This course develops the basic language of fibre bundles as geometric objects that look locally like products but may have nontrivial global structure. It begins with the notion of local triviality and bundle atlases, then explains how transition functions encode the data needed to glue local pieces into a global bundle. From there, the course studies sections as the natural objects living inside a bundle and uses them to formulate local-to-global questions that recur throughout geometry and topology. The central themes are classification, functoriality, and the interaction between local descriptions and global invariants. After establishing the elementary bundle formalism, the course moves to operations on vector bundles, bundle metrics, partitions of unity, and bundle maps, showing how these tools make the category of bundles workable and natural under pullback. It then develops classifying maps in simple cases, clutching constructions, and the sheaf of sections viewpoint, before turning to examples beyond vector bundles and the obstructions that prevent trivialization. The later chapters synthesize these ideas by comparing different ways of building bundles and recognizing when two presentations define the same object. By the end, the student should be able to move fluently between atlases, transition data, sections, and maps, and to understand how global bundle phenomena arise from local geometric information. # Introduction This opening chapter sets the direction for the course before the technical development begins. Fibre bundles are a language for spaces that look like products locally but may fail to be products globally. The central theme is that the failure of a global product structure is not mysterious: it is recorded by transition functions, cocycle identities, and the way local sections transform across overlaps. The course assumes familiarity with smooth manifolds, tangent and cotangent bundles, basic point-set topology, linear algebra, and partitions of unity. Many familiar geometric objects already have the shape of a bundle; the point of the first lectures is to isolate the structure they share and to make the local-to-global mechanism precise. ## Why Bundles Appear What goes wrong if every geometric family over a space is treated as a product? A product $B \times F$ has a fixed fibre $F$ over every point of $B$, but it also has a distinguished global identification of all fibres with the same model. Geometry often gives the first feature without the second. The tangent spaces $T_pM$ of a smooth manifold $M$ all have dimension $n$ when $M$ is $n$-dimensional, so each is abstractly isomorphic to $\mathbb R^n$. There is usually no preferred way to identify $T_pM$ with $T_qM$ for distinct points $p,q \in M$. This tension is the basic reason for introducing fibre bundles. [example: Tangent Spaces as a Local Product] Let $M$ be a smooth $n$-manifold and let $(U,\varphi)$ have coordinates $(x_1,\dots,x_n)$. For $p\in U$, the coordinate vector fields form a basis of $T_pM$, so every $v\in T_pM$ has a unique expression \begin{align*} v=\sum_{j=1}^n a^j\frac{\partial}{\partial x_j}\Big|_p. \end{align*} This gives a local product identification \begin{align*} \Phi_U:\pi^{-1}(U)\to U\times \mathbb R^n,\qquad \Phi_U(v)=\bigl(p,(a^1,\dots,a^n)^\top\bigr). \end{align*} Now let $(V,\psi)$ be another coordinate chart with coordinates $(y_1,\dots,y_n)$, and fix $p\in U\cap V$. On the overlap, each $y_i$ is a smooth function of the $x$-coordinates, so the chain rule gives \begin{align*} \frac{\partial}{\partial x_j}\Big|_p=\sum_{i=1}^n\frac{\partial y_i}{\partial x_j}(p)\frac{\partial}{\partial y_i}\Big|_p. \end{align*} Substituting this basis change into the expression for $v$ gives \begin{align*} v=\sum_{j=1}^n a^j\sum_{i=1}^n\frac{\partial y_i}{\partial x_j}(p)\frac{\partial}{\partial y_i}\Big|_p. \end{align*} Reordering the finite sums, \begin{align*} v=\sum_{i=1}^n\left(\sum_{j=1}^n\frac{\partial y_i}{\partial x_j}(p)a^j\right)\frac{\partial}{\partial y_i}\Big|_p. \end{align*} Thus, if $(a^1,\dots,a^n)^\top$ is the $x$-coordinate column of $v$ and $(b^1,\dots,b^n)^\top$ is the $y$-coordinate column, then \begin{align*} b^i=\sum_{j=1}^n\frac{\partial y_i}{\partial x_j}(p)a^j. \end{align*} Equivalently, \begin{align*} (b^1,\dots,b^n)^\top=\left(\frac{\partial y_i}{\partial x_j}(p)\right)_{i,j}(a^1,\dots,a^n)^\top. \end{align*} So $TM$ is locally identified with $U\times\mathbb R^n$, but two such identifications are glued on overlaps by the Jacobian matrix of the coordinate change, not by a single global choice of coordinates. [/example] The example also hints at the two layers of the subject. First we need a topological or smooth object whose fibres vary locally like a product. Then we need to understand how changing local product descriptions records the global geometry. [definition: Informal Fibre Bundle] A fibre bundle consists of a total space $E$, a base space $B$, a projection map $\pi:E\to B$, a model fibre $F$, an open cover $(U_i)_{i\in I}$ of $B$, and local identification maps \begin{align*} \phi_i:\pi^{-1}(U_i)\to U_i\times F \end{align*} such that $\operatorname{pr}_1\circ \phi_i=\pi|_{\pi^{-1}(U_i)}$ for every $i\in I$. [/definition] This is only an informal definition because the precise version must specify smoothness, compatibility with projections, and the regularity of the changes of local product coordinates. Chapter 1 turns this informal picture into the formal language of smooth fibre bundles, local trivializations, and bundle atlases. ## Local Data and Global Geometry How can local product charts determine a global object? The answer is that two local product descriptions over overlapping open sets must be compared, and these comparisons satisfy consistency identities on triple overlaps. These comparison maps are the transition functions of the bundle. [example: The Mobius Band as a Twisted Line Bundle] Realize the Mobius band as \begin{align*} E=([0,2\pi]\times \mathbb R)/\bigl((0,r)\sim(2\pi,-r)\bigr), \end{align*} with projection $\pi([\theta,r])=e^{i\theta}$. Let $U_0=S^1\setminus\{-1\}$ and $U_1=S^1\setminus\{1\}$. Then $U_0\cap U_1$ has two connected components \begin{align*} C_+=\{e^{i\theta}:0<\theta<\pi\} \end{align*} and \begin{align*} C_-=\{e^{i\theta}:\pi<\theta<2\pi\}. \end{align*} Choose the local product description over $U_1$ by \begin{align*} \phi_1([\theta,r])=(e^{i\theta},r),\qquad 0<\theta<2\pi. \end{align*} Over $U_0$, use the angle interval $-\pi<\alpha<\pi$ and set \begin{align*} \phi_0([\alpha,r])=(e^{i\alpha},r). \end{align*} On $C_+$, the same point has $0<\theta<\pi$ and $\alpha=\theta$, so \begin{align*} \phi_0\circ \phi_1^{-1}(e^{i\theta},r) =\phi_0([\theta,r]) =(e^{i\theta},r). \end{align*} On $C_-$, write $\alpha=\theta-2\pi$. Then $-\pi<\alpha<0$, $e^{i\alpha}=e^{i\theta}$, and the Mobius identification gives \begin{align*} [\theta,r]=[\alpha+2\pi,r]=[\alpha,-r]. \end{align*} Therefore \begin{align*} \phi_0\circ \phi_1^{-1}(e^{i\theta},r) =\phi_0([\theta,r]) =\phi_0([\alpha,-r]) =(e^{i\theta},-r). \end{align*} Thus the transition function $g_{01}:U_0\cap U_1\to \mathbb R^\times$ satisfies $g_{01}(p)=1$ for $p\in C_+$ and $g_{01}(p)=-1$ for $p\in C_-$. The fibres are locally copies of $\mathbb R$, but the sign change between the two overlap components records the half-twist that prevents the Mobius band from being the product $S^1\times\mathbb R$. [/example] The Mobius band is useful because the fibre is as simple as possible, yet the total space is not globally $S^1\times \mathbb R$. It also raises a new structural question: what extra algebraic conditions should be imposed when the fibre carries vector-space operations? This leads from general fibre bundles to vector bundles, where the transition functions must respect linear structure. [definition: Vector Bundle in Preview] A rank $k$ smooth vector bundle over a smooth manifold $B$ is a smooth fibre bundle $\pi:E\to B$ with model fibre $\mathbb R^k$, local trivialisations \begin{align*} \phi_i:\pi^{-1}(U_i)\to U_i\times \mathbb R^k \end{align*} over an open cover $(U_i)_{i\in I}$ of $B$, and transition maps \begin{align*} g_{ij}:U_i\cap U_j\to GL(k,\mathbb R) \end{align*} for all $i,j\in I$ with $U_i\cap U_j\ne\varnothing$, defined by \begin{align*} \phi_i\circ\phi_j^{-1}(b,v)=(b,g_{ij}(b)v). \end{align*} [/definition] The linearity condition makes vector bundles suitable for differential geometry. It lets us add sections, multiply them by smooth functions, form dual bundles, and construct tensor bundles. [example: Tautological Line Bundle Over Real Projective Space] Let $\mathbb{RP}^n$ be the set of lines through $0$ in $\mathbb R^{n+1}$, and let \begin{align*} \gamma^1=\{(\ell,v):\ell\in \mathbb{RP}^n,\ v\in \ell\} \end{align*} with projection $\pi(\ell,v)=\ell$. For each $i$, define \begin{align*} U_i=\{\ell\in \mathbb{RP}^n:\text{some, equivalently every, nonzero }z\in \ell\text{ has }z_i\ne 0\}. \end{align*} On $U_i$, every line $\ell$ has a unique representative $u_i(\ell)\in \ell$ whose $i$th coordinate is $1$: if $z=(z_0,\dots,z_n)\in \ell$ and $z_i\ne 0$, then \begin{align*} u_i(\ell)=\frac{1}{z_i}z. \end{align*} If $z'=cz$ is another nonzero representative of the same line, with $c\ne 0$, then $z_i'=cz_i$ and \begin{align*} \frac{1}{z_i'}z'=\frac{1}{cz_i}(cz)=\frac{1}{z_i}z, \end{align*} so $u_i(\ell)$ is independent of the chosen representative. Since $\ell$ is one-dimensional and $u_i(\ell)\ne 0$, every $v\in \ell$ has a unique form \begin{align*} v=a\,u_i(\ell) \end{align*} for some $a\in \mathbb R$. This gives the local product map \begin{align*} \phi_i:\pi^{-1}(U_i)\to U_i\times \mathbb R,\qquad \phi_i(\ell,v)=(\ell,a) \end{align*} where $v=a\,u_i(\ell)$. Now take $\ell\in U_i\cap U_j$. Write \begin{align*} u_i(\ell)=(u_i^0,\dots,u_i^n). \end{align*} Because $u_i(\ell)$ has $i$th coordinate $1$ and $\ell\in U_j$, its $j$th coordinate $u_i^j$ is nonzero. The representative with $j$th coordinate $1$ is therefore \begin{align*} u_j(\ell)=\frac{1}{u_i^j}u_i(\ell). \end{align*} Multiplying both sides by $u_i^j$ gives \begin{align*} u_i(\ell)=u_i^j\,u_j(\ell). \end{align*} If $\phi_i(\ell,v)=(\ell,a)$, then \begin{align*} v=a\,u_i(\ell). \end{align*} Substituting $u_i(\ell)=u_i^j\,u_j(\ell)$ gives \begin{align*} v=a\,u_i^j\,u_j(\ell). \end{align*} Thus the $j$-coordinate of the same vector is $u_i^j a$, and \begin{align*} \phi_j\circ \phi_i^{-1}(\ell,a)=(\ell,u_i^j a). \end{align*} The transition function on $U_i\cap U_j$ is multiplication by the nonzero scalar $u_i^j$, so the fibre over $\ell$ is the actual line $\ell\subset \mathbb R^{n+1}$, while the local identification with $\mathbb R$ changes by rescaling when the normalized representative of $\ell$ changes. [/example] This example separates two ideas that are often conflated. A fibre can be isomorphic to the model fibre without being supplied with a distinguished identification. Bundle theory keeps the local identifications explicit so that changes between them can be measured. ## Transition Functions and Cocycle Data Once local products have been chosen, the next problem is to decide when they fit together as a single global bundle. The comparison between two local product descriptions is not optional bookkeeping: it is the data that records the twisting of the bundle. [definition: Transition Function] Let $\pi:E\to B$ be a fibre bundle with local trivialisations $\phi_i:\pi^{-1}(U_i)\to U_i\times F$. For $U_i\cap U_j\ne\varnothing$, the transition function from the $j$-trivialisation to the $i$-trivialisation is the map \begin{align*} t_{ij}:U_i\cap U_j\to \operatorname{Homeo}(F) \end{align*} determined by \begin{align*} \phi_i\circ \phi_j^{-1}(b,v)=(b,t_{ij}(b)(v)) \end{align*} for all $b\in U_i\cap U_j$ and $v\in F$. [/definition] For vector bundles the same definition is used with $F=\mathbb R^k$ and $t_{ij}=g_{ij}$ taking values in $GL(k,\mathbb R)$. In computations, this means choosing local frames and writing how the $j$-frame is expressed in the $i$-frame on the overlap. [quotetheorem:6080] [citeproof:6080] These identities are the first real local-to-global test in the course. They say that transition functions cannot be chosen independently on each overlap; their values must agree with all possible ways of passing between local product charts. The hypotheses matter because the maps must come from actual local trivialisations over the same bundle: if three overlaps are assigned homeomorphisms with $t_{12}(b)=t_{23}(b)=\operatorname{id}_F$ but $t_{13}(b)\ne\operatorname{id}_F$ at a point in a triple overlap, then the two coordinate routes from chart $3$ to chart $1$ disagree and no single total space can realise those choices as transition functions. The theorem does not classify bundles, since changing local trivialisations changes the transition functions by a coboundary while leaving the bundle isomorphic. Its role is narrower and foundational: it identifies the compatibility condition that must be present before reconstruction or classification can begin. [example: Computing Tangent Bundle Transitions] Let $(U,\varphi)$ and $(V,\psi)$ be overlapping coordinate charts on an $n$-manifold $M$, with coordinate functions $x=(x_1,\dots,x_n)$ on $U$ and $y=(y_1,\dots,y_n)$ on $V$. Fix $p\in U\cap V$, and write $v\in T_pM$ in the $x$-coordinate basis as \begin{align*} v=\sum_{j=1}^n a^j\frac{\partial}{\partial x_j}\Big|_p. \end{align*} On $U\cap V$, each $y_i$ is a smooth function of the $x$-coordinates, so the chain rule gives \begin{align*} \frac{\partial}{\partial x_j}\Big|_p=\sum_{i=1}^n\frac{\partial y_i}{\partial x_j}(p)\frac{\partial}{\partial y_i}\Big|_p. \end{align*} Substituting this basis change into the expression for $v$ gives \begin{align*} v=\sum_{j=1}^n a^j\sum_{i=1}^n\frac{\partial y_i}{\partial x_j}(p)\frac{\partial}{\partial y_i}\Big|_p. \end{align*} Since both sums are finite, we may collect the coefficient of each $\frac{\partial}{\partial y_i}\big|_p$: \begin{align*} v=\sum_{i=1}^n\left(\sum_{j=1}^n\frac{\partial y_i}{\partial x_j}(p)a^j\right)\frac{\partial}{\partial y_i}\Big|_p. \end{align*} Thus, if $v=\sum_{i=1}^n b^i\frac{\partial}{\partial y_i}\big|_p$, uniqueness of coordinates in the basis $\left(\frac{\partial}{\partial y_i}\big|_p\right)_{i=1}^n$ gives \begin{align*} b^i=\sum_{j=1}^n\frac{\partial y_i}{\partial x_j}(p)a^j. \end{align*} Equivalently, \begin{align*} (b^1,\dots,b^n)^\top=\left(\frac{\partial y_i}{\partial x_j}(p)\right)_{i,j}(a^1,\dots,a^n)^\top. \end{align*} Therefore the transition map from the $x$-coordinate trivialisation to the $y$-coordinate trivialisation is \begin{align*} g_{UV}(p)=\left(\frac{\partial y_i}{\partial x_j}(p)\right)_{i,j}. \end{align*} Now take a triple overlap with coordinates $x,y,z$. The matrix product $g_{VW}(p)g_{UV}(p)$ has $(\alpha,j)$ entry \begin{align*} (g_{VW}(p)g_{UV}(p))_{\alpha j}=\sum_{i=1}^n\frac{\partial z_\alpha}{\partial y_i}(p)\frac{\partial y_i}{\partial x_j}(p). \end{align*} By the chain rule applied to the composite coordinate function $z_\alpha=z_\alpha(y(x))$, this sum is \begin{align*} \sum_{i=1}^n\frac{\partial z_\alpha}{\partial y_i}(p)\frac{\partial y_i}{\partial x_j}(p)=\frac{\partial z_\alpha}{\partial x_j}(p). \end{align*} The right-hand side is the $(\alpha,j)$ entry of $g_{UW}(p)$, so every matrix entry agrees: \begin{align*} g_{VW}(p)g_{UV}(p)=g_{UW}(p). \end{align*} Thus tangent bundle transition matrices satisfy the cocycle identity because Jacobian matrices compose in the same order as the corresponding coordinate changes. [/example] The tangent-bundle computation shows how cocycles arise from existing geometry. The next question reverses the direction: if the overlap maps are given first, when do they determine an actual bundle rather than only a formal list of coordinate changes? This reversal is needed for constructions such as the Mobius band and for later clutching arguments, where the bundle is built from prescribed gluing data rather than discovered inside an already known total space. [quotetheorem:6081] [citeproof:6081] This theorem explains why the Mobius band can be specified by signs on overlaps and why vector bundles are often built from transition matrices. The continuity hypothesis is not cosmetic: if $B=\mathbb R$, $F=\mathbb R$, and a two-set cover has an overlap on which $t_{12}(b)$ is the identity for rational $b$ and multiplication by $-1$ for irrational $b$, then the cocycle identities on that overlap can still hold after setting inverses accordingly, but the coordinate-change map $(b,v)\mapsto t_{12}(b)(v)$ is discontinuous and cannot be a transition map between topological bundle charts. The theorem constructs an existence result from compatible continuous gluing data; it does not say that two different cocycles give non-isomorphic bundles, nor does it choose a preferred representative among all cocycles obtained by changing local trivialisations. A practical construction therefore has three steps: choose a cover, write continuous overlap maps, and check the cocycle identities before taking the quotient; classification begins only after quotienting this data by changes of bundle charts. [illustration:overlapping-local-product-charts] [illustration:mobius-band-line-bundle-gluing] ## Sections as Geometric Objects Once a bundle is given, what are its natural functions? A section chooses one point in each fibre, so it is the bundle-theoretic analogue of a function whose value over $b\in B$ lies in a space depending on $b$. [definition: Section in Preview] Let $\pi:E\to B$ be a fibre bundle. A section of $\pi$ is a map $s:B\to E$ such that $\pi\circ s=\operatorname{id}_B$. [/definition] For a product bundle $B\times F\to B$, sections are the same as maps $B\to F$. For a non-product bundle, a section is still locally described by component functions, but these local descriptions must transform according to the transition functions. [example: Vector Fields as Sections] Let $M$ be a smooth $n$-manifold and let $\pi:TM\to M$ be its tangent bundle. A smooth vector field is a smooth section $X:M\to TM$, so for every $p\in M$ the value $X(p)$ lies in the fibre $T_pM$, because \begin{align*} \pi(X(p))=(\pi\circ X)(p)=p. \end{align*} On a coordinate chart $(U,\varphi)$ with coordinates $(x_1,\dots,x_n)$, the coordinate vectors $\left(\frac{\partial}{\partial x_i}\big|_p\right)_{i=1}^n$ form a basis of $T_pM$. Hence for each $p\in U$ there are unique [real numbers](/page/Real%20Numbers) $X_i(p)$ such that \begin{align*} X(p)=\sum_{i=1}^n X_i(p)\frac{\partial}{\partial x_i}\Big|_p. \end{align*} Thus the local coordinate description of $X$ on $U$ is the $n$-tuple of smooth functions \begin{align*} (X_1,\dots,X_n):U\to \mathbb R^n. \end{align*} Now let $(V,\psi)$ be another coordinate chart with coordinates $(y_1,\dots,y_n)$, and fix $p\in U\cap V$. Write the same tangent vector in the $y$-coordinate basis as \begin{align*} X(p)=\sum_{k=1}^n Y_k(p)\frac{\partial}{\partial y_k}\Big|_p. \end{align*} Since each $y_k$ is a smooth function of the $x$-coordinates on the overlap, the chain rule gives \begin{align*} \frac{\partial}{\partial x_i}\Big|_p=\sum_{k=1}^n\frac{\partial y_k}{\partial x_i}(p)\frac{\partial}{\partial y_k}\Big|_p. \end{align*} Substituting this basis change into the $x$-coordinate expression for $X(p)$ gives \begin{align*} X(p)=\sum_{i=1}^n X_i(p)\sum_{k=1}^n\frac{\partial y_k}{\partial x_i}(p)\frac{\partial}{\partial y_k}\Big|_p. \end{align*} Expanding the finite sums, \begin{align*} X(p)=\sum_{i=1}^n\sum_{k=1}^n X_i(p)\frac{\partial y_k}{\partial x_i}(p)\frac{\partial}{\partial y_k}\Big|_p. \end{align*} Collecting the coefficient of each $\frac{\partial}{\partial y_k}\big|_p$, \begin{align*} X(p)=\sum_{k=1}^n\left(\sum_{i=1}^n\frac{\partial y_k}{\partial x_i}(p)X_i(p)\right)\frac{\partial}{\partial y_k}\Big|_p. \end{align*} Comparing this with \begin{align*} X(p)=\sum_{k=1}^n Y_k(p)\frac{\partial}{\partial y_k}\Big|_p \end{align*} and using uniqueness of coordinates in the $y$-basis, for each $k$ we obtain \begin{align*} Y_k(p)=\sum_{i=1}^n\frac{\partial y_k}{\partial x_i}(p)X_i(p). \end{align*} Equivalently, in column-vector form, \begin{align*} (Y_1(p),\dots,Y_n(p))^\top=\left(\frac{\partial y_k}{\partial x_i}(p)\right)_{k,i}(X_1(p),\dots,X_n(p))^\top. \end{align*} So a vector field is locally represented by component functions, but on overlaps those component functions are related by the Jacobian matrix of the coordinate change; the geometric object is the section $X$, not any one coordinate $n$-tuple. [/example] Sections turn bundles into a practical tool. Instead of studying the total space in isolation, we often study the fields of fibrewise data that can be chosen smoothly over the base. ## Maps, Pullbacks, and Classification Questions How do bundles behave under maps of base spaces? Given a smooth map $f:B'\to B$ and a bundle $\pi:E\to B$, ordinary composition does not itself produce a bundle over $B'$, because the points of $E$ still project to $B$ rather than to $B'$. The pullback construction fixes this by pairing each point $b'\in B'$ with points of the fibre over $f(b')$, producing a bundle over $B'$ with the desired fibres. [definition: Pullback Bundle in Preview] Let $f:B'\to B$ be a map and let $\pi:E\to B$ be a fibre bundle. The pullback bundle $f^*E$ is the space \begin{align*} f^*E=\{(b',e)\in B'\times E: f(b')=\pi(e)\} \end{align*} with projection $f^*E\to B'$ given by $(b',e)\mapsto b'$. [/definition] Pullbacks are the functorial part of the theory. They allow bundles to be transported along maps, and they are essential for comparing constructions on different bases. [example: Pulling Back the Tautological Bundle] Let $g:M\to \mathbb{RP}^n$ be a smooth map, and let $\gamma^1\to \mathbb{RP}^n$ be the tautological line bundle \begin{align*} \gamma^1=\{(\ell,v):\ell\in \mathbb{RP}^n,\ v\in \ell\}, \end{align*} with projection \begin{align*} \pi_{\gamma^1}(\ell,v)=\ell. \end{align*} By the definition of pullback, the total space of $g^*\gamma^1$ is \begin{align*} g^*\gamma^1=\{(p,(\ell,v))\in M\times \gamma^1:g(p)=\pi_{\gamma^1}(\ell,v)\}. \end{align*} Since $\pi_{\gamma^1}(\ell,v)=\ell$, the condition inside this set is $g(p)=\ell$, so \begin{align*} g^*\gamma^1=\{(p,(\ell,v))\in M\times \gamma^1:g(p)=\ell\}. \end{align*} Membership in $\gamma^1$ means $v\in \ell$, and the equality $\ell=g(p)$ therefore gives $v\in g(p)$. Hence the same total space may be identified with \begin{align*} \{(p,v)\in M\times \mathbb R^{n+1}:v\in g(p)\}, \end{align*} where the projection to $M$ is \begin{align*} (p,v)\mapsto p. \end{align*} For a fixed point $p\in M$, the fibre over $p$ is the inverse image of $p$ under this projection: \begin{align*} (g^*\gamma^1)_p=\{(p,v)\in M\times \mathbb R^{n+1}:v\in g(p)\}. \end{align*} Forgetting the fixed first coordinate identifies this fibre with \begin{align*} g(p)\subset \mathbb R^{n+1}. \end{align*} Thus a map into projective space assigns to each point $p\in M$ a line through the origin, and the pullback tautological bundle is exactly the resulting family of lines over $M$. [/example] The pullback construction also frames classification problems. To classify bundles is to decide when two sets of local transition data produce equivalent global bundles, and to understand which invariants detect the difference. ## The Course Roadmap The remaining organisational problem is to decide which parts of the bundle language must be built first so that later constructions are not circular. The course begins with the local product structure because transition functions, sections, pullbacks, and classification questions all depend on having precise bundle charts. The first part of the course builds the formal definition of a smooth fibre bundle from this local product structure. We will define total space, base, fibre, projection, local product description, bundle chart, and bundle atlas, then prove that compatible local data gives a smooth total-space structure. The second part translates local product descriptions into transition functions. On double overlaps these functions compare charts, and on triple overlaps they satisfy cocycle identities. The reconstruction theorem of Chapter 2 then shows that suitable cocycle data can be glued into a bundle. The third part studies sections. Local component functions, extension questions, zero sets, and frames provide the bridge between abstract bundles and geometric fields on manifolds. The final part of this first course introduces bundle maps, pullbacks, vector bundle operations, and elementary clutching constructions. These tools prepare for characteristic classes, principal bundles, connections, and gauge-theoretic constructions in later courses. [remark: Guiding Principle] A fibre bundle is locally a product, but its global geometry is encoded by how the local products are identified on overlaps. Sections, pullbacks, and bundle maps are the basic operations that make this encoding useful. [/remark] The guiding principle in the preview is now made precise: a fibre bundle is built from local products whose overlaps carry the true geometry. In the next chapter, we introduce the language of local triviality, bundle atlases, and the basic operations that let us move between local and global descriptions. # 1. Local Triviality and Bundle Atlases After the introductory preview, this chapter sets up the language in which the rest of the course will describe geometry by local models and gluing data. The prerequisites are the basic theory of smooth manifolds, smooth maps, coordinate charts, products of manifolds, and the tangent bundle. A fibre bundle is a space that looks locally like a product, but may fail to be a product globally. The main point is that a bundle is not merely a map with similar-looking fibres: it comes with compatible local product descriptions, and those descriptions determine a smooth structure on the total space. We begin with smooth fibre bundles, pass to bundle atlases and their equivalence, and then isolate vector bundles by requiring the changes of local coordinates in the fibre direction to be linear. ## The Local Product Question The first problem is to decide what it should mean for a family of spaces parametrised by a manifold $B$ to vary smoothly. A projection map $\pi:E\to B$ records which point of the total space lies over which point of the base, but the fibres $E_b=\pi^{-1}(b)$ alone do not describe how nearby fibres fit together. The local triviality condition supplies the missing structure: near every base point, the family must be smoothly identifiable with a fixed model fibre. [definition: Smooth Fibre Bundle] Let $B$, $E$, and $F$ be smooth manifolds, and let $\pi:E\to B$ be a smooth surjective map. A smooth fibre bundle with fibre $F$ consists of $\pi:E\to B$ together with, for every $b\in B$, an open neighbourhood $U\subset B$ of $b$ and a diffeomorphism \begin{align*} \Phi_U:\pi^{-1}(U)\to U\times F \end{align*} such that $\operatorname{pr}_1\circ \Phi_U=\pi|_{\pi^{-1}(U)}$. [/definition] The condition $\operatorname{pr}_1\circ \Phi_U=\pi$ says that the diffeomorphism respects the base point. Thus $\Phi_U$ does not move a point of $E$ to an unrelated fibre; it rewrites the fibre over $b$ as the copy $\{b\}\times F$. This motivates giving the individual local product description a separate name, because later arguments manipulate these descriptions one [open set](/page/Open%20Set) at a time. [definition: Local Trivialisation] For a smooth fibre bundle $\pi:E\to B$ with fibre $F$, a local trivialisation over an open set $U\subset B$ is a diffeomorphism \begin{align*} \Phi_U:\pi^{-1}(U)\to U\times F \end{align*} satisfying $\operatorname{pr}_1\circ \Phi_U=\pi|_{\pi^{-1}(U)}$. [/definition] Once a local trivialization is chosen, each fibre over $U$ is identified with $F$ by the second component of $\Phi_U$. Different choices of local trivialization need not identify the same point of a fibre with the same element of $F$, and the comparison between choices becomes the transition data studied later. [example: Product Bundle] Let $B$ and $F$ be smooth manifolds, set $E=B\times F$, and let $\pi=\operatorname{pr}_1:B\times F\to B$. For an open set $U\subset B$, the inverse image of $U$ is \begin{align*} \pi^{-1}(U)=\{(b,y)\in B\times F:\pi(b,y)\in U\}. \end{align*} Since $\pi(b,y)=b$, this becomes \begin{align*} \pi^{-1}(U)=\{(b,y)\in B\times F:b\in U\}=U\times F. \end{align*} Define $\Phi_U:\pi^{-1}(U)\to U\times F$ by $\Phi_U(b,y)=(b,y)$. Under the identification $\pi^{-1}(U)=U\times F$, this is the identity map on $U\times F$, so its inverse is again $(b,y)\mapsto (b,y)$; hence $\Phi_U$ is a diffeomorphism. It remains to check that $\Phi_U$ preserves the base point. For every $(b,y)\in \pi^{-1}(U)=U\times F$, \begin{align*} (\operatorname{pr}_1\circ \Phi_U)(b,y)=\operatorname{pr}_1(b,y)=b. \end{align*} Also, \begin{align*} \pi|_{\pi^{-1}(U)}(b,y)=\pi(b,y)=b. \end{align*} Therefore $\operatorname{pr}_1\circ \Phi_U=\pi|_{\pi^{-1}(U)}$, so $\Phi_U$ is a local trivialization over $U$. The product bundle is the case where the local product descriptions are restrictions of one global product coordinate system. [/example] A bundle can be locally a product without being globally a product. The later transition functions measure the obstruction to extending the local identifications into a single global diffeomorphism $E\cong B\times F$ over $B$. [remark: Fibres Are Not Enough] A surjective smooth map $\pi:E\to B$ whose fibres are all diffeomorphic to $F$ is not automatically a fibre bundle. The definition requires local diffeomorphisms $\pi^{-1}(U)\cong U\times F$ that vary smoothly with the base point. This extra condition rules out pathological variation of fibres and is the feature that makes bundles usable in differential geometry. [/remark] ## Bundle Charts and Changes of Trivialization The next question is how two local product descriptions of the same bundle compare on an overlap. Since both descriptions preserve the base point, their composite fixes the base coordinate and acts only in the fibre coordinate. This observation turns local triviality into a calculus of fibre-coordinate changes. [definition: Bundle Chart] Let $\pi:E\to B$ be a smooth fibre bundle with fibre $F$. A bundle chart is a pair $(U,\Phi_U)$ where $U\subset B$ is open and $\Phi_U:\pi^{-1}(U)\to U\times F$ is a local trivialisation. [/definition] A collection of bundle charts is useful only when it covers the whole base. On intersections, the charts compare by smooth fibrewise diffeomorphisms. To use these overlap maps as transition data that can later be glued and classified, we need a definition that isolates the comparison map between two trivializations. [definition: Change of Trivialization] Let $(U,\Phi_U)$ and $(V,\Phi_V)$ be bundle charts for $\pi:E\to B$. On $U\cap V$, the change of trivialization is the diffeomorphism \begin{align*} \Phi_U\circ \Phi_V^{-1}:(U\cap V)\times F\to (U\cap V)\times F. \end{align*} It has the form \begin{align*} \Phi_U\circ \Phi_V^{-1}(b,y)=(b,g_{UV}(b)(y)) \end{align*} for a smooth map $g_{UV}:U\cap V\to \operatorname{Diff}(F)$ in the sense that $(b,y)\mapsto g_{UV}(b)(y)$ is smooth. [/definition] The formula says that the base coordinate is fixed and the fibre coordinate is transformed by a smoothly varying diffeomorphism of the model fibre. We need this definition because it turns the vague phrase "the local products agree on overlaps" into concrete data that can be computed, checked for compatibility, and later prescribed when constructing bundles from transition functions. The next example computes this data for tangent bundles, and Chapter 2 makes the same overlap maps the main object. [example: Tangent Bundle Changes From Coordinate Charts] Let $M$ be a smooth $n$-manifold, let $(U,\varphi)$ and $(V,\psi)$ be coordinate charts, and take $p\in U\cap V$. Write the corresponding tangent bundle trivializations as \begin{align*} \Phi_U(v)=(p,d\varphi_p(v)) \end{align*} for $v\in T_pM$, and \begin{align*} \Phi_V(v)=(p,d\psi_p(v)). \end{align*} We compute $\Phi_U\circ\Phi_V^{-1}$ at a point $(p,\xi)\in (U\cap V)\times\mathbb R^n$. By the definition of $\Phi_V$, the vector $v\in T_pM$ satisfying $\Phi_V(v)=(p,\xi)$ is characterized by \begin{align*} d\psi_p(v)=\xi. \end{align*} Since $\psi$ is a coordinate chart, $d\psi_p:T_pM\to\mathbb R^n$ is a linear isomorphism, and its inverse is $d(\psi^{-1})_{\psi(p)}$. Hence \begin{align*} v=(d\psi_p)^{-1}(\xi)=d(\psi^{-1})_{\psi(p)}(\xi). \end{align*} Substituting this vector into $\Phi_U$ gives \begin{align*} (\Phi_U\circ\Phi_V^{-1})(p,\xi)=\Phi_U\left(d(\psi^{-1})_{\psi(p)}(\xi)\right). \end{align*} Therefore \begin{align*} (\Phi_U\circ\Phi_V^{-1})(p,\xi)=\left(p,d\varphi_p\left(d(\psi^{-1})_{\psi(p)}(\xi)\right)\right). \end{align*} Equivalently, \begin{align*} (\Phi_U\circ\Phi_V^{-1})(p,\xi)=\left(p,\left(d\varphi_p\circ d(\psi^{-1})_{\psi(p)}\right)(\xi)\right). \end{align*} By the chain rule applied to $\varphi\circ\psi^{-1}$ at $\psi(p)$, \begin{align*} d\varphi_p\circ d(\psi^{-1})_{\psi(p)}=d(\varphi\circ\psi^{-1})_{\psi(p)}. \end{align*} Thus \begin{align*} (\Phi_U\circ\Phi_V^{-1})(p,\xi)=\left(p,d(\varphi\circ\psi^{-1})_{\psi(p)}(\xi)\right). \end{align*} Let $f=\varphi\circ\psi^{-1}$, and write $\xi=(\xi_1,\dots,\xi_n)$. If $d f_{\psi(p)}(\xi)=(\eta_1,\dots,\eta_n)$, then the $i$th component is \begin{align*} \eta_i=\sum_{j=1}^n \frac{\partial f_i}{\partial y_j}(\psi(p))\xi_j. \end{align*} This is exactly multiplication of the column vector $\xi$ by the Jacobian matrix $J(\varphi\circ\psi^{-1})_{\psi(p)}$, so \begin{align*} (\Phi_U\circ\Phi_V^{-1})(p,\xi)=\left(p,J(\varphi\circ\psi^{-1})_{\psi(p)}\xi\right). \end{align*} The base point remains $p$, while the fibre coordinate is multiplied by the Jacobian matrix of the coordinate change; tangent-vector coordinates therefore transform linearly even when $\varphi\circ\psi^{-1}$ is nonlinear. [/example] To build bundles systematically, we do not always start with a pre-existing smooth manifold $E$. Often we begin with local pieces $U_i\times F$ and gluing maps between them. The following definition records the compatibility requirements for those local pieces before we prove that they produce a bundle. [definition: Smooth Bundle Atlas] Let $B$ and $F$ be smooth manifolds. A smooth bundle atlas with fibre $F$ consists of an open cover $(U_i)_{i\in I}$ of $B$ and diffeomorphisms \begin{align*} \gamma_{ij}:(U_i\cap U_j)\times F\to (U_i\cap U_j)\times F \end{align*} of the form $\gamma_{ij}(b,y)=(b,g_{ij}(b)(y))$, where $g_{ij}:U_i\cap U_j\to \operatorname{Diff}(F)$ is smooth in the sense that $(b,y)\mapsto g_{ij}(b)(y)$ is smooth. These maps satisfy $\gamma_{ii}=\operatorname{id}_{U_i\times F}$, $\gamma_{ij}=\gamma_{ji}^{-1}$, and $\gamma_{ij}\circ \gamma_{jk}=\gamma_{ik}$ on every triple overlap $U_i\cap U_j\cap U_k$. [/definition] The last identity is the cocycle condition in chart form. It guarantees that passing from chart $k$ to chart $j$ and then to chart $i$ gives the same result as passing directly from chart $k$ to chart $i$. The construction theorem now asks whether these formal compatibility rules are strong enough to make an actual smooth total space. [quotetheorem:6082] [citeproof:6082] This theorem is the bridge from local data to geometry. The cocycle condition is essential: for instance, if $B=\mathbb R$, $F=\mathbb R$, $U_1=U_2=U_3=B$, with $\gamma_{12}$ and $\gamma_{23}$ the identity but $\gamma_{13}(b,y)=(b,y+1)$, then moving from chart $3$ to chart $1$ directly gives a different fibre coordinate from moving through chart $2$. The inverse condition is independent: if $\gamma_{12}(b,y)=(b,y+1)$ and $\gamma_{21}(b,y)=(b,y+1)$, then the two charts identify points in incompatible directions rather than undoing each other. Smoothness is also needed; taking a fibre coordinate change whose scalar factor has nonsmooth base-dependence, such as $b\mapsto 1+|b|$ near $0$, gives fibrewise diffeomorphisms but not a smooth bundle chart transition. Finally, the topology hypothesis cannot be omitted: quotients produced by gluing manifolds can fail to be Hausdorff, as in the line with two origins obtained by gluing two copies of $\mathbb R$ along $\mathbb R\setminus\{0\}$ by the identity. The theorem says that once the compatibility identities and the basic manifold-topology conditions hold, the local data determine a unique bundle up to the expected isomorphism. ## Equivalence of Bundle Atlases A single bundle has many possible atlases, because we can refine the cover or change the fibre coordinate in each chart. The right notion of sameness must ignore these choices while preserving the smooth structure and the projection to the base. [definition: Equivalent Bundle Atlases] Two smooth bundle atlases on $B$ with fibre $F$ are equivalent if their union is again a smooth bundle atlas. [/definition] Equivalently, every chart in one atlas is compatible with every chart in the other atlas on the corresponding overlap. This definition packages the usual maximal-atlas idea in the fibre bundle setting. Adding all charts compatible with a given atlas produces a maximal bundle atlas, and working with maximal atlases removes dependence on the first chosen cover. The next theorem states the parallel principle for a set already equipped with compatible prospective trivializations. [quotetheorem:6083] [citeproof:6083] The theorem is often used silently: once a total space is described by compatible local trivializations and the induced topology is a manifold topology, the smooth structure is not an additional choice. The compatibility hypothesis is doing real work. If two prospective charts on an overlap produce a map such as $x\mapsto |x|$ between open intervals, then the proposed coordinate system cannot define a smooth atlas on $E$ near $0$. If the changes fail the fibre-preserving form, for example $(b,y)\mapsto (y,b)$ on an overlap in $\mathbb R\times\mathbb R$, then the projection may not become the local product projection in both charts. The topology assumptions are also genuine: the line with two origins has compatible identity coordinate changes on the common punctured line, but the induced topology is not Hausdorff, so it is not a smooth manifold in the usual sense; an uncountable disjoint union of coordinate copies gives a locally Euclidean [Hausdorff space](/page/Hausdorff%20Space) that is not second-countable. Later gluing constructions rely on this principle whenever they build a bundle out of transition data. [example: Refining A Product Bundle Atlas] Let $E=B\times F$ with projection $\pi=\operatorname{pr}_1$, and let $(U_i)_{i\in I}$ be an open cover of $B$. The product chart over $U_i$ is \begin{align*} \Phi_i:\pi^{-1}(U_i)=U_i\times F\to U_i\times F,\qquad \Phi_i(b,y)=(b,y). \end{align*} On an overlap $U_i\cap U_j$, the inverse $\Phi_j^{-1}$ is also the identity map. Hence for every $(b,y)\in (U_i\cap U_j)\times F$, \begin{align*} (\Phi_i\circ \Phi_j^{-1})(b,y)=\Phi_i(b,y)=(b,y). \end{align*} Thus the original product atlas has identity transition maps. Now choose, for each $i$, a smooth fibre-coordinate change $h_i:U_i\to \operatorname{Diff}(F)$, meaning that $(b,y)\mapsto h_i(b)(y)$ is smooth and the corresponding inverse family is smooth. Define \begin{align*} \widetilde{\Phi}_i:U_i\times F\to U_i\times F,\qquad \widetilde{\Phi}_i(b,y)=(b,h_i(b)(y)). \end{align*} For fixed $b\in U_i$, the map $h_i(b):F\to F$ is a diffeomorphism, so the inverse is \begin{align*} \widetilde{\Phi}_i^{-1}(b,z)=\left(b,h_i(b)^{-1}(z)\right). \end{align*} Indeed, \begin{align*} \widetilde{\Phi}_i\left(b,h_i(b)^{-1}(z)\right)=\left(b,h_i(b)\left(h_i(b)^{-1}(z)\right)\right)=(b,z), \end{align*} and \begin{align*} \widetilde{\Phi}_i^{-1}\left(b,h_i(b)(y)\right)=\left(b,h_i(b)^{-1}\left(h_i(b)(y)\right)\right)=(b,y). \end{align*} Therefore each $\widetilde{\Phi}_i$ is a local product chart over $U_i$. On $U_i\cap U_j$, the transition from the altered $j$-chart to the altered $i$-chart is computed by first applying $\widetilde{\Phi}_j^{-1}$: \begin{align*} \widetilde{\Phi}_j^{-1}(b,y)=\left(b,h_j(b)^{-1}(y)\right). \end{align*} Applying $\widetilde{\Phi}_i$ to this point gives \begin{align*} (\widetilde{\Phi}_i\circ \widetilde{\Phi}_j^{-1})(b,y)=\widetilde{\Phi}_i\left(b,h_j(b)^{-1}(y)\right)=\left(b,h_i(b)\left(h_j(b)^{-1}(y)\right)\right). \end{align*} Equivalently, \begin{align*} (\widetilde{\Phi}_i\circ \widetilde{\Phi}_j^{-1})(b,y)=\left(b,\left(h_i(b)\circ h_j(b)^{-1}\right)(y)\right). \end{align*} So the altered transition function on $U_i\cap U_j$ is $h_i(b)\circ h_j(b)^{-1}$. The bundle is still the global product $B\times F$, but changing local fibre coordinates can make the transition functions non-identity. [/example] ## Vector Bundles And Linear Fibre Coordinates The final question in this chapter is what extra structure distinguishes vector bundles from general fibre bundles. A vector bundle has fibres that are vector spaces, and its local product descriptions identify each fibre with a fixed [vector space](/page/Vector%20Space) in a way that makes changes of trivialization linear. [definition: Smooth Vector Bundle] Let $B$ be a smooth manifold and let $k\in \mathbb N$. A smooth real vector bundle of rank $k$ over $B$ is a smooth fibre bundle $\pi:E\to B$ with fibre $\mathbb R^k$ such that every fibre $E_b$ is a real vector space and there exists a bundle atlas whose local trivializations restrict on each fibre to linear isomorphisms $E_b\to \mathbb R^k$, and whose changes of trivialization have the form \begin{align*} (b,v)\mapsto (b,A_{ij}(b)v) \end{align*} with $A_{ij}:U_i\cap U_j\to GL(k,\mathbb R)$ smooth. [/definition] The rank is the dimension of each fibre. The linearity condition ensures that changing charts does not alter vector-space operations on a fibre. To use this structure in calculations, we need to single out those bundle charts that identify fibres with $\mathbb R^k$ by linear maps. [definition: Vector Bundle Chart] Let $\pi:E\to B$ be a rank $k$ smooth real vector bundle. A vector bundle chart over $U\subset B$ is a local trivialization $\Phi_U:\pi^{-1}(U)\to U\times \mathbb R^k$ whose restriction to each fibre is a linear isomorphism $E_b\to \{b\}\times \mathbb R^k$ after identifying $\{b\}\times \mathbb R^k$ with $\mathbb R^k$. [/definition] Vector bundle charts are bundle charts with extra algebraic compatibility. The transition maps no longer take values in the full diffeomorphism group of $\mathbb R^k$; they take values in $GL(k,\mathbb R)$. [example: Tangent Bundle As A Vector Bundle] Let $M$ be a smooth $n$-manifold, and let $\pi:TM\to M$ be the tangent bundle projection, so $\pi^{-1}(p)=T_pM$ for each $p\in M$. For a coordinate chart $(U,\varphi)$, define \begin{align*} \Phi_U:\pi^{-1}(U)\to U\times \mathbb R^n,\qquad \Phi_U(v)=(p,d\varphi_p(v))\text{ for }v\in T_pM. \end{align*} Since $\varphi:U\to \varphi(U)\subset \mathbb R^n$ is a diffeomorphism onto an open subset, its derivative $d\varphi_p:T_pM\to \mathbb R^n$ is a linear isomorphism. Therefore the restriction of $\Phi_U$ to the fibre $T_pM$ is linear after identifying $\{p\}\times \mathbb R^n$ with $\mathbb R^n$. Now let $(V,\psi)$ be another coordinate chart, and take $(p,\xi)\in (U\cap V)\times \mathbb R^n$. To compute $\Phi_U\circ\Phi_V^{-1}$, we first find the unique vector $v\in T_pM$ with $\Phi_V(v)=(p,\xi)$. By the definition of $\Phi_V$, \begin{align*} d\psi_p(v)=\xi. \end{align*} Because $d\psi_p$ is a linear isomorphism, \begin{align*} v=(d\psi_p)^{-1}(\xi)=d(\psi^{-1})_{\psi(p)}(\xi). \end{align*} Applying $\Phi_U$ to this vector gives \begin{align*} (\Phi_U\circ\Phi_V^{-1})(p,\xi)=\Phi_U\left(d(\psi^{-1})_{\psi(p)}(\xi)\right). \end{align*} Using the definition of $\Phi_U$, this becomes \begin{align*} (\Phi_U\circ\Phi_V^{-1})(p,\xi)=\left(p,d\varphi_p\left(d(\psi^{-1})_{\psi(p)}(\xi)\right)\right). \end{align*} Equivalently, \begin{align*} (\Phi_U\circ\Phi_V^{-1})(p,\xi)=\left(p,\left(d\varphi_p\circ d(\psi^{-1})_{\psi(p)}\right)(\xi)\right). \end{align*} By the chain rule applied to the coordinate change $\varphi\circ\psi^{-1}$ at $\psi(p)$, \begin{align*} d\varphi_p\circ d(\psi^{-1})_{\psi(p)}=d(\varphi\circ\psi^{-1})_{\psi(p)}. \end{align*} Hence \begin{align*} (\Phi_U\circ\Phi_V^{-1})(p,\xi)=\left(p,d(\varphi\circ\psi^{-1})_{\psi(p)}(\xi)\right). \end{align*} Write $\varphi\circ\psi^{-1}=(f_1,\dots,f_n)$ and $\xi=(\xi_1,\dots,\xi_n)$. If $d(\varphi\circ\psi^{-1})_{\psi(p)}(\xi)=(\eta_1,\dots,\eta_n)$, then for each $i$, \begin{align*} \eta_i=\sum_{j=1}^n \frac{\partial f_i}{\partial y_j}(\psi(p))\xi_j. \end{align*} This is precisely multiplication by the Jacobian matrix $J(\varphi\circ\psi^{-1})_{\psi(p)}$, so \begin{align*} d(\varphi\circ\psi^{-1})_{\psi(p)}(\xi)=J(\varphi\circ\psi^{-1})_{\psi(p)}\xi. \end{align*} Therefore the transition map is \begin{align*} (\Phi_U\circ\Phi_V^{-1})(p,\xi)=\left(p,J(\varphi\circ\psi^{-1})_{\psi(p)}\xi\right). \end{align*} The coordinate change $\varphi\circ\psi^{-1}$ is a diffeomorphism, with inverse $\psi\circ\varphi^{-1}$, so its derivative at every point is an invertible [linear map](/page/Linear%20Map). Thus $J(\varphi\circ\psi^{-1})_{\psi(p)}\in GL(n,\mathbb R)$, and the tangent bundle transition maps are linear in the fibre coordinate. Hence $TM$ is a rank $n$ smooth real vector bundle. [/example] This example is the model for many geometric bundles: local coordinates give local bases, and Jacobian-type matrices describe how those bases change. The same pattern appears for cotangent bundles, tensor bundles, and later associated bundles. [example: Tautological Line Bundle Over Real Projective Space] Let $\mathbb{RP}^n$ be the space of lines $\ell\subset \mathbb R^{n+1}$ through the origin. The tautological line bundle is \begin{align*} \gamma_n=\{(\ell,v)\in \mathbb{RP}^n\times \mathbb R^{n+1}:v\in \ell\}, \end{align*} with projection $\pi(\ell,v)=\ell$. We show explicitly that the standard projective charts give local trivializations with linear one-dimensional transition maps. Fix $i\in\{0,\dots,n\}$ and let \begin{align*} U_i=\{[x_0:\cdots:x_n]\in \mathbb{RP}^n:x_i\ne 0\}. \end{align*} For $\ell=[x_0:\cdots:x_n]\in U_i$, define \begin{align*} s_i(\ell)=\left(\frac{x_0}{x_i},\dots,\frac{x_{i-1}}{x_i},1,\frac{x_{i+1}}{x_i},\dots,\frac{x_n}{x_i}\right)\in \mathbb R^{n+1}. \end{align*} This vector lies on the line $\ell$ and has $i$th coordinate $1$. If $w\in \ell$ also has $i$th coordinate $1$, then $w=\lambda s_i(\ell)$ for some $\lambda\ne 0$, because both vectors span $\ell$. Comparing the $i$th coordinate gives $1=\lambda\cdot 1$, so $\lambda=1$ and $w=s_i(\ell)$. Thus $s_i(\ell)$ is the unique representative of $\ell$ with $i$th coordinate $1$. Every $v\in \ell$ can be written uniquely as $v=t\,s_i(\ell)$ for a scalar $t\in\mathbb R$: existence holds because $s_i(\ell)$ spans $\ell$, and if $t\,s_i(\ell)=t'\,s_i(\ell)$, then comparing the $i$th coordinate gives $t=t'$. Define \begin{align*} \Phi_i:\pi^{-1}(U_i)\to U_i\times \mathbb R,\qquad \Phi_i(\ell,v)=(\ell,t)\quad\text{where }v=t\,s_i(\ell). \end{align*} Its inverse is \begin{align*} \Phi_i^{-1}:U_i\times \mathbb R\to \pi^{-1}(U_i),\qquad \Phi_i^{-1}(\ell,t)=(\ell,t\,s_i(\ell)). \end{align*} Indeed, if $v=t\,s_i(\ell)$, then \begin{align*} \Phi_i^{-1}(\Phi_i(\ell,v))=\Phi_i^{-1}(\ell,t)=(\ell,t\,s_i(\ell))=(\ell,v), \end{align*} and \begin{align*} \Phi_i(\Phi_i^{-1}(\ell,t))=\Phi_i(\ell,t\,s_i(\ell))=(\ell,t). \end{align*} Also, \begin{align*} (\operatorname{pr}_1\circ \Phi_i)(\ell,v)=\operatorname{pr}_1(\ell,t)=\ell=\pi(\ell,v), \end{align*} so $\Phi_i$ preserves the base point. Now take $\ell=[x_0:\cdots:x_n]\in U_i\cap U_j$. Since both $s_i(\ell)$ and $s_j(\ell)$ lie on the same line, and the $j$th coordinate of $s_i(\ell)$ is $\frac{x_j}{x_i}$, multiplying $s_i(\ell)$ by $\frac{x_i}{x_j}$ makes the $j$th coordinate equal to $1$: \begin{align*} \left(\frac{x_i}{x_j}s_i(\ell)\right)_j=\frac{x_i}{x_j}\cdot \frac{x_j}{x_i}=1. \end{align*} By uniqueness of the representative with $j$th coordinate $1$, \begin{align*} s_j(\ell)=\frac{x_i}{x_j}s_i(\ell). \end{align*} For $(\ell,t)\in (U_i\cap U_j)\times\mathbb R$, \begin{align*} \Phi_j^{-1}(\ell,t)=(\ell,t\,s_j(\ell)). \end{align*} Using $s_j(\ell)=\frac{x_i}{x_j}s_i(\ell)$, this becomes \begin{align*} t\,s_j(\ell)=t\frac{x_i}{x_j}s_i(\ell). \end{align*} Therefore \begin{align*} (\Phi_i\circ\Phi_j^{-1})(\ell,t)=\Phi_i\left(\ell,t\frac{x_i}{x_j}s_i(\ell)\right)=\left(\ell,\frac{x_i}{x_j}t\right). \end{align*} Since $x_i\ne 0$ and $x_j\ne 0$ on $U_i\cap U_j$, the scalar $\frac{x_i}{x_j}$ is nonzero, and it depends smoothly on the projective coordinates in the overlap chart. Thus the transition maps are multiplication by nonzero smooth scalars, so $\gamma_n$ is a smooth real line bundle. Its fibre over $\ell$ is exactly the line $\ell$ itself, which is why the bundle is called tautological. [/example] The tautological bundle also warns against confusing local and global product structure. Each chart sees a product $U_i\times \mathbb R$, but the transition scalars encode how the chosen representatives of projective lines change from chart to chart. [remark: What This Chapter Establishes] Local trivializations turn a projection into a bundle, bundle atlases record compatible systems of such trivializations, and vector bundles are the special case where fibre-coordinate changes are linear. The next chapter takes the transition maps $g_{ij}$ and $A_{ij}$ as primary data. The cocycle identities will become the algebraic mechanism by which local product pieces glue into global geometric objects. [/remark] Once a bundle has been covered by local products, the remaining problem is to understand how those pieces fit together. The next chapter isolates the transition functions and cocycle conditions that record this gluing data and determine the bundle up to isomorphism. # 2. Transition Functions and Cocycles This chapter turns the local descriptions from Chapter 1 into the algebraic data that records how they overlap. A bundle atlas is useful because each chart looks like a product, but the bundle itself need not be a product globally. The information lost when looking chart-by-chart is precisely the family of transition functions, and the compatibility conditions on these functions are the cocycle identities. The main point of the chapter is that, under smoothness hypotheses, this transition data is not merely a shadow of a bundle: it can be used to reconstruct the bundle. ## Transition Maps Valued in Structure Groups Suppose a smooth fibre bundle $\pi:E\to M$ with typical fibre $F$ is described over open sets $U_i\subset M$ by local trivialisations $\Phi_i:\pi^{-1}(U_i)\to U_i\times F$. The basic question is: if a point of the total space is written using two such local descriptions, what change of coordinates occurs in the fibre direction? [definition: Bundle Transition Function] Let $\pi:E\to M$ be a smooth fibre bundle with typical fibre $F$, and let $\Phi_i:\pi^{-1}(U_i)\to U_i\times F$ and $\Phi_j:\pi^{-1}(U_j)\to U_j\times F$ be local trivialisations. The transition function from $j$ to $i$ is the map \begin{align*} g_{ij}:U_i\cap U_j\to \operatorname{Diff}(F) \end{align*} determined by \begin{align*} \Phi_i\circ \Phi_j^{-1}(x,y)=(x,g_{ij}(x)(y)) \end{align*} for every $x\in U_i\cap U_j$ and $y\in F$. [/definition] The definition says that the base point $x$ is unchanged on overlaps, while the fibre coordinate is transformed by a diffeomorphism of $F$. Thus each overlap carries a map $U_i\cap U_j\to \operatorname{Diff}(F)$, interpreted through the smoothness of the associated map $(x,y)\mapsto g_{ij}(x)(y)$. [example: Product Bundle Transitions] Let $E=M\times F$ with projection $\pi(x,y)=x$, and on each open set $U_i\subset M$ use the product trivialisation $\Phi_i:\pi^{-1}(U_i)\to U_i\times F$ given by \begin{align*} \Phi_i(x,y)=(x,y). \end{align*} Its inverse sends the same pair back to the corresponding point of $E$: \begin{align*} \Phi_i^{-1}(x,y)=(x,y). \end{align*} Therefore, for $x\in U_i\cap U_j$ and $y\in F$, \begin{align*} (\Phi_i\circ \Phi_j^{-1})(x,y)=\Phi_i(x,y). \end{align*} Since $\Phi_i(x,y)=(x,y)$, we get \begin{align*} (\Phi_i\circ \Phi_j^{-1})(x,y)=(x,y). \end{align*} By the definition of the transition function, the same change of local description has the form \begin{align*} (\Phi_i\circ \Phi_j^{-1})(x,y)=(x,g_{ij}(x)(y)). \end{align*} Comparing the second coordinates in $(x,y)=(x,g_{ij}(x)(y))$ gives \begin{align*} g_{ij}(x)(y)=y. \end{align*} Thus $g_{ij}(x)=\operatorname{id}_F$ for every $x\in U_i\cap U_j$. The product bundle with its global product coordinates has identity transition functions; non-identity transitions record the failure of a chosen atlas to use one fibre coordinate system everywhere. [/example] The product example isolates the general role of transitions: they record fibre-coordinate changes, and an atlas with identity transitions is exactly the data of a global product trivialization. A product bundle may still acquire non-identity transition functions if we choose local descriptions that twist the fibre coordinates. Many geometric bundles have extra fibre structure, so the next refinement asks what happens when those coordinate changes must preserve vector addition and scalar multiplication. [definition: Vector Bundle Transition Function] Let $\pi:E\to M$ be a smooth vector bundle of rank $k$, with local vector-bundle descriptions $\Phi_i:\pi^{-1}(U_i)\to U_i\times \mathbb R^k$. The vector bundle transition function from $j$ to $i$ is the map \begin{align*} g_{ij}:U_i\cap U_j\to GL(k,\mathbb R) \end{align*} defined by \begin{align*} \Phi_i\circ \Phi_j^{-1}(x,v)=(x,g_{ij}(x)v). \end{align*} [/definition] The passage from $\operatorname{Diff}(F)$ to $GL(k,\mathbb R)$ is one of the first appearances of a structure group. The structure group records which transformations are permitted in the fibre coordinate, and reducing the structure group usually means adding geometric structure to the bundle. [example: Tangent Bundle Transition Functions] Let $M$ be a smooth $n$-manifold with coordinate charts $(U_i,\varphi_i)$ and $(U_j,\varphi_j)$. For $p\in U_i$, use the tangent-bundle trivialisation \begin{align*} \Phi_i:T U_i\to U_i\times \mathbb R^n,\quad \Phi_i(v_p)=\bigl(p,d(\varphi_i)_p(v_p)\bigr). \end{align*} If $(p,w)\in U_i\times \mathbb R^n$, then the inverse trivialisation is \begin{align*} \Phi_i^{-1}(p,w)=d(\varphi_i)_p^{-1}(w)=d(\varphi_i^{-1})_{\varphi_i(p)}(w). \end{align*} Now take $p\in U_i\cap U_j$ and $v\in \mathbb R^n$. Applying the inverse of the $j$-th trivialisation gives \begin{align*} \Phi_j^{-1}(p,v)=d(\varphi_j)_p^{-1}(v). \end{align*} Applying the $i$-th trivialisation to this tangent vector gives \begin{align*} (\Phi_i\circ \Phi_j^{-1})(p,v)=\Phi_i\bigl(d(\varphi_j)_p^{-1}(v)\bigr). \end{align*} Using the definition of $\Phi_i$, this becomes \begin{align*} (\Phi_i\circ \Phi_j^{-1})(p,v)=\bigl(p,d(\varphi_i)_p(d(\varphi_j)_p^{-1}(v))\bigr). \end{align*} By the *Chain Rule* applied to $\varphi_i\circ\varphi_j^{-1}$ at $\varphi_j(p)$, \begin{align*} d(\varphi_i\circ\varphi_j^{-1})_{\varphi_j(p)}=d(\varphi_i)_p\circ d(\varphi_j^{-1})_{\varphi_j(p)}. \end{align*} Since $d(\varphi_j^{-1})_{\varphi_j(p)}=d(\varphi_j)_p^{-1}$, we have \begin{align*} d(\varphi_i\circ\varphi_j^{-1})_{\varphi_j(p)}=d(\varphi_i)_p\circ d(\varphi_j)_p^{-1}. \end{align*} Therefore \begin{align*} (\Phi_i\circ \Phi_j^{-1})(p,v)=\bigl(p,d(\varphi_i\circ\varphi_j^{-1})_{\varphi_j(p)}(v)\bigr). \end{align*} Writing this linear map in the standard coordinates on $\mathbb R^n$ gives its Jacobian matrix, so the vector-bundle transition function is \begin{align*} g_{ij}(p)=J(\varphi_i\circ \varphi_j^{-1})_{\varphi_j(p)}. \end{align*} Thus tangent bundle transition functions are exactly the Jacobian matrices of coordinate changes, recovering the usual transformation law for tangent vectors. [/example] ## Cocycle Identities on Triple Overlaps Transition functions come from comparing local descriptions, so they cannot be arbitrary maps on pairwise overlaps. The next question is what consistency condition appears when three local descriptions are available at the same base point. [quotetheorem:6084] [citeproof:6084] The triple-overlap identity is the formal expression of coordinate independence: changing descriptions along a chain gives the same fibre coordinate as the direct change. The hypotheses matter because the maps must arise from one common system of local descriptions. For example, over a single base point, three proposed fibre-coordinate changes $a,b,c\in \operatorname{Diff}(F)$ with $a\circ b\ne c$ cannot be the changes from chart $k$ to chart $j$, then $j$ to $i$, and directly from $k$ to $i$ in any atlas. The theorem therefore gives a necessary compatibility condition; by itself it does not yet construct a total space or identify when two different compatible families describe the same bundle. This turns the theorem into a template for data that may be specified without first having a bundle, which leads to the notion of a cocycle. [definition: Smooth Cocycle] Let $(U_i)_{i\in I}$ be an open cover of a smooth manifold $M$, let $F$ be a smooth manifold, and let $G\subseteq \operatorname{Diff}(F)$ be a specified structure group. A smooth $G$-valued cocycle on the cover is a family of maps \begin{align*} g_{ij}:U_i\cap U_j\to G \end{align*} such that the associated map $(x,y)\mapsto g_{ij}(x)(y)$ is smooth and \begin{align*} g_{ii}(x)=\operatorname{id}_F \end{align*} and \begin{align*} g_{ij}(x)\circ g_{jk}(x)=g_{ik}(x) \end{align*} for all $x\in U_i\cap U_j\cap U_k$. [/definition] The inverse relation is encoded in the displayed identities by taking $k=i$. For vector bundles, a smooth $GL(k,\mathbb R)$-valued cocycle is a family of smooth maps $g_{ij}:U_i\cap U_j\to GL(k,\mathbb R)$ satisfying the same formulas with matrix multiplication. [example: Mobius Line Bundle Cocycle] Write $U_0\cap U_1=C_+\sqcup C_-$ as the disjoint union of its two connected components, and identify $GL(1,\mathbb R)$ with $\mathbb R^\times$ acting on a fibre coordinate $t\in\mathbb R$ by scalar multiplication. Define the identity transitions by \begin{align*} g_{00}(x)=1 \end{align*} and \begin{align*} g_{11}(x)=1. \end{align*} On the two overlap components, define \begin{align*} g_{01}(x)=1\quad\text{for }x\in C_+, \end{align*} and \begin{align*} g_{01}(x)=-1\quad\text{for }x\in C_-. \end{align*} Since $1^{-1}=1$ and $(-1)^{-1}=-1$, the inverse transition is \begin{align*} g_{10}(x)=g_{01}(x)^{-1}=g_{01}(x). \end{align*} The map $g_{01}$ is smooth because it is constant on each connected component of the overlap. The cocycle identities reduce to multiplication in $\mathbb R^\times$. If $x\in C_+$, then \begin{align*} g_{01}(x)g_{10}(x)=1\cdot 1=1=g_{00}(x). \end{align*} If $x\in C_-$, then \begin{align*} g_{01}(x)g_{10}(x)=(-1)(-1)=1=g_{00}(x). \end{align*} The same calculation with the indices reversed gives \begin{align*} g_{10}(x)g_{01}(x)=1=g_{11}(x). \end{align*} The remaining identities only multiply by an identity transition, for example \begin{align*} g_{00}(x)g_{01}(x)=1\cdot g_{01}(x)=g_{01}(x) \end{align*} and \begin{align*} g_{01}(x)g_{11}(x)=g_{01}(x)\cdot 1=g_{01}(x). \end{align*} Thus these functions form a smooth $GL(1,\mathbb R)$-valued cocycle. In local fibre coordinates, crossing the component $C_+$ sends $t$ to \begin{align*} 1\cdot t=t, \end{align*} while crossing $C_-$ sends $t$ to \begin{align*} (-1)t=-t. \end{align*} Going once around the circle crosses the two overlap components, so the total change in fibre coordinate is \begin{align*} (-1)\cdot 1\cdot t=-t. \end{align*} The sign reversal is the transition-data expression of the Mobius line bundle, whereas the product line bundle has identity transition $t\mapsto t$ on both overlap components. [/example] This example is the prototype for clutching data: on a cover with two large pieces, the whole bundle may be encoded by a transition function on their overlap. The cocycle identities become especially simple for two-set covers, but the same principle underlies the general reconstruction theorem. ## Reconstruction of Bundles from Cocycles We now reverse direction. Starting with open sets and transition functions satisfying the cocycle identities, can we manufacture a total space whose local descriptions have exactly those transitions? [quotetheorem:6085] [citeproof:6085] The Hausdorff and second-countable hypotheses prevent quotient pathologies and are automatic in many finite-cover constructions used in the course. A standard warning example is the line with two origins, obtained by gluing two copies of $\mathbb R$ along $\mathbb R\setminus\{0\}$: the two origins cannot be separated by disjoint open sets, so the quotient is not a smooth manifold. This example is not a valid smooth cocycle over an ordinary open-cover overlap, but it shows why quotient gluing arguments must include topological separation checks rather than only formal equivalence relations. The theorem also has a classification limitation. It constructs a bundle from a chosen cocycle, but a different cocycle may give an isomorphic bundle after changing local fibre coordinates. Thus cocycles are complete local data only up to the [equivalence relation](/page/Equivalence%20Relation) described next, not canonical names for bundles. [example: Complex Line Bundles from Unit Circle Transitions] Let $(U_i)$ be an open cover of $M$, and let $h_{ij}:U_i\cap U_j\to U(1)$ be smooth functions satisfying \begin{align*} h_{ij}(x)h_{jk}(x)=h_{ik}(x) \end{align*} for every $x\in U_i\cap U_j\cap U_k$. Since $U(1)\subset \mathbb C^\times=GL(1,\mathbb C)$, each $h_{ij}(x)$ acts on a fibre coordinate $z\in\mathbb C$ by scalar multiplication. Composition of two such scalar multiplications gives \begin{align*} h_{ij}(x)\bigl(h_{jk}(x)z\bigr)=\bigl(h_{ij}(x)h_{jk}(x)\bigr)z. \end{align*} Using the assumed cocycle identity, this becomes \begin{align*} \bigl(h_{ij}(x)h_{jk}(x)\bigr)z=h_{ik}(x)z. \end{align*} Thus the functions $h_{ij}$ form a smooth $GL(1,\mathbb C)$-valued cocycle, so by *[Bundle Reconstruction from a Smooth Cocycle](/theorems/6085)* they reconstruct a complex line bundle. In the local description over $U_i$, define the fibrewise Hermitian length by \begin{align*} \|z_i\|^2=z_i\overline{z_i}=|z_i|^2. \end{align*} On an overlap $U_i\cap U_j$, the two local fibre coordinates are related by \begin{align*} z_i=h_{ij}(x)z_j. \end{align*} Substituting this relation into the length formula gives \begin{align*} \|z_i\|^2=\bigl(h_{ij}(x)z_j\bigr)\overline{h_{ij}(x)z_j}. \end{align*} Since complex conjugation respects products, we have \begin{align*} \overline{h_{ij}(x)z_j}=\overline{h_{ij}(x)}\,\overline{z_j}. \end{align*} Therefore \begin{align*} \|z_i\|^2=h_{ij}(x)\overline{h_{ij}(x)}\,z_j\overline{z_j}. \end{align*} By the identity $a\overline a=|a|^2$ for complex numbers, \begin{align*} h_{ij}(x)\overline{h_{ij}(x)}\,z_j\overline{z_j}=|h_{ij}(x)|^2\|z_j\|^2. \end{align*} Because $h_{ij}(x)\in U(1)$, we have $|h_{ij}(x)|=1$, and hence \begin{align*} |h_{ij}(x)|^2\|z_j\|^2=1\cdot \|z_j\|^2=\|z_j\|^2. \end{align*} Thus the Hermitian length computed in the $i$-th trivialisation equals the Hermitian length computed in the $j$-th trivialisation. The reconstructed complex line bundle therefore carries a natural fibrewise Hermitian metric obtained by using the standard Hermitian metric in each local fibre coordinate. [/example] A reconstructed bundle depends on the chosen cocycle, but not on irrelevant changes of local coordinates. The obstruction is that two sets of transition functions may look different simply because each local product chart uses a different fibre coordinate. To decide whether the difference is only bookkeeping, one must ask whether there are local gauge transformations on the sets $U_i$ that convert one cocycle into the other on every overlap. [quotetheorem:6086] [citeproof:6086] This theorem is the cocycle version of the equivalent-atlas construction from Chapter 1. The maps $a_i$ are local gauge transformations: they rename the fibre coordinate over each $U_i$ while leaving the base point fixed. If no such family exists, the two cocycles may define genuinely non-isomorphic bundles; for instance, the product real line bundle on $S^1$ and the Mobius line bundle cannot be related by such local changes because doing so would remove the sign reversal around the circle. The theorem is therefore an isomorphism criterion for cocycles on a fixed cover, not a full classification theorem for all bundles over $M$; a full classification must also account for refinements of covers and the relevant cohomological equivalence relation. [example: Stereographic Cocycles for the Tangent Bundle of the Sphere] Let $S^2\subset \mathbb R^3$ have coordinates $(X,Y,Z)$, and use stereographic projection from the north and south poles: \begin{align*} \varphi_N(X,Y,Z)=\left(\frac{X}{1-Z},\frac{Y}{1-Z}\right) \end{align*} and \begin{align*} \varphi_S(X,Y,Z)=\left(\frac{X}{1+Z},\frac{Y}{1+Z}\right). \end{align*} Write $(u,v)=\varphi_S(p)$ and set $\rho=u^2+v^2$. On $U_N\cap U_S$, the point $p$ is neither pole, so $\rho>0$. The inverse southern stereographic chart is \begin{align*} \varphi_S^{-1}(u,v)=\left(\frac{2u}{\rho+1},\frac{2v}{\rho+1},\frac{1-\rho}{\rho+1}\right). \end{align*} Applying $\varphi_N$ to this point gives first coordinate \begin{align*} \frac{\frac{2u}{\rho+1}}{1-\frac{1-\rho}{\rho+1}}=\frac{\frac{2u}{\rho+1}}{\frac{\rho+1-(1-\rho)}{\rho+1}}=\frac{\frac{2u}{\rho+1}}{\frac{2\rho}{\rho+1}}=\frac{u}{\rho}. \end{align*} The same calculation for the second coordinate gives \begin{align*} \frac{\frac{2v}{\rho+1}}{1-\frac{1-\rho}{\rho+1}}=\frac{\frac{2v}{\rho+1}}{\frac{2\rho}{\rho+1}}=\frac{v}{\rho}. \end{align*} Therefore the coordinate change on the overlap is \begin{align*} F(u,v)=(\varphi_N\circ\varphi_S^{-1})(u,v)=\left(\frac{u}{u^2+v^2},\frac{v}{u^2+v^2}\right). \end{align*} For the tangent bundle, the transition function is the Jacobian matrix of this coordinate change, so \begin{align*} g_{NS}(p)=J F_{\varphi_S(p)}. \end{align*} The four entries of this Jacobian are obtained by the quotient rule: \begin{align*} \frac{\partial}{\partial u}\left(\frac{u}{u^2+v^2}\right)=\frac{(u^2+v^2)-u(2u)}{(u^2+v^2)^2}=\frac{v^2-u^2}{(u^2+v^2)^2}. \end{align*} \begin{align*} \frac{\partial}{\partial v}\left(\frac{u}{u^2+v^2}\right)=\frac{0\cdot (u^2+v^2)-u(2v)}{(u^2+v^2)^2}=\frac{-2uv}{(u^2+v^2)^2}. \end{align*} \begin{align*} \frac{\partial}{\partial u}\left(\frac{v}{u^2+v^2}\right)=\frac{0\cdot (u^2+v^2)-v(2u)}{(u^2+v^2)^2}=\frac{-2uv}{(u^2+v^2)^2}. \end{align*} \begin{align*} \frac{\partial}{\partial v}\left(\frac{v}{u^2+v^2}\right)=\frac{(u^2+v^2)-v(2v)}{(u^2+v^2)^2}=\frac{u^2-v^2}{(u^2+v^2)^2}. \end{align*} Thus $g_{NS}(p)$ is the $2\times 2$ matrix with entries \begin{align*} (g_{NS}(p))_{11}=\frac{v^2-u^2}{(u^2+v^2)^2}. \end{align*} \begin{align*} (g_{NS}(p))_{12}=\frac{-2uv}{(u^2+v^2)^2}. \end{align*} \begin{align*} (g_{NS}(p))_{21}=\frac{-2uv}{(u^2+v^2)^2}. \end{align*} \begin{align*} (g_{NS}(p))_{22}=\frac{u^2-v^2}{(u^2+v^2)^2}. \end{align*} Its determinant is \begin{align*} \det g_{NS}(p)=\frac{(v^2-u^2)(u^2-v^2)-(-2uv)(-2uv)}{(u^2+v^2)^4}. \end{align*} Since $(v^2-u^2)(u^2-v^2)=-(u^2-v^2)^2$, this becomes \begin{align*} \det g_{NS}(p)=\frac{-(u^2-v^2)^2-4u^2v^2}{(u^2+v^2)^4}. \end{align*} Expanding the square gives \begin{align*} -(u^2-v^2)^2-4u^2v^2=-(u^4-2u^2v^2+v^4)-4u^2v^2. \end{align*} Combining terms gives \begin{align*} -(u^4-2u^2v^2+v^4)-4u^2v^2=-(u^4+2u^2v^2+v^4)=-(u^2+v^2)^2. \end{align*} Therefore \begin{align*} \det g_{NS}(p)=\frac{-(u^2+v^2)^2}{(u^2+v^2)^4}=-\frac{1}{(u^2+v^2)^2}. \end{align*} Because $u^2+v^2>0$ on the overlap, this determinant is nonzero, so $g_{NS}(p)\in GL(2,\mathbb R)$. This matrix is the concrete transition datum for $TS^2$ in stereographic coordinates: it changes southern tangent-coordinate vectors into northern tangent-coordinate vectors. Its dependence on $(u,v)$ reflects the chosen coordinate systems on the annular overlap, but nonconstancy of one transition function by itself does not prove that the bundle is nontrivial. [/example] The chapter leaves us with a compact dictionary. A bundle atlas gives transition functions; transition functions satisfy cocycle identities; a smooth cocycle reconstructs a bundle; and changing local descriptions conjugates the cocycle by a family of local maps. Chapter 3 uses this dictionary to study sections, where local functions must satisfy compatibility equations determined by the same transition data. Transition functions turn local trivializations into algebraic data, but sections ask for something stricter: a compatible choice of points in every fibre. The next chapter studies how such local choices patch together, and when they fail to do so, by following the same overlap relations. # 3. Sections and Local-to-Global Problems This chapter belongs to the part of the course where fibre bundles are studied through the objects living over their base manifolds. Chapters 1 and 2 introduced local trivialisations and transition functions; here the goal is to understand sections, local frames, support, and the first local-to-global problems they create. The main prerequisite is the language of smooth manifolds and smooth fibre bundles, especially the fact that a bundle is locally a product but may not be globally one. The local-to-global theme is that sections are often easy to write in a local trivialisation, while their global existence is controlled by the transition data of Chapter 2 and by the topology of the bundle. ## Smooth Sections in a Trivialization How can a map into the total space be recorded by functions on the base? The point of a local trivialization is that, over a small open set, a section of a bundle is the same thing as a map into the model fibre. Smoothness of a section is therefore tested by looking at its component functions in every allowed trivialization. [definition: Smooth Section] Let $\rho:E\to M$ be a smooth fibre bundle. A smooth section of $\rho$ is a smooth map $s:M\to E$ such that \begin{align*} \rho\circ s=\operatorname{id}_M. \end{align*} For an open set $U\subset M$, a smooth local section over $U$ is a smooth map $s:U\to E$ such that $\rho\circ s=\operatorname{id}_U$. [/definition] The equation $\rho\circ s=\operatorname{id}$ says that $s(x)$ lies in the fibre $E_x=\rho^{-1}(x)$ for every $x$. Thus a section is a smoothly varying choice of one point in each fibre over its domain. Local sections are the natural objects because bundles are built locally, and global sections are the objects whose existence records a successful compatibility problem. To compare sections with ordinary functions, fix a local trivialization $\Phi:\rho^{-1}(U)\to U\times F$ of a smooth fibre bundle with fibre $F$. The obstruction is that a section takes values in the total space $E$, while ordinary calculus is done with maps into the model fibre $F$. A local trivialization removes the base coordinate, leaving the fibre coordinate that can be differentiated and compared on overlaps. [definition: Local Representative of a Section] Let $\rho:E\to M$ be a smooth fibre bundle with fibre $F$, let $\Phi:\rho^{-1}(U)\to U\times F$ be a local trivialization, and let $s:U\to E$ be a local section. The local representative of $s$ in the trivialization $\Phi$ is the map $s_\Phi:U\to F$ determined by \begin{align*} \Phi(s(x))=(x,s_\Phi(x)) \end{align*} for all $x\in U$. [/definition] The representative is where calculations live. For a vector bundle of rank $k$, the fibre is $\mathbb R^k$, so $s_\Phi$ is a $k$-tuple of smooth real-valued functions. A tempting but incomplete approach is to choose one convenient trivialization and check smoothness there. This fails because a section may pass through regions not covered by that trivialization, and on overlaps its component functions must transform smoothly according to the bundle atlas. The local test question is therefore sharper: if the representative is smooth in every trivialization of an atlas covering the domain, does that already force the original fibrewise map into $E$ to be smooth? [quotetheorem:6087] [citeproof:6087] This result is the bridge between geometric sections and analysis on open subsets of Euclidean space. Each hypothesis has a role. The condition $\rho\circ s=\operatorname{id}_M$ is what makes $s$ a fibrewise choice; without it, a smooth map $M\to E$ could move a point of the base into a fibre over a different point, so no representative $x\mapsto s_\Phi(x)$ over $x$ would be defined. The smooth bundle atlas matters because a continuous or incompatible family of trivializations could make component functions look smooth in one chart while destroying smoothness after changing charts. Finally, the atlas must cover the whole domain: checking a representative only on one open set says nothing about the behaviour of the map near points outside that set. Thus the theorem legitimises component calculations, but only after the section condition and the relevant smooth trivializations have fixed the geometric meaning of those components. [example: Vector Fields as Sections] Let $M$ be a smooth manifold and let $TM\to M$ be its tangent bundle. Write $\mathfrak{X}(M)=\Gamma(TM)$. A vector field $X\in\mathfrak{X}(M)$ assigns to each point $x\in M$ a tangent vector $X(x)\in T_xM$, and the bundle projection satisfies $\pi_{TM}(X(x))=x$, so $X$ is a section of $TM$. Fix a chart $(U,\varphi)$ with coordinate functions $(x_1,\dots,x_n)$. For each $x\in U$, the coordinate tangent vectors $\partial_{x_1}|_x,\dots,\partial_{x_n}|_x$ form a basis of $T_xM$. Hence there are unique real numbers $X_1(x),\dots,X_n(x)$ such that \begin{align*} X(x)=\sum_{i=1}^n X_i(x)\,\partial_{x_i}|_x. \end{align*} The tangent bundle trivialization determined by this coordinate frame is the map $\Phi:TM|_U\to U\times\mathbb R^n$ defined by \begin{align*} \Phi\left(\sum_{i=1}^n a_i\,\partial_{x_i}|_x\right)=(x,(a_1,\dots,a_n)). \end{align*} Applying this definition to $X(x)$ gives \begin{align*} \Phi(X(x))=\Phi\left(\sum_{i=1}^n X_i(x)\,\partial_{x_i}|_x\right). \end{align*} By the defining formula for $\Phi$, this equals \begin{align*} \Phi(X(x))=(x,(X_1(x),\dots,X_n(x))). \end{align*} Therefore the local representative of $X$ in this trivialization is \begin{align*} X_\Phi(x)=(X_1(x),\dots,X_n(x)). \end{align*} Thus the familiar coordinate expression \begin{align*} X|_U=\sum_{i=1}^n X_i\,\partial_{x_i} \end{align*} is exactly the fibre-coordinate form of the section in the coordinate-frame trivialization, and smoothness of the vector field is smoothness of these component functions in local coordinates. [/example] Vector fields also show why global questions are more delicate than local ones. Every coordinate chart supplies many local vector fields, but the existence of a vector field that never hits the zero vector is a stronger property. To formulate this obstruction for an arbitrary vector bundle, we isolate the notion of a nowhere-zero section. [definition: Nowhere-Zero Section] Let $\pi:E\to M$ be a vector bundle. A section $s:M\to E$ is nowhere-zero if \begin{align*} s(x)\ne 0_x \end{align*} for every $x\in M$, where $0_x$ denotes the zero vector in the fibre $E_x$. [/definition] A nowhere-zero section of $TM$ is a nonvanishing vector field. On $S^1$, the unit tangent field gives such a section. On $S^2$, the [hairy ball theorem](/theorems/2248) says no such section exists; this later becomes a prototype for characteristic-class obstructions. ## Support, Restriction, Extension, and Zero Loci A section is often constructed on a small open set first, then modified so that it becomes a global section. The problem is not just smoothness: we must control where the section is nonzero, how it restricts to smaller open sets, and whether it can be extended without disrupting the geometry elsewhere. [definition: Restriction of a Section] Let $\rho:E\to M$ be a smooth fibre bundle, let $s:U\to E$ be a local section over an open set $U\subset M$, and let $V\subset U$ be open. The restriction of $s$ to $V$ is the local section $s|_V:V\to E$ given by \begin{align*} (s|_V)(x)=s(x) \end{align*} for all $x\in V$. [/definition] Restriction loses no local information over $V$, but it forgets whether the original section could be continued outside $V$. The reverse direction, extension, needs extra hypotheses and is most flexible for vector bundles because fibres can be added and scaled. [definition: Support of a Section] Let $\pi:E\to M$ be a vector bundle and let $s:U\to E$ be a local section. The support of $s$ in $U$ is \begin{align*} \operatorname{supp}(s)=\overline{\{x\in U:s(x)\ne 0_x\}}^{\,U}. \end{align*} [/definition] The closure is taken in the domain $U$. Compactly supported sections are the right objects for local construction, because they can be extended by the zero section outside their support. This motivates the zero locus as the complementary bookkeeping device: instead of asking where a section is active, we ask exactly where it hits the zero vector. [definition: Zero Locus of a Section] Let $\pi:E\to M$ be a vector bundle and let $s:U\to E$ be a local section. The zero locus of $s$ is \begin{align*} Z(s)=\{x\in U:s(x)=0_x\}. \end{align*} [/definition] The zero locus is the place where a section fails to be nonvanishing. In later courses, transverse zero loci of sections become a mechanism for producing submanifolds and characteristic classes. Here the main point is more elementary: support and zero loci are complementary ways of tracking where a section carries information. [example: Sections of the Tautological Line Bundle] Let $\gamma^1\to\mathbb RP^n$ be the tautological line bundle, so the fibre over a point $[v]\in\mathbb RP^n$ is the line \begin{align*} \gamma^1_{[v]}=\operatorname{span}(v)\subset\mathbb R^{n+1}. \end{align*} The zero section is \begin{align*} z([v])=0\in \operatorname{span}(v), \end{align*} and this is well-defined because the zero vector belongs to every fibre. For each index $i$, let \begin{align*} U_i=\{[v_0:\cdots:v_n]\in\mathbb RP^n:v_i\ne 0\}. \end{align*} On $U_i$, every line has a unique representative whose $i$-th coordinate is $1$: \begin{align*} \widehat v_i =\frac{1}{v_i}(v_0,\dots,v_n) =\left(\frac{v_0}{v_i},\dots,\frac{v_{i-1}}{v_i},1,\frac{v_{i+1}}{v_i},\dots,\frac{v_n}{v_i}\right). \end{align*} This representative is independent of the chosen nonzero vector on the line, since if $w=\lambda v$ with $\lambda\ne 0$, then \begin{align*} \frac{w_j}{w_i} =\frac{\lambda v_j}{\lambda v_i} =\frac{v_j}{v_i} \end{align*} for every $j$. Define a local section $\sigma_i:U_i\to\gamma^1$ by \begin{align*} \sigma_i([v])=\widehat v_i. \end{align*} Because $\widehat v_i=(1/v_i)v$, the vector $\sigma_i([v])$ lies in $\operatorname{span}(v)=\gamma^1_{[v]}$, and because its $i$-th coordinate is $1$, it is never zero. In the affine coordinates on $U_i$ given by $y_j=v_j/v_i$ for $j\ne i$, the section is \begin{align*} \sigma_i([v]) =(y_0,\dots,y_{i-1},1,y_{i+1},\dots,y_n), \end{align*} so its coordinate functions are exactly the affine coordinate functions together with the constant function $1$. Thus the standard affine charts give explicit local nonzero sections. A global nonzero section would have to choose, for every line $[v]$, one nonzero vector in $\operatorname{span}(v)$ continuously and smoothly in these local coordinates, with the choices agreeing on overlaps of the charts. [/example] This example returns us to the extension problem: local sections are plentiful, but using them globally requires a controlled way to turn them off outside the region where they were defined. The naive extension by declaring $s=0$ outside $U$ usually fails to be smooth along the boundary of $U$; the values may jump, and even if the values match, derivatives need not match. The standard extension device inserts a bump function, preserving a prescribed section on a smaller set and making the result zero outside a slightly larger set. [quotetheorem:6088] [citeproof:6088] This theorem is one of the main reasons vector bundles have a rich sheaf of sections, but its hypotheses are not cosmetic. The compact containment $\overline V\subset U$ creates room for a bump function that is equal to $1$ near $\overline V$ and already zero before reaching the complement of $U$; without this buffer, cutting off at the boundary of $U$ can force nonsmooth behaviour. The vector bundle structure is also essential, because the construction multiplies a section by a scalar function and uses the zero vector $0_x$ in each fibre; an arbitrary smooth fibre bundle has no canonical scalar multiplication or zero section. The theorem preserves the original section only near $\overline V$, and it may introduce zeros wherever the cutoff function vanishes, so it is not a tool for preserving nonvanishing. Its purpose is localization: build a section in a trivializing neighbourhood, then convert it into a global test section whose support stays inside the chosen region. [example: Localising a Coordinate Vector Field] Let $(U,\varphi)$ be a coordinate chart on a smooth manifold $M$, with coordinate functions $(x_1,\dots,x_n)$, and let $V\subset U$ have compact closure in $U$. On $U$, the first coordinate vector field is the local section sending each $x\in U$ to $\partial_{x_1}|_x\in T_xM$, so $\pi_{TM}(\partial_{x_1}|_x)=x$. Choose a smooth function $\chi:M\to\mathbb R$ such that $\chi=1$ on a neighbourhood of $\overline V$ and $\operatorname{supp}(\chi)\subset U$. Define $X:M\to TM$ by \begin{align*} X(x)=\begin{cases}\chi(x)\partial_{x_1}|_x, & x\in U, 0_x, & x\notin U.\end{cases} \end{align*} If $x\in U$, then $\chi(x)\partial_{x_1}|_x\in T_xM$, and therefore \begin{align*} \pi_{TM}(X(x))=\pi_{TM}(\chi(x)\partial_{x_1}|_x)=x. \end{align*} If $x\notin U$, then $X(x)=0_x\in T_xM$, and therefore \begin{align*} \pi_{TM}(X(x))=\pi_{TM}(0_x)=x. \end{align*} Thus $\pi_{TM}\circ X=\operatorname{id}_M$, so $X$ is a section. On $U$, the coordinate-frame representative of $X$ is \begin{align*} X_\Phi(x)=(\chi(x),0,\dots,0), \end{align*} because \begin{align*} X(x)=\chi(x)\partial_{x_1}|_x+0\partial_{x_2}|_x+\cdots+0\partial_{x_n}|_x. \end{align*} The functions $\chi,0,\dots,0$ are smooth on $U$. Since $\operatorname{supp}(\chi)\subset U$, every point of $M\setminus U$ has a neighbourhood on which $\chi=0$, so near the boundary of $U$ the formula $\chi\partial_{x_1}$ is already the zero section. Hence the formula on $U$ and the zero formula on $M\setminus U$ glue to a smooth global vector field. Finally, if $X(x)\ne 0_x$, then the defining formula forces $\chi(x)\ne 0$, so \begin{align*} \operatorname{supp}(X)\subset \operatorname{supp}(\chi)\subset U. \end{align*} Since $\chi=1$ near $\overline V$, one also has \begin{align*} X(x)=\partial_{x_1}|_x \end{align*} on that neighbourhood. Thus a single coordinate basis vector has been converted into a globally defined test vector field, while its support remains inside the chosen coordinate chart. [/example] The [extension theorem](/theorems/59) has a useful limitation. It preserves a local section on a smaller region, but it may introduce zeros where the cutoff function vanishes. Thus it is a tool for building sections with compact support, not for solving global nonvanishing problems. ## Frames and Triviality by Global Frames When does a vector bundle look like a product for structural reasons rather than after choosing many local charts? A frame answers this by replacing transition data with sections. If we can choose a basis smoothly in every fibre at once, the bundle has no remaining gluing ambiguity. [definition: Local Frame] Let $\pi:E\to M$ be a rank $k$ smooth vector bundle. A local frame for $E$ over an open set $U\subset M$ is an ordered $k$-tuple of smooth local sections $(e_1,\dots,e_k)$ over $U$ such that \begin{align*} (e_1(x),\dots,e_k(x)) \end{align*} is a basis of the vector space $E_x$ for every $x\in U$. [/definition] A local frame is the vector bundle analogue of a coordinate frame on a tangent bundle. Once a frame is chosen, every section has component functions, and these functions determine the section over the frame domain. The next definition removes the word local and asks whether the same basis of sections can be chosen over all of $M$ at once. [definition: Global Frame] Let $\pi:E\to M$ be a rank $k$ smooth vector bundle. A global frame for $E$ is a local frame over $M$. [/definition] The difference between local and global frames is the difference between local triviality and product structure. Local frames always exist by the definition of a vector bundle; global frames may fail to exist. [example: Local Frames for the Tangent Bundle of the Sphere] Let $S^2\subset\mathbb R^3$, and set \begin{align*} U_N=S^2\setminus\{(0,0,1)\}, \qquad U_S=S^2\setminus\{(0,0,-1)\}. \end{align*} Stereographic projection from the north pole gives coordinates $(u,v)$ on $U_N$ whose inverse parametrization is \begin{align*} F_N(u,v)=\left(\frac{2u}{u^2+v^2+1},\frac{2v}{u^2+v^2+1},\frac{u^2+v^2-1}{u^2+v^2+1}\right). \end{align*} Writing $r^2=u^2+v^2$, this map lands in $S^2$ because \begin{align*} \|F_N(u,v)\|^2=\frac{4u^2+4v^2+(r^2-1)^2}{(r^2+1)^2}=\frac{4r^2+r^4-2r^2+1}{(r^2+1)^2}=\frac{r^4+2r^2+1}{(r^2+1)^2}=1. \end{align*} The coordinate vector fields on $U_N$ are defined by \begin{align*} e_1^N(F_N(u,v))=\frac{\partial F_N}{\partial u}(u,v), \qquad e_2^N(F_N(u,v))=\frac{\partial F_N}{\partial v}(u,v). \end{align*} Since $F_N(u,v)\cdot F_N(u,v)=1$, differentiating with respect to $u$ gives \begin{align*} 2F_N(u,v)\cdot \frac{\partial F_N}{\partial u}(u,v)=0. \end{align*} Differentiating with respect to $v$ gives \begin{align*} 2F_N(u,v)\cdot \frac{\partial F_N}{\partial v}(u,v)=0. \end{align*} Thus both coordinate vectors are orthogonal to the radius vector $F_N(u,v)$, so they lie in the tangent plane $T_{F_N(u,v)}S^2$. Because $F_N$ is the inverse of a coordinate chart, its differential \begin{align*} dF_N|_{(u,v)}:\mathbb R^2\to T_{F_N(u,v)}S^2 \end{align*} is a linear isomorphism. Therefore the images of the standard basis vectors, \begin{align*} \frac{\partial F_N}{\partial u}(u,v) \quad \text{and} \quad \frac{\partial F_N}{\partial v}(u,v), \end{align*} form a basis of $T_{F_N(u,v)}S^2$. Hence $(e_1^N,e_2^N)$ is a local frame for $TS^2$ over $U_N$. The same construction works over $U_S$. Stereographic projection from the south pole has inverse parametrization \begin{align*} F_S(u,v)=\left(\frac{2u}{u^2+v^2+1},\frac{2v}{u^2+v^2+1},\frac{1-u^2-v^2}{u^2+v^2+1}\right), \end{align*} and its coordinate vector fields are \begin{align*} e_1^S(F_S(u,v))=\frac{\partial F_S}{\partial u}(u,v), \qquad e_2^S(F_S(u,v))=\frac{\partial F_S}{\partial v}(u,v). \end{align*} As above, the identity $\|F_S(u,v)\|^2=1$ implies that these two vectors lie in $T_{F_S(u,v)}S^2$, and the differential of the coordinate parametrization identifies the standard basis of $\mathbb R^2$ with a basis of that tangent plane. Thus $(e_1^S,e_2^S)$ is a local frame for $TS^2$ over $U_S$. These two frames show explicitly that $TS^2$ is locally generated by smooth tangent vector fields. They cannot be combined into a global frame: if $(E_1,E_2)$ were a global frame of $TS^2$, then $E_1(x)$ would be nonzero for every $x\in S^2$, since it is the first vector in a basis of $T_xS^2$. That would give a nowhere-zero tangent vector field on $S^2$, contradicting the *Hairy Ball Theorem*. [/example] The next theorem is the central local-to-global criterion of the chapter. A naive hope would be that a few well-chosen sections, or even one nowhere-zero section, should make a vector bundle look like a product. That is true for line bundles, but in higher rank a product trivialization needs $k$ independent directions in every fibre, not merely one. The theorem identifies global frames as exactly the data needed to replace the vector bundle by a product bundle. [quotetheorem:6089] [citeproof:6089] This theorem turns a global frame into a coordinate system on the bundle, not on the base. The fibrewise basis condition is essential: if the sections become linearly dependent at even one point, the formula using their coefficients cannot be inverted on that fibre. In rank $k>1$, a single nowhere-zero section only selects a line subbundle candidate and leaves the complementary directions unspecified, so it does not by itself trivialise the whole bundle. The statement is also specific to vector bundles; for a general fibre bundle there is no linear combination of sections and no notion of a basis in each fibre. In transition-function language, a global frame is exactly a choice of local frames whose overlap matrices have been absorbed into one globally defined basis, so the remaining transition data is equivalent to the identity. [example: Line Bundles and Nonzero Sections] Let $L\to M$ be a smooth real line bundle, and let $s:M\to L$ be nowhere-zero. Since each fibre $L_x$ is one-dimensional and $s(x)\ne 0_x$, the one-element tuple $(s)$ is a basis of $L_x$ for every $x\in M$. Thus $(s)$ is a global frame. Define \begin{align*} \Phi:M\times\mathbb R\to L,\qquad \Phi(x,a)=a\,s(x). \end{align*} For each fixed $x$, the fibre map \begin{align*} \Phi_x:\mathbb R\to L_x,\qquad a\mapsto a\,s(x) \end{align*} is linear. It is injective because if $\Phi_x(a)=0_x$, then \begin{align*} a\,s(x)=0_x, \end{align*} and since $s(x)\ne 0_x$ in the one-dimensional vector space $L_x$, this forces $a=0$. It is surjective because every vector $\ell\in L_x$ is a scalar multiple of the basis vector $s(x)$, so there is a unique $a\in\mathbb R$ with \begin{align*} \ell=a\,s(x)=\Phi_x(a). \end{align*} Hence $\Phi$ is a fibrewise linear bijection over $M$. In a local trivialization where $s$ is represented by a nowhere-zero smooth function $f:U\to\mathbb R$, the map is represented by \begin{align*} (x,a)\mapsto (x,a f(x)), \end{align*} and its inverse is represented by \begin{align*} (x,b)\mapsto \left(x,\frac{b}{f(x)}\right), \end{align*} which is smooth because $f$ is smooth and $f(x)\ne 0$ on $U$. Therefore $\Phi$ is a smooth vector bundle isomorphism. Conversely, suppose $L$ is trivial, with a vector bundle isomorphism $\Psi:L\to M\times\mathbb R$ over $M$. Define \begin{align*} s(x)=\Psi^{-1}(x,1). \end{align*} Then $s(x)\in L_x$, and \begin{align*} \Psi(s(x))=(x,1)\ne (x,0)=\Psi(0_x), \end{align*} so injectivity of $\Psi$ on the fibre $L_x$ gives $s(x)\ne 0_x$. Thus a real line bundle is trivial exactly when it admits a nowhere-zero section. [/example] Frames also connect back to transition functions. On an overlap of two local frames, the change-of-frame matrix is precisely the transition matrix of the corresponding vector bundle trivializations. A global frame is the special case in which a single frame covers all of $M$, so there are no nontrivial overlap matrices to manage. [remark: Local Frames and Transition Matrices] Let $(e_1,\dots,e_k)$ and $(f_1,\dots,f_k)$ be local frames over the same open set $U$. There is a smooth map $A:U\to GL(k,\mathbb R)$ such that \begin{align*} f_j(x)=\sum_{i=1}^k A_{ij}(x)e_i(x). \end{align*} The columns of $A(x)$ are the coordinates of the $f_j(x)$ in the $e$-frame. This is the same matrix data that appears as a transition function between the two induced trivializations. [/remark] The chapter closes with the governing picture. Local sections and local frames exist because bundles are locally products; bump functions allow local vector-bundle constructions to be made global with controlled support; and global frames are exactly the extra data needed to make the whole vector bundle a product. Chapter 6 studies how pullbacks and bundle maps transport these section-level constructions between different bases, after Chapter 4 develops vector-bundle operations and Chapter 5 supplies metrics and partitions of unity. Sections are the first global objects we can build from transition data, and they reveal how fibrewise linear algebra interacts with the base manifold. The next chapter uses that viewpoint to construct new vector bundles from old ones through operations such as sums, tensor products, duals, and related fibrewise constructions. # 4. Operations on Vector Bundles Building on the transition-function viewpoint of Chapter 2 and the section language of Chapter 3, this chapter develops a calculus for building new vector bundles from old ones. The guiding question is: which familiar linear-algebra operations on vector spaces can be performed fibrewise, and how are the resulting bundles described by transition data? We also introduce the determinant line as the object that records orientation information, and we finish with subbundles, quotient bundles, and exact sequences. ## Fibrewise Linear Algebra from Transition Functions Suppose $E \to M$ and $F \to M$ are smooth vector bundles of ranks $r$ and $s$ over the same smooth manifold $M$. The problem is to construct new bundles whose fibre over $p \in M$ is obtained by applying a linear-algebra operation to $E_p$ and $F_p$, while keeping track of smooth variation in $p$. The transition-function method gives the clean answer: apply the same functor to the transition functions. [definition: Direct Sum Bundle] Let $\pi_E:E\to M$ and $\pi_F:F\to M$ be smooth vector bundles over $M$. The direct sum bundle $E\oplus F\to M$ is the smooth vector bundle whose fibre over $p\in M$ is \begin{align*} (E\oplus F)_p=E_p\oplus F_p. \end{align*} [/definition] This construction packages pairs of vectors [lying over](/theorems/2876) the same base point. In local trivializations of $E$ and $F$ over a common open set $U\subset M$, a point of $E\oplus F$ over $U$ is represented by $(p,v,w)$ with $v\in \mathbb R^r$ and $w\in \mathbb R^s$. [example: Tangent Plus Cotangent Bundle] For a smooth $n$-manifold $M$, the fibre of $TM\oplus T^*M\to M$ over $p\in M$ is \begin{align*} (TM\oplus T^*M)_p=T_pM\oplus T_p^*M. \end{align*} Thus a section assigns to each $p$ a pair $(X(p),\alpha(p))$ with $X(p)\in T_pM$ and $\alpha(p)\in T_p^*M$, so it is the same data as a vector field $X\in\mathfrak X(M)$ together with a $1$-form $\alpha\in\Omega^1(M)$. In coordinates $(x_1,\dots,x_n)$ on $U$, write \begin{align*} X|_U=\sum_{i=1}^n X_i\partial_{x_i}. \end{align*} \begin{align*} \alpha|_U=\sum_{i=1}^n \alpha_i dx_i. \end{align*} The corresponding section of the direct sum is \begin{align*} (X,\alpha)|_U=\left(\sum_{i=1}^n X_i\partial_{x_i},\sum_{i=1}^n \alpha_i dx_i\right). \end{align*} On an overlapping coordinate chart $(y_1,\dots,y_n)$, the tangent basis transforms by the chain rule: \begin{align*} \partial_{x_i}=\sum_{a=1}^n \frac{\partial y_a}{\partial x_i}\partial_{y_a}. \end{align*} Substituting this into the tangent component gives \begin{align*} X|_U=\sum_{i=1}^n X_i\sum_{a=1}^n \frac{\partial y_a}{\partial x_i}\partial_{y_a}. \end{align*} Reordering the finite sum gives \begin{align*} X|_U=\sum_{a=1}^n\left(\sum_{i=1}^n X_i\frac{\partial y_a}{\partial x_i}\right)\partial_{y_a}. \end{align*} So the $y$-coordinate components of $X$ are \begin{align*} Y_a=\sum_{i=1}^n X_i\frac{\partial y_a}{\partial x_i}. \end{align*} The cotangent basis transforms by applying the differential to $x_i=x_i(y_1,\dots,y_n)$: \begin{align*} dx_i=\sum_{a=1}^n \frac{\partial x_i}{\partial y_a}dy_a. \end{align*} Substituting this into the cotangent component gives \begin{align*} \alpha|_U=\sum_{i=1}^n \alpha_i\sum_{a=1}^n \frac{\partial x_i}{\partial y_a}dy_a. \end{align*} Reordering the finite sum gives \begin{align*} \alpha|_U=\sum_{a=1}^n\left(\sum_{i=1}^n \alpha_i\frac{\partial x_i}{\partial y_a}\right)dy_a. \end{align*} So the $y$-coordinate components of $\alpha$ are \begin{align*} \beta_a=\sum_{i=1}^n \alpha_i\frac{\partial x_i}{\partial y_a}. \end{align*} Therefore the tangent component transforms by the Jacobian matrix $\left(\frac{\partial y_a}{\partial x_i}\right)$, while the cotangent component transforms by the inverse Jacobian matrix $\left(\frac{\partial x_i}{\partial y_a}\right)$. Both matrices are evaluated at the same base point $p$, which is why the direct sum is $T_pM\oplus T_p^*M$ rather than a mixture of fibres over different points. [/example] The direct sum handles ordered pairs of fibre vectors, but it does not turn a vector and a covector into a bilinear quantity. For instance, trying to define a local tensor by multiplying coordinate components without changing the transition law gives an object that depends on the chosen frame. This failure motivates the [tensor product](/page/Tensor%20Product) bundle: tensoring must be done fibre by fibre, and the transition functions must act on both tensor factors. [definition: Tensor Product Bundle] Let $E\to M$ and $F\to M$ be smooth vector bundles. The tensor product bundle $E\otimes F\to M$ is the smooth vector bundle with total space \begin{align*} E\otimes F=\bigsqcup_{p\in M} E_p\otimes F_p \end{align*} and projection $E\otimes F\to M$ sending an element of $E_p\otimes F_p$ to $p$. [/definition] The tensor product bundle is the natural home for tensorial objects built from two varying vector spaces. It still does not describe a fibrewise linear test on a vector, because a functional on $E_p$ is contravariant: it must transform oppositely to vectors in order for evaluation to be well-defined. This is the obstruction that forces the dual bundle rather than another copy of $E$. [definition: Dual Bundle] Let $E\to M$ be a smooth vector bundle. The dual bundle $E^*\to M$ is the smooth vector bundle with total space \begin{align*} E^*=\bigsqcup_{p\in M}(E_p)^* \end{align*} and projection $E^*\to M$ sending an element of $(E_p)^*$ to $p$. [/definition] The dual construction reverses the direction of transition matrices. If vectors transform by a matrix $g_{ij}$, covectors transform by the inverse transpose, because evaluation must be independent of the chosen trivialization. [example: Dual Tautological Bundle] Let $U_i\subset \mathbb RP^n$ be the standard chart of lines $\ell=[x_0:\cdots:x_n]$ with $x_i\ne 0$. Over $U_i$, choose the local generator \begin{align*} e_i(\ell)=\frac{1}{x_i}(x_0,\dots,x_n)\in \ell, \end{align*} so the $i$-th coordinate of $e_i(\ell)$ is $1$. On $U_i\cap U_j$, the generator normalized in the $j$-th coordinate is \begin{align*} e_j(\ell)=\frac{1}{x_j}(x_0,\dots,x_n). \end{align*} Since \begin{align*} \frac{1}{x_j}(x_0,\dots,x_n)=\frac{x_i}{x_j}\frac{1}{x_i}(x_0,\dots,x_n), \end{align*} we have \begin{align*} e_j(\ell)=\frac{x_i}{x_j}e_i(\ell). \end{align*} If $v\in \ell$ has coordinate expressions $v=a_i e_i(\ell)=a_j e_j(\ell)$, then substituting the overlap relation gives \begin{align*} a_i e_i(\ell)=a_j\frac{x_i}{x_j}e_i(\ell). \end{align*} Because $e_i(\ell)\ne 0$, the coefficients satisfy \begin{align*} a_i=a_j\frac{x_i}{x_j}. \end{align*} Equivalently, \begin{align*} a_j=\frac{x_j}{x_i}a_i. \end{align*} Thus the vector coordinate changes by the scalar $\frac{x_j}{x_i}$. Let $\varepsilon_i(\ell)\in \ell^*$ be the dual generator defined by $\varepsilon_i(\ell)(e_i(\ell))=1$, and define $\varepsilon_j(\ell)$ similarly. Since $\ell^*$ is one-dimensional, write \begin{align*} \varepsilon_j(\ell)=c\,\varepsilon_i(\ell). \end{align*} Using $\varepsilon_j(\ell)(e_j(\ell))=1$ and $e_j(\ell)=\frac{x_i}{x_j}e_i(\ell)$ gives \begin{align*} 1=c\,\varepsilon_i(\ell)\left(\frac{x_i}{x_j}e_i(\ell)\right). \end{align*} By linearity of $\varepsilon_i(\ell)$, \begin{align*} c\,\varepsilon_i(\ell)\left(\frac{x_i}{x_j}e_i(\ell)\right)=c\frac{x_i}{x_j}\varepsilon_i(\ell)(e_i(\ell)). \end{align*} Since $\varepsilon_i(\ell)(e_i(\ell))=1$, this becomes \begin{align*} 1=c\frac{x_i}{x_j}. \end{align*} Therefore \begin{align*} c=\frac{x_j}{x_i}. \end{align*} So the dual generators transform by \begin{align*} \varepsilon_j(\ell)=\frac{x_j}{x_i}\varepsilon_i(\ell). \end{align*} If $\alpha\in \ell^*$ has coordinate expressions $\alpha=b_i\varepsilon_i(\ell)=b_j\varepsilon_j(\ell)$, then \begin{align*} b_i\varepsilon_i(\ell)=b_j\frac{x_j}{x_i}\varepsilon_i(\ell). \end{align*} Since $\varepsilon_i(\ell)\ne 0$, the coefficients satisfy \begin{align*} b_i=b_j\frac{x_j}{x_i}. \end{align*} Equivalently, \begin{align*} b_j=\frac{x_i}{x_j}b_i. \end{align*} The dual coordinate therefore changes by the reciprocal scalar to the vector coordinate. This reciprocal factor is forced by evaluation. In the $j$-chart, \begin{align*} \alpha(v)=b_j a_j. \end{align*} Substituting the coordinate-change formulae gives \begin{align*} b_j a_j=\left(\frac{x_i}{x_j}b_i\right)\left(\frac{x_j}{x_i}a_i\right). \end{align*} Multiplying the scalar factors, \begin{align*} \left(\frac{x_i}{x_j}b_i\right)\left(\frac{x_j}{x_i}a_i\right)=b_i a_i. \end{align*} Thus $\alpha(v)$ has the same value in both charts. If the dual coordinate changed by the same factor as the vector coordinate, the overlap computation would instead give \begin{align*} \alpha(v)=\left(\frac{x_j}{x_i}b_i\right)\left(\frac{x_j}{x_i}a_i\right). \end{align*} Multiplying those factors gives \begin{align*} \alpha(v)=\left(\frac{x_j}{x_i}\right)^2 b_i a_i, \end{align*} which depends on the chosen chart unless $\left(\frac{x_j}{x_i}\right)^2=1$. Thus the dual tautological line bundle is the rank-one model for the inverse-transpose transition rule. [/example] The tautological example isolates the lesson of the dual construction: the transformation law is determined by the invariant operation we want to preserve, here evaluation of a covector on a vector. The next invariant operation is alternating volume inside a single fibre. A naive choice of ordered local frames cannot be compared across overlaps unless the change-of-frame determinant is tracked, and exterior powers are exactly the functor that records this alternating information. [definition: Exterior Power Bundle] Let $E\to M$ be a smooth vector bundle of rank $r$. For $0\le k\le r$, the $k$-th exterior power bundle $\Lambda^k E\to M$ is the smooth vector bundle with total space \begin{align*} \Lambda^k E=\bigsqcup_{p\in M}\Lambda^k(E_p) \end{align*} and projection $\Lambda^k E\to M$ sending an element of $\Lambda^k(E_p)$ to $p$. [/definition] For $k=1$ this is $E$, and for $k=0$ it is the product line bundle $M\times \mathbb R$. Tensor and exterior powers describe multilinear objects, but many later geometric structures are not tensors themselves: they are smoothly varying linear maps between fibres. Treating such maps as unrelated maps $E_p\to F_p$ at each point would lose smoothness, which motivates the Hom bundle. [definition: Hom Bundle] Let $E\to M$ and $F\to M$ be smooth vector bundles. The Hom bundle $\operatorname{Hom}(E,F)\to M$ is the smooth vector bundle with total space \begin{align*} \operatorname{Hom}(E,F)=\bigsqcup_{p\in M}\operatorname{Hom}(E_p,F_p) \end{align*} and projection $\operatorname{Hom}(E,F)\to M$ sending a linear map $E_p\to F_p$ to $p$. [/definition] A section of $\operatorname{Hom}(E,F)$ is a smooth family of linear maps $E_p\to F_p$, in the sense of the section viewpoint from Chapter 3. Having defined the operation bundles fibre by fibre, we now need the practical rule that computes their transition functions from those of $E$ and $F$. [quotetheorem:6090] [citeproof:6090] This theorem is the main computational rule of the chapter. The hypothesis that $E$ and $F$ are described over the same cover is a bookkeeping convenience rather than a real restriction, because two covers can be replaced by their common refinement. It is still a necessary hypothesis for the displayed formulae as written. For instance, let $E$ be described on intervals $U_1,U_2\subset \mathbb R$ and $F$ on different intervals $V_1,V_2$; a symbol such as $g^E_{12}\oplus g^F_{12}$ has no meaning at a point lying in $U_1\cap U_2$ but not in $V_1\cap V_2$, and relabelling the covers does not repair the mismatch. Passing to the refinement by intersections $U_i\cap V_a$ creates overlaps where both transition functions are defined and the formula can be evaluated. The theorem does not say that these operation bundles are canonically trivial, nor that a choice of local frames has disappeared; it says that the dependence on frames is functorial and therefore controlled. This is the rule used below to turn the determinant into a line bundle and to interpret bundle maps as sections of Hom bundles. [example: Endomorphism Bundle] For a rank $r$ vector bundle $E\to M$, the endomorphism bundle is \begin{align*} \operatorname{End}(E)=\operatorname{Hom}(E,E). \end{align*} If $E$ has transition functions $g_{ij}:U_i\cap U_j\to GL(r)$, then the Hom transition rule from *Transition Functions for Standard Operations*, applied with $F=E$, gives \begin{align*} A\longmapsto g_{ij}A(g_{ij})^{-1}. \end{align*} Thus an endomorphism represented over $U_j$ by a matrix $A_j$ is represented over $U_i$ by \begin{align*} A_i=g_{ij}A_jg_{ij}^{-1}. \end{align*} The identity endomorphism is frame-independent. If $A_j=I_r$, then substituting into the transition formula gives \begin{align*} A_i=g_{ij}I_rg_{ij}^{-1}. \end{align*} Since $g_{ij}I_r=g_{ij}$, this becomes \begin{align*} A_i=g_{ij}g_{ij}^{-1}. \end{align*} Since $g_{ij}g_{ij}^{-1}=I_r$, we get \begin{align*} A_i=I_r. \end{align*} More generally, if $A_j=\lambda I_r$ for a scalar function $\lambda$ on the overlap, then \begin{align*} A_i=g_{ij}(\lambda I_r)g_{ij}^{-1}. \end{align*} Because scalar multiplication of matrices commutes with matrix multiplication, \begin{align*} A_i=\lambda g_{ij}I_rg_{ij}^{-1}. \end{align*} Using $g_{ij}I_r=g_{ij}$ gives \begin{align*} A_i=\lambda g_{ij}g_{ij}^{-1}. \end{align*} Therefore \begin{align*} A_i=\lambda I_r. \end{align*} So scalar endomorphisms agree across local frames and define global sections of $\operatorname{End}(E)$. A locally chosen projection matrix does not have this invariance unless it is compatible with the transition functions. If $P_j$ is a projection, so $P_j^2=P_j$, then its expression in the $i$-frame is \begin{align*} P_i=g_{ij}P_jg_{ij}^{-1}. \end{align*} For the same constant matrix $P$ to describe the projection in both frames, one would need \begin{align*} P=g_{ij}Pg_{ij}^{-1}. \end{align*} Multiplying this identity on the right by $g_{ij}$ gives \begin{align*} Pg_{ij}=g_{ij}P. \end{align*} Thus $P$ must commute with the transition matrix on the overlap. Since an arbitrary projection need not commute with $g_{ij}$, it need not glue to a global projection of $E$. The example separates endomorphisms whose formula is independent of frame, such as scalar multiples of the identity, from fibrewise linear choices that require extra structure to be globally defined. [/example] ## Determinant Lines and Orientations The next question is how a vector bundle remembers whether its fibres can be oriented in a way that varies smoothly over the base. For a single vector space, an orientation can be encoded by a choice of component in the set of ordered bases. For a vector bundle, the obstruction is measured by the signs of transition-function determinants. [definition: Determinant Line Bundle] Let $E\to M$ be a smooth vector bundle of rank $r$. The determinant line bundle of $E$ is \begin{align*} \det E=\Lambda^r E. \end{align*} [/definition] The determinant line converts rank $r$ transition matrices into scalar transition functions by taking determinants. Without passing to the top exterior power, the sign of a change of frame is mixed into a full matrix and is not visible as line-bundle transition data. Thus a high-rank orientation problem becomes a line-bundle problem, and the next result records the exact transition data. [quotetheorem:6091] [citeproof:6091] The determinant hypothesis is essential: lower exterior powers do not detect the sign of a full change of frame. For example, on a rank-two bundle the first exterior power is just the original bundle, while the top exterior power converts every transition matrix to the scalar $\det(g_{ij})$. The theorem does not by itself decide orientability; it only identifies the line bundle whose transition signs must be controlled. The determinant line is more than a convenient notation: it is the precise bundle whose nonvanishing sections encode orientations. To make that statement meaningful, we first phrase orientation as a compatibility condition on local frames rather than as an unrelated choice on each fibre. [definition: Orientation of a Vector Bundle] Let $E\to M$ be a smooth vector bundle of rank $r$. An orientation of $E$ is a choice of orientation of each fibre $E_p$ such that there is a bundle atlas whose transition functions all take values in $GL^+(r)=\{A\in GL(r):\det A>0\}$. [/definition] The definition says that local frames can be chosen so that all overlap changes preserve orientation. This is a global compatibility condition, and the determinant line is designed to detect exactly that compatibility; this motivates the determinant-line criterion for orientability. [quotetheorem:6092] [citeproof:6092] The nowhere-vanishing condition is essential: a section of $\det E$ with zeros cannot choose an orientation at those fibres. In the product line bundle $M\times \mathbb R\to M$, the section $p\mapsto (p,f(p))$ with $f(p_0)=0$ gives no sign at $p_0$, so it cannot distinguish the positive ray from the negative ray in that fibre. Thus even for an orientable line bundle, a section with zeros is weaker data than an orientation. The theorem does not provide a preferred orientation, since multiplying a nonvanishing determinant section by a negative function reverses the choice on each connected component. It does, however, turn orientability into the existence problem for a line-bundle section, which is the form used later in topology through orientation local systems, the first Stiefel-Whitney class, and densities. This criterion is the bundle-theoretic version of the familiar statement that a manifold is orientable exactly when its tangent bundle is orientable. [example: Determinant Bundle of the Tangent Bundle] For an $n$-dimensional smooth manifold $M$, the determinant line of the tangent bundle is \begin{align*} \det(TM)=\Lambda^n TM. \end{align*} In a coordinate chart $(x_1,\dots,x_n)$ on $U\subset M$, the vector fields \begin{align*} \partial_{x_1},\dots,\partial_{x_n} \end{align*} form a local frame for $TM|_U$, so \begin{align*} \partial_{x_1}\wedge\cdots\wedge \partial_{x_n} \end{align*} is a local generator of $\Lambda^n TM|_U$. Therefore a section of $\Lambda^n TM$ is locally of the form \begin{align*} \xi|_U=f\,\partial_{x_1}\wedge\cdots\wedge \partial_{x_n} \end{align*} for a smooth function $f$ on $U$, and $\xi$ is nowhere vanishing exactly when $f(p)\ne 0$ in every such local expression. On an overlapping coordinate chart $(y_1,\dots,y_n)$, the chain rule gives \begin{align*} \partial_{x_i}=\sum_{a=1}^n\frac{\partial y_a}{\partial x_i}\partial_{y_a}. \end{align*} Substituting this into the top wedge gives \begin{align*} \partial_{x_1}\wedge\cdots\wedge\partial_{x_n} = \left(\sum_{a_1=1}^n\frac{\partial y_{a_1}}{\partial x_1}\partial_{y_{a_1}}\right) \wedge\cdots\wedge \left(\sum_{a_n=1}^n\frac{\partial y_{a_n}}{\partial x_n}\partial_{y_{a_n}}\right). \end{align*} By multilinearity of the wedge product, \begin{align*} \partial_{x_1}\wedge\cdots\wedge\partial_{x_n} = \sum_{a_1,\dots,a_n=1}^n \left(\prod_{i=1}^n\frac{\partial y_{a_i}}{\partial x_i}\right) \partial_{y_{a_1}}\wedge\cdots\wedge\partial_{y_{a_n}}. \end{align*} Every term with two equal indices $a_i=a_j$ is zero because the wedge product is alternating. The remaining terms are indexed by permutations $\sigma\in S_n$, and \begin{align*} \partial_{y_{\sigma(1)}}\wedge\cdots\wedge\partial_{y_{\sigma(n)}} = \operatorname{sgn}(\sigma)\, \partial_{y_1}\wedge\cdots\wedge\partial_{y_n}. \end{align*} Hence \begin{align*} \partial_{x_1}\wedge\cdots\wedge\partial_{x_n} = \left(\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma) \prod_{i=1}^n\frac{\partial y_{\sigma(i)}}{\partial x_i}\right) \partial_{y_1}\wedge\cdots\wedge\partial_{y_n}. \end{align*} The coefficient is the determinant of the Jacobian matrix $\left(\frac{\partial y_a}{\partial x_i}\right)$, so \begin{align*} \partial_{x_1}\wedge\cdots\wedge\partial_{x_n} = \det\left(\frac{\partial y_a}{\partial x_i}\right) \partial_{y_1}\wedge\cdots\wedge\partial_{y_n}. \end{align*} The dual line bundle satisfies the canonical fibrewise identification \begin{align*} (\Lambda^n T_pM)^*\cong \Lambda^n T_p^*M. \end{align*} In the same coordinates, the dual generator is \begin{align*} dx_1\wedge\cdots\wedge dx_n. \end{align*} Since \begin{align*} dx_i=\sum_{a=1}^n\frac{\partial x_i}{\partial y_a}dy_a, \end{align*} the same multilinear expansion gives \begin{align*} dx_1\wedge\cdots\wedge dx_n = \det\left(\frac{\partial x_i}{\partial y_a}\right) dy_1\wedge\cdots\wedge dy_n. \end{align*} Because the matrices $\left(\frac{\partial y_a}{\partial x_i}\right)$ and $\left(\frac{\partial x_i}{\partial y_a}\right)$ are inverse Jacobians, \begin{align*} \det\left(\frac{\partial x_i}{\partial y_a}\right) = \det\left(\frac{\partial y_a}{\partial x_i}\right)^{-1}. \end{align*} Thus a nowhere-vanishing section of $\Lambda^n TM$ records a consistent choice of positive ray in each determinant fibre, and a nowhere-vanishing section of $\Lambda^n T^*M$ records the dual positive ray. Such a section of $\Lambda^n T^*M$ is a volume form. Consequently, orientability of $M$ can be detected either by a globally nonzero top-degree form or by choosing coordinate charts whose overlap Jacobian determinants are positive. [/example] ## Subbundles, Quotients, and Constant Rank We now ask when a family of vector subspaces $F_p\subset E_p$ forms a vector bundle in its own right. The pointwise condition alone is not enough: the dimension must be locally constant and the subspaces must vary smoothly. This is the bundle analogue of the constant rank condition for smooth maps. [definition: Vector Subbundle] Let $E\to M$ be a smooth vector bundle. A vector subbundle of $E$ is a smooth vector bundle $F\to M$ together with a smooth bundle map $\iota:F\to E$ over $\operatorname{id}_M$ such that each fibre map $\iota_p:F_p\to E_p$ is injective. [/definition] After identifying $F_p$ with its image in $E_p$, we usually write $F_p\subset E_p$. The definition requires the inclusion to be smooth as a bundle map, so the subspaces are not allowed to jump as the base point moves. [example: Normal Bundle of an Embedded Submanifold] Let $N\subset M$ be an embedded $k$-dimensional submanifold of an $m$-dimensional manifold, and choose a Riemannian metric $g$ on $M$. The tangent bundle $TN\to N$ sits inside $TM|_N\to N$ by the inclusion $T_pN\subset T_pM$, and the normal bundle is the fibrewise orthogonal complement \begin{align*} \nu(N)_p=\{v\in T_pM:g_p(v,w)=0 \text{ for all } w\in T_pN\}. \end{align*} Choose adapted local coordinates $(x_1,\dots,x_m)$ near $p\in N$ such that \begin{align*} N\cap U=\{x_{k+1}=\cdots=x_m=0\}. \end{align*} Along $N\cap U$, set $e_i=\partial_{x_i}$. Then $e_1,\dots,e_k$ span $TN$, while $e_1,\dots,e_m$ span $TM|_{N\cap U}$. Define \begin{align*} A_{ij}=g(e_i,e_j),\qquad 1\le i,j\le k, \end{align*} and, for $k+1\le \alpha\le m$, \begin{align*} b_{j\alpha}=g(e_\alpha,e_j),\qquad 1\le j\le k. \end{align*} The matrix $A=(A_{ij})$ is positive definite because $g$ is positive definite on $T_qN$, so $A$ is invertible at each $q\in N\cap U$. For each $\alpha>k$, define smooth coefficients $c_{i\alpha}$ by \begin{align*} \sum_{i=1}^k A_{ji}c_{i\alpha}=b_{j\alpha},\qquad 1\le j\le k, \end{align*} and set \begin{align*} n_\alpha=e_\alpha-\sum_{i=1}^k c_{i\alpha}e_i. \end{align*} The coefficients are smooth because the entries of $A^{-1}$ and $b$ are smooth. For $1\le j\le k$, bilinearity of $g$ gives \begin{align*} g(n_\alpha,e_j)=g(e_\alpha,e_j)-\sum_{i=1}^k c_{i\alpha}g(e_i,e_j). \end{align*} Using the definitions of $A_{ji}$ and $b_{j\alpha}$, this is \begin{align*} g(n_\alpha,e_j)=b_{j\alpha}-\sum_{i=1}^k A_{ji}c_{i\alpha}. \end{align*} By the defining equation for the coefficients $c_{i\alpha}$, \begin{align*} g(n_\alpha,e_j)=0. \end{align*} Thus each $n_\alpha$ lies in $\nu(N)$. Conversely, let \begin{align*} v=\sum_{i=1}^k t_i e_i+\sum_{\alpha=k+1}^m s_\alpha e_\alpha \end{align*} satisfy $v\in \nu(N)_q$. Orthogonality to each $e_j$ gives \begin{align*} 0=g(v,e_j). \end{align*} Expanding by bilinearity, \begin{align*} 0=\sum_{i=1}^k A_{ji}t_i+\sum_{\alpha=k+1}^m s_\alpha b_{j\alpha}. \end{align*} Substitute $b_{j\alpha}=\sum_{i=1}^k A_{ji}c_{i\alpha}$: \begin{align*} 0=\sum_{i=1}^k A_{ji}t_i+\sum_{\alpha=k+1}^m s_\alpha\sum_{i=1}^k A_{ji}c_{i\alpha}. \end{align*} Reordering the finite sums gives \begin{align*} 0=\sum_{i=1}^k A_{ji}\left(t_i+\sum_{\alpha=k+1}^m s_\alpha c_{i\alpha}\right). \end{align*} Since $A$ is invertible, the vector in parentheses is zero, so \begin{align*} t_i=-\sum_{\alpha=k+1}^m s_\alpha c_{i\alpha}. \end{align*} Substituting this expression for $t_i$ into $v$ gives \begin{align*} v=\sum_{i=1}^k\left(-\sum_{\alpha=k+1}^m s_\alpha c_{i\alpha}\right)e_i+\sum_{\alpha=k+1}^m s_\alpha e_\alpha. \end{align*} Reordering the finite sums, \begin{align*} v=\sum_{\alpha=k+1}^m s_\alpha\left(e_\alpha-\sum_{i=1}^k c_{i\alpha}e_i\right). \end{align*} Therefore \begin{align*} v=\sum_{\alpha=k+1}^m s_\alpha n_\alpha. \end{align*} So $n_{k+1},\dots,n_m$ span $\nu(N)$ locally. They are linearly independent because, in the frame $e_1,\dots,e_m$, the coefficient of $e_\alpha$ in $n_\alpha$ is $1$, while the coefficient of $e_\alpha$ in $n_\beta$ is $0$ for $\beta\ne\alpha$. Hence $\nu(N)$ is locally framed by smooth sections, and its fibre at $p$ is exactly the metric normal vector space to $N$ in $M$. [/example] The normal bundle illustrates smooth local spanning by independent sections. This motivates the practical criterion below: to prove that a fibrewise collection of subspaces is a subbundle, it is enough to find local smooth frames of constant size. [quotetheorem:6093] [citeproof:6093] This criterion explains why constant rank appears in the statement. If the dimension of $S_p$ changes, no local product model $U\times \mathbb R^k$ can describe the total space near the jump; for a smooth function $f:M\to \mathbb R$, the subsets $\ker(df_p)\subset T_pM$ fail to form a fixed-rank subbundle at critical points where the rank of $df_p$ changes. The theorem does not say that every fibrewise vector subspace assignment is smooth, even when the rank is constant; the local spanning sections are the smoothness data. This criterion is the practical bridge to kernels, images, and quotients of bundle maps. [example: Kernel of a Constant Rank Bundle Map] Let $\Phi:E\to F$ be a smooth bundle map over $\operatorname{id}_M$, and suppose every fibre map $\Phi_p:E_p\to F_p$ has the same rank $k$. We show that the fibrewise kernel \begin{align*} (\ker\Phi)_p=\ker(\Phi_p) \end{align*} is locally spanned by smooth independent sections, so it is a smooth rank $r-k$ subbundle of $E$ by the local constant rank criterion. Fix $p\in M$, choose local frames $e_1,\dots,e_r$ for $E$ and $f_1,\dots,f_s$ for $F$ near $p$, and write \begin{align*} \Phi(e_i)=\sum_{a=1}^s B_{ai}f_a. \end{align*} The functions $B_{ai}$ are smooth, so $\Phi$ is represented by a smooth $s\times r$ matrix $B(q)$. Since $B(p)$ has rank $k$, after reordering the frames we may assume the first $k$ rows and first $k$ columns form an invertible block $A(p)$. Shrinking the neighbourhood if necessary, $A(q)$ remains invertible. Write the remaining blocks as follows: $C$ is the first $k$ rows and last $r-k$ columns, $D$ is the last $s-k$ rows and first $k$ columns, and $H$ is the last $s-k$ rows and last $r-k$ columns. For a vector in $E_q$ with coordinates $(u,w)\in \mathbb R^k\oplus \mathbb R^{r-k}$, the equation $\Phi_q(u,w)=0$ is equivalent to the pair of equations \begin{align*} Au+Cw=0. \end{align*} \begin{align*} Du+Hw=0. \end{align*} The first equation gives \begin{align*} u=-A^{-1}Cw. \end{align*} Substituting this into the second equation gives \begin{align*} D(-A^{-1}Cw)+Hw=(H-DA^{-1}C)w. \end{align*} It remains to identify the last coefficient. Perform the smooth row operation that replaces the lower block row by the lower block row minus $DA^{-1}$ times the upper block row. This row operation is invertible, so it preserves rank, and it changes the four blocks to $A$, $C$, $0$, and $H-DA^{-1}C$. The upper block row has rank $k$ because $A$ is invertible. Since the whole matrix has rank exactly $k$, the lower-right block must vanish: \begin{align*} H-DA^{-1}C=0. \end{align*} Therefore the second kernel equation is automatically satisfied once the first one is satisfied, and every element of $\ker(\Phi_q)$ has the form \begin{align*} (-A^{-1}Cw,w). \end{align*} For $j=1,\dots,r-k$, let $\epsilon_j$ be the $j$-th standard basis vector of $\mathbb R^{r-k}$ and define \begin{align*} s_j(q)=\sum_{i=1}^k\left(-(A^{-1}C)_{ij}(q)\right)e_i(q)+e_{k+j}(q). \end{align*} The coefficients are smooth because $A^{-1}$ and $C$ are smooth. The displayed kernel formula shows that $s_1(q),\dots,s_{r-k}(q)$ span $\ker(\Phi_q)$, and they are linearly independent because their last $r-k$ coordinates are $\epsilon_1,\dots,\epsilon_{r-k}$. Thus the constant-rank hypothesis turns the pointwise kernels into a smoothly varying family of subspaces, namely a smooth rank $r-k$ subbundle of $E$. [/example] The kernel example shows that subbundles arise naturally from maps, not only from geometric inclusions chosen in advance. Having identified a smooth family of subspaces inside $E$, the next operation is to keep the directions transverse to that family while identifying vectors that differ by an element of the subbundle. Doing this for an arbitrary fibrewise collection of subspaces can produce fibres of different dimensions and therefore no vector bundle, so the smooth subbundle hypothesis is the condition that makes quotienting locally uniform. [definition: Quotient Bundle] Let $F\subset E$ be a smooth vector subbundle. The quotient bundle $E/F\to M$ is the smooth vector bundle with total space \begin{align*} E/F=\bigsqcup_{p\in M}E_p/F_p \end{align*} and projection $E/F\to M$ sending a quotient class in $E_p/F_p$ to $p$. [/definition] The quotient bundle is characterised by the projection $E\to E/F$, which is fibrewise the ordinary quotient map. To record subbundles and quotients in one object, this motivates the language of short exact sequences. [definition: Short Exact Sequence of Vector Bundles] A short exact sequence of smooth vector bundles over $M$ is a diagram \begin{align*} 0\to F\xrightarrow{\iota}E\xrightarrow{\pi}Q\to 0 \end{align*} of smooth bundle maps over $\operatorname{id}_M$ such that for every $p\in M$ the sequence \begin{align*} 0\to F_p\xrightarrow{\iota_p}E_p\xrightarrow{\pi_p}Q_p\to 0 \end{align*} is exact. [/definition] Exactness means that $F$ is identified with the kernel of $\pi$, and $Q$ is fibrewise the corresponding quotient. The quotient fibres are easy to describe abstractly, but calculations often require choosing actual vectors in $E$ to represent them. A metric supplies a controlled way to choose those representatives: take the orthogonal complement of $F$ inside $E$. [quotetheorem:6094] [citeproof:6094] The metric hypothesis is doing real work: without a chosen complement, a quotient $E/F$ is not naturally a subbundle of $E$, even though it is fibrewise isomorphic to any complement. The theorem does not say the splitting is canonical; different metrics generally produce different orthogonal complements. Its point is that, once a smooth bundle metric is available, there is no further smooth obstruction to splitting a short exact sequence, and this is why quotient bundles can often be replaced by explicit complementary subbundles in geometry. [example: Tangent Bundle Along an Embedded Submanifold] For an embedded submanifold $N\subset M$, the inclusion $T_pN\subset T_pM$ gives a short exact sequence over $N$, \begin{align*} 0\to TN\to TM|_N\to TM|_N/TN\to 0, \end{align*} where the last map sends $v\in T_pM$ to its quotient class $[v]\in T_pM/T_pN$. Exactness at $TN$ says the inclusion $T_pN\to T_pM$ is injective, and exactness at $TM|_N$ says precisely that \begin{align*} \ker\bigl(T_pM\to T_pM/T_pN\bigr)=T_pN. \end{align*} Choose a Riemannian metric $g$ on $M$ and define \begin{align*} \nu(N)_p=\{w\in T_pM:g_p(w,u)=0\text{ for every }u\in T_pN\}. \end{align*} For each $v\in T_pM$, finite-dimensional [inner product](/page/Inner%20Product) linear algebra gives a unique decomposition \begin{align*} v=t+n,\qquad t\in T_pN,\quad n\in \nu(N)_p. \end{align*} If $v'=v+u$ represents the same quotient class, with $u\in T_pN$, then \begin{align*} v'=v+u=(t+n)+u=(t+u)+n. \end{align*} Since $t+u\in T_pN$ and $n\in \nu(N)_p$, the orthogonal component of $v'$ is the same $n$. Thus the map \begin{align*} T_pM/T_pN\to \nu(N)_p,\qquad [v]\mapsto n \end{align*} is well-defined. This map is linear: if $v_1=t_1+n_1$ and $v_2=t_2+n_2$, then \begin{align*} av_1+bv_2=a(t_1+n_1)+b(t_2+n_2) =(at_1+bt_2)+(an_1+bn_2), \end{align*} with $at_1+bt_2\in T_pN$ and $an_1+bn_2\in \nu(N)_p$. Hence \begin{align*} a[v_1]+b[v_2]\longmapsto an_1+bn_2. \end{align*} It is injective because $[v]$ maps to $0$ exactly when $v=t+0\in T_pN$, so $[v]=0$. It is surjective because every $n\in \nu(N)_p$ is the image of $[n]\in T_pM/T_pN$. Therefore \begin{align*} TM|_N/TN\cong \nu(N) \end{align*} after the metric has been chosen, and the corresponding fibrewise decomposition is \begin{align*} TM|_N\cong TN\oplus \nu(N). \end{align*} The quotient bundle $TM|_N/TN$ exists before choosing $g$, while the displayed identification with actual normal vectors depends on the metric through the choice of orthogonal complement. [/example] ## How Operations Interact with Sections The constructions above are not only operations on total spaces; they also organise the algebra of sections. The final question of the chapter is how local frames and sections behave under these operations, since this is what later chapters use when studying pullbacks, connections, and characteristic classes. [quotetheorem:6095] [citeproof:6095] The smoothness hypotheses on both sides are essential. A merely fibrewise assignment $p\mapsto A_p\in \operatorname{Hom}(E_p,F_p)$ can fail to be a section if its matrix entries are not smooth in local frames, and a map of total spaces that is not fibrewise linear does not define a point of the Hom fibre. The theorem does not identify arbitrary smooth maps $E\to F$ with Hom sections; it identifies precisely the bundle maps covering $\operatorname{id}_M$ and linear on each fibre. This is why metrics, projections, almost complex structures, and bundle morphisms can all be studied as sections of associated operation bundles. [example: Bundle Metrics as Sections] Let $E\to M$ be a rank $r$ real vector bundle, and let $h$ be a bundle metric, so each fibre has a positive-definite symmetric [bilinear form](/page/Bilinear%20Form) $h_p:E_p\times E_p\to \mathbb R$ depending smoothly on $p$. In a local frame $e_1,\dots,e_r$ over $U$, define \begin{align*} h_{ab}(p)=h_p(e_a(p),e_b(p)). \end{align*} If \begin{align*} v=\sum_{a=1}^r v^a e_a(p) \end{align*} and \begin{align*} w=\sum_{b=1}^r w^b e_b(p), \end{align*} then bilinearity gives \begin{align*} h_p(v,w)=\sum_{a=1}^r\sum_{b=1}^r v^a w^b h_p(e_a(p),e_b(p)). \end{align*} Using the definition of $h_{ab}(p)$, this becomes \begin{align*} h_p(v,w)=\sum_{a=1}^r\sum_{b=1}^r h_{ab}(p)v^a w^b. \end{align*} Symmetry gives \begin{align*} h_{ab}(p)=h_p(e_a(p),e_b(p))=h_p(e_b(p),e_a(p))=h_{ba}(p), \end{align*} so $H(p)=(h_{ab}(p))$ is symmetric. Positive definiteness says that for every nonzero coordinate vector $v=(v^1,\dots,v^r)$, \begin{align*} v^\top H(p)v=\sum_{a=1}^r\sum_{b=1}^r h_{ab}(p)v^a v^b. \end{align*} The right-hand side is \begin{align*} h_p\left(\sum_{a=1}^r v^a e_a(p),\sum_{b=1}^r v^b e_b(p)\right)>0. \end{align*} Now let $e'_1,\dots,e'_r$ be another local frame on an overlap, related by \begin{align*} e'_i=\sum_{a=1}^r e_a g_{ai}. \end{align*} The metric coefficients in the primed frame are \begin{align*} h'_{ij}=h(e'_i,e'_j). \end{align*} Substituting the change of frame gives \begin{align*} h'_{ij}=h\left(\sum_{a=1}^r e_a g_{ai},\sum_{b=1}^r e_b g_{bj}\right). \end{align*} By bilinearity, \begin{align*} h'_{ij}=\sum_{a=1}^r\sum_{b=1}^r g_{ai}g_{bj}h(e_a,e_b). \end{align*} Using $h(e_a,e_b)=h_{ab}$, this is \begin{align*} h'_{ij}=\sum_{a=1}^r\sum_{b=1}^r g_{ai}h_{ab}g_{bj}. \end{align*} In matrix form, \begin{align*} H'=g^\top H g. \end{align*} This is the transformation law for a symmetric covariant $2$-tensor, so a bundle metric is a section of $\operatorname{Sym}^2(E^*)$ whose local matrices are positive definite. Conversely, a section of $\operatorname{Sym}^2(E^*)$ is represented in each local frame by a symmetric matrix $H(p)$ transforming by \begin{align*} H'=g^\top H g. \end{align*} If $H(p)$ is positive definite in one frame, then for every nonzero vector $u$ in the primed coordinates, \begin{align*} u^\top H'(p)u=u^\top g(p)^\top H(p)g(p)u. \end{align*} Associativity of matrix multiplication gives \begin{align*} u^\top g(p)^\top H(p)g(p)u=(g(p)u)^\top H(p)(g(p)u). \end{align*} Since $g(p)$ is invertible, $u\ne 0$ implies $g(p)u\ne 0$, so the last expression is positive. Thus positive definiteness is independent of the chosen frame. On a paracompact smooth manifold, choose local frames and put the Euclidean metric $I_r$ in each one. If $(\rho_i)$ is a smooth [partition of unity](/page/Partition%20of%20Unity) subordinate to these trivializing sets, define \begin{align*} h_p(v,w)=\sum_i \rho_i(p)h^{(i)}_p(v,w). \end{align*} The sum is locally finite, so it defines a smooth symmetric bilinear form. For $v\ne 0$, \begin{align*} h_p(v,v)=\sum_i \rho_i(p)h^{(i)}_p(v,v). \end{align*} Each term is nonnegative, and at least one $\rho_i(p)$ is positive; for that index, $h^{(i)}_p(v,v)>0$. Hence \begin{align*} h_p(v,v)>0. \end{align*} Thus partitions of unity glue local Euclidean metrics into a global bundle metric, and the tensor-section viewpoint records exactly the transformation law that makes this construction independent of coordinates. [/example] The chapter's constructions give the basic algebra of vector bundles. Chapter 6 applies the same principle to pullback bundles, and Chapter 10 applies it to associated bundles: do the construction locally, then verify that the transition data transform by the corresponding functorial rule. The fibrewise operations of the previous chapter are useful only because local formulas can be assembled globally. The next chapter adds the missing global structure by introducing bundle metrics and partitions of unity, which let local data be averaged, compared, and glued into smooth global objects. # 5. Bundle Metrics and Partitions of Unity This chapter turns the local description of vector bundles into global geometric structure. Chapters 2 and 3 showed that a bundle is assembled from transition functions and that sections are global objects written locally as compatible component functions. We now ask how to build additional structure on a bundle when it is specified only locally: inner products on fibres, complements to subbundles, projections, and glued bundle maps. The technical device that makes this possible is the partition of unity, which lets finite-dimensional linear constructions vary smoothly over a manifold. ## Smooth Fibre Metrics on Vector Bundles A vector bundle has vector spaces as fibres, but a bare vector space has no preferred notion of length or angle. Many later constructions, including [orthogonal projection](/theorems/437), adjoints of bundle maps, and Riemannian geometry itself, require a smoothly varying inner product on the fibres. The first question is therefore: when can local fibrewise inner products be assembled into a global one? [definition: Smooth Fibre Metric] Let $\pi:E\to M$ be a smooth real vector bundle of rank $k$. A smooth fibre metric on $E$ is a smooth section \begin{align*} g \in \Gamma(\operatorname{Sym}^2(E^*)) \end{align*} such that, for every $p\in M$, the bilinear form $g_p:E_p\times E_p\to \mathbb R$ is symmetric and positive definite. [/definition] The section viewpoint is useful because it separates the algebraic condition in each fibre from the smoothness condition over the base. In a local trivialization $E|_U\cong U\times \mathbb R^k$, a fibre metric is represented by a smooth map $G:U\to \operatorname{Sym}^2_+(\mathbb R^k)$, where $G(p)$ is a positive definite symmetric matrix. [example: Product Bundle Metric] On the product vector bundle $M\times \mathbb R^k\to M$, write $v=(v_1,\dots,v_k)$ and $w=(w_1,\dots,w_k)$ in the standard coordinates on $\mathbb R^k$. The formula \begin{align*} g_p(v,w)=\sum_{i=1}^k v_iw_i \end{align*} has coefficient matrix $I_k$ in the global product frame, independent of $p$, so its coefficient functions are smooth. It is bilinear because, for $a,b\in\mathbb R$ and $u,v,w\in\mathbb R^k$, \begin{align*} g_p(au+bv,w)=\sum_{i=1}^k(au_i+bv_i)w_i=a\sum_{i=1}^k u_iw_i+b\sum_{i=1}^k v_iw_i=a g_p(u,w)+b g_p(v,w). \end{align*} The same expansion in the second argument gives linearity in $w$. It is symmetric since \begin{align*} g_p(v,w)=\sum_{i=1}^k v_iw_i=\sum_{i=1}^k w_iv_i=g_p(w,v), \end{align*} and it is positive definite because \begin{align*} g_p(v,v)=\sum_{i=1}^k v_i^2, \end{align*} which is positive exactly when $v\ne 0$. Thus the constant formula defines a smooth fibre metric. More generally, if $G:M\to \operatorname{Sym}^2_+(\mathbb R^k)$ is smooth, define \begin{align*} g_p(v,w)=v^\top G(p)w. \end{align*} Writing $G(p)=(G_{ij}(p))$, this means \begin{align*} g_p(v,w)=\sum_{i=1}^k\sum_{j=1}^k v_iG_{ij}(p)w_j. \end{align*} The functions $G_{ij}$ are smooth, so $g$ is a smooth section of $\operatorname{Sym}^2(E^*)$ in the global product frame. Since $G(p)$ is symmetric, $G_{ij}(p)=G_{ji}(p)$, and therefore \begin{align*} g_p(w,v)=\sum_{i=1}^k\sum_{j=1}^k w_iG_{ij}(p)v_j=\sum_{j=1}^k\sum_{i=1}^k v_jG_{ji}(p)w_i=\sum_{j=1}^k\sum_{i=1}^k v_jG_{ij}(p)w_i=g_p(v,w). \end{align*} Because each $G(p)$ is positive definite, $g_p(v,v)=v^\top G(p)v>0$ for every nonzero $v\in\mathbb R^k$. This shows that the local problem is not the existence of inner products on individual fibres; the issue is choosing them smoothly as $p$ varies. [/example] The product example identifies the local building blocks: every trivializing neighbourhood carries a preferred Euclidean metric after a choice of trivialization. A naive attempt to glue these metrics would demand agreement on overlaps, but this is too strict: on a rank-one bundle, two local trivializations can give local metrics represented by $1$ and by a positive function $a(p)^2$ on the overlap, and these coincide only when $a(p)=1$. The remaining question is whether these incompatible local choices can be combined without requiring equality on overlaps. The next theorem answers this by using a partition of unity to average the local metrics while preserving positive definiteness. [quotetheorem:6096] [citeproof:6096] This theorem is the first major instance of a common principle: on paracompact manifolds, local smooth linear data can often be glued if the target condition is convex. The paracompactness hypothesis is not cosmetic; on a non-paracompact smooth manifold such as the Prüfer surface, smooth partitions of unity subordinate to arbitrary covers can fail to exist, and the tangent bundle need not admit a Riemannian metric because a Riemannian metric would make the surface metrizable. The theorem also does not choose a canonical metric, since different covers, trivializations, and partitions of unity generally produce different answers. What it gives is an existence tool strong enough for later topology and geometry: once a metric exists, vector-bundle exact sequences can be split, subbundles have complements, and characteristic-class arguments can compare bundles through these auxiliary splittings. [example: Riemannian Metrics] A Riemannian metric on a smooth manifold $M$ is a smooth fibre metric on the tangent bundle $TM\to M$. In the notation of fibre metrics, it is a smooth section \begin{align*} g\in \Gamma(\operatorname{Sym}^2(T^*M)) \end{align*} such that, for every $p\in M$, the bilinear form $g_p:T_pM\times T_pM\to\mathbb R$ is symmetric and positive definite. In a coordinate chart $(U,x^1,\dots,x^n)$, the tangent bundle has local frame $\partial/\partial x^1,\dots,\partial/\partial x^n$. If \begin{align*} v=\sum_{i=1}^n v^i\frac{\partial}{\partial x^i}\Big|_p \end{align*} and \begin{align*} w=\sum_{j=1}^n w^j\frac{\partial}{\partial x^j}\Big|_p, \end{align*} then bilinearity gives \begin{align*} g_p(v,w)=\sum_{i=1}^n\sum_{j=1}^n v^i w^j\,g_p\left(\frac{\partial}{\partial x^i}\Big|_p,\frac{\partial}{\partial x^j}\Big|_p\right). \end{align*} Writing \begin{align*} g_{ij}(p)=g_p\left(\frac{\partial}{\partial x^i}\Big|_p,\frac{\partial}{\partial x^j}\Big|_p\right), \end{align*} this becomes \begin{align*} g_p(v,w)=\sum_{i=1}^n\sum_{j=1}^n v^i g_{ij}(p)w^j. \end{align*} Smoothness of $g$ means that the coefficient functions $g_{ij}:U\to\mathbb R$ are smooth. Symmetry gives $g_{ij}(p)=g_{ji}(p)$ for all $i,j$, and positive definiteness says that, for every nonzero $v\in T_pM$, \begin{align*} g_p(v,v)=\sum_{i=1}^n\sum_{j=1}^n v^i g_{ij}(p)v^j>0. \end{align*} Since $TM\to M$ is a smooth real vector bundle of finite rank, *[Existence of Smooth Bundle Metrics](/theorems/6096)* applied to $E=TM$ gives a smooth fibre metric on $TM$ whenever $M$ is paracompact. Thus every paracompact smooth manifold admits a Riemannian metric, and locally it is represented by a smooth positive definite symmetric matrix of functions $(g_{ij})$. [/example] A metric also transports structure to related bundles. For instance, a metric on $E$ identifies $E$ with $E^*$ by sending $v\in E_p$ to the functional $w\mapsto g_p(v,w)$, and the smoothness of this identification follows from the smoothness of $g$ in local frames. [remark: Complex Bundles] For a complex vector bundle, the analogous object is a Hermitian fibre metric: a smooth section of the Hermitian forms on $E$ whose restriction to each fibre is positive definite, complex-linear in the first argument and conjugate-linear in the second. The same partition-of-unity proof works after choosing the standard Hermitian form on each local trivialization. [/remark] ## Orthogonal Complements and Projections A subbundle is locally a smoothly varying family of subspaces. Linear algebra says that a subspace of an [inner product space](/page/Inner%20Product%20Space) has an orthogonal complement, but a bundle statement also requires those complements to fit together smoothly. The question is how the fibrewise construction can be promoted to a smooth subbundle. [illustration:orthogonal-complement-subbundle] [definition: Orthogonal Complement Subbundle] Let $E\to M$ be a smooth vector bundle with smooth fibre metric $g$, and let $F\subset E$ be a smooth vector subbundle. The orthogonal complement of $F$ in $E$ is the subset \begin{align*} F^\perp=\{v\in E: g_{\pi(v)}(v,w)=0\text{ for all }w\in F_{\pi(v)}\}. \end{align*} [/definition] The definition gives the expected fibrewise subspaces, but a subset of the total space is not yet known to be a smooth subbundle. To use $F^\perp$ in bundle constructions, we need local smooth frames for it and a smooth direct-sum decomposition. The next theorem supplies exactly this smoothness statement. [quotetheorem:6097] [citeproof:6097] This result shows why metrics are so powerful: they turn the noncanonical problem of finding a complement into a canonical fibrewise operation. The hypotheses are essential: if $E=M\times \mathbb R$ over $M=\mathbb R$ and the proposed fibre family is $F_x=0$ for $x\le 0$ and $F_x=\mathbb R$ for $x>0$, then the fibrewise orthogonal complements jump rank at $0$ and do not form a smooth subbundle. Smoothness of the metric is also essential, not only convenient for the construction. For example, over $M=(-1,1)$ take $E=M\times\mathbb R^2$, let $F$ be the constant line spanned by $(1,0)$, and define a positive definite fibrewise inner product $G_x$ by \begin{align*} G_x(e_1,e_1)=1,\qquad G_x(e_1,e_2)=G_x(e_2,e_1)=|x|/2,\qquad G_x(e_2,e_2)=1. \end{align*} The fibrewise orthogonal complement is the line spanned by $(-|x|/2,1)$, whose slope is not a smooth function of $x$ at $0$; hence it is not a smooth line subbundle. The theorem does not say that an arbitrary fibrewise choice of complements is smooth; it says that the particular complement defined from a smooth metric and a smooth subbundle is smooth. Once this is known, the associated projection maps are smooth bundle morphisms and can be used to split geometric constructions locally and globally. [definition: Orthogonal Projection] Let $E\to M$ be a smooth vector bundle with smooth fibre metric $g$, and let $F\subset E$ be a smooth vector subbundle. The orthogonal projection onto $F$ is the bundle map \begin{align*} P_F:E\to E \end{align*} whose restriction to each fibre $E_p=F_p\oplus F_p^\perp$ sends $u+w$ to $u$ for $u\in F_p$ and $w\in F_p^\perp$. [/definition] The projection is a bundle endomorphism over $\operatorname{id}_M$, and it satisfies $P_F^2=P_F$. This raises a broader complement question that no longer mentions a preselected metric: given only the subbundle $F\subset E$, can we find some smooth partner $G$? The metric existence theorem and the smoothness of $F^\perp$ combine to give the answer. [quotetheorem:6098] [citeproof:6098] This result is often used in the contrapositive direction: when a short exact sequence of vector bundles appears, a choice of metric supplies a splitting. The paracompactness assumption enters through the existence of the metric; on non-paracompact smooth manifolds such as the Prüfer surface, even the tangent bundle may fail to admit a Riemannian metric, so this proof of complements breaks down. The smooth subbundle assumption is also essential: a family of subspaces whose dimension jumps, such as $F_x=0$ for $x\le 0$ and $F_x=\mathbb R$ for $x>0$ inside the product line bundle over $\mathbb R$, has no smooth complementary subbundle compatible with a constant-rank direct-sum decomposition. The theorem does not say that the complement is unique or natural; different metrics can give different complements. Its geometric force is that topological questions about extensions of bundles can often be reduced to the existence of smooth splittings, even when no preferred splitting exists. [example: Subbundle Of A Product Bundle] Let $F\subset M\times \mathbb R^n$ be a rank $r$ smooth subbundle, and equip $M\times\mathbb R^n$ with the standard product metric. By the [Orthogonal Complement Subbundle Theorem](/theorems/6097), \begin{align*} F_p^\perp=\{v\in \mathbb R^n: v\cdot w=0\text{ for all }w\in F_p\} \end{align*} forms a smooth subbundle of rank $n-r$. On a neighbourhood $U\subset M$, suppose $F$ is spanned by the columns of a smooth full-rank matrix \begin{align*} A(p)=\begin{bmatrix} a_1(p)&\cdots&a_r(p)\end{bmatrix}, \end{align*} so $F_p=\operatorname{im}A(p)$. For $x\in\mathbb R^r$, \begin{align*} x^\top A(p)^\top A(p)x=(A(p)x)^\top(A(p)x)=\|A(p)x\|^2. \end{align*} Since $A(p)$ has rank $r$, $A(p)x=0$ implies $x=0$, so $x^\top A(p)^\top A(p)x>0$ for $x\ne 0$. Hence $A(p)^\top A(p)$ is invertible. For $y\in\mathbb R^n$, write the projection of $y$ onto $F_p$ as $A(p)c$ for some $c\in\mathbb R^r$. The error term $y-A(p)c$ must be orthogonal to every column of $A(p)$, which is equivalent to \begin{align*} A(p)^\top(y-A(p)c)=0. \end{align*} Expanding gives \begin{align*} A(p)^\top y-A(p)^\top A(p)c=0, \end{align*} so \begin{align*} c=(A(p)^\top A(p))^{-1}A(p)^\top y. \end{align*} Therefore \begin{align*} P(p)y=A(p)c=A(p)(A(p)^\top A(p))^{-1}A(p)^\top y, \end{align*} and the projection matrix is \begin{align*} P(p)=A(p)(A(p)^\top A(p))^{-1}A(p)^\top. \end{align*} Its entries are smooth functions of $p$ because the entries of $A(p)$ are smooth and the inverse of the matrix $A(p)^\top A(p)$ is smooth wherever its determinant is nonzero. Thus the orthogonal projection onto $F$ is a smooth bundle map, and its kernel is the smooth orthogonal complement $F^\perp$. [/example] ## Using Partitions Of Unity To Glue Local Bundle Data The previous sections used partitions of unity to build metrics, but the same mechanism applies more broadly. The guiding problem is this: given local sections, bundle maps, or fibrewise tensors that agree only approximately or are defined on different open sets, when can they be combined into a global smooth object? [definition: Locally Finite Smooth Partition Of Unity] Let $M$ be a smooth manifold and let $(U_i)_{i\in I}$ be an open cover of $M$. A smooth partition of unity subordinate to $(U_i)_{i\in I}$ is a family of smooth functions $\rho_i:M\to [0,1]$ such that $\operatorname{supp}\rho_i\subset U_i$, the family $(\operatorname{supp}\rho_i)_{i\in I}$ is locally finite, and \begin{align*} \sum_{i\in I}\rho_i(p)=1 \end{align*} for every $p\in M$. [/definition] Local finiteness is the condition that makes the infinite sum behave like a finite sum near every point. This is precisely what is needed when local sections are multiplied by the weights $\rho_i$ and summed. The next theorem records the gluing construction in the vector-bundle form used throughout the chapter. [quotetheorem:6099] [citeproof:6099] This theorem glues sections without requiring the local sections to agree on overlaps. Local finiteness is essential: on $\mathbb R$, if infinitely many bump functions are all nonzero near $0$, the pointwise sum of the corresponding weighted local sections may be an infinite sum near $0$ and need not define a smooth section. The support condition is equally important, because extending $\rho_i s_i$ by zero across the boundary of $U_i$ is smooth only when $\rho_i$ has already vanished near that boundary; without this, a local section can acquire a jump discontinuity after extension. The theorem also does not produce a common extension of the $s_i$ on overlaps. The price is that the resulting section is an average rather than a common extension, so it is best suited to affine or convex constructions rather than rigid uniqueness problems. This is the mechanism behind the bundle-map example next: a fibrewise linear map is itself a section of a Hom-bundle, so local maps can be averaged exactly as local metrics were. Later global constructions use the same pattern whenever the desired object is defined by linear equations or by an open convex condition preserved under weighted sums. [example: Gluing Local Bundle Maps] Let $E\to M$ and $F\to M$ be smooth vector bundles, and let $A_i:E|_{U_i}\to F|_{U_i}$ be smooth bundle maps over $\operatorname{id}_{U_i}$. Given a locally finite smooth partition of unity $(\rho_i)$ with $\operatorname{supp}\rho_i\subset U_i$, define \begin{align*} A_p(v)=\sum_i \rho_i(p)(A_i)_p(v), \end{align*} where $\rho_i A_i$ is extended by zero outside $U_i$. For each fixed $p$, local finiteness leaves only finitely many nonzero terms. If $a,b\in\mathbb R$ and $v,w\in E_p$, then the linearity of each fibre map $(A_i)_p:E_p\to F_p$ gives \begin{align*} A_p(av+bw)=\sum_i \rho_i(p)\bigl(a(A_i)_p(v)+b(A_i)_p(w)\bigr). \end{align*} Distributing the finite sum gives \begin{align*} A_p(av+bw)=a\sum_i \rho_i(p)(A_i)_p(v)+b\sum_i \rho_i(p)(A_i)_p(w). \end{align*} By the definition of $A_p$, this is \begin{align*} A_p(av+bw)=aA_p(v)+bA_p(w). \end{align*} Thus each $A_p:E_p\to F_p$ is linear. To check smoothness, choose a neighbourhood $V$ meeting only finitely many supports $\operatorname{supp}\rho_i$, and choose smooth local frames $e_1,\dots,e_r$ for $E$ and $f_1,\dots,f_s$ for $F$ over $V$. On $V\cap U_i$, write \begin{align*} A_i(e_\alpha)=\sum_{\beta=1}^s a_{i\beta\alpha}f_\beta, \end{align*} where each coefficient $a_{i\beta\alpha}$ is smooth on $V\cap U_i$. Since $\operatorname{supp}\rho_i\subset U_i$, the product $\rho_i a_{i\beta\alpha}$ extends smoothly by zero to all of $V$. Therefore \begin{align*} A(e_\alpha)=\sum_i \rho_i A_i(e_\alpha). \end{align*} Substituting the frame expansion gives \begin{align*} A(e_\alpha)=\sum_i \rho_i\sum_{\beta=1}^s a_{i\beta\alpha}f_\beta. \end{align*} Reordering the finite sum on $V$ gives \begin{align*} A(e_\alpha)=\sum_{\beta=1}^s\left(\sum_i \rho_i a_{i\beta\alpha}\right)f_\beta. \end{align*} The coefficient $\sum_i \rho_i a_{i\beta\alpha}$ is a finite sum of smooth functions on $V$, so it is smooth. Hence $A:E\to F$ is a smooth bundle map. The metric construction is the same weighted-sum operation in the vector bundle $\operatorname{Sym}^2(E^*)$. If $g_i$ are local fibre metrics, then \begin{align*} g_p(v,v)=\sum_i \rho_i(p)(g_i)_p(v,v). \end{align*} For $v\ne 0$, every term is nonnegative, and at least one coefficient $\rho_i(p)$ is positive because $\sum_i\rho_i(p)=1$. For that index, positive definiteness gives $(g_i)_p(v,v)>0$, so the whole sum is positive. Thus partition-of-unity averaging preserves the positive definiteness needed in the metric proof. [/example] Partitions of unity also let us isolate local behaviour. Multiplying a section by a bump function turns a local section into a global section supported inside a prescribed coordinate neighbourhood, a technique used repeatedly when proving local-to-global statements. [remark: What Cannot Be Glued By Averaging] Averaging works naturally in vector spaces and convex subsets of vector spaces. It does not directly glue local trivializations, local frames, or transition functions valued in nonlinear groups such as $GL(k,\mathbb R)$, because a weighted average of invertible matrices need not be invertible. For such data, compatibility on overlaps or a different construction is required. [/remark] This chapter's main pattern is now in place. Smooth bundle metrics exist because positive definite forms can be averaged; orthogonal complements exist because the metric turns complement-finding into fibrewise linear algebra; and partitions of unity provide the bridge from local formulas to global smooth bundle data. Bundle metrics and partitions of unity show that many local arguments can be made global without losing smoothness. The next chapter uses that same idea for morphisms of bundles, explaining how pullbacks and bundle maps transport structure along maps of the base manifold. # 6. Bundle Maps, Pullbacks, and Naturality These notes develop the part of vector-bundle theory where bundles are allowed to move along maps of their base manifolds. The prerequisites are the local construction of smooth fibre bundles from Chapters 1 and 2, vector bundles, local frames, and sections from Chapter 3. Earlier chapters built bundles from local data and studied sections as maps into a fixed total space; this chapter asks how bundles themselves move under maps of bases. The central construction is the pullback bundle, which transports a bundle on $N$ to a bundle on $M$ along a smooth map $f:M\to N$ and thereby explains why later geometric constructions, such as induced metrics and characteristic classes, behave functorially. ## Bundle Morphisms Over Fixed and Moving Bases The first problem is to say what it means for a map between total spaces to respect the bundle projections. A map of total spaces alone need not send fibres to fibres, so it may ignore the geometry encoded by the bundle structure. Bundle morphisms impose compatibility with the base, and in the vector-bundle case they also impose linearity on each fibre. [definition: Bundle Morphism Over A Smooth Map] Let $\pi_E:E\to M$ and $\pi_F:F\to N$ be smooth fibre bundles. A bundle morphism from $E$ to $F$ over a smooth map $f:M\to N$ is a smooth map $\Phi:E\to F$ such that \begin{align*} \pi_F\circ \Phi=f\circ \pi_E. \end{align*} [/definition] The equation says that if $e\in E_x$, then $\Phi(e)\in F_{f(x)}$. This captures fibre preservation, but it does not yet express the linear structure present in vector bundles. To compare vector bundles rather than only their underlying fibre bundles, the fibre maps must be linear. [definition: Vector Bundle Morphism] Let $\pi_E:E\to M$ and $\pi_F:F\to N$ be smooth vector bundles. A vector bundle morphism from $E$ to $F$ over $f:M\to N$ is a bundle morphism $\Phi:E\to F$ over $f$ such that for every $x\in M$, the restricted map \begin{align*} \Phi_x:E_x\to F_{f(x)} \end{align*} is linear. [/definition] This linearity condition is fibrewise, while smoothness is a condition on the total space. For computations, we need a local test that separates the base map from the fibrewise linear part. In adapted bundle coordinates, the fibrewise linear part should be a smooth matrix-valued function. [quotetheorem:6100] [citeproof:6100] The local form makes bundle morphisms computable: they are base maps together with compatible families of matrices. The smoothness of $f$ is essential; if the first component is not smooth, then no choice of smooth fibre matrix can make $(x,v)\mapsto(f(x),A(x)v)$ a smooth map of total spaces. Fibrewise linearity is also essential, since without it the second component could be any smooth fibre map rather than a matrix-valued one. The theorem is local rather than global: the matrices $A$ depend on the chosen trivializations and must transform on overlaps according to the transition functions of the two bundles. This overlap behaviour motivates treating familiar differentials as bundle morphisms rather than as unrelated pointwise linear maps. [example: Tangent Map As A Bundle Morphism] Let $f:M\to N$ be smooth, and let $\pi_M:TM\to M$ and $\pi_N:TN\to N$ be the tangent bundle projections. The tangent map sends a tangent vector at $p$ to its pushforward at $f(p)$: \begin{align*} df(p,u)=df_p(u). \end{align*} For $u\in T_pM$, the vector $df_p(u)$ lies in $T_{f(p)}N$, so \begin{align*} (\pi_N\circ df)(p,u)=\pi_N(df_p(u))=f(p)=f(\pi_M(p,u))=(f\circ \pi_M)(p,u). \end{align*} Thus $df:TM\to TN$ is a bundle morphism over $f$. For each fixed $p\in M$, the map $df_p:T_pM\to T_{f(p)}N$ is linear by the definition of the differential, so $df$ is a vector bundle morphism. Now choose charts $(U,\varphi)$ on $M$ and $(V,\psi)$ on $N$ with $f(U)\subset V$, and write \begin{align*} F=\psi\circ f\circ \varphi^{-1}:\varphi(U)\to \psi(V). \end{align*} The induced tangent coordinates identify $(p,u)\in TU$ with $(x,\xi)$, where $x=\varphi(p)$ and $\xi=d\varphi_p(u)\in \mathbb R^m$. By the chain rule, \begin{align*} d\psi_{f(p)}(df_p(u))=d(\psi\circ f)_p(u). \end{align*} Applying the chain rule again to $\psi\circ f=F\circ \varphi$ gives \begin{align*} d(\psi\circ f)_p(u)=dF_{\varphi(p)}(d\varphi_p(u))=dF_x(\xi). \end{align*} If $F=(F^1,\dots,F^n)$ and $\xi=(\xi^1,\dots,\xi^m)$, then for each component $1\le a\le n$, \begin{align*} (dF_x(\xi))^a=\sum_{j=1}^m \frac{\partial F^a}{\partial x^j}(x)\xi^j. \end{align*} Equivalently, \begin{align*} dF_x(\xi)=\left(\frac{\partial F^a}{\partial x^j}(x)\right)_{a,j}\xi. \end{align*} Hence, in the tangent bundle charts induced by $\varphi$ and $\psi$, the local matrix of $df$ is the Jacobian matrix of $\psi\circ f\circ \varphi^{-1}$. The important feature is that $df_p$ has target $T_{f(p)}N$, which changes with $p$; tangent maps therefore naturally live over non-identity base maps. [/example] When the base is fixed, the set of bundle maps forms the natural morphisms in the category of vector bundles over $M$. When the base varies, morphisms behave like arrows in a larger category whose objects are bundles over all manifolds. [remark: Isomorphisms] A vector bundle morphism $\Phi:E\to F$ over a diffeomorphism $f:M\to N$ is a vector bundle isomorphism if $\Phi$ is a diffeomorphism and every fibre map $\Phi_x:E_x\to F_{f(x)}$ is a linear isomorphism. Over the identity map on $M$, this is the usual notion of an isomorphism of vector bundles over the same base. [/remark] ## Pullback Bundles And Their Universal Property The next question is how to compare a bundle over $N$ with geometry on a different manifold $M$. A smooth map $f:M\to N$ lets each point $x\in M$ select the fibre of a bundle $E\to N$ over $f(x)$. The pullback bundle packages these selected fibres into a new bundle over $M$. [definition: Pullback Bundle] Let $\pi:E\to N$ be a smooth fibre bundle and let $f:M\to N$ be smooth. The pullback set is \begin{align*} f^*E=\{(x,e)\in M\times E: f(x)=\pi(e)\}. \end{align*} The projection is $\pi_{f^*E}:f^*E\to M$, given by $\pi_{f^*E}(x,e)=x$. [/definition] The fibre over $x$ is naturally the old fibre $E_{f(x)}$, with the element $e\in E_{f(x)}$ recorded together with the new base point $x$. The remaining work is to put a smooth bundle structure on this set, since it is initially only a subset of $M\times E$. [quotetheorem:6101] [citeproof:6101] The construction follows the same gluing philosophy as transition functions, but with no new cocycle invented: all transition data are read off from the old bundle along $f$. The hypothesis that $f$ is smooth is needed because the transition functions of the pullback are obtained by composing the old transition functions with $f$; a merely continuous base map would not produce a smooth bundle in general. The theorem constructs a bundle over $M$, not a canonical subbundle of $E$, because different points of $M$ may map to the same point of $N$ and must still carry distinct base labels. This distinction is what makes restriction to a submanifold fit the same construction without treating it as a separate operation. [example: Restriction To A Submanifold] Let $i:S\hookrightarrow M$ be the inclusion of an embedded submanifold, and let $\pi:E\to M$ be a smooth vector bundle. By the definition of pullback, \begin{align*} i^*E=\{(s,e)\in S\times E:i(s)=\pi(e)\}. \end{align*} Since $i(s)=s$ as a point of $M$, this becomes \begin{align*} i^*E=\{(s,e)\in S\times E:s=\pi(e)\}. \end{align*} Equivalently, \begin{align*} i^*E=\{(s,e)\in S\times E:e\in E_s\}. \end{align*} The usual restricted total space is $E|_S=\pi^{-1}(S)$. Define \begin{align*} \Theta:E|_S\to i^*E,\qquad \Theta(e)=(\pi(e),e). \end{align*} This is well-defined because $e\in E|_S$ implies $\pi(e)\in S$, and then \begin{align*} i(\pi(e))=\pi(e). \end{align*} Define also \begin{align*} \Lambda:i^*E\to E|_S,\qquad \Lambda(s,e)=e. \end{align*} If $(s,e)\in i^*E$, then $\pi(e)=s\in S$, so $e\in E|_S$ and $\Lambda$ is well-defined. The two maps are inverse because \begin{align*} (\Lambda\circ\Theta)(e)=\Lambda(\pi(e),e)=e \end{align*} and, for $(s,e)\in i^*E$, \begin{align*} (\Theta\circ\Lambda)(s,e)=\Theta(e)=(\pi(e),e)=(s,e). \end{align*} Thus the fibre over $s\in S$ is \begin{align*} (i^*E)_s=\{(s,e):e\in E_s\}\cong E_s. \end{align*} If $\tau:E|_U\to U\times \mathbb R^r$ is a local trivialization over an open set $U\subset M$, then its restriction gives a chart over $S\cap U$: \begin{align*} E|_{S\cap U}\to (S\cap U)\times \mathbb R^r,\qquad e\mapsto \bigl(\pi(e),\operatorname{pr}_{\mathbb R^r}\tau(e)\bigr). \end{align*} Under the identification $\Theta:E|_S\cong i^*E$, this is exactly the pullback chart obtained from $\tau$ by replacing the base point $x\in U$ with $s\in S\cap U$. Hence restriction to a submanifold is the pullback along the inclusion map, not a separate construction. [/example] The construction gives explicit charts, while many later arguments need a coordinate-free recognition principle. If a space $P$ over $M$ carries a map into $E$ whose base point is always $f(q(p))$, then the map should factor through $f^*E$. This is the universal property of the pullback. [quotetheorem:6102] [citeproof:6102] This property says that any family of points in fibres of $E$ parametrized by $P$ and lying above $f\circ q$ factors uniquely through the pullback. The compatibility equation is necessary: without $\pi\circ\Psi=f\circ q$, the pair $(q(p),\Psi(p))$ may fail to lie in the defining set of $f^*E$. The result does not assert that every map $P\to f^*E$ is a section, only that maps into $E$ with the correct base behaviour lift uniquely to the fibre product. This recognition principle is what lets pullbacks classify geometric data such as line subbundles produced by maps to projective space. [example: Pullback Of The Tautological Line Bundle] Let $\gamma^1\to \mathbb{R}P^n$ be the tautological line bundle and let $f:M\to \mathbb{R}P^n$ be smooth. Recall that a point of $\mathbb{R}P^n$ is a line $\ell\subset \mathbb R^{n+1}$, and the tautological line bundle has total space \begin{align*} \gamma^1=\{(\ell,v)\in \mathbb{R}P^n\times \mathbb R^{n+1}:v\in \ell\}, \end{align*} with projection $\pi(\ell,v)=\ell$. By the definition of pullback, \begin{align*} f^*\gamma^1=\{(x,(\ell,v))\in M\times \gamma^1:f(x)=\pi(\ell,v)\}. \end{align*} Since $\pi(\ell,v)=\ell$, this is \begin{align*} f^*\gamma^1=\{(x,(\ell,v))\in M\times \gamma^1:f(x)=\ell\}. \end{align*} Because membership in $\gamma^1$ also means $v\in \ell$, the condition $f(x)=\ell$ gives the equivalent description \begin{align*} f^*\gamma^1=\{(x,(f(x),v)):v\in f(x)\}. \end{align*} Thus the fibre over $x\in M$ is \begin{align*} (f^*\gamma^1)_x=\{(x,(f(x),v)):v\in f(x)\}. \end{align*} For this fixed $x$, define \begin{align*} \Theta_x:(f^*\gamma^1)_x\to f(x),\qquad \Theta_x(x,(f(x),v))=v. \end{align*} If $a,b\in \mathbb R$ and $v,w\in f(x)$, then \begin{align*} \Theta_x\bigl(a(x,(f(x),v))+b(x,(f(x),w))\bigr)=\Theta_x(x,(f(x),av+bw)). \end{align*} Since $f(x)$ is a line, $av+bw\in f(x)$, and therefore \begin{align*} \Theta_x(x,(f(x),av+bw))=av+bw=a\Theta_x(x,(f(x),v))+b\Theta_x(x,(f(x),w)). \end{align*} So $\Theta_x$ is linear. It is bijective because the inverse map is \begin{align*} f(x)\to (f^*\gamma^1)_x,\qquad v\mapsto (x,(f(x),v)). \end{align*} Hence the pulled-back fibre is canonically the line in $\mathbb R^{n+1}$ represented by $f(x)$. Under these fibrewise identifications, $f^*\gamma^1$ sits inside the product bundle $M\times \mathbb R^{n+1}$ as \begin{align*} \{(x,v)\in M\times \mathbb R^{n+1}:v\in f(x)\}. \end{align*} Since $f$ is smooth and $\gamma^1$ is a smooth line bundle, the pullback construction gives this subset a smooth line-bundle structure over $M$. Thus a smooth map to projective space selects one smoothly varying line in $\mathbb R^{n+1}$ over each point of $M$. [/example] ## Pulling Back Sections, Frames, Metrics, And Transition Data Once the pullback bundle exists, the next problem is to transport the structures carried by a bundle. Sections, frames, metrics, and transition functions should all pull back by substituting $f(x)$ for the old base variable. This section records the resulting formulas and the reason they are natural. A section of $E\to N$ gives a section of $f^*E\to M$ by evaluating it at the image point $f(x)$. [definition: Pullback Section] Let $s:N\to E$ be a smooth section of $\pi:E\to N$ and let $f:M\to N$ be smooth. The pullback section $f^*s:M\to f^*E$ is \begin{align*} (f^*s)(x)=(x,s(f(x))). \end{align*} [/definition] The section condition follows from $\pi(s(f(x)))=f(x)$. Smoothness follows from smoothness of $s\circ f$ and the construction of the pullback bundle. [example: Pulling Back A Nowhere-Zero Section] Let $E\to N$ be a line bundle with a nowhere-zero section $s$, and let $f:M\to N$ be smooth. The pullback section is \begin{align*} (f^*s)(x)=(x,s(f(x))). \end{align*} This lies in the fibre $(f^*E)_x$ because \begin{align*} \pi(s(f(x)))=f(x), \end{align*} since $s$ is a section of $E\to N$. Since $s$ is nowhere zero, for every $x\in M$ one has \begin{align*} s(f(x))\neq 0\in E_{f(x)}. \end{align*} Under the fibre identification \begin{align*} (f^*E)_x=\{(x,e):e\in E_{f(x)}\}, \end{align*} the zero vector in $(f^*E)_x$ is $(x,0_{E_{f(x)}})$. Therefore \begin{align*} (f^*s)(x)=(x,s(f(x)))\neq (x,0_{E_{f(x)}}), \end{align*} so $f^*s$ is nowhere zero. The corresponding product presentation is obtained fibre by fibre. Define \begin{align*} \Theta:M\times \mathbb R\to f^*E,\qquad \Theta(x,a)=(x,a\,s(f(x))). \end{align*} For each fixed $x$, the vector $s(f(x))$ is a nonzero vector in the one-dimensional vector space $E_{f(x)}$, so it is a basis. Hence every element $(x,e)\in (f^*E)_x$ has a unique scalar $a\in \mathbb R$ with \begin{align*} e=a\,s(f(x)). \end{align*} Thus $\Theta_x:\mathbb R\to (f^*E)_x$ is a linear isomorphism, with inverse sending $(x,e)$ to the unique coefficient $a$ in this expansion. Hence a nowhere-zero section trivializes a line bundle, and pulling the section back gives the corresponding trivialization of $f^*E$. [/example] A single section detects one vector in each fibre, while a frame detects an entire basis. To pass from pulled-back sections to pulled-back coordinates, we need to know that a basis of sections remains a basis after evaluating at $f(x)$. This is the frame version of pullback. [quotetheorem:6103] [citeproof:6103] The local-frame hypothesis is essential here: a pulled-back list of sections is a frame only because each original list was already a basis in every fibre over its domain. If the original sections merely span at some points, or if one of them becomes dependent somewhere, the pullback may fail to give coordinates. Smoothness also matters because the pulled-back sections must vary smoothly with $x$ through the map $f$. The theorem gives a practical test for pullbacks: a local computation in $E$ over $f(U)$ becomes a local computation in $f^*E$ over $U$. For example, if a line bundle has a nonzero local generator on $V\subset M$, then its pullback has the generator $x\mapsto s(f(x))$ on $f^{-1}(V)$. This explains why the exact transition functions of $f^*E$ should be obtained by substituting $f(x)$ into the transition functions of $E$. The frame statement gives local bases after pullback, but computations with bundles usually happen through transition matrices. We therefore need the corresponding cocycle formula: it should identify the overlap matrix for $f^*E$ directly from the overlap matrix for $E$, and it should make clear why pullback preserves the cocycle identities. [quotetheorem:6104] [citeproof:6104] This formula proves that cocycle identities are preserved under pullback, because composition with $f$ preserves products and identities of matrices. The smoothness of the original transition functions and of $f$ is needed to keep the pulled-back transition functions smooth; if either ingredient is not smooth, the cocycle may define only a topological bundle. The theorem does not say that the pulled-back bundle is less twisted, since composition with $f$ may preserve, kill, or duplicate the original twisting. It does give a practical test for that question, which is especially visible when $f$ is a covering map. [example: Pullback Of The Mobius Line Bundle To A Double Cover] Write $S^1=\mathbb R/\mathbb Z$, so the double cover is \begin{align*} p([t])=[2t]. \end{align*} Use the standard quotient model of the Mobius line bundle \begin{align*} L=([0,1]\times \mathbb R)/((1,a)\sim(0,-a)), \end{align*} equivalently on $\mathbb R\times \mathbb R$ with \begin{align*} (u+1,a)\sim(u,-a). \end{align*} Thus going once around the base multiplies the fibre coordinate by $-1$. The pullback fibre over $[t]\in S^1$ is $L_{[2t]}$. Define \begin{align*} \Theta:S^1\times \mathbb R\to p^*L,\qquad \Theta([t],a)=([t],[2t,a]). \end{align*} This is well-defined in the base variable because replacing $t$ by $t+1$ gives \begin{align*} [2(t+1),a]=[2t+2,a]\sim[2t+1,-a]\sim[2t,a]. \end{align*} The two applications of the Mobius identification contribute the factors \begin{align*} (-1)(-1)=1, \end{align*} so the fibre coordinate returns to $a$ rather than to $-a$. For each fixed $[t]$, the map \begin{align*} \Theta_{[t]}:\mathbb R\to (p^*L)_{[t]},\qquad a\mapsto ([t],[2t,a]) \end{align*} is linear, since \begin{align*} \Theta_{[t]}(\alpha a+\beta b)=([t],[2t,\alpha a+\beta b]) =\alpha([t],[2t,a])+\beta([t],[2t,b]). \end{align*} It is bijective because every element of $(p^*L)_{[t]}$ has the form $([t],[2t,a])$ for a unique $a\in\mathbb R$. In the quotient coordinates above, $\Theta$ and its fibrewise inverse are given by the displayed coordinate formulas, so they are smooth. Therefore $p^*L\cong S^1\times\mathbb R$ as a line bundle. Pulling the Mobius bundle back along the double cover trivializes it because the original sign change is encountered twice. [/example] Transition functions describe coordinate changes, while metrics describe fibrewise measurement. The preceding example concerns twisting, but geometric applications also require lengths and angles to pull back. Since the fibre of $f^*E$ over $x$ is the fibre of $E$ over $f(x)$, the next definition transports a metric by evaluating it at $f(x)$. [definition: Pullback Bundle Metric] Let $E\to N$ be a smooth real vector bundle with smooth bundle metric $h$, and let $f:M\to N$ be smooth. The pullback metric $f^*h$ on $f^*E$ is defined fibrewise by \begin{align*} (f^*h)_x((x,e_1),(x,e_2))=h_{f(x)}(e_1,e_2). \end{align*} [/definition] This is positive definite because $h_{f(x)}$ is positive definite on $E_{f(x)}$. In a pulled-back local frame, its matrix is the original metric matrix composed with $f$. [example: Pullback Of A Riemannian Metric] Let $g$ be a Riemannian metric on $N$ and let $f:M\to N$ be an immersion. The differential defines a vector bundle morphism from $TM$ to $f^*TN$ by sending each $u\in T_xM$ to $(x,df_xu)$. Use this map to define \begin{align*} (f^\sharp g)_x(u,v)=g_{f(x)}(df_xu,df_xv),\qquad u,v\in T_xM. \end{align*} For $a,b\in \mathbb R$ and $u_1,u_2,v\in T_xM$, linearity of $df_x$ and bilinearity of $g_{f(x)}$ give \begin{align*} (f^\sharp g)_x(au_1+bu_2,v)=g_{f(x)}(a\,df_xu_1+b\,df_xu_2,df_xv)=a(f^\sharp g)_x(u_1,v)+b(f^\sharp g)_x(u_2,v). \end{align*} The same argument in the second variable proves bilinearity. Symmetry follows from symmetry of $g_{f(x)}$: \begin{align*} (f^\sharp g)_x(u,v)=g_{f(x)}(df_xu,df_xv)=g_{f(x)}(df_xv,df_xu)=(f^\sharp g)_x(v,u). \end{align*} If $u\neq 0$, then $df_xu\neq 0$ because $f$ is an immersion, so positive definiteness of $g_{f(x)}$ gives \begin{align*} (f^\sharp g)_x(u,u)=g_{f(x)}(df_xu,df_xu)>0. \end{align*} Also, \begin{align*} (f^\sharp g)_x(0,0)=g_{f(x)}(0,0)=0. \end{align*} In local coordinates $x^1,\dots,x^m$ on $M$ and $y^1,\dots,y^n$ on $N$, write $f^\alpha=y^\alpha\circ f$ and \begin{align*} g=\sum_{\alpha,\beta=1}^n g_{\alpha\beta}(y)\,dy^\alpha dy^\beta. \end{align*} For each coordinate vector, \begin{align*} df_x\left(\frac{\partial}{\partial x^i}\right)=\sum_{\alpha=1}^n \frac{\partial f^\alpha}{\partial x^i}(x)\frac{\partial}{\partial y^\alpha}\bigg|_{f(x)}. \end{align*} Therefore \begin{align*} (f^\sharp g)_{ij}(x)=\sum_{\alpha,\beta=1}^n g_{\alpha\beta}(f(x))\frac{\partial f^\alpha}{\partial x^i}(x)\frac{\partial f^\beta}{\partial x^j}(x). \end{align*} The functions $g_{\alpha\beta}\circ f$ and $\partial f^\alpha/\partial x^i$ are smooth, so the coefficients $(f^\sharp g)_{ij}$ are smooth. Hence $f^\sharp g$ is a Riemannian metric on $M$: it measures tangent vectors on $M$ by first pushing them forward into $T_{f(x)}N$ and then using $g$ there. [/example] The notation $f^\sharp g$ distinguishes the induced metric on $TM$ from the pullback bundle metric on $f^*TN$. The two are related by the bundle morphism $df:TM\to f^*TN$. ## Naturality Of Vector-Bundle Operations The final problem is to verify that pullback commutes with the standard operations on vector bundles. This matters because it allows constructions made from bundles to be transported along maps without rebuilding them from scratch. Without naturality, a construction such as $E\oplus F$ would have two competing pullbacks, $f^*(E\oplus F)$ and $f^*E\oplus f^*F$, and later formulae involving metrics, characteristic classes, or associated bundles would depend on arbitrary choices of identification. The next theorem removes that ambiguity for the first algebraic operations. [quotetheorem:6105] [citeproof:6105] Direct sums and tensor products are the first algebraic operations on bundles. The fibrewise hypotheses in the theorem are essential: the two summands or tensor factors must lie over the same point of $N$, since otherwise there is no canonical fibrewise sum or tensor product to pull back. The theorem gives canonical isomorphisms, not literal equalities of total spaces, so the notation $f^*(E\oplus F)=f^*E\oplus f^*F$ is shorthand for an identified pair of bundles. The same naturality question arises for duals and exterior powers, which produce covectors and alternating tensors and are needed for differential forms and characteristic classes. [quotetheorem:6106] [citeproof:6106] These naturality isomorphisms justify writing pullbacks as if they commute with algebraic bundle operations. The bound $0\le k\le r$ records that exterior powers above the rank vanish under the usual convention, while the dual statement depends on the fibrewise identification $(f^*E)_x=E_{f(x)}$. The theorem does not choose bases of covectors or alternating tensors; it identifies the corresponding bundles by their universal fibrewise constructions. This is the mechanism behind pulling back differential forms and comparing characteristic classes along smooth maps. [example: Normal Bundles Under Embeddings] Let $i:S\hookrightarrow M$ be an embedded submanifold and let $g$ be a Riemannian metric on $M$. The restricted tangent bundle is the pullback bundle \begin{align*} i^*TM=\{(s,v)\in S\times TM:i(s)=\pi_{TM}(v)\}. \end{align*} Since $i(s)=s$, an element of the fibre $(i^*TM)_s$ has the form $(s,v)$ with $v\in T_sM$. The differential gives a vector bundle morphism \begin{align*} di:TS\to i^*TM,\qquad di(s,w)=(s,di_s(w)), \end{align*} because \begin{align*} \pi_{i^*TM}(s,di_s(w))=s=\pi_{TS}(s,w). \end{align*} For each $s\in S$, define \begin{align*} \nu_s(S)=\{v\in T_sM:g_s(v,di_s(w))=0 \text{ for all } w\in T_sS\}. \end{align*} This is a vector subspace of $T_sM$. Indeed, if $v_1,v_2\in \nu_s(S)$ and $a,b\in\mathbb R$, then for every $w\in T_sS$, bilinearity of $g_s$ gives \begin{align*} g_s(av_1+bv_2,di_s(w))=a\,g_s(v_1,di_s(w))+b\,g_s(v_2,di_s(w)). \end{align*} Since $v_1,v_2\in \nu_s(S)$, the two terms on the right are $a\cdot 0$ and $b\cdot 0$, so \begin{align*} g_s(av_1+bv_2,di_s(w))=a\cdot 0+b\cdot 0=0. \end{align*} Thus $av_1+bv_2\in \nu_s(S)$. Since $i$ is an embedding, $di_s:T_sS\to T_sM$ is injective, so $di_s(T_sS)$ is a $\dim S$-dimensional subspace of $T_sM$. With respect to the positive-definite inner product $g_s$, its orthogonal complement has dimension $\dim M-\dim S$, and \begin{align*} T_sM=di_s(T_sS)\oplus \nu_s(S). \end{align*} The normal bundle is therefore \begin{align*} \nu(S)=\{(s,v)\in i^*TM:v\in \nu_s(S)\}. \end{align*} Equivalently, \begin{align*} \nu(S)=\{(s,v)\in i^*TM:g_s(v,di_s(w))=0 \text{ for all } w\in T_sS\}. \end{align*} It is naturally a subbundle of the pullback bundle $i^*TM\to S$, whose fibre over $s$ is $T_sM$ with the base point recorded as $s$; it is not a subbundle of $TM\to M$ over all of $M$, because its fibres are defined only along points of $S$. [/example] The examples above show that pullback is compatible with algebraic operations and with restriction to submanifolds. A final coherence question remains: if the base is changed in two stages, the resulting bundle should agree with the bundle obtained from the composite base map. Without this statement, pulling a bundle back from $P$ to $N$ and then to $M$ could produce notation and transition data unrelated to the direct pullback along $g\circ f$, which would make covering-space computations and induced submanifold geometry depend on the chosen route. Functoriality says that the two-stage and one-stage constructions carry the same geometric information. [quotetheorem:6107] [citeproof:6107] The smoothness hypotheses on $f$ and $g$ ensure that the induced transition functions remain smooth after one or two pullbacks. The theorem is canonical but still an isomorphism statement: the two total spaces are presented with different parenthesized data, and the displayed map identifies them by forgetting the intermediate base label. This is the final coherence needed for the chapter's guiding principle: maps of bases induce pullback operations on bundles, and these operations respect the fibrewise algebra. Associated bundles in Chapter 10 and later courses on connections and characteristic classes rely on this naturality to compare geometric objects across different manifolds. Pullbacks and bundle maps make bundle theory functorial: geometry on one base can be compared systematically with geometry on another. The next chapter exploits this naturality to classify certain bundles by maps into universal parameter spaces, beginning with the simplest cases. # 7. Classifying Maps in Elementary Cases Chapters 1 through 6 developed vector bundles through local trivialisations, transition functions, sections, pullbacks, and bundle maps. This chapter turns those constructions into elementary classification results: some bundles can be recovered from maps into a universal parameter space. The guiding case is a rank-$k$ subbundle of a product bundle $X \times \mathbb R^n$, whose fibre over $x$ is a $k$-plane in $\mathbb R^n$ and hence determines a map from $X$ to a Grassmannian. The second theme is finite generation by global sections. A finite list of sections gives a map from a product bundle onto the given bundle, and under constant-rank hypotheses its kernel becomes a subbundle. The chapter ends with homotopy invariance of pullbacks, which explains why classification by maps is naturally insensitive to deforming the classifying map through a homotopy. ## Grassmannian Tautological Bundles and Subbundles of Product Bundles How can a family of vector subspaces of a fixed vector space be recorded as a single geometric object? If each fibre $E_x \subset \mathbb R^n$ is a $k$-plane and varies smoothly with $x$, then the point $x$ should determine the corresponding point of the Grassmannian. The classification statement in this section makes that sentence precise by comparing subbundles of $X \times \mathbb R^n$ with maps into $\operatorname{Gr}(k,n)$. The Grassmannian is the parameter space for possible fibres. Before any universal bundle can be built, we need a base space whose points name the $k$-planes that may occur as fibres of a subbundle. [definition: Grassmannian] Let $\operatorname{Gr}(k,n)$ be the smooth manifold whose points are $k$-dimensional linear subspaces $P \subset \mathbb R^n$. [/definition] The definition gives the base of the universal example: it records which plane is chosen, but not the vectors belonging to that plane. The next construction is needed because pullback classification requires an actual vector bundle over the parameter space, not only the parameter space itself. [definition: Tautological Bundle] The tautological rank-$k$ bundle over $\operatorname{Gr}(k,n)$ is the subset \begin{align*} \gamma_k^n := \{(P,v) \in \operatorname{Gr}(k,n) \times \mathbb R^n : v \in P\} \end{align*} with projection $\pi: \gamma_k^n \to \operatorname{Gr}(k,n)$ given by $\pi(P,v)=P$. [/definition] This bundle is tautological because the fibre over $P$ is the plane $P$ itself. Local coordinates on $\operatorname{Gr}(k,n)$, obtained by representing nearby planes as graphs of linear maps $P \to P^\perp$, give local trivializations of $\gamma_k^n$. [example: Tautological Line Over Projective Space] For $k=1$, a point of $\operatorname{Gr}(1,n)$ is a one-dimensional subspace $\ell\subset \mathbb R^n$, which is the same datum as a point of $\mathbb RP^{n-1}$. The tautological bundle is \begin{align*} \gamma_1^n=\{(\ell,v)\in \mathbb RP^{n-1}\times \mathbb R^n : v\in \ell\}, \end{align*} so its fibre over $\ell$ is \begin{align*} \pi^{-1}(\ell) &=\{(\ell,v):v\in \ell\}. \end{align*} After identifying this fibre with its second coordinate, the fibre is exactly the line $\ell$. Consider the standard chart $U_1\subset \mathbb RP^{n-1}$ consisting of lines whose vectors have nonzero first coordinate. If $\ell\in U_1$ and $w=(w_1,\dots,w_n)\in \ell$ has $w_1\ne 0$, then \begin{align*} \frac{1}{w_1}w &=\left(1,\frac{w_2}{w_1},\dots,\frac{w_n}{w_1}\right). \end{align*} Thus $\ell$ has a spanning vector of the form $(1,a_2,\dots,a_n)$. This vector is unique: if both $(1,a_2,\dots,a_n)$ and $(1,b_2,\dots,b_n)$ span $\ell$, then for some nonzero scalar $\lambda$, \begin{align*} (1,b_2,\dots,b_n) &=\lambda(1,a_2,\dots,a_n). \end{align*} Comparing first coordinates gives $1=\lambda$, and then comparing the remaining coordinates gives $b_i=a_i$ for every $i=2,\dots,n$. The local trivialization over $U_1$ sends \begin{align*} (\ell,t(1,a_2,\dots,a_n))\longmapsto (\ell,t), \end{align*} where $(1,a_2,\dots,a_n)$ is the unique normalized spanning vector of $\ell$. Its inverse is \begin{align*} (\ell,t)\longmapsto (\ell,t(1,a_2,\dots,a_n)). \end{align*} These two formulas compose to the identity in both directions, so over this chart the tautological line bundle is identified with $U_1\times \mathbb R$. [/example] The example shows the intended universal behaviour in the special case of lines: pulling back the tautological line over projective space gives a line over each point of the source. This motivates the following theorem, which is needed to pass from the model line example to an exact classification of all rank-$k$ subbundles of a fixed product bundle. [quotetheorem:6108] [citeproof:6108] The theorem says that a subbundle is the same information as its fibrewise position inside the ambient product bundle. The hypotheses matter: the fibres must have constant dimension $k$ and must vary smoothly, since an arbitrary assignment $x\mapsto P_x\subset \mathbb R^n$ need not define either a smooth map to the Grassmannian or a vector subbundle. For instance, a family of kernels whose dimension jumps cannot be locally trivial as a vector bundle, because local triviality forces the fibre dimension to be locally constant. The theorem also does not classify all rank-$k$ bundles on $X$ at once, only those already realised as subbundles of a specified product bundle $X\times \mathbb R^n$. The next section explains how finite sets of global sections can place more bundles into this framework by presenting them as quotients of product bundles with Grassmannian-classified kernels. [example: A Line Subbundle From A Projective Map] Write a point of $S^1$ as $e^{i\theta}$, with $\theta$ understood modulo $2\pi$, and define \begin{align*} f(e^{i\theta})=\operatorname{span}\left(\cos\frac{\theta}{2},\sin\frac{\theta}{2}\right)\subset \mathbb R^2. \end{align*} This is well-defined as a map to $\mathbb RP^1$: replacing $\theta$ by $\theta+2\pi$ changes the spanning vector to \begin{align*} \left(\cos\frac{\theta+2\pi}{2},\sin\frac{\theta+2\pi}{2}\right)=\left(-\cos\frac{\theta}{2},-\sin\frac{\theta}{2}\right). \end{align*} Multiplying a nonzero vector by $-1$ does not change the one-dimensional subspace it spans, so the unoriented line is unchanged. The pullback $f^*\gamma_1^2$ has fibre over $e^{i\theta}$ equal to the line selected by $f$: \begin{align*} (f^*\gamma_1^2)_{e^{i\theta}}=\left\{a\left(\cos\frac{\theta}{2},\sin\frac{\theta}{2}\right):a\in \mathbb R\right\}. \end{align*} On the arc with coordinate $\theta\in(-\pi,\pi)$, choose the local generator \begin{align*} u_0(\theta)=\left(\cos\frac{\theta}{2},\sin\frac{\theta}{2}\right). \end{align*} On the arc with coordinate $\phi\in(0,2\pi)$, choose the local generator \begin{align*} u_1(\phi)=\left(\cos\frac{\phi}{2},\sin\frac{\phi}{2}\right). \end{align*} On the lower overlap, the same point has $\phi=\theta+2\pi$ with $\theta\in(-\pi,0)$, and therefore \begin{align*} u_1(\theta+2\pi)=-u_0(\theta). \end{align*} Thus the transition function on this overlap is multiplication by $-1$. Equivalently, as $\theta$ runs from $0$ to $2\pi$, the lifted generator runs from $(1,0)$ to $(-1,0)$, so the line bundle closes with a sign reversal; this is the Möbius line bundle. [/example] The line-bundle example shows why projective space appears naturally: a line subbundle of $X\times \mathbb R^n$ is determined by a line in $\mathbb R^n$ at every point of $X$. For higher rank, projective space is replaced by the Grassmannian of $k$-planes. ## Bundles Generated By Finitely Many Global Sections When is a vector bundle controlled by finitely many sections? A collection of global sections of a rank-$k$ bundle $E\to X$ gives, at each point, vectors in the fibre $E_x$. If these vectors span every fibre, then the bundle is a quotient of a product bundle, and if the number of generators is large enough, it can also be studied through a complementary subbundle of a product bundle. The first construction turns sections into a bundle map from a product bundle. It is the bundle analogue of a surjective linear map from a free vector space onto a finitely generated vector space. [definition: Fibrewise Generating Sections] Let $p:E\to X$ be a rank-$k$ vector bundle. Sections $s_1,\dots,s_n\in \Gamma(X,E)$ fibrewise generate $E$ if for every $x\in X$, the vectors $s_1(x),\dots,s_n(x)$ span the vector space $E_x$. [/definition] Fibrewise generation is stronger than having many sections locally: it asks for one fixed finite list defined on all of $X$. The associated map packages the list into a single morphism of vector bundles, so it is the object whose kernel and image can be studied by local linear algebra. [definition: Evaluation Map Of Sections] Let $s_1,\dots,s_n\in \Gamma(X,E)$ be global sections of a vector bundle $E\to X$. The evaluation map is the vector bundle map $\operatorname{ev}_s:X\times \mathbb R^n \longrightarrow E$ defined by \begin{align*} \operatorname{ev}_s(x,a_1,\dots,a_n) = \sum_{i=1}^n a_i s_i(x). \end{align*} [/definition] The rank of this map at $x$ is the dimension of the span of $s_1(x),\dots,s_n(x)$. This prepares the next theorem: if the sections generate every fibre, then the evaluation map has constant rank, and its kernel should vary as a vector bundle rather than as an arbitrary family of subspaces. [quotetheorem:6109] [citeproof:6109] This result reduces a finitely generated bundle to two pieces of linear algebra varying smoothly over $X$: a product source and a kernel subbundle. The fibrewise generating hypothesis is essential, because without it the evaluation map may fail to be surjective over points where the chosen sections do not span the fibre. Constant rank is equally important for the kernel statement: for example, the map $\mathbb R\times \mathbb R\to \mathbb R\times \mathbb R$ over $\mathbb R$ given on fibres by multiplication by $x$ has kernel dimension $1$ over $x=0$ and dimension $0$ over $x\ne 0$, so its kernel is not a vector subbundle of the product line bundle. Thus the theorem is not merely a pointwise quotient statement; it is a smooth constant-rank statement ensuring that the quotient data varies locally trivially. The kernel can now be classified by a Grassmannian map, and the original bundle is recovered as the quotient by that kernel. [example: Two Sections Generating A Line Bundle] Let $L\to X$ be a line bundle generated by global sections $s_1,s_2$. The evaluation map is the bundle map $\operatorname{ev}_s:X\times \mathbb R^2\to L$ defined by \begin{align*} \operatorname{ev}_s(x,a,b)=a\,s_1(x)+b\,s_2(x). \end{align*} For a fixed $x\in X$, the fibre map $(\operatorname{ev}_s)_x:\mathbb R^2\to L_x$ sends $(a,b)$ to $a\,s_1(x)+b\,s_2(x)$. Since $s_1(x)$ and $s_2(x)$ span $L_x$, this fibre map is surjective. Since $L_x$ is one-dimensional, rank-nullity gives \begin{align*} \dim \ker((\operatorname{ev}_s)_x)=\dim \mathbb R^2-\dim \operatorname{im}((\operatorname{ev}_s)_x)=2-1=1. \end{align*} Thus $K_x:=\ker((\operatorname{ev}_s)_x)$ is a line in $\mathbb R^2$ for every $x$, and $K=\ker(\operatorname{ev}_s)$ is a rank-$1$ subbundle of $X\times \mathbb R^2$ by the [Globally Generated Vector Bundles Are Quotients of Trivial Bundles](/theorems/6109) theorem. Each fibre $K_x\subset \mathbb R^2$ therefore determines a point of $\operatorname{Gr}(1,2)=\mathbb RP^1$, so the kernel defines the classifying map $g:X\to \mathbb RP^1$ given by $g(x)=K_x$. The quotient map identifies $(x,a,b)$ and $(x,a',b')$ exactly when $(a'-a,b'-b)\in K_x$. Because $\operatorname{ev}_s$ vanishes on $K$, it induces a fibrewise linear map $\overline{\operatorname{ev}}_s:(X\times \mathbb R^2)/K\to L$ by \begin{align*} \overline{\operatorname{ev}}_s([x,a,b])=a\,s_1(x)+b\,s_2(x). \end{align*} This is well-defined: if $(a'-a,b'-b)\in K_x$, then \begin{align*} a's_1(x)+b's_2(x)=a\,s_1(x)+b\,s_2(x)+(a'-a)s_1(x)+(b'-b)s_2(x)=a\,s_1(x)+b\,s_2(x). \end{align*} It is surjective on each fibre because $s_1(x),s_2(x)$ span $L_x$. It is injective on each fibre because $a\,s_1(x)+b\,s_2(x)=0$ means $(a,b)\in K_x$, so $[x,a,b]=[x,0,0]$ in the quotient. Hence \begin{align*} L\cong (X\times \mathbb R^2)/K. \end{align*} The two generating sections encode the line bundle by producing a kernel line subbundle of the product plane bundle, and that kernel is recorded by the map $g:X\to \mathbb RP^1$. [/example] The quotient description explains how finite generation turns an arbitrary bundle into a construction from a product bundle and a subbundle. The next question is when finite generation is available without being assumed; compactness and partitions of unity provide the standard answer for smooth bundles. [quotetheorem:6110] [citeproof:6110] Combined with the previous theorem, compactness lets us present any vector bundle over $X$ as a quotient of a product bundle by a Grassmannian-classified kernel. Compactness is the finiteness input: it turns the local frames from a trivializing cover into a finite list after passing to a finite subcover. On a noncompact base, the same partition-of-unity construction may produce infinitely many global sections, and finite generation is not automatic unless another finiteness hypothesis is supplied. The theorem therefore should not be read as saying that every smooth vector bundle over every smooth manifold has a finite global generating set by this argument. This is the first version of the classifying-space philosophy: under suitable finiteness assumptions, global bundles can be encoded by maps into spaces of linear data. ## Homotopy Invariance Of Pullback Bundles If a bundle is classified by a map, what happens when the map is deformed continuously or smoothly? A classification theory should not distinguish two classifying maps joined by a homotopy when the corresponding pullback bundles are isomorphic. The theorem in this section proves that pullbacks of a fixed bundle are invariant under homotopy of the pulling map, under the usual smooth hypotheses. We begin by naming the geometric situation. A homotopy is a map from a cylinder $X\times I$, and restricting it to the two ends gives the maps whose pullbacks we want to compare. [definition: Smooth Homotopy] Let $X$ and $Y$ be smooth manifolds, and let $I=[0,1]$. A smooth homotopy from $f_0:X\to Y$ to $f_1:X\to Y$ is a smooth map $F:X\times I\to Y$ such that $F(x,0)=f_0(x)$ and $F(x,1)=f_1(x)$ for all $x\in X$. [/definition] The pullback $F^*E$ is a bundle over the cylinder. To compare the two endpoint pullbacks, we need a way to move vectors in this cylinder bundle from time $0$ to time $1$ while preserving linear structure in each fibre. [quotetheorem:6111] [citeproof:6111] The isomorphism uses more structure than the statement remembers: a concrete construction depends on choices such as a connection, but its existence is independent of those choices. The paracompactness hypothesis is what guarantees that the required connection can be assembled from local data using a partition of unity; without such a hypothesis, the transport argument may have no global connection to start from. The smoothness hypothesis also matches the smooth category: a merely continuous homotopy does not by itself provide a smooth bundle over $X\times I$ to which smooth parallel transport can be applied. For classification problems, existence of some endpoint isomorphism is the important point, while the particular isomorphism produced by transport is not canonical. [example: Constant Homotopy Gives The Identity Pullback] Let $f:X\to Y$ be smooth, and let $F:X\times I\to Y$ be the constant homotopy $F(x,t)=f(x)$. By the definition of pullback, \begin{align*} F^*E=\{((x,t),e)\in (X\times I)\times E : q(e)=F(x,t)\}. \end{align*} Since $F(x,t)=f(x)$, this becomes \begin{align*} F^*E=\{((x,t),e)\in (X\times I)\times E : q(e)=f(x)\}. \end{align*} Also, \begin{align*} f^*E=\{(x,e)\in X\times E : q(e)=f(x)\}. \end{align*} Define $\Phi:F^*E\to f^*E\times I$ by \begin{align*} \Phi((x,t),e)=((x,e),t). \end{align*} This is well-defined because $((x,t),e)\in F^*E$ implies $q(e)=f(x)$, so $(x,e)\in f^*E$. Define $\Psi:f^*E\times I\to F^*E$ by \begin{align*} \Psi((x,e),t)=((x,t),e). \end{align*} This is well-defined because $(x,e)\in f^*E$ implies $q(e)=f(x)=F(x,t)$, so $((x,t),e)\in F^*E$. The two maps are inverse to each other. For $((x,t),e)\in F^*E$, \begin{align*} (\Psi\circ\Phi)((x,t),e)=\Psi((x,e),t)=((x,t),e). \end{align*} For $((x,e),t)\in f^*E\times I$, \begin{align*} (\Phi\circ\Psi)((x,e),t)=\Phi((x,t),e)=((x,e),t). \end{align*} Thus $F^*E$ is the bundle $f^*E$ carried constantly along the interval direction. At the endpoint $t=0$, the fibre over $(x,0)$ is \begin{align*} (F^*E)_{(x,0)}=\{((x,0),e):q(e)=f(x)\}. \end{align*} At the endpoint $t=1$, the fibre over $(x,1)$ is \begin{align*} (F^*E)_{(x,1)}=\{((x,1),e):q(e)=f(x)\}. \end{align*} Under the endpoint identifications $((x,0),e)\leftrightarrow (x,e)$ and $((x,1),e)\leftrightarrow (x,e)$, the transported vector is unchanged: \begin{align*} (x,e)\mapsto (x,e). \end{align*} So for the constant homotopy, the endpoint isomorphism $f^*E\to f^*E$ is the identity on every fibre. [/example] More interesting examples come from deforming a classifying map. The resulting pullback subbundles may sit differently inside a product bundle, while their abstract vector bundle type stays fixed. [example: Rotating A Classifying Map] Let $X=S^1$, let $f_0:S^1\to \mathbb RP^2$ classify a line subbundle of $S^1\times \mathbb R^3$, and write \begin{align*} f_0(x)=\ell_x\subset \mathbb R^3. \end{align*} Let $A:I\to SO(3)$ be a smooth path, with $A(t)=A_t$. The rotation $A_t$ acts on the line $\ell_x$ by \begin{align*} A_t\ell_x=\{A_t v:v\in \ell_x\}. \end{align*} This is again a line. Indeed, if $v\ne 0$ spans $\ell_x$, then every element of $\ell_x$ is $\lambda v$ for some $\lambda\in\mathbb R$, and linearity of $A_t$ gives \begin{align*} A_t(\lambda v)=\lambda A_t v. \end{align*} Therefore \begin{align*} A_t\ell_x=\{\lambda A_t v:\lambda\in\mathbb R\}=\operatorname{span}(A_t v). \end{align*} Since $A_t\in SO(3)$ is invertible, $A_t v\ne 0$, so $\operatorname{span}(A_t v)$ is a point of $\mathbb RP^2$. Define \begin{align*} f_t(x)=A_t f_0(x)=A_t\ell_x. \end{align*} Equivalently, $F:S^1\times I\to \mathbb RP^2$ is given by \begin{align*} F(x,t)=A_t\ell_x. \end{align*} The map $F$ is smooth because $x\mapsto \ell_x$ is smooth, $t\mapsto A_t$ is smooth, and the standard action $SO(3)\times\mathbb RP^2\to\mathbb RP^2$ is smooth. For each $t$, the pullback of the tautological line bundle is the subbundle \begin{align*} f_t^*\gamma_1^3=\{(x,w)\in S^1\times\mathbb R^3:w\in f_t(x)\}. \end{align*} Using $f_t(x)=A_t\ell_x$, this becomes \begin{align*} f_t^*\gamma_1^3=\{(x,w)\in S^1\times\mathbb R^3:w\in A_t\ell_x\}. \end{align*} Its fibre over $x$ is therefore \begin{align*} (f_t^*\gamma_1^3)_x=A_t\ell_x=\{A_t v:v\in\ell_x\}. \end{align*} Thus the fibre at time $t$ is obtained by rotating the original line $\ell_x$ inside $\mathbb R^3$. Since $F$ is a smooth homotopy from $f_0$ to $f_1$, *Homotopy Invariance Of Pullbacks* gives an isomorphism of line bundles \begin{align*} f_0^*\gamma_1^3\cong f_1^*\gamma_1^3. \end{align*} The embedded line over $x$ has changed from $\ell_x$ to $A_1\ell_x$, but the abstract pullback line bundle has not changed up to isomorphism. [/example] The theorem also explains why the Grassmannian classification above is naturally a classification by homotopy classes once the ambient dimension is sufficiently flexible. If two maps $X\to \operatorname{Gr}(k,n)$ are homotopic, their pullbacks of $\gamma_k^n$ are isomorphic rank-$k$ bundles. [remark: Homotopy Classes And Bundle Isomorphism] Homotopy invariance gives a well-defined assignment \begin{align*} [X,\operatorname{Gr}(k,n)] \longrightarrow \{\text{isomorphism classes of rank-}k\text{ bundles over }X\} \end{align*} by sending the homotopy class of $f$ to the isomorphism class of $f^*\gamma_k^n$. This is weaker than the earlier exact correspondence between embedded subbundles $E\subset X\times \mathbb R^n$ and actual maps $X\to \operatorname{Gr}(k,n)$. Passing to homotopy classes deliberately forgets the precise embedded position of the subbundle inside the product bundle and keeps only the abstract isomorphism type forced by homotopy invariance. [/remark] This chapter has moved from local transition data to classifying maps in the first elementary cases. A subbundle of a product bundle gives a map to a Grassmannian, a finitely generated bundle gives a quotient of a product bundle with Grassmannian-classified kernel, and homotopy of classifying maps produces isomorphic pullbacks. Later courses enlarge the Grassmannian to a stable classifying space so that arbitrary vector bundles over suitable bases are represented without fixing a finite ambient product bundle in advance. Classification by maps into a universal space shows that some bundles are determined by homotopy-theoretic data rather than by explicit charts. The next chapter turns to clutching constructions, where bundles are assembled from two product pieces glued along a boundary, giving a concrete model for this classification idea. # 8. Clutching Constructions Chapter 2 showed that transition functions encode a bundle once a cover has been chosen, and Chapter 7 showed how pullbacks from parameter spaces classify elementary families. This chapter studies the special case where the base is obtained by gluing two large pieces along a common boundary. In that situation most of the transition data collapses to a single map on the overlap, called a clutching function, and the topology of the bundle becomes a homotopy problem. The prerequisites for this chapter are vector bundles and bundle isomorphisms, transition functions for a chosen cover, homotopy of maps, and the degree or [winding number](/page/Winding%20Number) of a map $S^1\to S^1$. We will invoke the standard deformation retractions $GL(k,\mathbb C)\simeq U(k)$ and $GL^+(2,\mathbb R)\simeq SO(2)$ when needed. ## Gluing Bundles Over Two Closed Pieces Suppose a space or manifold is built from two pieces $A$ and $B$ whose intersection is a common boundary region. What data is needed to build a bundle over $A \cup B$ if bundles over $A$ and $B$ are already known? The answer is that the only missing datum is an identification of the two restricted bundles over $A \cap B$. This is the two-piece version of reconstruction from transition cocycles: instead of many charts and many transition functions, there is one overlap and one gluing isomorphism. [definition: Closed Bundle-Gluing Cover] Fix either the topological category or the smooth category. A closed bundle-gluing cover of a space or manifold $X$ is a pair of closed subspaces or submanifolds-with-boundary $A,B\subset X$ such that $X=A\cup B$, the intersection $C=A\cap B$ has neighbourhoods $U_A\subset A$ and $U_B\subset B$ identified with collar neighbourhoods $C\times[0,\varepsilon)$ on the two sides, and these collars glue to a neighbourhood $U\subset X$ of $C$. In the smooth category the collars and gluing maps are required to be smooth. [/definition] This cover condition is the local model that prevents gluing from being only a set-theoretic quotient. It says that near the overlap the base itself already looks like two product collars glued along their common boundary, so a fibrewise gluing map can be checked in product coordinates. Once this controlled overlap has been fixed, the remaining information is no longer about the shape of the base but about how the two already-existing bundles are identified above $C$; the next definition packages exactly that identification as the datum to be glued. [definition: Gluing Datum For Two Pieces] Fix either the topological category or the smooth category. Let $X=A\cup B$, where $A,B\subset X$ form a closed bundle-gluing cover with $C=A\cap B$. A gluing datum for rank $k$ real vector bundles consists of rank $k$ real vector bundles $E_A\to A$ and $E_B\to B$ in the chosen category, together with a vector bundle isomorphism in that category \begin{align*} \varphi:E_A|_{C}\longrightarrow E_B|_{C}. \end{align*} [/definition] The isomorphism $\varphi$ tells us when a vector over a point of $A\cap B$ in the $A$-bundle is to be regarded as the same vector as one in the $B$-bundle. This raises the basic existence question for gluing data: does the quotient of the two total spaces automatically carry a vector bundle structure, or can the [quotient topology](/page/Quotient%20Topology) destroy local product charts? The next theorem gives the construction needed for every clutching example below. [quotetheorem:6112] [citeproof:6112] The cover hypothesis is not decorative. For example, if two closed disks are attached along a topologist's sine-curve subset rather than along a collared boundary arc, the resulting quotient near the attachment can have no product neighbourhood model; gluing two product bundles across that subset need not produce local bundle charts. The theorem also does not say that the glued bundle is determined by the abstract restrictions alone; the specific isomorphism $\varphi$ is part of the data, and changing it can change the resulting bundle. In the manifold examples below this pathology is avoided because hemispheres meet along a smooth equator with collar neighbourhoods on both sides. This theorem is most useful when $E_A$ and $E_B$ are product bundles. Then all of the interesting information is carried by the overlap isomorphism. [example: Gluing Two Product Line Bundles] Let $X=A\cup B$, set $C=A\cap B$, and take $E_A=A\times\mathbb R$ and $E_B=B\times\mathbb R$. A gluing isomorphism over $C$ must preserve the base point and be linear on each one-dimensional fibre, so for each $x\in C$ it sends $(x,t)$ to $(x,g(x)t)$ for a unique scalar $g(x)\in\mathbb R^*$. Thus the gluing map has the form \begin{align*} \varphi(x,t)=(x,g(x)t), \end{align*} where $g:C\to\mathbb R^*$ is continuous or smooth in the chosen category. Changing the trivialization on $A$ by a nowhere-zero function $a:A\to\mathbb R^*$ replaces the fibre coordinate $t$ by $t'=a(x)t$, and changing the trivialization on $B$ by $b:B\to\mathbb R^*$ replaces the fibre coordinate $u$ by $u'=b(x)u$. On the overlap, the original gluing relation is \begin{align*} u=g(x)t. \end{align*} Substituting $u=g(x)t$ into $u'=b(x)u$ gives \begin{align*} u'=b(x)g(x)t. \end{align*} Since $t'=a(x)t$, we have $t=a(x)^{-1}t'$, and therefore \begin{align*} u'=b(x)g(x)a(x)^{-1}t'. \end{align*} So the new clutching function is \begin{align*} g'(x)=b(x)g(x)a(x)^{-1}. \end{align*} The glued line bundle is a product exactly when there are nowhere-zero functions $a$ on $A$ and $b$ on $B$ such that \begin{align*} b(x)g(x)a(x)^{-1}=1 \end{align*} for every $x\in C$, equivalently \begin{align*} g(x)=b(x)^{-1}a(x). \end{align*} In that case the new gluing relation is $u'=t'$, so the two product trivializations agree across $C$ and assemble into one global product trivialization over $X$. [/example] This example is the prototype for the rest of the chapter. A change of trivialization on either side multiplies the clutching function by functions extending across the two pieces, so the bundle is controlled not by a raw function but by its equivalence class under such changes. ## Clutching Functions On Spheres How does this construction simplify when the base is a sphere? The sphere $S^n$ is the union of two closed hemispheres, and each hemisphere is contractible, so every vector bundle over a hemisphere is isomorphic to a product bundle. The only remaining datum is therefore a map from the equator to a general linear group. [definition: Clutching Function] For the decomposition \begin{align*} S^n=D^n_+\cup D^n_-,\qquad D^n_+\cap D^n_-=S^{n-1}, \end{align*} a rank $k$ real clutching function is a continuous or smooth map \begin{align*} g:S^{n-1}\longrightarrow GL(k,\mathbb R). \end{align*} [/definition] The associated bundle $E_g\to S^n$ is obtained by gluing $D^n_+\times \mathbb R^k$ to $D^n_-\times \mathbb R^k$ through \begin{align*} (x,v)_+\sim (x,g(x)v)_-\qquad x\in S^{n-1}. \end{align*} For complex vector bundles the same construction uses $GL(k,\mathbb C)$ and fibres $\mathbb C^k$. This construction produces bundles from maps, but a classification requires knowing when two different maps produce the same bundle. The next theorem separates the real quotient by hemisphere trivializations from the ordinary homotopy classification available for complex bundles; the case $n=1$ has an extra endpoint effect because $S^0$ is disconnected. [illustration:hemisphere-clutching-construction] The preceding construction gives a bundle for every clutching function, and the next question is when this process identifies two clutching functions. The classification theorem answers this by measuring exactly how changing the two hemisphere trivializations modifies the equatorial map. [quotetheorem:6113] [citeproof:6113] The theorem converts a classification of bundles into a classification of maps, but for real bundles it is not the unquotiented homotopy set of maps. The quotient by constant changes matters even when the equator is connected: for instance, rank-one real clutching over $S^2$ uses maps $S^1\to\mathbb R^*$, whose ordinary homotopy classes remember whether the loop lies in the positive or negative component, but multiplication by the constant $-1$ identifies those two classes and all real line bundles over $S^2$ are products. If the pieces were not contractible, nontrivial bundles on the pieces would remain part of the gluing data. If $n=1$, the equator $S^0$ has two points, and changes of trivialization on the two intervals can alter the two endpoint values in a way that is not captured by ordinary homotopy of maps $S^0\to GL(k,\mathbb R)$. This is why the circle case is treated separately below rather than read directly from $[S^0,GL(k,\mathbb R)]$. [remark: Smooth And Continuous Clutching] For smooth vector bundles over smooth spheres, smooth clutching functions are used. The resulting smooth classification agrees with the continuous classification in these examples because continuous maps into the general linear group can be approximated and homotoped to smooth maps. [/remark] The rank-one cases already detect familiar geometry. A real line clutching map lands in $GL(1,\mathbb R)=\mathbb R^*$, while a complex line clutching map lands in $GL(1,\mathbb C)=\mathbb C^*$. ## Rank One Examples In Low Dimensions What does the clutching classification say in the first dimensions where non-product line bundles appear? For $S^1$ the equator is $S^0$, so a real clutching function records two nonzero real numbers up to homotopy. For $S^2$ the equator is $S^1$, so complex line bundles are governed by the winding number of a loop in $\mathbb C^*$. [example: Mobius Bundle From A Clutching Map] Write $S^1=D^1_+\cup D^1_-$, so the overlap is $S^0=\{-1,1\}$. A real line clutching map is a function $g:S^0\to\mathbb R^*$, and the glued bundle is obtained from $D^1_+\times\mathbb R$ and $D^1_-\times\mathbb R$ by imposing, for each endpoint $x\in\{-1,1\}$, the relation \begin{align*} (x,t)_+\sim (x,g(x)t)_-. \end{align*} If the trivialization on $D^1_-$ is changed by a nowhere-zero function $b:D^1_-\to\mathbb R^*$ while the trivialization on $D^1_+$ is kept fixed, the new clutching value at each endpoint is \begin{align*} g'(x)=b(x)g(x). \end{align*} Suppose first that $g(-1)$ and $g(1)$ have the same sign. Then $1/g(-1)$ and $1/g(1)$ also have the same sign, so they lie in the same path component of $\mathbb R^*$. Since $D^1_-$ is an interval, choose a continuous nowhere-zero function $b:D^1_-\to\mathbb R^*$ with \begin{align*} b(-1)=\frac{1}{g(-1)}. \end{align*} and \begin{align*} b(1)=\frac{1}{g(1)}. \end{align*} For this choice, \begin{align*} g'(-1)=\frac{1}{g(-1)}g(-1)=1. \end{align*} and \begin{align*} g'(1)=\frac{1}{g(1)}g(1)=1. \end{align*} Thus both endpoint identifications become $(x,t)_+\sim(x,t)_-$, so the two interval product trivializations glue to a product line bundle over $S^1$. Now suppose that $g(-1)$ and $g(1)$ have opposite signs. Then $1/g(-1)$ and $-1/g(1)$ have the same sign, so choose a continuous nowhere-zero function $b:D^1_-\to\mathbb R^*$ with \begin{align*} b(-1)=\frac{1}{g(-1)}. \end{align*} and \begin{align*} b(1)=-\frac{1}{g(1)}. \end{align*} Then \begin{align*} g'(-1)=\frac{1}{g(-1)}g(-1)=1. \end{align*} and \begin{align*} g'(1)=-\frac{1}{g(1)}g(1)=-1. \end{align*} The normalized gluing therefore preserves the fibre coordinate at one endpoint, \begin{align*} (-1,t)_+\sim(-1,t)_-, \end{align*} and reverses it at the other endpoint, \begin{align*} (1,t)_+\sim(1,-t)_-. \end{align*} This is the Mobius line bundle: after one trip around the circle, a nonzero fibre vector returns with the opposite orientation. [/example] This is the smallest clutching example, and it shows that the Mobius bundle comes from gluing two product line bundles with a negative transition on one end relative to the other. The classification question is whether any further real line bundles over $S^1$ can occur, or whether this relative sign is the only invariant. [quotetheorem:6114] [citeproof:6114] This theorem is the first place where the disconnected equator matters. A map $S^0\to\mathbb R^*$ has two independent signs, but the bundle does not remember both signs separately because the two interval trivializations may be changed. What survives is the relative sign encountered after going once around the circle, so the apparent four sign choices collapse to two bundle classes. This also explains why the earlier homotopy-classification theorem was stated only for $n\ge 2$. A useful limitation is that this argument is special to a two-point equator. If one tried to classify real line bundles over $S^2$ by the unquotiented homotopy set $[S^1,\mathbb R^*]$, one would falsely obtain two classes, depending on whether the clutching loop lies in $\mathbb R_{>0}$ or $\mathbb R_{<0}$. Multiplication by a negative constant on one hemisphere identifies these two choices, so the extra sign is a choice of trivialization rather than a bundle invariant. The same bookkeeping issue will reappear in a more useful form for complex line bundles over $S^2$, where the target $\mathbb C^*$ is connected up to its winding data and the surviving invariant is an integer rather than a relative endpoint sign. Complex line bundles over $S^2$ are the next basic family. Here the classification is no longer finite, because loops in $\mathbb C^*$ have an integer winding number around the origin. [definition: Degree Clutching Map] For $m\in\mathbb Z$, the degree $m$ clutching map for a complex line bundle over $S^2$ is \begin{align*} g_m:S^1\longrightarrow \mathbb C^*,\qquad g_m(z)=z^m. \end{align*} [/definition] The integer $m$ measures how many times the equatorial transition function winds around $0\in\mathbb C$. The remaining classification problem is to decide whether two clutching loops with the same winding can still produce different complex line bundles. For rank one over $S^2$, the equator contains all the gluing data, and homotopy of loops in $\mathbb C^*$ is exactly measured by this winding number. [quotetheorem:6115] [citeproof:6115] The hypotheses here are doing real work. The target is $\mathbb C^*$ because the fibres have complex rank one; replacing it by $GL(k,\mathbb C)$ for higher rank bundles gives a different homotopy problem, and replacing $S^2$ by a general base space usually requires more than one clutching map. The puncture at $0$ is also essential: if the transition function were allowed to land in $\mathbb C$, the loop $z\mapsto z$ would contract through the homotopy $z\mapsto tz$ as $t$ moves to $0$, but that homotopy passes through non-invertible fibre maps and therefore cannot define bundle transition functions. Thus the integer invariant is not a generic feature of all line-bundle gluings, but a consequence of the equator $S^1$ and the topology of the nonzero complex numbers. This distinction points forward to characteristic classes. Winding number is a clutching invariant that later becomes the first Chern class of a complex line bundle over $S^2$. The tangent-bundle example below is the real rank-two analogue: the invariant is again a degree, but now the loop comes from comparing tangent frames rather than from multiplying complex line fibres. This integer classification links the clutching picture to the standard line bundles on complex projective space. The next example identifies the tautological line bundle on $\mathbb C P^1$ as a generator, showing that the abstract winding number is visible in projective coordinate changes. [example: The Tautological Complex Line Bundle On Projective Space] Identify $\mathbb C P^1$ with the two affine charts $U_0=\{[Z_0:Z_1]:Z_0\ne 0\}$ and $U_1=\{[Z_0:Z_1]:Z_1\ne 0\}$, with coordinates \begin{align*} w=\frac{Z_1}{Z_0} \end{align*} on $U_0$ and \begin{align*} \eta=\frac{Z_0}{Z_1} \end{align*} on $U_1$. On the overlap $U_0\cap U_1$, these coordinates satisfy \begin{align*} \eta=\frac{1}{w}. \end{align*} The tautological line over $[Z_0:Z_1]$ is the one-dimensional subspace of $\mathbb C^2$ spanned by $(Z_0,Z_1)$, so over $U_0$ use the frame $(1,w)$ and over $U_1$ use the frame $(\eta,1)$. On the overlap, substituting $\eta=1/w$ gives \begin{align*} (\eta,1)=\left(\frac{1}{w},1\right). \end{align*} Factoring out $1/w$ gives \begin{align*} \left(\frac{1}{w},1\right)=\frac{1}{w}(1,w). \end{align*} Thus \begin{align*} (\eta,1)=\frac{1}{w}(1,w). \end{align*} If a vector in the tautological line is written as $v=a(1,w)$ in the $U_0$ frame and as $v=b(\eta,1)$ in the $U_1$ frame, then \begin{align*} a(1,w)=b(\eta,1). \end{align*} Using the frame relation above, this becomes \begin{align*} a(1,w)=\frac{b}{w}(1,w). \end{align*} Since $(1,w)$ is a nonzero vector, equality of these scalar multiples gives \begin{align*} a=\frac{b}{w}. \end{align*} Multiplying by $w$ gives \begin{align*} b=wa. \end{align*} Therefore, with the clutching convention that sends the $U_0$ fibre coordinate $a$ to the $U_1$ fibre coordinate $b$, the equatorial transition function is \begin{align*} g(w)=w. \end{align*} Restricting to the equator $|w|=1$ and writing $w=z\in S^1$, this is \begin{align*} g(z)=z. \end{align*} If the opposite chart-order convention is used, the transition function is the inverse loop \begin{align*} g^{-1}(z)=z^{-1}. \end{align*} The loop $z\mapsto z$ has winding number $1$, while $z\mapsto z^{-1}$ has winding number $-1$. Hence the tautological complex line bundle over $\mathbb C P^1\cong S^2$ represents a generator of the integer classification of complex line bundles over $S^2$. [/example] This example connects clutching to the transition-function viewpoint from earlier chapters. The same bundle can be described either by projective coordinate charts or by a single equatorial clutching map. ## The Tangent Bundle Of The Two Sphere How can clutching recognise the tangent bundle of a sphere? The two hemispheres of $S^2$ are disks, so the tangent bundle restricted to either hemisphere is trivializable. The transition along the equator records how a tangent frame chosen in the northern hemisphere is compared with a tangent frame chosen in the southern hemisphere. [example: Tangent Bundle Of The Two Sphere] Use stereographic coordinates $z$ on the northern hemisphere and $w$ on the southern hemisphere, identifying each oriented tangent plane with $\mathbb C$ by the corresponding coordinate frame. On the equatorial overlap the coordinates satisfy \begin{align*} w=\frac{1}{z} \end{align*} and hence, equivalently, \begin{align*} z=\frac{1}{w}. \end{align*} Differentiating the second identity gives \begin{align*} \frac{dz}{dw}=-\frac{1}{w^2}. \end{align*} On the equator $|z|=1$ we have $w=1/z$, so \begin{align*} \frac{1}{w^2}=\frac{1}{(1/z)^2}=z^2, \end{align*} and therefore \begin{align*} \frac{dz}{dw}=-z^2. \end{align*} Thus a southern tangent coordinate $v$ is written in the northern frame as \begin{align*} v\longmapsto -z^2v. \end{align*} Writing $z=e^{i\theta}$, the clutching factor is \begin{align*} -z^2=-e^{2i\theta}=e^{i\pi}e^{2i\theta}=e^{i(2\theta+\pi)}. \end{align*} As $\theta$ runs from $0$ to $2\pi$, the angle $2\theta+\pi$ changes by \begin{align*} (2\cdot 2\pi+\pi)-(2\cdot 0+\pi)=4\pi. \end{align*} After the standard deformation retraction $GL^+(2,\mathbb R)\simeq SO(2)$, the constant factor $-1=e^{i\pi}$ does not change the degree, so the clutching loop is homotopic in $SO(2)\cong S^1$ to \begin{align*} z\longmapsto z^2. \end{align*} The tangent bundle $TS^2$ therefore has clutching degree $2$, recording that the equatorial comparison of tangent frames rotates through two full turns. [/example] [illustration:tangent-bundle-clutching-s2] The tangent bundle example is a useful warning: even when both pieces of the base are geometrically simple, the boundary comparison of frames can carry global curvature and characteristic-class information. To make this computation independent of the chosen picture, the result should be stated as a clutching degree rather than as a property of a particular drawing of frames. [quotetheorem:6116] [citeproof:6116] The orientation and rank-two hypotheses are essential in this statement. They allow the clutching loop to be reduced from $GL^+(2,\mathbb R)$ to $SO(2)$, where degree is defined; without an orientation-preserving comparison, the loop could move through the other component of $GL(2,\mathbb R)$ and degree in $SO(2)$ would no longer be the right invariant. The result is also special to $S^2$: the tangent bundle of $S^1$ is a product bundle, while higher-dimensional spheres have clutching maps landing in $SO(n)$ rather than in the circle group $SO(2)$. Thus the number $2$ is not a universal tangent-bundle clutching degree, but the particular equatorial frame-rotation invariant for the two-sphere. Clutching therefore packages a major theme of the course into one construction: local product structure is inexpensive on contractible pieces, while global geometry enters through the compatibility data on overlaps. This point of view prepares the transition to Chapter 11's elementary obstructions, and in later courses it becomes the bridge from transition functions to characteristic classes and $K$-theory. Clutching makes the overlap data visible in its simplest form: a bundle is encoded by how product pieces are attached along a common boundary. The next chapter shifts from building bundles to studying their sections, interpreting them as smooth objects whose local descriptions must satisfy the same compatibility conditions. # 9. Vector Bundles and Sheaves of Sections After the constructions and obstructions of Chapters 7 and 8, this chapter shifts from building vector bundles by local product charts to studying them through their smooth sections. It uses the construction of vector bundles from transition functions in Chapter 2, partitions of unity and metrics from Chapter 5, and local frames from Chapter 3. The algebra here is a compact dictionary, not a new prerequisite: product bundles give free modules of sections, subbundles give direct summands, and local frames give local bases. The central question is how much of the geometry remains visible after one records sections and restriction maps. For this course, the answer needed is concrete: section data remembers local product structure and gives another way to recognize familiar bundle constructions. ## Algebraic Language Used in This Chapter The ring $C^\infty(M)$ consists of smooth real-valued functions on $M$, with pointwise addition and multiplication. A module over this ring is like a vector space in which the scalars are smooth functions rather than real numbers. Thus a section can be multiplied by a different scalar at each point of the base. This is exactly the operation needed for vector bundles, because a smooth function $f$ and a section $s$ determine a new section $fs$ by multiplying the vector $s(p)$ in the fibre $E_p$ by the number $f(p)$. A free $C^\infty(M)$-module of rank $k$ is the module $(C^\infty(M))^k$. It has the standard basis of $k$ coordinate sections, just as the product bundle $M\times\mathbb R^k$ has its constant coordinate frame. A module is projective, in the only form needed here, if it is a direct summand of a free module: there is another module $Q$ such that \begin{align*} P\oplus Q\cong (C^\infty(M))^n. \end{align*} Geometrically, this means that the object behaves like the sections of a subbundle sitting inside a product bundle, with a complementary subbundle filling out the remaining directions. An idempotent is an algebraic way to describe such a summand. A matrix $P$ with $P^2=P$ is a projection: applying it twice is the same as applying it once. If the entries of $P$ are smooth functions on $M$, then $P(p)$ projects $\mathbb R^n$ onto a subspace for each point $p$. When the rank is locally constant, these image subspaces vary smoothly and form a vector bundle. Finally, a sheaf is only a bookkeeping device for local data in this chapter. It records which sections exist on each open set and how they restrict to smaller open sets. Saying that two descriptions are equivalent here means that one can translate from bundles to section data and back without losing the original object up to natural isomorphism; no additional category theory is needed for the uses below. ## Sections as a Module over Smooth Functions A vector bundle $\pi:E\to M$ has fibres that are vector spaces, but its global sections are not usually a vector space over a field large enough to remember the base. The correct scalar ring is $C^\infty(M)$, since a smooth function can scale a section fibrewise. This turns the space of sections into an algebraic object whose module structure remembers both pointwise linear algebra and variation over $M$. [definition: Smooth Section Module] Let $\pi:E\to M$ be a smooth real vector bundle. The smooth section module of $E$ is \begin{align*} \Gamma(M,E)=\{s:M\to E\mid \pi\circ s=\operatorname{id}_M,\ s\text{ is smooth}\}. \end{align*} For $f\in C^\infty(M)$ and $s\in \Gamma(M,E)$, the scalar multiple $fs\in \Gamma(M,E)$ is defined by \begin{align*} (fs)(p)=f(p)s(p),\qquad p\in M. \end{align*} [/definition] The module operation is fibrewise, so addition and scalar multiplication are tested after projecting to each fibre $E_p$. This viewpoint explains why sections of a product bundle look like vector-valued functions, which gives the first model for the general theory. [example: Sections of a Product Bundle] Let $E=M\times \mathbb R^k$ with projection $\pi(p,v)=p$. If $s\in \Gamma(M,E)$, then $\pi(s(p))=p$ for every $p\in M$, so there is a unique vector $u(p)\in \mathbb R^k$ such that \begin{align*} s(p)=(p,u(p)). \end{align*} Writing $u(p)=(u_1(p),\dots,u_k(p))$, the section is \begin{align*} s(p)=(p,(u_1(p),\dots,u_k(p))). \end{align*} Since this is the product bundle, $s$ is smooth exactly when the map $u:M\to \mathbb R^k$ is smooth, and this is equivalent to $u_i\in C^\infty(M)$ for each $i=1,\dots,k$. Define the correspondence by sending $s$ to its component tuple: \begin{align*} s\longmapsto (u_1,\dots,u_k). \end{align*} Its inverse sends a tuple of smooth functions to the section \begin{align*} (u_1,\dots,u_k)\longmapsto \bigl(p\mapsto (p,(u_1(p),\dots,u_k(p)))). \end{align*} These two assignments undo each other pointwise, so they identify $\Gamma(M,M\times\mathbb R^k)$ with $(C^\infty(M))^k$. Under this identification, if $f\in C^\infty(M)$ and $s(p)=(p,(u_1(p),\dots,u_k(p)))$, then fibrewise scalar multiplication gives \begin{align*} (fs)(p)=(p,f(p)(u_1(p),\dots,u_k(p))). \end{align*} Expanding the scalar product in $\mathbb R^k$ gives \begin{align*} (fs)(p)=(p,(f(p)u_1(p),\dots,f(p)u_k(p))). \end{align*} Thus multiplication by $f$ corresponds to \begin{align*} f\cdot (u_1,\dots,u_k)=(fu_1,\dots,fu_k). \end{align*} Hence $\Gamma(M,M\times\mathbb R^k)\cong (C^\infty(M))^k$ as $C^\infty(M)$-modules, giving the local algebraic model for sections of every rank $k$ vector bundle. [/example] The preceding example suggests that a non-product vector bundle should still become algebraically free after restricting to a small enough open set. To make this precise, sections must first be restricted to open subsets, so that local product charts can be read as local statements about modules. [definition: Restriction of Sections] Let $\pi:E\to M$ be a smooth vector bundle and let $U\subset M$ be open. The restriction map \begin{align*} \rho^M_U:\Gamma(M,E)\to \Gamma(U,E|_U) \end{align*} is defined by $\rho^M_U(s)=s|_U$. [/definition] Restriction is compatible with restriction of smooth functions, so the same bundle gives modules over the rings $C^\infty(U)$ for all open $U\subset M$. The algebraic question is whether the local product charts of a vector bundle are visible purely from sections. On a sufficiently small open set, a trivialization should provide finitely many sections that generate every other local section uniquely by smooth function coefficients. [quotetheorem:6117] [citeproof:6117] This theorem translates local product structure into local freeness, and the local product hypothesis is doing real work: the proof uses a product chart to manufacture a local frame. The conclusion is deliberately local. It does not say that $\Gamma(M,E)$ is globally free; global freeness would amount to a global frame, and many vector bundles do not admit one. The obstruction is not that sections fail to exist locally, but that the transition functions may prevent the locally chosen basis sections from agreeing on overlaps. The tautological line bundle gives the standard warning that local bases need not glue to a single global basis. [example: Nonfree Behaviour from the Tautological Line Bundle] Let $\gamma^1\to \mathbb{RP}^1$ be the tautological real line bundle, whose fibre over a line $\ell\subset \mathbb R^2$ is $\ell$ itself. We show that $\Gamma(\mathbb{RP}^1,\gamma^1)$ is not free of rank $1$ by proving that no global generator can exist. If $s$ were a global generator, then $s(\ell)\neq 0$ for every $\ell$. Indeed, if $s(\ell_0)=0$, choose a nonzero vector $y\in \ell_0$ and define a section $t$ near $\ell_0$ by orthogonally projecting $y$ onto nearby lines. Then $t(\ell_0)=y\neq 0$, but every multiple $fs$ satisfies \begin{align*} (fs)(\ell_0)=f(\ell_0)s(\ell_0)=f(\ell_0)\cdot 0=0, \end{align*} so $s$ could not generate $t$. Now identify $\mathbb{RP}^1$ as the quotient of $S^1$ by $x\sim -x$. Since $s([x])\in [x]$ and $s([x])\neq 0$, there is a unique scalar $a(x)\in \mathbb R\setminus\{0\}$ such that \begin{align*} s([x])=a(x)x. \end{align*} Because $[x]=[-x]$, the same vector also satisfies \begin{align*} s([-x])=a(-x)(-x). \end{align*} Thus \begin{align*} a(x)x=a(-x)(-x), \end{align*} and since $x\neq 0$, this gives \begin{align*} a(-x)=-a(x). \end{align*} The function $a:S^1\to \mathbb R\setminus\{0\}$ is continuous, so its sign is constant because $S^1$ is connected. But $a(-x)=-a(x)$ forces $a(x)$ and $a(-x)$ to have opposite signs, a contradiction. Therefore $\gamma^1$ has no nowhere-zero global section and $\Gamma(\mathbb{RP}^1,\gamma^1)$ is not free of rank $1$. Locally, however, $\gamma^1$ is a line bundle, so its section module is locally free of rank $1$; the obstruction is exactly that the local generators reverse sign around the circle and cannot glue to one global generator. [/example] ## Sheaves of Sections and Local Freeness The module of global sections is powerful, but local statements require a structure that remembers all open sets at once. The guiding problem is to encode restriction, gluing, and local bases in one object. Sheaves provide exactly this bookkeeping. [definition: Sheaf of Sections] Let $\pi:E\to M$ be a smooth vector bundle. The sheaf of smooth sections of $E$ is the assignment \begin{align*} U\longmapsto \Gamma(U,E|_U) \end{align*} for each open set $U\subset M$, together with the restriction maps $\rho^U_V:\Gamma(U,E|_U)\to\Gamma(V,E|_V)$ for open subsets $V\subset U$. [/definition] Because smooth maps can be checked locally and glued from compatible local data, this assignment satisfies the sheaf axioms. The next algebraic property needed is the sheaf-level version of having a local frame, since this is the condition that should correspond to vector bundles. [definition: Locally Free Sheaf of Modules] Let $M$ be a smooth manifold and let $\mathcal{C}^\infty_M$ be the sheaf $U\mapsto C^\infty(U)$. A sheaf $\mathcal F$ of $\mathcal{C}^\infty_M$-modules is locally free of rank $k$ if every point $p\in M$ has an open neighbourhood $U\subset M$ such that \begin{align*} \mathcal F|_U \cong (\mathcal{C}^\infty_M|_U)^k \end{align*} as sheaves of $\mathcal{C}^\infty_M|_U$-modules. [/definition] This definition is the sheaf-theoretic version of having a local frame. The isomorphism supplies $k$ local sections that generate all smaller-open-set sections uniquely over smooth functions, so the section sheaf of a vector bundle should satisfy it. [quotetheorem:6118] [citeproof:6118] The theorem shows that vector bundles give locally free sheaves, with the rank fixed by the rank of the original bundle. This is a local statement, not a global product-bundle statement: a locally free sheaf of rank $k$ need not be isomorphic to $(\mathcal{C}^\infty_M)^k$ on all of $M$, just as a vector bundle can have local frames without admitting a global frame. Both parts matter. If the rank were allowed to vary from point to point, there would be no single fibre dimension for the reconstructed object. If local freeness failed, then near some point there would be no stable local basis of sections, so the data would not look like a product bundle in a neighbourhood of that point. The converse problem is now forced: if a sheaf has local bases of fixed rank, then changes between those bases should be transition matrices, and those matrices should reconstruct a vector bundle. [quotetheorem:6119] [citeproof:6119] This result shows that vector bundles and locally free sheaves of fixed rank are two languages for the same local data. The fixed-rank local basis condition is essential because the transition functions must take values in one group $GL(k,\mathbb R)$ on every overlap. A sheaf that is only locally generated, or whose minimal number of generators jumps, does not produce invertible transition matrices of a fixed size. For example, data that needs one local generator near one point but no generator on nearby regions cannot describe the sections of a rank-one bundle, because a line bundle has one-dimensional fibres everywhere and local bases on every sufficiently small neighbourhood. Such data may still be useful algebraically, but it is not the section sheaf of a vector bundle. The geometric language emphasizes total spaces and fibres, while the sheaf language emphasizes sections and restrictions. The next section asks what happens when we forget the sheaf and keep only the global module. ## Projective Modules from Compact Manifolds Local freeness is not yet the same as projectivity of the global module. The next question is why compactness of the base allows local finite data to become a finite global algebraic statement. Partitions of unity are the bridge: they let local frames be patched into finitely many global generators. [definition: Finitely Generated Projective Module] Let $R$ be a commutative ring. An $R$-module $P$ is finitely generated projective if there exists $n\in\mathbb N$ and an $R$-module $Q$ such that \begin{align*} P\oplus Q\cong R^n. \end{align*} [/definition] Projective modules are the algebraic analogue of direct summands of product bundles. Local frames show that section modules look free in small neighbourhoods, but projectivity is a global finite statement. The obstruction is that infinitely many local generators would not give a finitely generated module; compactness and partitions of unity are what turn local frames into finite global algebraic data. [quotetheorem:6120] [citeproof:6120] The statement reflects both hypotheses: compactness supplies a finite cover, and smoothness supplies partitions of unity and fibrewise inner products. Without compactness, the section module is still locally free, but finite generation may fail. The direct-summand picture is worth isolating because it is the geometric form of projectivity. [example: A Direct Summand from a Subbundle] Let $E\subset M\times\mathbb R^n$ be a smooth rank $k$ subbundle, and write the chosen smooth fibrewise inner product on $\{p\}\times\mathbb R^n$ as $\langle -,-\rangle_p$. Define the orthogonal complement fibre by fibre: \begin{align*} E^\perp_p=\{v\in \mathbb R^n\mid \langle v,w\rangle_p=0\text{ for every }w\in E_p\}. \end{align*} Locally choose a smooth frame $s_1,\dots,s_k$ for $E$. Then $v\in E^\perp_p$ exactly when $\langle v,s_i(p)\rangle_p=0$ for every $i=1,\dots,k$. Completing $s_1,\dots,s_k$ locally to a smooth frame of the product bundle and applying fibrewise Gram-Schmidt produces a smooth local frame whose last $n-k$ vectors span $E^\perp_p$, so $E^\perp$ is a smooth rank $n-k$ subbundle. For each $p\in M$, finite-dimensional inner product linear algebra gives \begin{align*} \mathbb R^n=E_p\oplus E^\perp_p. \end{align*} Indeed, $E_p\cap E^\perp_p=\{0\}$, and \begin{align*} \dim E_p+\dim E^\perp_p=k+(n-k)=n. \end{align*} Thus the product bundle splits fibrewise as \begin{align*} M\times\mathbb R^n=E\oplus E^\perp. \end{align*} Taking sections gives a $C^\infty(M)$-linear map \begin{align*} \Phi:\Gamma(M,E)\oplus\Gamma(M,E^\perp)\to \Gamma(M,M\times\mathbb R^n) \end{align*} defined by $\Phi(s,t)=s+t$. Linearity follows because, for $f\in C^\infty(M)$, \begin{align*} \Phi(f(s,t))=fs+ft=f(s+t)=f\Phi(s,t). \end{align*} The inverse sends a product-bundle section $u$ to its fibrewise orthogonal components: \begin{align*} \Psi(u)=(\operatorname{pr}_E u,\operatorname{pr}_{E^\perp}u). \end{align*} The projections are smooth because, in a local smooth orthonormal frame adapted to $E\oplus E^\perp$, they are represented by constant diagonal projection matrices. Hence \begin{align*} \Phi(\Psi(u))=\operatorname{pr}_E u+\operatorname{pr}_{E^\perp}u=u, \end{align*} and, if $s(p)\in E_p$ and $t(p)\in E^\perp_p$ for every $p$, \begin{align*} \Psi(\Phi(s,t))=\Psi(s+t)=(s,t). \end{align*} Therefore \begin{align*} \Gamma(M,E)\oplus\Gamma(M,E^\perp)\cong \Gamma(M,M\times\mathbb R^n). \end{align*} A section of $M\times\mathbb R^n$ is uniquely a smooth map of the form \begin{align*} p\longmapsto (p,(u_1(p),\dots,u_n(p))) \end{align*} with $u_i\in C^\infty(M)$. Hence \begin{align*} \Gamma(M,M\times\mathbb R^n)\cong (C^\infty(M))^n. \end{align*} Combining the two identifications gives \begin{align*} \Gamma(M,E)\oplus\Gamma(M,E^\perp)\cong (C^\infty(M))^n. \end{align*} Thus $\Gamma(M,E)$ is projective, with complementary module $\Gamma(M,E^\perp)$; geometrically, projectivity says that the section module of a subbundle is a direct summand of the free module of sections of a product bundle. [/example] ## Idempotents and Subbundles of Product Bundles Projective modules can be represented by idempotent endomorphisms of free modules. The geometric version is a smooth field of projection matrices. The problem is to translate an algebraic equation $P^2=P$ into a family of vector subspaces varying smoothly over $M$. [definition: Smooth Idempotent Matrix] Let $M$ be a smooth manifold. A smooth idempotent matrix of size $n$ on $M$ is a smooth map \begin{align*} P:M\to M_n(\mathbb R) \end{align*} such that $P(p)^2=P(p)$ for every $p\in M$. [/definition] Here $M_n(\mathbb R)$ denotes the set of $n\times n$ real matrices, viewed as a finite-dimensional real vector space. Thus smoothness of $P$ means smoothness of each matrix entry as a real-valued function on $M$. At each point, an idempotent matrix is a projection onto its image. Pointwise images alone do not automatically form a bundle, because their dimensions and local coordinates could vary badly from point to point. The obstruction is local triviality. Even if every $\operatorname{im}P(p)$ is a vector subspace of $\mathbb R^n$, these subspaces must have a fixed dimension near each point and admit smoothly varying local frames. Smoothness of the matrix entries controls the variation of the projection in coordinates, while constant rank prevents the fibre dimension from jumping. Under exactly these conditions, the images can be organized as the fibres of a smooth vector bundle over $M$. [quotetheorem:6121] [citeproof:6121] The constant-rank hypothesis is essential. If the rank of $P(p)$ jumps, the dimensions of the images $\operatorname{im}P(p)$ jump as well, so the set of images cannot be locally isomorphic to $U\times\mathbb R^k$ for any fixed $k$. For example, on the disconnected manifold $M=(-\infty,0)\cup(0,\infty)$ the smooth idempotent $P(x)=0$ on the negative component and $P(x)=1$ on the positive component has fibres of ranks $0$ and $1$; more seriously, on any space where such a jump occurs at an accumulation point, no vector bundle structure of fixed rank can exist across the jump. The associated module is the image of the idempotent endomorphism of the free module $(C^\infty(M))^n$. Thus an idempotent matrix is simultaneously a projection of a product bundle and a projection of a free module. The tautological line bundle gives a concrete case where this algebraic projector is visible in coordinates. [example: Projector onto the Tautological Line over Projective Space] Let $M=\mathbb{RP}^{m}$, and write a point as $[x]$, the line spanned by a nonzero vector $x\in\mathbb R^{m+1}_{0}=\mathbb R^{m+1}\setminus\{0\}$. The tautological line bundle $\gamma^1\to\mathbb{RP}^{m}$ has fibre $\gamma^1_{[x]}=[x]\subset\mathbb R^{m+1}$. Define \begin{align*} P([x])v=\frac{\langle v,x\rangle}{\langle x,x\rangle}x,\qquad v\in\mathbb R^{m+1}. \end{align*} In coordinates this is the matrix \begin{align*} P([x])=\frac{x\otimes x}{|x|^2}, \end{align*} whose $(i,j)$ entry is \begin{align*} P_{ij}([x])=\frac{x_i x_j}{x_0^2+\cdots+x_m^2}. \end{align*} The formula is independent of the chosen nonzero representative of the line. If $\lambda\ne 0$, then the numerator changes by \begin{align*} (\lambda x)\otimes(\lambda x)=\lambda^2(x\otimes x), \end{align*} and the denominator changes by \begin{align*} |\lambda x|^2=\lambda^2|x|^2. \end{align*} Therefore \begin{align*} \frac{(\lambda x)\otimes(\lambda x)}{|\lambda x|^2}=\frac{\lambda^2(x\otimes x)}{\lambda^2|x|^2}=\frac{x\otimes x}{|x|^2}. \end{align*} Thus $P$ is well defined on $\mathbb{RP}^m$. Its entries are smooth on $\mathbb R^{m+1}_{0}$ and are invariant under rescaling $x\mapsto \lambda x$, so they define smooth functions on projective space in the standard projective charts. Now verify the idempotent equation. For $v\in\mathbb R^{m+1}$, first applying $P([x])$ gives \begin{align*} P([x])v=\frac{\langle v,x\rangle}{\langle x,x\rangle}x. \end{align*} Applying $P([x])$ again gives \begin{align*} P([x])^2v=P([x])\left(\frac{\langle v,x\rangle}{\langle x,x\rangle}x\right). \end{align*} Using bilinearity of the Euclidean inner product, \begin{align*} P([x])^2v=\frac{\left\langle \frac{\langle v,x\rangle}{\langle x,x\rangle}x,x\right\rangle}{\langle x,x\rangle}x=\frac{\frac{\langle v,x\rangle}{\langle x,x\rangle}\langle x,x\rangle}{\langle x,x\rangle}x. \end{align*} Cancelling the common nonzero factor $\langle x,x\rangle=|x|^2$ gives \begin{align*} P([x])^2v=\frac{\langle v,x\rangle}{\langle x,x\rangle}x=P([x])v. \end{align*} Hence $P([x])^2=P([x])$ for every $[x]\in\mathbb{RP}^m$. Its image is exactly the tautological line. For every $v\in\mathbb R^{m+1}$, \begin{align*} P([x])v=\frac{\langle v,x\rangle}{\langle x,x\rangle}x\in [x], \end{align*} so $\operatorname{im}P([x])\subseteq [x]$. Conversely, \begin{align*} P([x])x=\frac{\langle x,x\rangle}{\langle x,x\rangle}x=x. \end{align*} If $w\in [x]$, then $w=cx$ for some $c\in\mathbb R$, and linearity gives \begin{align*} P([x])w=P([x])(cx)=cP([x])x=cx=w. \end{align*} Thus every vector in $[x]$ lies in $\operatorname{im}P([x])$, and therefore \begin{align*} \operatorname{im}P([x])=[x]=\gamma^1_{[x]}. \end{align*} The smooth idempotent matrix $P$ therefore realizes the tautological line bundle as the image subbundle of the product bundle $\mathbb{RP}^m\times\mathbb R^{m+1}$. [/example] ## The Section-Algebra Dictionary The chapter culminates in a structural dictionary rather than in a new algebraic theory. We have seen that vector bundles produce modules of sections, that subbundles of product bundles give direct summands, and that smooth idempotent matrices produce image subbundles. The useful takeaway is concrete: section modules, projection matrices, and local frames are three ways of recording the same geometric data when enough compactness and smoothness hypotheses are present. In computations, this gives a modest workflow. Start with a bundle or a subbundle, choose local frames or finitely many global sections when compactness supplies them, record the associated projection data when the bundle sits inside a product bundle, and remember that a global product frame is a stronger condition than projectivity. A bundle can have a section module that is a direct summand of a free module without having a global basis. Thus this perspective explains why the Mobius line bundle and tautological line bundles have local bases of sections without having one global product frame, while keeping the main course focused on bundles, sections, transition functions, and projections. [remark: Compactness and finite generators] The compactness assumption is the version used in this course because it makes finite generation automatic from finite covers. For noncompact bases, section spaces still carry useful local information, but a finite global list of generators need not exist. For the present course, compactness is simply the condition that turns local frames into finite global algebraic data. [/remark] This perspective also clarifies why local product charts, transition functions, sections, sheaves, and modules are not separate theories in this course. They are different presentations of the same object: a locally free family of vector spaces varying smoothly over the base. Sections convert a bundle into a sheaf-like object of local smooth data, and this viewpoint connects geometry with algebraic structure. The next chapter moves beyond vector bundles to show that the same local-to-global language also describes more general fibre bundles, where the fibres need not be linear. # 10. First Examples Beyond Vector Bundles This chapter develops the first non-vector-bundle examples in the course. The prerequisites are the local trivialisation and transition-function language from Chapters 1 and 2, together with the gluing principle used throughout the vector-bundle chapters. The goal is to separate the bundle idea from linear algebra: the fibre may be any smooth manifold $F$, and the transition maps may be diffeomorphisms rather than linear isomorphisms. The same local-to-global principle still governs the theory, but the examples now include sphere bundles, covering spaces, mapping tori, and projectivizations built from transition data. ## General Smooth Fibres and Structure Groups What remains of the vector bundle story when the fibre has no addition or scalar multiplication? The answer is that local triviality and transition functions survive unchanged, while linear algebra is replaced by the geometry of a fixed smooth manifold $F$ and a group acting on it by diffeomorphisms. [definition: Smooth Fibre Bundle With Fibre] Let $F$ be a smooth manifold. A smooth fibre bundle with fibre $F$ consists of smooth manifolds $E$ and $B$, a smooth surjective map $\rho:E\to B$, an open cover $(U_i)_{i\in I}$ of $B$, and diffeomorphisms \begin{align*} \varphi_i:\rho^{-1}(U_i)\to U_i\times F \end{align*} such that $\rho=\operatorname{pr}_1\circ \varphi_i$ for every $i\in I$. [/definition] The maps $\varphi_i$ are local trivializations, and the definition records the local product shape of the bundle. To use such a bundle globally, one must compare the fibre coordinates supplied by different trivializations on overlaps. On an overlap, two trivializations assign two possibly different fibre coordinates to the same point of $E$. Since the fibre has no preferred linear structure, the change of fibre coordinate is no longer a matrix; it is a diffeomorphism of the model fibre depending on the base point. This comparison must be packaged as actual transition data: for each pair of trivializations and each base point in their overlap, we need the diffeomorphism of $F$ that converts one fibre coordinate into the other. The following definition names this overlap map and records the formula that determines it from the two local trivializations. [definition: Transition Functions For General Fibre] Let $\rho:E\to B$ be a smooth fibre bundle with fibre $F$ and local trivializations $(\varphi_i)_{i\in I}$. The transition function on $U_i\cap U_j$ is the map \begin{align*} g_{ij}:U_i\cap U_j\to \operatorname{Diff}(F) \end{align*} determined by \begin{align*} \varphi_i\circ \varphi_j^{-1}(b,y)=(b,g_{ij}(b)(y)). \end{align*} [/definition] Unlike in a vector bundle, the target $\operatorname{Diff}(F)$ is usually infinite-dimensional. For this introductory course, smoothness of $g_{ij}$ means that the associated evaluation map \begin{align*} (U_i\cap U_j)\times F&\to F, & (b,y)&\mapsto g_{ij}(b)(y) \end{align*} is smooth. The next problem is to express geometric restrictions on these transition functions, which leads to the idea of a prescribed structure group. [definition: Bundle With Structure Group] Let $G$ be a Lie group acting smoothly on a smooth manifold $F$ by diffeomorphisms, with action homomorphism $\alpha:G\to \operatorname{Diff}(F)$. A smooth fibre bundle $\rho:E\to B$ with fibre $F$ has structure group $G$ if it admits local trivializations whose transition functions take values in $\alpha(G)\subseteq \operatorname{Diff}(F)$ and are represented by smooth maps to $G$. [/definition] Reducing the structure group records geometric information on the fibre. For instance, a vector bundle of rank $k$ is a fibre bundle with fibre $\mathbb R^k$ and structure group $GL(k,\mathbb R)$, while a bundle of oriented orthonormal frames has transition maps in $SO(k)$ after a metric and orientation have been chosen. [example: Product Bundle With Nonlinear Fibre] Let $B$ and $F$ be smooth manifolds, and consider the projection \begin{align*} \operatorname{pr}_1:B\times F&\to B, & \operatorname{pr}_1(b,y)&=b. \end{align*} This is a smooth fibre bundle with fibre $F$: take the single open set $U=B$ and the single trivialization \begin{align*} \varphi:\operatorname{pr}_1^{-1}(B)=B\times F&\to B\times F, & \varphi(b,y)&=(b,y). \end{align*} Then \begin{align*} (\operatorname{pr}_1\circ \varphi)(b,y)=\operatorname{pr}_1(b,y)=b, \end{align*} so the required compatibility with the projection holds. Because there is only one chart, the only transition function is on $B\cap B=B$. For $(b,y)\in B\times F$, \begin{align*} (\varphi\circ \varphi^{-1})(b,y)=(b,y)=(b,\operatorname{id}_F(y)), \end{align*} so $g_{BB}(b)=\operatorname{id}_F$ for every $b\in B$. If $F=S^1$, the product $B\times S^1$ has fibre circles that return to themselves by the identity map along every loop in $B$. Thus any twisting in a non-product circle bundle must appear in non-identity transition functions, not in the local product form itself. [/example] The product example is the local model for every bundle in the chapter, but local product charts by themselves do not determine how the products match on overlaps. If the overlap maps fail to agree on triple intersections, a point glued through $U_i$, then $U_j$, then $U_k$ can acquire two incompatible representatives. The global construction problem is therefore how to glue such local products using a smooth cocycle of diffeomorphisms, and the next theorem gives the general answer. [quotetheorem:6123] [citeproof:6123] The theorem is the main construction principle for the chapter: a bundle may be specified without first drawing a total space, provided the atlas topology on the quotient is Hausdorff and second-countable. The smoothness hypothesis is essential, since a merely set-theoretic cocycle can glue topological fibres while producing nonsmooth changes of chart. The cocycle identity is equally essential: without it, the proposed equivalence relation is not compatible with passage across triple overlaps. The theorem does not classify all bundles by a single cocycle until a cover has been chosen, but it explains why classification problems naturally become questions about cocycles modulo changes of trivialization. [example: Circle Bundle From Two Arcs] Choose the two overlap components so that crossing one component corresponds to entering the second arc and crossing the other corresponds to returning to the first arc after one circuit around the base. With fibre $S^1=\{z\in \mathbb C:|z|=1\}$, identity gluing on both components identifies fibre coordinates by \begin{align*} z&\longmapsto z \end{align*} at both crossings. Thus one circuit around the base sends every fibre point $z$ back to the same point $z$, so the quotient is \begin{align*} (S^1\times[0,1])/{(z,1)\sim(z,0)}\cong S^1\times S^1. \end{align*} If one overlap uses the identity and the other uses complex conjugation, then the return map on the fibre after one circuit is \begin{align*} z&\longmapsto \overline z. \end{align*} The resulting total space is therefore \begin{align*} (S^1\times[0,1])/{(z,1)\sim(\overline z,0)}. \end{align*} Writing $z=e^{i\theta}$ gives \begin{align*} \overline z=\overline{e^{i\theta}}=e^{-i\theta}, \end{align*} so the two boundary circles of the cylinder are glued with opposite angular orientation. This is the standard quotient model of the Klein bottle. The torus case preserves the fibre orientation after going once around the base, while the Klein bottle case reverses it. [/example] [illustration:circle-bundle-two-arc-cover] ## Sphere Bundles and Covering Maps Which familiar geometric objects become fibre bundles once we forget vector-space operations? Two useful families are sphere bundles, which keep only directions or unit vectors, and covering maps, whose fibre is a discrete set. [definition: Sphere Bundle] Let $E\to B$ be a smooth vector bundle of rank $k$ equipped with a smoothly varying fibrewise inner product. The unit sphere bundle is \begin{align*} S(E)=\{v\in E: |v|=1\}, \end{align*} with projection $S(E)\to B$ induced by the projection of $E$. [/definition] The fibre of $S(E)$ is $S^{k-1}$ rather than $\mathbb R^k$, so the definition keeps directions in each fibre and discards the zero vector. The natural question is whether this subset of the vector bundle is itself locally trivial, and the next theorem answers that question using orthonormal frames. [quotetheorem:6124] [citeproof:6124] This theorem is the first example where a vector bundle canonically produces a non-vector fibre bundle. The metric hypothesis is what makes the phrase "unit vector" meaningful in every fibre and what forces the transition maps between orthonormal frames to preserve $S^{k-1}$. If arbitrary $GL(k,\mathbb R)$-frames were used, a change of frame would generally send a unit vector to a non-unit vector, so it would not define a transition map of the sphere fibre. The theorem does not say that the sphere bundle is a vector bundle; for $k>1$ its fibre $S^{k-1}$ has no compatible vector-space structure. It also illustrates reduction of structure group: the metric changes $GL(k,\mathbb R)$-valued transition maps into $O(k)$-valued transition maps on unit spheres. [example: Tangent Unit Sphere Bundle] Let $(M,g)$ be a Riemannian manifold of dimension $n$, and let $\pi:TM\to M$ be the tangent bundle projection. The unit tangent sphere bundle is \begin{align*} SM=\{v\in TM:g_{\pi(v)}(v,v)=1\}. \end{align*} Choose an open set $U\subset M$ with a smooth local orthonormal frame $e_1,\dots,e_n$ for $TM|_U$. Every $v\in T_bM$ has a unique expression \begin{align*} v=x_1e_1(b)+\cdots+x_ne_n(b). \end{align*} By bilinearity of $g_b$, \begin{align*} g_b(v,v)=g_b\left(\sum_{a=1}^n x_ae_a(b),\sum_{c=1}^n x_ce_c(b)\right). \end{align*} Expanding both sums gives \begin{align*} g_b(v,v)=\sum_{a=1}^n\sum_{c=1}^n x_ax_c\,g_b(e_a(b),e_c(b)). \end{align*} Since the frame is orthonormal, $g_b(e_a(b),e_c(b))=\delta_{ac}$, so \begin{align*} g_b(v,v)=\sum_{a=1}^n\sum_{c=1}^n x_ax_c\,\delta_{ac}. \end{align*} The Kronecker delta kills all terms with $a\ne c$ and leaves the terms with $a=c$, hence \begin{align*} g_b(v,v)=\sum_{a=1}^n x_a^2. \end{align*} Thus $v\in SM$ exactly when $(x_1,\dots,x_n)\in S^{n-1}$. The corresponding local trivialization sends \begin{align*} v=\sum_{a=1}^n x_ae_a(\pi(v)) \end{align*} to \begin{align*} (\pi(v),(x_1,\dots,x_n))\in U\times S^{n-1}. \end{align*} On an overlap $U\cap V$, let $f_1,\dots,f_n$ be another smooth local orthonormal frame. Write \begin{align*} f_c(b)=\sum_{a=1}^n A_{ac}(b)e_a(b). \end{align*} For each $b\in U\cap V$, orthonormality of the $f$-frame gives \begin{align*} \delta_{cd}=g_b(f_c(b),f_d(b)). \end{align*} Substituting the expressions for $f_c(b)$ and $f_d(b)$ gives \begin{align*} \delta_{cd}=g_b\left(\sum_{a=1}^n A_{ac}(b)e_a(b),\sum_{r=1}^n A_{rd}(b)e_r(b)\right). \end{align*} By bilinearity, \begin{align*} \delta_{cd}=\sum_{a=1}^n\sum_{r=1}^n A_{ac}(b)A_{rd}(b)g_b(e_a(b),e_r(b)). \end{align*} Using $g_b(e_a(b),e_r(b))=\delta_{ar}$ gives \begin{align*} \delta_{cd}=\sum_{a=1}^n\sum_{r=1}^n A_{ac}(b)A_{rd}(b)\delta_{ar}. \end{align*} Therefore \begin{align*} \delta_{cd}=\sum_{a=1}^n A_{ac}(b)A_{ad}(b). \end{align*} This is exactly the $(c,d)$-entry of $A(b)^TA(b)=I$, so $A(b)\in O(n)$. If $v=\sum_{c=1}^n y_cf_c(b)$, then substituting the $f$-frame in terms of the $e$-frame gives \begin{align*} v=\sum_{c=1}^n y_c\sum_{a=1}^n A_{ac}(b)e_a(b). \end{align*} Reordering the finite sums gives \begin{align*} v=\sum_{a=1}^n\left(\sum_{c=1}^n A_{ac}(b)y_c\right)e_a(b). \end{align*} Hence the coordinate change on the sphere fibre is \begin{align*} y\longmapsto A(b)y. \end{align*} The entries $A_{ac}(b)$ are smooth because they are the coordinate functions of one smooth frame in another smooth frame, and the calculation above shows $A(b)\in O(n)$. Therefore $SM\to M$ is locally $U\times S^{n-1}\to U$, with transition maps given pointwise by elements of $O(n)$. The fibre over $b$ is the set of unit directions in $T_bM$, not the whole tangent vector space. [/example] The tangent unit sphere bundle has positive-dimensional fibre. To see how the same language handles discrete fibres, we next need the local sheet decomposition used in the definition of a covering map. [definition: Covering Map] A smooth map $p:E\to B$ is a covering map if every point $b\in B$ has an open neighbourhood $U\subset B$ such that \begin{align*} p^{-1}(U)=\bigsqcup_{\alpha\in A} V_\alpha \end{align*} and each restriction $p|_{V_\alpha}:V_\alpha\to U$ is a diffeomorphism. [/definition] The set $A$ indexes the sheets over $U$. A covering is locally a disjoint union of copies of the base, but a fibre bundle description requires one fixed model fibre and compatible local trivializations. When there are exactly $n$ sheets everywhere, choosing a local ordering of the sheets identifies each fibre with the same discrete set $\{1,\dots,n\}$. [quotetheorem:6125] [citeproof:6125] This reformulation packages covering theory into the language of transition data. The $n$-sheeted hypothesis fixes one model fibre, so every local trivialization uses the same finite set $\{1,\dots,n\}$. If the number of sheets varied, no single discrete fibre could model all fibres; for infinite coverings one can still obtain a bundle with an infinite discrete fibre when the cardinality is locally constant, but the structure group becomes a permutation group of that fibre rather than $S_n$. Monodromy of a covering is then the special case of bundle holonomy where the structure group is the finite symmetric group, and classification over connected bases is governed by the corresponding permutation representation of the fundamental group. [example: Double Cover Of The Circle] Let $p:S^1\to S^1$ be $p(z)=z^2$. Choose an arc \begin{align*} U=\{e^{i\theta}:a<\theta<b\} \end{align*} with $b-a<2\pi$, so each point of $U$ has a unique angle $\theta\in(a,b)$. If $w=e^{i\theta}\in U$, then the two displayed square roots of $w$ are \begin{align*} z_1(\theta)=e^{i\theta/2} \end{align*} and \begin{align*} z_2(\theta)=-e^{i\theta/2}=e^{i(\theta/2+\pi)}. \end{align*} Indeed, \begin{align*} p(z_1(\theta))=(e^{i\theta/2})^2=e^{i\theta}=w \end{align*} and \begin{align*} p(z_2(\theta))=(-e^{i\theta/2})^2=(-1)^2(e^{i\theta/2})^2=e^{i\theta}=w. \end{align*} Conversely, if $z=e^{i\phi}$ and $z^2=e^{i\theta}$, then \begin{align*} e^{2i\phi}=e^{i\theta}. \end{align*} Hence $2\phi=\theta+2\pi m$ for some $m\in\mathbb Z$, so \begin{align*} z=e^{i\phi}=e^{i\theta/2}e^{i\pi m}. \end{align*} If $m$ is even, then $e^{i\pi m}=1$, and if $m$ is odd, then $e^{i\pi m}=-1$. Thus the only two preimages of $e^{i\theta}$ are $e^{i\theta/2}$ and $-e^{i\theta/2}$. Therefore $p^{-1}(U)$ is the disjoint union of the two sheets \begin{align*} V_1=\{e^{i\theta/2}:a<\theta<b\} \end{align*} and \begin{align*} V_2=\{-e^{i\theta/2}:a<\theta<b\}. \end{align*} The local trivialization over $U$ sends \begin{align*} e^{i\theta/2}\longmapsto (e^{i\theta},1) \end{align*} and \begin{align*} -e^{i\theta/2}\longmapsto (e^{i\theta},2). \end{align*} To see the monodromy, lift the loop $\gamma(t)=e^{2\pi i t}$ starting at $1\in S^1$. The lift beginning at $1$ is \begin{align*} \widetilde\gamma(t)=e^{\pi i t}, \end{align*} because \begin{align*} p(\widetilde\gamma(t))=(e^{\pi i t})^2=e^{2\pi i t}=\gamma(t). \end{align*} At $t=0$ this lift satisfies \begin{align*} \widetilde\gamma(0)=e^0=1, \end{align*} while at $t=1$ it satisfies \begin{align*} \widetilde\gamma(1)=e^{\pi i}=-1. \end{align*} Thus one circuit around the base carries the first square-root sheet to the second sheet, and the same calculation starting at $-1$ carries the second sheet back to the first. The transition permutation is therefore the transposition $(1\ 2)$, the non-identity element of $S_2$. [/example] ## Mapping Tori Over The Circle How can a single diffeomorphism of a fibre produce a bundle over $S^1$? The construction identifies the two ends of a cylinder $F\times[0,1]$ using the diffeomorphism, so the twisting is concentrated in the return map around the circle. [definition: Mapping Torus] Let $F$ be a smooth manifold and let $f:F\to F$ be a diffeomorphism. The mapping torus of $f$ is \begin{align*} T_f=(F\times[0,1])/{(y,1)\sim(f(y),0)}. \end{align*} [/definition] The quotient comes with a natural map to $S^1$ by sending the class of $(y,t)$ to $e^{2\pi i t}$. This definition would fail to produce a smooth bundle if the endpoint identification were not made by a diffeomorphism, because the two collar neighbourhoods at $t=0$ and $t=1$ would not have compatible smooth fibre coordinates. The next theorem is needed to verify that this quotient map is locally a product projection, including near the identified endpoints. [quotetheorem:6126] [citeproof:6126] Mapping tori are the basic bundles over the circle. The diffeomorphism hypothesis is necessary because the overlap chart crossing the cut must have a smooth inverse. The construction records only the [conjugacy class](/page/Conjugacy%20Class) of the return diffeomorphism after changing trivializations, so it is a geometric version of monodromy rather than a choice of preferred coordinates on $F$. Over a one-dimensional base, transition data is therefore governed by the diffeomorphism obtained after one circuit around the circle. [example: Torus And Klein Bottle As Mapping Tori] Take $F=S^1$. For the identity map $\operatorname{id}_{S^1}:S^1\to S^1$, the mapping torus is \begin{align*} T_{\operatorname{id}}=(S^1\times[0,1])/{(z,1)\sim(\operatorname{id}_{S^1}(z),0)}=(S^1\times[0,1])/{(z,1)\sim(z,0)}. \end{align*} Define \begin{align*} \Phi:T_{\operatorname{id}}\to S^1\times S^1,\qquad \Phi([z,t])=(z,e^{2\pi i t}). \end{align*} This is well-defined at the glued endpoints because \begin{align*} \Phi([z,1])=(z,e^{2\pi i})=(z,1) \end{align*} and \begin{align*} \Phi([z,0])=(z,e^0)=(z,1). \end{align*} Thus the endpoint identification in $T_{\operatorname{id}}$ is exactly the usual identification that turns the interval coordinate into a circle coordinate, so $T_{\operatorname{id}}\cong S^1\times S^1$, the torus. For the conjugation map $c:S^1\to S^1$, $c(z)=\overline z$, the mapping torus is \begin{align*} T_c=(S^1\times[0,1])/{(z,1)\sim(\overline z,0)}. \end{align*} Writing $z=e^{i\theta}$ gives \begin{align*} \overline z=\overline{e^{i\theta}}=e^{-i\theta}. \end{align*} Hence the endpoint gluing is \begin{align*} (e^{i\theta},1)\sim(e^{-i\theta},0). \end{align*} In angular coordinates this identifies the top boundary circle by the map $\theta\mapsto -\theta \pmod{2\pi}$, which reverses orientation because its derivative with respect to $\theta$ is $-1$. Therefore \begin{align*} T_c=(S^1\times[0,1])/{(e^{i\theta},1)\sim(e^{-i\theta},0)} \end{align*} is the standard quotient model of the Klein bottle. The torus case glues the fibre back by the identity, while the Klein bottle case glues the fibre back by an orientation-reversing diffeomorphism. [/example] The previous example matches the two-arc transition-data construction for circle bundles over $S^1$. The mapping-torus description is often more economical because all twisting is recorded in a single diffeomorphism of the fibre. [illustration:mapping-torus-construction] ## Associated Fibre Constructions Without Connections When a bundle is given by transition functions, what other bundles can be built from the same cocycle? The guiding principle is functoriality: if the structure group acts on another space, the same transition functions glue local products with that new fibre. [definition: Associated Fibre Bundle] Let a bundle over $B$ be described by transition functions $g_{ij}:U_i\cap U_j\to G$, and let $G$ act smoothly on a manifold $F'$. The associated fibre bundle with fibre $F'$ has total space \begin{align*} E' = \bigsqcup_i U_i\times F'/{\sim}, \end{align*} where \begin{align*} (i,b,y')\sim (j,b,g_{ji}(b)y'). \end{align*} Its projection is the map $\rho':E'\to B$ defined by $\rho'([i,b,y'])=b$. [/definition] No connection, horizontal lift, or curvature is involved in this construction. The point to check is more basic: the equivalence relation must glue the products $U_i\times F'$ into a smooth bundle, and the overlap maps on the new fibre must come from the given action of $G$ on $F'$. This raises the verification problem for associated bundles: when does the quotient formula actually produce a smooth fibre bundle with the expected transition functions? The result below supplies that justification, turning the construction from a set-theoretic recipe into a legitimate smooth bundle construction. [quotetheorem:6127] [citeproof:6127] This theorem explains why many geometric bundles appear in families. The smooth action hypothesis is necessary: without it the new overlap maps on $F'$ need not be smooth, even if the original cocycle is smooth as a $G$-valued map. The construction is also functorial rather than connection-dependent; it uses only transition data and does not choose horizontal subspaces or parallel transport. A single vector bundle cocycle can therefore act not only on $\mathbb R^k$, but also on spheres, projective spaces, exterior powers, and other spaces naturally attached to $\mathbb R^k$. [example: Projectivization Of A Vector Bundle] Let $E\to B$ be a rank $k$ real vector bundle with transition functions \begin{align*} g_{ij}:U_i\cap U_j\to GL(k,\mathbb R). \end{align*} Write a point of $\mathbb RP^{k-1}$ as a line $[v]=\mathbb R v$ with $v\in\mathbb R^k\setminus\{0\}$. For $g\in GL(k,\mathbb R)$, define \begin{align*} g\cdot [v]=[gv]. \end{align*} This is well-defined: if $[v]=[w]$, then $w=\lambda v$ for some $\lambda\in\mathbb R^\times$, and hence \begin{align*} gw=g(\lambda v)=\lambda gv, \end{align*} so $[gw]=[gv]$. Also $gv\ne 0$ because $g$ is invertible and $v\ne 0$, so $[gv]$ is again a point of $\mathbb RP^{k-1}$. The identity matrix acts by \begin{align*} I\cdot [v]=[Iv]=[v], \end{align*} and for $g,h\in GL(k,\mathbb R)$, \begin{align*} g\cdot(h\cdot [v])=g\cdot[hv]=[g(hv)]=[(gh)v]=(gh)\cdot[v]. \end{align*} Thus $GL(k,\mathbb R)$ acts on $\mathbb RP^{k-1}$ by diffeomorphisms. Using this action, glue the local products $U_i\times \mathbb RP^{k-1}$ by \begin{align*} (i,b,[v])\sim (j,b,[g_{ji}(b)v]). \end{align*} The resulting associated bundle is \begin{align*} \mathbb P(E)=\bigsqcup_i U_i\times \mathbb RP^{k-1}/\sim . \end{align*} On an overlap, the change of projective fibre coordinate is \begin{align*} [b,v]_j\longmapsto [b,g_{ij}(b)v]_i, \end{align*} which is exactly the original vector-bundle transition map followed by passage from a nonzero vector to its span. Therefore $\mathbb P(E)\to B$ is a smooth fibre bundle with fibre $\mathbb RP^{k-1}$. For a fixed $b\in B$, choose $i$ with $b\in U_i$. The trivialization of $E$ over $U_i$ identifies $E_b$ with $\mathbb R^k$, so a line $L\subset E_b$ corresponds to a line $[v]\in\mathbb RP^{k-1}$. If another chart $U_j$ is used, the same vector coordinates are changed by $g_{ji}(b)$, and the projective gluing identifies $[v]$ with $[g_{ji}(b)v]$. Hence the description is independent of the chosen local trivialization, and the fibre of $\mathbb P(E)$ over $b$ is precisely the set of lines in the vector space $E_b$. [/example] Projectivization keeps the directions in each fibre while identifying nonzero scalar multiples. It is therefore related to the sphere bundle, but it quotients the unit sphere fibre by the antipodal relation. [example: Unit Sphere Bundle From Vector Bundle Transition Functions] Let $E\to B$ have orthonormal local trivializations \begin{align*} \psi_i:E|_{U_i}\to U_i\times \mathbb R^k \end{align*} with transition functions determined by \begin{align*} \psi_i\circ \psi_j^{-1}(b,y)=(b,g_{ij}(b)y), \end{align*} where $g_{ij}(b)\in O(k)$. If $y\in S^{k-1}$, then \begin{align*} \|g_{ij}(b)y\|^2=(g_{ij}(b)y)^\top(g_{ij}(b)y). \end{align*} Associativity of matrix multiplication gives \begin{align*} (g_{ij}(b)y)^\top(g_{ij}(b)y)=y^Tg_{ij}(b)^Tg_{ij}(b)y. \end{align*} Since $g_{ij}(b)\in O(k)$, we have $g_{ij}(b)^Tg_{ij}(b)=I$, hence \begin{align*} y^Tg_{ij}(b)^Tg_{ij}(b)y=y^TIy. \end{align*} Since $Iy=y$, this becomes \begin{align*} y^TIy=y^Ty=\|y\|^2=1. \end{align*} Thus each transition map restricts to a diffeomorphism $S^{k-1}\to S^{k-1}$. The associated bundle from the standard $O(k)$-action on $S^{k-1}$ is \begin{align*} E_S=\bigsqcup_i U_i\times S^{k-1}/\sim, \end{align*} where \begin{align*} (i,b,y)\sim (j,b,g_{ji}(b)y). \end{align*} Define \begin{align*} \Phi:E_S\to S(E),\qquad \Phi([i,b,y])=\psi_i^{-1}(b,y). \end{align*} This lands in $S(E)$ because $\psi_i$ is orthonormal: if $y=(y_1,\dots,y_k)\in S^{k-1}$, then \begin{align*} \|\psi_i^{-1}(b,y)\|_b^2=y_1^2+\cdots+y_k^2=1. \end{align*} The map $\Phi$ is well-defined. If $(i,b,y)\sim(j,b,g_{ji}(b)y)$, then \begin{align*} \psi_i\circ\psi_j^{-1}(b,g_{ji}(b)y)=(b,g_{ij}(b)g_{ji}(b)y). \end{align*} The inverse transition identity gives $g_{ij}(b)g_{ji}(b)=I$, so \begin{align*} (b,g_{ij}(b)g_{ji}(b)y)=(b,Iy)=(b,y). \end{align*} Applying $\psi_i^{-1}$ to both sides gives \begin{align*} \psi_j^{-1}(b,g_{ji}(b)y)=\psi_i^{-1}(b,y). \end{align*} Conversely, every unit vector $v\in S(E)_b$ has, in any orthonormal trivialization, a unique coordinate vector $y\in S^{k-1}$ satisfying \begin{align*} \psi_i(v)=(b,y). \end{align*} Therefore $\Phi$ is bijective on each fibre. In the chart over $U_i$, both $E_S$ and $S(E)$ are identified with $U_i\times S^{k-1}$, and $\Phi$ is the identity in these coordinates. Hence the associated bundle obtained from the standard $O(k)$-action on $S^{k-1}$ is the unit sphere bundle $S(E)\to B$: its fibre over $b$ is the set of unit vectors in $E_b$. [/example] The constructions in this chapter all express the same idea at different levels of generality. Vector bundles use linear transition maps; sphere bundles, coverings, and mapping tori replace linear maps with diffeomorphisms of other fibres; associated bundles reuse existing transition data through a new [group action](/page/Group%20Action). This is the point at which fibre bundles become a flexible language for geometry rather than only a generalization of vector bundles. The examples in the previous chapter show that bundle theory is not limited to vector spaces in each fibre. The next chapter returns to vector bundles to isolate the first obstructions to triviality, showing how global invariants prevent the local product description from extending everywhere. # 11. Obstructions to Triviality in Examples After Chapter 10 separated fibre bundles from linear algebra, this chapter returns to vector bundles and collects the first obstructions to being a product. The guiding theme is that local bundle charts can always be chosen, but global choices such as a nowhere-zero section or a coherent orientation may fail to exist. These failures are already visible in examples developed earlier: the tangent bundle of the sphere, the Mobius line bundle, and transition functions whose determinants change sign. ## Nonzero Sections and Splitting Off Product Line Summands When a vector bundle is a product, it has many global sections that never vanish: in $M \times \mathbb R^k$, any constant nonzero vector gives one. The first question is whether the converse holds in a partial form. A single nowhere-zero section cannot identify a rank-$k$ bundle with a product by itself, but it should give one globally defined line inside every fibre. [definition: Nowhere-Zero Section] Let $\pi:E\to M$ be a smooth vector bundle. A smooth section $s:M\to E$ is nowhere-zero if $s(x)\ne 0$ in $E_x$ for every $x\in M$. [/definition] A nowhere-zero section is stronger than a section that is merely not identically zero: it picks a nonzero vector in every fibre. The point of the next definition is to package the line generated by this vector as a subbundle rather than just a set of fibrewise lines. [definition: Product Line Subbundle Generated By A Section] Let $\pi:E\to M$ be a smooth vector bundle and let $s:M\to E$ be a nowhere-zero smooth section. The line subbundle generated by $s$ is the rank-$1$ vector subbundle $L_s\subset E$ with fibres \begin{align*} (L_s)_x=\operatorname{span}(s(x))\subset E_x. \end{align*} [/definition] The map $M\times \mathbb R\to L_s$ given by $(x,t)\mapsto t s(x)$ is a bundle isomorphism, so $L_s$ is a product line bundle. To split $E$ as this line plus something else, we need a complementary subbundle; for smooth vector bundles this is supplied by a bundle metric. [quotetheorem:6128] [citeproof:6128] The nowhere-zero hypothesis is essential because a section that vanishes at even one point does not span a line there; for example, the zero section of any bundle would otherwise produce a false splitting. Smoothness is also part of the content: a discontinuous choice of a nonzero vector in the fibres of a product bundle need not define a smooth subbundle. The paracompactness assumption is what permits the bundle metric used in the proof, through partitions of unity on the base. The theorem does not say that $E$ is a product bundle, only that one product line summand has been separated; the remaining summand $F$ may still carry all the global twisting. Products pass this test in the most direct way, so the next example fixes the model case before we look for failures. [example: Product Bundle Has Many Splittings] Let $E=M\times \mathbb R^k$, and let $e_1=(1,0,\dots,0)$. The section $s:M\to E$ defined by $s(x)=(x,e_1)$ is nowhere-zero because, in the fibre $E_x=\{x\}\times \mathbb R^k$, the vector $e_1$ is not the zero vector. For each $x\in M$, the line generated by $s(x)$ is \begin{align*} (L_s)_x=\{(x,t e_1):t\in \mathbb R\}. \end{align*} Since $\{t e_1:t\in \mathbb R\}=\operatorname{span}(e_1)$, this gives \begin{align*} L_s=M\times \operatorname{span}(e_1). \end{align*} The complementary coordinate subspace is $\operatorname{span}(e_2,\dots,e_k)$. If $v=(v_1,\dots,v_k)\in \mathbb R^k$, then \begin{align*} v=(v_1,0,\dots,0)+(0,v_2,\dots,v_k). \end{align*} The first summand is $v_1e_1\in \operatorname{span}(e_1)$, and the second is $v_2e_2+\cdots+v_ke_k\in \operatorname{span}(e_2,\dots,e_k)$. If a vector lies in both subspaces, then it has first coordinate possibly nonzero and all other coordinates zero, and also first coordinate zero; hence every coordinate is zero. Therefore \begin{align*} \mathbb R^k=\operatorname{span}(e_1)\oplus \operatorname{span}(e_2,\dots,e_k). \end{align*} Taking the product with $M$ gives the bundle decomposition \begin{align*} M\times \mathbb R^k\cong M\times \operatorname{span}(e_1)\oplus M\times \operatorname{span}(e_2,\dots,e_k). \end{align*} The first factor is the product line bundle $\underline{\mathbb R}$, and the second factor is the product rank-$(k-1)$ bundle $\underline{\mathbb R}^{k-1}$, so \begin{align*} M\times \mathbb R^k\cong \underline{\mathbb R}\oplus \underline{\mathbb R}^{k-1}. \end{align*} Thus in a product bundle, a constant nonzero vector selects a product line summand, and the remaining coordinate directions give an explicit complementary product bundle. [/example] The product example shows what a global nonvanishing choice accomplishes. The next problem is whether a natural bundle can fail this test; tangent bundles provide the first geometric obstruction because sections of $TM$ are vector fields. [quotetheorem:2248] [citeproof:2248] The base and the smoothness matter here: on $S^1$ there is a nowhere-zero tangent field, while on $S^2$ the topology forces a zero. The tangent-bundle hypothesis also matters, since a product rank-$2$ bundle over $S^2$ has many nowhere-zero sections even though it has the same base. The theorem does not classify all rank-$2$ bundles over $S^2$, and it does not prove that every failure of product form is caused by the absence of a line summand. It identifies a specific global choice that cannot be made, and this style of obstruction will reappear in transition-function language as a failure to choose consistent signs or matrices across overlaps. ## Orientability of Line Bundles and Determinant Bundles The next obstruction asks whether the fibres of a real vector bundle can be oriented consistently. Local bundle charts orient each fibre once we orient $\mathbb R^k$, but transition functions may reverse orientation on overlaps. The determinant records the part of a transition matrix that matters for this question. [definition: Orientation Of A Real Vector Bundle] Let $E\to M$ be a rank-$k$ real vector bundle. An orientation of $E$ is a choice of orientation of each fibre $E_x$ such that around every point of $M$ there is a local frame $(e_1,\dots,e_k)$ whose ordered basis is positively oriented in every fibre over its domain. [/definition] The local-frame condition is the smooth compatibility requirement; without it, arbitrary fibrewise choices would ignore the topology of the bundle. To express the condition using transition functions, we fix a bundle atlas and inspect determinants. [quotetheorem:6129] [citeproof:6129] The possibility of changing to an equivalent atlas is essential: a poorly chosen atlas for a product bundle may contain negative determinant changes of frame, but those signs do not represent a real obstruction if they can be removed. The positivity condition is also stronger than asking for nonzero determinants, since every vector-bundle transition matrix already lies in $GL(k,\mathbb R)$; the Mobius line bundle shows what fails when signs cannot be made positive on all overlaps. The theorem does not say that an orientable bundle is a product bundle, only that its transition functions can be chosen in the identity component of $GL(k,\mathbb R)$. For rank one, the determinant is the transition function itself, so orientability becomes especially concrete. [quotetheorem:6130] [citeproof:6130] The rank-one hypothesis is the sharp point of this result: in higher rank, orientability does not force product form, as the tangent bundle $TS^2$ is orientable but still has no product line summand. For a line bundle, choosing positive transition functions is enough to glue local positive unit vectors into a nowhere-zero global section, and that section trivializes the line bundle. The theorem does not say that every line bundle has a preferred product frame, since a chosen isomorphism still depends on a chosen nowhere-zero section. For higher rank bundles the determinant criterion is still only about a line worth of data, so the natural question is whether this line can be separated from the original bundle and studied as an ordinary line bundle in its own right. [definition: Determinant Bundle] Let $E\to M$ be a rank-$k$ real vector bundle. The determinant bundle of $E$ is the real line bundle \begin{align*} \det E=\Lambda^k E\to M, \end{align*} whose fibre at $x\in M$ is $\Lambda^k E_x$. [/definition] The transition function of $\det E$ associated to a transition matrix $g_{ij}$ is the scalar function $\det g_{ij}$. Thus the determinant bundle turns the orientability problem for $E$ into the product-form problem for a line bundle. [quotetheorem:6092] [citeproof:6092] The determinant bundle is necessary because individual matrix entries are not invariant under changes of frame; two atlases may use very different matrices while inducing the same scalar determinant transitions. Without passing to $\Lambda^k E$, an orientation reversal can be hidden inside a complicated change of basis, whereas the determinant line records exactly whether top exterior volume changes sign. The theorem does not say that $\det E$ determines $E$, since nonisomorphic higher-rank bundles may have the same determinant line. It packages only the sign obstruction, and we can now read standard examples directly from their transition data. [example: Orientability Of The Mobius Bundle] Cover $S^1$ by arcs $U_0,U_1$ such that $U_0\cap U_1=C_+\sqcup C_-$ has two connected components. Describe the Mobius line bundle by a transition function $g_{01}:U_0\cap U_1\to \mathbb R^\times$ with $g_{01}(x)=+1$ for $x\in C_+$ and $g_{01}(x)=-1$ for $x\in C_-$. For a line bundle, changing the local frame over $U_i$ multiplies that frame by a smooth nowhere-zero function $h_i:U_i\to \mathbb R^\times$. Since each $U_i$ is connected and $\mathbb R^\times=(-\infty,0)\cup(0,\infty)$ has two connected components, the sign of $h_i$ is constant on $U_i$. Under this frame change, the transition function becomes \begin{align*} g'_{01}(x)=h_1(x)^{-1}g_{01}(x)h_0(x). \end{align*} Taking signs gives \begin{align*} \operatorname{sgn}(g'_{01}(x))=\operatorname{sgn}(h_1(x))^{-1}\operatorname{sgn}(g_{01}(x))\operatorname{sgn}(h_0(x)). \end{align*} The factor $\operatorname{sgn}(h_1(x))^{-1}\operatorname{sgn}(h_0(x))$ is constant on $U_0\cap U_1$, because each $\operatorname{sgn}(h_i)$ is constant on $U_i$. Therefore multiplying $g_{01}$ by this factor either preserves both signs or reverses both signs at once. Since $g_{01}$ is positive on $C_+$ and negative on $C_-$, the modified transition function $g'_{01}$ still has opposite signs on the two components. Thus no equivalent atlas can make all transition functions positive, so the Mobius line bundle is not orientable by the determinant criterion for orientability. Because the Mobius bundle has rank $1$, \begin{align*} \det L=\Lambda^1 L=L, \end{align*} so its determinant bundle is the same nonorientable line bundle. [/example] The Mobius example shows that a local choice of positive generator can return with the opposite sign after going once around the base. In rank $k$, the same phenomenon is detected by the sign of the determinant rather than by each matrix entry. ## Elementary Invariants from Clutching and Transition Functions Transition functions do more than reconstruct a bundle: they can expose features that cannot be removed by changing bundle charts. The simplest invariant is discrete, asking whether determinant signs can be made positive. Clutching descriptions make this visible because all the global information is concentrated in a map on the equator or overlap. [definition: Clutching Function] Let $X=A\cup B$ with $A$ and $B$ open and let $A\cap B$ deformation retract onto a space $C$. A rank-$k$ real vector bundle obtained by gluing the product bundles $A\times \mathbb R^k$ and $B\times \mathbb R^k$ along a smooth map $g:C\to GL(k,\mathbb R)$ is said to have clutching function $g$. [/definition] The clutching function is not unique, since changing bundle charts conjugates or modifies it by maps from the two pieces. For bundles over $S^1$, however, cutting the circle to an interval leaves a single gluing matrix, and the determinant sign of that matrix answers the orientability question. [quotetheorem:6131] [citeproof:6131] The assumption that the bundle is described by a single clutching matrix after cutting $S^1$ is doing real work: over a more complicated base, one loop sign cannot encode all transition data. Invertibility is also essential, since a singular gluing matrix would not identify fibres and would not define a vector bundle. In this special circle setting, the sign records the component of the clutching matrix: a positive determinant matrix lies in the identity component of $GL(k,\mathbb R)$ and gives a product bundle after the allowed change of clutching data, while a negative determinant gives the orientation-reversing obstruction. Thus the theorem is the rank-$k$ version of the Mobius sign obstruction, and it shows why the determinant is the right invariant for orientation: many different matrices may define equivalent local data, but the sign of whether a loop reverses orientation is unchanged by orientation-preserving changes of frame. [example: Determinant Transition Signs] Let a rank-$2$ bundle over $S^1$ be formed by gluing $(0,v)\sim (1,Av)$ on $[0,1]\times \mathbb R^2$. First take $A=\operatorname{diag}(1,-1)$. For a $2\times 2$ diagonal matrix, the determinant is the product of the diagonal entries, so \begin{align*} \det A=(1)(-1)=-1. \end{align*} Since the clutching determinant is negative, the [Orientability Criterion for Real Vector Bundles over the Circle by the Clutching Determinant](/theorems/6131) says that the resulting rank-$2$ bundle is nonorientable. Now take $A=\operatorname{diag}(-1,-1)$. Again multiplying the diagonal entries gives \begin{align*} \det A=(-1)(-1)=1. \end{align*} Thus this gluing preserves orientation, so the determinant-sign test gives no orientability obstruction. Fibrewise, $e_1$ is sent to $-e_1$ and $e_2$ is sent to $-e_2$, but the ordered area orientation is multiplied by $\det A=1$, so the two coordinate sign reversals cancel at the level of orientation. [/example] The second matrix illustrates an important limitation: elementary invariants detect only the features they are designed to see. A bundle may pass the determinant-sign test and still fail to be a product for other reasons. [remark: What These Obstructions Measure] A nowhere-zero section measures whether a product line summand can be split off. Orientability measures whether the determinant line bundle is isomorphic to a product line. Clutching signs measure whether going around a gluing loop reverses orientation. These are different questions, and none of them by itself classifies all vector bundles. [/remark] The examples in this chapter should be read as prototypes for later invariants and as preparation for the final synthesis in Chapter 12. Instead of trying to identify a bundle with a product directly, we ask for a global structure that every product bundle has, translate its existence into sections or transition functions, and then show that the requested global structure cannot be chosen. The obstructions found in the examples show that local triviality is only the beginning of the story. The final chapter collects the course's constructions into a synthesis, using transition functions, sections, pullbacks, and gluing methods to recognize when a bundle is built globally from familiar local pieces. # 12. Synthesis: Building and Recognizing Bundles This closing chapter gathers the constructions developed throughout the course into a practical toolkit for building and recognizing bundles. The prerequisites are Chapters 1 through 11: local product charts, transition functions, sections, pullbacks, bundle maps, bundle metrics, clutching, sheaves, and elementary obstructions. Here the emphasis is on using those ingredients to build bundles and decide when two constructions give the same bundle. The guiding principle is that a bundle can be recognized either geometrically, through its total space and maps, or algebraically, through transition data satisfying a cocycle condition. ## From Local Data to Global Bundles When a construction begins with charts on the base, the first question is whether the local pieces glue to an honest global bundle. The answer is not merely set-theoretic: the gluing must produce a smooth total space, a smooth projection, and local product charts whose changes agree with the prescribed data. [quotetheorem:6132] [citeproof:6132] The hypotheses are doing separate jobs. Without the identity and cocycle equations, the proposed relation may fail to be an equivalence relation: on a triple overlap, gluing first through $U_j$ and then through $U_k$ can give a different fibre point from gluing directly through $U_k$. Without smoothness of the induced maps on $(U_i\cap U_j)\times F$, the quotient may still be a topological bundle candidate, but its chart changes need not be smooth. The theorem is also not a bare assertion about an arbitrary quotient topology: the smooth bundle structure is the atlas topology determined by the local trivializations, and Hausdorffness and second-countability are explicit topological hypotheses. Finally, the theorem does not classify the resulting bundle; different cocycles can produce isomorphic bundles after a change of local frame, which is the recognition problem addressed next. [quotetheorem:6133] [citeproof:6133] The displayed formula is the main computational test for whether two transition descriptions define isomorphic bundles, but the fixed cover and chosen charts are part of the statement. If the cover is changed, the cocycles must first be compared on a common refinement; otherwise the symbols $g_{ij}$ and $g'_{ij}$ do not live on the same domains. If the cocycle identity fails, gluing around a triple overlap gives path-dependent identifications and no bundle is produced by the construction theorem. If the frame changes $h_i$ are only continuous, the resulting comparison is not a smooth bundle isomorphism. The theorem also does not give a canonical choice of cocycle for a bundle; it gives a dictionary between geometric data and cocycles modulo smooth frame changes. [example: Product Bundle From A Coboundary] Let $E$ be the vector bundle obtained by gluing the pieces $U_i\times \mathbb R^k$ using the cocycle $g_{ij}=h_i^{-1}h_j$. Define a map from the $i$th local piece to the product bundle by \begin{align*} \Phi_i(x,v)=(x,h_i(x)v). \end{align*} To check that these formulas agree with the gluing relation, suppose $(x,v,i)$ is identified with $(x,g_{ji}(x)v,j)$. Since $g_{ji}=h_j^{-1}h_i$, we have \begin{align*} h_j(x)g_{ji}(x)v=h_j(x)\bigl(h_j(x)^{-1}h_i(x)\bigr)v=\bigl(h_j(x)h_j(x)^{-1}\bigr)h_i(x)v=h_i(x)v. \end{align*} Therefore \begin{align*} \Phi_j(x,g_{ji}(x)v)=(x,h_j(x)g_{ji}(x)v)=(x,h_i(x)v)=\Phi_i(x,v). \end{align*} Thus the local formulas descend to a smooth bundle map $\Phi:E\to M\times\mathbb R^k$. The inverse is defined in local coordinates as follows: for $x\in U_i$, set \begin{align*} \Psi(x,w)=[x,h_i(x)^{-1}w,i]. \end{align*} If $x\in U_i\cap U_j$, then \begin{align*} g_{ji}(x)h_i(x)^{-1}w=\bigl(h_j(x)^{-1}h_i(x)\bigr)h_i(x)^{-1}w=h_j(x)^{-1}w. \end{align*} Hence $[x,h_i(x)^{-1}w,i]=[x,h_j(x)^{-1}w,j]$, so $\Psi$ is independent of the chosen index. The two composites are \begin{align*} \Phi(\Psi(x,w))=(x,h_i(x)h_i(x)^{-1}w)=(x,w) \end{align*} and \begin{align*} \Psi(\Phi([x,v,i]))=\Psi(x,h_i(x)v)=[x,h_i(x)^{-1}h_i(x)v,i]=[x,v,i]. \end{align*} Therefore $E\cong M\times\mathbb R^k$. The calculation shows that a cocycle of the form $h_i^{-1}h_j$ is completely removed by the frame change $h_i$ on each chart. [/example] This example isolates the basic obstruction to being a product bundle: the transition functions must be removable by changing frames. The same checklist applies not only to vector bundles but also to bundles constructed functorially from a given one. [example: Projectivization And Sphere Bundle] Let $E\to M$ be a rank-$k$ smooth vector bundle with transition functions $g_{ij}:U_i\cap U_j\to GL(k,\mathbb R)$. For $x\in U_i\cap U_j$, define the induced projective transition map by \begin{align*} \overline g_{ij}(x)([v])=[g_{ij}(x)v]. \end{align*} This is well-defined: if $[v]=[v']$, then $v'=av$ for some $a\in\mathbb R^\times$, and linearity gives \begin{align*} [g_{ij}(x)v']=[g_{ij}(x)(av)]=[a\,g_{ij}(x)v]=[g_{ij}(x)v]. \end{align*} The induced maps satisfy the identity condition because \begin{align*} \overline g_{ii}(x)([v])=[g_{ii}(x)v]=[v]. \end{align*} On a triple overlap $U_i\cap U_j\cap U_k$, they satisfy the cocycle condition because \begin{align*} \overline g_{ij}(x)\bigl(\overline g_{jk}(x)([v])\bigr)=\overline g_{ij}(x)([g_{jk}(x)v])=[g_{ij}(x)g_{jk}(x)v]=[g_{ik}(x)v]=\overline g_{ik}(x)([v]). \end{align*} Thus the local pieces $U_i\times\mathbb{RP}^{k-1}$ glue to the projective bundle $P(E)\to M$. Now suppose $E$ has a bundle metric and the local frames are orthonormal, so each $g_{ij}(x)$ is orthogonal. For $v\in S^{k-1}$, orthogonality gives \begin{align*} \|g_{ij}(x)v\|^2=\langle g_{ij}(x)v,g_{ij}(x)v\rangle=\langle v,v\rangle=1. \end{align*} Hence $g_{ij}(x)$ maps $S^{k-1}$ to itself. The same transition matrices satisfy \begin{align*} g_{ii}(x)v=v. \end{align*} On triple overlaps they also satisfy \begin{align*} g_{ij}(x)\bigl(g_{jk}(x)v\bigr)=\bigl(g_{ij}(x)g_{jk}(x)\bigr)v=g_{ik}(x)v. \end{align*} Therefore the pieces $U_i\times S^{k-1}$ glue to the sphere bundle $S(E)\to M$. Projectivization keeps only the lines in the fibres, while the sphere bundle keeps the unit vectors determined by the chosen metric. [/example] ## Comparing Bundles By Their Constructions Once two bundles have been built, the next problem is to recognize whether they are the same up to bundle isomorphism. The course gives several interchangeable viewpoints: compare transition cocycles, construct explicit bundle maps, compare enough sections, or use pullbacks to reduce a bundle over one space to a bundle over another. [definition: Vector Bundle Isomorphism] Let $E\to M$ and $E'\to M$ be smooth rank-$k$ vector bundles over the same base. A vector bundle isomorphism from $E$ to $E'$ is a diffeomorphism $\Phi:E\to E'$ such that the diagram over $M$ commutes and each fibre map \begin{align*} \Phi_x:E_x\to E'_x \end{align*} is a linear isomorphism. [/definition] The definition packages the geometric meaning of equality for bundles: points in different total spaces may be identified, but the base point and fibrewise linear structure must be preserved. This raises the practical question of how to verify such an isomorphism from transition matrices alone; the next result turns the geometric definition into a local matrix equation. [quotetheorem:6134] [citeproof:6134] This theorem is often the fastest route in calculations, but its hypotheses prevent several false comparisons. The two cocycles must be written on the same cover; after refining a cover, the comparison maps must also be refined. The matrices $h_i(x)$ must be invertible at every point, since a rank drop would produce a fibrewise linear map that is not an isomorphism. Smoothness matters as well: a continuous family of invertible matrices can give a topological bundle isomorphism without giving a smooth one. The test proves isomorphism over the fixed base $M$ and does not compare bundles after a non-identity base map. For bundles over spaces assembled from two pieces, the natural question is whether the comparison data can be compressed to the common boundary; clutching provides exactly that compression. [definition: Clutching Construction] Let $A$ and $B$ be smooth manifolds with boundary, let $C$ be a smooth manifold, and identify $C$ diffeomorphically with an open-and-closed union of boundary components of both $A$ and $B$. Suppose the glued space $M=A\cup_C B$ has the smooth structure determined by collars of these boundary components. A rank-$k$ clutching function is a smooth map \begin{align*} \gamma:C\to GL(k,\mathbb R). \end{align*} [/definition] Given a clutching function $\gamma$, the associated vector bundle is obtained from $(A\times \mathbb R^k)\sqcup (B\times \mathbb R^k)$ by identifying $(x,v)\in C\times \mathbb R^k$ on the $B$ side with $(x,\gamma(x)v)$ on the $A$ side. Clutching is transition-function gluing with one overlap after replacing the boundary by a collar neighbourhood. The next issue is classification: if two boundary maps can be deformed into one another, or if local frames on the two halves are changed, should the glued bundle change? The homotopy test gives the answer used in low-dimensional examples. [quotetheorem:6135] [citeproof:6135] The contractibility of the two pieces is what lets all bundle information be pushed to the common boundary: if a piece itself carries a non-product bundle, its internal transition data would remain even after the boundary map is fixed. Collars matter because the clutching map must define smooth transition data on an overlap, not only a set-theoretic boundary identification. Smooth frame changes are also part of the equivalence relation; a nonsmooth boundary change may preserve fibres as sets while failing to produce a smooth bundle isomorphism. Homotopy classification records the vector bundle obtained from this particular two-piece decomposition, but it does not choose a preferred isomorphism and it does not classify extra structures such as a chosen connection or metric. Sections now give a more geometric recognition tool. [quotetheorem:6136] [citeproof:6136] The criterion requires exactly $k$ smooth sections because a rank-$k$ fibre needs $k$ independent vectors to define coordinates. Fewer sections may describe a subbundle but cannot determine every fibre vector; more sections may generate the fibres while still containing relations that do not identify the bundle with $M\times\mathbb R^k$. The independence must hold at every point: for instance, a single section of a line bundle that vanishes at one point cannot give a product coordinate there. Smoothness is also essential, since a pointwise basis varying nonsmoothly gives no smooth inverse frame map. The theorem detects productness, but it does not classify non-product bundles or choose a unique product structure when several global frames exist. Many constructions ask for moving a bundle along a map rather than identifying it with a product, which motivates pullbacks. [definition: Pullback Bundle] Let $f:N\to M$ be a smooth map and let $\pi:E\to M$ be a smooth fibre bundle. The pullback bundle $f^*E\to N$ is \begin{align*} f^*E=\{(y,e)\in N\times E: f(y)=\pi(e)\}, \end{align*} with projection $(y,e)\mapsto y$. [/definition] In a local product chart of $E$ over $U\subset M$, the pullback is a product over $f^{-1}(U)$ by sending $(y,e)$ to $(y,v)$ whenever $e$ has fibre coordinate $v$. The construction should be characterized without choosing charts, so the next theorem states the universal property that makes pullback the canonical solution to this transport problem. [quotetheorem:6102] [citeproof:6102] The covering condition is the hypothesis that makes the formula meaningful. If $F$ does not cover $f$, then $\pi(F(p))$ need not equal $f(q(p))$, so the pair $(q(p),F(p))$ may not lie in the fibre product defining $f^*E$. Smoothness of $f$ and $F$ ensures that the pullback total space and the induced map are smooth; dropping it changes the category of the statement. The universal property characterizes pullbacks up to unique bundle isomorphism over $N$, but it does not say that $f^*E$ is product, nor that different maps $f,g:N\to M$ give the same pullback unless an additional comparison, such as homotopy invariance under suitable hypotheses, is supplied. ## Worked Classification Problems In Low Dimension The low-dimensional examples show how the abstract tests become concrete invariants. In each case the base is covered by simple pieces, so the bundle is controlled by a clutching function on a low-dimensional overlap. [example: Real Line Bundles Over The Circle] Let $S^1=U\cup V$, where $U$ and $V$ are arcs and $U\cap V=C_0\sqcup C_1$ has two connected components. A real line bundle on this cover is described by a transition function \begin{align*} g_{UV}:U\cap V\to GL(1,\mathbb R)=\mathbb R^\times . \end{align*} Since each $C_r$ is connected and $\mathbb R^\times=(-\infty,0)\sqcup(0,\infty)$ has two connected components, the sign of $g_{UV}$ is constant on each $C_r$. Write \begin{align*} \sigma_r=\operatorname{sgn}(g_{UV}|_{C_r})\in\{+1,-1\}. \end{align*} Changing the local frames over $U$ and $V$ by smooth nowhere-zero functions $a:U\to\mathbb R^\times$ and $b:V\to\mathbb R^\times$ replaces the transition function by \begin{align*} g'_{UV}(x)=a(x)g_{UV}(x)b(x)^{-1}. \end{align*} Because $U$ and $V$ are connected, $\operatorname{sgn}(a)$ and $\operatorname{sgn}(b)$ are constant. Also $\operatorname{sgn}(b^{-1})=\operatorname{sgn}(b)$, since $b(x)$ and $b(x)^{-1}$ are either both positive or both negative. Hence on each component $C_r$, \begin{align*} \sigma'_r=\operatorname{sgn}(a)\sigma_r\operatorname{sgn}(b). \end{align*} Multiplying the two component signs gives \begin{align*} \sigma'_0\sigma'_1=\bigl(\operatorname{sgn}(a)\sigma_0\operatorname{sgn}(b)\bigr)\bigl(\operatorname{sgn}(a)\sigma_1\operatorname{sgn}(b)\bigr). \end{align*} Reordering the factors in $\{+1,-1\}$ gives \begin{align*} \sigma'_0\sigma'_1=\operatorname{sgn}(a)^2\operatorname{sgn}(b)^2\sigma_0\sigma_1. \end{align*} Since $\operatorname{sgn}(a)^2=\operatorname{sgn}(b)^2=1$, this becomes \begin{align*} \sigma'_0\sigma'_1=\sigma_0\sigma_1. \end{align*} Thus the product $\sigma_0\sigma_1$ is unchanged by changing local frames. If $\sigma_0\sigma_1=+1$, the two signs can be made both positive by changing one local frame by a negative constant when necessary, so the gluing has no sign reversal and gives the product line bundle. If $\sigma_0\sigma_1=-1$, one sign reversal remains after every frame change; a nonzero vector transported once around the circle returns as its negative, which is the Mobius line bundle. A nowhere-zero global section would choose a nonzero vector continuously in every fibre and return to the same value after one loop, so this sign reversal is exactly the obstruction to being a product line bundle. [/example] This example is the first appearance of a characteristic obstruction in elementary form: a line bundle over $S^1$ is classified by its orientation monodromy. For complex line bundles over $S^2$, the analogous invariant is not a sign but a winding number. [example: Complex Line Bundles Over The Two-Sphere] Write $S^2=D^2_+\cup D^2_-$ with common boundary the equator $S^1$, and identify $GL(1,\mathbb C)$ with $\mathbb C^\times$ by letting a nonzero complex number act by multiplication on $\mathbb C$. A complex line bundle built from a clutching function $\gamma:S^1\to \mathbb C^\times$ is obtained by identifying boundary points by $(x,v)_-\sim (x,\gamma(x)v)_+$. Because $\gamma(x)\neq 0$, decompose it as \begin{align*} \gamma(x)=|\gamma(x)|\frac{\gamma(x)}{|\gamma(x)|}. \end{align*} Here $|\gamma(x)|\in (0,\infty)$ and $\gamma(x)/|\gamma(x)|\in U(1)$. The formula \begin{align*} H(x,t)=\bigl((1-t)|\gamma(x)|+t\bigr)\frac{\gamma(x)}{|\gamma(x)|} \end{align*} defines a homotopy in $\mathbb C^\times$, since $(1-t)|\gamma(x)|+t>0$ for every $0\leq t\leq 1$. Thus every clutching map is homotopic to its normalized $U(1)$-valued map $x\mapsto \gamma(x)/|\gamma(x)|$, so its homotopy class is measured by the degree of this circle map. For the model map $\gamma_n:S^1\to U(1)$ given by $\gamma_n(z)=z^n$, parametrize $S^1$ by $z=e^{i\theta}$. Then \begin{align*} \gamma_n(e^{i\theta})=(e^{i\theta})^n=e^{in\theta}. \end{align*} As $\theta$ runs from $0$ to $2\pi$, the image angle runs from $0$ to $2\pi n$, so $\deg(\gamma_n)=n$. By the Classification of Complex Line Bundles over $S^2$ theorem and the degree classification of maps $S^1\to U(1)$, the clutching maps $\gamma_m$ and $\gamma_n$ define isomorphic complex line bundles only when $m=n$. Tensor product multiplies clutching functions. Indeed, on the overlap the first bundle identifies $v_-$ with $\gamma_m(z)v_+$, and the second identifies $w_-$ with $\gamma_n(z)w_+$. Therefore the tensor product identifies $v_-\otimes w_-$ with $(\gamma_m(z)v_+)\otimes(\gamma_n(z)w_+)$. By bilinearity of tensor product, \begin{align*} (\gamma_m(z)v_+)\otimes(\gamma_n(z)w_+)=\gamma_m(z)\gamma_n(z)(v_+\otimes w_+). \end{align*} For the model maps, \begin{align*} \gamma_m(z)\gamma_n(z)=z^m z^n=z^{m+n}=\gamma_{m+n}(z). \end{align*} Thus the tensor product of the complex line bundles of degrees $m$ and $n$ has degree $m+n$. [/example] The tangent bundle of the sphere is a rank-two real example of the same method. Its clutching function records how tangent frames rotate along the equator when the northern and southern frames are compared. [example: Clutching Function For The Tangent Bundle Of The Two-Sphere] Parametrize the equator by $z=e^{i\theta}$, and identify $SO(2)$ with $U(1)$ by letting $e^{i\alpha}$ act on $\mathbb R^2\cong\mathbb C$ as rotation through angle $\alpha$. Using stereographic coordinates from the two poles, the overlap coordinate change is \begin{align*} w=\frac{1}{z}. \end{align*} For a tangent vector $\xi\in T_z\mathbb C\cong\mathbb C$, its derivative is \begin{align*} d\left(\frac{1}{z}\right)_z(\xi)=-z^{-2}\xi. \end{align*} On the equator this gives \begin{align*} -z^{-2}\xi=-e^{-2i\theta}\xi. \end{align*} Since $-1=e^{i\pi}$, the same factor is \begin{align*} -e^{-2i\theta}=e^{i\pi}e^{-2i\theta}=e^{i(\pi-2\theta)}. \end{align*} Thus one standard comparison convention gives the clutching map \begin{align*} \gamma(e^{i\theta})=R_{\pi-2\theta}. \end{align*} As $\theta$ increases from $0$ to $2\pi$, the angle $\pi-2\theta$ changes from $\pi$ to $\pi-4\pi$, so the map winds twice around $SO(2)\cong S^1$ with negative sign in this convention. If the northern and southern collar comparison is reversed, the clutching function is replaced by $\gamma^{-1}$, whose angle changes by $+4\pi$ and hence has degree $2$. With the orientation convention used for the Euler number of $TS^2$, the tangent bundle therefore has clutching degree $2$, recording that tangent frames rotate twice while one travels once around the equator. [/example] These computations also clarify why auxiliary choices usually do not affect the answer. Different covers, collars, or local frames change the clutching function by the allowed operations above, so the resulting invariant is the homotopy class or degree rather than a particular formula. ## Splittings And Associated Constructions The last recognition principle is that some bundle constructions require extra choices but produce canonical isomorphism classes once the choices are varied. Bundle metrics are the standard device for turning exact sequences into direct sums and for reducing transition functions to orthogonal or unitary groups. [quotetheorem:6137] [citeproof:6137] The smooth vector bundle and metric hypotheses are substantial. The metric is obtained from partitions of unity, so paracompactness of the smooth base is being used; on spaces without such partitions, a global smooth complement need not be available. Smoothness is also needed for the complement to be a smooth subbundle rather than only a fibrewise linear complement. The vector bundle setting matters because fibrewise inner products and ranks make the orthogonal-complement argument work; short exact sequences of sheaves or more general module objects need not split. The result also gives existence rather than a preferred splitting: different metrics can produce different complements. A different problem arises when a construction changes the fibre itself, such as passing from a vector space to a tensor power, a projective space, or another representation space; transition functions must then be transported through the corresponding group action. [quotetheorem:6138] [citeproof:6138] The representation theorem explains why many familiar bundles are obtained by applying a fibrewise construction to transition functions. Projectivization is a slightly nonlinear version of the same idea, and it asks which part of a linear transition map still matters after vectors are replaced by lines. [example: Projectivization Depends On Scalar Classes] Let $E\to M$ be a rank-$k$ vector bundle with transition functions $g_{ij}:U_i\cap U_j\to GL(k,\mathbb R)$. The induced projective transition function sends a line $[v]\in \mathbb{RP}^{k-1}$, with $v\neq 0$, to \begin{align*} \overline g_{ij}(x)([v])=[g_{ij}(x)v]. \end{align*} Now replace $g_{ij}$ by \begin{align*} g'_{ij}(x)=\lambda_{ij}(x)g_{ij}(x), \end{align*} where $\lambda_{ij}:U_i\cap U_j\to \mathbb R^\times$ is nowhere zero. The induced projective transition function is \begin{align*} \overline g'_{ij}(x)([v])=[g'_{ij}(x)v]. \end{align*} Substituting the definition of $g'_{ij}$ gives \begin{align*} \overline g'_{ij}(x)([v])=[\lambda_{ij}(x)g_{ij}(x)v]. \end{align*} Since $\lambda_{ij}(x)\in\mathbb R^\times$, the vectors $\lambda_{ij}(x)g_{ij}(x)v$ and $g_{ij}(x)v$ span the same one-dimensional subspace of $\mathbb R^k$. Therefore \begin{align*} [\lambda_{ij}(x)g_{ij}(x)v]=[g_{ij}(x)v]. \end{align*} Hence \begin{align*} \overline g'_{ij}(x)([v])=\overline g_{ij}(x)([v]). \end{align*} Thus multiplying transition matrices by nowhere-zero scalar functions changes the parametrization of nonzero fibre vectors, but it does not change the induced transition data on lines. The projective bundle $P(E)$ therefore records the scalar classes of the transition functions, not lengths or chosen representatives of nonzero vectors. [/example] The synthesis of the course is therefore the working dictionary promised in the introduction. Local product charts produce cocycles; cocycles glue bundles; frame changes identify cocycles that describe isomorphic bundles; sections test whether a vector bundle is a product; pullbacks transport bundles; clutching functions reduce many classifications to homotopy classes; and associated constructions build new bundles functorially from old ones. ## Beyond This Course This course stops at the point where transition data, sections, pullbacks, clutching functions, and associated constructions have become usable tools. The next layer is to make the structure group itself the main object. Principal bundles replace a chosen fibre vector space by a group acting freely and transitively on each fibre, and many vector bundles are recovered from principal frame bundles by applying a representation. This is the natural setting for gauge transformations, reductions of structure group, and associated bundles with nonlinear fibres. Connections add the geometric operation that is missing from bare transition data: they let one compare nearby fibres and differentiate sections. Curvature then measures the failure of this comparison to be path independent, a theme that continues in Androma's [Cambridge III Riemannian Geometry](/page/Cambridge%20III%20Riemannian%20Geometry) notes. Characteristic classes turn that curvature, or the underlying transition cocycles, into cohomological obstructions, using the language developed in [Differential Forms I: Exterior Calculus](/page/Differential%20Forms%20I%3A%20Exterior%20Calculus), [Differential Forms and de Rham Cohomology](/page/Differential%20Forms%20and%20de%20Rham%20Cohomology), and [Differential Forms II: Manifolds and Cohomology](/page/Differential%20Forms%20II%3A%20Manifolds%20and%20Cohomology). In this way the first examples in the course, such as the Mobius bundle and the tangent bundle of a sphere, are early forms of the same obstruction theory that later produces Stiefel-Whitney classes, Euler classes, Chern classes, and Pontryagin classes. The clutching viewpoint also points toward classifying spaces and K-theory. Instead of classifying one bundle by one transition function, classifying spaces organize all bundles with a fixed structure group up to homotopy, while K-theory records stable differences of vector bundles. The elementary operations on bundles in this course are the concrete precursors of those stable algebraic constructions. ## References - Dale Husemoller, *Fibre Bundles*. - Norman Steenrod, *The Topology of Fibre Bundles*. - John Milnor and James Stasheff, *Characteristic Classes*. - Raoul Bott and Loring Tu, *Differential Forms in Algebraic Topology*. - Androma, [Differential Forms I: Exterior Calculus](/page/Differential%20Forms%20I%3A%20Exterior%20Calculus). - Androma, [Differential Forms and de Rham Cohomology](/page/Differential%20Forms%20and%20de%20Rham%20Cohomology). - Androma, [Differential Forms II: Manifolds and Cohomology](/page/Differential%20Forms%20II%3A%20Manifolds%20and%20Cohomology). - Androma, [Cambridge III Riemannian Geometry](/page/Cambridge%20III%20Riemannian%20Geometry).