}\left\langle p_1(TM),[M]\right\rangle. \end{align*} Hence \begin{align*} \sigma(M)=\frac{1}{3}\left\langle p_1(TM),[M]\right\rangle. \end{align*} Multiplying both sides by $3$ gives \begin{align*} 3\sigma(M)=\left\langle p_1(TM),[M]\right\rangle. \end{align*} Equivalently, \begin{align*} \left\langle p_1(TM),[M]\right\rangle=3\sigma(M). \end{align*} For $\mathbb{CP}^2$, the middle cohomology is generated by the hyperplane class $H\in H^2(\mathbb{CP}^2;\mathbb Z)$ with $\langle H^2,[\mathbb{CP}^2]\rangle=1$. Therefore the intersection form sends $H$ paired with itself to $1$, so its matrix in the basis $(H)$ is $(1)$. This form has one positive eigenvalue and no negative eigenvalues, so \begin{align*} \sigma(\mathbb{CP}^2)=1-0=1. \end{align*} Substituting into the four-dimensional signature formula gives \begin{align*} \left\langle p_1(T\mathbb{CP}^2),[\mathbb{CP}^2]\right\rangle=3\sigma(\mathbb{CP}^2). \end{align*} Since $\sigma(\mathbb{CP}^2)=1$, \begin{align*} \left\langle p_1(T\mathbb{CP}^2),[\mathbb{CP}^2]\right\rangle=3\cdot 1. \end{align*} Thus \begin{align*} \left\langle p_1(T\mathbb{CP}^2),[\mathbb{CP}^2]\right\rangle=3. \end{align*} In real dimension four, the signature determines the first Pontryagin number by this factor of $3$. [/example] The four-dimensional formula gives a practical way to compute the first Pontryagin number when the intersection form is known. It also explains why the comparison between $S^2\times S^2$ and connected sums involving $\mathbb{CP}^2$ is a natural test case. [example: Comparing $S^2\times S^2$ and $\mathbb{CP}^2\#(-\mathbb{CP}^2)$] For $S^2\times S^2$, multiplicativity of Euler characteristic for finite CW complexes gives \begin{align*} \chi(S^2\times S^2)=\chi(S^2)\chi(S^2)=2\cdot 2=4. \end{align*} For $\mathbb{CP}^2\#(-\mathbb{CP}^2)$, the connected-sum formula in real dimension at least $2$ gives \begin{align*} \chi(\mathbb{CP}^2\#(-\mathbb{CP}^2))=\chi(\mathbb{CP}^2)+\chi(-\mathbb{CP}^2)-2=3+3-2=4. \end{align*} The orientation reversal in $-\mathbb{CP}^2$ changes the sign of the fundamental class, not the cell count, so its Euler characteristic is still $3$. Let $a,b\in H^2(S^2\times S^2;\mathbb Z)$ be the pullbacks of the fundamental degree-two classes from the two factors. Since $a^2=0$, $b^2=0$, and $\langle ab,[S^2\times S^2]\rangle=1$, the intersection form satisfies \begin{align*} q(xa+yb)=\langle (xa+yb)^2,[S^2\times S^2]\rangle. \end{align*} Expanding the square gives \begin{align*} (xa+yb)^2=x^2a^2+2xyab+y^2b^2. \end{align*} Using $a^2=0$ and $b^2=0$, this becomes \begin{align*} (xa+yb)^2=2xyab. \end{align*} Therefore \begin{align*} q(xa+yb)=\langle 2xyab,[S^2\times S^2]\rangle=2xy. \end{align*} In the basis $(a,b)$, the associated [bilinear form](/page/Bilinear%20Form) has matrix with diagonal entries $0,0$ and off-diagonal entries $1,1$. The vector $a+b$ has eigenvalue $1$, and the vector $a-b$ has eigenvalue $-1$, so \begin{align*} \sigma(S^2\times S^2)=1-1=0. \end{align*} For $\mathbb{CP}^2\#(-\mathbb{CP}^2)$, the intersection form is the orthogonal sum of the forms on the two summands. The generator of $H^2(\mathbb{CP}^2;\mathbb Z)$ has square $1$, while reversing orientation changes the square to $-1$ on the second summand. Hence, in a basis $(u,v)$, \begin{align*} q(xu+yv)=x^2-y^2. \end{align*} The corresponding bilinear form has diagonal entries $1,-1$ and off-diagonal entries $0,0$, so it has one positive eigenvalue and one negative eigenvalue. Therefore \begin{align*} \sigma(\mathbb{CP}^2\#(-\mathbb{CP}^2))=1-1=0. \end{align*} By the four-dimensional case of the *Hirzebruch Signature Theorem*, \begin{align*} \left\langle p_1(TM),[M]\right\rangle=3\sigma(M). \end{align*} Substituting $\sigma(S^2\times S^2)=0$ gives \begin{align*} \left\langle p_1(T(S^2\times S^2)),[S^2\times S^2]\right\rangle=3\cdot 0=0. \end{align*} Substituting $\sigma(\mathbb{CP}^2\#(-\mathbb{CP}^2))=0$ gives \begin{align*} \left\langle p_1(T(\mathbb{CP}^2\#(-\mathbb{CP}^2))),[\mathbb{CP}^2\#(-\mathbb{CP}^2)]\right\rangle=3\cdot 0=0. \end{align*} The two intersection forms are nevertheless not equivalent over $\mathbb Z$. The hyperbolic form $q(x,y)=2xy$ is even because $2xy$ is divisible by $2$ for every pair of integers $x,y$, while the diagonal form $q(x,y)=x^2-y^2$ is odd because $q(1,0)=1$. Thus these manifolds have the same Euler characteristic, signature, and first Pontryagin number, but their integral intersection forms record information that those characteristic numbers do not see. [/example] Characteristic numbers therefore sit between cohomology classes and classification theorems. They are computable, stable under natural equivalences, and powerful enough to detect many geometric distinctions, but they compress the full cohomology ring and bundle data into finitely many evaluations. # 7. Transgression And Secondary Forms This chapter studies what remains after the main Chern-Weil construction has passed to cohomology. Chapters 1 and 2 showed that an invariant polynomial applied to curvature gives a closed differential form whose cohomology class is independent of the connection. Here we keep track of the actual difference between two representatives, and the answer is a differential form of one lower degree: a transgression form. The main examples are Chern-Simons forms, which turn characteristic classes into secondary geometric quantities depending on a chosen connection or on a path of connections. ## Difference Forms For Two Connections The Chern-Weil theorem says that $P(F_A)$ and $P(F_{A'})$ define the same de Rham cohomology class when $A$ and $A'$ are connections on the same principal bundle. The problem in this section is to identify a canonical primitive for their difference, at least after choosing a path joining the two connections. Let $P \to M$ be a principal $G$-bundle, let $\mathfrak g$ be the Lie algebra of $G$, and let $P_0 \in I^k(G)$ be a homogeneous Ad-invariant polynomial of degree $k$, using the notation of Chapter 2. If $A$ is a connection form on the principal bundle, its curvature is denoted $F_A$. [definition: Chern-Weil Form] Let $P \in I^k(\mathfrak g)$ and let $A$ be a connection on a principal $G$-bundle $\nu \to M$. The Chern-Weil form associated to $P$ and $A$ is the differential form \begin{align*} P(F_A) \in \Omega^{2k}(M) \end{align*} obtained by applying the symmetric $k$-linear polarisation of $P$ to $F_A,\dots,F_A$. [/definition] This form is closed by the Bianchi identity and Ad-invariance. The point now is not just that its cohomology class is independent of $A$, but that varying $A$ produces a derivative which is already exact. [quotetheorem:9790] [citeproof:9790] This theorem is the local calculation behind connection-independence in de Rham cohomology, but each hypothesis is doing real work. Smoothness of $A_t$ is needed because the formula differentiates both the connection and its curvature; a merely continuous path of connections has no well-defined $\dot A_t$ and hence no candidate transgression integrand. Ad-invariance of $P$ is also essential: without it, the terms produced by moving $d_{A_t}$ past $P$ do not collapse to an ordinary exterior derivative, so $P(F_{A_t})$ need not be closed or connection-independent. Finally, the Bianchi identity is the mechanism that removes the curvature-derivative terms, and the integrated transgression formula below is precisely the global form of this infinitesimal cancellation. [example: Trace Polynomial Variation] Let $A_t$ be the connection matrix of $\nabla^t$ in a local frame, so \begin{align*} F_t=dA_t+A_t\wedge A_t \end{align*} and $\dot A_t=\frac{dA_t}{dt}$. Differentiating the curvature entry by entry gives \begin{align*} \dot F_t=d\dot A_t+\dot A_t\wedge A_t+A_t\wedge \dot A_t. \end{align*} This is the covariant exterior derivative $D_t\dot A_t$, where $D_t\eta=d\eta+A_t\wedge\eta-(-1)^{|\eta|}\eta\wedge A_t$. Now compute the derivative of the trace polynomial. Since $F_t$ and $\dot F_t$ are both matrix-valued two-forms, cyclically moving factors inside the trace introduces no sign, so \begin{align*} \frac{d}{dt}\operatorname{tr}(F_t^k)=\sum_{j=0}^{k-1}\operatorname{tr}(F_t^j\wedge \dot F_t\wedge F_t^{k-1-j})=k\operatorname{tr}(\dot F_t\wedge F_t^{k-1}). \end{align*} Substituting $\dot F_t=D_t\dot A_t$ gives \begin{align*} \frac{d}{dt}\operatorname{tr}(F_t^k)=k\operatorname{tr}(D_t\dot A_t\wedge F_t^{k-1}). \end{align*} The Bianchi identity is $D_tF_t=0$, hence $D_t(F_t^{k-1})=0$ by the graded Leibniz rule. Therefore \begin{align*} D_t(\dot A_t\wedge F_t^{k-1})=D_t\dot A_t\wedge F_t^{k-1}. \end{align*} Taking the trace kills the connection commutator terms in $D_t$, so \begin{align*} d\operatorname{tr}(\dot A_t\wedge F_t^{k-1})=\operatorname{tr}(D_t\dot A_t\wedge F_t^{k-1}). \end{align*} Combining the last two displayed identities yields \begin{align*} \frac{d}{dt}\operatorname{tr}(F_t^k)=k\,d\operatorname{tr}(\dot A_t\wedge F_t^{k-1}). \end{align*} Thus, along the path of connections, the infinitesimal change of the $2k$-form representative is exact, with primitive built from the variation $\dot A_t$ and the curvature $F_t$. [/example] The trace example is the model for the general formula, but it also shows why the construction needs the path rather than only the two endpoints. The algebraic invariant polynomial supplies the characteristic form in degree $2k$, while the path derivative supplies the missing one-form input needed to build something in degree $2k-1$. Integrating that infinitesimal correction is the mechanism that turns equality of cohomology classes into an explicit primitive for the difference of forms. ## Transgression Along Paths Of Connections The next question is how to package the integrated variation into a form depending on the endpoints and the chosen path. This construction is called transgression because it turns a relation between closed forms in degree $2k$ into a form in degree $2k-1$ whose exterior derivative records their difference. [definition: Transgression Form Along A Path] Let $P \in I^k(\mathfrak g)$ and let $A_t$, $0\le t\le 1$, be a smooth path of connections on a principal $G$-bundle $\nu \to M$. The transgression form associated to the path is \begin{align*} T_P(A_t) := k\int_0^1 P(\dot A_t,F_{A_t},\dots,F_{A_t})\,dt \in \Omega^{2k-1}(M). \end{align*} [/definition] The definition uses the polarised form associated to $P$, with one input equal to $\dot A_t$ and the other $k-1$ inputs equal to the curvature. The required theorem is that this candidate is not merely natural-looking: its exterior derivative is the exact endpoint difference that measures connection-independence. [quotetheorem:9791] [citeproof:9791] The formula explains why the de Rham class of $P(F_A)$ is independent of the connection: any two representatives joined by a smooth path differ by an exact form. Each hypothesis is needed for that conclusion in the displayed form. If the path is not smooth, for instance a path that is only continuous and has a corner or worse as a map into the [affine space](/page/Affine%20Space) of connections, the derivative $\dot A_t$ is not available at every parameter value, so the integral defining $T_P(A_t)$ is not the object stated in the theorem. If the two endpoint connections live on different principal bundles, $P(F_{A_1})-P(F_{A_0})$ is not a difference of forms built from a common adjoint bundle unless an isomorphism has first identified the bundles and their curvature data. If $P$ is not Ad-invariant, the infinitesimal variation contains extra commutator terms; for a matrix group, a non-cyclic polynomial in matrix entries is changed by conjugating the curvature, so the corresponding form is not a characteristic form with connection-independent cohomology class. The theorem also does not assert that the primitive is canonical: changing the path by a loop of connections can change $T_P(A_t)$ by a closed $(2k-1)$-form with nonzero de Rham class, and this residual path dependence is the source of the secondary invariants used below. [remark: Dependence On The Path] Different paths from $A_0$ to $A_1$ can produce transgression forms differing by a closed $(2k-1)$-form, and sometimes by a form with nonzero cohomology class. Thus transgression is a secondary construction: the primary characteristic class is fixed in cohomology, while the transgression remembers auxiliary geometric data. [/remark] This dependence is useful rather than accidental. In applications, the chosen path is often the affine path between two connections, and then the transgression form becomes a concrete formula in the two connection forms. [example: Affine Path Of Connections] Let $A_0$ and $A_1$ be connections on the same principal bundle, and write $\alpha=A_1-A_0$, so $\alpha\in\Omega^1(M;\operatorname{ad}\nu)$. For the affine path $A_t=A_0+t\alpha$, differentiation in $t$ gives $\dot A_t=\alpha$. Using the curvature convention $F_A=dA+\frac12[A\wedge A]$, bilinearity of the bracket gives \begin{align*} F_{A_t}=d(A_0+t\alpha)+\frac12[A_0+t\alpha\wedge A_0+t\alpha]. \end{align*} Expanding the two terms separately, \begin{align*} d(A_0+t\alpha)=dA_0+t\,d\alpha. \end{align*} Also, \begin{align*} [A_0+t\alpha\wedge A_0+t\alpha]=[A_0\wedge A_0]+t[A_0\wedge\alpha]+t[\alpha\wedge A_0]+t^2[\alpha\wedge\alpha]. \end{align*} Since $A_0$ and $\alpha$ are both one-forms, the two mixed bracket terms combine into $2[A_0\wedge\alpha]$ under the usual graded bracket convention. Therefore \begin{align*} F_{A_t}=dA_0+\frac12[A_0\wedge A_0]+t\bigl(d\alpha+[A_0\wedge\alpha]\bigr)+\frac{t^2}{2}[\alpha\wedge\alpha]. \end{align*} The first two terms are $F_{A_0}$, and the expression in parentheses is $d_{A_0}\alpha$, so \begin{align*} F_{A_t}=F_{A_0}+t\,d_{A_0}\alpha+\frac{t^2}{2}[\alpha\wedge\alpha]. \end{align*} Substituting $\dot A_t=\alpha$ into the path transgression definition gives \begin{align*} T_P(A_0,A_1)=k\int_0^1P(\alpha,F_{A_t},\dots,F_{A_t})\,dt. \end{align*} By the [infinitesimal Chern-Weil variation formula](/theorems/9790), \begin{align*} k\,dP(\alpha,F_{A_t},\dots,F_{A_t})=\frac{d}{dt}P(F_{A_t}). \end{align*} Exterior differentiation commutes with integration over the parameter $t$, hence \begin{align*} dT_P(A_0,A_1)=\int_0^1\frac{d}{dt}P(F_{A_t})\,dt. \end{align*} The [fundamental theorem of calculus](/theorems/632) then gives \begin{align*} dT_P(A_0,A_1)=P(F_{A_1})-P(F_{A_0}). \end{align*} Thus the affine path turns the endpoint difference of Chern-Weil forms into an explicit primitive built from $\alpha=A_1-A_0$ and the curvature along the path. [/example] The affine formula is especially compact when one endpoint is the zero connection in a chosen local product chart. That specialisation leads to the classical Chern-Simons form. ## Chern-Simons Forms Chern-Simons theory asks for a differential form whose exterior derivative is a prescribed Chern-Weil form. The construction is not intrinsic on every bundle without choices, but once a connection is compared with a reference connection, or once a local product chart is fixed, the transgression formula gives an explicit secondary characteristic form. [definition: Chern-Simons Form Relative To A Reference Connection] Let $P \in I^k(\mathfrak g)$, and let $A_0$ and $A_1$ be two connections on a principal $G$-bundle. The Chern-Simons form of $A_1$ relative to $A_0$ is \begin{align*} \operatorname{CS}_P(A_0,A_1):=T_P(A_t)\in\Omega^{2k-1}(M), \end{align*} where $A_t$ is the affine path $A_t=A_0+t(A_1-A_0)$. [/definition] The reference connection fixes the ambiguity in choosing a path, but the form is useful only if its exterior derivative measures something intrinsic about the endpoints. The obstruction is that Chern-Weil forms are closed rather than usually exact, so a primitive can exist only for the difference between two representatives on the same bundle. The differential identity below is the mechanism that turns the relative transgression form into a controlled primitive. [quotetheorem:9792] [citeproof:9792] The theorem is the core identity of the chapter: a characteristic form becomes exact after choosing a reference connection and passing to the corresponding secondary form. The reference connection is not decoration; if $A_0$ is replaced by another connection $A_0'$, the resulting Chern-Simons form changes by the transgression between $A_0$ and $A_0'$, and this change can carry nonzero closed components. The same-principal-bundle hypothesis is also essential: without a common bundle, the difference $A_1-A_0$ is not an $\operatorname{ad}\nu$-valued one-form, so the affine path $A_0+t(A_1-A_0)$ is not defined. The invariant-polynomial hypothesis cannot be weakened to an arbitrary polynomial expression in curvature, since conjugating a connection by a gauge transformation would then change the polynomial by commutator terms rather than preserve the characteristic form. The theorem therefore does not attach an intrinsic form only to $A_1$ on an arbitrary bundle; it attaches a relative form to the pair $(A_0,A_1)$ and to the affine-path convention. The more general path-dependent construction above records exactly what changes when a different homotopy of connections is chosen. [example: U One Chern-Simons Three-Form] Let $G=U(1)$ and let $A\in\Omega^1(M;i\mathbb R)$ be a connection one-form on a product $U(1)$-bundle. The Lie algebra is abelian, so the bracket term in the curvature vanishes and \begin{align*} F_A=dA. \end{align*} Relative to the zero connection, the affine path is $A_t=tA$, so $\dot A_t=A$ and \begin{align*} F_{A_t}=d(tA)=t\,dA. \end{align*} For the degree two invariant polynomial $P(X)=cX^2$, the transgression formula gives \begin{align*} \operatorname{CS}_P(A)=2\int_0^1 P(A,F_{A_t})\,dt. \end{align*} Since the polarised form is $P(X,Y)=c\,X\wedge Y$ in the abelian case, \begin{align*} \operatorname{CS}_P(A)=2c\int_0^1 A\wedge t\,dA\,dt. \end{align*} The form $A\wedge dA$ is independent of $t$, hence \begin{align*} 2c\int_0^1 A\wedge t\,dA\,dt=2c\left(\int_0^1 t\,dt\right)A\wedge dA. \end{align*} Because $\int_0^1 t\,dt=\frac12$, this becomes \begin{align*} \operatorname{CS}_P(A)=c\,A\wedge dA. \end{align*} Taking the exterior derivative and using the graded Leibniz rule with $\deg A=1$, \begin{align*} d\operatorname{CS}_P(A)=c\,d(A\wedge dA)=c\bigl(dA\wedge dA-A\wedge d^2A\bigr). \end{align*} Since $d^2A=0$, \begin{align*} d\operatorname{CS}_P(A)=c\,dA\wedge dA. \end{align*} Finally, $F_A=dA$, so \begin{align*} P(F_A)=c\,F_A\wedge F_A=c\,dA\wedge dA. \end{align*} Thus $d\operatorname{CS}_P(A)=P(F_A)$, and on a three-manifold the integral of $A\wedge dA$ is the abelian Chern-Simons functional once the normalising constant $c$ is fixed by the characteristic class convention. [/example] The abelian example has no cubic term because the Lie bracket vanishes. Nonabelian groups introduce an additional interaction term coming from the $A\wedge A$ part of the curvature. [example: Matrix Chern-Simons Three-Form] Let $G\subset GL(n,\mathbb C)$ be a matrix Lie group, let $A\in\Omega^1(M;\mathfrak g)$, and take $P(X)=\operatorname{tr}(X^2)$. Use the matrix curvature convention \begin{align*} F_A=dA+A\wedge A, \end{align*} where multiplication means wedge product of forms together with matrix multiplication. Along the affine path $A_t=tA$, one has $dA_t=t\,dA$ and \begin{align*} A_t\wedge A_t=(tA)\wedge(tA)=t^2A\wedge A. \end{align*} Thus \begin{align*} F_{A_t}=dA_t+A_t\wedge A_t=t\,dA+t^2A\wedge A. \end{align*} For $P(X)=\operatorname{tr}(X^2)$, the polarised form is $P(X,Y)=\operatorname{tr}(X\wedge Y)$, so the transgression integral from the zero connection to $A$ gives \begin{align*} \operatorname{CS}_P(A)=2\int_0^1\operatorname{tr}(A\wedge F_{A_t})\,dt. \end{align*} Substituting the displayed expression for $F_{A_t}$ gives \begin{align*} \operatorname{CS}_P(A)=2\int_0^1\operatorname{tr}(A\wedge(t\,dA+t^2A\wedge A))\,dt. \end{align*} By linearity of wedge product, trace, and integration, \begin{align*} \operatorname{CS}_P(A)=2\left(\int_0^1t\,dt\right)\operatorname{tr}(A\wedge dA)+2\left(\int_0^1t^2\,dt\right)\operatorname{tr}(A\wedge A\wedge A). \end{align*} Since $\int_0^1t\,dt=\frac12$ and $\int_0^1t^2\,dt=\frac13$, this becomes \begin{align*} \operatorname{CS}_P(A)=\operatorname{tr}(A\wedge dA)+\frac23\operatorname{tr}(A\wedge A\wedge A). \end{align*} Equivalently, \begin{align*} \operatorname{CS}_P(A)=\operatorname{tr}\left(A\wedge dA+\frac23A\wedge A\wedge A\right). \end{align*} Now compute its exterior derivative. Since $\deg A=1$ and $d^2A=0$, the graded Leibniz rule gives \begin{align*} d\operatorname{tr}(A\wedge dA)=\operatorname{tr}(dA\wedge dA-A\wedge d^2A)=\operatorname{tr}(dA\wedge dA). \end{align*} For the cubic term, \begin{align*} d(A\wedge A\wedge A)=dA\wedge A\wedge A-A\wedge dA\wedge A+A\wedge A\wedge dA. \end{align*} Inside the trace, cyclically moving a form of degree $p$ past forms of total degree $q$ contributes the sign $(-1)^{pq}$. Hence \begin{align*} \operatorname{tr}(A\wedge dA\wedge A)=-\operatorname{tr}(dA\wedge A\wedge A). \end{align*} Also, \begin{align*} \operatorname{tr}(A\wedge A\wedge dA)=\operatorname{tr}(dA\wedge A\wedge A), \end{align*} because moving $dA$ of degree $2$ past the two one-forms contributes sign $(-1)^4=1$. Therefore \begin{align*} d\operatorname{tr}(A\wedge A\wedge A)=3\operatorname{tr}(dA\wedge A\wedge A). \end{align*} Combining the two derivative computations, \begin{align*} d\operatorname{CS}_P(A)=\operatorname{tr}(dA\wedge dA)+2\operatorname{tr}(dA\wedge A\wedge A). \end{align*} Finally, \begin{align*} F_A\wedge F_A=(dA+A\wedge A)\wedge(dA+A\wedge A). \end{align*} Expanding the product gives \begin{align*} F_A\wedge F_A=dA\wedge dA+dA\wedge A\wedge A+A\wedge A\wedge dA+A\wedge A\wedge A\wedge A. \end{align*} Taking traces, the third term satisfies \begin{align*} \operatorname{tr}(A\wedge A\wedge dA)=\operatorname{tr}(dA\wedge A\wedge A). \end{align*} The quartic term vanishes under the trace, since cyclically moving the first $A$ past the other three one-forms gives \begin{align*} \operatorname{tr}(A\wedge A\wedge A\wedge A)=-\operatorname{tr}(A\wedge A\wedge A\wedge A). \end{align*} Thus \begin{align*} \operatorname{tr}(F_A\wedge F_A)=\operatorname{tr}(dA\wedge dA)+2\operatorname{tr}(dA\wedge A\wedge A). \end{align*} Comparing with the previous expression gives \begin{align*} d\operatorname{CS}_P(A)=\operatorname{tr}(F_A\wedge F_A). \end{align*} So the three-form $\operatorname{tr}(A\wedge dA+\frac23A\wedge A\wedge A)$ is the explicit Chern-Simons primitive for the degree-two trace Chern-Weil form. [/example] This three-dimensional formula is the one most often called the Chern-Simons form. Higher degree invariant polynomials produce higher odd-degree secondary forms by the same transgression integral. ## Gauge Transformations And Winding Terms A final question is how Chern-Simons forms change under gauge transformations. Primary Chern-Weil forms are gauge-invariant, but secondary forms can change by exact terms and by topological terms measuring the homotopy class of the gauge transformation. [definition: Gauge Transform Of A Connection] Let $\nu\to M$ be a principal $G$-bundle and let $g:\nu\to\nu$ be a gauge transformation covering $\operatorname{id}_M$. In a local product chart, $g$ is represented by a smooth map $g:M\to G$, and it defines a map from local connection one-forms to local connection one-forms \begin{align*} \mathcal G_g:\Omega^1(M;\mathfrak g)&\longrightarrow \Omega^1(M;\mathfrak g). \end{align*} Its value on $A$ is \begin{align*} \mathcal G_g(A)=A^g:=g^{-1}Ag+g^{-1}dg. \end{align*} [/definition] The curvature transforms by conjugation, so Ad-invariant polynomials satisfy $P(F_{A^g})=P(F_A)$. Chern-Simons forms therefore have exterior derivatives that agree after gauge transformation, and their difference is closed; the theorem below identifies the exact part and the topological remainder in degree three. [quotetheorem:9793] [citeproof:9793] This result explains why the Chern-Simons functional is not an ordinary gauge-invariant functional on all connections. The matrix-group hypothesis is what makes the displayed local calculation meaningful: it uses ordinary matrix multiplication to form $A\wedge A$ and $\theta\wedge\theta$, and a general Lie group requires a representation or an invariant bilinear form before the trace expression has any meaning. The trace-polynomial hypothesis is equally specific. If the degree-two polynomial is replaced by a non-invariant bilinear expression, cyclic cancellation fails under conjugation, so the transformed formula acquires extra commutator terms rather than the stated exact term plus cubic remainder. The curvature convention is also part of the statement: using $F_A=dA+\frac12[A\wedge A]$ while keeping the displayed three-form changes the coefficient of the cubic term, so the constants in the gauge law no longer match. The degree-three restriction matters because higher Chern-Simons forms have longer gauge transformation formulae with several mixed terms rather than only one exact term and one cubic winding form. A concrete failure of gauge invariance occurs already when $A=0$ on the product bundle over $S^3$ and $g:S^3\to SU(2)$ has degree $1$: then $A^g=\theta$ and the integral of the difference is a nonzero multiple of $\int_{S^3}\operatorname{tr}(\theta\wedge\theta\wedge\theta)$. With suitable normalisation, the exponential of the functional can still be gauge-invariant because this winding term has integral periods. [example: SU Two Maurer-Cartan Form And Winding Number] Let $\Theta=h^{-1}dh$ be the Maurer-Cartan form on $SU(2)$, so for a map $g:S^3\to SU(2)$ one has $\theta=g^{-1}dg=g^*\Theta$. Choose the standard generators $\tau_j=i\sigma_j$ of $\mathfrak{su}(2)$, where $\sigma_j$ are the Pauli matrices, and write \begin{align*} \Theta=\tau_1\alpha_1+\tau_2\alpha_2+\tau_3\alpha_3. \end{align*} They satisfy $\operatorname{tr}(\tau_i\tau_j)=-2\delta_{ij}$ and \begin{align*} \operatorname{tr}(\tau_i\tau_j\tau_k)=2\varepsilon_{ijk}. \end{align*} Therefore \begin{align*} \operatorname{tr}(\Theta\wedge\Theta\wedge\Theta)=\sum_{i,j,k}\operatorname{tr}(\tau_i\tau_j\tau_k)\,\alpha_i\wedge\alpha_j\wedge\alpha_k. \end{align*} Substituting $\operatorname{tr}(\tau_i\tau_j\tau_k)=2\varepsilon_{ijk}$ gives \begin{align*} \operatorname{tr}(\Theta\wedge\Theta\wedge\Theta)=2\sum_{i,j,k}\varepsilon_{ijk}\,\alpha_i\wedge\alpha_j\wedge\alpha_k. \end{align*} Each nonzero term in the sum is $\alpha_1\wedge\alpha_2\wedge\alpha_3$, because the sign of the permutation in $\varepsilon_{ijk}$ is cancelled by the same sign from reordering the wedge product. There are $6$ such permutations, hence \begin{align*} \operatorname{tr}(\Theta\wedge\Theta\wedge\Theta)=12\,\alpha_1\wedge\alpha_2\wedge\alpha_3. \end{align*} With the orientation for which $\alpha_1\wedge\alpha_2\wedge\alpha_3$ is positive, the Haar-volume normalisation on $SU(2)\cong S^3$ gives \begin{align*} \int_{SU(2)}\alpha_1\wedge\alpha_2\wedge\alpha_3=2\pi^2. \end{align*} Thus for the identity map, \begin{align*} \frac{1}{24\pi^2}\int_{SU(2)}\operatorname{tr}(\Theta\wedge\Theta\wedge\Theta)=\frac{1}{24\pi^2}\cdot 12\cdot 2\pi^2=1. \end{align*} For a general smooth $g:S^3\to SU(2)$, \begin{align*} \frac{1}{24\pi^2}\int_{S^3}\operatorname{tr}(\theta\wedge\theta\wedge\theta)=\int_{S^3}g^*\left(\frac{1}{24\pi^2}\operatorname{tr}(\Theta\wedge\Theta\wedge\Theta)\right). \end{align*} Since the displayed normalised $3$-form integrates to $1$ on $SU(2)\cong S^3$, integration of its pullback over $S^3$ gives the degree of $g$. Hence \begin{align*} \operatorname{wind}(g)=\frac{1}{24\pi^2}\int_{S^3}\operatorname{tr}(\theta\wedge\theta\wedge\theta)=\deg(g)\in\mathbb Z. \end{align*} This is exactly the topological term in the gauge transformation law: it depends only on the homotopy class of $g$, not on the connection being transformed. [/example] The appearance of the [winding number](/page/Winding%20Number) is the first sign that secondary characteristic forms connect curvature calculations with global topology beyond de Rham representatives. Later applications use these forms to detect geometric structures whose primary characteristic classes vanish, or to refine characteristic classes in the presence of boundary data. # 8. Chern-Simons Invariants After Chapter 7 identified transgression forms as primitives for differences of Chern-Weil representatives, this chapter studies what remains when the curvature term itself vanishes, so that the primary characteristic form is zero but the connection can still carry global information. The main prerequisites are principal bundles, connection and curvature forms, invariant polynomials, de Rham cohomology, holonomy, and the basic Chern-Weil construction from the preceding chapters. The resulting Chern-Simons invariants are secondary: they are built from transgression forms, depend on a choice of connection, and become well-defined modulo periods in the cases most useful for topology and three-dimensional geometry. ## Flat Connections and Secondary Classes The guiding question is this: if a characteristic form $P(F_A)$ vanishes because $A$ is flat, can the connection still determine a non-zero invariant? Transgression gives a way to compare a connection with a reference connection, and on a closed manifold of one dimension lower than the characteristic form, the transgression form itself can be integrated. [definition: Flat Principal Connection] Let $P \to M$ be a principal $G$-bundle with principal connection form $A \in \Omega^1(P;\mathfrak g)$. Let $F_A^{\mathrm{prin}} \in \Omega^2(P;\mathfrak g)$ be its curvature form, and let $F_A \in \Omega^2(M;\operatorname{ad} P)$ be the corresponding adjoint-valued curvature form on the base. The connection $A$ is flat if $F_A = 0$. [/definition] Flatness is a first-order condition on the parallel transport of the connection. It implies that local parallel frames exist, but it does not force the global holonomy representation to be trivial. The missing information is global: parallel transport around loops may still record a non-zero holonomy class. This is the first sign that curvature forms alone cannot capture all of the geometry of a connection. [example: Flat Line Bundles Over The Circle] Let $G=U(1)$ and write the product bundle over $S^1=\mathbb R/2\pi\mathbb Z$ in the coordinate $\theta$. For the connection \begin{align*} A=i\alpha\,d\theta \end{align*} the group $U(1)$ is abelian, so the curvature is $F_A=dA$. Since $\alpha$ is constant, \begin{align*} F_A=d(i\alpha\,d\theta)=i\alpha\,d(d\theta)=0. \end{align*} Thus $A$ is flat. With the convention that the holonomy of a $U(1)$-connection around an oriented loop is $\exp\!\left(\int A\right)$, the positively oriented generator of $\pi_1(S^1)$ gives \begin{align*} \operatorname{Hol}_{S^1}(A)=\exp\!\left(\int_0^{2\pi} i\alpha\,d\theta\right)=\exp(2\pi i\alpha). \end{align*} If $n\in\mathbb Z$, the gauge transformation $u_n(\theta)=e^{in\theta}$ satisfies \begin{align*} u_n^{-1}du_n=e^{-in\theta}\,d(e^{in\theta})=e^{-in\theta}(in e^{in\theta}\,d\theta)=in\,d\theta. \end{align*} Under the usual abelian gauge action this shifts $i\alpha\,d\theta$ by an integral multiple of $i\,d\theta$, so $\alpha$ changes by an integer. Conversely, \begin{align*} \exp(2\pi i\alpha)=\exp(2\pi i\beta) \end{align*} holds exactly when $\alpha-\beta\in\mathbb Z$. Hence the flat connections in this family are distinguished up to gauge equivalence by the class of $\alpha$ modulo integers, even though all of them have zero curvature. [/example] The example shows the basic theme: curvature can vanish while holonomy remains visible. Chern-Weil forms miss this information because they are built directly from curvature. To turn the observation into an invariant, we compare two connections by a transgression form whose exterior derivative measures the difference between the two Chern-Weil representatives. The secondary form is therefore not a new primary characteristic class; it is the primitive left behind after a primary class has been represented in two different ways. [definition: Chern-Simons Form] Let $P \in (S^k\mathfrak g^*)^G$ be an invariant symmetric $k$-linear polynomial, viewed as a map \begin{align*} P:\mathfrak g^k \to \mathbb R. \end{align*} Let $\operatorname{Conn}(E)$ denote the space of connections on a principal $G$-bundle $E\to M$. The Chern-Simons transgression associated to $P$ is the assignment \begin{align*} \operatorname{CS}_P:\operatorname{Conn}(E)\times\operatorname{Conn}(E)\to \Omega^{2k-1}(M). \end{align*} For $A_0,A_1\in \operatorname{Conn}(E)$, let $A_t=A_0+t(A_1-A_0)$ and let $F_t$ be its curvature. The value of the transgression form is \begin{align*} \operatorname{CS}_P(A_0,A_1)=k\int_0^1 P(A_1-A_0,F_t,\dots,F_t)\,dt \in \Omega^{2k-1}(M). \end{align*} [/definition] Here $A_1-A_0$ is interpreted using the affine structure on $\operatorname{Conn}(E)$, so it is an element of $\Omega^1(M;\operatorname{ad}E)$. The polynomial $P$ is extended to adjoint-valued differential forms by wedging the form parts and applying the symmetric multilinear map to the Lie algebra parts, with the usual graded signs. With this convention the formula defines a candidate primitive, but the construction is useful only if its derivative is exactly the difference of characteristic forms. The next theorem supplies that identity and is the calculation that makes Chern-Simons theory a secondary version of Chern-Weil theory. [quotetheorem:9792] [citeproof:9792] The hypotheses in the transgression formula are doing real work. The invariant polynomial condition is what allows the covariant exterior derivative to pass through $P$ without producing extra commutator terms; for a non-invariant multilinear form the displayed derivative would acquire correction terms and would no longer be the difference of Chern-Weil representatives. The path $A_t=A_0+t(A_1-A_0)$ also matters only through transgression: the theorem does not say that $\operatorname{CS}_P(A_0,A_1)$ is itself canonical as a form, only that its exterior derivative is canonical. This is why the next step is to integrate in the special odd-dimensional situation, where exact changes disappear on closed manifolds but global periods can remain. When $A_0$ is fixed, the form $\operatorname{CS}_P(A_0,A)$ is a primitive for $P(F_A)-P(F_{A_0})$. If both endpoint connections are flat, the Chern-Simons form is closed, so on a closed manifold its integral becomes the natural numerical object to study. [definition: Chern-Simons Invariant Of A Flat Connection] Let $E \to M$ be a principal $G$-bundle over a closed oriented $(2k-1)$-manifold, let $A_0$ be a fixed reference connection on $E$, and let $\mathcal A_{\mathrm{flat}}(E)$ be the set of flat connections on $E$. The relative Chern-Simons invariant associated to $P$ and $A_0$ is the function \begin{align*} \operatorname{cs}_P(-;A_0):\mathcal A_{\mathrm{flat}}(E) \to \mathbb R. \end{align*} For $A\in \mathcal A_{\mathrm{flat}}(E)$, its value is \begin{align*} \operatorname{cs}_P(A;A_0)=\int_M \operatorname{CS}_P(A_0,A). \end{align*} [/definition] This real number is a relative invariant: before adding integrality data it depends on the fixed reference connection and on the transgression convention. It is still useful because its change under controlled modifications can be measured by characteristic periods. The next section explains the precise form of that control and separates two issues that should not be conflated: gauge transformation of a flat connection, and the independent choice of reference data used to define the relative integral. [example: The Basic Three-Dimensional Formula] For a compact matrix Lie group, take \begin{align*} P(X,Y)=c\operatorname{tr}(XY) \end{align*} with $c$ chosen so that the corresponding degree-four Chern-Weil class is integral. For a matrix-valued curvature form $F_A$, the associated Chern-Weil form is \begin{align*} P(F_A,F_A)=c\operatorname{tr}(F_A\wedge F_A). \end{align*} On the product bundle, compare $A$ with the product connection $0$ by the path $A_t=tA$. Its curvature is \begin{align*} F_t=d(tA)+(tA)\wedge(tA)=t\,dA+t^2A\wedge A. \end{align*} For $k=2$, the transgression formula gives \begin{align*} \operatorname{CS}(A)=2\int_0^1 P(A,F_t)\,dt. \end{align*} Substituting $P(X,Y)=c\operatorname{tr}(XY)$ and the displayed expression for $F_t$ gives \begin{align*} \operatorname{CS}(A)=2c\int_0^1 \operatorname{tr}\bigl(A\wedge(t\,dA+t^2A\wedge A)\bigr)\,dt. \end{align*} By bilinearity of the wedge product and [linearity of the trace](/theorems/7814), \begin{align*} \operatorname{tr}\bigl(A\wedge(t\,dA+t^2A\wedge A)\bigr)=t\,\operatorname{tr}(A\wedge dA)+t^2\,\operatorname{tr}(A\wedge A\wedge A). \end{align*} Therefore \begin{align*} \operatorname{CS}(A)=2c\left(\int_0^1 t\,dt\right)\operatorname{tr}(A\wedge dA)+2c\left(\int_0^1 t^2\,dt\right)\operatorname{tr}(A\wedge A\wedge A). \end{align*} Since $\int_0^1t\,dt=1/2$ and $\int_0^1t^2\,dt=1/3$, this becomes \begin{align*} \operatorname{CS}(A)=c\operatorname{tr}\left(A\wedge dA+\frac{2}{3}A\wedge A\wedge A\right). \end{align*} If $A$ is flat, then \begin{align*} F_A=dA+A\wedge A=0. \end{align*} Hence \begin{align*} dA=-A\wedge A. \end{align*} Substituting this into the first term gives \begin{align*} A\wedge dA=A\wedge(-A\wedge A)=-A\wedge A\wedge A. \end{align*} Thus, for a flat connection, \begin{align*} \operatorname{CS}(A)=c\operatorname{tr}\left(-A\wedge A\wedge A+\frac{2}{3}A\wedge A\wedge A\right). \end{align*} Combining the scalar coefficients $-1+2/3=-1/3$ yields \begin{align*} \operatorname{CS}(A)=-\frac{c}{3}\operatorname{tr}(A\wedge A\wedge A). \end{align*} So in dimension three, the Chern-Simons form for a flat matrix connection is a cubic trace expression in the connection one-form, even though the curvature form itself is zero. [/example] ## Gauge Equivalence and Integral Periods A secondary invariant is useful only if its dependence on gauge transformations is controlled. The problem is that a gauge transformation need not preserve the Chern-Simons integral as a real number; instead it shifts it by a period determined by the universal characteristic class. [definition: Gauge Equivalent Connections] Let $E \to M$ be a principal $G$-bundle. Two connections $A$ and $A'$ are gauge equivalent if there exists a bundle automorphism $u:E\to E$ covering $\operatorname{id}_M$ such that $A'=u^*A$. [/definition] The set of all such bundle automorphisms is the gauge group and is denoted $\mathcal G(E)$. Gauge equivalence preserves curvature up to the adjoint action, so Chern-Weil forms are unchanged for invariant polynomials. Since Chern-Simons forms are primitives rather than closed characteristic forms, the next issue is to compute the residual change in the integral under a gauge transformation. [quotetheorem:9794] [citeproof:9794] The theorem says that the real-valued integral contains an unavoidable integer ambiguity under gauge transformations once the auxiliary data is part of a compatible integral construction. Each hypothesis prevents a specific failure: if $M$ has boundary, Stokes' theorem leaves a boundary correction; if $M$ is not oriented, the integral is not defined as a signed period; if the endpoint connections are not flat, curvature terms survive and the expression is not only a secondary invariant. Integrality is also essential, since rescaling $P$ by an irrational constant would rescale the period lattice and destroy the conclusion modulo $\mathbb Z$. The caveat about reference data is mathematical, not cosmetic: fixing an arbitrary connection $A_0$ and transforming only $A$ can leave real secondary terms depending on $A_0$, so the reduced invariant below is defined only after choosing compatible integral data. [definition: Reduced Chern-Simons Invariant] Under the hypotheses of the compatible gauge-change theorem and for the fixed compatible integral reference datum, the reduced Chern-Simons invariant is the map on gauge equivalence classes of flat connections \begin{align*} \overline{\operatorname{cs}}_P:\mathcal A_{\mathrm{flat}}(E)/\mathcal G(E) \to \mathbb R/\mathbb Z. \end{align*} It is defined by \begin{align*} \overline{\operatorname{cs}}_P([A])=\operatorname{cs}_P(A) \mod \mathbb Z. \end{align*} [/definition] The reduced invariant is independent of the representative of the gauge equivalence class for the chosen compatible datum. Its dependence on auxiliary reference data is a separate issue: arbitrary reference connections can change the relative real integral continuously, while fully reference-free formulations use a differential character, a bounding construction, or explicitly fixed trivialisation data. [remark: Normalisation Matters] The phrase "modulo integers" depends on choosing $P$ so that the corresponding Chern-Weil class is integral. If $P$ is rescaled by a real number, the period lattice is rescaled by the same factor. [/remark] The normalisation remark is not only a convention about coefficients; it is the mechanism that turns winding number into an integer shift. In dimension three this becomes concrete because $S^3$ is itself the compact Lie group $SU(2)$, and gauge transformations $S^3\to SU(2)$ are classified by their degree. The next example is the calibration point: it shows how the standard second Chern class normalisation makes a pure gauge change the unreduced integral by exactly that degree, so that the reduced invariant forgets the choice of representative. [example: Product $SU(2)$-Bundle Over $S^3$] Let $M=S^3$ and consider the product $SU(2)$-bundle. Use the convention that a gauge transformation sends the product connection $0$ to \begin{align*} A^u=u^{-1}du. \end{align*} Write $\omega=u^{-1}du$. Since $d(u^{-1}u)=d1=0$, we have \begin{align*} d(u^{-1})u+u^{-1}du=0. \end{align*} Multiplying on the right by $u^{-1}$ gives \begin{align*} d(u^{-1})=-u^{-1}du\,u^{-1}. \end{align*} Therefore \begin{align*} d\omega=d(u^{-1}du)=d(u^{-1})\wedge du+u^{-1}d^2u. \end{align*} Since $d^2u=0$, this becomes \begin{align*} d\omega=(-u^{-1}du\,u^{-1})\wedge du=-u^{-1}du\wedge u^{-1}du=-\omega\wedge\omega. \end{align*} Thus \begin{align*} F_{\omega}=d\omega+\omega\wedge\omega=-\omega\wedge\omega+\omega\wedge\omega=0, \end{align*} so $\omega$ is flat. Now normalise \begin{align*} P(X,Y)=-\frac{1}{8\pi^2}\operatorname{tr}(XY), \end{align*} so here $c=-1/(8\pi^2)$. For a three-dimensional matrix connection, the Chern-Simons form computed from the path $A_t=tA$ is \begin{align*} \operatorname{CS}(A)=c\operatorname{tr}\left(A\wedge dA+\frac{2}{3}A\wedge A\wedge A\right). \end{align*} Substituting $A=\omega$ and $d\omega=-\omega\wedge\omega$ gives \begin{align*} \operatorname{CS}(\omega)=-\frac{1}{8\pi^2}\operatorname{tr}\left(\omega\wedge(-\omega\wedge\omega)+\frac{2}{3}\omega\wedge\omega\wedge\omega\right). \end{align*} Inside the trace, \begin{align*} \omega\wedge(-\omega\wedge\omega)+\frac{2}{3}\omega\wedge\omega\wedge\omega=-\omega\wedge\omega\wedge\omega+\frac{2}{3}\omega\wedge\omega\wedge\omega=-\frac{1}{3}\omega\wedge\omega\wedge\omega. \end{align*} Hence \begin{align*} \operatorname{CS}(\omega)=\frac{1}{24\pi^2}\operatorname{tr}(\omega\wedge\omega\wedge\omega). \end{align*} With the standard orientation of $SU(2)\cong S^3$, the degree of $u:S^3\to SU(2)$ is computed by \begin{align*} \deg(u)=\frac{1}{24\pi^2}\int_{S^3}\operatorname{tr}\bigl((u^{-1}du)\wedge(u^{-1}du)\wedge(u^{-1}du)\bigr). \end{align*} Therefore \begin{align*} \int_{S^3}\operatorname{CS}(u^{-1}du)=\deg(u). \end{align*} Since the product connection $0$ has zero Chern-Simons integral, the unreduced change from $0$ to the pure gauge $u^{-1}du$ is exactly $\deg(u)$. If the pure gauge form is instead written as $u\,du^{-1}$, the same degree form is computed using the opposite Maurer-Cartan form, which reverses the sign of the displayed integral under this convention. Reversing the sign convention for the second Chern representative also changes $P$ to $-P$, so the Chern-Simons integral is multiplied by $-1$. In either case the change is $\pm\deg(u)\in\mathbb Z$, so all pure gauges represent the same reduced invariant in $\mathbb R/\mathbb Z$. [/example] ## Three-Manifold Examples and Mapping Tori The practical difficulty is that the definition of $\overline{\operatorname{cs}}_P$ is an integral of a transgression form, while flat bundles in topology are usually presented by holonomy representations. The question in dimension three is therefore how to pass from representation-theoretic data, such as a homomorphism $\pi_1(M)\to G$, to a computable number in $\mathbb R/\mathbb Z$. Degree-two invariant polynomials produce three-forms, so this is the first dimension where the secondary invariant can be seen directly. The examples in this section show the main computational inputs: winding number for pure gauges, linking forms for lens spaces, and monodromy data for mapping tori. [definition: Flat Connection From A Representation] Let $M$ be a connected manifold and let $G$ be a Lie group. A representation $\rho:\pi_1(M)\to G$ determines a flat principal $G$-bundle \begin{align*} \widetilde M\times_{\rho}G \to M, \end{align*} where $\pi_1(M)$ acts on $\widetilde M\times G$ by $\gamma\cdot(x,g)=(\gamma x,\rho(\gamma)g)$. The flat connection associated to $\rho$ is the unique connection on $\widetilde M\times_{\rho}G$ whose pullback to $\widetilde M\times G\to \widetilde M$ is the product flat connection. [/definition] The product flat connection is invariant under the displayed action of $\pi_1(M)$, which is why the pulled-back description descends to the quotient bundle. This construction turns the computation of flat connections into a problem about group representations. More precisely, it gives a map from representations modulo conjugacy to isomorphism classes of flat principal $G$-bundles with flat connection, with the usual qualification that disconnected bases and reducible representations may require stabiliser data. The Chern-Simons invariant can then distinguish some components and special points that have the same primary characteristic forms. [example: Lens Space Flat $SU(2)$-Connections] Let $L(p,q)$ have $\pi_1(L(p,q))\cong \mathbb Z/p\mathbb Z$, and let $\gamma$ be a generator. A representation $\rho:\pi_1(L(p,q))\to SU(2)$ must satisfy $\rho(\gamma)^p=1$. Every element of $SU(2)$ is conjugate to a diagonal matrix of the form \begin{align*} \operatorname{diag}(e^{i\phi},e^{-i\phi}) \end{align*} and the condition $\rho(\gamma)^p=1$ gives \begin{align*} \operatorname{diag}(e^{ip\phi},e^{-ip\phi})=I. \end{align*} Thus $e^{ip\phi}=1$, so $p\phi\in 2\pi\mathbb Z$, and after conjugating we may write \begin{align*} \rho_m(\gamma)=\operatorname{diag}(e^{2\pi i m/p},e^{-2\pi i m/p}) \end{align*} for some $m\in\mathbb Z/p\mathbb Z$. The values $m$ and $-m$ give conjugate representations because the standard Weyl element in $SU(2)$ interchanges the two diagonal entries, sending $\operatorname{diag}(z,z^{-1})$ to $\operatorname{diag}(z^{-1},z)$. Let $q^*$ satisfy $qq^*\equiv 1\pmod p$. With the usual orientation convention for $L(p,q)$, the torsion linking form on a generator $x\in H_1(L(p,q);\mathbb Z)$ is \begin{align*} \lambda(x,x)\equiv \frac{q^*}{p}\pmod 1. \end{align*} By bilinearity of $\lambda$, the class $mx$ satisfies \begin{align*} \lambda(mx,mx)\equiv m^2\lambda(x,x)\pmod 1. \end{align*} Substituting the displayed value of $\lambda(x,x)$ gives \begin{align*} \lambda(mx,mx)\equiv \frac{q^*m^2}{p}\pmod 1. \end{align*} For the standard $SU(2)$ Chern-Simons normalisation used here, the flat connection $A_m$ associated to $\rho_m$ has reduced value equal to the negative of this quadratic linking value: \begin{align*} \overline{\operatorname{cs}}(A_m)\equiv -\frac{q^*m^2}{p}\pmod 1. \end{align*} This representative is independent of the chosen inverse $q^*$ modulo $p$: if $q_1^*-q_2^*=kp$, then \begin{align*} -\frac{q_1^*m^2}{p}+\frac{q_2^*m^2}{p}=-km^2\in\mathbb Z. \end{align*} It is also unchanged when $m$ is replaced by $-m$, since $(-m)^2=m^2$. Changing the orientation or changing the sign convention for the Chern-Weil representative multiplies the displayed representative by $-1$, so lens spaces give a finite arithmetic family of reduced Chern-Simons values governed by the torsion linking form. [/example] Lens spaces show that reduced Chern-Simons values need not vanish even for flat connections on rational homology spheres. Mapping tori give a different source of examples, where the invariant records monodromy. [definition: Mapping Torus] Let $X$ be a smooth manifold and let $f:X\to X$ be a diffeomorphism. The mapping torus of $f$ is \begin{align*} T_f = X\times[0,1]/(x,1)\sim(f(x),0). \end{align*} [/definition] A mapping torus comes with a built-in circle direction, so a flat bundle over it can be described by data on $X$ together with compatibility under the monodromy. This makes the Chern-Simons invariant a transgression measurement of how the monodromy acts on flat data. [example: Circle Bundles Over Surfaces] Let $\Sigma$ be a closed oriented surface and let $\pi:M\to \Sigma$ be an oriented circle bundle with Euler number $e\in\mathbb Z$, with $e\ne 0$. Let $f\in H_1(M;\mathbb Z)$ denote the oriented fibre class. The Gysin sequence gives that the fibre direction is torsion with relation \begin{align*} e f=0, \end{align*} so the torsion subgroup generated by $f$ has order $|e|$, apart from the free homology pulled back from the base. With the orientation convention fixed by the circle bundle, the torsion linking form satisfies \begin{align*} \lambda(f,f)\equiv \frac{1}{e}\pmod 1. \end{align*} A flat abelian connection whose fibre holonomy is $\exp(2\pi i r/e)$ corresponds to the torsion holonomy class $r f$. Since the linking form is bilinear on torsion homology, \begin{align*} \lambda(rf,rf)\equiv r\,\lambda(f,rf)\pmod 1. \end{align*} Applying bilinearity once more gives \begin{align*} \lambda(f,rf)\equiv r\,\lambda(f,f)\pmod 1. \end{align*} Therefore \begin{align*} \lambda(rf,rf)\equiv r^2\lambda(f,f)\pmod 1. \end{align*} Substituting $\lambda(f,f)\equiv 1/e$ gives \begin{align*} \lambda(rf,rf)\equiv \frac{r^2}{e}\pmod 1. \end{align*} For the common abelian normalisation in which the reduced Chern-Simons value is one half of the torsion linking pairing on the holonomy class, we obtain \begin{align*} \overline{\operatorname{cs}}(A_r)\equiv \frac{1}{2}\lambda(rf,rf)\pmod 1. \end{align*} Using the displayed value of $\lambda(rf,rf)$, \begin{align*} \overline{\operatorname{cs}}(A_r)\equiv \frac{r^2}{2e}\pmod 1. \end{align*} Thus the Euler number controls the denominator of the reduced invariant: the larger the torsion order of the fibre class, the finer the possible quadratic Chern-Simons values. [/example] The examples show the same pattern in three settings: a pure gauge on $S^3$, representations of a lens space group, and fibre holonomy in a circle bundle. They also show why the auxiliary data has to be specified: the integer ambiguity belongs to the gauge transformation once the integral refinement has been fixed, while changing the reference construction is a different operation. This motivates the following theorem, which states the quantisation result with that distinction built in. [quotetheorem:9795] [citeproof:9795] Each hypothesis in the theorem marks a boundary of the construction. Closedness removes boundary transgression terms, while orientability is needed to integrate the $(2k-1)$-form over $M$ with a sign. Flatness is what makes the invariant secondary rather than a mixture of secondary and primary curvature data; without flatness, the endpoint Chern-Weil form contributes real curvature information. Integrality is what turns the possible change under gauge into an element of $\mathbb Z$ rather than an arbitrary real period. The theorem is therefore a quantisation statement for gauge ambiguity, not a statement that all auxiliary choices have disappeared. This is the role of Chern-Simons theory in the course: primary characteristic classes live in even-dimensional cohomology and are represented by curvature forms, while Chern-Simons invariants live naturally in odd dimensions and remember the period ambiguity left by choosing primitives. # 9. Local Index-Theoretic Characteristic Forms This chapter explains why the characteristic forms constructed by Chern-Weil theory are the local densities that occur in index theorems. Chapters 1 through 6 attached de Rham classes and characteristic numbers to invariant polynomials in curvature. We now package those same constructions into the Todd class, the $\hat A$-class, and the $L$-class, because these are the combinations of curvature that measure holomorphic Euler characteristic, Dirac index, and signature. The central theme is that global analytic invariants can be computed by integrating differential forms made from curvature. The index theorems stated here are used as motivation and reference points: their proofs belong to analysis and [elliptic operator](/page/Elliptic%20Operator) theory, but the characteristic forms appearing in their statements are exactly the objects developed in this course. ## Curvature Roots and the Main Characteristic Forms What should replace the elementary symmetric polynomial formulas for Chern and Pontryagin classes when an index theorem asks for a more complicated expression such as \begin{align*} \frac{x}{1-e^{-x}}? \end{align*} The answer is to treat curvature formally through its roots, apply a [power series](/page/Power%20Series) to each root, and then reassemble the result as a symmetric polynomial. For a complex vector bundle $E \to M$ with connection $\nabla^E$ and curvature $F^E \in \Omega^2(M;\operatorname{End}(E))$, the exponential of the curvature is a matrix-valued differential form. Taking the trace gives the Chern character form, which is additive under direct sum and multiplicative under tensor product at the level of cohomology. [definition: Chern Character Form] Let $E \to M$ be a complex vector bundle with connection $\nabla^E$ and curvature $F^E$. The Chern character form of $(E,\nabla^E)$ is \begin{align*} \operatorname{ch}(E,\nabla^E) := \operatorname{tr}\left(\exp\left(\frac{F^E}{2\pi i}\right)\right) \in \Omega^{\mathrm{even}}(M;\mathbb C). \end{align*} [/definition] Its degree-$2k$ component is \begin{align*} \frac{1}{k!}\operatorname{tr}\left(\left(\frac{F^E}{2\pi i}\right)^k\right). \end{align*} The definition packages the trace polynomials already studied in Chern-Weil theory, and the next question is whether this package has the same two properties as the ordinary characteristic forms: closedness and independence of the chosen connection. [quotetheorem:9796] [citeproof:9796] The theorem turns the Chern character into a reliable cohomology class rather than a connection-dependent form. Each hypothesis controls a specific failure mode. If the expression is not built from an invariant polynomial, conjugating a local curvature matrix by a change of frame can change the local formula; for example, a selected diagonal entry of $F^E$ is not globally defined, while its trace is. If connection independence is omitted, the same bundle can receive different forms from different choices of connection, as happens for the connection one-form itself on a line bundle. If the splitting-principle descent is ignored, a formula depending on an ordered root such as $x_1$ would change after permuting the two line summands of a split rank-two bundle. The theorem also does not say that the Chern character by itself is the correct index density. For instance, on $\mathbb{CP}^1$ the class $\operatorname{ch}(\mathcal O(k))=1+kH$ integrates to $k$, while the holomorphic Euler characteristic is $k+1$; the missing contribution comes from the tangent geometry of the base. This obstruction is the reason for introducing the Todd class. [definition: Todd Class] Let $E \to M$ be a complex vector bundle with formal Chern roots $x_1,\dots,x_r$. The Todd class of $E$ is the characteristic class $\operatorname{Td}(E) \in H^{\mathrm{even}}(M;\mathbb Q)$ defined by \begin{align*} \operatorname{Td}(E) := \prod_{j=1}^{r} \frac{x_j}{1-e^{-x_j}}. \end{align*} The Todd form $\operatorname{Td}(E,\nabla^E) \in \Omega^{\mathrm{even}}(M;\mathbb C)$ is the Chern-Weil representative obtained by applying the same symmetric power series to the curvature roots of $F^E/2\pi i$. [/definition] The first terms of the Todd class are useful in computations: \begin{align*} \operatorname{Td}(E)=1 + \frac{1}{2}c_1(E) + \frac{1}{12}\bigl(c_1(E)^2+c_2(E)\bigr) + \frac{1}{24}c_1(E)c_2(E) + \cdots . \end{align*} It is the correction factor that converts Chern character data into holomorphic Euler characteristics. [example: Todd Class Of Complex Projective Space] Let $H\in H^2(\mathbb{CP}^n;\mathbb Z)$ be the hyperplane class, so $H^{n+1}=0$ and only powers $1,H,\dots,H^n$ can contribute. The Euler sequence \begin{align*} 0\to \mathcal O \to \mathcal O(1)^{\oplus(n+1)}\to T\mathbb{CP}^n\to 0 \end{align*} gives, by multiplicativity of total Chern classes in short exact sequences, \begin{align*} c(T\mathbb{CP}^n)c(\mathcal O)=c(\mathcal O(1)^{\oplus(n+1)}). \end{align*} Since $c(\mathcal O)=1$ and $c(\mathcal O(1))=1+H$, the Whitney product formula gives \begin{align*} c(\mathcal O(1)^{\oplus(n+1)})=(1+H)^{n+1}. \end{align*} Thus \begin{align*} c(T\mathbb{CP}^n)=(1+H)^{n+1} \end{align*} in the [quotient ring](/page/Quotient%20Ring) where $H^{n+1}=0$. The Todd class is multiplicative under direct sums, and $\operatorname{Td}(\mathcal O)=1$. Applying the Euler sequence to the multiplicative sequence $Q(z)=z/(1-e^{-z})$ therefore gives \begin{align*} \operatorname{Td}(T\mathbb{CP}^n)\operatorname{Td}(\mathcal O)=\operatorname{Td}(\mathcal O(1)^{\oplus(n+1)}). \end{align*} Since $\mathcal O(1)$ has Chern root $H$, \begin{align*} \operatorname{Td}(\mathcal O(1)^{\oplus(n+1)})=\left(\frac{H}{1-e^{-H}}\right)^{n+1}. \end{align*} Hence \begin{align*} \operatorname{Td}(T\mathbb{CP}^n)=\left(\frac{H}{1-e^{-H}}\right)^{n+1}, \end{align*} with all terms involving $H^{n+1}$ or higher set equal to zero. For $n=1$, the relation is $H^2=0$. The power series begins \begin{align*} \frac{H}{1-e^{-H}}=1+\frac{1}{2}H+\frac{1}{12}H^2+\cdots . \end{align*} After imposing $H^2=0$, this becomes \begin{align*} \frac{H}{1-e^{-H}}=1+\frac{1}{2}H. \end{align*} Therefore \begin{align*} \operatorname{Td}(T\mathbb{CP}^1)=\left(1+\frac{1}{2}H\right)^2. \end{align*} Expanding and using $H^2=0$ gives \begin{align*} \left(1+\frac{1}{2}H\right)^2=1+H+\frac{1}{4}H^2=1+H. \end{align*} The top-degree component is $H$, and since $\int_{\mathbb{CP}^1}H=1$, the Todd genus is \begin{align*} \int_{\mathbb{CP}^1}\operatorname{Td}(T\mathbb{CP}^1)=1. \end{align*} This agrees with $\chi(\mathbb{CP}^1,\mathcal O_{\mathbb{CP}^1})=1$, so the tangent-bundle correction supplies exactly the missing constant term in the Riemann-Roch computation. [/example] The Todd example shows how a complex tangent bundle feeds into holomorphic index formulae. Spin Dirac operators create a different obstruction: the relevant tangent-bundle correction cannot be recovered from a raw Pontryagin class chosen in isolation. Already in degree four the Dirac density is $-p_1/24$, while the next correction involves the specific combination $7p_1^2-4p_2$; the coefficients are forced by the power series attached to the Clifford-theoretic operator. For real oriented bundles the natural input is therefore Pontryagin-root data, and we need an even power series in the curvature roots so that the answer is expressible in Pontryagin classes. [definition: A-Hat Class] Let $E \to M$ be a real oriented vector bundle with formal Pontryagin roots $x_1^2,\dots,x_m^2$. The $\hat A$-class of $E$ is the rational characteristic class $\hat A(E)$ whose nonzero components have degrees divisible by $4$, defined by \begin{align*} \hat A(E) := \prod_{j=1}^{m} \frac{x_j/2}{\sinh(x_j/2)}. \end{align*} The $\hat A$-form $\hat A(E,\nabla^E) \in \Omega^{4*}(M;\mathbb R)$ is the corresponding Chern-Weil representative formed from the curvature of a metric connection on $E$. [/definition] The first nonzero terms are \begin{align*} \hat A(E) = 1 - \frac{1}{24}p_1(E) + \frac{1}{5760}\bigl(7p_1(E)^2-4p_2(E)\bigr)+\cdots . \end{align*} This class is the curvature density in the spin Dirac index theorem. [example: A-Hat Class Vanishing On Spheres] Write $S_n$ for the $n$-sphere, with $n>0$. The outward unit normal embedding $S_n\subset \mathbb R^{n+1}$ gives the stable tangent-bundle isomorphism \begin{align*} TS_n\oplus \mathbb R \cong \mathbb R^{n+1}. \end{align*} Pontryagin classes are unchanged after adding a trivial real line bundle, and the trivial bundle has total Pontryagin class $1$. Hence \begin{align*} p(TS_n)=p(TS_n\oplus \mathbb R)=p(\mathbb R^{n+1})=1. \end{align*} Writing $p(TS_n)=1+p_1(TS_n)+p_2(TS_n)+\cdots$, comparison of homogeneous degree-$4j$ terms gives \begin{align*} p_j(TS_n)=0 \end{align*} for every $j\ge 1$. The $\hat A$-class is a universal polynomial in the Pontryagin classes whose constant term is $1$: \begin{align*} \hat A(TS_n)=1-\frac{1}{24}p_1(TS_n)+\frac{1}{5760}\bigl(7p_1(TS_n)^2-4p_2(TS_n)\bigr)+\cdots . \end{align*} Substituting $p_j(TS_n)=0$ for every $j\ge 1$ leaves \begin{align*} \hat A(TS_n)=1. \end{align*} Thus every positive-degree component of $\hat A(TS_n)$ is zero. When the $\hat A$-genus is defined, the top-degree component has positive degree because $n>0$, so \begin{align*} \int_{S_n}\hat A(TS_n)=0. \end{align*} The vanishing comes from stable triviality of the sphere tangent bundle, not from a special curvature choice. [/example] The $\hat A$-class detects the Dirac operator, but a different elliptic operator controls the middle-dimensional intersection form. The same obstruction appears again: ordinary Pontryagin classes supply the raw variables, but the signature is not computed by an arbitrary Pontryagin polynomial. In dimension four the correct density is $p_1/3$, and in dimension eight the correction is $(7p_2-p_1^2)/45$, so the coefficients are part of the theorem rather than decorative normalisation. To prepare for the signature theorem, we introduce the multiplicative sequence whose degree-four term is proportional to $p_1$. [definition: L-Class] Let $E \to M$ be a real oriented vector bundle with formal Pontryagin roots $x_1^2,\dots,x_m^2$. The Hirzebruch $L$-class of $E$ is the rational characteristic class $L(E)$ whose nonzero components have degrees divisible by $4$, defined by \begin{align*} L(E) := \prod_{j=1}^{m} \frac{x_j}{\tanh x_j}. \end{align*} The $L$-form $L(E,\nabla^E) \in \Omega^{4*}(M;\mathbb R)$ is its Chern-Weil representative for a metric connection on $E$. [/definition] Its expansion begins as \begin{align*} L(E)=1+\frac{1}{3}p_1(E)+\frac{1}{45}\bigl(7p_2(E)-p_1(E)^2\bigr)+\cdots . \end{align*} Thus on a closed oriented four-manifold $M$, the degree-four signature density is $\frac{1}{3}p_1(TM)$. [example: Signature Density In Dimension Four] Let $M$ be a closed oriented smooth four-manifold with a Riemannian metric and Levi-Civita connection. In dimension four, only the degree-four term of the $L$-class can contribute to the integral, and the displayed expansion \begin{align*} L(TM)=1+\frac{1}{3}p_1(TM)+\cdots \end{align*} therefore gives \begin{align*} L(TM)_{\mathrm{top}}=\frac{1}{3}p_1(TM,\nabla^{TM}). \end{align*} By the *Hirzebruch Signature Theorem*, \begin{align*} \sigma(M)=\int_M L(TM)_{\mathrm{top}}. \end{align*} Substituting the degree-four component gives \begin{align*} \sigma(M)=\int_M \frac{1}{3}p_1(TM,\nabla^{TM}). \end{align*} For $M=\mathbb{CP}^2$, let $H\in H^2(\mathbb{CP}^2;\mathbb Z)$ be the hyperplane class. From the Euler-sequence computation, \begin{align*} c(T\mathbb{CP}^2)=(1+H)^3 \end{align*} with $H^3=0$. Expanding in degrees at most four gives \begin{align*} (1+H)^3=1+3H+3H^2+H^3=1+3H+3H^2. \end{align*} Hence \begin{align*} c_1(T\mathbb{CP}^2)=3H \end{align*} and \begin{align*} c_2(T\mathbb{CP}^2)=3H^2. \end{align*} For the underlying real bundle of a complex rank-two bundle, the first Pontryagin class is \begin{align*} p_1=c_1^2-2c_2. \end{align*} Thus \begin{align*} p_1(T\mathbb{CP}^2)=(3H)^2-2(3H^2)=9H^2-6H^2=3H^2. \end{align*} Therefore \begin{align*} \sigma(\mathbb{CP}^2)=\int_{\mathbb{CP}^2}\frac{1}{3}\,3H^2. \end{align*} Since $\int_{\mathbb{CP}^2}H^2=1$, this becomes \begin{align*} \sigma(\mathbb{CP}^2)=\int_{\mathbb{CP}^2}H^2=1. \end{align*} The curvature density $\frac{1}{3}p_1$ recovers the one-dimensional positive intersection form on $H^2(\mathbb{CP}^2;\mathbb R)$. [/example] ## Multiplicative Sequences From Power Series Why do these different looking characteristic classes all have the same formal shape? The common mechanism is a multiplicative sequence: a formal power series determines a characteristic class by being evaluated on formal roots, and direct sums become products. [definition: Multiplicative Sequence] Let $Q(z) \in R[[z]]$ be a formal power series with constant term $1$, where $R$ is a commutative coefficient ring. For a complex vector bundle $E$ with formal Chern roots $x_1,\dots,x_r$, the multiplicative sequence associated to $Q$ is \begin{align*} K_Q(E):=\prod_{j=1}^{r} Q(x_j). \end{align*} For a real oriented vector bundle with formal Pontryagin roots $x_1^2,\dots,x_m^2$, the corresponding even multiplicative sequence is defined by a power series in the variables $x_j$ whose product is symmetric in $x_1^2,\dots,x_m^2$. [/definition] The condition $Q(0)=1$ normalises the characteristic class of a rank-zero summand. A root-wise formula would be unusable for characteristic classes unless it were insensitive to the splitting presentation and compatible with Whitney sum: adding the roots of two bundles should multiply the resulting classes. The definition has therefore introduced a candidate construction, but the construction still has to pass two tests before it can be used geometrically. It must descend from formal roots to genuine characteristic classes, and it must satisfy the Whitney product rule. The following result supplies exactly these checks, which are later needed for virtual bundles and index-theoretic multiplicative genera. [quotetheorem:9797] [citeproof:9797] The theorem is useful because direct sum is the operation by which tangent bundles, twisting bundles, and virtual bundles are assembled. The hypothesis $Q(0)=1$ is not cosmetic: it makes adding a zero-root summand harmless, so the class depends on the bundle rather than on a chosen presentation with extra rank-one zero-root factors. If instead $Q(0)=a\ne 1$, then adjoining such a summand would multiply the class by $a$; the same geometric bundle presented with an extra zero root would receive a different value. The theorem is also limited to symmetric root expressions. A nonsymmetric expression such as $x_1$ for a rank-two bundle has no invariant meaning before an ordering of roots is chosen, and such an ordering is not part of the geometry. The Todd, $\hat A$, and $L$ classes correspond to three specific power series: \begin{align*} Q_{\operatorname{Td}}(z)=\frac{z}{1-e^{-z}}, \qquad Q_{\hat A}(z)=\frac{z/2}{\sinh(z/2)}, \qquad Q_L(z)=\frac{z}{\tanh z}. \end{align*} These series encode different elliptic operators. The geometric input is curvature; the analytic output is an index. [remark: Formal Roots As Bookkeeping] Formal roots need not be actual cohomology classes on $M$. They are a controlled notation justified by the splitting principle: symmetric expressions in the roots descend to genuine characteristic classes. This is why formulas such as $\prod_j x_j/(1-e^{-x_j})$ are meaningful even when the bundle does not split. [/remark] The root notation explains the algebraic identity behind the class, but index formulae require differential forms that can be integrated over the manifold. We therefore need the [Chern-Weil representative of a multiplicative sequence](/theorems/9798), not only its abstract cohomology class. [quotetheorem:9798] [citeproof:9798] This theorem is the local bridge needed for index formulae: it turns a formal characteristic class into an actual closed differential form whose top-degree component can be integrated. The coefficient ring is required to support the formal coefficients of $Q$ and the resulting characteristic classes; in the standard Todd, $\hat A$, and $L$ examples this is why rational or complex coefficients appear. The constant-term condition fixes the degree-zero normalisation, while symmetry ensures that the expression is a polynomial in invariant Chern-Weil data rather than in labelled roots. For example, applying a non-symmetric function to a single chosen curvature root would change under permutation of the roots and would not define a global form on the bundle. The theorem also remains a Chern-Weil statement only; it constructs candidate local densities but does not prove that any of them computes an analytic index. ## Index Theorem Statements As Curvature Formulae Which curvature forms actually occur in geometry, rather than just being formal constructions? The answer is supplied by index theory: the Todd class, $\hat A$-class, and $L$-class are precisely the local densities for natural elliptic operators. This chapter is a synthesis point rather than another repetition of the closedness and transgression calculations. Earlier chapters explained why characteristic forms are closed and how their representatives change under a connection; here those constructions are organized by the analytic problems they solve. The point is to identify which characteristic form belongs to which geometric operator, and what numerical invariant its integral computes. The holomorphic case begins with a compact complex manifold. The analytic invariant is the alternating sum of sheaf cohomology dimensions, while the topological expression is an integral involving the Chern character and Todd class. [definition: Holomorphic Euler Characteristic] Let $X$ be a compact complex manifold and let $E \to X$ be a holomorphic vector bundle. Write $H^q(X,E)$ for the $q$-th sheaf cohomology group of holomorphic sections of $E$. The holomorphic Euler characteristic of $E$ is the integer \begin{align*} \chi(X,E):=\sum_{q\ge 0}(-1)^q\dim_{\mathbb C} H^q(X,E) \in \mathbb Z. \end{align*} [/definition] This invariant is analytic in its definition because it is built from holomorphic cohomology groups. The problem is that these dimensions are not visibly computable from curvature, while Chern-Weil theory supplies differential forms such as the Chern character and Todd class. Hirzebruch-Riemann-Roch identifies the precise top-degree characteristic form whose integral recovers the alternating holomorphic Euler characteristic. [quotetheorem:3866] [remark: Proof Status] This theorem is quoted here as an index theorem rather than proved in this course note. The surrounding discussion records how the statement is used: it identifies the characteristic-class expression whose top-degree integral computes the holomorphic Euler characteristic. [/remark] This proof status matters because Hirzebruch-Riemann-Roch is being used as a bridge from Chern-Weil forms to analytic invariants, not as a local consequence of the preceding curvature computations. Compactness ensures that the cohomology groups are finite-dimensional and that the integral has no boundary correction. Without compactness, the invariant may fail even to be finite: for $X=\mathbb C$ and the structure sheaf, $H^0(X,\mathcal O_X)$ is the infinite-dimensional vector space of entire functions. Holomorphicity is also essential because the groups $H^q(X,E)$ and the coupled Dolbeault operator use the holomorphic structure, not just the underlying smooth complex vector bundle; a smooth complex line bundle over a complex manifold with no specified $\bar\partial$-operator does not determine sheaf cohomology groups. The Todd factor uses the complex tangent bundle of the manifold, so the formula is not a statement about an arbitrary real tangent bundle; for instance, an oriented smooth four-manifold without an almost complex structure has no holomorphic tangent bundle $T_{\mathrm{hol}}X$ to insert into the right-hand side. What the theorem computes is the alternating Euler characteristic, not the individual cohomology dimensions: on $\mathbb{CP}^1$, both $\mathcal O(-1)$ and the zero sheaf have Euler characteristic $0$, but their cohomology groups are different. [example: Riemann-Roch For Line Bundles On The Riemann Sphere] Let $X=\mathbb{CP}^1$ and let $E=\mathcal O(k)$. If $H\in H^2(\mathbb{CP}^1;\mathbb Z)$ is the hyperplane class, then $H^2=0$ and $c_1(\mathcal O(k))=kH$. Hence the Chern character is \begin{align*} \operatorname{ch}(\mathcal O(k))=e^{kH}=1+kH+\frac{(kH)^2}{2!}+\cdots=1+kH. \end{align*} From the Todd-class computation for $\mathbb{CP}^1$, \begin{align*} \operatorname{Td}(T\mathbb{CP}^1)=1+H. \end{align*} Therefore \begin{align*} \operatorname{ch}(\mathcal O(k))\operatorname{Td}(T\mathbb{CP}^1)=(1+kH)(1+H). \end{align*} Expanding the product gives \begin{align*} (1+kH)(1+H)=1+H+kH+kH^2. \end{align*} Since $H^2=0$, this becomes \begin{align*} (1+kH)(1+H)=1+(k+1)H. \end{align*} The top-degree component is therefore $(k+1)H$. Substituting into Hirzebruch-Riemann-Roch and using $\int_{\mathbb{CP}^1}H=1$ gives \begin{align*} \chi(\mathbb{CP}^1,\mathcal O(k))=\int_{\mathbb{CP}^1}(k+1)H=k+1. \end{align*} This matches the standard cohomology calculation: for $k\ge 0$, $\dim_{\mathbb C}H^0(\mathbb{CP}^1,\mathcal O(k))=k+1$ and $H^1=0$; for $k=-1$, both groups vanish; and for $k\le -2$, $H^0=0$ while $\dim_{\mathbb C}H^1(\mathbb{CP}^1,\mathcal O(k))=-k-1$, so in every case $\dim H^0-\dim H^1=k+1$. [/example] The Riemann-Roch example shows how a complex elliptic operator selects the Todd class. This motivates the next definition: for spin geometry the basic elliptic operator is built from Clifford multiplication, and twisting it by a vector bundle is the mechanism that brings the Chern character into the formula. [definition: Twisted Dirac Operator] Let $M$ be a closed Riemannian spin manifold with spinor bundle decomposed as $S=S_+\oplus S_-$ in even dimension, and let $E\to M$ be a complex vector bundle with connection. The twisted Dirac operator is the elliptic differential operator \begin{align*} D_{E,+} : \Gamma(S_+\otimes E) \to \Gamma(S_-\otimes E) \end{align*} obtained by composing the covariant derivative on $S\otimes E$ with Clifford multiplication. [/definition] Its index is the dimension of the kernel minus the dimension of the cokernel. This integer is defined analytically from solutions of a differential equation, so it is not evident why it should be computable by integrating characteristic forms. The local index formula identifies the curvature density selected by the twisted spin Dirac operator: the tangent-bundle $\hat A$-class multiplied by the Chern character of the twisting bundle. The obstruction is the apparent mismatch between analysis and topology: kernels of differential operators depend on metrics and connections, while characteristic numbers are cohomological. The index theorem is needed precisely to show that this analytic integer is forced by the top-degree part of $\hat A(TM)\operatorname{ch}(E)$. [quotetheorem:9799] This theorem is stated as analytic input for the course. Closedness of $M$ gives a Fredholm operator without boundary conditions; on a compact manifold with boundary, such as the disk, the Dirac operator requires boundary conditions and the index formula acquires boundary contributions such as the Atiyah-Patodi-Singer eta term. Even dimension is used here because the spinor bundle splits chirally, giving an operator from the positive spinors to the negative spinors whose index is defined in this form; on an odd-dimensional closed spin manifold the ordinary Dirac operator is self-adjoint on an ungraded spinor bundle, so this chiral index is not the invariant being computed. The spin hypothesis supplies the spinor bundle and Clifford module structure; a closed oriented manifold need not be spin, with $\mathbb{CP}^2$ as the standard example since $w_2(T\mathbb{CP}^2)\ne 0$, and the displayed spin Dirac operator is then not defined. The twisting bundle must also carry the connection used in the coupled operator and in $\operatorname{ch}(E)$; changing the twisting data changes the operator and its index. Its proof uses elliptic operator theory, heat kernels, and the local index theorem; the heat-kernel calculation is what explains why the curvature expression $\hat A(TM)\operatorname{ch}(E)$ appears. [example: Untwisted Dirac Index On A Sphere] Write $S_{2m}$ for the $2m$-sphere with $m\ge 1$, equip it with a spin structure, and take $E=S_{2m}\times \mathbb C$ with the flat product connection. The curvature of this line bundle is zero, so its Chern character is \begin{align*} \operatorname{ch}(E)=\operatorname{tr}(e^0)=1. \end{align*} By the *[Atiyah-Singer Index Theorem for Twisted Dirac Operators](/theorems/9799)*, \begin{align*} \operatorname{ind}(D_+)=\int_{S_{2m}}\bigl(\hat A(TS_{2m})\operatorname{ch}(E)\bigr)_{\mathrm{top}}. \end{align*} Substituting $\operatorname{ch}(E)=1$ gives \begin{align*} \operatorname{ind}(D_+)=\int_{S_{2m}}\hat A(TS_{2m})_{\mathrm{top}}. \end{align*} The outward normal line gives the stable bundle identity \begin{align*} TS_{2m}\oplus \mathbb R \cong S_{2m}\times \mathbb R^{2m+1}. \end{align*} Pontryagin classes are unchanged after adding a product real line bundle, and a product real vector bundle has total Pontryagin class $1$. Hence \begin{align*} p(TS_{2m})=p(TS_{2m}\oplus \mathbb R)=p(S_{2m}\times \mathbb R^{2m+1})=1. \end{align*} Writing \begin{align*} p(TS_{2m})=1+p_1(TS_{2m})+p_2(TS_{2m})+\cdots \end{align*} and comparing homogeneous components gives \begin{align*} p_j(TS_{2m})=0 \end{align*} for every $j\ge 1$. The $\hat A$-class is a universal polynomial in the Pontryagin classes with constant term $1$: \begin{align*} \hat A(TS_{2m})=1-\frac{1}{24}p_1(TS_{2m})+\frac{1}{5760}\bigl(7p_1(TS_{2m})^2-4p_2(TS_{2m})\bigr)+\cdots . \end{align*} Substituting $p_j(TS_{2m})=0$ for every $j\ge 1$ leaves \begin{align*} \hat A(TS_{2m})=1. \end{align*} Since $m\ge 1$, the top degree $2m$ is positive, while the class $1$ has degree zero. Therefore \begin{align*} \hat A(TS_{2m})_{\mathrm{top}}=0. \end{align*} Thus \begin{align*} \operatorname{ind}(D_+)=\int_{S_{2m}}0=0. \end{align*} The vanishing is forced by the stable product-bundle identity for the sphere tangent bundle: the untwisted Dirac density has no positive-degree component to integrate. [/example] The Dirac example closes the spin part of the chapter. The signature theorem starts from a different kind of global invariant, one that does not require a spin structure or a twisting bundle. Its input is the middle-dimensional cup-product pairing of an oriented manifold, and the first issue is to turn that pairing into a single integer. [definition: Signature] Let $M$ be a closed oriented smooth manifold of dimension $4k$. The signature $\sigma(M)$ is the signature of the symmetric bilinear form \begin{align*} H^{2k}(M;\mathbb R)\times H^{2k}(M;\mathbb R) \to \mathbb R. \end{align*} It sends a pair $(\alpha,\beta)$ to \begin{align*} \int_M \alpha\wedge\beta. \end{align*} Here $H^{2k}(M;\mathbb R)$ denotes the de Rham cohomology group in degree $2k$. [/definition] The signature is topological, yet the signature operator represents it as an elliptic index. The comparison problem is to understand which Chern-Weil form can recover this purely cohomological integer by integration. For the signature operator, the selected local density is the $L$-form, giving the analogue of the Todd density for Dolbeault operators and the $\hat A$ density for Dirac operators. [quotetheorem:9789] This theorem is quoted without proof in this course. Closedness is needed so that the intersection pairing is nondegenerate without boundary terms; for example, the compact oriented manifold $[0,1]\times S^{4k-1}$ has boundary, and the ordinary closed-manifold intersection pairing is not the invariant appearing in the signature theorem. Orientation is needed both to define the fundamental class used in integration and to make the middle-dimensional pairing meaningful with a sign; reversing orientation changes the sign of the fundamental class and hence the sign of both sides. The dimension condition $\dim M=4k$ ensures that the middle-degree cup-product pairing is symmetric; in dimensions $4k+2$ it is skew-symmetric, so a closed oriented surface has a symplectic intersection form on $H^1$ and no signature in this sense. The theorem computes the signature of this pairing, not the full cohomology ring or the individual Betti numbers: $S^4$ and $S^2\times S^2$ both have signature $0$, but their middle cohomology ranks differ. [remark: Local Versus Global Content] The right-hand sides of these index formulae are integrals of local curvature densities, while the left-hand sides are dimensions or signatures defined globally. The equality is not a formal consequence of Chern-Weil theory alone: Chern-Weil theory constructs closed forms and cohomology classes, whereas index theory identifies which closed forms compute analytic indices. [/remark] These statements complete the bridge from characteristic forms to curvature methods in index theory. The forms $\operatorname{Td}$, $\hat A$, and $L$ are not arbitrary refinements of Chern and Pontryagin classes; they are the canonical multiplicative sequences selected by the Dolbeault, Dirac, and signature operators. # 10. Geometric Computations In Representative Bundles This chapter turns the general Chern-Weil machine into computations on representative bundles. The point is to see how curvature forms encode integral characteristic classes in examples where the topology is already visible: the Hopf fibration, projective space, and Grassmannians. The previous chapters developed characteristic forms, characteristic numbers, and secondary transgression forms abstractly; here we choose canonical geometric connections and calculate the resulting de Rham representatives. ## Hermitian Line Bundles And Curvature Representatives For a complex line bundle, the Chern-Weil construction becomes especially concrete. The structure group reduces to $U(1)$, whose Lie algebra is $i\mathbb R$, so the curvature of a unitary connection is an ordinary imaginary-valued $2$-form. The guiding question is how the topological class $c_1(L)$ is read off from this curvature. [definition: Hermitian Line Bundle] A Hermitian line bundle over a smooth manifold $M$ is a complex line bundle $L \to M$ equipped with maps \begin{align*} h_p:L_p\times L_p\longrightarrow \mathbb C \end{align*} for each $p\in M$, varying smoothly with $p$, such that $h_p$ is positive-definite, complex-linear in the first argument, and conjugate-linear in the second argument. [/definition] A Hermitian metric gives a preferred reduction of the frame bundle to $U(1)$, but curvature computations also require connections compatible with that reduction. The next definition isolates the connections whose local forms live in $i\mathbb R$, which is the condition needed for a real Chern-Weil representative after normalisation. [definition: Unitary Connection On A Hermitian Line Bundle] Let $L \to M$ be a Hermitian line bundle. A unitary connection on $L$ is a connection \begin{align*} \nabla : \Gamma(L) \longrightarrow \Omega^1(M;L) \end{align*} such that, for all smooth sections $s,t \in \Gamma(L)$ and all vector fields $X \in \mathfrak X(M)$, \begin{align*} X(h(s,t)) = h(\nabla_X s,t) + h(s,\nabla_X t). \end{align*} [/definition] A local unitary frame $e$ writes every section as $s = fe$, and the connection has the form $\nabla e = A\otimes e$ for a local $1$-form $A \in \Omega^1(U;i\mathbb R)$. Since $U(1)$ is abelian, the curvature is locally $F_\nabla=dA$; the next theorem explains the precise normalisation that turns this curvature into the first Chern class. [quotetheorem:9800] [citeproof:9800] This theorem is the computational entry point for all examples in this chapter: choose a geometric unitary connection, compute its curvature, and normalise by $i/(2\pi)$. The unitary hypothesis is essential for the displayed form to be real; for instance, an arbitrary complex connection on a product line bundle may have connection form $A=x\,dy$ on a complex coordinate chart, and then $dA=dx\wedge dy$ is real-valued rather than imaginary-valued. The Chern-Weil normalisation in the theorem therefore no longer produces a real form from such a connection. The theorem also does not say that the particular $2$-form is unique, since changing the connection changes the representative by an exact form while preserving the de Rham class. The normalisation is fixed by requiring integral periods, so the first useful test case is a degree-one bundle over the Riemann sphere. [example: Curvature Period On A Degree One Line Bundle] Let $L=\mathcal O(1)\to \mathbb{CP}^1$ carry the Chern connection induced by the Fubini-Study Hermitian metric. In the affine chart $\mathbb C\subset \mathbb{CP}^1$ with coordinate $z$, the first Chern form is \begin{align*} \omega_{FS}=\frac{i}{2\pi}\frac{dz\wedge d\bar z}{(1+|z|^2)^2}. \end{align*} We compute its period over $\mathbb{CP}^1$ by using polar coordinates $z=re^{i\varphi}$ on the affine chart; the missing point at infinity does not change the integral, and the displayed form extends smoothly across it. Since \begin{align*} dz=e^{i\varphi}dr+ir e^{i\varphi}d\varphi \end{align*} and \begin{align*} d\bar z=e^{-i\varphi}dr-ir e^{-i\varphi}d\varphi, \end{align*} their wedge product is \begin{align*} dz\wedge d\bar z=(dr+ir\,d\varphi)\wedge(dr-ir\,d\varphi). \end{align*} Expanding the wedge product and using $dr\wedge dr=0$ and $d\varphi\wedge dr=-dr\wedge d\varphi$ gives \begin{align*} dz\wedge d\bar z=-ir\,dr\wedge d\varphi+ir\,d\varphi\wedge dr. \end{align*} Therefore \begin{align*} dz\wedge d\bar z=-2ir\,dr\wedge d\varphi. \end{align*} Substituting this into $\omega_{FS}$ gives \begin{align*} \omega_{FS}=\frac{r}{\pi(1+r^2)^2}\,dr\wedge d\varphi. \end{align*} Thus \begin{align*} \int_{\mathbb{CP}^1}\omega_{FS}=\int_0^{2\pi}\int_0^\infty \frac{r}{\pi(1+r^2)^2}\,dr\,d\varphi. \end{align*} The angular integral contributes $2\pi$, so \begin{align*} \int_{\mathbb{CP}^1}\omega_{FS}=2\int_0^\infty \frac{r}{(1+r^2)^2}\,dr. \end{align*} With $u=1+r^2$, so that $du=2r\,dr$, the radial integral becomes \begin{align*} 2\int_0^\infty \frac{r}{(1+r^2)^2}\,dr=\int_1^\infty u^{-2}\,du. \end{align*} Since \begin{align*} \int_1^\infty u^{-2}\,du=\left[-u^{-1}\right]_1^\infty=1, \end{align*} we obtain \begin{align*} \int_{\mathbb{CP}^1}\omega_{FS}=1. \end{align*} Hence $\omega_{FS}$ represents the positive integral generator of $H^2(\mathbb{CP}^1;\mathbb Z)$ under the de Rham comparison, fixing the sign convention used below for Hopf and tautological bundles. [/example] The example also indicates why curvature is more than local differential data. Its integral over a closed surface detects the obstruction to finding a nowhere-vanishing global section. ## Hopf Fibration And Monopole Curvature The Hopf fibration is the model calculation where a principal circle bundle, a complex line bundle, and a curvature form on a sphere all meet. The problem is to compute a $2$-form on $S^2$ whose integral is the Chern number of the associated line bundle. This is the geometric origin of the Dirac monopole curvature. [definition: Hopf Principal Circle Bundle] The Hopf principal circle bundle is the map \begin{align*} \pi:S^3 &\longrightarrow \mathbb{CP}^1, & \pi(z_0,z_1)&=[z_0:z_1], \end{align*} where $S^3=\{(z_0,z_1)\in \mathbb C^2: |z_0|^2+|z_1|^2=1\}$. The circle action is the map \begin{align*} U(1)\times S^3&\longrightarrow S^3, & (\lambda,(z_0,z_1))&\longmapsto (\lambda z_0,\lambda z_1). \end{align*} [/definition] Identifying $\mathbb{CP}^1$ with $S^2$ turns this into a circle bundle over the sphere, but a Chern-Weil computation still needs a chosen connection. The Hermitian geometry of $S^3\subset\mathbb C^2$ supplies one by declaring horizontal vectors to be orthogonal to the circle direction. [definition: Standard Hopf Connection] On $S^3 \subset \mathbb C^2$, the standard Hopf connection is the $i\mathbb R$-valued $1$-form \begin{align*} \alpha:TS^3\longrightarrow i\mathbb R. \end{align*} It is defined by restricting the following ambient $1$-form to tangent vectors of $S^3$: \begin{align*} \alpha = \bar z_0\,dz_0 + \bar z_1\,dz_1. \end{align*} [/definition] The form $\alpha$ evaluates to $i$ on the fundamental vector field of the circle action, and its kernel gives the horizontal distribution. Its curvature descends to the base because it is horizontal and invariant; the next theorem computes the descended form and identifies the associated Chern class. [quotetheorem:9801] [citeproof:9801] The sign is a useful diagnostic: the dual Hopf line bundle has Chern class $+u$. The theorem depends on using the Hopf connection and the scalar $U(1)$ action; replacing the connection by one with an added basic imaginary $1$-form changes the descended curvature form, although not its cohomology class after normalisation. A concrete contrast is the product circle bundle $S^2\times U(1)\to S^2$, whose product connection has zero curvature and therefore cannot represent the generator of $H^2(S^2;\mathbb Z)$. The theorem identifies the Chern class and a canonical curvature representative for this bundle, but it does not classify all possible monopole gauge potentials. Physicists often call the same curvature, after multiplying by a real constant, the field strength of a unit magnetic monopole, and this language is best seen through local gauge potentials. [example: Dirac Monopole On The Two-Sphere] Cover $S^2\cong \mathbb{CP}^1$ by northern and southern charts, and use spherical coordinates $(\theta,\varphi)$ on their overlap, where $0<\theta<\pi$. Define the local $i\mathbb R$-valued potentials \begin{align*} A_N=\frac{i}{2}(1-\cos\theta)\,d\varphi \end{align*} and \begin{align*} A_S=-\frac{i}{2}(1+\cos\theta)\,d\varphi. \end{align*} On the overlap, \begin{align*} A_N-A_S=\frac{i}{2}(1-\cos\theta)\,d\varphi+\frac{i}{2}(1+\cos\theta)\,d\varphi. \end{align*} Combining the scalar factors gives \begin{align*} A_N-A_S=\frac{i}{2}(2)\,d\varphi=i\,d\varphi. \end{align*} If $g=e^{i\varphi}$, then \begin{align*} g^{-1}dg=e^{-i\varphi}d(e^{i\varphi})=e^{-i\varphi}(i e^{i\varphi}\,d\varphi)=i\,d\varphi, \end{align*} so the two potentials differ by the logarithmic derivative of the transition function $e^{i\varphi}$. Now compute the curvature on each chart. Since $d(d\varphi)=0$, \begin{align*} dA_N=\frac{i}{2}d(1-\cos\theta)\wedge d\varphi. \end{align*} Because $d(1-\cos\theta)=\sin\theta\,d\theta$, this becomes \begin{align*} dA_N=\frac{i}{2}\sin\theta\,d\theta\wedge d\varphi. \end{align*} Similarly, \begin{align*} dA_S=-\frac{i}{2}d(1+\cos\theta)\wedge d\varphi. \end{align*} Since $d(1+\cos\theta)=-\sin\theta\,d\theta$, we get \begin{align*} dA_S=\frac{i}{2}\sin\theta\,d\theta\wedge d\varphi. \end{align*} Thus both local potentials define the same global curvature form \begin{align*} F=\frac{i}{2}\sin\theta\,d\theta\wedge d\varphi. \end{align*} Using the orientation for which $\theta\in[0,\pi]$ and $\varphi\in[0,2\pi]$ give the standard area form $\sin\theta\,d\theta\wedge d\varphi$, its Chern-Weil period is \begin{align*} \frac{i}{2\pi}\int_{S^2}F=\frac{i}{2\pi}\int_0^{2\pi}\int_0^\pi \frac{i}{2}\sin\theta\,d\theta\,d\varphi. \end{align*} Since $i^2=-1$, this is \begin{align*} -\frac{1}{4\pi}\int_0^{2\pi}\int_0^\pi \sin\theta\,d\theta\,d\varphi. \end{align*} The inner integral is \begin{align*} \int_0^\pi \sin\theta\,d\theta=\left[-\cos\theta\right]_0^\pi=2, \end{align*} and the outer integral contributes \begin{align*} \int_0^{2\pi}d\varphi=2\pi. \end{align*} Therefore \begin{align*} \frac{i}{2\pi}\int_{S^2}F=-\frac{1}{4\pi}(2)(2\pi)=-1. \end{align*} The monopole curvature therefore represents the negative generator in the tautological sign convention, matching the first Chern class of $\mathcal O(-1)$. [/example] This local-potential description shows why no single smooth potential exists on the whole sphere. The obstruction is exactly the nonzero first Chern class. ## Grassmannians And Tautological Bundles The Hopf calculation is the rank-one case of a broader universal construction. Over a Grassmannian, every point is itself a subspace, so there is a canonical vector bundle whose fibre is that subspace. The question is how to compute its Chern classes in a way compatible with the Schubert-cell description of the Grassmannian. [definition: Complex Grassmannian] For integers $1\le k\le n$, the complex Grassmannian $\operatorname{Gr}(k,n)$ is the smooth complex manifold whose points are $k$-dimensional complex linear subspaces $V\subset \mathbb C^n$. [/definition] The Grassmannian is not only a parameter space; it carries the family of subspaces it parameterises. To express characteristic classes universally, we need both the selected subspace and the quotient left after removing it from the fixed ambient vector space. [definition: Tautological And Quotient Bundles On A Grassmannian] The tautological bundle $S\to \operatorname{Gr}(k,n)$ is the rank $k$ complex vector bundle with fibre \begin{align*} S_V=V\subset \mathbb C^n \end{align*} at $V\in \operatorname{Gr}(k,n)$. The universal quotient bundle $Q\to \operatorname{Gr}(k,n)$ is the rank $n-k$ complex vector bundle with fibre $Q_V=\mathbb C^n/V$. [/definition] These two bundles are linked by the universal short exact sequence \begin{align*} 0\longrightarrow S\longrightarrow \underline{\mathbb C}^n\longrightarrow Q\longrightarrow 0, \end{align*} where $\underline{\mathbb C}^n$ is the product rank $n$ bundle. The next theorem asks what the Whitney product formula forces when the middle term has no positive Chern classes. [quotetheorem:9802] [citeproof:9802] The relation $c(S)c(Q)=1$ determines how the two universal bundles interact, but the exact-sequence hypothesis is doing real work. If $E$ and $F$ are unrelated bundles over the same space, there is no reason for $c(E)c(F)$ to be $1$; for example, two copies of $\mathcal O(1)$ over $\mathbb{CP}^1$ have product total Chern class $(1+u)^2$, not $1$. The theorem also does not determine the individual classes geometrically inside the cohomology ring, because it only gives formal inverse relations. To identify these classes inside $H^*(\operatorname{Gr}(k,n);\mathbb Z)$, we need cycles that generate the cohomology ring and that arise from the geometry of subspaces. Schubert calculus supplies exactly such cycles by measuring how a varying $k$-plane meets a fixed flag. The next result records the convention used here and explains which special Schubert classes are represented by the Chern classes of the universal quotient bundle. [quotetheorem:9803] [citeproof:9803] The theorem packages the enumerative geometry of the Grassmannian into vector-bundle language, but it depends on the stated flag convention and the universal quotient bundle. With the incidence condition $\dim(V\cap F_{n-k+j})\ge 1$, the case $\operatorname{Gr}(1,n+1)$ and $j=1$ would be automatic rather than the hyperplane Schubert condition, so it would not represent $c_1(Q)$. The result identifies the special Chern classes, but it does not by itself give all multiplication rules among Schubert classes; those require the Pieri and Littlewood-Richardson rules. In computations, it lets us replace incidence conditions by Chern classes and then use formal algebra in the cohomology ring. [example: The Tautological Bundle Over Projective Space] Since $\mathbb{CP}^n=\operatorname{Gr}(1,n+1)$, the tautological bundle $S$ is the line bundle $\mathcal O(-1)$, and the universal quotient bundle $Q$ has rank $n$. Let $u=c_1(\mathcal O(1))$. Since $\mathcal O(-1)$ is dual to $\mathcal O(1)$, we have $c_1(S)=-u$, so \begin{align*} c(S)=1-u. \end{align*} By *Chern Relation For Tautological And Quotient Bundles*, the universal exact sequence gives \begin{align*} c(S)c(Q)=1. \end{align*} Thus $c(Q)$ is the multiplicative inverse of $1-u$ in \begin{align*} H^*(\mathbb{CP}^n;\mathbb Z)=\mathbb Z[u]/(u^{n+1}). \end{align*} To compute this inverse, multiply: \begin{align*} (1-u)(1+u+u^2+\cdots+u^n)=1+u+u^2+\cdots+u^n-u-u^2-\cdots-u^n-u^{n+1}. \end{align*} All intermediate powers cancel in pairs, leaving \begin{align*} (1-u)(1+u+u^2+\cdots+u^n)=1-u^{n+1}. \end{align*} Since $u^{n+1}=0$ in $\mathbb Z[u]/(u^{n+1})$, this becomes \begin{align*} (1-u)(1+u+u^2+\cdots+u^n)=1. \end{align*} Therefore \begin{align*} c(Q)=1+u+u^2+\cdots+u^n. \end{align*} Comparing homogeneous degrees gives $c_j(Q)=u^j$ for $1\le j\le n$, so the $j$th Chern class of the universal quotient bundle is the special Schubert class of codimension $j$ in the projective-space convention. [/example] The projective-space case recovers the Hopf computation in degree two and extends it to all Chern classes of the universal quotient bundle. The finite truncation comes from the cohomological relation $u^{n+1}=0$. [example: Universal Quotient Bundle On $\operatorname{Gr}(2,4)$] On $\operatorname{Gr}(2,4)$, both the tautological bundle $S$ and the universal quotient bundle $Q$ have rank $2$. Write \begin{align*} c(Q)=1+\sigma_1+\sigma_2 \end{align*} and \begin{align*} c(S)=1+s_1+s_2, \end{align*} where $s_j=c_j(S)$ and $\sigma_j=c_j(Q)$. By *Chern Relation For Tautological And Quotient Bundles*, the universal exact sequence gives \begin{align*} (1+s_1+s_2)(1+\sigma_1+\sigma_2)=1. \end{align*} Comparing degree-$2$ terms gives \begin{align*} s_1+\sigma_1=0, \end{align*} so \begin{align*} s_1=-\sigma_1. \end{align*} Comparing degree-$4$ terms gives \begin{align*} s_2+s_1\sigma_1+\sigma_2=0. \end{align*} Substituting $s_1=-\sigma_1$ yields \begin{align*} s_2+(-\sigma_1)\sigma_1+\sigma_2=0. \end{align*} Thus \begin{align*} s_2-\sigma_1^2+\sigma_2=0, \end{align*} and hence \begin{align*} s_2=\sigma_1^2-\sigma_2. \end{align*} Therefore \begin{align*} c(S)=1-\sigma_1+(\sigma_1^2-\sigma_2). \end{align*} This expresses the first two Chern classes of the tautological $2$-plane bundle in terms of the special Schubert classes of the quotient bundle. [/example] ## Universal Connections And Curvature Formulae The final computation explains why the preceding characteristic forms are canonical rather than arbitrary representatives. Grassmannians carry natural Hermitian geometry inherited from the fixed Hermitian [inner product](/page/Inner%20Product) on $\mathbb C^n$. If a section is differentiated inside the ambient product bundle, the derivative usually points partly outside the tautological subspace, so projection is the operation that turns ambient differentiation into a connection on the moving fibres. [Orthogonal projection](/theorems/437) onto the tautological subspace defines a universal connection, and its curvature has a compact formula in terms of the projection operator. [definition: Projection Model Of The Tautological Bundle] Identify $\operatorname{Gr}(k,n)$ with the set of Hermitian orthogonal projections $P\in \operatorname{End}(\mathbb C^n)$ satisfying $P^2=P$, $P^*=P$, and $\operatorname{rank}P=k$. The tautological bundle is realised as \begin{align*} S=\{(P,v)\in \operatorname{Gr}(k,n)\times \mathbb C^n: Pv=v\}. \end{align*} [/definition] In this model, a section of $S$ is a function $s$ into $\mathbb C^n$ satisfying $Ps=s$. Ordinary differentiation may leave the moving subspace, so $ds$ is generally a vector in the ambient product bundle rather than a one-form with values in $S$. To use the tautological bundle in Chern-Weil theory, one needs an actual connection whose output remains tangent to the moving fibre over each projection $P$. The natural candidate is to differentiate in the ambient trivial bundle and then project back into the tautological subspace; the following definition packages that operation as a connection. [definition: Universal Connection On The Tautological Bundle] The universal connection on $S\to\operatorname{Gr}(k,n)$ is the map \begin{align*} \nabla : \Gamma(S) \longrightarrow \Omega^1(\operatorname{Gr}(k,n);S) \end{align*} defined by \begin{align*} \nabla s = P\,ds \end{align*} for every smooth section $s$ of $S$, where $d$ is the exterior derivative in the product bundle $\operatorname{Gr}(k,n)\times\mathbb C^n$. [/definition] The definition makes the connection canonical, but Chern-Weil computation requires its curvature. The projection identity $P^2=P$ is strong enough to reduce the curvature to a compact formula involving only $P$ and $dP$. [quotetheorem:9804] [citeproof:9804] Substituting this curvature into invariant polynomials gives differential-form representatives for the Chern classes already described by Schubert calculus. The projection model is essential here: if $P$ is replaced by an arbitrary endomorphism-valued function that is not an orthogonal projection, the identities $P^2=P$ and $P^*=P$ fail, and the formula $P\,dP\wedge dP\,P$ need not be the curvature of a connection on a moving subbundle. Even for a genuine subbundle, choosing a non-projected ambient derivative gives a connection whose curvature differs by exact Chern-Weil transgression terms. The theorem gives the curvature of this canonical universal connection, but it does not say that every connection on $S$ has the same curvature form. This is the promised meeting point of geometry and topology. [example: Chern-Weil Forms From The Projection Formula] For the tautological bundle $S\to\operatorname{Gr}(k,n)$ in the projection model, *Curvature Formula For The Universal Connection* gives \begin{align*} F_\nabla=P\,dP\wedge dP\,P. \end{align*} Substituting this curvature into the line-bundle normalization of *[First Chern Form Of A Hermitian Line Bundle](/theorems/9800)*, with trace replacing the scalar curvature in rank $k$, gives \begin{align*} c_1(\nabla)=\frac{i}{2\pi}\operatorname{tr}(F_\nabla). \end{align*} Using the displayed curvature formula, this becomes \begin{align*} c_1(\nabla)=\frac{i}{2\pi}\operatorname{tr}(P\,dP\wedge dP\,P). \end{align*} Since $P$ has degree $0$ as a matrix-valued form, cyclicity of the trace also gives \begin{align*} \operatorname{tr}(P\,dP\wedge dP\,P)=\operatorname{tr}(P^2\,dP\wedge dP). \end{align*} Because $P^2=P$, the same form may be written as \begin{align*} c_1(\nabla)=\frac{i}{2\pi}\operatorname{tr}(P\,dP\wedge dP). \end{align*} For all Chern forms at once, set \begin{align*} B=\frac{i}{2\pi}F_\nabla. \end{align*} The total Chern form is defined by expanding the determinant \begin{align*} \det(I+tB)=1+c_1(\nabla)t+c_2(\nabla)t^2+\cdots+c_k(\nabla)t^k. \end{align*} The coefficient of $t$ is the trace: \begin{align*} c_1(\nabla)=\operatorname{tr}(B)=\frac{i}{2\pi}\operatorname{tr}(F_\nabla). \end{align*} The coefficient of $t^2$ is the second elementary symmetric polynomial in the curvature eigenvalues, equivalently \begin{align*} c_2(\nabla)=\frac{1}{2}\left(\operatorname{tr}(B)\wedge\operatorname{tr}(B)-\operatorname{tr}(B\wedge B)\right). \end{align*} Substituting $B=\frac{i}{2\pi}F_\nabla$ gives \begin{align*} c_2(\nabla)=\frac{1}{2}\left(\frac{i}{2\pi}\right)^2\left(\operatorname{tr}(F_\nabla)\wedge\operatorname{tr}(F_\nabla)-\operatorname{tr}(F_\nabla\wedge F_\nabla)\right). \end{align*} The same determinant expansion gives each higher $c_j(\nabla)$ as the $j$th elementary symmetric polynomial in $\frac{i}{2\pi}F_\nabla$. Thus the projection formula gives explicit closed differential-form representatives for the Chern classes of $S$. By *Chern Relation For Tautological And Quotient Bundles* and *Schubert-Compatible Chern Classes*, their de Rham cohomology classes are the same classes determined algebraically from the universal exact sequence and the Schubert basis. [/example] The representative-bundle computations in this chapter form a template for later applications. A canonical connection supplies curvature, Chern-Weil theory turns curvature into closed forms, and comparison with integral generators identifies the characteristic classes. # 11. Four-Dimensional Geometry And Instanton Examples This chapter brings the Chern-Weil constructions, characteristic-number pairings, and index-theoretic densities of the course into the special geometry of oriented Riemannian four-manifolds. In real dimension four, the Hodge star splits two-forms into self-dual and anti-self-dual parts, so curvature itself decomposes into two geometrically meaningful halves. The main theme is that analytic quantities such as Yang-Mills energy are constrained by characteristic numbers, and that the same curvature algebra produces the Euler characteristic, signature, and instanton number. ## Yang-Mills Energy and the Second Chern Number The first question is how much curvature a connection on an $SU(2)$-bundle over a compact four-manifold is forced to have by topology. Chern-Weil theory gives a closed four-form representing the second Chern class, while Riemannian geometry gives an $L^2$ norm of curvature. The special feature in dimension four is that both are obtained by integrating four-forms. Let $X$ be a compact oriented Riemannian four-manifold, and let $E \to X$ be a rank-$2$ complex vector bundle with structure group $SU(2)$. A unitary connection $A$ on $E$ has curvature $F_A \in \Omega^2(X;\mathfrak{su}(E))$. Before asking for the best possible curvature bound, we need the analytic quantity that records the size of this curvature. [definition: Yang-Mills Energy] Let $E \to X$ be a Hermitian vector bundle over a compact Riemannian manifold $X$, and let $\mathcal A(E)$ denote the affine space of compatible connections on $E$. The Yang-Mills energy is the functional \begin{align*} YM: \mathcal A(E) \longrightarrow \mathbb R_{\ge 0}. \end{align*} It is defined on a connection $A$ by \begin{align*} YM(A)=\int_X |F_A|^2\,d\operatorname{vol}_g. \end{align*} [/definition] This energy measures the total squared curvature of the connection. In four dimensions it is scale-invariant: under a conformal change of metric, the pointwise norm of a two-form and the volume form scale in opposite ways. To turn this analytic number into a topological lower bound, we next need the characteristic number that Chern-Weil theory computes from an $SU(2)$ curvature form. [definition: Second Chern Number] Let $E \to X$ be a rank-$2$ complex vector bundle over a closed oriented four-manifold $X$. The second Chern number of $E$ is \begin{align*} c_2(E)[X] := \int_X c_2(E), \end{align*} where the integrand is any closed $4$-form representing the real image of the cohomology class $c_2(E)$. [/definition] For an $SU(2)$-bundle the first Chern class vanishes, so $c_2(E)$ is the first nonzero Chern class. The definition is topological, but the course needs a curvature formula because the Yang-Mills energy is also expressed through $F_A$. This motivates the following theorem, which fixes the precise normalisation of the instanton number. [quotetheorem:9805] [citeproof:9805] This formula is the bridge between curvature and topology, but its hypotheses are doing real work. Closedness gives a compact fundamental class without boundary, orientation fixes the sign of that fundamental class, and the $SU(2)$ assumption gives trace-free curvature and $c_1(E)=0$. For a general $U(2)$-bundle, the trace-square Chern-Weil form computes the degree-four component of the Chern character, \begin{align*} \frac{1}{2}(c_1^2-2c_2) \end{align*} rather than only $c_2$, so the trace-square integral no longer isolates the second Chern number without accounting for the $c_1^2$ contribution. The determinant formula for the total Chern class still has degree-four Chern component $c_2$. If the orientation of $X$ is reversed, the integral changes sign because the fundamental class changes sign. On a noncompact four-manifold such as $\mathbb R^4$, the integral may converge only after imposing decay conditions on $F_A$, and without those conditions it is not a characteristic number of a compact fundamental class. The theorem does not classify bundles or connections with a given value of $k$; it only says that every connection on the same closed $SU(2)$-bundle gives the same characteristic number after integration. The simplest test case is a connection with no curvature at all, which shows how topology obstructs flatness. [example: Flat Bundles Have Zero Instanton Number] Let $A$ be a flat $SU(2)$-connection on $E \to X$. Flatness means that its curvature two-form vanishes: \begin{align*} F_A=0. \end{align*} By the *Chern-Weil Formula For The Instanton Number*, \begin{align*} c_2(E)[X]=-\frac{1}{8\pi^2}\int_X \operatorname{tr}(F_A\wedge F_A). \end{align*} Substituting $F_A=0$ gives \begin{align*} F_A\wedge F_A=0\wedge 0=0. \end{align*} Hence \begin{align*} \operatorname{tr}(F_A\wedge F_A)=\operatorname{tr}(0)=0. \end{align*} Therefore \begin{align*} c_2(E)[X]=-\frac{1}{8\pi^2}\int_X 0=0. \end{align*} So any $SU(2)$-bundle with $c_2(E)[X]\ne 0$ cannot admit a flat $SU(2)$-connection; on such a bundle, every connection must have nonzero curvature somewhere. [/example] ## Self-Dual and Anti-Self-Dual Curvature The central four-dimensional question is whether curvature can sit entirely in one half of the two-form space. On an oriented Riemannian four-manifold, the Hodge star is the bundle operator \begin{align*} *: \Lambda^2T^*X \longrightarrow \Lambda^2T^*X, \end{align*} and hence the induced operator $*: \Omega^2(X) \to \Omega^2(X)$ on smooth two-forms. In real dimension four it satisfies $*^2=1$ on two-forms, giving a splitting that has no analogue in most other dimensions. The preceding section expressed $c_2(E)[X]$ as an integral of $F_A\wedge F_A$, but it has not yet related that integral to $|F_A|^2$. The missing ingredient is the decomposition of two-forms into the two eigenspaces of this explicitly defined Hodge star operator. [definition: Self-Dual And Anti-Self-Dual Two-Forms] Let $(X,g)$ be an oriented Riemannian four-manifold. A two-form $\omega \in \Omega^2(X)$ is self-dual if $*\omega=\omega$, and anti-self-dual if $*\omega=-\omega$. [/definition] The corresponding subbundles are denoted $\Lambda^2_+$ and $\Lambda^2_-$. Every two-form decomposes uniquely as \begin{align*} \omega = \omega^+ + \omega^-, \qquad \omega^+ = \frac{1}{2}(\omega+*\omega), \qquad \omega^- = \frac{1}{2}(\omega-*\omega). \end{align*} For a connection $A$, the curvature decomposes as $F_A=F_A^+ + F_A^-$ with $F_A^\pm \in \Omega^2(X;\mathfrak{su}(E)\otimes \Lambda^2_\pm)$. Once curvature has this splitting, a natural way to minimise energy is to require one summand to vanish. [definition: Instanton] Let $E \to X$ be an $SU(2)$-bundle over an oriented Riemannian four-manifold. A connection $A$ is anti-self-dual if \begin{align*} F_A^+ = 0, \end{align*} and self-dual if \begin{align*} F_A^- = 0. \end{align*} [/definition] A connection satisfying either of these equations is called an instanton. The sign convention varies across gauge theory sources. In these notes, with \begin{align*} c_2(E)[X] = -\frac{1}{8\pi^2}\int_X \operatorname{tr}(F_A \wedge F_A) \end{align*} and the usual negative definite trace pairing on $\mathfrak{su}(2)$, the self-dual equation $F_A^-=0$ is the model equation when $c_2(E)[X]>0$. This motivates the following theorem, because it identifies the first-order instanton equation with equality in the topological Yang-Mills bound. [quotetheorem:9806] [citeproof:9806] This theorem explains why instantons are absolute minima of Yang-Mills energy in a fixed topological class, subject to the stated conventions. Compactness is needed so both the energy and Chern-Weil integral are finite global quantities, orientation fixes the sign of $k$, and the $SU(2)$ trace normalisation determines whether the self-dual or anti-self-dual equation gives equality for positive $k$. Reversing the orientation swaps $\Lambda^2_+$ and $\Lambda^2_-$ and changes the sign of the fundamental class, so the same connection may be described by the opposite equation after orientation reversal. For a different structure group, a different trace pairing, or the reversed orientation on $X$, the displayed equality condition must be rechecked rather than copied unchanged. The result also does not assert that an instanton exists in every topological class; it only says that if equality is achieved, then one Hodge component of curvature must vanish. The basic example shows that the lower bound is sometimes attained, not merely formal. [example: BPST Instanton On The Four-Sphere] With the orientation and trace convention used in these notes, the basic BPST connection is an $SU(2)$-connection $A$ on a bundle $E\to S^4$ with \begin{align*} c_2(E)[S^4]=1. \end{align*} Under stereographic coordinates $S^4\setminus\{\infty\}\cong \mathbb R^4$ and the standard identification $\mathbb R^4\cong\mathbb H$, the usual BPST formula depends on a scale parameter $\lambda>0$; this parameter changes the pointwise concentration of the curvature but not the characteristic number. The curvature is self-dual in the convention of these notes, so \begin{align*} F_A^-=0. \end{align*} By the *Chern-Weil Formula For The Instanton Number*, \begin{align*} c_2(E)[S^4]=-\frac{1}{8\pi^2}\int_{S^4}\operatorname{tr}(F_A\wedge F_A). \end{align*} Substituting $c_2(E)[S^4]=1$ gives \begin{align*} 1=-\frac{1}{8\pi^2}\int_{S^4}\operatorname{tr}(F_A\wedge F_A). \end{align*} Multiplying both sides by $-8\pi^2$ gives \begin{align*} \int_{S^4}\operatorname{tr}(F_A\wedge F_A)=-8\pi^2. \end{align*} Equivalently, \begin{align*} -\frac{1}{8\pi^2}\int_{S^4}\operatorname{tr}(F_A\wedge F_A)=1. \end{align*} Now $k=c_2(E)[S^4]=1>0$, and the curvature satisfies $F_A^-=0$. Therefore the equality case of the *[Yang-Mills Energy Bound By The Second Chern Number](/theorems/9806)* applies: \begin{align*} YM(A)=8\pi^2|k|. \end{align*} Since $|k|=|1|=1$, this becomes \begin{align*} YM(A)=8\pi^2. \end{align*} If the orientation of $S^4$ is reversed, then the Hodge star on two-forms changes sign, so the same curvature obeys $*_{\mathrm{new}}F_A=-F_A$ instead of $*F_A=F_A$. Thus the same local BPST formula is anti-self-dual for the reversed orientation, which explains the opposite convention in many gauge theory texts. This example shows that the lower bound $YM(A)\ge 8\pi^2|c_2(E)[S^4]|$ is sharp. [/example] ## Pontryagin and Euler Densities in Dimension Four The final question is how the same four-dimensional curvature algebra controls the topology of the tangent bundle. For an oriented Riemannian four-manifold, the Levi-Civita connection on $TX$ produces Pontryagin and Euler forms whose integrals recover characteristic numbers of $X$ itself. The instanton discussion used the curvature of an auxiliary $SU(2)$-bundle. For the tangent bundle, the relevant connection is the Levi-Civita connection and the curvature takes values in $\mathfrak{so}(4)$. The trace-square Chern-Weil form gives the Pontryagin density, which is the curvature input for the signature. This trace-square construction detects the Pontryagin class but it does not recover the Euler characteristic, because the Euler class is governed by the skew-symmetric determinant data encoded by the Pfaffian rather than by the trace of $\Omega\wedge\Omega$. [definition: Pontryagin Density] Let $(X,g)$ be an oriented Riemannian four-manifold with Levi-Civita curvature matrix $\Omega \in \Omega^2(X;\mathfrak{so}(4))$. The first Pontryagin density is the Chern-Weil four-form \begin{align*} p_1: \Omega^2(X;\mathfrak{so}(4)) \longrightarrow \Omega^4(X). \end{align*} It is defined by \begin{align*} p_1(\Omega) := -\frac{1}{8\pi^2}\operatorname{tr}(\Omega\wedge\Omega). \end{align*} [/definition] Integrating this form gives the first Pontryagin number $p_1(TX)[X]$. In dimension four, the signature of the intersection form is the topological invariant controlled by this number. The next formula is the four-dimensional form of Hirzebruch's signature theorem. [quotetheorem:9807] [citeproof:9807] The signature formula accounts for the difference between the positive and negative parts of the symmetric middle-dimensional intersection form through the Pontryagin form. Compactness is needed to integrate the Chern-Weil form over the fundamental class; on a noncompact four-manifold such as the total space of a gravitational instanton, the integral of $p_1(\Omega)$ may diverge or depend on decay conditions and eta-invariant corrections at infinity, so it need not define the ordinary signature by itself. If $X$ has boundary, the Atiyah-Patodi-Singer signature formula includes a boundary eta term, and the interior integral alone does not usually equal $\sigma(X)$. Orientation fixes both the fundamental class and the sign convention for the intersection form: reversing the orientation changes the sign of $\sigma(X)$ and also changes the sign of the Pontryagin number pairing $p_1(TX)[X]$. The formula does not determine the pointwise curvature of a metric, nor does it imply that $p_1(\Omega)$ is pointwise positive or negative; it only fixes its total integral. A second topological invariant, the Euler characteristic, is not captured by the trace-square polynomial alone. The missing algebraic operation is the Pfaffian: it is the invariant polynomial on skew-symmetric curvature matrices whose Chern-Weil form represents the Euler class, so it records orientation-sensitive determinant data that trace-square cannot see. This motivates the following definition. [definition: Euler Density] Let $(X,g)$ be an oriented Riemannian four-manifold with Levi-Civita curvature matrix $\Omega \in \Omega^2(X;\mathfrak{so}(4))$. The Euler density is the Chern-Weil four-form \begin{align*} e: \Omega^2(X;\mathfrak{so}(4)) \longrightarrow \Omega^4(X). \end{align*} It is defined by \begin{align*} e(\Omega) := \operatorname{Pf}\left(\frac{\Omega}{2\pi}\right). \end{align*} [/definition] The Euler density integrates to the Euler characteristic. In dimension four, the same density can also be written using the curvature decomposition into Weyl, Ricci, and scalar curvature pieces. This motivates the following theorem, which records both the characteristic-class statement and the explicit Riemannian density. [quotetheorem:9808] [citeproof:9808] Together, the signature and Euler formulas show that curvature components cannot vary independently on a fixed compact four-manifold. Compactness again turns the local density into a global number; on a noncompact manifold the curvature density can fail to be integrable, and even when it is integrable its value may depend on the geometry at infinity rather than only on the ordinary Euler characteristic. If $X$ has boundary, Chern-Gauss-Bonnet acquires an additional boundary term built from the induced metric and second fundamental form, so the interior integral alone does not usually equal $\chi(X)$. The Levi-Civita connection matters because its curvature is the Chern-Weil input for the characteristic classes of $TX$. The theorem does not say that the displayed integrand is pointwise topological: different metrics can redistribute the Weyl, Ricci, and scalar curvature terms while leaving the total integral equal to $\chi(X)$. In later computations, the practical strategy is often to compute characteristic classes algebraically and then use these curvature formulas as consistency checks. The tangent bundle of complex projective space is the standard computation where Chern classes, Pontryagin classes, Euler characteristic, and signature all meet. [example: Tangent Bundle Of Complex Projective Plane] For $\mathbb{CP}^2$ with its standard orientation, let $h\in H^2(\mathbb{CP}^2;\mathbb Z)$ be the hyperplane class normalized by \begin{align*} \int_{\mathbb{CP}^2}h^2=1. \end{align*} The Euler sequence gives the total Chern class of the complex tangent bundle as \begin{align*} c(T\mathbb{CP}^2)=(1+h)^3 \end{align*} through degree four. Expanding the binomial gives \begin{align*} (1+h)^3=1+3h+3h^2+h^3. \end{align*} Since $\mathbb{CP}^2$ has real dimension $4$, the degree-six class $h^3$ vanishes in cohomology, so the part relevant to $T\mathbb{CP}^2$ is \begin{align*} c(T\mathbb{CP}^2)=1+3h+3h^2. \end{align*} Matching this with \begin{align*} c(T\mathbb{CP}^2)=1+c_1(T\mathbb{CP}^2)+c_2(T\mathbb{CP}^2) \end{align*} gives \begin{align*} c_1(T\mathbb{CP}^2)=3h. \end{align*} It also gives \begin{align*} c_2(T\mathbb{CP}^2)=3h^2. \end{align*} For a complex surface with its complex orientation, the Euler class of the underlying real tangent bundle is the top Chern class, so \begin{align*} \chi(\mathbb{CP}^2)=\int_{\mathbb{CP}^2}c_2(T\mathbb{CP}^2). \end{align*} Substituting $c_2(T\mathbb{CP}^2)=3h^2$ gives \begin{align*} \chi(\mathbb{CP}^2)=\int_{\mathbb{CP}^2}3h^2. \end{align*} By linearity of integration, \begin{align*} \int_{\mathbb{CP}^2}3h^2=3\int_{\mathbb{CP}^2}h^2. \end{align*} Using $\int_{\mathbb{CP}^2}h^2=1$, we obtain \begin{align*} \chi(\mathbb{CP}^2)=3. \end{align*} For the underlying real bundle of a complex rank-$2$ bundle, the first Pontryagin class satisfies the standard relation \begin{align*} p_1=c_1^2-2c_2. \end{align*} Here \begin{align*} c_1^2=(3h)^2=9h^2. \end{align*} Also, \begin{align*} 2c_2=2(3h^2)=6h^2. \end{align*} Therefore \begin{align*} p_1(T\mathbb{CP}^2)=9h^2-6h^2=3h^2. \end{align*} By the *Signature Formula In Dimension Four*, \begin{align*} \sigma(\mathbb{CP}^2)=\frac{1}{3}\int_{\mathbb{CP}^2}p_1(T\mathbb{CP}^2). \end{align*} Substituting $p_1(T\mathbb{CP}^2)=3h^2$ gives \begin{align*} \sigma(\mathbb{CP}^2)=\frac{1}{3}\int_{\mathbb{CP}^2}3h^2. \end{align*} Since $\int_{\mathbb{CP}^2}3h^2=3$, this becomes \begin{align*} \sigma(\mathbb{CP}^2)=\frac{1}{3}\cdot 3=1. \end{align*} Thus $\mathbb{CP}^2$ has Euler characteristic $3$ and signature $1$, with both numbers coming from the same generators $c_2=3h^2$ and $p_1=3h^2$ in degree four. [/example] The projective plane has positive signature and small Euler characteristic, making it a useful calibration point for the formulas. A K3 surface gives a very different benchmark: its first Chern class vanishes, but its second Chern number is large enough to force a large negative signature. [example: K3 Characteristic Numbers] Let $X$ be a K3 surface. By definition, $X$ is a compact complex surface with \begin{align*} c_1(TX)=0 \end{align*} and \begin{align*} \int_X c_2(TX)=24. \end{align*} For a complex surface with its complex orientation, the Euler class of the underlying real tangent bundle is the top Chern class, so \begin{align*} \chi(X)=\int_X c_2(TX). \end{align*} Substituting the K3 value gives \begin{align*} \chi(X)=24. \end{align*} For the underlying real bundle of a complex rank-$2$ bundle, the first Pontryagin class satisfies \begin{align*} p_1=c_1^2-2c_2. \end{align*} Since $c_1(TX)=0$, we have \begin{align*} c_1(TX)^2=0^2=0. \end{align*} Thus \begin{align*} p_1(TX)=0-2c_2(TX)=-2c_2(TX). \end{align*} Integrating over $X$ gives \begin{align*} \int_X p_1(TX)=\int_X -2c_2(TX). \end{align*} By linearity of integration, \begin{align*} \int_X -2c_2(TX)=-2\int_X c_2(TX). \end{align*} Using $\int_X c_2(TX)=24$, this becomes \begin{align*} \int_X p_1(TX)=-2\cdot 24=-48. \end{align*} By the *Signature Formula In Dimension Four*, \begin{align*} \sigma(X)=\frac{1}{3}\int_X p_1(TX). \end{align*} Substituting $\int_X p_1(TX)=-48$ gives \begin{align*} \sigma(X)=\frac{1}{3}(-48)=-16. \end{align*} Thus a K3 surface has Euler characteristic $24$ and signature $-16$, with the negative signature coming from the relation $p_1=-2c_2$ when $c_1=0$. [/example] The chapter closes the course's passage from abstract characteristic classes to concrete curvature identities. In four dimensions, Chern-Weil forms are not merely representatives of cohomology classes: their pointwise decomposition interacts with the Hodge star, producing sharp energy bounds and rigid topological formulas. # 12. Synthesis: Characteristic Classes As Obstructions And Measurements Chapters 1 through 11 built characteristic classes from invariant polynomials, curvature, splitting principles, transgression, index densities, and representative-bundle computations. This final chapter gathers those constructions into a working synthesis: characteristic classes are simultaneously obstructions to simplifying a bundle, measurements extracted from geometric choices, and computational devices controlled by functoriality. The aim is to make precise how the same class can block a section, record curvature, distinguish bundles, or appear as the boundary term between two geometric representatives. ## Obstructions To Product Structures And Reduction Of Structure Group The first global question about a bundle is whether it can be simplified: can it be made into a product bundle, can it admit a nowhere-zero section, or can its structure group be reduced? Characteristic classes answer these questions by converting a geometric problem into a cohomological obstruction. The important point is not that every obstruction is a characteristic class, but that the standard characteristic classes were designed to detect exactly these failures in the most common geometric situations. A complex line bundle is the basic model. Its transition functions take values in $U(1)$, so the first Chern class measures the twisting which prevents a global non-vanishing frame from existing. [quotetheorem:9809] [citeproof:9809] This theorem is the cleanest instance of the obstruction philosophy, but its hypotheses matter. Paracompactness is what allows the usual classification of principal bundles by maps into $BU(1)$ and ensures that the local-to-global constructions of line bundles behave well; without it, principal-bundle classification can fail to be represented by ordinary homotopy classes of classifying maps. The conclusion is also special to complex rank one: for higher-rank complex bundles, the total Chern class is a powerful invariant but it does not classify all bundles in general. A concrete example occurs over $S^6$: complex rank-two bundles are classified by the clutching group $\pi_5(U(2))\cong \mathbb Z/2$, while \begin{align*} H^2(S^6;\mathbb Z)=H^4(S^6;\mathbb Z)=0, \end{align*} so both the product bundle and the non-product rank-two bundle have the same total Chern class $1$. Finally, $c_1(L)$ classifies the isomorphism class of the underlying line bundle, not a chosen local trivialisation, Hermitian metric, connection, or unitary frame; those extra choices may change representatives without changing the class. A line bundle has a global nowhere-zero section exactly when it is a product line bundle, so $c_1(L)$ is the complete obstruction to being a product in complex rank one. [example: Line Bundles On Complex Projective Space] On $\mathbb{C}P^n$ with $n\ge 1$, let $h\in H^2(\mathbb{C}P^n;\mathbb Z)$ be the positive generator, normalized by \begin{align*} c_1(\mathcal O(1))=h. \end{align*} For $k\ge 0$, the tensor power $\mathcal O(k)=\mathcal O(1)^{\otimes k}$ has first Chern class \begin{align*} c_1(\mathcal O(k))=c_1(\mathcal O(1)^{\otimes k})=kc_1(\mathcal O(1))=kh, \end{align*} where the middle equality is repeated use of the tensor product rule $c_1(L\otimes M)=c_1(L)+c_1(M)$. For negative $k$, write $k=-m$ with $m>0$. Since $\mathcal O(-1)=\mathcal O(1)^*$ and $\mathcal O(1)\otimes \mathcal O(1)^*\cong \mathcal O$, the tensor product rule gives \begin{align*} 0=c_1(\mathcal O)=c_1(\mathcal O(1))+c_1(\mathcal O(1)^*)=h+c_1(\mathcal O(-1)). \end{align*} Thus $c_1(\mathcal O(-1))=-h$, and applying the nonnegative tensor-power calculation to $\mathcal O(-1)^{\otimes m}$ gives \begin{align*} c_1(\mathcal O(k))=c_1(\mathcal O(-1)^{\otimes m})=m(-h)=kh. \end{align*} If $k\ne \ell$, then \begin{align*} c_1(\mathcal O(k))-c_1(\mathcal O(\ell))=(k-\ell)h\ne 0 \end{align*} because $H^2(\mathbb{C}P^n;\mathbb Z)\cong \mathbb Z h$ is torsion-free. By *First Chern Class Classifies Complex Line Bundles*, the line bundles $\mathcal O(k)$ and $\mathcal O(\ell)$ are therefore non-isomorphic. [/example] For real oriented bundles, the analogous obstruction is the Euler class. Instead of measuring the twisting of a complex line, it measures whether a real oriented bundle can avoid the zero section. [quotetheorem:2284] [citeproof:2284] The theorem gives a necessary condition, and the orientation hypothesis is not cosmetic. The integral Euler class is defined for oriented real bundles because its construction uses an oriented Thom class; without an orientation, the corresponding obstruction naturally lives with twisted coefficients or appears only after reducing coefficients, for example through Stiefel-Whitney classes. The vanishing of $e(E)$ is therefore not a universal classification theorem for sections in every setting: additional obstruction groups may appear when the base has high dimension relative to the rank, and special low-rank or nonorientable cases require separate invariants. It is especially concrete for tangent bundles, where a nowhere-zero section is a vector field without zeros. [example: The Hairy Ball Obstruction] For the oriented tangent bundle $TS^2\to S^2$, *Gauss-Bonnet As Euler Class Evaluation* gives \begin{align*} \int_{S^2} e(TS^2)=\chi(S^2) \end{align*} and the cell decomposition of $S^2$ with one $0$-cell and one $2$-cell gives \begin{align*} \chi(S^2)=1+1=2. \end{align*} Hence \begin{align*} \int_{S^2} e(TS^2)=2. \end{align*} If $e(TS^2)=0$ in $H^2(S^2;\mathbb Z)$, then its evaluation on the fundamental class would be \begin{align*} \int_{S^2} e(TS^2)=0, \end{align*} contradicting the value $2$. Thus $e(TS^2)\neq 0$, so *Euler Class Obstructs A Nowhere-Zero Section* implies that $S^2$ admits no nowhere-zero tangent vector field. For comparison, identify $S^1$ with $\{z\in \mathbb C:|z|=1\}$. The vector field $V_z=iz$ is tangent because the curve $\gamma_z(t)=ze^{it}$ lies in $S^1$ and satisfies \begin{align*} \gamma_z'(0)=iz. \end{align*} It is nowhere zero because $|iz|=|z|=1$. Therefore $V$ is a global frame for the real line bundle $TS^1$, so \begin{align*} TS^1\cong S^1\times \mathbb R. \end{align*} The contrast is exactly the obstruction detected above: $TS^2$ cannot have such a global tangent frame, while $TS^1$ does. [/example] Reduction of structure group is a broader version of the same problem. A reduction from $G$ to a subgroup $H$ means that the transition functions can be chosen to land in $H$; the next theorem is needed because it translates that geometric factorisation into a test on universal cohomology classes. To state the obstruction test precisely, the reduction has to be treated as additional bundle data rather than as a smaller-looking collection of local transition functions. The issue is that transition functions depend on choices of local trivialisations, while the existence of a reduction is an invariant property of the bundle itself. The definition therefore records a genuine principal $H$-bundle whose extension along $H\subset G$ recovers the original principal $G$-bundle. [definition: Reduction Of Structure Group] Let $P \to M$ be a principal $G$-bundle and let $H \subset G$ be a Lie subgroup. A reduction of structure group of $P$ to $H$ is a principal $H$-bundle $Q \to M$ together with an isomorphism of principal $G$-bundles \begin{align*} Q \times_H G \cong P. \end{align*} [/definition] The definition packages many familiar structures: Hermitian metrics, orientations, and spin structures all arise by replacing a large structure group with a smaller one. What cohomological condition must hold if such a replacement exists? Characteristic classes must then come from the classifying space of the smaller group, which is the obstruction test in the next theorem. [quotetheorem:9810] [citeproof:9810] This theorem is a template rather than a single computation, and it is only as strong as the chosen cohomology theory and the universal classes being tested. The classifying-space hypothesis is the bridge from geometry to cohomology: the reduction is being tested through an actual lift of the classifying map $f:M\to BG$ to $BH$, not through an arbitrary formal relation among classes. On bases where principal bundles are not controlled by homotopy classes of maps into classifying spaces, this argument no longer captures the full reduction problem. The compatibility of the classifying data also matters: even if a class on $M$ lies in the image of some map from $H^*(BH;R)$, it must arise from a single map $g:M\to BH$ whose composite with $i$ classifies the given bundle $P$. The factorisation condition is necessary for a reduction, but it need not be sufficient: characteristic classes can vanish while a reduction is still prevented by torsion, unstable homotopy, or other non-cohomological data. For instance, asking for a reduction of a rank-two complex bundle to the identity group is the same as asking for a global frame. Over $S^6$, the non-product rank-two complex bundle coming from the nonzero element of $\pi_5(U(2))\cong \mathbb Z/2$ has $c_1=c_2=0$ because the relevant cohomology groups vanish, but it still has no global complex frame. Thus Stiefel-Whitney classes obstruct orientability and spin structures, $c_1$ obstructs an $SU(n)$-reduction of a unitary bundle, and Euler classes obstruct splitting off a real line, but the absence of these particular obstructions is not a blanket existence theorem without the relevant classification result. [example: Vanishing First Chern Class For An SU Bundle] Let $E\to M$ be a rank $n$ Hermitian vector bundle with unitary frame bundle $P_{U(n)}(E)$. A reduction to $SU(n)$ means that, after choosing local unitary frames, the transition functions can be written as maps $g_{ij}:U_i\cap U_j\to SU(n)$. Taking determinants gives \begin{align*} \det(g_{ij})=1. \end{align*} But the determinant line bundle $\det E=\bigwedge^n E$ has transition functions $\det(g_{ij})$, so these transition functions are all $1$. Hence the local determinant frames glue to a global nowhere-zero section of $\det E$, and $\det E$ is a product line bundle. Now compute the first Chern class. By the splitting principle, after an injective pullback we may write $E$ as a sum of line bundles $L_1\oplus\cdots\oplus L_n$. Then \begin{align*} c(E)=\prod_{a=1}^n c(L_a)=\prod_{a=1}^n(1+c_1(L_a)). \end{align*} The degree-two part is therefore \begin{align*} c_1(E)=c_1(L_1)+\cdots+c_1(L_n). \end{align*} The determinant line pulls back to \begin{align*} \det E\cong L_1\otimes\cdots\otimes L_n. \end{align*} Repeated use of the tensor product rule for line bundles gives \begin{align*} c_1(\det E)=c_1(L_1)+\cdots+c_1(L_n). \end{align*} Thus $c_1(E)=c_1(\det E)$. Since a product line bundle has first Chern class $0$, an $SU(n)$-reduction forces \begin{align*} c_1(E)=c_1(\det E)=0. \end{align*} Conversely, on a space where complex line bundles are classified by $H^2(-;\mathbb Z)$, *First Chern Class Classifies Complex Line Bundles* implies that $c_1(\det E)=0$ exactly when $\det E$ is a product line bundle. In that setting, the obstruction measured by $c_1(E)$ is precisely the obstruction to choosing unitary frames whose determinant is globally normalized to $1$. [/example] ## Comparing Curvature Representatives Under Geometric Choices The second synthesis question is how a topological invariant can arise from choices such as a connection, a metric, or a local frame. Chern-Weil theory answers this by producing differential forms depending on the geometric choice while proving that their de Rham cohomology classes do not depend on the choice. Thus the form is geometric, but its cohomology class is topological. [definition: Chern-Weil Representative] Let $P\to M$ be a principal $G$-bundle with connection $A$ and curvature $F_A$. Let $\mathbb K$ be either $\mathbb R$ or $\mathbb C$, and let \begin{align*} P_0:\mathfrak g^k\to \mathbb K \end{align*} be a symmetric $k$-linear map invariant under the adjoint action of $G$. The Chern-Weil representative associated to $A$ is the closed differential form \begin{align*} P_0(F_A)\in \Omega^{2k}(M;\mathbb K). \end{align*} [/definition] The invariant polynomial may also be recorded as an element of $(\operatorname{Sym}^k\mathfrak g^*)^G=I^k(\mathfrak g;\mathbb K)$. This notation is useful because Chern-Weil theory depends on invariance of the polynomial, not on a particular basis for $\mathfrak g$. The notation hides a useful distinction. The connection $A$ changes the actual form $P_0(F_A)$, so the next point is to prove that changing $A$ changes only the representative and not the de Rham class. [quotetheorem:9811] [citeproof:9811] This theorem is the technical reason curvature can compute topology, and each hypothesis is doing work. The connections must be genuine connections on the same principal bundle so that the affine path $A_t$ and its curvature forms live in a common setting. The polynomial must be invariant under the adjoint action: this invariance is what combines with the Bianchi identity to make the derivative along the path exact rather than merely another differential form. The conclusion is also a de Rham statement; after passing to real cohomology, integral torsion information is lost, so Chern-Weil representatives cannot by themselves detect torsion characteristic classes. This raises the special case of tangent bundles, where the same real cohomology class can be represented by a curvature expression built from a Riemannian metric. [example: First Chern Form Of A Hermitian Line Bundle] Let $L\to M$ be a Hermitian line bundle with unitary connection $\nabla$. Choose a unitary local frame $s_i$ over $U_i$, and write \begin{align*} \nabla s_i=A_i\otimes s_i \end{align*} with $A_i$ an imaginary-valued $1$-form. On $U_i\cap U_j$, suppose $s_j=g_{ij}s_i$ with $g_{ij}:U_i\cap U_j\to U(1)$. Then \begin{align*} \nabla s_j=\nabla(g_{ij}s_i)=dg_{ij}\otimes s_i+g_{ij}A_i\otimes s_i=(dg_{ij}g_{ij}^{-1}+A_i)\otimes s_j. \end{align*} Since $U(1)$ is abelian, $dg_{ij}g_{ij}^{-1}=g_{ij}^{-1}dg_{ij}$, so \begin{align*} A_j=A_i+g_{ij}^{-1}dg_{ij}. \end{align*} Define the local curvature form by \begin{align*} F_i=dA_i. \end{align*} On an overlap, \begin{align*} F_j=dA_j=dA_i+d(g_{ij}^{-1}dg_{ij}). \end{align*} Using $d(g_{ij}^{-1})=-g_{ij}^{-2}dg_{ij}$, the second term is \begin{align*} d(g_{ij}^{-1}dg_{ij})=d(g_{ij}^{-1})\wedge dg_{ij}=-g_{ij}^{-2}dg_{ij}\wedge dg_{ij}=0. \end{align*} Therefore \begin{align*} F_j=F_i. \end{align*} The forms $F_i$ glue to a global imaginary-valued closed $2$-form $F_A$, closed because locally \begin{align*} dF_A=dF_i=d^2A_i=0. \end{align*} For the standard Chern-Weil normalization of a unitary line bundle, the invariant polynomial on $\mathfrak u(1)$ is \begin{align*} X\mapsto \frac{i}{2\pi}X. \end{align*} Thus the first Chern form is \begin{align*} \frac{i}{2\pi}F_A. \end{align*} Since $F_A$ is imaginary-valued, this is a real closed $2$-form, and its de Rham cohomology class is the image of $c_1(L)$ in $H^2_{\mathrm{dR}}(M)$. [/example] The line-bundle calculation shows how local forms glue into a global representative. What does the same principle give for an oriented tangent bundle in even dimension? The next theorem answers this by identifying the Pfaffian curvature form with the Euler class and the Euler characteristic. [quotetheorem:9812] [citeproof:9812] This formulation separates the analytic representative from the topological number, and the geometric assumptions cannot be dropped without changing the statement. Closedness ensures that integration over $M$ pairs the top-degree form with a fundamental class without boundary correction terms; for manifolds with boundary, Gauss-Bonnet includes boundary contributions involving the induced geometry. Orientability supplies the fundamental class and the integral Euler class used in the displayed evaluation. Even dimensionality is also essential for this Pfaffian curvature representative: it produces a top-degree form only for even-dimensional tangent bundles. For a closed odd-dimensional oriented manifold, the Euler characteristic vanishes, and the top Euler class is correspondingly constrained; over coefficient fields of characteristic not equal to $2$, the Euler class of an odd-rank oriented real bundle vanishes because it is $2$-torsion. In noncompact settings, convergence of the integral and boundary-at-infinity terms become extra analytic questions. [example: Round Two-Sphere] For the round sphere $S^2$ of radius $1$, the Gaussian curvature is $K=1$, and in dimension two the Euler form of the Levi-Civita connection is \begin{align*} e(\nabla^{TS^2})=\frac{1}{2\pi}K\,d\operatorname{area}. \end{align*} Substituting $K=1$ gives \begin{align*} e(\nabla^{TS^2})=\frac{1}{2\pi}\,d\operatorname{area}. \end{align*} Using spherical coordinates $(\theta,\phi)$ with $0\leq \theta\leq \pi$ and $0\leq \phi\leq 2\pi$, the area form on the unit sphere is $\sin\theta\,d\theta\,d\phi$, so \begin{align*} \operatorname{area}(S^2)=\int_0^{2\pi}\int_0^\pi \sin\theta\,d\theta\,d\phi. \end{align*} The inner integral is \begin{align*} \int_0^\pi \sin\theta\,d\theta=[-\cos\theta]_0^\pi=-\cos\pi+\cos 0=1+1=2. \end{align*} Therefore \begin{align*} \operatorname{area}(S^2)=\int_0^{2\pi}2\,d\phi=2[\phi]_0^{2\pi}=4\pi. \end{align*} Hence the Euler-form integral is \begin{align*} \int_{S^2}e(\nabla^{TS^2})=\int_{S^2}\frac{1}{2\pi}\,d\operatorname{area}=\frac{1}{2\pi}\operatorname{area}(S^2)=\frac{1}{2\pi}\cdot 4\pi=2. \end{align*} A cell decomposition of $S^2$ with one $0$-cell and one $2$-cell gives \begin{align*} \chi(S^2)=1+1=2. \end{align*} Thus the curvature integral equals the Euler characteristic in this case. [/example] The sphere calculation uses nonzero curvature to produce the Euler form. A different question arises when the curvature itself vanishes: primary Chern-Weil forms then disappear, so any remaining information must be carried by holonomy or secondary transgression data. [definition: Flat Connection] Let $P\to M$ be a principal $G$-bundle with connection $A$. The connection is flat if its curvature satisfies \begin{align*} F_A=0 \quad \text{in } \Omega^2(M;\operatorname{ad}P). \end{align*} [/definition] If $F_A=0$, every positive-degree Chern-Weil form $P_0(F_A)$ vanishes. This makes it necessary to compare a flat connection with a reference connection, because the difference between representatives is where secondary classes enter. [quotetheorem:9813] [citeproof:9813] This statement deliberately treats the simplified case with a chosen integral null datum rather than the full differential-character theory. Flatness matters because it kills the positive-degree Chern-Weil form $P_0(F_A)$, so the homotopy formula no longer compares two nonzero primary representatives; instead it says that the reference representative $P_0(F_{A_0})$ is the exterior derivative of a transgression form. The flat bundle may still carry nontrivial holonomy, and the secondary class records part of that remaining global information. The period lattice $\Lambda$ is needed because Chern-Weil forms produce real cohomology classes, while characteristic classes often have integral or otherwise discrete periods. Passing to $\mathbb R/\Lambda$ records the ambiguity caused by changing representatives by forms with periods in $\Lambda$. The integral trivialisation is extra data: the vanishing of the primary class identifies the obstruction as zero, but a specific null-homotopy or cochain-level trivialisation is what lets the transgression be turned into a well-defined mod-period class. If the primary class is torsion rather than zero, its image in real cohomology vanishes, but there need not be an actual integral trivialisation. In that situation secondary invariants are usually formulated using differential characters or related exact sequences, where torsion primary data and real transgression data are kept together. The theorem above isolates the case needed for the present synthesis: a chosen trivialisation lets the Chern-Simons class be represented by integrating transgression forms over closed cycles. ## Building Computations From Splitting, Functoriality, And Transgression The final computational question is how to avoid recomputing characteristic classes from curvature every time. The course has developed three reusable tools: split bundles into line-bundle pieces when possible, pull classes through natural maps, and compare representatives through transgression. Together they turn difficult bundle calculations into algebra in cohomology. [quotetheorem:7041] [citeproof:7041] The splitting principle is not a claim that every bundle is itself a sum of line bundles. The injectivity of $\pi^*$ is the point that makes the method legitimate: an identity proved on $Y$ descends to $X$ only because two classes on $X$ with the same pullback must already have been equal. Without injectivity, pullback can erase information; for example, the map $\{*\}\to S^2$ pulls the generator of $H^2(S^2;\mathbb Z)$ to zero, so equality after such a pullback says nothing about equality on $S^2$. Without passing to a flag bundle or another space with this injective pullback property, a formal expression in Chern roots would have no reason to describe the original bundle. The theorem also does not produce canonical line subbundles of $E$ on $X$; the line bundles $L_i$ live after pullback, and the Chern roots are bookkeeping devices for symmetric polynomials in characteristic classes. A concrete warning is $T\mathbb{C}P^2$: if it split as $\mathcal O(a)\oplus \mathcal O(b)$ on $\mathbb{C}P^2$, then its Chern classes would force $a+b=3$ and $ab=3$, which has no integer solution. The splitting-space construction creates line-bundle summands after pullback, where computations with roots are legitimate and then descend by injectivity. [example: Pontryagin Class Of An Oriented Rank-Two Bundle] Let $E\to X$ be an oriented real rank-two bundle, and choose a bundle metric. Rotation by $\pi/2$ in each oriented Euclidean fiber defines a complex structure $J$ on $E$, so $E$ becomes a complex line bundle $L$. With this convention, the Euler class of the oriented real bundle is the first Chern class of the associated complex line bundle: \begin{align*} e(E)=c_1(L). \end{align*} Write $x=c_1(L)$. The conjugate line bundle $\overline L$ has transition functions $\overline{g_{ij}}$ when $L$ has unitary transition functions $g_{ij}$. Since $g_{ij}\in U(1)$, one has $\overline{g_{ij}}=g_{ij}^{-1}$, so $\overline L\cong L^*$ and therefore \begin{align*} c_1(\overline L)=c_1(L^*)=-c_1(L)=-x. \end{align*} The complexification of the underlying real bundle splits as \begin{align*} E\otimes_{\mathbb R}\mathbb C\cong L\oplus \overline L. \end{align*} Hence its total Chern class is \begin{align*} c(E\otimes_{\mathbb R}\mathbb C)=c(L)c(\overline L)=(1+x)(1-x). \end{align*} Multiplying the two factors gives \begin{align*} (1+x)(1-x)=1-x^2. \end{align*} Thus \begin{align*} c_2(E\otimes_{\mathbb R}\mathbb C)=-x^2. \end{align*} By the defining convention for the first Pontryagin class, \begin{align*} p_1(E)=-c_2(E\otimes_{\mathbb R}\mathbb C). \end{align*} Substituting the computed value of $c_2$ gives \begin{align*} p_1(E)=-(-x^2)=x^2. \end{align*} Since $x=e(E)$, this becomes \begin{align*} p_1(E)=e(E)^2. \end{align*} So in real rank two, the first Pontryagin class contains no new degree-four information beyond the square of the Euler class. [/example] The rank-two computation shows the strength of checking identities after converting to line-bundle data. The next reusable rule is needed when the bundle is obtained by pullback, restriction, inclusion of a submanifold, or a classifying construction. [quotetheorem:9814] [citeproof:9814] Naturality is what makes characteristic classes measurements rather than labels attached by hand. The statement is about pullback because pullback preserves the entire bundle construction: transition functions, classifying maps, and connections all transport functorially along $f$. It is not a rule for comparing arbitrary bundles of the same rank over unrelated spaces, and it does not say that equality of characteristic classes forces an isomorphism. For Euler classes, the orientation compatibility is part of the data: reversing the orientation changes the sign of the Euler class, so the pulled-back orientation is needed for the displayed identity. [example: Detecting Non-Isomorphic Bundles Of Equal Rank] Let $X=\mathbb{C}P^2$, and let $h\in H^2(X;\mathbb Z)$ be the hyperplane class, so $c_1(\mathcal O(1))=h$. The two bundles \begin{align*} E=\mathcal O\oplus \mathcal O \end{align*} and \begin{align*} F=\mathcal O(1)\oplus \mathcal O(-1) \end{align*} both have complex rank $2$, because each is a direct sum of two complex line bundles. For the product line bundle $\mathcal O$, the total Chern class is \begin{align*} c(\mathcal O)=1. \end{align*} Using the Whitney product formula for direct sums gives \begin{align*} c(E)=c(\mathcal O\oplus\mathcal O)=c(\mathcal O)c(\mathcal O). \end{align*} Substituting $c(\mathcal O)=1$ gives \begin{align*} c(E)=1\cdot 1=1. \end{align*} For $F$, the line-bundle normalization gives \begin{align*} c(\mathcal O(1))=1+c_1(\mathcal O(1))=1+h. \end{align*} Since $\mathcal O(-1)=\mathcal O(1)^*$, the dual line bundle has first Chern class \begin{align*} c_1(\mathcal O(-1))=-c_1(\mathcal O(1))=-h. \end{align*} Hence \begin{align*} c(\mathcal O(-1))=1-h. \end{align*} Applying the Whitney product formula again, \begin{align*} c(F)=c(\mathcal O(1)\oplus\mathcal O(-1))=c(\mathcal O(1))c(\mathcal O(-1)). \end{align*} Substituting the two line-bundle classes gives \begin{align*} c(F)=(1+h)(1-h). \end{align*} Multiplying the factors, \begin{align*} (1+h)(1-h)=1-h+h-h^2=1-h^2. \end{align*} Therefore \begin{align*} c(F)=1-h^2. \end{align*} In $H^*(\mathbb{C}P^2;\mathbb Z)\cong \mathbb Z[h]/(h^3)$, the class $h^2$ is the nonzero generator of $H^4(\mathbb{C}P^2;\mathbb Z)$. Thus \begin{align*} c(E)=1\neq 1-h^2=c(F). \end{align*} Isomorphic complex vector bundles have the same total Chern class, so $E$ and $F$ are not isomorphic. The rank alone sees only that both bundles have two complex-dimensional fibers, while the total Chern class detects different global twisting. [/example] The previous example distinguishes bundles by using the cohomology ring of projective space. For the tangent bundle of $\mathbb{C}P^n$, the obstruction is that tangent vectors are not given directly as copies of the hyperplane line bundle, so the Whitney formula cannot be applied without first expressing the tangent bundle in terms of simpler bundles. The Euler sequence supplies that expression, converting tangent-bundle geometry into Chern class algebra. [quotetheorem:9815] [citeproof:9815] This formula packages both local geometry and global topology. The bundle is the holomorphic tangent bundle, so its transition functions are complex-linear and its Chern classes are the ordinary Chern classes of the underlying complex vector bundle. The Euler sequence is the geometric input: it expresses tangent directions on projective space as the quotient of $(n+1)$ copies of the hyperplane bundle by the product line, which is why the Whitney product formula reduces the calculation to powers of $1+h$. The displayed expression must be read inside $H^*(\mathbb{C}P^n;\mathbb Z)\cong \mathbb Z[h]/(h^{n+1})$. Terms of degree above $2n$ vanish because $h^{n+1}=0$, so $(1+h)^{n+1}$ is a compact way of writing the truncated total class. In particular, it proves that $T^{1,0}\mathbb{C}P^n$ is not usually a product bundle, since a product complex vector bundle has total Chern class $1$. The same computation will also feed later numerical checks, such as extracting Euler characteristics, top Chern classes, and restrictions of tangent bundles to projective subspaces. [example: Non-Product Tangent Bundle Of Projective Space] For $\mathbb{C}P^n$ with $n\ge 1$, let $h=c_1(\mathcal O(1))\in H^2(\mathbb{C}P^n;\mathbb Z)$. By *Chern Class Of The Tangent Bundle Of Complex Projective Space*, \begin{align*} c(T^{1,0}\mathbb{C}P^n)=(1+h)^{n+1} \end{align*} in $H^*(\mathbb{C}P^n;\mathbb Z)\cong \mathbb Z[h]/(h^{n+1})$. Expanding the binomial up to degree two gives \begin{align*} (1+h)^{n+1}=1+(n+1)h+\binom{n+1}{2}h^2+\cdots \end{align*} so the degree-two part is \begin{align*} c_1(T^{1,0}\mathbb{C}P^n)=(n+1)h. \end{align*} Since $H^2(\mathbb{C}P^n;\mathbb Z)\cong \mathbb Z h$, the equality $(n+1)h=0$ would imply $n+1=0$ in $\mathbb Z$. But $n\ge 1$, so $n+1\neq 0$, and therefore \begin{align*} c_1(T^{1,0}\mathbb{C}P^n)\neq 0. \end{align*} If $T^{1,0}\mathbb{C}P^n$ were a product complex vector bundle of rank $n$, then \begin{align*} T^{1,0}\mathbb{C}P^n\cong \mathbb{C}P^n\times \mathbb C^n\cong \mathcal O^{\oplus n}. \end{align*} For the product line bundle $\mathcal O$, one has $c(\mathcal O)=1$, and the Whitney product formula gives \begin{align*} c(\mathcal O^{\oplus n})=\underbrace{c(\mathcal O)\cdots c(\mathcal O)}_{n\text{ factors}}=\underbrace{1\cdots 1}_{n\text{ factors}}=1. \end{align*} Thus a product rank-$n$ bundle would have first Chern class $0$, contradicting $c_1(T^{1,0}\mathbb{C}P^n)=(n+1)h\neq 0$. Hence $T^{1,0}\mathbb{C}P^n$ is not a product complex vector bundle. This obstruction is not visible from rank or local triviality alone. Every rank-$n$ complex vector bundle is locally isomorphic to $U\times \mathbb C^n$ over a sufficiently small open set $U$, and the product bundle $\mathbb{C}P^n\times \mathbb C^n$ has the same rank as $T^{1,0}\mathbb{C}P^n$. The nonzero class $(n+1)h$ records exactly that these local product frames cannot be chosen compatibly on all overlaps to form a global product frame. [/example] The projective-space computation used splitting and functoriality. What remains when the problem is not to compute a class, but to compare two differential forms representing the same class? The required object is a transgression form whose exterior derivative is their difference. [definition: Chern-Simons Transgression Form] Let $A_0$ and $A_1$ be connections on a principal $G$-bundle $P\to M$. Let $\mathbb K$ be either $\mathbb R$ or $\mathbb C$, and let \begin{align*} P_0:\mathfrak g^k\to \mathbb K \end{align*} be a symmetric $k$-linear map invariant under the adjoint action of $G$. A Chern-Simons transgression form for $P_0$ is a differential form $\operatorname{CS}_{P_0}(A_0,A_1)\in \Omega^{2k-1}(M;\mathbb K)$ satisfying \begin{align*} d\operatorname{CS}_{P_0}(A_0,A_1)=P_0(F_{A_1})-P_0(F_{A_0}). \end{align*} [/definition] The same invariant polynomial may be denoted as an element of $I^k(\mathfrak g;\mathbb K)$. The definition itself only uses the differential equation for the transgression form; the invariant-polynomial notation records why such forms arise from Chern-Weil representatives. This definition records the property needed for applications. For actual calculations, the path $A_t=A_0+t(A_1-A_0)$ gives an explicit integral formula. [example: The Trace-Square Chern-Simons Form] Let $P_0(X)=\operatorname{tr}(X^2)$, and compare the zero connection with a connection form $A$ on a product bundle. Its curvature is \begin{align*}F_A=dA+A\wedge A.\end{align*} For this invariant polynomial, the Chern-Simons transgression form is \begin{align*}\operatorname{CS}_{P_0}(A)=\operatorname{tr}\left(A\wedge dA+\frac{2}{3}A\wedge A\wedge A\right).\end{align*} We compute its exterior derivative using the graded Leibniz rule. Since $d^2A=0$, \begin{align*}d\operatorname{tr}(A\wedge dA)=\operatorname{tr}(dA\wedge dA-A\wedge d^2A)=\operatorname{tr}(dA\wedge dA).\end{align*} For the cubic term, \begin{align*}d\operatorname{tr}(A\wedge A\wedge A)=\operatorname{tr}(dA\wedge A\wedge A-A\wedge dA\wedge A+A\wedge A\wedge dA).\end{align*} The graded cyclic property of trace gives \begin{align*}\operatorname{tr}(A\wedge dA\wedge A)=-\operatorname{tr}(dA\wedge A\wedge A)\end{align*} and \begin{align*}\operatorname{tr}(A\wedge A\wedge dA)=\operatorname{tr}(dA\wedge A\wedge A).\end{align*} Therefore \begin{align*}d\operatorname{tr}(A\wedge A\wedge A)=3\operatorname{tr}(dA\wedge A\wedge A).\end{align*} Multiplying by $\frac{2}{3}$ and adding the two pieces, \begin{align*}d\operatorname{CS}_{P_0}(A)=\operatorname{tr}(dA\wedge dA)+2\operatorname{tr}(dA\wedge A\wedge A).\end{align*} Now expand the Chern-Weil form: \begin{align*}\operatorname{tr}(F_A\wedge F_A)=\operatorname{tr}\left((dA+A\wedge A)\wedge(dA+A\wedge A)\right).\end{align*} Distributing the product gives \begin{align*}\operatorname{tr}(F_A\wedge F_A)=\operatorname{tr}(dA\wedge dA)+\operatorname{tr}(dA\wedge A\wedge A)+\operatorname{tr}(A\wedge A\wedge dA)+\operatorname{tr}(A\wedge A\wedge A\wedge A).\end{align*} Again by graded cyclicity, \begin{align*}\operatorname{tr}(A\wedge A\wedge dA)=\operatorname{tr}(dA\wedge A\wedge A).\end{align*} Also, \begin{align*}\operatorname{tr}(A\wedge A\wedge A\wedge A)=-\operatorname{tr}(A\wedge A\wedge A\wedge A),\end{align*} because moving the first degree-one factor past the other three degree-one factors contributes the sign $(-1)^3=-1$. Over real or complex coefficients this forces \begin{align*}\operatorname{tr}(A\wedge A\wedge A\wedge A)=0.\end{align*} Hence \begin{align*}\operatorname{tr}(F_A\wedge F_A)=\operatorname{tr}(dA\wedge dA)+2\operatorname{tr}(dA\wedge A\wedge A)=d\operatorname{CS}_{P_0}(A).\end{align*} Thus this Chern-Simons form transgresses the trace-square Chern-Weil representative: its exterior derivative is exactly $\operatorname{tr}(F_A\wedge F_A)$. [/example] The synthesis is that characteristic classes are not merely invariants to list. They organize the passage between local geometry and global topology: obstruction classes prevent simplification, curvature representatives measure geometric choices, functorial formulas make calculations transportable, and transgression records the difference between representatives. This is the common mechanism behind Chern, Euler, Pontryagin, and Chern-Simons constructions throughout the course. ## Beyond and Connections The natural next direction is index theory. Characteristic classes become numerical invariants when paired with fundamental classes, and the Atiyah-Singer index theorem explains why those characteristic numbers also compute analytic indices of elliptic operators. In this sense, the Chern character, Todd class, $\hat A$-class, and $L$-class are not isolated constructions: they are the characteristic-class language in which topological obstructions, elliptic operators, and manifold signatures meet. Another continuation is gauge theory and secondary invariants. Chern-Simons forms measure the difference between characteristic forms attached to different connections, so they are especially important when the primary characteristic class vanishes or when boundary terms carry the geometry. This viewpoint connects the material here to flat bundles, moduli spaces of connections, anomalies in field theory, and geometric refinements of cohomology. Finally, the obstruction-theoretic interpretation remains useful beyond smooth manifolds. Euler and Stiefel-Whitney classes control vector fields and orientability phenomena, Chern classes organize complex bundles and projective geometry, and Pontryagin classes enter the topology of real manifolds. These links explain why characteristic classes recur across algebraic topology, differential geometry, analytic indices, and mathematical physics. Within Androma, the closest linked continuations are [Fibre Bundles I: Bundles, Sections, and Transition Data](/page/Fibre%20Bundles%20I%3A%20Bundles%2C%20Sections%2C%20and%20Transition%20Data), [Fibre Bundles II: Principal Bundles and Connections](/page/Fibre%20Bundles%20II%3A%20Principal%20Bundles%20and%20Connections), [Differential Forms and de Rham Cohomology](/page/Differential%20Forms%20and%20de%20Rham%20Cohomology), and [Lie Groups I: Foundations](/page/Lie%20Groups%20I%3A%20Foundations); these provide the surrounding language in which characteristic classes are used to compare local geometric data with global topological constraints. ## References - J. W. Milnor and J. D. Stasheff, *Characteristic Classes*, Princeton University Press, 1974. - R. Bott and L. W. Tu, *Differential Forms in Algebraic Topology*, Springer, 1982. - S. S. Chern and J. Simons, "Characteristic Forms and Geometric Invariants," *Annals of Mathematics*, 1974. - M. F. Atiyah and I. M. Singer, "The Index of Elliptic Operators I," *Annals of Mathematics*, 1963. - F. Hirzebruch, *Topological Methods in Algebraic Geometry*, Springer, 1966. - Androma internal continuations: [Fibre Bundles I: Bundles, Sections, and Transition Data](/page/Fibre%20Bundles%20I%3A%20Bundles%2C%20Sections%2C%20and%20Transition%20Data); [Fibre Bundles II: Principal Bundles and Connections](/page/Fibre%20Bundles%20II%3A%20Principal%20Bundles%20and%20Connections); [Differential Forms and de Rham Cohomology](/page/Differential%20Forms%20and%20de%20Rham%20Cohomology); [Lie Groups I: Foundations](/page/Lie%20Groups%20I%3A%20Foundations).