This course develops the foundational language and core techniques of symplectic geometry, the study of even-dimensional spaces equipped with a closed, nondegenerate 2-form. It begins at the linear level, where symplectic structures have no local invariants, and then moves to manifolds, where the same algebraic ideas control geometry, dynamics, and topology. Along the way, the course connects the abstract theory to concrete examples such as cotangent bundles, Hamiltonian systems, and geometric structures arising from physics and mechanics. The main themes are the passage from linear symplectic algebra to global geometry, the special role of Hamiltonian vector fields and Poisson brackets, and the rigidity phenomena that distinguish symplectic geometry from ordinary differential topology. After establishing local normal forms through Darboux’s theorem and Moser’s method, the course studies symplectic and Hamiltonian isotopies, contact-type hypersurfaces, moment maps, and symplectic reduction. These ideas are then applied to toric and other global examples, before concluding with deeper topics such as rigidity beyond Darboux and integrable systems with action-angle coordinates. Each chapter builds on the previous one, moving from basic structures to powerful global tools and culminating in the interplay between symmetry, reduction, and dynamics. # Introduction Symplectic geometry is the geometry of spaces that carry a nondegenerate closed $2$-form. It grew from Hamiltonian mechanics, where the same structure records position, momentum, and the evolution of a physical system, but its modern form is a central part of differential geometry and topology. This course develops the subject from its linear algebraic core to Hamiltonian dynamics, moment maps, and symplectic reduction. The guiding theme is the tension between local flexibility and global rigidity: locally every symplectic manifold has the same model, while globally volume, cohomology, and holomorphic-curve phenomena impose strong constraints. The prerequisites are smooth manifolds, tangent and cotangent bundles, differential forms, de Rham cohomology, basic Lie groups and Lie algebras, multilinear algebra, and elementary classical mechanics. The notes will use these tools without rebuilding them from first principles, but the first lectures will recall the pieces of linear algebra and differential forms that symplectic geometry uses in a distinctive way. ## The Geometry Behind Hamiltonian Mechanics What geometric structure is preserved by Hamiltonian evolution, and why is it different from the metric geometry of lengths and angles? Classical phase space has coordinates $(q_1,\dots,q_n,p_1,\dots,p_n)$, where $q_i$ records position and $p_i$ records momentum. The fundamental object is not a Riemannian metric but the $2$-form \begin{align*} \omega_0 = \sum_{i=1}^n dq_i \wedge dp_i. \end{align*} This form pairs position directions with momentum directions and gives the coordinate-free language for Hamiltonian equations. [motivation] ### Phase Space Rather Than Configuration Space A mechanical system with $n$ degrees of freedom is usually described by a configuration manifold $Q$. The actual state of the system also includes momentum, so the natural phase space is the cotangent bundle $T^*Q$. A point of $T^*Q$ is a pair $(q,p)$ with $q \in Q$ and $p \in T_q^*Q$, and the cotangent bundle carries a canonical $2$-form independent of coordinate choices. ### Conservation As Geometry Hamiltonian evolution is generated by a function $H:T^*Q \to \mathbb R$, the Hamiltonian. Instead of measuring the gradient of $H$ using a metric, symplectic geometry converts $dH$ into a vector field by contracting with the symplectic form. The resulting flow preserves the symplectic form, and hence preserves the associated phase-space volume. ### Why Closed Nondegenerate Two-Forms Nondegeneracy says that every covector can be represented by contraction with a unique vector. Closedness is the infinitesimal condition that makes the local normal form and Hamiltonian conservation laws work. Together these two requirements produce a geometry with no local curvature invariant but many global constraints. [/motivation] The first use of the theory is therefore explanatory: it gives a coordinate-free account of Hamiltonian mechanics. The same language also creates questions with no mechanical origin, such as which manifolds admit symplectic forms and which embeddings preserve symplectic structure. ## Linear Algebra As The Local Model What must be understood before putting a symplectic form on a manifold? At each point of a symplectic manifold there is a tangent space equipped with a nondegenerate alternating [bilinear form](/page/Bilinear%20Form). The hypotheses cannot be weakened without changing the geometry: a degenerate alternating form leaves some nonzero directions invisible, while a symmetric nondegenerate form measures lengths and signs rather than paired position-momentum directions. The first chapter studies this purely linear situation because every local and infinitesimal argument later depends on it. [definition: Symplectic Vector Space] A symplectic [vector space](/page/Vector%20Space) is a finite-dimensional real vector space $V$ equipped with a bilinear form $\omega:V \times V \to \mathbb R$ such that: 1. $\omega(v,w)=-\omega(w,v)$ for all $v,w \in V$; 2. if $v \in V$ satisfies $\omega(v,w)=0$ for all $w \in V$, then $v=0$. [/definition] The alternating condition forces every vector to pair to zero with itself, so the information in $\omega$ is entirely in how different directions interact. Nondegeneracy is the replacement for invertibility: it lets the form identify vectors with covectors. This identification is the linear operation behind Hamiltonian vector fields. [example: Standard Symplectic Vector Space] Let $V=\mathbb R^{2n}$ with coordinate functions $(q_1,\dots,q_n,p_1,\dots,p_n)$, and define \begin{align*} \omega_0=\sum_{i=1}^n dq_i\wedge dp_i. \end{align*} We verify directly that $\omega_0$ is alternating and nondegenerate. For tangent vectors \begin{align*} u=\sum_{i=1}^n a_i\frac{\partial}{\partial q_i}+b_i\frac{\partial}{\partial p_i} \end{align*} and \begin{align*} v=\sum_{i=1}^n c_i\frac{\partial}{\partial q_i}+d_i\frac{\partial}{\partial p_i}, \end{align*} the identity $(\alpha\wedge\beta)(u,v)=\alpha(u)\beta(v)-\alpha(v)\beta(u)$ gives \begin{align*} (dq_i\wedge dp_i)(u,v)=dq_i(u)dp_i(v)-dq_i(v)dp_i(u)=a_i d_i-c_i b_i. \end{align*} Therefore \begin{align*} \omega_0(u,v)=\sum_{i=1}^n(a_i d_i-b_i c_i). \end{align*} Interchanging $u$ and $v$ gives \begin{align*} \omega_0(v,u)=\sum_{i=1}^n(c_i b_i-d_i a_i)=-\sum_{i=1}^n(a_i d_i-b_i c_i)=-\omega_0(u,v), \end{align*} so $\omega_0$ is alternating. To prove nondegeneracy, suppose $\omega_0(u,v)=0$ for every $v\in V$. Taking $v=\partial/\partial p_j$ gives $c_i=0$ for all $i$, $d_j=1$, and $d_i=0$ for $i\ne j$, hence \begin{align*} 0=\omega_0\left(u,\frac{\partial}{\partial p_j}\right)=a_j. \end{align*} Taking $v=\partial/\partial q_j$ gives $c_j=1$, $c_i=0$ for $i\ne j$, and $d_i=0$ for all $i$, hence \begin{align*} 0=\omega_0\left(u,\frac{\partial}{\partial q_j}\right)=-b_j. \end{align*} Thus $a_j=b_j=0$ for every $j$, so $u=0$. Hence $(\mathbb R^{2n},\omega_0)$ is a symplectic vector space, and the formula shows explicitly that each $q_i$-direction pairs only with its corresponding $p_i$-direction. [/example] The standard model shows what the desired coordinates should look like, but it does not yet explain whether an arbitrary nondegenerate alternating form can be put into that shape. The next theorem solves that classification problem and gives the normal form used in almost every later local argument. [quotetheorem:3295] [citeproof:3295] This theorem explains why symplectic geometry has no pointwise eigenvalue-type invariant for the form itself. Both hypotheses are doing real work: if the form is alternating but degenerate, for instance the zero form on $\mathbb R^2$, no basis can make it look like $dq \wedge dp$ because some nonzero vector pairs to zero with every vector. Alternation also matters, since a symmetric nondegenerate form is classified by signature rather than by a universal standard symplectic matrix. The theorem is only a linear classification at one vector space; it does not choose bases smoothly over a manifold, nor does it compare different global symplectic forms. The difficulty of the subject begins when these pointwise normal forms have to be chosen coherently over a manifold and compared with global topology. ## From Forms On Vector Spaces To Forms On Manifolds How does a bilinear form on each tangent space become a geometric structure on a manifold? A differential $2$-form assigns an alternating bilinear form to every tangent space, and smoothness relates these assignments from point to point. Symplectic geometry asks for this family to be nondegenerate everywhere and closed under exterior differentiation. [definition: Symplectic Manifold] A symplectic manifold is a pair $(M,\omega)$, where $M$ is a [smooth manifold](/page/Smooth%20Manifold) and $\omega \in \Omega^2(M)$ satisfies: 1. $d\omega=0$; 2. for every $p \in M$, the bilinear form $\omega_p:T_pM \times T_pM \to \mathbb R$ is nondegenerate. [/definition] The definition immediately forces $M$ to be even-dimensional. It also supplies a canonical orientation, since powers of the form give a nowhere-vanishing top-degree form. [example: Cotangent Bundles] Let $Q$ be an $n$-dimensional smooth manifold. On a coordinate chart with coordinates $(q_1,\dots,q_n)$, the induced coordinates on $T^*Q$ are $(q_1,\dots,q_n,p_1,\dots,p_n)$, where a covector is written as $\sum_{i=1}^n p_i\,dq_i$. The tautological $1$-form is \begin{align*} \lambda=\sum_{i=1}^n p_i\,dq_i. \end{align*} Using $d(f\alpha)=df\wedge\alpha+f\,d\alpha$ and $d(dq_i)=0$, we get \begin{align*} d\lambda=\sum_{i=1}^n d(p_i\,dq_i)=\sum_{i=1}^n dp_i\wedge dq_i=-\sum_{i=1}^n dq_i\wedge dp_i. \end{align*} Thus, with the convention $\omega=-d\lambda$, the cotangent bundle carries the local form \begin{align*} \omega=\sum_{i=1}^n dq_i\wedge dp_i. \end{align*} This form is closed because \begin{align*} d\omega=\sum_{i=1}^n d(dq_i\wedge dp_i)=\sum_{i=1}^n \bigl(d(dq_i)\wedge dp_i-dq_i\wedge d(dp_i)\bigr)=0. \end{align*} For tangent vectors $u=\sum_i a_i\partial/\partial q_i+b_i\partial/\partial p_i$ and $v=\sum_i c_i\partial/\partial q_i+d_i\partial/\partial p_i$, the wedge formula gives \begin{align*} \omega(u,v)=\sum_{i=1}^n(a_i d_i-b_i c_i). \end{align*} If $\omega(u,v)=0$ for every $v$, then taking $v=\partial/\partial p_j$ gives $a_j=0$, and taking $v=\partial/\partial q_j$ gives $-b_j=0$. Hence all $a_j$ and $b_j$ vanish, so $u=0$. Therefore the canonical cotangent-bundle form is symplectic, and locally it is exactly the standard position-momentum form. [/example] Cotangent bundles show that symplectic manifolds occur naturally, but they also raise a basic structural question: what geometric data does any symplectic form automatically provide? The next theorem answers this by turning the form into a canonical top-degree form, which is the source of orientation and volume in the subject. [quotetheorem:10035] [citeproof:10035] The volume form is the first global invariant that appears in the course. Nondegeneracy and even dimension are essential here: on $\mathbb R^3$ a $2$-form cannot have a nonzero third-half wedge power, and on $\mathbb R^2$ the degenerate form $0\,dx \wedge dy$ produces no volume element. More generally, a closed $2$-form may exist on an odd-dimensional or degenerate space, but it cannot define a symplectic orientation by this construction. The theorem does not say that the symplectic volume determines the symplectic form, or even the symplectic manifold, because many inequivalent forms can have the same total volume. Its main role here is to identify the first place where symplectic geometry stops being only a pointwise linear-algebraic condition: a symplectic form canonically orients the whole manifold and gives a positive volume density. Later compactness obstructions and cohomological constraints will use this volume form as their global input. ## Local Flexibility And Global Rigidity Cotangent bundles give a large class of symplectic manifolds whose canonical form is locally the standard position-momentum form. For an arbitrary symplectic manifold, however, this has not yet been shown: the definition only gives a closed nondegenerate $2$-form on each tangent space. The next local question is whether the linear normal form can be made smoothly in coordinates around every point. Darboux's theorem answers this in the strongest possible way: every symplectic manifold is locally symplectomorphic to the standard model. This is the first result that justifies the slogan that symplectic geometry has no local curvature-type invariant, and it is a major contrast with Riemannian geometry, where curvature gives local invariants. [quotetheorem:10036] [citeproof:10036] Darboux's theorem says that local symplectic geometry is governed by normal forms rather than curvature. Closedness is essential: a nondegenerate $2$-form with $d\omega \ne 0$ need not be locally equivalent to $\sum_i dq_i \wedge dp_i$, since pullback commutes with exterior differentiation and the standard form is closed. Nondegeneracy is also essential, because the rank of a $2$-form is preserved by coordinate changes; a degenerate form cannot be transformed into the full-rank standard form. The theorem is strictly local: it does not say that two symplectic manifolds are globally symplectomorphic, or that two forms on the same manifold are globally equivalent. The rest of the course investigates why this local flexibility does not make the subject globally featureless. [example: Volume Is Not The Whole Story] Let $\Omega=\omega_0^n/n!$ be the [symplectic volume form](/theorems/10035) from *Symplectic Volume Form*. If $\varphi:U\to \mathbb R^{2n}$ is a symplectic embedding, then $\varphi^*\omega_0=\omega_0$, so \begin{align*} \varphi^*\Omega=\varphi^*\left(\frac{\omega_0^n}{n!}\right)=\frac{(\varphi^*\omega_0)^n}{n!}=\frac{\omega_0^n}{n!}=\Omega. \end{align*} Therefore, for any compact domain $U$ on which the change-of-variables formula applies, \begin{align*} \operatorname{Vol}(\varphi(U))=\int_{\varphi(U)}\Omega=\int_U \varphi^*\Omega=\int_U\Omega=\operatorname{Vol}(U). \end{align*} Thus volume is a necessary obstruction: a symplectic embedding cannot place a domain into another domain of smaller symplectic volume. The obstruction is not complete. Let \begin{align*} B^{2n}(R)=\{(q,p)\in\mathbb R^{2n}:q_1^2+p_1^2+\cdots+q_n^2+p_n^2<R^2\} \end{align*} and \begin{align*} Z^{2n}(r)=\{(q,p)\in\mathbb R^{2n}:q_1^2+p_1^2<r^2\}. \end{align*} For $n\ge 2$, the cylinder $Z^{2n}(r)=B^2(r)\times\mathbb R^{2n-2}$ has infinite ordinary volume, while $B^{2n}(R)$ has finite volume $\pi^nR^{2n}/n!$. Hence volume alone gives no contradiction to embedding $B^{2n}(R)$ into $Z^{2n}(r)$, even when $R>r$. The *Gromov non-squeezing theorem* says that such a symplectic embedding exists only if $R\le r$. The missing invariant is therefore not total volume but the two-dimensional symplectic radius seen in each conjugate $(q_i,p_i)$-plane, which is a first instance of global rigidity beyond Darboux local normal form. [/example] This rigidity-flexibility contrast will reappear in several forms. Darboux's theorem removes local invariants, Moser's method shows when families of forms are equivalent, and non-squeezing reveals constraints invisible to ordinary volume comparison. ## Hamiltonian Vector Fields And Flows How does a function generate motion on a symplectic manifold? In Riemannian geometry a function produces a gradient vector field using the metric. In symplectic geometry a function produces a Hamiltonian vector field using the nondegenerate $2$-form. [definition: Hamiltonian Vector Field] Let $(M,\omega)$ be a symplectic manifold and let $H:M \to \mathbb R$ be a smooth function. The Hamiltonian vector field $X_H \in \mathfrak X(M)$ is defined by \begin{align*} \iota_{X_H}\omega = dH. \end{align*} [/definition] The sign convention for $X_H$ varies between texts; this course uses the convention displayed above unless a later comparison states otherwise. Nondegeneracy of $\omega$ gives existence and uniqueness of $X_H$ at each point. [example: Hamilton Equations In Standard Coordinates] On $(\mathbb R^{2n},\omega_0)$, take \begin{align*} \omega_0=\sum_{i=1}^n dq_i\wedge dp_i \end{align*} and let $H:\mathbb R^{2n}\to\mathbb R$ be smooth. Write the Hamiltonian vector field in coordinates as \begin{align*} X_H=\sum_{i=1}^n a_i\frac{\partial}{\partial q_i}+b_i\frac{\partial}{\partial p_i}. \end{align*} For an arbitrary tangent vector \begin{align*} Y=\sum_{i=1}^n c_i\frac{\partial}{\partial q_i}+d_i\frac{\partial}{\partial p_i}, \end{align*} the wedge formula gives \begin{align*} (dq_i\wedge dp_i)(X_H,Y)=dq_i(X_H)dp_i(Y)-dq_i(Y)dp_i(X_H)=a_i d_i-c_i b_i. \end{align*} Therefore \begin{align*} (\iota_{X_H}\omega_0)(Y)=\omega_0(X_H,Y)=\sum_{i=1}^n(a_i d_i-b_i c_i). \end{align*} On the other hand, \begin{align*} dH=\sum_{i=1}^n \frac{\partial H}{\partial q_i}\,dq_i+\frac{\partial H}{\partial p_i}\,dp_i, \end{align*} so evaluating on the same vector $Y$ gives \begin{align*} dH(Y)=\sum_{i=1}^n\frac{\partial H}{\partial q_i}c_i+\frac{\partial H}{\partial p_i}d_i. \end{align*} The defining equation $\iota_{X_H}\omega_0=dH$ means these two expressions are equal for every choice of the coefficients $c_i,d_i$: \begin{align*} \sum_{i=1}^n(a_i d_i-b_i c_i)=\sum_{i=1}^n\frac{\partial H}{\partial q_i}c_i+\frac{\partial H}{\partial p_i}d_i. \end{align*} Comparing the coefficient of $d_i$ gives \begin{align*} a_i=\frac{\partial H}{\partial p_i}. \end{align*} Comparing the coefficient of $c_i$ gives \begin{align*} -b_i=\frac{\partial H}{\partial q_i}. \end{align*} Hence \begin{align*} b_i=-\frac{\partial H}{\partial q_i}. \end{align*} If $\gamma(t)=(q_1(t),\dots,q_n(t),p_1(t),\dots,p_n(t))$ is an integral curve of $X_H$, then \begin{align*} \dot\gamma(t)=\sum_{i=1}^n \dot q_i(t)\frac{\partial}{\partial q_i}+\dot p_i(t)\frac{\partial}{\partial p_i}=X_H(\gamma(t)). \end{align*} Equating the coordinate coefficients gives Hamilton's equations: \begin{align*} \dot q_i=\frac{\partial H}{\partial p_i} \end{align*} and \begin{align*} \dot p_i=-\frac{\partial H}{\partial q_i}. \end{align*} Thus the coordinate form of the abstract equation $\iota_{X_H}\omega_0=dH$ is exactly the usual position-momentum system from Hamiltonian mechanics. [/example] The coordinate computation recovers the usual equations of motion, but the geometric definition is useful only if the resulting flow respects the structure that produced it. A Hamiltonian trajectory is built from $\omega$ by solving $\iota_{X_H}\omega=dH$; if its flow distorted $\omega$, the dynamics would immediately leave behind the geometry used to define it. The key structural question is therefore whether Hamiltonian time evolution preserves the symplectic form, so that the motion is symplectic and not merely smooth. [quotetheorem:6844] [citeproof:6844] This theorem is the geometric conservation law behind Hamiltonian mechanics. The closedness of $\omega$ is used exactly in Cartan's formula: for a nondegenerate but non-closed $2$-form, the extra term $\iota_{X_H}d\omega$ need not vanish, so the same calculation does not prove preservation. The sign convention for $X_H$ changes the direction of the Hamiltonian vector field and the signs in Hamilton's equations, but either consistent convention gives a flow preserving $\omega$ after the corresponding definition is fixed. The local-flow hypothesis is also a real limitation: on a noncompact manifold the vector field may fail to be complete, so $\varphi_t$ may not exist for all $t \in \mathbb R$. Later lectures will refine this conservation law into moment maps, Noether-type statements, and reduction procedures. ## The Route Through The Course What sequence of ideas turns the definition of a symplectic manifold into a working geometric toolkit? The course begins with linear symplectic algebra, because it supplies the normal forms and subspace vocabulary used everywhere else. It then moves to manifolds, local coordinates, and the Moser argument, which explains why closedness is the correct differential condition. The middle part of the course studies Hamiltonian vector fields, Poisson brackets, symplectomorphisms, and group actions. Moment maps turn infinitesimal symmetries into functions, and symplectic reduction constructs new symplectic manifolds from level sets modulo symmetry groups. These topics connect geometry with mechanics and representation theory. The final part introduces global phenomena. Lagrangian submanifolds, symplectic capacities, and Gromov non-squeezing show that symplectic embeddings are much more constrained than volume-preserving embeddings. By the end of the course, the reader should be able to move between the linear algebra, differential-form, dynamical, and topological viewpoints that make symplectic geometry a single subject. The course has now shown why symplectic embeddings are far more rigid than volume alone would suggest. To understand that rigidity from the ground up, we first isolate the linear algebra of nondegenerate skew-symmetric forms and the maps that preserve them. # 1. Symplectic Linear Algebra Symplectic geometry begins with a piece of linear algebra: a vector space equipped with a nondegenerate skew-symmetric bilinear form. This chapter isolates that algebraic core before any manifold enters the story. The main questions are how such forms can look, which linear maps preserve them, and how subspaces behave under the symplectic complement operation. These answers become the pointwise model for symplectic manifolds, Darboux charts, Hamiltonian vector fields, and reduction. ## Skew Forms and the Standard Symplectic Vector Space What should replace an [inner product](/page/Inner%20Product) if the geometry is meant to measure oriented two-dimensional area rather than length? A skew-symmetric bilinear form is the linear object that records signed area in every two-plane. The symplectic condition is the requirement that this area pairing has no invisible directions. [definition: Skew-Symmetric Bilinear Form] Let $V$ be a real vector space. A skew-symmetric bilinear form on $V$ is a map $\omega: V \times V \to \mathbb R$ such that, for all $u,v,w \in V$ and $a,b \in \mathbb R$, the map is linear in each argument and satisfies \begin{align*} \omega(u,v) = -\omega(v,u). \end{align*} [/definition] Skew-symmetry forces $\omega(v,v)=0$ for every $v\in V$, so a symplectic form never measures the length of a single vector. Instead, it pairs two independent directions. This raises the basic obstruction question: when can a skew form be strong enough to detect every vector? [definition: Symplectic Vector Space] A symplectic vector space is a pair $(V,\omega)$, where $V$ is a finite-dimensional real vector space and $\omega:V\times V\to\mathbb R$ is a skew-symmetric bilinear form satisfying the nondegeneracy condition \begin{align*} \{v\in V: \omega(v,w)=0 \text{ for all } w\in V\}=\{0\}. \end{align*} [/definition] The definition can be recast as a statement about the [linear map](/page/Linear%20Map) $V\to V^*$ sending $v$ to $\omega(v,\cdot)$. Nondegeneracy says this map is an isomorphism. Since skew-symmetric matrices have special determinant behaviour, the first structural question is whether every dimension can support such an isomorphism. [quotetheorem:3295] [citeproof:3295] This parity restriction uses both finite-dimensionality and nondegeneracy. If nondegeneracy is dropped, odd-dimensional spaces carry skew forms, for instance the zero form on $\mathbb R^3$; what fails is that every vector is then invisible to the pairing. The theorem also does not say that every even-dimensional skew form is symplectic, since the zero form on $\mathbb R^{2n}$ is still degenerate. It only identifies the first obstruction, and the next step is to build the basic nondegenerate model by pairing each position coordinate with a corresponding momentum coordinate. [example: Standard Symplectic Form On Real Space] Let $\mathbb R^{2n}$ have coordinates $(q_1,\dots,q_n,p_1,\dots,p_n)$, and write a vector as $u=(a,b)$ with $a=(a_1,\dots,a_n)$ in the $q$-coordinates and $b=(b_1,\dots,b_n)$ in the $p$-coordinates. Define \begin{align*} \omega_0=\sum_{i=1}^n dq_i\wedge dp_i. \end{align*} For $u=(a,b)$ and $v=(c,d)$, the coordinate covectors satisfy $dq_i(u)=a_i$, $dp_i(u)=b_i$, $dq_i(v)=c_i$, and $dp_i(v)=d_i$. Hence \begin{align*} (dq_i\wedge dp_i)(u,v)=dq_i(u)dp_i(v)-dq_i(v)dp_i(u)=a_i d_i-c_i b_i. \end{align*} Summing over $i$ gives \begin{align*} \omega_0(u,v)=\sum_{i=1}^n(a_i d_i-b_i c_i). \end{align*} Interchanging $u$ and $v$ gives \begin{align*} \omega_0(v,u)=\sum_{i=1}^n(c_i b_i-d_i a_i)=-\sum_{i=1}^n(a_i d_i-b_i c_i)=-\omega_0(u,v), \end{align*} so $\omega_0$ is skew-symmetric. To check nondegeneracy, suppose $\omega_0(u,v)=0$ for every $v\in\mathbb R^{2n}$. Taking $v=(0,e_j)$, where $e_j$ is the $j$th standard vector in $\mathbb R^n$, gives \begin{align*} 0=\omega_0((a,b),(0,e_j))=\sum_{i=1}^n(a_i(e_j)_i-b_i0_i)=a_j. \end{align*} Taking $v=(e_j,0)$ gives \begin{align*} 0=\omega_0((a,b),(e_j,0))=\sum_{i=1}^n(a_i0_i-b_i(e_j)_i)=-b_j. \end{align*} Thus $a_j=0$ and $b_j=0$ for every $j$, so $u=0$. Therefore $\omega_0$ is nondegenerate, and $(\mathbb R^{2n},\omega_0)$ is the standard symplectic vector space. [/example] In the ordered basis $(e_1,\dots,e_n,f_1,\dots,f_n)$, where $e_i$ is the $q_i$-direction and $f_i$ is the $p_i$-direction, the standard form has matrix \begin{align*} J=\begin{pmatrix}0&I_n;-I_n&0\end{pmatrix}, \end{align*} with $\omega_0(u,v)=u^\top Jv$ in column coordinates. The next section asks whether this example is merely a model or the universal local linear form. ## Symplectic Bases and Linear Symplectomorphisms Once the standard model is known, the next problem is whether every symplectic vector space can be put into that model by choosing a suitable basis. For inner product spaces, orthonormal bases solve this problem. For symplectic spaces, the corresponding adapted basis consists of paired vectors whose only nonzero pairings are between matching partners. [definition: Symplectic Basis] Let $(V,\omega)$ be a symplectic vector space with $\dim V=2n$. A symplectic basis of $V$ is an ordered basis $(e_1,\dots,e_n,f_1,\dots,f_n)$ such that, for all $1\le i,j\le n$, \begin{align*} \omega(e_i,e_j)=0, \end{align*} \begin{align*} \omega(f_i,f_j)=0, \end{align*} and \begin{align*} \omega(e_i,f_j)=\delta_{ij}. \end{align*} [/definition] A symplectic basis turns an abstract symplectic vector space into the standard model. The existence theorem is the linear algebraic ancestor of Darboux's theorem, though it is much easier because no derivatives or closedness condition are involved. [quotetheorem:3295] [citeproof:3295] The theorem depends essentially on nondegeneracy: for a degenerate skew form, the kernel dimension is an invariant that cannot be removed by changing basis, so the standard symplectic matrix cannot be reached. It also does not choose a canonical basis; many different symplectic bases exist, and the freedom among them is itself important. Thus the form has no linear invariant beyond dimension only after symplectic bases are allowed. Having chosen such coordinates, the next question is which changes of coordinates preserve the same symplectic measurements. [definition: Linear Symplectomorphism] Let $(V,\omega_V)$ and $(W,\omega_W)$ be symplectic vector spaces. A linear symplectomorphism is a linear isomorphism $A:V\to W$ such that \begin{align*} \omega_W(Av,Aw)=\omega_V(v,w) \end{align*} for all $v,w\in V$. [/definition] Linear symplectomorphisms are the maps that preserve the signed area pairing, not lengths or angles. To compute with them, we need the coordinate version of this preservation law in the standard basis. [definition: Symplectic Group] The real symplectic group is \begin{align*} Sp(2n,\mathbb R)=\{A\in GL(2n,\mathbb R):A^\top J A=J\}, \end{align*} where \begin{align*} J=\begin{pmatrix}0&I_n;-I_n&0\end{pmatrix}. \end{align*} [/definition] The equation $A^\top J A=J$ is the matrix form of $\omega_0(Au,Av)=\omega_0(u,v)$. It also implies a volume-preservation statement, but it is more restrictive than having determinant one. [example: Two-Dimensional Symplectic Matrices] When $n=1$, the standard symplectic matrix is \begin{align*} J=\begin{pmatrix}0&1;-1&0\end{pmatrix}. \end{align*} Write \begin{align*} A=\begin{pmatrix}a&b;c&d\end{pmatrix}. \end{align*} First multiply $J$ by $A$: \begin{align*} JA=\begin{pmatrix}0&1;-1&0\end{pmatrix}\begin{pmatrix}a&b;c&d\end{pmatrix}=\begin{pmatrix}c&d;-a&-b\end{pmatrix}. \end{align*} Since \begin{align*} A^\top=\begin{pmatrix}a&c;b&d\end{pmatrix}, \end{align*} we get \begin{align*} A^\top JA=\begin{pmatrix}a&c;b&d\end{pmatrix}\begin{pmatrix}c&d;-a&-b\end{pmatrix}=\begin{pmatrix}ac-ca&ad-bc;bc-ad&bd-db\end{pmatrix}. \end{align*} The diagonal entries are zero, and $bc-ad=-(ad-bc)$, so \begin{align*} A^\top JA=\begin{pmatrix}0&ad-bc;-(ad-bc)&0\end{pmatrix}=(ad-bc)\begin{pmatrix}0&1;-1&0\end{pmatrix}=(ad-bc)J. \end{align*} Therefore $A^\top JA=J$ holds exactly when $ad-bc=1$, which is exactly $\det A=1$. Hence \begin{align*} Sp(2,\mathbb R)=SL(2,\mathbb R). \end{align*} This equality is special to dimension $2$: in dimensions $2n\ge 4$, the matrix equation $A^\top JA=J$ contains more relations than the single determinant condition $\det A=1$. [/example] The two-dimensional computation shows that symplectic matrices form a Lie group of coordinate changes. For later Hamiltonian dynamics, we need its tangent space at the identity, since infinitesimal symplectic motions are the linear models of Hamiltonian vector fields. [quotetheorem:10038] [citeproof:10038] The path hypothesis is needed because this is a tangent-space statement, not a statement about arbitrary symplectic matrices themselves. For example, a fixed matrix $A\in Sp(2n,\mathbb R)$ need not satisfy $A^\top J+JA=0$; the displayed identity belongs to its infinitesimal velocity $B$ at the identity. Conversely, not every determinant-zero or trace-zero matrix is allowed: the condition is exactly that $JB$ be symmetric. The [symmetric matrix](/page/Symmetric%20Matrix) $JB$ will later be interpreted as a quadratic Hamiltonian, giving the linear prototype for the relation between symplectic vector fields and Hamiltonian functions. ## Symplectic Complements and Distinguished Subspaces Subspaces of a symplectic vector space behave differently from subspaces of an [inner product space](/page/Inner%20Product%20Space) because every vector is orthogonal to itself. In $(\mathbb R^2,dq\wedge dp)$, the line $Q=\mathbb R\times\{0\}$ satisfies $Q\subset Q^\omega$, so a subspace can lie inside its own complement. This already shows that the Euclidean picture $V=W\oplus W^\perp$ cannot be copied: the relevant complement may intersect the original subspace nontrivially, or even coincide with it. The right replacement for orthogonal complement is still available, but it has different inclusion and dimension properties. These properties organise the main types of subspaces used in constraint geometry and reduction. [definition: Symplectic Complement] Let $(V,\omega)$ be a symplectic vector space and let $W\subset V$ be a linear subspace. The symplectic complement of $W$ is \begin{align*} W^\omega=\{v\in V:\omega(v,w)=0\text{ for all }w\in W\}. \end{align*} [/definition] The complement records all directions that pair to zero with $W$. Unlike the Euclidean complement, it may intersect $W$ nontrivially and may even contain $W$, so the next question is which dimension identities survive without a direct-sum decomposition. [quotetheorem:10039] [citeproof:10039] The identity $\dim W+\dim W^\omega=\dim V$ uses nondegeneracy of the ambient form; for a degenerate form the map $V\to V^*$ has a kernel and the annihilator argument loses rank. The formula replaces the familiar orthogonal direct-sum formula, but it does not say $V=W\oplus W^\omega$: in $(\mathbb R^2,dq\wedge dp)$, the line $Q=\mathbb R\times\{0\}$ has $Q^\omega=Q$. What survives is a dimension count and a double-complement operation, not a canonical complementary subspace. The possible relative positions of $W$ and $W^\omega$ therefore lead to the standard classes of subspaces. [definition: Isotropic Coisotropic Symplectic And Lagrangian Subspaces] Let $(V,\omega)$ be a symplectic vector space and let $W\subset V$ be a linear subspace. The subspace $W$ is isotropic if $W\subset W^\omega$; coisotropic if $W^\omega\subset W$; symplectic if $W\cap W^\omega=\{0\}$; and Lagrangian if $W=W^\omega$. [/definition] These four classes measure how much of the symplectic form remains visible on $W$. Isotropic subspaces carry no internal symplectic area, symplectic subspaces carry a nondegenerate restricted form, coisotropic subspaces contain their null directions, and Lagrangian subspaces are maximal isotropic. The next result turns these inclusions into sharp dimension bounds. [quotetheorem:10040] [citeproof:10040] The dimension bounds depend on the ambient space being symplectic and finite-dimensional; without the complement formula, the sharp half-dimensional conclusion need not follow. The converse also has a precise hypothesis: an $n$-dimensional subspace is not automatically Lagrangian unless it is first isotropic, as a generic $n$-plane in $\mathbb R^{2n}$ need not have vanishing restricted form. Thus Lagrangian means both maximal size and zero internal symplectic area. The simplest examples are the coordinate position and momentum subspaces of the standard model, and they also show how different symplectic complements are from Euclidean complements. [example: Coordinate Lagrangian Subspaces] In $(\mathbb R^{2n},\omega_0)$, write vectors as pairs $(a,b)$ with $a,b\in\mathbb R^n$, so \begin{align*} \omega_0((a,b),(c,d))=\sum_{i=1}^n(a_i d_i-b_i c_i). \end{align*} Let \begin{align*} Q=\mathbb R^n\times\{0\}, \qquad P=\{0\}\times\mathbb R^n. \end{align*} If $u=(a,0)$ and $v=(c,0)$ lie in $Q$, then \begin{align*} \omega_0(u,v)=\omega_0((a,0),(c,0))=\sum_{i=1}^n(a_i0_i-0_i c_i)=0. \end{align*} Thus $Q$ is isotropic. Since $\dim Q=n$ inside the $2n$-dimensional symplectic vector space $\mathbb R^{2n}$, $Q$ is Lagrangian by the maximal-dimensional isotropic criterion. The complement can also be computed explicitly. A vector $(x,y)$ lies in $Q^\omega$ exactly when $\omega_0((x,y),(c,0))=0$ for every $c\in\mathbb R^n$. But \begin{align*} \omega_0((x,y),(c,0))=\sum_{i=1}^n(x_i0_i-y_i c_i)=-\sum_{i=1}^n y_i c_i=-y\cdot c. \end{align*} Taking $c=e_j$ gives $0=-y_j$ for each $j$, so $y=0$. Hence $Q^\omega=Q$. Similarly, if $u=(0,b)$ and $v=(0,d)$ lie in $P$, then \begin{align*} \omega_0(u,v)=\omega_0((0,b),(0,d))=\sum_{i=1}^n(0_i d_i-b_i0_i)=0. \end{align*} Thus $P$ is isotropic and has dimension $n$, so $P$ is Lagrangian. For the complement, $(x,y)\in P^\omega$ exactly when $\omega_0((x,y),(0,d))=0$ for every $d\in\mathbb R^n$, and \begin{align*} \omega_0((x,y),(0,d))=\sum_{i=1}^n(x_i d_i-y_i0_i)=x\cdot d. \end{align*} Taking $d=e_j$ gives $x_j=0$ for each $j$, so $x=0$ and $P^\omega=P$. Therefore the symplectic complement of the position space is the position space itself, not the momentum space, even though $\mathbb R^{2n}=Q\oplus P$ as vector spaces. The half-dimensional condition is sharp: if $(a,b)\notin Q$, then $b\ne0$, and with $(-b,0)\in Q$ we get \begin{align*} \omega_0((a,b),(-b,0))=\sum_{i=1}^n(a_i0_i-b_i(-b_i))=\sum_{i=1}^n b_i^2=|b|^2\ne0. \end{align*} So adjoining any such vector to $Q$ creates a subspace on which the restricted symplectic form is no longer zero. [/example] Graphs give a less coordinate-axis-bound family of Lagrangian subspaces and explain why symmetric matrices appear naturally in linear symplectic geometry. [example: Graphs Of Symmetric Linear Maps] Let $S:\mathbb R^n\to\mathbb R^n$ be linear, and consider its graph \begin{align*} \Gamma_S=\{(q,Sq):q\in\mathbb R^n\}\subset \mathbb R^n\oplus\mathbb R^n. \end{align*} Write $u=(q,Sq)$ and $v=(r,Sr)$. Since $\omega_0((a,b),(c,d))=\sum_{i=1}^n(a_i d_i-b_i c_i)$, we have \begin{align*} \omega_0((q,Sq),(r,Sr))=\sum_{i=1}^n(q_i(Sr)_i-(Sq)_i r_i). \end{align*} The first sum is $q\cdot Sr$, and the second sum is $Sq\cdot r$, so \begin{align*} \omega_0((q,Sq),(r,Sr))=q\cdot Sr-Sq\cdot r. \end{align*} Let $A$ be the matrix of $S$ in the standard basis, so $Sq=Aq$. Then \begin{align*} q\cdot Sr=q^\top A r. \end{align*} Also, \begin{align*} Sq\cdot r=(Aq)^\top r=q^\top A^\top r. \end{align*} Therefore \begin{align*} \omega_0((q,Sq),(r,Sr))=q^\top(A-A^\top)r. \end{align*} Thus $\Gamma_S$ is isotropic exactly when $q^\top(A-A^\top)r=0$ for all $q,r\in\mathbb R^n$. Taking $q=e_i$ and $r=e_j$ gives \begin{align*} 0=e_i^\top(A-A^\top)e_j=A_{ij}-A_{ji}. \end{align*} Hence $A_{ij}=A_{ji}$ for every $i,j$, so $A$ is symmetric. Conversely, if $A$ is symmetric, then $A-A^\top=0$, and the displayed formula gives $\omega_0((q,Sq),(r,Sr))=0$ for all $q,r$, so $\Gamma_S$ is isotropic. The map $q\mapsto(q,Sq)$ is injective because $(q,Sq)=(0,0)$ forces $q=0$, so $\dim\Gamma_S=n$. By *Dimension Bounds For Isotropic And Lagrangian Subspaces*, an isotropic $n$-dimensional subspace of $\mathbb R^{2n}$ is Lagrangian. Therefore $\Gamma_S$ is Lagrangian exactly when the matrix of $S$ is symmetric. [/example] The graph example previews a recurring theme: Lagrangian geometry converts geometric questions into questions about functions, differentials, and symmetric second derivatives. In later chapters, this linear picture becomes the local model for Lagrangian submanifolds inside symplectic manifolds. After working out the algebraic normal forms and the linear notion of a Lagrangian subspace, we can ask how that structure appears on tangent spaces of a manifold. The next chapter globalizes the linear picture by requiring a smoothly varying symplectic form on each tangent space. # 2. From Linear Algebra to Manifolds The first chapter treated symplectic geometry as linear algebra: a vector space equipped with a skew-symmetric nondegenerate bilinear form. This chapter globalises that structure to smooth manifolds by putting such a form on every tangent space in a way that varies smoothly. The new condition is closedness, which is invisible in a single vector space but decisive on a manifold because it connects symplectic geometry to de Rham cohomology, volume, and [Stokes' theorem](/theorems/1530). The guiding question is how much of the linear model survives after the vector space is replaced by a curved space. Nondegeneracy remains pointwise linear algebra, while closedness is a differential condition. Together they force every symplectic manifold to be even-dimensional, orientable, and equipped with a canonical volume form. ## Symplectic Forms on Smooth Manifolds The problem is to turn the standard form on $\mathbb R^{2n}$ into an intrinsic object that does not depend on a preferred coordinate system. A differential $2$-form assigns a skew-symmetric bilinear form to each tangent space, so it is the right global container for the linear algebra from Chapter 1. The extra requirement is that the form be closed, which makes the geometry compatible across nearby tangent spaces. [definition: Symplectic Manifold] A symplectic manifold is a pair $(M, \omega)$ where $M$ is a smooth manifold of dimension $2n$ and $\omega \in \Omega^2(M)$ satisfies: 1. $d\omega = 0$. 2. For every $p \in M$, the bilinear form $\omega_p: T_pM \times T_pM \to \mathbb R$ is nondegenerate. [/definition] The first condition is global and differential; the second condition is pointwise and algebraic. At each point $p$, the tangent space $(T_pM, \omega_p)$ is a symplectic vector space, so all the dimension restrictions and symplectic-basis statements from linear algebra apply fibre by fibre. [example: The Standard Symplectic Space] Let $M=\mathbb R^{2n}$ with coordinates $(q_1,\dots,q_n,p_1,\dots,p_n)$, and define \begin{align*} \omega_0=\sum_{i=1}^n dq_i\wedge dp_i. \end{align*} Each coordinate $1$-form is exact, since $dq_i=d(q_i)$ and $dp_i=d(p_i)$, so $d(dq_i)=0$ and $d(dp_i)=0$. Using the graded Leibniz rule for $d$, \begin{align*} d(dq_i\wedge dp_i)=d(dq_i)\wedge dp_i-dq_i\wedge d(dp_i)=0\wedge dp_i-dq_i\wedge 0=0. \end{align*} Therefore \begin{align*} d\omega_0=d\left(\sum_{i=1}^n dq_i\wedge dp_i\right)=\sum_{i=1}^n d(dq_i\wedge dp_i)=0. \end{align*} It remains to check nondegeneracy at each point. Let \begin{align*} v=\sum_{i=1}^n a_i\frac{\partial}{\partial q_i}+\sum_{i=1}^n b_i\frac{\partial}{\partial p_i}. \end{align*} For each $j$, \begin{align*} \omega_0\left(v,\frac{\partial}{\partial p_j}\right)=\sum_{i=1}^n\left(dq_i(v)\,dp_i\left(\frac{\partial}{\partial p_j}\right)-dq_i\left(\frac{\partial}{\partial p_j}\right)\,dp_i(v)\right)=\sum_{i=1}^n a_i\delta_{ij}=a_j. \end{align*} Similarly, \begin{align*} \omega_0\left(v,\frac{\partial}{\partial q_j}\right)=\sum_{i=1}^n\left(dq_i(v)\,dp_i\left(\frac{\partial}{\partial q_j}\right)-dq_i\left(\frac{\partial}{\partial q_j}\right)\,dp_i(v)\right)=\sum_{i=1}^n(0-\delta_{ij}b_i)=-b_j. \end{align*} If $\omega_0(v,w)=0$ for every tangent vector $w$, then taking $w=\partial/\partial p_j$ gives $a_j=0$, and taking $w=\partial/\partial q_j$ gives $b_j=0$, for every $j$. Hence $v=0$, so $\omega_0$ is nondegenerate. Thus $(\mathbb R^{2n},\omega_0)$ is the standard model of a symplectic manifold. [/example] This example shows that the linear model is already a manifold example, not merely a tangent-space approximation. The next natural source comes from mechanics: positions live on a manifold $Q$, and momenta live in cotangent fibres. [example: Cotangent Bundles] Let $Q$ be a smooth $n$-manifold and let $\pi:T^*Q\to Q$ be its cotangent bundle. At a covector $\alpha\in T_q^*Q$, the tautological $1$-form is intrinsically defined by \begin{align*} \lambda_\alpha(V)=\alpha(d\pi_\alpha(V)) \end{align*} for $V\in T_\alpha(T^*Q)$. In local coordinates $(q_1,\dots,q_n)$ on $Q$, write a covector as $\alpha=\sum_i p_i\,dq_i|_q$. Since $d\pi(\partial/\partial q_i)=\partial/\partial q_i$ and $d\pi(\partial/\partial p_i)=0$, this gives \begin{align*} \lambda\left(\frac{\partial}{\partial q_j}\right)=p_j \end{align*} and \begin{align*} \lambda\left(\frac{\partial}{\partial p_j}\right)=0. \end{align*} Thus, in the induced coordinates $(q_1,\dots,q_n,p_1,\dots,p_n)$ on $T^*Q$, \begin{align*} \lambda=\sum_{i=1}^n p_i\,dq_i. \end{align*} Define the canonical $2$-form by $\omega=-d\lambda$. Using the Leibniz rule and $d(dq_i)=0$, \begin{align*} d(p_i\,dq_i)=dp_i\wedge dq_i+p_i\,d(dq_i)=dp_i\wedge dq_i. \end{align*} Therefore \begin{align*} \omega=-\sum_{i=1}^n dp_i\wedge dq_i=\sum_{i=1}^n dq_i\wedge dp_i. \end{align*} Since $d^2=0$, \begin{align*} d\omega=d(-d\lambda)=-d^2\lambda=0. \end{align*} It remains to check nondegeneracy. Let \begin{align*} v=\sum_{i=1}^n a_i\frac{\partial}{\partial q_i}+\sum_{i=1}^n b_i\frac{\partial}{\partial p_i}. \end{align*} For each $j$, \begin{align*} \omega\left(v,\frac{\partial}{\partial p_j}\right)=\sum_{i=1}^n dq_i(v)\,dp_i\left(\frac{\partial}{\partial p_j}\right)-\sum_{i=1}^n dq_i\left(\frac{\partial}{\partial p_j}\right)\,dp_i(v)=\sum_{i=1}^n a_i\delta_{ij}=a_j. \end{align*} Similarly, \begin{align*} \omega\left(v,\frac{\partial}{\partial q_j}\right)=\sum_{i=1}^n dq_i(v)\,dp_i\left(\frac{\partial}{\partial q_j}\right)-\sum_{i=1}^n dq_i\left(\frac{\partial}{\partial q_j}\right)\,dp_i(v)=-\sum_{i=1}^n \delta_{ij}b_i=-b_j. \end{align*} If $\omega(v,w)=0$ for every tangent vector $w$, then these two identities give $a_j=0$ and $b_j=0$ for every $j$, hence $v=0$. Thus $\omega$ is closed and nondegenerate, so $T^*Q$ carries a canonical symplectic structure. [/example] Cotangent bundles are the basic phase spaces of Hamiltonian mechanics. They also explain why many symplectic forms are locally written as $\sum_i dq_i\wedge dp_i$, even when the underlying manifold has no global coordinate system. ## Orientation and Symplectic Volume A general smooth manifold may or may not be orientable, and even when it is orientable there may be many choices of orientation. A symplectic form removes this ambiguity: wedging the form with itself the maximal possible number of times produces a nowhere-vanishing top-degree form. This is the first global consequence of nondegeneracy. [quotetheorem:10035] [citeproof:10035] The factorial normalisation is not cosmetic: it cancels the $n!$ repeated terms that occur when wedging a sum of $n$ commuting degree-$2$ pieces with itself. Thus the canonical orientation is represented by \begin{align*} \frac{\omega^n}{n!}. \end{align*} Nondegeneracy is essential here: if a closed $2$-form has a kernel at some point, its top wedge can vanish there and no orientation form is produced. Even-dimensionality is also forced by the same linear algebra, since a skew-symmetric nondegenerate form cannot exist on an odd-dimensional tangent space. The theorem gives an orientation and a volume density, but it does not say that every oriented even-dimensional manifold is symplectic, nor does it classify symplectic forms on a fixed manifold. Its role is to convert the pointwise linear condition into a global invariant that will interact with integration and Stokes' theorem. [definition: Symplectic Volume] Let $(M,\omega)$ be a symplectic manifold of dimension $2n$. If $M$ is compact, its symplectic volume is \begin{align*} \operatorname{Vol}(M,\omega) := \int_M \frac{\omega^n}{n!}. \end{align*} [/definition] This definition depends on the orientation already determined by $\omega$. For noncompact manifolds the same top form is still the symplectic volume density, but the total integral may be infinite or undefined without support conditions. [example: Volume in the Standard Model] Let $\alpha_i=dq_i\wedge dp_i$, so $\omega_0=\alpha_1+\cdots+\alpha_n$. Each $\alpha_i$ has degree $2$, hence $\alpha_i\wedge\alpha_j=\alpha_j\wedge\alpha_i$, and \begin{align*} \alpha_i\wedge\alpha_i=dq_i\wedge dp_i\wedge dq_i\wedge dp_i=0 \end{align*} because the factor $dq_i$ is repeated. Therefore, in the expansion of $\omega_0^n$, every term with a repeated index vanishes, and the only surviving terms are the $n!$ permutations of $\alpha_1\wedge\cdots\wedge\alpha_n$. Thus \begin{align*} \omega_0^n=n!\,\alpha_1\wedge\cdots\wedge\alpha_n \end{align*} and therefore \begin{align*} \frac{\omega_0^n}{n!}=dq_1\wedge dp_1\wedge\cdots\wedge dq_n\wedge dp_n. \end{align*} Now let \begin{align*} B=\prod_{i=1}^n [a_i,b_i]\times[c_i,d_i] \end{align*} with the orientation determined by $dq_1\wedge dp_1\wedge\cdots\wedge dq_n\wedge dp_n$. Since the coefficient of the volume form is the constant function $1$, iterated integration gives \begin{align*} \int_B \frac{\omega_0^n}{n!}=\int_{a_1}^{b_1}\int_{c_1}^{d_1}\cdots\int_{a_n}^{b_n}\int_{c_n}^{d_n}1\,dp_n\,dq_n\cdots dp_1\,dq_1. \end{align*} Evaluating the innermost integral first gives $\int_{c_n}^{d_n}1\,dp_n=d_n-c_n$, and repeating this for each coordinate interval yields \begin{align*} \int_B \frac{\omega_0^n}{n!}=\prod_{i=1}^n (b_i-a_i)(d_i-c_i). \end{align*} Thus the symplectic volume of a coordinate rectangle in the standard model is exactly the product of its side lengths in the $q_i$ and $p_i$ directions. [/example] The volume formula is the bridge from local linear algebra to global topology. Because it gives a positive total volume on compact symplectic manifolds, it can contradict exactness through Stokes' theorem. ## Exact Symplectic Forms and the Compact Obstruction Closed forms are allowed to be exact, and exact symplectic forms occur constantly on noncompact spaces such as cotangent bundles. The question is whether a compact manifold can carry such a form. The answer is no, and the proof is one of the first places where the closedness condition has a visible global effect. [definition: Exact Symplectic Form] A symplectic form $\omega$ on a smooth manifold $M$ is exact if there exists a $1$-form $\lambda\in\Omega^1(M)$ such that \begin{align*} \omega = d\lambda. \end{align*} [/definition] Exactness is a cohomological condition: it says that the de Rham class $[\omega]\in H^2_{\mathrm{dR}}(M)$ vanishes. The canonical form on a cotangent bundle is exact up to the sign convention chosen for the tautological form, so exactness is natural on open phase spaces. The next result asks whether the same phenomenon can occur on closed compact spaces, where Stokes' theorem forces every exact top-degree form to integrate to zero. [quotetheorem:10041] [citeproof:10041] This theorem is often the quickest way to rule out proposed symplectic structures. The compactness hypothesis is doing real work: cotangent bundles such as $T^*Q$ carry exact canonical symplectic forms, but they are typically noncompact. The absence of boundary is equally important, since compact domains with boundary in $\mathbb R^{2n}$ inherit the exact standard form from the ambient space. Thus the obstruction is not a local statement about exact symplectic forms; it is a closed-manifold statement where Stokes' theorem has no boundary term to absorb the positive symplectic volume. [example: The Obstruction on a Closed Surface] Let $\Sigma$ be a compact connected oriented surface without boundary, and let $\omega$ be an area form compatible with the chosen orientation. Since $\omega$ is a $2$-form on a $2$-manifold, $d\omega$ is a $3$-form on $\Sigma$, and $\Omega^3(\Sigma)=0$. Hence \begin{align*} d\omega=0. \end{align*} Because an area form is nowhere vanishing, $\omega_p$ is a nonzero alternating bilinear form on the $2$-dimensional vector space $T_p\Sigma$ for every $p\in\Sigma$, and a nonzero alternating form in dimension $2$ is nondegenerate. Thus $\omega$ is symplectic. Suppose, for contradiction, that $\omega=d\lambda$ for some $1$-form $\lambda\in\Omega^1(\Sigma)$. Then *Stokes' theorem* gives \begin{align*} \int_\Sigma \omega=\int_\Sigma d\lambda=\int_{\partial\Sigma}\lambda. \end{align*} Since $\Sigma$ has no boundary, $\partial\Sigma=\varnothing$, so \begin{align*} \int_{\partial\Sigma}\lambda=\int_\varnothing \lambda=0. \end{align*} Therefore $\int_\Sigma\omega=0$. But $\omega$ is compatible with the orientation, so it is a positive area density and \begin{align*} \int_\Sigma\omega>0. \end{align*} This contradiction shows that no area form on a closed oriented surface is exact. [/example] The surface case is the two-dimensional version of the general obstruction. It also previews the role of cohomology: on a compact symplectic manifold the class $[\omega]$ cannot vanish, and in fact $[\omega]^n$ evaluates nonzero on the fundamental class. ## Symplectomorphisms and Local Coordinates Once symplectic manifolds have been defined, the correct maps between them are the diffeomorphisms preserving the $2$-form. This is stricter than preserving orientation or volume, because a symplectic form records pairings of tangent directions, not just a top-degree density. A volume-preserving diffeomorphism may preserve $\omega^n/n!$ while changing the individual pairings $\omega_p(v,w)$, so it need not be symplectic. The central question is which coordinate changes preserve the geometry. [definition: Symplectomorphism] Let $(M,\omega)$ and $(N,\eta)$ be symplectic manifolds. A symplectomorphism from $(M,\omega)$ to $(N,\eta)$ is a diffeomorphism $\varphi:M\to N$ such that \begin{align*} \varphi^*\eta = \omega. \end{align*} [/definition] This definition is the manifold version of a linear symplectomorphism. The derivative $d\varphi_p:T_pM\to T_{\varphi(p)}N$ is a linear symplectomorphism at every point, because pullback equality says that tangent-vector pairings are preserved. [example: Linear Symplectic Maps as Global Symplectomorphisms] Since $A$ is invertible and linear, it is a diffeomorphism of $\mathbb R^{2n}$. Write tangent vectors in the coordinate splitting as $v=(a,b)$ and $w=(c,d)$, where $a,b,c,d\in\mathbb R^n$. For the standard form $\omega_0=\sum_{i=1}^n dq_i\wedge dp_i$, \begin{align*} \omega_0(v,w)=\sum_{i=1}^n\left(dq_i(v)dp_i(w)-dq_i(w)dp_i(v)\right)=\sum_{i=1}^n(a_i d_i-c_i b_i). \end{align*} Let $J$ be the standard symplectic matrix, characterized by $J_{q_i p_j}=\delta_{ij}$, $J_{p_i q_j}=-\delta_{ij}$, and all $q_iq_j$ and $p_ip_j$ entries equal to $0$. Then the previous identity is exactly \begin{align*} \omega_0(v,w)=v^\top Jw. \end{align*} For a linear map $A$, the pullback is evaluated by applying $A$ to tangent vectors: \begin{align*} (A^*\omega_0)(v,w)=\omega_0(Av,Aw)=(Av)^\top J(Aw)=v^\top A^\top JAw. \end{align*} Thus $A^*\omega_0=\omega_0$ holds exactly when \begin{align*} v^\top A^\top JAw=v^\top Jw \end{align*} for every pair of tangent vectors $v,w$. Equivalently, \begin{align*} v^\top(A^\top JA-J)w=0 \end{align*} for every $v,w$. Taking $v$ and $w$ to be standard coordinate basis vectors shows that every entry of $A^\top JA-J$ is zero, so \begin{align*} A^\top JA=J. \end{align*} Conversely, if $A^\top JA=J$, then the displayed pullback formula gives $(A^*\omega_0)(v,w)=\omega_0(v,w)$ for all $v,w$, hence $A^*\omega_0=\omega_0$. Therefore an invertible linear map is a global symplectomorphism of $(\mathbb R^{2n},\omega_0)$ exactly when it satisfies $A^\top JA=J$. [/example] The local-coordinate problem asks for the nonlinear analogue of a symplectic basis. In full generality this is Darboux's theorem, which will be treated later; at this stage the important point is that local coordinates are not arbitrary labels but tests of whether the form has been normalised to the standard model. [definition: Darboux Coordinate Chart] Let $(M,\omega)$ be a symplectic manifold of dimension $2n$. A Darboux coordinate chart on an [open set](/page/Open%20Set) $U\subset M$ is a coordinate chart $\varphi:U\to\varphi(U)\subset\mathbb R^{2n}$ with coordinates $(q_1,\dots,q_n,p_1,\dots,p_n)$ such that \begin{align*} \omega\big|_U = \sum_{i=1}^n dq_i\wedge dp_i. \end{align*} [/definition] The existence of such charts is a deep local theorem, not part of the definition. The definition gives the language needed to state that theorem and to compare symplectic geometry with other geometric structures. [remark: Contrast with Riemannian Geometry] A Riemannian metric has local invariants such as curvature, so it cannot usually be made equal to the [Euclidean metric](/page/Euclidean%20Metric) on an open neighbourhood. A symplectic form has no analogous local curvature invariant once Darboux's theorem is available. The geometric information in symplectic topology is therefore concentrated in global features such as volume, cohomology classes, submanifolds, and Hamiltonian dynamics. [/remark] This contrast is a major theme of the course. Symplectic geometry is locally flexible but globally rigid, and the compact exact obstruction is the first example of that rigidity. ## Core Families of Symplectic Manifolds The definitions above are useful only if they produce a substantial supply of examples. The main families in the foundations course come from linear spaces, cotangent bundles, surfaces, and complex projective spaces. Each family emphasises a different source of the symplectic form: coordinates, canonical mechanics, orientation, and complex geometry. [example: Oriented Surfaces with Area Forms] Let $\Sigma$ be an oriented smooth surface, and let $\omega\in\Omega^2(\Sigma)$ be an area form compatible with the orientation. Since $\dim \Sigma=2$, there are no nonzero $3$-forms on $\Sigma$, so $\Omega^3(\Sigma)=0$. Hence \begin{align*} d\omega\in\Omega^3(\Sigma)=0, \end{align*} and therefore $d\omega=0$. It remains to check nondegeneracy. Fix $p\in\Sigma$, and choose a positively oriented basis $(e_1,e_2)$ for $T_p\Sigma$. Because $\omega$ is a compatible area form, \begin{align*} c:=\omega_p(e_1,e_2)>0. \end{align*} Let $v=a e_1+b e_2\in T_p\Sigma$, and suppose $\omega_p(v,w)=0$ for every $w\in T_p\Sigma$. Taking $w=e_1$ gives \begin{align*} 0=\omega_p(v,e_1)=\omega_p(ae_1+be_2,e_1)=a\omega_p(e_1,e_1)+b\omega_p(e_2,e_1)=0-bc=-bc. \end{align*} Since $c>0$, this implies $b=0$. Taking $w=e_2$ gives \begin{align*} 0=\omega_p(v,e_2)=\omega_p(ae_1+be_2,e_2)=a\omega_p(e_1,e_2)+b\omega_p(e_2,e_2)=ac+0=ac. \end{align*} Again $c>0$, so $a=0$. Thus $v=0$, proving that $\omega_p$ is nondegenerate at every point $p$. Therefore $(\Sigma,\omega)$ is a symplectic manifold. [/example] Surfaces are special because closedness is automatic for degree reasons. In higher dimensions closedness is a genuine equation, and nondegeneracy alone is not enough to define a symplectic structure. [example: Complex Projective Space] On the affine chart $U_0=\{[Z_0:\cdots:Z_n]\mid Z_0\ne 0\}$, write $z_j=Z_j/Z_0$ for $1\le j\le n$ and set \begin{align*} \rho=\log(1+|z_1|^2+\cdots+|z_n|^2). \end{align*} The Fubini-Study form is locally \begin{align*} \omega_{\mathrm{FS}}=\frac{i}{2}\partial\bar\partial\rho. \end{align*} Since $\omega_{\mathrm{FS}}$ is obtained by applying the [exterior derivative](/theorems/1525) operators to the local potential $\rho$, the identities $\partial^2=0$, $\bar\partial^2=0$, and $\partial\bar\partial=-\bar\partial\partial$ give $d\omega_{\mathrm{FS}}=0$. Let $r=1+\sum_{\ell=1}^n |z_\ell|^2$. Differentiating $\rho=\log r$ gives \begin{align*} \frac{\partial\rho}{\partial z_j}=\frac{\bar z_j}{r}. \end{align*} Differentiating again, \begin{align*} \frac{\partial^2\rho}{\partial z_j\partial \bar z_k}=\frac{\delta_{jk}r-\bar z_j z_k}{r^2}. \end{align*} Thus \begin{align*} \omega_{\mathrm{FS}}=\frac{i}{2}\sum_{j,k=1}^n\frac{\delta_{jk}r-\bar z_j z_k}{r^2}\,dz_j\wedge d\bar z_k. \end{align*} To check nondegeneracy, take $\xi=(\xi_1,\dots,\xi_n)\in\mathbb C^n$. The Hermitian matrix associated to the displayed form satisfies \begin{align*} \sum_{j,k=1}^n\frac{\delta_{jk}r-\bar z_j z_k}{r^2}\xi_j\overline{\xi_k}=\frac{r|\xi|^2-\left|\sum_{j=1}^n\bar z_j\xi_j\right|^2}{r^2}. \end{align*} Since $r=1+|z|^2$, the numerator is \begin{align*} |\xi|^2+|z|^2|\xi|^2-|\langle z,\xi\rangle|^2. \end{align*} By Cauchy-Schwarz, $|\langle z,\xi\rangle|^2\le |z|^2|\xi|^2$, so this numerator is at least $|\xi|^2$, and it is positive whenever $\xi\ne 0$. Hence the associated Hermitian form is positive definite, so the real $2$-form $\omega_{\mathrm{FS}}$ is nondegenerate. The same formula on the other affine charts agrees on overlaps, so these local forms assemble into a global closed nondegenerate real $2$-form on $\mathbb{C}P^n$. Since $\mathbb{C}P^n$ is compact, $(\mathbb{C}P^n,\omega_{\mathrm{FS}})$ is a compact symplectic manifold; by *No Compact Exact Symplectic Manifolds*, $\omega_{\mathrm{FS}}$ cannot be exact. [/example] This example is important because it is compact and not built from a cotangent bundle. It shows that symplectic manifolds arise naturally in complex and algebraic geometry, not only in mechanics. [remark: First Cohomological Signal] For any compact symplectic manifold $(M,\omega)$ of dimension $2n$, the class $[\omega]^n\in H^{2n}_{\mathrm{dR}}(M)$ is nonzero because it integrates to $n!\operatorname{Vol}(M,\omega)$. In particular, $[\omega]\ne 0$. This does not classify symplectic forms, but it gives a necessary cohomological condition for compact examples. [/remark] The chapter has moved from the tangent-space model to a global object with orientation, volume, maps, and examples. The next step is to understand the strongest local normal-form statement: near any point, every symplectic manifold looks like $(\mathbb R^{2n},\omega_0)$ in suitable Darboux coordinates. We have now passed from tangent-space models to genuine symplectic manifolds, with examples, maps, and orientation properties in hand. The next step is to see the canonical local model in its strongest form: every symplectic manifold is locally standard in Darboux coordinates. # 3. Cotangent Bundles and Canonical Geometry This chapter studies the first large family of symplectic manifolds that appears in the course: cotangent bundles. The point is not only that $T^*Q$ carries a natural symplectic form, but that this form is forced by the geometry of covectors and is independent of coordinates. Cotangent bundles also provide the basic phase spaces of classical mechanics, and their Lagrangian submanifolds already contain many of the guiding examples for the rest of the course. ## The Tautological Form on a Cotangent Bundle The central question is how to build a differential form on $T^*Q$ using no choices on the base manifold $Q$. A point of $T^*Q$ is already a covector on $Q$, so the most economical construction is to let that covector evaluate tangent vectors after projecting them down to $Q$. Let $\tau:T^*Q\to Q$ be the cotangent bundle projection. If $\beta\in T_q^*Q$, then a tangent vector $v\in T_\beta(T^*Q)$ projects to $d\tau_\beta(v)\in T_qQ$, and $\beta$ can evaluate this projected vector. [definition: Tautological One Form] Let $Q$ be a smooth manifold and let $\tau:T^*Q\to Q$ be the cotangent bundle projection. The tautological one-form on $T^*Q$ is the differential one-form $\lambda\in\Omega^1(T^*Q)$ defined by \begin{align*} \lambda_\beta(v)=\beta(d\tau_\beta(v)) \end{align*} for every $\beta\in T_q^*Q$ and every $v\in T_\beta(T^*Q)$. [/definition] This definition says that $\lambda$ records the momentum component of a tangent vector after forgetting its vertical part. To connect the intrinsic definition with the standard phase-space formula, we need its expression in coordinates induced from a chart on $Q$. Choose a chart $(U,x)$ on $Q$ with coordinates $(q_1,\dots,q_n)$. A covector over $q\in U$ is written $\sum_{i=1}^n p_i\,dq_i|_q$, so $(q_1,\dots,q_n,p_1,\dots,p_n)$ are induced coordinates on $T^*U$; the following theorem computes the form in these coordinates so that the later symplectic formula has a fixed sign convention. [quotetheorem:10042] [citeproof:10042] The coordinate formula reveals why cotangent bundles model position-momentum phase spaces: $q_i$ are position coordinates and $p_i$ are covector-valued momentum coordinates. Its content is local, however: the expression $\sum_i p_i\,dq_i$ depends on cotangent coordinates induced from a chart, while the object it describes is the globally defined one-form $\lambda$. A formula such as $\sum_i p_i\,dq_i$ written in arbitrary coordinates on $T^*Q$ would have no intrinsic meaning unless those coordinates came from a base chart and the dual fibre coordinates. The next problem is to pass from this canonical one-form to a canonical two-form, since symplectic geometry is governed by closed nondegenerate two-forms. [definition: Canonical Symplectic Form] Let $Q$ be a smooth manifold and let $\lambda\in\Omega^1(T^*Q)$ be the tautological one-form. The canonical symplectic form on $T^*Q$ is the differential two-form $\omega\in\Omega^2(T^*Q)$ defined by \begin{align*} \omega=-d\lambda. \end{align*} [/definition] The sign convention gives the standard local expression $\omega=\sum_i dq_i\wedge dp_i$, matching the convention from symplectic linear algebra. Closedness is immediate from $d^2=0$, but closedness alone is not enough for symplectic geometry: a closed two-form may still have a kernel. The real question is whether the tautological construction has avoided this degeneracy globally, rather than only in the familiar coordinate formula. This is the point at which the cotangent bundle construction must be checked to produce a genuine symplectic manifold. [quotetheorem:6843] [citeproof:6843] The theorem explains the phrase "canonical geometry": no Riemannian metric, connection, or coordinate system is needed. The smooth-manifold hypothesis is still doing work, since the proof uses coordinate charts on $Q$ and the [smooth vector bundle](/page/Smooth%20Vector%20Bundle) structure of $T^*Q$; for a singular configuration space there is no ordinary cotangent bundle carrying this argument without additional machinery. The result also says something special about cotangent bundles, not about arbitrary even-dimensional manifolds: a closed nondegenerate two-form is extra structure in general, whereas here it is forced by the tautological one-form. The canonical symplectic form is exact, with primitive $-\lambda$, so $T^*Q$ is the basic local source of exact symplectic manifolds. [example: Cotangent Bundle of the Circle] Write the standard angular one-form on $S^1$ as $d q$ locally; equivalently, under $S^1\subset\mathbb R^2$, it is the global one-form $-y\,dx+x\,dy$ restricted to the circle. Every covector over a point of $S^1$ is therefore uniquely of the form $p\,dq$, so the map \begin{align*} S^1\times\mathbb R\to T^*S^1,\quad (q,p)\mapsto p\,dq|_q \end{align*} trivializes the cotangent bundle. In these coordinates, the tautological one-form is \begin{align*} \lambda=p\,dq \end{align*} by *Coordinate Formula for the Tautological Form*. Hence the canonical symplectic form is computed from $\omega=-d\lambda$: \begin{align*} d\lambda=d(p\,dq)=dp\wedge dq+p\,d(dq) \end{align*} and since $d(dq)=0$, \begin{align*} d\lambda=dp\wedge dq. \end{align*} Using antisymmetry of the wedge product, $dp\wedge dq=-dq\wedge dp$, so \begin{align*} \omega=-d\lambda=-dp\wedge dq=dq\wedge dp. \end{align*} Thus $T^*S^1$ is the cylinder $S^1\times\mathbb R$ with its standard area form $dq\wedge dp$, showing that the compact configuration space $S^1$ has a noncompact exact symplectic phase space. [/example] ## Coordinate-Free Naturality The local formula gives confidence, but the construction should also behave correctly under changes of configuration space. The guiding question is: if a diffeomorphism $f:Q\to Q'$ identifies two manifolds, how does it identify their cotangent phase spaces and canonical forms? A diffeomorphism pushes tangent vectors forward, so covectors are transported in the opposite direction. This contravariance is the reason cotangent bundles are functorial in a slightly reversed way. [definition: Cotangent Lift] Let $f:Q\to Q'$ be a diffeomorphism. The cotangent lift of $f$ is the diffeomorphism \begin{align*} T^*f:T^*Q'\to T^*Q \end{align*} defined by \begin{align*} T^*f(\alpha_{q'})=\alpha_{q'}\circ df_q \end{align*} where $q\in Q$, $q'=f(q)$, and $\alpha_{q'}\in T_{q'}^*Q'$. [/definition] The direction $T^*Q'\to T^*Q$ is important: covectors pull back. This raises the naturality question for the forms themselves: does the cotangent lift preserve the tautological one-form, and hence the canonical symplectic form? [quotetheorem:10043] [citeproof:10043] Naturality means that the canonical symplectic structure is intrinsic to the smooth manifold $Q$. The diffeomorphism hypothesis is essential in this statement: if $f:Q\to Q'$ is not invertible, then the assignment $\alpha_{q'}\mapsto \alpha_{q'}\circ df_q$ is not a globally defined diffeomorphism $T^*Q'\to T^*Q$ covering an inverse map. For example, an inclusion $i:N\hookrightarrow Q$ pulls covectors on $Q$ back to covectors on $N$, but it does not identify the two cotangent bundles and need not preserve dimensions. Thus the theorem is a coordinate-change result, not a statement that every smooth map gives a symplectomorphism. A change of coordinates on $Q$ produces a symplectomorphism between the corresponding coordinate descriptions of $T^*Q$. [example: Phase Space of a Particle on a Manifold] Let $Q$ be the configuration manifold of a particle, so a phase-space point is a covector $p\in T_q^*Q$ together with its base point $q$. Thus the intrinsic phase space is $T^*Q$, and the pair is written $(q,p)$ only after choosing notation for the base point and the covector over it. Suppose $f:Q\to Q'$ is a diffeomorphism and write $q'=f(q)$. A momentum covector $p'\in T_{q'}^*Q'$ is transported to a covector on $Q$ by pullback: \begin{align*} (T^*f)(p')=p'\circ df_q\in T_q^*Q. \end{align*} For a tangent vector $v\in T_qQ$, this transported momentum evaluates by \begin{align*} ((T^*f)(p'))(v)=(p'\circ df_q)(v)=p'(df_q(v)). \end{align*} So the changed momentum is exactly the covector that gives the same numerical pairing after the velocity vector is expressed in the new configuration variables. By *Naturality of the Tautological Form*, the cotangent lift satisfies \begin{align*} (T^*f)^*\omega_Q=\omega_{Q'}. \end{align*} Therefore changing configuration variables by a diffeomorphism does not change the canonical symplectic geometry of phase space. Hamiltonian mechanics on $T^*Q$ is coordinate-independent before any Hamiltonian function is chosen. [/example] The same construction also explains why using tangent rather than cotangent bundles would require extra structure. To identify velocities with momenta, a Riemannian metric or Lagrangian is needed; the cotangent bundle carries the symplectic form without that extra choice. [remark: Exactness and Compactness] The canonical symplectic form on $T^*Q$ is exact because $\omega=-d\lambda$. This does not contradict the compact exact obstruction from the previous chapter when $Q$ is compact, since $T^*Q$ is noncompact unless $Q$ has dimension $0$. [/remark] ## Graphs of One-Forms as Lagrangians We now ask how submanifolds of $T^*Q$ reflect differential forms on $Q$. The most direct construction takes a one-form $\alpha\in\Omega^1(Q)$ and forms its graph inside the cotangent bundle. The question is when that graph is Lagrangian for the canonical symplectic form. [definition: Graph of a One Form] Let $Q$ be a smooth manifold and let $\alpha\in\Omega^1(Q)$. The graph of $\alpha$ is the embedded submanifold \begin{align*} \Gamma_\alpha=\{\alpha_q\in T_q^*Q:q\in Q\}\subset T^*Q. \end{align*} [/definition] The graph has dimension $\dim Q$, half the dimension of $T^*Q$, so it has the right size to be Lagrangian. To test the vanishing of $\omega$ on the graph, we first need to understand how $\lambda$ pulls back along the section $\alpha:Q\to T^*Q$. [quotetheorem:10044] [citeproof:10044] This identity converts a symplectic condition on the graph into a differential condition on the original one-form. The section hypothesis is essential: it gives $\tau\circ\alpha=\operatorname{id}_Q$, so the projection in the definition of $\lambda$ returns the original tangent vector. For a general map $F:M\to T^*Q$ whose projection to $Q$ is not the identity, $F^*\lambda$ is instead the pulled-back covector field along $\tau\circ F$, not a one-form originally living on $M$ by the same formula. The identity also does not say that the graph is Lagrangian; it only identifies the restriction of the primitive, and the exterior derivative of that identity is what detects the obstruction. The next theorem makes this conversion precise. [quotetheorem:10045] [citeproof:10045] Closed one-forms therefore produce Lagrangian submanifolds, and closedness is not a cosmetic condition. On $\mathbb R^2$ with coordinates $(x,y)$, the one-form $\alpha=x\,dy$ has $d\alpha=dx\wedge dy\ne 0$, so its graph in $T^*\mathbb R^2$ has nonzero restricted symplectic form and is not Lagrangian. The criterion is also limited to Lagrangians that are graphs over all of $Q$; cotangent fibres and conormal bundles are Lagrangian but fail to project diffeomorphically to the base. The most concrete subclass of closed graphs comes from functions, and the following result records why exact graphs are the basic generating-function examples: they carry a preferred primitive for the restricted tautological form. [quotetheorem:10046] [citeproof:10046] The distinction between closed and exact is global. The theorem above gives a stronger conclusion than the closed graph criterion: not only is $\Gamma_{dS}$ Lagrangian, but the restricted primitive is globally the differential of a function. It does not say that every Lagrangian graph has such a global [generating function](/page/Generating%20Function), because a closed one-form may fail to be exact when its de Rham cohomology class is nonzero. Locally every closed one-form is exact by the Poincare lemma, but globally the obstruction lies in $H^1_{\mathrm{dR}}(Q)$. [example: Closed Nonexact Graph on the Circle] Let $\theta=-y\,dx+x\,dy$ be the standard angular one-form on $S^1\subset \mathbb R^2$. On every angular coordinate arc, $\theta=dq$, so for a constant $c\in\mathbb R$ the one-form $\alpha=c\theta$ is locally $c\,dq$. Its exterior derivative vanishes because \begin{align*} d\alpha=d(c\,dq)=dc\wedge dq+c\,d(dq)=0\wedge dq+c\cdot 0=0. \end{align*} Thus $\alpha$ is closed, and by *Graph Criterion for Closed Lagrangians* its graph is Lagrangian in $T^*S^1$. Under the trivialization $T^*S^1\cong S^1\times\mathbb R$ determined by $\theta$, a covector over $q$ has the form $p\,\theta_q$. The graph condition $\alpha_q=c\theta_q=p\,\theta_q$ gives $p=c$, so \begin{align*} \Gamma_\alpha=\{(q,p)\in S^1\times\mathbb R:p=c\}. \end{align*} This graph is exact exactly when $\alpha$ is exact. To test exactness, parametrize the circle by $\gamma(t)=(\cos t,\sin t)$ for $0\le t\le 2\pi$. Along $\gamma$, \begin{align*} \gamma^*\theta=-\sin t\,d(\cos t)+\cos t\,d(\sin t)=\sin^2 t\,dt+\cos^2 t\,dt=dt. \end{align*} Therefore \begin{align*} \int_{S^1}\alpha=\int_0^{2\pi}c\,dt=2\pi c. \end{align*} If $\alpha=dS$ for a globally defined smooth function $S:S^1\to\mathbb R$, then the integral over the closed loop would be $S(\gamma(2\pi))-S(\gamma(0))=0$, so $2\pi c=0$ and hence $c=0$. Conversely, when $c=0$, $\alpha=0=d0$ is exact. Thus the circle $p=c$ is a closed Lagrangian graph for every $c$, but it is an exact graph only for $c=0$. [/example] ## Conormal Bundles Graphs of one-forms are Lagrangians that project diffeomorphically to $Q$. Cotangent bundles also contain Lagrangians associated to submanifolds of $Q$, where the projection drops rank. The guiding construction assigns to a submanifold all covectors that vanish along its tangent directions. [definition: Conormal Bundle] Let $N\subset Q$ be an embedded submanifold. The conormal bundle of $N$ in $Q$ is \begin{align*} N^*N=\{\beta\in T_q^*Q:q\in N,\ \beta(v)=0\text{ for all }v\in T_qN\}\subset T^*Q. \end{align*} [/definition] The notation $N^*N$ records that this is a bundle over $N$, not over all of $Q$. If $\dim Q=n$ and $\dim N=k$, then the fibre over each point has dimension $n-k$, so the total dimension is $n$; the next question is whether the canonical symplectic form vanishes on it. [quotetheorem:10047] [citeproof:10047] This gives a second basic source of Lagrangians, complementary to graphs. The embedded-submanifold hypothesis matters because the proof uses adapted coordinates and a smooth conormal [vector bundle](/page/Vector%20Bundle) of constant rank. For a singular subset, such as two curves crossing in a surface, the annihilator directions can jump at the crossing and need not form a smooth submanifold of $T^*Q$. For an immersed submanifold with self-intersections, the conormal construction may give an immersed Lagrangian rather than an embedded one unless the immersion has suitable global behaviour. A graph is transverse to the cotangent fibres, while a conormal bundle contains entire covector directions over points of $N$. [example: Conormal Bundle of a Point] Let $Q$ be an $n$-manifold and let $N=\{q_0\}$. Since the only tangent vector to a point is the zero vector, \begin{align*} T_{q_0}N=\{0\}. \end{align*} If $\beta\in T_{q_0}^*Q$, then the only condition for $\beta$ to lie in the conormal bundle is $\beta(0)=0$, which holds by linearity. Hence \begin{align*} N^*N=\{\beta\in T_{q_0}^*Q:\beta(v)=0\text{ for every }v\in T_{q_0}N\}=T_{q_0}^*Q. \end{align*} To see the Lagrangian condition explicitly, choose local coordinates $(q_1,\dots,q_n)$ centered at $q_0$. In the induced cotangent coordinates, the fibre $T_{q_0}^*Q$ is cut out by \begin{align*} q_1=\cdots=q_n=0. \end{align*} A tangent vector to this fibre has the form $X=\sum_{i=1}^n b_i\partial_{p_i}$. Since $dq_i(\partial_{p_j})=0$ for every $i,j$, \begin{align*} \lambda(X)=\sum_{i=1}^n p_i\,dq_i\left(\sum_{j=1}^n b_j\partial_{p_j}\right)=\sum_{i=1}^n\sum_{j=1}^n p_i b_j\,dq_i(\partial_{p_j})=0. \end{align*} Therefore the restriction of $\lambda$ to the fibre is zero, so the restriction of $\omega=-d\lambda$ is also zero: \begin{align*} \omega|_{T_{q_0}^*Q}=-d(\lambda|_{T_{q_0}^*Q})=-d0=0. \end{align*} The fibre has dimension $n$, while $\dim T^*Q=2n$, so it is an $n$-dimensional isotropic submanifold of a $2n$-dimensional symplectic manifold. Hence $T_{q_0}^*Q$ is Lagrangian. [/example] The point case shows that conormals include the vertical Lagrangians given by cotangent fibres. For a positive-dimensional submanifold, the same annihilator condition leaves only the momentum directions normal to the constraint. [example: Conormal Bundle of a Hypersurface] Let $N\subset Q$ be a hypersurface locally defined by $r=0$ with $dr\ne 0$ along $N$. Choose local coordinates $(x_1,\dots,x_{n-1},r)$ near a point of $N$, so tangent vectors to $N$ have the form \begin{align*} v=\sum_{i=1}^{n-1} c_i\partial_{x_i}. \end{align*} A covector over $q\in N$ can be written uniquely as \begin{align*} \beta=\sum_{i=1}^{n-1}\eta_i\,dx_i|_q+a\,dr|_q. \end{align*} For each $j\le n-1$, evaluating on $\partial_{x_j}$ gives \begin{align*} \beta(\partial_{x_j})=\sum_{i=1}^{n-1}\eta_i\,dx_i(\partial_{x_j})+a\,dr(\partial_{x_j})=\eta_j+a\cdot 0=\eta_j. \end{align*} Thus $\beta$ annihilates every vector in $T_qN$ exactly when $\eta_1=\cdots=\eta_{n-1}=0$, which means \begin{align*} \beta=a\,dr|_q. \end{align*} Therefore the conormal bundle is locally \begin{align*} N^*N=\{(q,p):r(q)=0,\ p=a\,dr_q\text{ for some }a\in\mathbb R\}. \end{align*} In the induced cotangent coordinates $(x_1,\dots,x_{n-1},r,\eta_1,\dots,\eta_{n-1},a)$, this same subset is cut out by \begin{align*} r=0,\qquad \eta_1=\cdots=\eta_{n-1}=0. \end{align*} So a codimension-one constraint in configuration space leaves exactly one free momentum coordinate, namely the coefficient $a$ of the normal covector $dr$. [/example] The cotangent bundle is therefore more than a supply of examples: it is the universal local model for symplectic geometry. The next step in the course is Darboux's theorem, which says that every symplectic manifold locally looks like this canonical model, even though its global geometry may be very different. The cotangent-bundle chapter showed a canonical source of the standard local form, but it has not yet shown that arbitrary symplectic manifolds have the same local model. The next step is Darboux's theorem, which proves precisely that local universality and sets up the later study of Hamiltonian vector fields and Poisson brackets. # 4. Darboux's Theorem and Moser's Method This chapter proves the local flexibility theorem at the heart of symplectic geometry. The previous chapters built symplectic manifolds from closed nondegenerate $2$-forms and compared them with cotangent bundles and linear models; now we show that every symplectic manifold has the same local model near each point. The mechanism is Moser's method: instead of changing coordinates directly, we build an isotopy whose time-dependent vector field compensates for a moving family of forms. ## Darboux Coordinates and the Meaning of Local Flatness What should a coordinate chart remember about a symplectic form? For a Riemannian metric, curvature gives local invariants, so no coordinate change can usually turn the metric into the Euclidean metric on an open set. Darboux's theorem says that symplectic forms behave in the opposite way: after choosing suitable coordinates, the form has the same expression as the constant linear model. We first fix the standard model that all local symplectic forms will be compared with. This is the manifold version of the linear normal form from symplectic linear algebra. [definition: Standard Symplectic Form] On $\mathbb R^{2n}$ with coordinates $(q_1,\dots,q_n,p_1,\dots,p_n)$, the standard symplectic form is \begin{align*} \omega_0=\sum_{i=1}^n dq_i\wedge dp_i. \end{align*} [/definition] This form is closed because its coefficients are constant, and it is nondegenerate because its value at each point is the standard symplectic form on the vector space $\mathbb R^{2n}$. Darboux coordinates are coordinates in which an arbitrary symplectic form becomes this model. [definition: Darboux Chart] Let $(M,\omega)$ be a symplectic manifold of dimension $2n$. A Darboux chart around $x_0\in M$ is a coordinate chart $(U,\varphi)$, where $x_0\in U$ and $\varphi:U\to\varphi(U)\subseteq\mathbb R^{2n}$ is a diffeomorphism onto an open subset, whose coordinate functions are $(q_1,\dots,q_n,p_1,\dots,p_n)$ and such that \begin{align*} \omega|_U=\sum_{i=1}^n dq_i\wedge dp_i. \end{align*} [/definition] The central point is that these charts always exist. The theorem is local, but its conclusion is stronger than a pointwise linear normal form: the coefficients are normalized throughout a neighbourhood. [quotetheorem:10036] [citeproof:10036] Darboux's theorem removes local invariants of the symplectic $2$-form itself, and each hypothesis is doing work. Closedness is necessary because pullbacks commute with $d$; for instance, a non-closed nondegenerate $2$-form on an open subset of $\mathbb R^{4}$ cannot be transformed into $\omega_0$, since $d\omega_0=0$. Nondegeneracy is also necessary: the zero $2$-form, or any $2$-form with a kernel at a point, cannot become $\omega_0$ under a diffeomorphism because rank is preserved. Finally, the even-dimensional condition is forced by linear algebra, since every skew-symmetric bilinear form on an odd-dimensional vector space has nontrivial kernel. The theorem therefore does not say that all geometric phenomena are local; instead, it pushes the subject toward global invariants, boundary behaviour, distinguished submanifolds, and dynamics. [example: Local Normal Coordinates Near A Point] Let $(M,\omega)$ be a $2n$-dimensional symplectic manifold and let $x_0\in M$. Choose local coordinates near $x_0$ with $x_0$ sent to $0$, and write $B=\omega_{x_0}$ on $T_{x_0}M$. By the [symplectic basis theorem](/theorems/10037), there is a basis $e_1,\dots,e_n,f_1,\dots,f_n$ of $T_{x_0}M$ such that \begin{align*} B(e_i,e_j)=0,\quad B(f_i,f_j)=0,\quad B(e_i,f_j)=\delta_{ij}. \end{align*} Using the corresponding linear coordinates $(q_1,\dots,q_n,p_1,\dots,p_n)$ at the origin, the standard form satisfies \begin{align*} (dq_i\wedge dp_i)(e_j,f_k)=dq_i(e_j)dp_i(f_k)-dq_i(f_k)dp_i(e_j)=\delta_{ij}\delta_{ik}. \end{align*} Summing over $i$ gives \begin{align*} \left(\sum_{i=1}^n dq_i\wedge dp_i\right)(e_j,f_k)=\delta_{jk}, \end{align*} and the same formula gives zero on pairs $(e_j,e_k)$ and $(f_j,f_k)$. Hence this linear change of coordinates makes \begin{align*} \omega_{x_0}=\sum_{i=1}^n dq_i\wedge dp_i \end{align*} at the single tangent space $T_{x_0}M$. The *[Darboux Theorem](/theorems/10036)* strengthens this pointwise statement: after shrinking the neighbourhood and changing coordinates again, there are coordinates $(Q_1,\dots,Q_n,P_1,\dots,P_n)$ in which \begin{align*} \omega=\sum_{i=1}^n dQ_i\wedge dP_i \end{align*} on the whole neighbourhood. Thus the linear algebra step fixes only the value of $\omega$ at $x_0$, while the Moser step in Darboux's proof removes the nearby coefficient variation and identifies the neighbourhood with an open subset of $(\mathbb R^{2n},\omega_0)$. [/example] For surfaces this theorem has a familiar interpretation. A symplectic form is just an area form, so Darboux's theorem says that every area form can locally be written as $dq\wedge dp$ after a coordinate change preserving orientation. [example: Darboux Coordinates On A Symplectic Surface] Let $(\Sigma,\omega)$ be a symplectic surface and let $x_0\in\Sigma$. Choose an oriented coordinate chart $(x,y)$ near $x_0$. Since $dx,dy$ form a basis of the cotangent space in this chart, every local $2$-form has the shape \begin{align*} \omega=f(x,y)\,dx\wedge dy \end{align*} for a smooth function $f$. Nondegeneracy of $\omega_{x_0}$ means $\omega_{x_0}(\partial_x,\partial_y)\neq 0$, and here \begin{align*} \omega_{x_0}(\partial_x,\partial_y)=f(x_0)(dx\wedge dy)(\partial_x,\partial_y)=f(x_0)(1\cdot 1-0\cdot 0)=f(x_0). \end{align*} Because the chart is oriented compatibly with the area form, we may take $f(x_0)>0$; after shrinking the chart, continuity gives $f>0$ throughout the neighbourhood. By *Darboux Theorem*, there are local coordinates $(q,p)$ centred near $x_0$ such that \begin{align*} \omega=dq\wedge dp. \end{align*} Thus the function $f(x,y)$, although it may vary in the original oriented chart, can be absorbed by a change of coordinates. In dimension two, Darboux coordinates are exactly local area coordinates, so symplectic surfaces have no curvature-like local invariant coming from the $2$-form alone. [/example] ## Moser's Path Method How can a coordinate change be found when two symplectic forms are connected by a path? Moser's method converts a path of forms into a path of diffeomorphisms. The guiding equation says that the change in the form should be cancelled by dragging the manifold along a time-dependent vector field. The method needs a time-dependent vector field and its flow. We use the standard convention that $\psi_t$ is generated by $X_t$ when \begin{align*} \frac{d}{dt}\psi_t(x)&=X_t(\psi_t(x)), & \psi_0&=\operatorname{id}. \end{align*} [definition: Moser Isotopy] Let $M$ be a smooth manifold and let $(\omega_t)_{t\in[0,1]}$ be a smooth family of symplectic forms. A Moser isotopy for $(\omega_t)$ is a smooth family of diffeomorphisms $\psi_t:M\to M$ with $\psi_0=\operatorname{id}$ such that \begin{align*} \psi_t^*\omega_t=\omega_0 \end{align*} for all $t\in[0,1]$. [/definition] The definition packages the desired outcome, but it does not tell us when such an isotopy exists. The next theorem identifies the two conditions that make the construction work on a compact manifold: the forms must remain symplectic along the path, and their cohomology class must stay fixed so that the infinitesimal change has a primitive. [quotetheorem:10048] [citeproof:10048] The theorem explains why the de Rham cohomology class of a symplectic form is the right global datum in this problem. The fixed-cohomology hypothesis is necessary for isotopies starting at the identity: if $\psi_t$ is such an isotopy and $\psi_1^*\omega_1=\omega_0$, then $[\omega_0]=[\omega_1]$ because isotopic maps act identically on de Rham cohomology. Compactness is a separate analytic hypothesis: it ensures that the time-dependent vector field produced by Moser's equation integrates for the whole interval $[0,1]$. On a noncompact manifold a smooth vector field can have finite-time escape, such as $X=x^2\partial_x$ on $\mathbb R$, whose positive-time flow from $x>0$ blows up at time $1/x$. Thus Moser stability is both topological and analytic: the primitive exists because the cohomology class is fixed, and the global isotopy exists because the flow cannot leave a compact manifold. [example: Cohomologous Area Forms On A Compact Surface] Let $\Sigma$ be a compact connected oriented surface, and let $\omega_0$ and $\omega_1$ be area forms with equal total area: \begin{align*} \int_\Sigma \omega_0=\int_\Sigma \omega_1. \end{align*} Choose a positive local oriented area form $\rho$. On each coordinate patch we can write $\omega_0=f_0\rho$ and $\omega_1=f_1\rho$ with $f_0>0$ and $f_1>0$. For $t\in[0,1]$, \begin{align*} \omega_t=(1-t)\omega_0+t\omega_1=((1-t)f_0+t f_1)\rho. \end{align*} Since $(1-t)f_0+t f_1>0$, each $\omega_t$ is again an area form. Also $d\omega_t=0$, because $\omega_t$ is a $2$-form on a surface and there are no nonzero $3$-forms. On a compact connected oriented surface, the top-degree de Rham class is determined by integration over $\Sigma$. Thus the equal-area assumption gives \begin{align*} [\omega_0]=[\omega_1]\in H^2_{\mathrm{dR}}(\Sigma). \end{align*} Because $\omega_t=(1-t)\omega_0+t\omega_1$, all classes are the same: \begin{align*} [\omega_t]=(1-t)[\omega_0]+t[\omega_1]=[\omega_0]. \end{align*} By *[Moser Stability Theorem](/theorems/10048)*, there is an isotopy $\psi_t$ with $\psi_0=\operatorname{id}$ and \begin{align*} \psi_t^*\omega_t=\omega_0. \end{align*} Taking $t=1$ and writing $\psi=\psi_1$ gives \begin{align*} \psi^*\omega_1=\omega_0. \end{align*} Conversely, if such a diffeomorphism isotopic to the identity exists, then \begin{align*} \int_\Sigma \omega_0=\int_\Sigma \psi^*\omega_1=\int_{\psi(\Sigma)}\omega_1=\int_\Sigma \omega_1. \end{align*} Thus on a compact connected oriented surface, total area is exactly the invariant of an area form up to isotopy. [/example] This example uses compactness to avoid any question about whether the vector field can be integrated for the whole interval. In local normal form problems the same differential equation is used, but the domain is allowed to shrink so that the relevant flow remains defined. [quotetheorem:10049] [citeproof:10049] This local lemma is often the practical form of the method: set up a path, solve a contraction equation, and shrink the domain if necessary. Its hypotheses also mark the method's limitations. If $\dot\omega_t$ is not exact, then the equation $\dot\omega_t=d\sigma_t$ has no global primitive on the chosen open set, and the Moser derivative computation cannot be made to vanish; on $T^2$, the path $\omega_t=(1+t)\omega_0$ has $\dot\omega_t=\omega_0$, which is closed but not exact. If the path leaves the symplectic cone, the contraction equation may stop determining a vector field; for example, in dimension two the path $(1-2t)\,dq\wedge dp$ becomes degenerate at $t=1/2$. The difficulty is therefore not usually solving the vector field equation once the hypotheses hold, but arranging exactness, nondegeneracy, and any boundary or submanifold conditions needed for the flow to preserve the feature under study. ## Relative Forms and Fixed Submanifolds What if the coordinate change must leave a submanifold fixed? Darboux's theorem normalizes the ambient form near a point, but many geometric problems come with a distinguished submanifold, such as a Lagrangian, a hypersurface, or a fixed zero set. The relative version of Moser's method adds vanishing conditions to the primitive so that the vector field vanishes along the submanifold. The needed input is a relative Poincare lemma. It says that if a closed form vanishes along a submanifold in the right sense, then it has a primitive that also vanishes there. For a Moser argument that fixes the submanifold pointwise, the required vanishing is stronger than vanishing after pullback to the submanifold: the primitive must vanish as a covector on the ambient tangent spaces over the submanifold. [quotetheorem:10050] [citeproof:10050] This lemma is the relative replacement for the local Poincare lemma in Darboux's proof, and the restriction hypothesis is necessary. If a form is exact with primitive vanishing on $N$, then its restriction to $N$ must vanish after pulling back to $N$; hence a closed form with nonzero restriction cannot have such a relative primitive. For a concrete example, take $M=\mathbb R^2$ and $N=M$ with $\alpha=dx\wedge dy$: then $\alpha|_N\neq 0$, so no $1$-form $\beta$ with $\beta|_N=0$ can satisfy $d\beta=\alpha$ in the relative sense. The next step is to feed the vanishing primitive into Moser's equation, because a primitive with suitable vanishing makes the resulting vector field fix the chosen submanifold during the isotopy. [quotetheorem:10051] [citeproof:10051] The relative lemma is a template rather than a single-purpose result, and its hypotheses are exactly what let the isotopy preserve the submanifold. Agreement of the ambient forms along $N$ is stronger than equality after restriction to $TN$; the stronger condition is what gives a primitive that vanishes as an ambient covector and hence a Moser vector field vanishing on $N$. Symplecticity of the whole path is also necessary for this proof: if some $\omega_t$ becomes degenerate, the equation $\iota_{X_t}\omega_t=-\sigma$ may no longer determine $X_t$. The conclusion is local near $N$, not global on $M$, because the flow and the chosen primitives may only exist after shrinking neighbourhoods. The next normal forms apply this local mechanism to submanifolds whose first-order symplectic data determine the whole neighbourhood model. ## Lagrangian Neighbourhoods What is the local model near a Lagrangian submanifold? Since a Lagrangian has half the ambient dimension and the symplectic form vanishes on its tangent spaces, ordinary Darboux coordinates at each point do not describe how the neighbourhoods glue along the whole submanifold. The correct model is the cotangent bundle, with the Lagrangian identified with the zero section. We recall the canonical form on a cotangent bundle. It is the basic example in which the base is Lagrangian. [definition: Tautological One-Form] Let $L$ be a smooth manifold and let $\pi:T^*L\to L$ be the cotangent bundle projection. The tautological $1$-form $\lambda_{\mathrm{can}}$ on $T^*L$ is defined by \begin{align*} (\lambda_{\mathrm{can}})_\xi(v)=\xi(d\pi_\xi(v)) \end{align*} for $\xi\in T^*L$ and $v\in T_\xi(T^*L)$. [/definition] The tautological form records the covector sitting over each point of $L$, so its exterior derivative measures how base and fibre directions pair. With the sign convention below, the resulting symplectic form has the same local expression as the standard Darboux form. [definition: Canonical Cotangent Symplectic Form] The canonical symplectic form on $T^*L$ is \begin{align*} \omega_{\mathrm{can}}=-d\lambda_{\mathrm{can}}. \end{align*} [/definition] The canonical form completes the cotangent model: in local coordinates $(q_1,\dots,q_n)$ on $L$ and fibre coordinates $(p_1,\dots,p_n)$, one has $\lambda_{\mathrm{can}}=\sum_i p_i dq_i$ and therefore $\omega_{\mathrm{can}}=\sum_i dq_i\wedge dp_i$. This calculation shows that the zero section is Lagrangian and motivates the neighbourhood theorem: every embedded Lagrangian should have this same cotangent-bundle model, not merely the literal zero section. [quotetheorem:10052] [citeproof:10052] This theorem is the symplectic analogue of a [tubular neighbourhood theorem](/theorems/2276), but the normal bundle is not arbitrary. The symplectic form identifies normal directions to a Lagrangian with covectors on the Lagrangian, which is why $T^*L$ appears canonically. Embeddedness matters because the theorem describes an actual neighbourhood of $L$ inside $M$; an immersed Lagrangian with self-intersections has different local branches meeting at the same ambient point, so no single neighbourhood of the abstract source manifold can model the image without recording the self-intersection data. The conclusion is also only local near the zero section: two cotangent bundles can contain neighbourhoods of their zero sections that are symplectomorphic in this sense even though their larger global symplectic geometry, such as fibrewise behaviour at infinity, is different. [example: Neighbourhoods Of The Zero Section In A Cotangent Bundle] Let $s:L\to T^*L$ be the zero section, so $s(x)=0_x\in T_x^*L$. For $v\in T_xL$, the definition of the tautological form gives \begin{align*} (s^*\lambda_{\mathrm{can}})_x(v)=(\lambda_{\mathrm{can}})_{0_x}(ds_xv)=0_x(d\pi_{0_x}(ds_xv))=0. \end{align*} Hence $s^*\lambda_{\mathrm{can}}=0$, and therefore \begin{align*} s^*\omega_{\mathrm{can}}=s^*(-d\lambda_{\mathrm{can}})=-d(s^*\lambda_{\mathrm{can}})=-d0=0. \end{align*} The zero section has dimension $n$, while $T^*L$ has dimension $2n$, so the zero section is Lagrangian. In this model example, the *[Lagrangian Neighbourhood Theorem](/theorems/10052)* may be taken with $U=V$ and $\Phi=\operatorname{id}$, because \begin{align*} \operatorname{id}^*\omega_{\mathrm{can}}=\omega_{\mathrm{can}}. \end{align*} More generally, if $L\subset(M,\omega)$ is Lagrangian, the *Lagrangian Neighbourhood Theorem* identifies a sufficiently small neighbourhood of $L$ in $M$ with a neighbourhood of the zero section in $(T^*L,\omega_{\mathrm{can}})$. Under this identification, a nearby submanifold transverse to the cotangent fibres is the graph of a unique $1$-form $\alpha\in\Omega^1(L)$. If $a_\alpha:L\to T^*L$ is the graph map $a_\alpha(x)=\alpha_x$, then for $v\in T_xL$, \begin{align*} (a_\alpha^*\lambda_{\mathrm{can}})_x(v)=(\lambda_{\mathrm{can}})_{\alpha_x}(d(a_\alpha)_xv)=\alpha_x(d\pi_{\alpha_x}(d(a_\alpha)_xv)). \end{align*} Since $\pi\circ a_\alpha=\operatorname{id}_L$, we have $d\pi_{\alpha_x}(d(a_\alpha)_xv)=v$, so \begin{align*} (a_\alpha^*\lambda_{\mathrm{can}})_x(v)=\alpha_x(v). \end{align*} Thus $a_\alpha^*\lambda_{\mathrm{can}}=\alpha$, and \begin{align*} a_\alpha^*\omega_{\mathrm{can}}=a_\alpha^*(-d\lambda_{\mathrm{can}})=-d(a_\alpha^*\lambda_{\mathrm{can}})=-d\alpha. \end{align*} Therefore the graph is Lagrangian exactly when $d\alpha=0$. The cotangent neighbourhood model turns small fibre-transverse Lagrangian submanifolds near $L$ into graphs of closed $1$-forms on $L$. [/example] Graphs in the cotangent model make deformations of Lagrangians concrete. If $\alpha$ is a $1$-form on $L$, its graph is a submanifold of $T^*L$, and the condition for it to be Lagrangian can be read from $d\alpha$. [example: Graphs Of Closed One-Forms] Let $\alpha\in\Omega^1(L)$, and define the graph map $a_\alpha:L\to T^*L$ by $a_\alpha(x)=\alpha_x$. Its image is $\Gamma_\alpha=\{\alpha_x:x\in L\}$. For $v\in T_xL$, the definition of the tautological form gives \begin{align*} (a_\alpha^*\lambda_{\mathrm{can}})_x(v)=(\lambda_{\mathrm{can}})_{\alpha_x}(d(a_\alpha)_xv)=\alpha_x(d\pi_{\alpha_x}(d(a_\alpha)_xv)). \end{align*} Since $\pi\circ a_\alpha=\operatorname{id}_L$, differentiating gives \begin{align*} d\pi_{\alpha_x}(d(a_\alpha)_xv)=d(\pi\circ a_\alpha)_x(v)=d(\operatorname{id}_L)_x(v)=v. \end{align*} Substituting this into the previous formula yields \begin{align*} (a_\alpha^*\lambda_{\mathrm{can}})_x(v)=\alpha_x(v). \end{align*} Because this holds for every $x\in L$ and every $v\in T_xL$, we have \begin{align*} a_\alpha^*\lambda_{\mathrm{can}}=\alpha. \end{align*} Using $\omega_{\mathrm{can}}=-d\lambda_{\mathrm{can}}$ and the identity $a_\alpha^*(d\eta)=d(a_\alpha^*\eta)$ for pullbacks and exterior derivatives, \begin{align*} a_\alpha^*\omega_{\mathrm{can}}=a_\alpha^*(-d\lambda_{\mathrm{can}})=-d(a_\alpha^*\lambda_{\mathrm{can}})=-d\alpha. \end{align*} The graph $\Gamma_\alpha$ has dimension $\dim L$, while $T^*L$ has dimension $2\dim L$. Therefore $\Gamma_\alpha$ is Lagrangian exactly when the restriction of $\omega_{\mathrm{can}}$ to $\Gamma_\alpha$ vanishes, which under the parametrization $a_\alpha$ is exactly the condition \begin{align*} a_\alpha^*\omega_{\mathrm{can}}=0. \end{align*} By the computation above, this is equivalent to \begin{align*} -d\alpha=0, \end{align*} and hence to $d\alpha=0$. Thus graphs of closed $1$-forms are precisely the fibre-transverse Lagrangian submanifolds in the cotangent model. [/example] ## Hypersurface Neighbourhoods and Characteristic Directions What data determine a symplectic form near a hypersurface? A hypersurface in a symplectic manifold cannot be symplectic, because it has odd dimension. Instead, the restriction of the symplectic form has a one-dimensional kernel, called the characteristic direction, and neighbourhood normal forms keep track of this presymplectic data. The first definition records the structure induced on a hypersurface by the ambient symplectic form. [definition: Characteristic Line Field] Let $(M,\omega)$ be a symplectic manifold and let $H\subset M$ be an embedded hypersurface. The characteristic line field on $H$ is \begin{align*} \mathcal C_H=\ker(\omega|_H)\subset TH, \end{align*} where $\omega|_H$ denotes the pullback of $\omega$ to $H$. [/definition] The kernel has rank one because $TH$ is a codimension-one subspace of a symplectic vector space. Its integral curves are the characteristic curves of the hypersurface, and they become central in Hamiltonian dynamics and contact-type boundary theory. [example: Energy Hypersurfaces In Hamiltonian Mechanics] Let $H_0:M\to\mathbb R$ be a smooth Hamiltonian on a symplectic manifold $(M,\omega)$, and suppose $c$ is a regular value. Then for each $x\in\Sigma:=H_0^{-1}(c)$, the differential $(dH_0)_x$ is nonzero, so $\Sigma$ is a hypersurface and \begin{align*} T_x\Sigma=\ker (dH_0)_x. \end{align*} The Hamiltonian vector field is defined by \begin{align*} \iota_{X_{H_0}}\omega=-dH_0. \end{align*} Evaluating this identity on $X_{H_0}$ gives \begin{align*} \omega(X_{H_0},X_{H_0})=-dH_0(X_{H_0}). \end{align*} Since $\omega$ is alternating, $\omega(X_{H_0},X_{H_0})=0$, and hence \begin{align*} dH_0(X_{H_0})=0. \end{align*} Thus $X_{H_0}(x)\in\ker(dH_0)_x=T_x\Sigma$, so the Hamiltonian vector field is tangent to the energy hypersurface. Now let $v\in T_x\Sigma$. Since $T_x\Sigma=\ker(dH_0)_x$, we have $(dH_0)_x(v)=0$. Using the Hamiltonian equation at $x$, \begin{align*} \omega_x(X_{H_0}(x),v)=-(dH_0)_x(v). \end{align*} Therefore \begin{align*} \omega_x(X_{H_0}(x),v)=0. \end{align*} Because this holds for every $v\in T_x\Sigma$, the tangent vector $X_{H_0}(x)$ lies in $\ker(\omega|_\Sigma)_x$. Since $(dH_0)_x\neq 0$ and $\omega_x$ is nondegenerate, $X_{H_0}(x)\neq 0$, so it spans the one-dimensional characteristic line at $x$. Hence the characteristic curves on $\Sigma$ are exactly the Hamiltonian trajectories with energy $c$. [/example] The example shows that the restricted form on a hypersurface controls a genuine dynamical direction rather than disappearing under Darboux coordinates. The normal form question is therefore relative: if two hypersurfaces have the same restricted closed $2$-form under a given identification, does the ambient symplectic geometry near them also agree? [quotetheorem:10053] [citeproof:10053] The theorem is weaker than Darboux's theorem in one respect and stronger in another. It does not erase the characteristic foliation on the hypersurface, but it says that no additional transverse symplectic data remain once the restricted form is fixed. The hypothesis on the restricted forms is necessary: if $F$ extends $f$ and satisfies $F^*\omega_1=\omega_0$, then pulling back to $H_0$ gives $f^*(\omega_1|_{H_1})=\omega_0|_{H_0}$. For example, on the same hypersurface $H\subset M$, two ambient forms whose restrictions have different kernels induce different characteristic line fields, so no extension fixing the given identification can be symplectic. The result is also local after shrinking; it does not assert that global collar neighbourhoods, return maps, or Hamiltonian dynamics away from the hypersurface are determined by the restricted $2$-form. [remark: No Local Symplectic Invariants Versus Relative Invariants] Darboux's theorem says that points have no local symplectic invariants. Relative problems can still have local data because the submanifold is part of the structure being preserved. For Lagrangians, the local model is always the zero section in $T^*L$; for hypersurfaces, the restricted closed $2$-form and its characteristic line field are the data that survive. [/remark] ## The Role Of Moser's Method In The Course Why is Moser's method more than a proof of Darboux's theorem? It is a general strategy for proving that two geometric structures are equivalent: connect them by a path, solve a time-dependent equation, and integrate the resulting vector field. The method will reappear whenever a symplectic form varies in a fixed cohomology class or when a normal form is determined by first-order data along a submanifold. The algebraic heart of the method is the isomorphism between vector fields and $1$-forms induced by a symplectic form. The analytic heart is the existence of the time-dependent flow, which is automatic on compact manifolds and local after shrinking. The topological heart is exactness: without a primitive for $\dot\omega_t$, there is no reason for the derivative of the pulled-back form to vanish. [example: Why The Cohomology Hypothesis Is Necessary] Let $M$ be a compact symplectic manifold, and suppose $\psi:M\to M$ is a diffeomorphism isotopic to the identity. An isotopy from $\operatorname{id}_M$ to $\psi$ is a homotopy between the two maps, so by *[Homotopy Invariance of de Rham Cohomology](/theorems/3585)* the induced maps on cohomology agree: \begin{align*} \psi^*=(\operatorname{id}_M)^*=\operatorname{id}_{H^*_{\mathrm{dR}}(M)}. \end{align*} If $\psi^*\omega_1=\omega_0$, then passing to de Rham classes gives \begin{align*} [\omega_0]=[\psi^*\omega_1]. \end{align*} Pullback commutes with exterior derivative, so it descends to cohomology classes, and therefore \begin{align*} [\psi^*\omega_1]=\psi^*[\omega_1]. \end{align*} Using that $\psi^*$ is the identity on cohomology, \begin{align*} [\omega_0]=\psi^*[\omega_1]=[\omega_1]. \end{align*} Thus an isotopy starting at the identity can only carry $\omega_1$ to $\omega_0$ when the two forms have the same de Rham cohomology class. Moser stability therefore cannot hold for arbitrary paths of symplectic forms whose cohomology class changes. On a compact connected oriented surface, $H^2_{\mathrm{dR}}(\Sigma)$ is determined by integration over $\Sigma$, so the condition becomes \begin{align*} [\omega_0]=[\omega_1]\quad\Longleftrightarrow\quad \int_\Sigma\omega_0=\int_\Sigma\omega_1. \end{align*} In higher dimensions the same obstruction is the full degree-two class $[\omega]\in H^2_{\mathrm{dR}}(M)$, not just a single total-area number. [/example] The chapter's normal forms set up the Hamiltonian chapters that follow. Since Darboux coordinates remove local invariants of the two-form itself, the next structures to study are distinguished functions and vector fields: Hamiltonian flows in Chapter 5, symplectic and Hamiltonian isotopies in Chapter 6, moment maps in Chapter 8, and quotient constructions in Chapter 9. Darboux coordinates eliminate local invariants of the symplectic form, but they leave behind the functions and flows built from it. This chapter turns the nondegeneracy of ω into a mechanism that associates vector fields and dynamics to Hamiltonians. # 5. Hamiltonian Vector Fields and Poisson Brackets Hamiltonian dynamics is where the nondegeneracy of a symplectic form begins to act as a machine: a function determines a vector field, and the vector field determines a flow. Chapters 2 through 4 built symplectic manifolds, symplectomorphisms, and Darboux local normal forms; this chapter uses that structure to turn energy functions into motion. The prerequisites are the definition of a symplectic manifold, the musical-map viewpoint on nondegenerate bilinear forms, exterior differentiation, Lie derivatives, pullbacks of differential forms, and the local flow theorem for smooth vector fields. The main questions are how a Hamiltonian function $H$ produces its vector field $X_H$, why the resulting flow preserves both $H$ and $\omega$, and how Poisson brackets package conservation laws into algebra. ## From One-Forms To Vector Fields On a Riemannian manifold, a function produces a gradient vector field by using the metric to convert $dH$ into a vector. A symplectic manifold has no metric and no preferred notion of length, but nondegeneracy of $\omega$ gives a different conversion between tangent and cotangent spaces. The first problem is to identify this conversion and fix the sign convention for Hamiltonian vector fields. [definition: Symplectic Musical Map] Let $(M,\omega)$ be a symplectic manifold. The symplectic musical map is the vector bundle map \begin{align*} \omega^\flat : TM \longrightarrow T^*M, \quad \omega^\flat(v)(w)=\omega(v,w) \end{align*} for $v,w \in T_pM$ and $p \in M$. [/definition] The adjective musical is borrowed from Riemannian geometry, but the symplectic map is skew rather than symmetric. Nondegeneracy of $\omega$ says exactly that $\omega^\flat_p:T_pM\to T_p^*M$ is an isomorphism for every $p\in M$. If the two-form were degenerate, this conversion could fail in two ways: a one-form such as $dH_p$ might lie outside the image of $v\mapsto\omega_p(v,\cdot)$, or it might have several preimages differing by vectors in the kernel of $\omega_p$. Nondegeneracy rules out both nonexistence and nonuniqueness, so it leads to the defining object of Hamiltonian dynamics. [definition: Hamiltonian Vector Field] Let $(M,\omega)$ be a symplectic manifold and let $H\in C^\infty(M)$. The Hamiltonian vector field of $H$ is the unique vector field $X_H\in\mathfrak X(M)$ satisfying \begin{align*} \omega(X_H,\cdot)=dH. \end{align*} [/definition] This course uses the convention $\omega(X_H,\cdot)=dH$. Some books put a minus sign on the right-hand side; every formula involving the Poisson bracket changes sign under that alternative convention, so the convention must be held fixed throughout. [example: Hamiltonian Vector Field In Standard Coordinates] On $\mathbb R^{2n}$ with coordinates $(q_1,\dots,q_n,p_1,\dots,p_n)$ and \begin{align*} \omega_0=\sum_{i=1}^n dq_i\wedge dp_i, \end{align*} let \begin{align*} X_H=\sum_{i=1}^n a_i\partial_{q_i}+\sum_{i=1}^n b_i\partial_{p_i}. \end{align*} For an arbitrary tangent vector \begin{align*} Y=\sum_{i=1}^n u_i\partial_{q_i}+\sum_{i=1}^n v_i\partial_{p_i}, \end{align*} the wedge product satisfies $(dq_i\wedge dp_i)(X_H,Y)=dq_i(X_H)dp_i(Y)-dp_i(X_H)dq_i(Y)$. Since $dq_i(X_H)=a_i$, $dp_i(Y)=v_i$, $dp_i(X_H)=b_i$, and $dq_i(Y)=u_i$, we get \begin{align*} \omega_0(X_H,Y)=\sum_{i=1}^n(a_i v_i-b_i u_i). \end{align*} On the other hand, \begin{align*} dH(Y)=\sum_{i=1}^n\frac{\partial H}{\partial q_i}u_i+\sum_{i=1}^n\frac{\partial H}{\partial p_i}v_i. \end{align*} The defining equation $\omega_0(X_H,\cdot)=dH$ therefore gives, for every choice of the coefficients $u_i$ and $v_i$, \begin{align*} \sum_{i=1}^n(a_i v_i-b_i u_i)=\sum_{i=1}^n\frac{\partial H}{\partial q_i}u_i+\sum_{i=1}^n\frac{\partial H}{\partial p_i}v_i. \end{align*} Comparing the coefficient of $v_i$ gives $a_i=\partial H/\partial p_i$, and comparing the coefficient of $u_i$ gives $-b_i=\partial H/\partial q_i$, hence $b_i=-\partial H/\partial q_i$. Thus \begin{align*} X_H=\sum_{i=1}^n\frac{\partial H}{\partial p_i}\partial_{q_i}-\sum_{i=1}^n\frac{\partial H}{\partial q_i}\partial_{p_i}. \end{align*} An integral curve $(q(t),p(t))$ of $X_H$ has velocity \begin{align*} \frac{d}{dt}(q(t),p(t))=\sum_{i=1}^n\dot q_i(t)\partial_{q_i}+\sum_{i=1}^n\dot p_i(t)\partial_{p_i}, \end{align*} so matching this velocity with $X_H$ gives Hamilton's equations \begin{align*} \dot q_i=\frac{\partial H}{\partial p_i}, \quad \dot p_i=-\frac{\partial H}{\partial q_i}. \end{align*} This coordinate calculation shows exactly how the abstract equation $\omega_0(X_H,\cdot)=dH$ recovers the canonical equations from mechanics. [/example] The coordinate formula also shows that a constant Hamiltonian produces the zero vector field. More generally, $X_H$ only depends on $dH$, so adding a constant to $H$ does not change the dynamics; this observation will later explain why the Poisson algebra naturally uses functions modulo constants. [remark: Constants In The Hamiltonian] If $c\in\mathbb R$, then $X_{H+c}=X_H$. The assignment $H\mapsto X_H$ therefore factors through $C^\infty(M)/\mathbb R$ when $M$ is connected, and through functions modulo locally constant functions on a disconnected manifold. [/remark] The free particle is the simplest phase-space model: the Hamiltonian depends only on momentum, so the system should move without force. This example fixes the interpretation of the $q$- and $p$-coordinates before adding a restoring force. [example: Free Particle On Euclidean Space] On $T^*\mathbb R^n\cong\mathbb R^{2n}$ with canonical coordinates $(q,p)=(q_1,\dots,q_n,p_1,\dots,p_n)$, take \begin{align*} H(q,p)=\frac{1}{2}|p|^2=\frac{1}{2}\sum_{i=1}^n p_i^2. \end{align*} For each $i$, the partial derivatives are \begin{align*} \frac{\partial H}{\partial p_i}=p_i. \end{align*} Since $H$ has no dependence on any $q_i$, \begin{align*} \frac{\partial H}{\partial q_i}=0. \end{align*} Hamilton's equations therefore become \begin{align*} \dot q_i=p_i,\quad \dot p_i=0. \end{align*} The equations $\dot p_i=0$ imply $p_i(t)=p_i(0)$ for every $i$. Substituting this into $\dot q_i=p_i$ gives $\dot q_i(t)=p_i(0)$, hence \begin{align*} q_i(t)=q_i(0)+t\,p_i(0). \end{align*} Thus $p(t)=p(0)$ and $q(t)=q(0)+t\,p(0)$, so the Hamiltonian describes straight-line motion in configuration space with constant velocity equal to the initial momentum. [/example] The next model adds a quadratic potential. It is the first case where the Hamiltonian level sets are compact curves and the flow is periodic. [example: Harmonic Oscillator] On $\mathbb R^2$ with coordinates $(q,p)$ and $\omega_0=dq\wedge dp$, take \begin{align*} H(q,p)=\frac{1}{2}(p^2+q^2). \end{align*} The partial derivatives are \begin{align*} \frac{\partial H}{\partial p}=p \end{align*} and \begin{align*} \frac{\partial H}{\partial q}=q. \end{align*} Using the standard-coordinate formula for Hamiltonian vector fields, this gives \begin{align*} X_H=\frac{\partial H}{\partial p}\partial_q-\frac{\partial H}{\partial q}\partial_p=p\partial_q-q\partial_p. \end{align*} Therefore an integral curve $(q(t),p(t))$ satisfies \begin{align*} \dot q=p, \quad \dot p=-q. \end{align*} Along such a curve, the squared radius has derivative \begin{align*} \frac{d}{dt}(p(t)^2+q(t)^2)=2p(t)\dot p(t)+2q(t)\dot q(t). \end{align*} Substituting $\dot p(t)=-q(t)$ and $\dot q(t)=p(t)$ gives \begin{align*} 2p(t)(-q(t))+2q(t)p(t)=-2p(t)q(t)+2q(t)p(t)=0. \end{align*} Hence $p(t)^2+q(t)^2$ is constant, so the trajectories lie on circles centered at the origin. The vector field $(\dot q,\dot p)=(p,-q)$ is tangent to these circles, and the Hamiltonian \begin{align*} H(q,p)=\frac{1}{2}(p^2+q^2) \end{align*} is exactly one half of the squared radius. [/example] ## Hamiltonian Flows And Symplectic Invariance A vector field gives local flows when the usual existence theorem for ODEs applies. The symplectic question is more rigid: once $X_H$ is produced from $H$, what geometric structure is transported by its flow? The answer is that [Hamiltonian flows preserve the symplectic form](/theorems/6844), so they are symplectomorphisms wherever they are defined. [definition: Hamiltonian Flow] Let $(M,\omega)$ be a symplectic manifold and let $H\in C^\infty(M)$. A Hamiltonian flow for $H$ is a smooth map \begin{align*} \varphi:\mathcal D\longrightarrow M \end{align*} defined on an open subset $\mathcal D\subset \mathbb R\times M$ containing $\{0\}\times M$, with time maps $\varphi_t:U_t\to M$ where $U_t=\{p\in M:(t,p)\in\mathcal D\}$, such that \begin{align*} \frac{d}{dt}\varphi_t(p)=X_H(\varphi_t(p)), \quad \varphi_0(p)=p \end{align*} whenever both sides are defined. [/definition] The definition turns the Hamiltonian vector field into actual motion. A priori, however, the function that generates the motion could still vary along the resulting trajectories; an arbitrary vector field need not preserve a chosen smooth function. For Hamiltonian dynamics to deserve the language of fixed energy, the vector field constructed from $H$ must be tangent to the level sets of $H$. The next issue is therefore whether skew-symmetry of the symplectic form forces this tangency. [quotetheorem:6842] [citeproof:6842] Conservation of $H$ is a scalar invariant, and the skew-symmetry of $\omega$ is the essential hypothesis behind it: it forces every vector to pair to zero with itself. If the same construction is attempted with a non-skew nondegenerate pairing, conservation can fail. On $\mathbb R$ with the symmetric pairing $B(v,w)=vw$, the equation $B(X_H,\cdot)=dH$ for $H(x)=x^2/2$ gives $X_H=x\,\partial_x$, and along its flow \begin{align*} \frac{d}{dt}H(x(t))=dH(X_H)=x(t)^2, \end{align*} which is nonzero away from $0$. The theorem does not say that every level set of $H$ is a smooth hypersurface; singular points of $H$ may produce critical energy levels where the geometry of the flow changes. It also does not imply that trajectories are periodic or complete, because local existence of the Hamiltonian flow is still an ODE statement depending on the vector field and the manifold. The stronger geometric question is whether the flow preserves the differential form from which the vector field was constructed. The next theorem answers this using Cartan's formula and closedness of $\omega$. [quotetheorem:6844] [citeproof:6844] This theorem explains why Hamiltonian dynamics belongs to symplectic geometry rather than to general dynamics. The closedness condition $d\omega=0$ is not decorative: it removes the $\iota_{X_H}d\omega$ term in Cartan's formula, so without it the same vector field construction would not generally give symplectic flows. The result is also one-directional. Every Hamiltonian flow is a symplectic isotopy, but a symplectic isotopy need not be generated by a globally defined Hamiltonian function; global obstructions appear especially on manifolds with nontrivial first cohomology. Thus Hamiltonian flows form a distinguished subclass of symplectic motions, not all possible symplectic motions. [example: Symplectic Invariance Of The Harmonic Oscillator] For the harmonic oscillator $H(q,p)=\frac12(p^2+q^2)$, the equations are $\dot q=p$ and $\dot p=-q$. Starting from $(q_0,p_0)$, define \begin{align*} q(t)=q_0\cos t+p_0\sin t,\quad p(t)=-q_0\sin t+p_0\cos t. \end{align*} Then $q(0)=q_0$ and $p(0)=p_0$. Differentiating the first component gives \begin{align*} \dot q(t)=-q_0\sin t+p_0\cos t=p(t). \end{align*} Differentiating the second component gives \begin{align*} \dot p(t)=-q_0\cos t-p_0\sin t=-q(t). \end{align*} Thus the displayed formula is the time-$t$ Hamiltonian flow. Write the time-$t$ map as \begin{align*} \varphi_t(q,p)=(Q,P)=(q\cos t+p\sin t,-q\sin t+p\cos t). \end{align*} Its differential on coordinate one-forms is \begin{align*} dQ=\cos t\,dq+\sin t\,dp,\quad dP=-\sin t\,dq+\cos t\,dp. \end{align*} Using bilinearity and skew-symmetry of the wedge product, \begin{align*} dQ\wedge dP=(\cos t\,dq+\sin t\,dp)\wedge(-\sin t\,dq+\cos t\,dp). \end{align*} The terms with $dq\wedge dq$ and $dp\wedge dp$ vanish, so \begin{align*} dQ\wedge dP=(\cos^2 t)\,dq\wedge dp-(\sin^2 t)\,dp\wedge dq. \end{align*} Since $dp\wedge dq=-dq\wedge dp$, this becomes \begin{align*} dQ\wedge dP=(\cos^2 t+\sin^2 t)\,dq\wedge dp=dq\wedge dp. \end{align*} Therefore $\varphi_t^*(dq\wedge dp)=dq\wedge dp$. Equivalently, the linear rotation has determinant $\cos^2 t+\sin^2 t=1$, so this explicit flow calculation is the two-dimensional instance of the *Hamiltonian Flow Preservation Theorem*. [/example] Hamiltonian preservation gives a structural invariant, but phase portraits also depend on the geometry of the level sets of $H$. The pendulum is the standard example where regular energy levels and a singular separatrix coexist. [example: Pendulum Away From The Separatrix] On $T^*S^1$ with angle $q$ and momentum $p$, take \begin{align*} H(q,p)=\frac12p^2+1-\cos q. \end{align*} The partial derivatives are \begin{align*} \frac{\partial H}{\partial p}=p. \end{align*} Since $\frac{d}{dq}(-\cos q)=\sin q$, we also have \begin{align*} \frac{\partial H}{\partial q}=\sin q. \end{align*} The standard-coordinate formula for Hamiltonian vector fields gives \begin{align*} X_H=\frac{\partial H}{\partial p}\partial_q-\frac{\partial H}{\partial q}\partial_p=p\partial_q-\sin q\,\partial_p. \end{align*} Thus an integral curve satisfies \begin{align*} \dot q=p,\quad \dot p=-\sin q. \end{align*} The critical points of $H$ occur where $dH=0$, meaning $p=0$ and $\sin q=0$. Hence $q=0$ or $q=\pi$ modulo $2\pi$. Their energies are \begin{align*} H(0,0)=\frac12\cdot 0^2+1-\cos 0=0 \end{align*} and \begin{align*} H(\pi,0)=\frac12\cdot 0^2+1-\cos \pi=2. \end{align*} So every level $H=E$ with $0<E<2$ or $E>2$ is regular. On the level $H=E$, the equation is \begin{align*} \frac12p^2+1-\cos q=E. \end{align*} Solving for $p^2$ gives \begin{align*} p^2=2(E-1+\cos q). \end{align*} If $0<E<2$, then points on the level must satisfy \begin{align*} E-1+\cos q\ge 0. \end{align*} Equivalently, \begin{align*} \cos q\ge 1-E. \end{align*} Since $1-E>-1$, the allowed angles form a proper interval around $q=0$ modulo $2\pi$. At the endpoints of this interval, $p=0$; between them, the two branches \begin{align*} p=\sqrt{2(E-1+\cos q)} \end{align*} and \begin{align*} p=-\sqrt{2(E-1+\cos q)} \end{align*} join at the turning points. These two branches form a closed oval, so the motion oscillates around the stable equilibrium $(0,0)$. If $E>2$, then for every $q$, \begin{align*} E-1+\cos q\ge E-2>0. \end{align*} Thus $p$ is never zero on the level set. The two branches $p>0$ and $p<0$ are smooth graphs over the whole circle $S^1$, so each branch wraps once around the cylinder $T^*S^1$. Since $\dot q=p$, the branch with $p>0$ rotates in the increasing $q$ direction, while the branch with $p<0$ rotates in the decreasing $q$ direction. At the energy $E=2$, the level equation becomes \begin{align*} p^2=2(1+\cos q). \end{align*} Using $1+\cos q=2\cos^2(q/2)$, this is \begin{align*} p^2=4\cos^2(q/2). \end{align*} This level passes through $(\pi,0)$, the unstable equilibrium. It is the separatrix: below it the regular trajectories are oscillating closed ovals, while above it the regular trajectories are rotating curves wrapping around the cylinder. [/example] ## Poisson Brackets And First Integrals The Hamiltonian vector field turns a function into an infinitesimal symmetry. The next problem is to measure how one function changes along the Hamiltonian flow of another. The resulting operation is the Poisson bracket, which records both dynamics and a [Lie algebra](/page/Lie%20Algebra) structure on functions modulo constants. [definition: Poisson Bracket] Let $(M,\omega)$ be a symplectic manifold. The Poisson bracket is the map \begin{align*} \{\cdot,\cdot\}:C^\infty(M)\times C^\infty(M)\longrightarrow C^\infty(M) \end{align*} defined, for $F,H\in C^\infty(M)$, by \begin{align*} \{F,H\}:=dF(X_H)=\omega(X_F,X_H). \end{align*} [/definition] The second equality follows from the defining relation $dF=\omega(X_F,\cdot)$ evaluated on $X_H$. With the convention used here, reversing the two vector fields would introduce a minus sign. To compute brackets in examples, the abstract definition must be translated into canonical coordinates. [example: Coordinate Formula For The Poisson Bracket] On $\mathbb R^{2n}$ with $\omega_0=\sum_{i=1}^n dq_i\wedge dp_i$, the standard-coordinate Hamiltonian vector field of $H$ is \begin{align*} X_H=\sum_{k=1}^n\frac{\partial H}{\partial p_k}\partial_{q_k}-\sum_{k=1}^n\frac{\partial H}{\partial q_k}\partial_{p_k}. \end{align*} Since \begin{align*} dF=\sum_{k=1}^n\frac{\partial F}{\partial q_k}dq_k+\sum_{k=1}^n\frac{\partial F}{\partial p_k}dp_k, \end{align*} the definition $\{F,H\}=dF(X_H)$ gives \begin{align*} \{F,H\}=\sum_{k=1}^n\frac{\partial F}{\partial q_k}dq_k(X_H)+\sum_{k=1}^n\frac{\partial F}{\partial p_k}dp_k(X_H). \end{align*} Now $dq_k(X_H)=\partial H/\partial p_k$ and $dp_k(X_H)=-\partial H/\partial q_k$, so \begin{align*} \{F,H\}=\sum_{k=1}^n\left(\frac{\partial F}{\partial q_k}\frac{\partial H}{\partial p_k}-\frac{\partial F}{\partial p_k}\frac{\partial H}{\partial q_k}\right). \end{align*} For the coordinate functions, take $F=q_i$ and $H=p_j$. Then \begin{align*} \frac{\partial q_i}{\partial q_k}=\delta_{ik},\quad \frac{\partial q_i}{\partial p_k}=0,\quad \frac{\partial p_j}{\partial p_k}=\delta_{jk},\quad \frac{\partial p_j}{\partial q_k}=0. \end{align*} Substitution into the coordinate formula gives \begin{align*} \{q_i,p_j\}=\sum_{k=1}^n(\delta_{ik}\delta_{jk}-0)=\delta_{ij}. \end{align*} Similarly, \begin{align*} \{q_i,q_j\}=\sum_{k=1}^n(\delta_{ik}\cdot 0-0\cdot \delta_{jk})=0 \end{align*} and \begin{align*} \{p_i,p_j\}=\sum_{k=1}^n(0\cdot \delta_{jk}-\delta_{ik}\cdot 0)=0. \end{align*} Thus the coordinate functions pair nontrivially only across matching $q_i$- and $p_i$-directions, exactly reflecting the form $\omega_0=\sum_i dq_i\wedge dp_i$. [/example] The coordinate formula makes the bracket computable, but the bracket is not merely a calculation device. The example only verifies the basic coordinate relations in one model case; it does not show that the same operation has stable algebraic meaning on an arbitrary symplectic manifold. For observables to be manipulated as infinitesimal symmetries, the bracket must behave like a Lie bracket and like differentiation in each argument where products appear. The main obstruction is the Jacobi identity: nested Hamiltonian evolutions must close coherently instead of producing an operation outside the same class. The next result records exactly the algebraic structure needed for that interpretation. [quotetheorem:1333] [citeproof:1333] The theorem says that $C^\infty(M)$ is a Lie algebra under $\{\cdot,\cdot\}$, with constants in the centre. Each listed property has a separate role. Bilinearity makes the bracket an algebraic operation on observables rather than a nonlinear rule depending on representatives. Skew-symmetry is forced by the alternating form $\omega$ and gives $\{H,H\}=0$, which is the infinitesimal form of energy conservation. The Leibniz rule says that, for fixed $H$, the operator $F\mapsto\{F,H\}=X_HF$ is a derivation, so the bracket differentiates products as a vector field should. The Jacobi identity is the property that uses the full symplectic hypothesis, not only nondegeneracy. Closedness enters when Cartan's formula gives $\mathcal L_{X_F}\omega=0$; without $d\omega=0$, Hamiltonian vector fields need not preserve the two-form and the commutator calculation can acquire an extra term involving $d\omega$. A concrete failure occurs on $\mathbb R^4$ with coordinates $(x,y,z,w)$ and the nondegenerate two-form \begin{align*} \omega=dx\wedge dy+dz\wedge dw+x\,dy\wedge dz. \end{align*} Here $d\omega=dx\wedge dy\wedge dz\neq 0$. The same rule $\{F,H\}=dF(X_H)$ gives \begin{align*} \{x,y\}=1,\quad \{z,w\}=1,\quad \{x,w\}=x, \end{align*} with the remaining coordinate brackets determined by skew-symmetry and the displayed formula. Then \begin{align*} \{x,\{y,w\}\}+\{y,\{w,x\}\}+\{w,\{x,y\}\} =0+\{y,-x\}+0=1, \end{align*} so Jacobi fails. In the symplectic setting, nondegeneracy supplies the Hamiltonian vector fields and closedness is the extra condition that forces the Jacobi identity. Since constants have zero Hamiltonian vector field, the next step is to remove them and state the precise relationship between functions and Hamiltonian vector fields. [quotetheorem:10054] [citeproof:10054] The connectedness hypothesis is exactly what turns $dH=0$ into $H$ being constant. On a disconnected symplectic manifold, a function that is constant on each [connected component](/page/Connected%20Component) has zero Hamiltonian vector field but need not represent a single global constant, so quotienting only by $\mathbb R$ would not give injectivity. The quotient by constants is also necessary: before quotienting, the map $H\mapsto X_H$ has kernel containing every constant function, so injectivity cannot hold even on a connected manifold. The bracket identity depends on the sign convention fixed at the start of the chapter. With $\omega(X_H,\cdot)=dH$ and $\{F,H\}=dF(X_H)$, the compatibility with vector field commutators is \begin{align*} [X_F,X_H]=-X_{\{F,H\}}. \end{align*} If the convention were changed to $\omega(X_H,\cdot)=-dH$, or if the Poisson bracket were defined with the vector fields in the opposite order, the sign in this formula would change. Thus the theorem is not a convention-free statement about all bracket normalisations; it is the precise compatibility statement for the convention used in these notes. The theorem also does not say that every symplectic vector field is Hamiltonian; it identifies only the image of the map $H\mapsto X_H$, and global topology can prevent a symplectic vector field from having a global Hamiltonian function. For dynamics, the corresponding question is which functions remain constant under a chosen Hamiltonian flow. This leads to first integrals and the bracket criterion for conservation. [definition: First Integral] Let $(M,\omega)$ be a symplectic manifold and let $H\in C^\infty(M)$. A function $F\in C^\infty(M)$ is a first integral of the Hamiltonian system generated by $H$ if \begin{align*} F\circ\varphi_t=F \end{align*} wherever the Hamiltonian flow $(\varphi_t)$ of $H$ is defined. [/definition] The definition is global along the flow, while the Poisson bracket tests the corresponding infinitesimal statement. This is the simplest Noether-type conservation criterion in the course: symmetry of $F$ under the $H$-flow is equivalent to vanishing bracket with $H$. [quotetheorem:10055] [citeproof:10055] This criterion includes [conservation of energy](/theorems/1335) by taking $F=H$, since skew-symmetry gives $\{H,H\}=0$. The hypotheses are global over the region where the flow is being considered, not pointwise along a favourite solution. On $(\mathbb R^2,dq\wedge dp)$ with $H=p$, the Hamiltonian vector field is $\partial_q$. For $F(q,p)=pq$, the trajectory with $p=0$ has $F=0$ for all time, but $\{F,H\}=p$ is not the zero function on $\mathbb R^2$. A pointwise bracket check is also insufficient: for $F(q,p)=q^2$, the bracket $\{F,H\}=2q$ vanishes at points with $q=0$, yet the trajectory $q(t)=t$ starting from $(0,p)$ has $F(t,p)=t^2$ and is not conserved. Conversely, if $\{F,H\}=0$ on an open invariant region, then $F$ is conserved for trajectories staying in that region, but the statement gives no information elsewhere. For instance, choosing a smooth function $\rho:\mathbb R\to\mathbb R$ with $\rho(p)=0$ for $|p|<1$ and $\rho(p)=1$ for $|p|>2$, the function $F(q,p)=\rho(p)q$ is conserved on the invariant strip $|p|<1$ for the flow of $H=p$, while $\{F,H\}=\rho(p)$ is nonzero outside that strip. The criterion therefore separates quantities determined by the Hamiltonian from independent conserved quantities, which is the entry point to integrable systems. [example: Angular Momentum For A Central Potential] On $T^*\mathbb R^2$ with coordinates $(q_1,q_2,p_1,p_2)$, let \begin{align*} H(q,p)=\frac12(p_1^2+p_2^2)+V(q_1^2+q_2^2), \quad L=q_1p_2-q_2p_1. \end{align*} Write $r^2=q_1^2+q_2^2$. The partial derivatives of $L$ are \begin{align*} \frac{\partial L}{\partial q_1}=p_2, \quad \frac{\partial L}{\partial q_2}=-p_1, \quad \frac{\partial L}{\partial p_1}=-q_2, \quad \frac{\partial L}{\partial p_2}=q_1. \end{align*} For $H$, the kinetic term gives \begin{align*} \frac{\partial H}{\partial p_1}=p_1, \quad \frac{\partial H}{\partial p_2}=p_2. \end{align*} By the chain rule applied to $V(r^2)$, \begin{align*} \frac{\partial H}{\partial q_1}=2q_1V'(r^2), \quad \frac{\partial H}{\partial q_2}=2q_2V'(r^2). \end{align*} Using the coordinate formula for the Poisson bracket, \begin{align*} \{L,H\}=\frac{\partial L}{\partial q_1}\frac{\partial H}{\partial p_1}-\frac{\partial L}{\partial p_1}\frac{\partial H}{\partial q_1}+\frac{\partial L}{\partial q_2}\frac{\partial H}{\partial p_2}-\frac{\partial L}{\partial p_2}\frac{\partial H}{\partial q_2}. \end{align*} Substituting the four derivatives of $L$ and the four derivatives of $H$ gives \begin{align*} \{L,H\}=p_2p_1-(-q_2)(2q_1V'(r^2))+(-p_1)p_2-q_1(2q_2V'(r^2)). \end{align*} Grouping the kinetic and potential contributions, \begin{align*} \{L,H\}=p_1p_2-p_1p_2+2q_1q_2V'(r^2)-2q_1q_2V'(r^2). \end{align*} Thus \begin{align*} \{L,H\}=0. \end{align*} By the *[Noether Type Conservation Criterion](/theorems/10055)*, $L$ is conserved along the Hamiltonian flow of $H$. This is the Hamiltonian form of angular momentum conservation for a rotationally symmetric system. [/example] Poisson brackets therefore unify three viewpoints. They describe rates of change along Hamiltonian flows, encode the Lie algebra of Hamiltonian vector fields, and give an efficient test for conserved quantities. Hamiltonian flows give a distinguished class of symplectomorphisms, but they are only part of the story of time-dependent symmetries. The next chapter studies isotopies more generally and measures when a symplectic isotopy can actually be realized by Hamiltonian data. # 6. Symplectic and Hamiltonian Isotopies This chapter studies symmetries of a symplectic manifold that vary with time. Chapter 2 defined symplectomorphisms as maps preserving the form, and Chapter 5 showed that Hamiltonian flows give examples; here we ask more generally how such maps arise as time-dependent flows. The prerequisites are the language of smooth flows, differential forms, Cartan's formula for Lie derivatives, and de Rham cohomology. The main distinction is between symplectic motion, generated by closed $1$-forms, and Hamiltonian motion, generated by exact $1$-forms. This distinction is invisible locally but is measured globally by cohomology through the flux construction. ## Time-Dependent Vector Fields and Symplectic Isotopies A single vector field gives an autonomous flow, but Hamiltonian mechanics and many deformation arguments use vector fields depending on time. The first question is how to encode a path of diffeomorphisms by its instantaneous velocity. [definition: Time-Dependent Vector Field] Let $M$ be a smooth manifold. A time-dependent vector field on $M$ is a smooth map \begin{align*} X &: [0,1] \times M \to TM \end{align*} such that $X_t(p) := X(t,p)$ belongs to $T_pM$ for every $t \in [0,1]$ and $p \in M$. [/definition] Given $X_t$, an integral curve is a path $\gamma:[0,1]\to M$ satisfying $\dot\gamma(t)=X_t(\gamma(t))$. When the time-dependent flow exists for all $t\in[0,1]$, it gives a path $\varphi_t$ of diffeomorphisms with $\varphi_0=\operatorname{id}_M$ and \begin{align*} \frac{d}{dt}\varphi_t(p)=X_t(\varphi_t(p)). \end{align*} This motivates isolating the paths of diffeomorphisms that begin at the identity, since these are the deformations generated by time-dependent vector fields. The resulting notion is the ambient smooth category in which symplectic and Hamiltonian paths will later be special cases. [definition: Isotopy] An isotopy of a smooth manifold $M$ is a smooth map \begin{align*} \varphi &: [0,1]\times M \to M \end{align*} such that, writing $\varphi_t(p):=\varphi(t,p)$, each map $\varphi_t:M\to M$ is a diffeomorphism and $\varphi_0=\operatorname{id}_M$. [/definition] An isotopy records positions, while a vector field records velocities. To compare a diffeomorphism path with a symplectic or Hamiltonian equation, we need the vector field recovered from the path itself. The correct velocity is measured at the current point of the moving manifold, so the inverse map $\varphi_t^{-1}$ enters the formula. [definition: Generating Vector Field of an Isotopy] Let $(\varphi_t)_{t\in[0,1]}$ be an isotopy of $M$. Its generating time-dependent vector field is \begin{align*} X_t = \frac{d\varphi_t}{dt}\circ \varphi_t^{-1}. \end{align*} [/definition] This convention places $X_t$ on the target point $\varphi_t(p)$, so the differential equation for the isotopy is $\dot\varphi_t=X_t\circ\varphi_t$. The next refinement asks when the moving diffeomorphisms preserve the symplectic structure at every instant, rather than only preserving smooth structure. [definition: Symplectic Isotopy] Let $(M,\omega)$ be a symplectic manifold. A symplectic isotopy is a smooth map \begin{align*} \varphi &: [0,1]\times M \to M \end{align*} such that, writing $\varphi_t(p):=\varphi(t,p)$, the family $(\varphi_t)_{t\in[0,1]}$ is an isotopy and $\varphi_t^*\omega=\omega$ for every $t\in[0,1]$. [/definition] The definition is global in time, but computations and constructions require an infinitesimal test. Differentiating $\varphi_t^*\omega$ should turn preservation of the two-form into a condition on the generating vector field. Since $d\omega=0$, Cartan's formula reduces that condition to closedness of a $1$-form. [quotetheorem:10056] [citeproof:10056] This criterion turns the geometric condition of preserving a two-form into a cohomological condition on a family of one-forms. The hypothesis $d\omega=0$ is essential: for a nondegenerate two-form that is not closed, Cartan's formula contains the extra term $\iota_{X_t}d\omega$, so closedness of $\iota_{X_t}\omega$ no longer controls preservation of the form. The isotopy hypothesis is also doing work, because the theorem tests a path through its instantaneous velocity and does not characterize an isolated diffeomorphism with no chosen path from the identity. Thus the result explains local Hamiltonian behaviour on contractible coordinate charts while pointing toward the global obstruction measured by cohomology. [example: Rotations of the Symplectic Sphere] For the usual rotation-invariant area form on the unit sphere, use coordinates $(\theta,z)$ away from the poles, where \begin{align*} (x,y,z)=\left(\sqrt{1-z^2}\cos\theta,\sqrt{1-z^2}\sin\theta,z\right). \end{align*} With the outward orientation, the area form is \begin{align*} \omega=d\theta\wedge dz. \end{align*} The rotation through angle $t$ is $\varphi_t(\theta,z)=(\theta+t,z)$, so $\varphi_0=\operatorname{id}$ and \begin{align*} \frac{d}{dt}\varphi_t(\theta,z)=\partial_\theta\big|_{(\theta+t,z)}. \end{align*} Thus the generating vector field is $X=\partial_\theta$. Now compute the contraction with the area form. Since $d\theta(\partial_\theta)=1$ and $dz(\partial_\theta)=0$, \begin{align*} \iota_{\partial_\theta}(d\theta\wedge dz)=d\theta(\partial_\theta)\,dz-dz(\partial_\theta)\,d\theta=dz. \end{align*} Hence $\iota_X\omega=dz$, so \begin{align*} d(\iota_X\omega)=d(dz)=0. \end{align*} Therefore the rotation flow is symplectic by *Symplectic Vector Field Criterion*. Moreover, \begin{align*} \iota_X\omega=dz=dH \end{align*} with $H=z+C$ for any constant $C$, so the same isotopy is Hamiltonian by *[Hamiltonian Criterion by Exactness](/theorems/10057)*. Choosing $C$ so that the mean value of $H$ over $S^2$ is zero gives the normalized Hamiltonian $H=z$, up to the sign convention for the rotation direction. [/example] The sphere example is cohomologically simple. The next section studies what survives when $H^1(M;\mathbb R)$ is nonzero. ## Flux and the Difference Between Symplectic and Hamiltonian Motion A symplectic isotopy supplies a whole path of closed $1$-forms $\iota_{X_t}\omega$. The natural invariant is obtained by adding their cohomology classes over time, and it records the net cohomological displacement of the isotopy. [definition: Flux of a Symplectic Isotopy] Let $(M,\omega)$ be a symplectic manifold and let $(\varphi_t)_{t\in[0,1]}$ be a symplectic isotopy generated by $X_t$. The flux of $(\varphi_t)$ is the de Rham cohomology class \begin{align*} \operatorname{Flux}(\varphi_t)=\left[\int_0^1 \iota_{X_t}\omega\,dt\right]\in H^1(M;\mathbb R). \end{align*} [/definition] The integral is taken as a smooth $1$-form: on a tangent vector $v\in T_pM$, it evaluates to \begin{align*} \int_0^1\omega_p(X_t(p),v)\,dt. \end{align*} Closedness at each time makes the time integral closed, so the cohomology class is defined. To identify which symplectic isotopies come from Hamiltonian mechanics, we now compare this closed form condition with exactness at each time. [quotetheorem:10057] [citeproof:10057] Exactness at every time implies vanishing flux, but this criterion deliberately says nothing about whether two Hamiltonian paths have the same endpoint or whether a symplectic path can be deformed into a Hamiltonian one. The smooth family hypothesis is necessary: choosing primitives for each closed form separately does not produce a Hamiltonian isotopy unless those primitives vary smoothly in time. A basic limitation is visible on $T^2$: the constant form $a\,dp-b\,dq$ is closed but not exact when $(a,b)\ne(0,0)$, so the associated translation is symplectic but not Hamiltonian. The preceding criterion is pointwise in time, while flux is invariant under homotopy of the full path, which creates the need for a global statement on the universal cover of the symplectomorphism group. [quotetheorem:10058] [proofunderconstruction:10058] The theorem is stated here because it is the organizing principle for the distinction between symplectic and Hamiltonian isotopies. Closedness of $M$ keeps the standard flux theory in its cleanest form: on noncompact manifolds, support conditions must be added or the time-integrated form may not capture the intended compactly supported Hamiltonian group. Passing to the universal cover is also essential, since flux depends on the homotopy class of the path with fixed endpoints rather than only on the endpoint symplectomorphism; different paths to the same endpoint can differ by a loop. Thus the theorem does not yet give a complete invariant on $\operatorname{Symp}_0(M,\omega)$ itself, and its deeper global consequences depend on the flux group, which lies beyond this introductory chapter. [example: Translation on the Symplectic Torus] Let $T^2=\mathbb R^2/\mathbb Z^2$ with coordinates $q,p$ modulo $1$ and symplectic form $\omega=dq\wedge dp$. For fixed [real numbers](/page/Real%20Numbers) $a,b$, the translation isotopy is \begin{align*} \varphi_t(q,p)=(q+ta,p+tb). \end{align*} Its velocity at the point $\varphi_t(q,p)$ is \begin{align*} \frac{d}{dt}\varphi_t(q,p)=a\partial_q+b\partial_p. \end{align*} Thus its generating vector field is the time-independent field $X=a\partial_q+b\partial_p$. For any tangent vector $Y$, contraction with a wedge product gives \begin{align*} \iota_X(dq\wedge dp)(Y)=(dq\wedge dp)(X,Y). \end{align*} Writing $Y=u\partial_q+v\partial_p$, we have $dq(X)=a$, $dp(X)=b$, $dq(Y)=u$, and $dp(Y)=v$, so \begin{align*} (dq\wedge dp)(X,Y)=dq(X)dp(Y)-dp(X)dq(Y)=av-bu. \end{align*} The $1$-form $a\,dp-b\,dq$ evaluates on the same vector $Y$ as \begin{align*} (a\,dp-b\,dq)(Y)=a v-bu. \end{align*} Therefore \begin{align*} \iota_X\omega=a\,dp-b\,dq. \end{align*} Since $a$ and $b$ are constants, \begin{align*} d(\iota_X\omega)=d(a\,dp-b\,dq)=a\,d(dp)-b\,d(dq)=0. \end{align*} So the translation isotopy is symplectic by *Symplectic Vector Field Criterion*. Because $X$ is independent of $t$, its flux is \begin{align*} \operatorname{Flux}(\varphi_t)=\left[\int_0^1(a\,dp-b\,dq)\,dt\right]=[a\,dp-b\,dq]. \end{align*} To see when this class vanishes, integrate over the coordinate circles $\gamma_q(s)=(s,0)$ and $\gamma_p(s)=(0,s)$ for $0\le s\le 1$. Along $\gamma_q$, $dq=ds$ and $dp=0$, hence \begin{align*} \int_{\gamma_q}(a\,dp-b\,dq)=\int_0^1(-b)\,ds=-b. \end{align*} Along $\gamma_p$, $dq=0$ and $dp=ds$, hence \begin{align*} \int_{\gamma_p}(a\,dp-b\,dq)=\int_0^1 a\,ds=a. \end{align*} An exact $1$-form has integral $0$ around every closed loop, so $[a\,dp-b\,dq]\ne 0$ whenever $(a,b)\ne(0,0)$. Thus nontrivial translations of the symplectic torus are symplectic isotopies with nonzero flux, and they are not Hamiltonian by *Hamiltonian Criterion by Exactness*. [/example] The torus example is the basic warning that preserving $\omega$ is not the same as being generated by an energy function. Hamiltonian motion is a subgroup-level condition, not merely a pointwise preservation condition. ## Compactly Supported Hamiltonians and the Hamiltonian Group On noncompact manifolds, flows may fail to exist for all time unless the vector field has controlled support. Compactly supported Hamiltonians provide the standard class of globally defined Hamiltonian isotopies. [definition: Compactly Supported Time-Dependent Hamiltonian] Let $(M,\omega)$ be a symplectic manifold. A compactly supported time-dependent Hamiltonian is a smooth function \begin{align*} H:[0,1]\times M\to\mathbb R \end{align*} such that there is a compact set $K\subset M$ with $\operatorname{supp}(H_t)\subset K$ for every $t\in[0,1]$. [/definition] The uniform compact support condition keeps the Hamiltonian vector fields inside a fixed compact region. Outside that region the vector field vanishes, so the flow is the identity there. This support convention motivates the next definition of Hamiltonian isotopy on noncompact manifolds, because it packages exactness and global flow control into one notion. [definition: Hamiltonian Isotopy] Let $(M,\omega)$ be a symplectic manifold. A Hamiltonian isotopy is an isotopy $(\varphi_t)_{t\in[0,1]}$ generated by a time-dependent vector field $X_t$ for which there exists a smooth family $H_t:M\to\mathbb R$ satisfying \begin{align*} \iota_{X_t}\omega=dH_t. \end{align*} If $M$ is noncompact, the Hamiltonians are required to be compactly supported unless another support condition is stated. [/definition] Hamiltonian isotopies are symplectic isotopies by the closedness criterion, since [exact forms are closed](/theorems/3565). Their endpoints are the actual symmetries usually used in geometry, because composing endpoint maps corresponds to performing motions successively. This motivates defining a Hamiltonian group from time-one maps rather than from the entire space of paths. [definition: Hamiltonian Group] Let $(M,\omega)$ be a symplectic manifold. The Hamiltonian group is the subset \begin{align*} \operatorname{Ham}(M,\omega) := \{\varphi_1\in \operatorname{Diff}(M): (\varphi_t)_{t\in[0,1]} \text{ is a Hamiltonian isotopy with } \varphi_0=\operatorname{id}_M\}. \end{align*} [/definition] This definition is endpoint-based: different Hamiltonians can generate the same element of $\operatorname{Ham}(M,\omega)$. That creates a real well-definedness problem for the word "group": composing two endpoint maps requires a new Hamiltonian path whose time-one map is the composition, and taking an inverse requires reversing a Hamiltonian motion without leaving the allowed support class. Until these closure operations are checked, the endpoint set is only a collection of diffeomorphisms, not a transformation group. [quotetheorem:10059] [citeproof:10059] The formulas in the proof are often more useful than the abstract subgroup statement. The support hypothesis in the noncompact case is part of the statement: without it, a Hamiltonian vector field may fail to have a globally defined time-one flow, and the endpoint set would not be a well-defined group of diffeomorphisms. The theorem also does not say that every symplectomorphism isotopic to the identity is Hamiltonian; the torus translations above give symplectic endpoints in the identity component that are excluded by nonzero flux. What the result provides is closure of the Hamiltonian endpoints under the operations needed for a transformation group, preparing for later constructions such as moment maps and symplectic reduction. [example: Compactly Supported Hamiltonian Bump Flow] Let $(\mathbb R^{2n},\omega_0)$ have coordinates $(q_1,\dots,q_n,p_1,\dots,p_n)$ and \begin{align*} \omega_0=\sum_{i=1}^n dq_i\wedge dp_i. \end{align*} Choose $\rho\in C_c^\infty(\mathbb R^{2n})$ with $\rho=1$ on $B(0,1)$ and $\operatorname{supp}(\rho)\subset B(0,2)$, and define \begin{align*} H(q,p)=\rho(q,p)p_1. \end{align*} Write a vector field in the form \begin{align*} X=\sum_{i=1}^n A_i\partial_{q_i}+\sum_{i=1}^n B_i\partial_{p_i}. \end{align*} Since $dq_i(X)=A_i$ and $dp_i(X)=B_i$, contraction with the $i$th summand gives \begin{align*} \iota_X(dq_i\wedge dp_i)=A_i\,dp_i-B_i\,dq_i. \end{align*} Therefore \begin{align*} \iota_X\omega_0=\sum_{i=1}^n A_i\,dp_i-\sum_{i=1}^n B_i\,dq_i. \end{align*} The differential of $H=\rho p_1$ is \begin{align*} dH=\rho\,dp_1+p_1\,d\rho. \end{align*} Expanding $d\rho$ in the coordinates gives \begin{align*} dH=\rho\,dp_1+p_1\sum_{i=1}^n \frac{\partial\rho}{\partial q_i}\,dq_i+p_1\sum_{i=1}^n \frac{\partial\rho}{\partial p_i}\,dp_i. \end{align*} The Hamiltonian convention in this chapter is $\iota_{X_H}\omega_0=dH$, so matching the $dq_i$ and $dp_i$ coefficients gives \begin{align*} A_i=\frac{\partial H}{\partial p_i} \end{align*} and \begin{align*} -B_i=\frac{\partial H}{\partial q_i}. \end{align*} Thus \begin{align*} X_H=\left(\rho+p_1\frac{\partial\rho}{\partial p_1}\right)\partial_{q_1}+\sum_{i=2}^n p_1\frac{\partial\rho}{\partial p_i}\partial_{q_i}-\sum_{i=1}^n p_1\frac{\partial\rho}{\partial q_i}\partial_{p_i}. \end{align*} On $B(0,1)$, the function $\rho$ is identically $1$, so every first derivative of $\rho$ is $0$ there. Substituting these values into the formula for $X_H$ gives \begin{align*} X_H=\partial_{q_1} \end{align*} on $B(0,1)$. Outside $B(0,2)$, the function $\rho$ is identically $0$ on a neighborhood, so its first derivatives are also $0$ there, and the same formula gives \begin{align*} X_H=0. \end{align*} Hence the flow has the desired local translation field in the inner ball, is the identity outside the compact support region, and changes only in the annular transition region where derivatives of $\rho$ may be nonzero. [/example] This example captures a recurring technique: build local Hamiltonian motion by writing down the desired linear Hamiltonian, then multiply by a bump function to keep the support compact. The transition region changes the vector field, but the resulting flow remains Hamiltonian because it is still generated by a global compactly supported function. [remark: Normalization on Closed Manifolds] On a closed connected symplectic manifold, adding a time-dependent constant $c(t)$ to $H_t$ does not change $dH_t$ and hence does not change the Hamiltonian vector field. A common normalization is \begin{align*} \int_M H_t\,\omega^n=0 \end{align*} for every $t$, or equivalently zero mean with respect to the symplectic volume form after including the conventional factor $1/n!$. [/remark] The chapter's main lesson is that symplectic isotopies are governed infinitesimally by closed $1$-forms, while Hamiltonian isotopies are governed by exact $1$-forms. Flux packages the cohomological obstruction to being Hamiltonian and prepares the way for the later study of moment maps, reduction, and global symplectic rigidity. Once Hamiltonian isotopies are understood, the next natural question is what happens on hypersurfaces cut out by energy constraints. Such level sets often inherit characteristic directions, and when the geometry is favorable these become contact-type structures with Reeb dynamics. # 7. Contact-Type Hypersurfaces and Energy Levels Symplectic geometry turns Hamiltonian mechanics into geometry by replacing trajectories in phase space with intrinsic objects determined by $\omega$. The previous chapters developed Hamiltonian vector fields and symplectic invariance; this chapter asks what remains when the Hamiltonian is restricted to a single regular energy level. The central point is that a hypersurface in a symplectic manifold carries a canonical one-dimensional direction field, and under a contact-type hypothesis that direction field becomes Reeb dynamics. ## Regular Energy Hypersurfaces and Characteristic Line Fields Suppose a Hamiltonian $H:M\to \mathbb R$ is conserved along its own flow. The physical motion at fixed energy $c$ is constrained to $\Sigma=H^{-1}(c)$, so the first geometric question is: what structure does the ambient symplectic form induce on this odd-dimensional hypersurface? [definition: Regular Energy Hypersurface] Let $(M^{2n},\omega)$ be a symplectic manifold and let $H:M\to\mathbb R$ be smooth. A level set $\Sigma=H^{-1}(c)$ is a regular energy hypersurface if $c$ is a regular value of $H$. [/definition] Regularity means that $\Sigma$ is a smooth submanifold of codimension $1$. Since $\dim \Sigma=2n-1$, the restricted two-form $\omega|_{\Sigma}$ cannot be nondegenerate. The useful structure is therefore not a symplectic form on $\Sigma$, but the direction in which the restricted form vanishes. This degeneracy is not a defect to be ignored: it is the only intrinsic direction on the energy surface selected by the ambient symplectic form. To compare constrained Hamiltonian motion with geometry on $\Sigma$, we need a pointwise name for the vectors tangent to $\Sigma$ whose $\omega$-pairing with every other tangent vector to $\Sigma$ is zero. [definition: Characteristic Line Field] Let $\Sigma\subset (M^{2n},\omega)$ be a smooth hypersurface. The characteristic line field of $\Sigma$ is the rank-one distribution \begin{align*} \mathcal L_{\Sigma,x}:=\ker(\omega|_{T_x\Sigma})=\{v\in T_x\Sigma: \omega_x(v,w)=0\text{ for all }w\in T_x\Sigma\}. \end{align*} [/definition] The definition is pointwise linear algebra, but it is not yet automatic from the notation that these kernels assemble smoothly or have dimension one everywhere. A pointwise kernel could, for a general restricted form, jump in rank and fail to define a smooth distribution. For hypersurfaces in a symplectic manifold, the codimension-one condition forces exactly one null direction, which is the regularity needed before speaking about characteristic curves. [quotetheorem:10060] [citeproof:10060] This theorem explains why hypersurfaces are the first odd-dimensional spaces encountered inside symplectic geometry: they inherit one direction rather than a symplectic form. The hypersurface hypothesis is essential here. For a submanifold of higher codimension, the kernel of the restricted two-form need not be one-dimensional and may even change rank from point to point; for an arbitrary odd-dimensional submanifold, being odd-dimensional only forces some degeneracy, not a canonical line field. Thus the theorem is not a general statement about all odd-dimensional submanifolds, and it does not provide a preferred parametrisation of the characteristic curves. For regular level sets, we also have a distinguished vector field $X_H$, so the next question is whether the characteristic direction is the actual Hamiltonian direction. [quotetheorem:10061] [citeproof:10061] The theorem says that on a regular energy shell the Hamiltonian vector field is determined up to reparametrisation by the hypersurface itself. Regularity is not a cosmetic assumption: if $c$ is a critical value, $H^{-1}(c)$ may fail to be a smooth hypersurface and $X_H$ may vanish at critical points. For example, for $H(q,p)=\frac{1}{2}(q^2+p^2)$ on $\mathbb R^2$, the level $H^{-1}(0)$ is the single point $(0,0)$ and $X_H(0,0)=0$, so there is no characteristic line field to span. Different Hamiltonians defining the same regular hypersurface give the same characteristic leaves, although the speed along the leaves may change. [example: Harmonic Oscillator Energy Spheres] On $(\mathbb R^{2n},\omega_0=\sum_{i=1}^n dq_i\wedge dp_i)$, let \begin{align*} H(q,p)=\frac{1}{2}\sum_{i=1}^n(q_i^2+p_i^2). \end{align*} For $c>0$, the equation $H(q,p)=c$ is equivalent to \begin{align*} \sum_{i=1}^n(q_i^2+p_i^2)=2c, \end{align*} so $H^{-1}(c)$ is the sphere $S^{2n-1}_{\sqrt{2c}}$. Write \begin{align*} X_H=\sum_{i=1}^n\left(a_i\frac{\partial}{\partial q_i}+b_i\frac{\partial}{\partial p_i}\right). \end{align*} Since $\omega_0(X_H,\cdot)=dH$, we compare the two one-forms \begin{align*} \omega_0(X_H,\cdot)=\sum_{i=1}^n(a_i\,dp_i-b_i\,dq_i) \end{align*} and \begin{align*} dH=\sum_{i=1}^n(q_i\,dq_i+p_i\,dp_i). \end{align*} Matching the coefficients of $dq_i$ and $dp_i$ gives $-b_i=q_i$ and $a_i=p_i$, hence \begin{align*} X_H=\sum_{i=1}^n\left(p_i\frac{\partial}{\partial q_i}-q_i\frac{\partial}{\partial p_i}\right). \end{align*} The flow equations are therefore \begin{align*} \dot q_i=p_i,\qquad \dot p_i=-q_i. \end{align*} Differentiating the first equation gives $\ddot q_i=\dot p_i=-q_i$, and the initial data $q_i(0),p_i(0)$ determine \begin{align*} q_i(t)=q_i(0)\cos t+p_i(0)\sin t. \end{align*} Then $\dot q_i(t)=-q_i(0)\sin t+p_i(0)\cos t$, so \begin{align*} p_i(t)=p_i(0)\cos t-q_i(0)\sin t. \end{align*} Equivalently, with $z_i=q_i+ip_i$, this is $z_i(t)=e^{-it}z_i(0)$ for every $i$. Thus the same circle action rotates all complex coordinates with the same angular speed, and by *Hamiltonian Vector Field Spans The Characteristic Line* these circles are exactly the characteristic leaves on $S^{2n-1}_{\sqrt{2c}}$. The energy value $c$ changes only the radius $\sqrt{2c}$, not the Hopf-circle pattern of the foliation. [/example] ## Liouville Vector Fields and Contact-Type Hypersurfaces The characteristic line field gives unparametrised dynamics, but it does not yet choose a preferred one-form or a preferred time scale. The next question is: when does the restricted symplectic geometry on a hypersurface come from a contact form? [definition: Liouville Vector Field] Let $(M,\omega)$ be a symplectic manifold and let $U\subset M$ be open. A smooth vector field $Y\in\Gamma(TU)$ is a Liouville vector field if \begin{align*} \mathcal L_Y\omega=\omega \end{align*} on $U$. [/definition] Cartan's formula turns this condition into an exactness statement. If $\lambda:=\iota_Y\omega$, then $d\lambda=\omega$ because $d\omega=0$, so a Liouville vector field supplies a local primitive for the symplectic form. This motivates the next definition: to make a contact form on a hypersurface, the Liouville field must cross the hypersurface so that its primitive has a nonzero component in the missing normal direction. [definition: Contact-Type Hypersurface] Let $(M,\omega)$ be a symplectic manifold and let $\Sigma\subset M$ be a smooth hypersurface. The hypersurface $\Sigma$ is of contact type if there is a Liouville vector field $Y$ defined near $\Sigma$ that is transverse to $\Sigma$. [/definition] Transversality is the condition that prevents the primitive from degenerating on $\Sigma$. The induced one-form is the restriction of $\iota_Y\omega$, but we still need to prove that it satisfies the contact nonvanishing condition and that it keeps the same characteristic line field. This is the content of the restriction lemma. [quotetheorem:10062] [citeproof:10062] The lemma converts an odd-dimensional hypersurface into a contact manifold, while keeping the same characteristic curves. The transversality condition is exactly what makes the characteristic direction visible to the primitive: if $Y$ were tangent to $\Sigma$, then for $v\in\ker(\omega|_{T\Sigma})$ we would have $\alpha(v)=\omega(Y,v)=0$, so $\alpha$ could not normalise that line as a Reeb direction. Contact type is also a genuine extra hypothesis, not a consequence of being a hypersurface; some hypersurfaces admit no transverse Liouville primitive in a neighbourhood. Its main effect, when it holds, is a normalisation: the Reeb vector field is not merely tangent to the characteristic line; it is the characteristic vector field with a distinguished speed. [example: Star-Shaped Hypersurfaces In Euclidean Symplectic Space] On $\mathbb R^{2n}$ with coordinates $(q,p)$ and $\omega_0=\sum_{i=1}^n dq_i\wedge dp_i$, consider \begin{align*} Y=\frac{1}{2}\sum_{i=1}^n\left(q_i\frac{\partial}{\partial q_i}+p_i\frac{\partial}{\partial p_i}\right). \end{align*} For each $i$, contraction with $dq_i\wedge dp_i$ gives \begin{align*} \iota_Y(dq_i\wedge dp_i)=dq_i(Y)\,dp_i-dp_i(Y)\,dq_i=\frac{1}{2}q_i\,dp_i-\frac{1}{2}p_i\,dq_i. \end{align*} Summing over $i$, \begin{align*} \iota_Y\omega_0=\frac{1}{2}\sum_{i=1}^n(q_i\,dp_i-p_i\,dq_i). \end{align*} Taking exterior derivatives term by term, \begin{align*} d(q_i\,dp_i-p_i\,dq_i)=dq_i\wedge dp_i-dp_i\wedge dq_i=2\,dq_i\wedge dp_i. \end{align*} Therefore \begin{align*} d(\iota_Y\omega_0)=\sum_{i=1}^n dq_i\wedge dp_i=\omega_0. \end{align*} Since $d\omega_0=0$, Cartan's formula gives \begin{align*} \mathcal L_Y\omega_0=d(\iota_Y\omega_0)+\iota_Y(d\omega_0)=\omega_0. \end{align*} Thus $Y$ is a Liouville vector field. If $\Sigma$ is smooth and star-shaped with respect to the origin, the radial direction has a nonzero normal component at every point of $\Sigma$, so $Y$ is transverse to $\Sigma$. By *[Contact-Type Restriction Lemma](/theorems/10062)*, the restriction \begin{align*} \alpha=(\iota_Y\omega_0)|_{\Sigma} \end{align*} is a contact form. Using the contraction computed above, this contact form is \begin{align*} \alpha=\frac{1}{2}\sum_{i=1}^n(q_i\,dp_i-p_i\,dq_i)\bigg|_{\Sigma}. \end{align*} For the round sphere $S^{2n-1}_r$, the radial vector field is everywhere normal to the sphere and hence transverse, so the same formula gives the standard contact form on $S^{2n-1}_r$. [/example] Contact type is not automatic for every hypersurface. It is an additional compatibility between the hypersurface and a primitive of the ambient symplectic form near it, and the transverse Liouville field is the geometric witness for that compatibility. ## Reeb Dynamics As Hamiltonian Dynamics On Restricted Energy Shells We now compare the two vector fields present on a contact-type regular energy shell: the Hamiltonian vector field $X_H$ coming from a function and the Reeb vector field $R_\alpha$ coming from the induced contact form. Since both span the same line, the only possible difference is a positive or negative scalar factor. [definition: Reeb Vector Field] Let $(\Sigma^{2n-1},\alpha)$ be a contact manifold. The Reeb vector field is the unique smooth vector field $R_\alpha\in\Gamma(T\Sigma)$ satisfying \begin{align*} \alpha(R_\alpha)&=1, & \iota_{R_\alpha}d\alpha&=0. \end{align*} [/definition] The uniqueness in this definition comes from the contact condition: $d\alpha$ is nondegenerate on $\ker\alpha$, leaving exactly one transverse direction normalised by $\alpha(R_\alpha)=1$. On a contact-type energy shell, the Hamiltonian vector field already lies in the same one-dimensional kernel of $d\alpha$, so the remaining problem is to compute the scalar imposed by the Reeb normalisation. [quotetheorem:10063] [citeproof:10063] This result is the bridge between symplectic and contact dynamics, but each hypothesis is doing work. Regularity ensures that the energy level is a hypersurface and that $X_H$ does not vanish there, while $dH(Y)\ne0$ is exactly the transversality condition needed to divide by the normal derivative. The theorem identifies the same unparametrised orbits, not the same time parametrisation unless $dH(Y)$ is constant along the relevant trajectory. If the opposite Hamiltonian convention is used, the scalar relating $R_\alpha$ and $X_H$ changes sign, while the characteristic leaves remain the same. [remark: Sign Conventions For Hamiltonian Vector Fields] Some texts define $X_H$ by $\iota_{X_H}\omega=dH$, while others use $\iota_{X_H}\omega=-dH$. The characteristic foliation is unchanged by this choice. Formulae comparing $X_H$ with $R_\alpha$ must keep the sign convention fixed throughout the course. [/remark] The most important family of contact-type energy shells in mechanics occurs in cotangent bundles. The Liouville form is already present there, so the contact form on an energy level is the restriction of the canonical one-form. [example: Unit Cotangent Bundle And Geodesic Flow] In local coordinates $(q^1,\dots,q^m,p_1,\dots,p_m)$ on $T^*Q$, the canonical one-form is \begin{align*} \lambda=\sum_{i=1}^m p_i\,dq^i. \end{align*} With the convention $\omega=-d\lambda$, we have \begin{align*} d\lambda=\sum_{i=1}^m dp_i\wedge dq^i \end{align*} and therefore \begin{align*} \omega=\sum_{i=1}^m dq^i\wedge dp_i. \end{align*} Let \begin{align*} Y=-\sum_{i=1}^m p_i\frac{\partial}{\partial p_i} \end{align*} be the inward fibre-radial vector field. For each $i$, \begin{align*} \iota_Y(dq^i\wedge dp_i)=dq^i(Y)\,dp_i-dp_i(Y)\,dq^i=0\cdot dp_i-(-p_i)\,dq^i=p_i\,dq^i. \end{align*} Summing over $i$ gives \begin{align*} \iota_Y\omega=\sum_{i=1}^m p_i\,dq^i=\lambda. \end{align*} Since $d(\iota_Y\omega)=-d\lambda=\omega$ and $d\omega=0$, this sign convention makes $Y$ the Liouville field whose induced primitive is $\lambda$. The kinetic energy Hamiltonian is \begin{align*} H(q,p)=\frac{1}{2}|p|_{g_q^{-1}}^2=\frac{1}{2}\sum_{i,j=1}^m g^{ij}(q)p_i p_j. \end{align*} Differentiating in the fibre directions gives \begin{align*} \frac{\partial H}{\partial p_k}=\frac{1}{2}\sum_{j=1}^m g^{kj}(q)p_j+\frac{1}{2}\sum_{i=1}^m g^{ik}(q)p_i=\sum_{j=1}^m g^{kj}(q)p_j, \end{align*} using the symmetry $g^{ij}=g^{ji}$. Hence \begin{align*} dH(Y)=-\sum_{k=1}^m p_k\frac{\partial H}{\partial p_k}=-\sum_{k,j=1}^m g^{kj}(q)p_kp_j=-|p|_{g_q^{-1}}^2=-2H. \end{align*} On $S^*Q=H^{-1}(1/2)$ this becomes $dH(Y)=-1$, so $S^*Q$ is a regular energy hypersurface and $Y$ is transverse to it. The induced contact form is therefore \begin{align*} \alpha=(\iota_Y\omega)|_{S^*Q}=\lambda|_{S^*Q}. \end{align*} The Hamiltonian vector field of $H$ is the co-geodesic vector field, and the reparametrisation factor is constant because $dH(Y)=-1$ on $S^*Q$. Thus the Reeb vector field of $\lambda|_{S^*Q}$ is the same vector field as the co-geodesic flow. Under the metric identification $T^*Q\cong TQ$, this flow projects to the usual geodesic flow on the unit tangent bundle. [/example] The cotangent example shows how contact-type hypersurfaces encode familiar mechanical flows without changing their orbits. The harmonic oscillator gives a finite-dimensional model where the same comparison can be computed on a round sphere and the reparametrisation factor is constant. [example: Harmonic Oscillator As A Reeb Flow] On $\mathbb R^{2n}$ with $\omega_0=\sum_{i=1}^n dq_i\wedge dp_i$ and \begin{align*} H(q,p)=\frac{1}{2}\sum_{i=1}^n(q_i^2+p_i^2), \end{align*} the positive energy level $H^{-1}(c)$ satisfies $\sum_i(q_i^2+p_i^2)=2c$, so it is the sphere $S^{2n-1}_{\sqrt{2c}}$. Let \begin{align*} Y=\frac{1}{2}\sum_{i=1}^n\left(q_i\frac{\partial}{\partial q_i}+p_i\frac{\partial}{\partial p_i}\right). \end{align*} Since \begin{align*} dH=\sum_{i=1}^n(q_i\,dq_i+p_i\,dp_i), \end{align*} we get \begin{align*} dH(Y)=\sum_{i=1}^n\left(q_i\cdot\frac{q_i}{2}+p_i\cdot\frac{p_i}{2}\right)=\frac{1}{2}\sum_{i=1}^n(q_i^2+p_i^2)=H. \end{align*} Thus on $H^{-1}(c)$, $dH(Y)=c\ne0$, so $Y$ is transverse to the energy sphere. For each $i$, \begin{align*} \iota_Y(dq_i\wedge dp_i)=dq_i(Y)\,dp_i-dp_i(Y)\,dq_i=\frac{q_i}{2}\,dp_i-\frac{p_i}{2}\,dq_i. \end{align*} Summing over $i$ gives the induced contact form \begin{align*} \alpha=(\iota_Y\omega_0)|_{H^{-1}(c)}=\frac{1}{2}\sum_{i=1}^n(q_i\,dp_i-p_i\,dq_i)\big|_{H^{-1}(c)}. \end{align*} By *Reeb Hamiltonian Reparametrisation On A Contact-Type Energy Shell*, \begin{align*} R_\alpha=-\frac{1}{dH(Y)}X_H=-\frac{1}{c}X_H \end{align*} on $H^{-1}(c)$. The scalar $-1/c$ is constant on the whole sphere, so the Reeb flow has exactly the same unparametrised closed curves as the harmonic oscillator flow. Hence the Hopf circles on $S^{2n-1}_{\sqrt{2c}}$ are simultaneously the fixed-energy Hamiltonian trajectories and the Reeb orbits of the induced contact form. [/example] The chapter's conclusion is conceptual: hypersurfaces inherit characteristic directions from symplectic geometry, while contact-type hypersurfaces upgrade those directions to Reeb dynamics. This is why contact geometry appears naturally at the boundary of symplectic geometry and on regular energy surfaces in Hamiltonian mechanics. Energy hypersurfaces show that symplectic geometry often packages dynamics into boundary behavior. From there it is natural to study symmetries that preserve the symplectic form and to ask when their infinitesimal generators admit moment maps. # 8. Lie Group Actions and Moment Maps Symmetries of a symplectic manifold should interact with the Hamiltonian formalism rather than merely preserve the underlying smooth manifold. This chapter studies Lie group actions by symplectomorphisms, asks when their infinitesimal vector fields are Hamiltonian, and packages the corresponding Hamiltonians into a moment map. The main theme is that a moment map is both a collection of conserved quantities and a bridge from group actions to symplectic reduction. ## Symplectic Actions and Infinitesimal Generators The first question is what it means for a Lie group symmetry to respect the symplectic structure. A smooth [group action](/page/Group%20Action) already gives diffeomorphisms of the manifold; the symplectic condition asks whether those diffeomorphisms preserve the closed nondegenerate $2$-form. [definition: Symplectic Group Action] Let $(M,\omega)$ be a symplectic manifold and let $G$ be a Lie group. A smooth left action $G \times M \to M$, written $(g,p) \mapsto g \cdot p$, is a symplectic action if, for every $g \in G$, the diffeomorphism \begin{align*} \Phi_g:M &\to M, & \Phi_g(p)&=g\cdot p \end{align*} is a symplectomorphism, so $\Phi_g^*\omega=\omega$. [/definition] This definition turns the global action into a family of symplectomorphisms. To test whether such a symmetry is Hamiltonian, we need the vector field obtained by differentiating the orbit of a point in a specified Lie algebra direction. [definition: Infinitesimal Generator] Let $G$ act smoothly on a manifold $M$, and let $\mathfrak g=T_eG$ be its Lie algebra. For $\xi\in\mathfrak g$, the infinitesimal generator is the vector field $\xi_M\in\mathfrak X(M)$ defined by \begin{align*} (\xi_M)_p=\frac{d}{dt}\Big|_{t=0}\exp(t\xi)\cdot p. \end{align*} [/definition] The infinitesimal generator records the first-order motion along group orbits. For a symplectic action, the next question is what preservation of $\theta$ becomes after differentiating at the identity, because that infinitesimal condition is what can be compared with Hamiltonian vector fields. [quotetheorem:10064] [citeproof:10064] The theorem uses both parts of the phrase "symplectic action". Smoothness of the action is needed to differentiate the curve $t\mapsto \Phi_{\exp(t\xi)}$, and preservation of $\theta$ is needed before taking the derivative; a smooth action that does not preserve $\theta$ need not satisfy $\mathcal L_{\xi_M}\theta=0$. The conclusion is also only infinitesimal: it says that $\iota_{\xi_M}\theta$ is closed, not that it has a global Hamiltonian primitive. A concrete failure of exactness occurs on the symplectic torus $M=\mathbb R^2/\mathbb Z^2$ with $\theta=dx\wedge dy$. The circle action by translation in the $x$-direction is symplectic, its infinitesimal generator is $\partial_x$, and $\iota_{\partial_x}\theta=dy$, which is closed but not exact on $T^2$. This example explains why the next definition is stronger than symplecticity: Hamiltonian actions require the closed forms from the theorem to be exact, and the moment maps in the next section package chosen primitives in a linear way. [definition: Hamiltonian Group Action] Let $(M,\theta)$ be a symplectic manifold with a symplectic action of a Lie group $G$. The action is Hamiltonian if for every $\xi\in\mathfrak g$ there exists a smooth function $H_\xi:M\to\mathbb R$ such that \begin{align*} dH_\xi=\iota_{\xi_M}\theta, \end{align*} and the assignment \begin{align*} \mathfrak g&\to C^\infty(M), & \xi&\mapsto H_\xi \end{align*} is linear. [/definition] The sign convention here fixes $\xi_M$ as the Hamiltonian vector field of $H_\xi$ under the convention $dH=\iota_{X_H}\theta$. Other texts use the opposite sign, so formulas involving moment maps should always be read with the convention stated in the course. [example: Rotation Of The Complex Plane] Let $S^1$ act on $\mathbb C$ by $e^{it}\cdot z=e^{it}z$, and write $z=x+iy$ with $\theta=dx\wedge dy$. For the Lie algebra element $1\in\mathfrak{s}^1\cong\mathbb R$, the orbit through $z$ is \begin{align*} e^{it}z=(x\cos t-y\sin t)+i(x\sin t+y\cos t). \end{align*} Differentiating at $t=0$ gives \begin{align*} X_z=\frac{d}{dt}\Big|_{t=0}e^{it}z=-y\partial_x+x\partial_y. \end{align*} Now \begin{align*} \iota_X(dx\wedge dy)=dx(X)\,dy-dy(X)\,dx. \end{align*} Since $dx(X)=-y$ and $dy(X)=x$, this becomes \begin{align*} \iota_X\theta=-y\,dy-x\,dx=-(x\,dx+y\,dy). \end{align*} On the other hand, \begin{align*} d\left(-\frac12|z|^2\right)=d\left(-\frac12(x^2+y^2)\right)=-x\,dx-y\,dy. \end{align*} Thus $dH=\iota_X\theta$ for \begin{align*} H(z)=-\frac12|z|^2. \end{align*} The Hamiltonian is quadratic because the infinitesimal rotation field is linear in the coordinates, and contracting it with the constant symplectic form produces the differential of a quadratic function. [/example] ## Moment Maps as Lie Algebra Dual-Valued Hamiltonians A Hamiltonian action initially gives one Hamiltonian function for each Lie algebra direction. The next problem is to assemble all these functions into one geometric object, so that evaluating it on $\xi\in\mathfrak g$ recovers the Hamiltonian for the vector field generated by $\xi$. [definition: Moment Map] Let $(M,\theta)$ be a symplectic manifold with a symplectic action of a Lie group $G$. A moment map for the action is a smooth map \begin{align*} \mu:M\to\mathfrak g^* \end{align*} such that, for every $\xi\in\mathfrak g$, the component function \begin{align*} \mu^\xi:M&\to\mathbb R, & \mu^\xi(p)&=\mu(p)(\xi) \end{align*} satisfies \begin{align*} d\mu^\xi=\iota_{\xi_M}\theta. \end{align*} [/definition] Thus a moment map is a Lie algebra dual-valued Hamiltonian. The definition leads to an existence problem: for which symplectic actions do the closed $1$-forms $\iota_{\xi_M}\theta$ admit primitives that can be assembled into a smooth $\mathfrak g^*$-valued map? [quotetheorem:10065] [citeproof:10065] This criterion separates the local and global issues. The symplectic hypothesis is needed because the proof begins with closedness of $\iota_{\xi_M}\theta$; without it, there may be no local primitive to assemble. The vanishing of the de Rham cohomology classes is a genuine global condition: the translation action of $S^1$ on $(T^2,dx\wedge dy)$ gives the closed form $dy$, so the action is symplectic but has no moment map. The theorem does not make the moment map unique. Even on a connected manifold, each primitive is determined only up to a constant, and changing all constants linearly translates $\mu$ by an element of $\mathfrak g^*$. It also does not assert equivariance; the abelian translation action of $\mathbb R^2$ on $(\mathbb R^2,dx\wedge dy)$ has a moment map $\mu(x,y)(a,b)=ay-bx$, but this map is not invariant under translations even though the coadjoint action is the identity action. This limitation leads directly to the equivariance obstruction studied next. [remark: Non-Uniqueness Of Moment Maps] If $M$ is connected and $\mu$ is a moment map, then another moment map with the same sign convention has the form $\mu+c$ for some constant element $c\in\mathfrak g^*$. On a disconnected manifold, the constant may vary by connected component. [/remark] The standard linear examples are the model cases for the general definition. They also explain why quadratic functions appear throughout Hamiltonian mechanics. [example: Diagonal Circle Action On Complex Space] Let $z_j=x_j+iy_j$ and let $S^1$ act by \begin{align*} e^{it}\cdot(z_1,\dots,z_n)=(e^{ia_1t}z_1,\dots,e^{ia_nt}z_n). \end{align*} For the Lie algebra element $1\in\mathfrak s^1\cong\mathbb R$, the $j$th coordinate of the orbit is \begin{align*} e^{ia_jt}z_j=(x_j\cos(a_jt)-y_j\sin(a_jt))+i(x_j\sin(a_jt)+y_j\cos(a_jt)). \end{align*} Differentiating at $t=0$ gives the infinitesimal generator \begin{align*} X=\sum_{j=1}^n\left(-a_jy_j\partial_{x_j}+a_jx_j\partial_{y_j}\right). \end{align*} With $\theta=\sum_{j=1}^n dx_j\wedge dy_j$, contraction in the $j$th coordinate gives \begin{align*} \iota_X(dx_j\wedge dy_j)=dx_j(X)\,dy_j-dy_j(X)\,dx_j. \end{align*} Since $dx_j(X)=-a_jy_j$ and $dy_j(X)=a_jx_j$, we get \begin{align*} \iota_X\theta=\sum_{j=1}^n\left(-a_jy_j\,dy_j-a_jx_j\,dx_j\right). \end{align*} Now \begin{align*} d\left(-\frac12\sum_{j=1}^n a_j|z_j|^2\right)=d\left(-\frac12\sum_{j=1}^n a_j(x_j^2+y_j^2)\right)=\sum_{j=1}^n\left(-a_jx_j\,dx_j-a_jy_j\,dy_j\right). \end{align*} Thus $d\mu=\iota_X\theta$ for \begin{align*} \mu(z)=-\frac12\sum_{j=1}^n a_j|z_j|^2. \end{align*} Under the identification $\mathfrak s^{1*}\cong\mathbb R$, this is the moment map for the generator $1$; for a Lie algebra element $\lambda\in\mathbb R$, the generator is $\lambda X$ and the corresponding component is $\lambda\mu$. The weights $a_j$ therefore appear exactly as the coefficients of the quadratic coordinate terms in the moment map. [/example] ## Equivariance and the Coadjoint Action A moment map is tied to the Lie algebra, but a group action contains global symmetry data. The next question is whether the moment map itself respects the group action, and the correct action on the target is the coadjoint action on $\mathfrak g^*$. [definition: Coadjoint Action] Let $G$ be a Lie group with Lie algebra $\mathfrak g$. The coadjoint action of $G$ on $\mathfrak g^*$ is the left action defined by \begin{align*} (g\cdot\alpha)(\xi)=\alpha(\operatorname{Ad}_{g^{-1}}\xi) \end{align*} for $g\in G$, $\alpha\in\mathfrak g^*$, and $\xi\in\mathfrak g$. [/definition] The coadjoint action is forced by the way Lie algebra directions transform under conjugation. This makes it possible to ask for a moment map that intertwines the given action on $M$ with the induced action on the space of momentum values. [definition: Equivariant Moment Map] Let $(M,\theta)$ be a Hamiltonian $G$-manifold with moment map $\mu:M\to\mathfrak g^*$. The moment map is equivariant if \begin{align*} \mu(g\cdot p)=g\cdot\mu(p) \end{align*} for every $g\in G$ and $p\in M$, where $G$ acts on $\mathfrak g^*$ by the coadjoint action. [/definition] Equivariance says that applying a symmetry before measuring momentum gives the same answer as transforming the measured momentum. Not every moment map has this property: adding constants to Hamiltonian functions can change how the components transform, and the remaining failure may be invisible in each separate Hamiltonian equation. The obstruction is whether these failures fit together as a constant Lie algebra cocycle rather than disappearing under a normalization. [quotetheorem:10066] [citeproof:10066] Connectedness of $M$ is what turns the failure of equivariance into a constant rather than a locally constant function on different components. Connectedness of $G$ ensures that the infinitesimal cocycle controls the identity component of the group action; for a disconnected group, a moment map can be equivariant for the identity component while still failing to respect a discrete symmetry. The Hamiltonian hypothesis is also essential, since the formula uses the component functions $\mu^\xi$ and cannot even be formed for a merely symplectic non-Hamiltonian action such as translation on the symplectic torus. The standard non-equivariant example is the translation action of the [abelian group](/page/Abelian%20Group) $\mathbb R^2$ on $(\mathbb R^2,dx\wedge dy)$. For $\xi=(a,b)$, the infinitesimal generator is $a\partial_x+b\partial_y$, and $\mu^\xi=ay-bx$ is a Hamiltonian. Since the coadjoint action is the identity action, equivariance would require $\mu$ to be translation-invariant, but translating $(x,y)$ changes $\mu$ by a constant depending on the translation. The resulting Lie algebra cocycle is \begin{align*} \sigma((a,b),(c,d))=ad-bc, \end{align*} which is a nonzero constant bilinear form; for this abelian Lie algebra it is not removed by adding a constant to $\mu$. For compact connected groups, averaging or normalization often removes the obstruction, while non-compact actions and central extensions can retain it as meaningful geometry. [example: Rotation Action On The Cotangent Bundle Of The Sphere] Let $SO(3)$ act on $S^2\subset\mathbb R^3$ by rotations, and lift the action to $T^*S^2$. Using the Euclidean metric, write a point of $T^*S^2$ as $(q,p)$ with $q\in S^2$, $p\in T_qS^2$, so $q\cdot q=1$ and $q\cdot p=0$. Let $\alpha$ be the canonical $1$-form, so for a tangent vector $(u,v)\in T_{(q,p)}T^*S^2$, \begin{align*} \alpha_{(q,p)}(u,v)=p\cdot u. \end{align*} With the canonical symplectic form $\theta=-d\alpha$, we show that the moment map is \begin{align*} \mu(q,p)=q\times p\in\mathfrak{so}(3)^*\cong\mathbb R^3. \end{align*} Identify $\mathfrak{so}(3)$ with $\mathbb R^3$ by sending $a\in\mathbb R^3$ to the infinitesimal rotation $q\mapsto a\times q$. The infinitesimal generator on $T^*S^2$ has base component $a\times q$, so \begin{align*} \alpha_{(q,p)}(a_{T^*S^2})=p\cdot(a\times q). \end{align*} Using the scalar triple product identity, \begin{align*} p\cdot(a\times q)=a\cdot(q\times p). \end{align*} Thus the $a$-component of $\mu$ is \begin{align*} \mu^a(q,p)=\mu(q,p)\cdot a=(q\times p)\cdot a=p\cdot(a\times q)=\alpha(a_{T^*S^2}). \end{align*} The lifted action preserves the canonical $1$-form: for $R\in SO(3)$, \begin{align*} \alpha_{(Rq,Rp)}(Ru,Rv)=Rp\cdot Ru=p\cdot u=\alpha_{(q,p)}(u,v). \end{align*} Differentiating this invariance in the infinitesimal direction $a$ gives \begin{align*} \mathcal L_{a_{T^*S^2}}\alpha=0. \end{align*} Cartan's formula gives \begin{align*} \mathcal L_{a_{T^*S^2}}\alpha=d(\iota_{a_{T^*S^2}}\alpha)+\iota_{a_{T^*S^2}}d\alpha. \end{align*} Since $\iota_{a_{T^*S^2}}\alpha=\alpha(a_{T^*S^2})=\mu^a$ and $\theta=-d\alpha$, this becomes \begin{align*} 0=d\mu^a-\iota_{a_{T^*S^2}}\theta. \end{align*} Hence \begin{align*} d\mu^a=\iota_{a_{T^*S^2}}\theta. \end{align*} This is exactly the moment map identity for every $a\in\mathfrak{so}(3)$. Finally, for $R\in SO(3)$, \begin{align*} \mu(Rq,Rp)=Rq\times Rp=R(q\times p). \end{align*} Under the identification $\mathfrak{so}(3)^*\cong\mathbb R^3$, this is the coadjoint action of $SO(3)$, so angular momentum is an equivariant moment map. [/example] ## Noether Theorem for Hamiltonian Group Actions Moment maps matter dynamically because their components are conserved along Hamiltonian flows with compatible symmetry. The final problem of the chapter is to make precise the principle that continuous symmetries produce conserved quantities. [quotetheorem:10067] [citeproof:10067] This result is the symplectic form of conservation of momentum, and the $G$-invariance assumption is the point where the conservation law enters. If $H$ is not invariant, then $dH(\xi_M)$ need not vanish, and the same computation in the proof says that $\mu^\xi$ changes at rate $-dH(\xi_M)$ along the Hamiltonian flow. On $(\mathbb R^2,dx\wedge dy)$ with the translation action in the $x$-direction, the moment map component is $y$; for the Hamiltonian $H=x$, the Hamiltonian vector field satisfies $dH=\iota_{X_H}\theta$, so $X_H=-\partial_y$ and $y$ is not conserved. The theorem also does not say that every conserved quantity comes from a group action, nor that the whole map $\mu$ is constant as an equivariant object under all symmetries. It gives conservation component by component for infinitesimal generators of a Hamiltonian symmetry group. The equivariance discussion above explains how those components transform under the symmetry group itself, and the reduction theory that follows uses the conserved level sets $\mu^{-1}(\alpha)$ as spaces on which quotienting by symmetry can produce new symplectic manifolds. [example: Torus Actions On Complex Projective Space] Let $T^{n+1}$ act on $\mathbb C^{n+1}$ by \begin{align*} (e^{it_0},\dots,e^{it_n})\cdot(z_0,\dots,z_n)=(e^{it_0}z_0,\dots,e^{it_n}z_n). \end{align*} This action descends to $\mathbb CP^n$ because phase rotations send the complex line spanned by $(z_0,\dots,z_n)$ to another complex line. If $e^{it_0}=\cdots=e^{it_n}=e^{it}$, then \begin{align*} [e^{it}z_0:\cdots:e^{it}z_n]=[z_0:\cdots:z_n], \end{align*} so the diagonal circle acts as the identity on $\mathbb CP^n$ and the effective torus is $T^{n+1}/S^1$. For $[z_0:\dots:z_n]\in\mathbb CP^n$, define \begin{align*} \rho_j([z])=\frac{|z_j|^2}{\sum_{k=0}^n |z_k|^2}. \end{align*} This is well-defined on [projective space](/page/Projective%20Space): if $\lambda\in\mathbb C^\times$, then \begin{align*} \frac{|\lambda z_j|^2}{\sum_{k=0}^n |\lambda z_k|^2}=\frac{|\lambda|^2|z_j|^2}{|\lambda|^2\sum_{k=0}^n |z_k|^2}=\frac{|z_j|^2}{\sum_{k=0}^n |z_k|^2}. \end{align*} For the standard normalization of the Fubini--Study form, the $T^{n+1}$-moment map is, up to the global sign convention used for Hamiltonian vector fields, \begin{align*} \widetilde\mu([z])=(\rho_0([z]),\dots,\rho_n([z]))\in(\mathbb R^{n+1})^*. \end{align*} The diagonal direction $(1,\dots,1)$ gives the component \begin{align*} \widetilde\mu^{(1,\dots,1)}([z])=\sum_{j=0}^n \rho_j([z])=\frac{\sum_{j=0}^n |z_j|^2}{\sum_{k=0}^n |z_k|^2}=1. \end{align*} Its differential is zero, which matches the fact that the diagonal circle has zero infinitesimal action on $\mathbb CP^n$. Thus the effective moment map is obtained by removing the common diagonal component: \begin{align*} \mu([z])=\left(\rho_0([z])-\frac1{n+1},\dots,\rho_n([z])-\frac1{n+1}\right). \end{align*} This lies in the hyperplane $\sum_j u_j=0$, since \begin{align*} \sum_{j=0}^n\left(\rho_j([z])-\frac1{n+1}\right)=\sum_{j=0}^n\rho_j([z])-1=1-1=0. \end{align*} Before this translation, the image consists exactly of all $(\rho_0,\dots,\rho_n)$ with $\rho_j\ge 0$ and $\sum_j\rho_j=1$: the inequalities and sum condition follow from the displayed formula, and conversely any such point is obtained from \begin{align*} [z_0:\dots:z_n]=[\sqrt{\rho_0}:\dots:\sqrt{\rho_n}]. \end{align*} Therefore the moment map image is the standard simplex, translated into the dual of the effective torus, giving the first basic example of a moment polytope. [/example] The chapter has moved from symmetry of a symplectic form to conserved quantities and equivariant Lie algebra dual-valued maps. In the next stage of the course, level sets of moment maps become the input for symplectic reduction, where quotienting by symmetry produces new symplectic manifolds. The moment map records symmetry in a form that can be analyzed on level sets, so it becomes the right input for taking quotients. In the next chapter we turn that data into new symplectic spaces by symplectic reduction. # 9. Symplectic Reduction Chapter 8 introduced Hamiltonian group actions and moment maps as a way of recording infinitesimal symmetry by functions. This chapter assumes the course material on smooth manifolds, Lie group actions, symplectic forms, Hamiltonian vector fields, and the regular level set theorem; it also uses the coadjoint action from the moment-map chapter. The main question is what happens after imposing conserved quantities and then identifying states that differ only by symmetry. The result is symplectic reduction: a construction that turns a constrained quotient into a new symplectic manifold, with Hamiltonian dynamics descending to the quotient. ## Level Sets of Moment Maps and Quotient Spaces A Hamiltonian symmetry gives conserved quantities, but a conserved level set is usually not itself a symplectic manifold. The first problem is to understand the geometry of a level set $J^{-1}(\mu)$ and to identify the directions in which the symplectic form degenerates after restriction. [definition: Regular Value of a Moment Map] Let $(M,\omega)$ be a symplectic manifold, let $G$ be a Lie group with Lie algebra $\mathfrak g$, and let $J:M\to \mathfrak g^*$ be a smooth map. A point $\mu\in \mathfrak g^*$ is a regular value of $J$ if, for every $x\in J^{-1}(\mu)$, the total derivative \begin{align*} DJ_x:T_xM\to T_\mu\mathfrak g^*\cong \mathfrak g^* \end{align*} is surjective. [/definition] Regular values matter because they make the constraint set a manifold rather than a singular space. When $\mu$ is regular, the regular level set theorem gives \begin{align*} T_xJ^{-1}(\mu)=\ker DJ_x. \end{align*} The next calculation connects this kernel to the infinitesimal action. [quotetheorem:10068] [citeproof:10068] This theorem says that the constraint directions are already encoded by the orbit directions, but it is only a tangent-space statement on a smooth level set. The Hamiltonian hypothesis enters through the identity $dJ^\xi=\iota_{\xi_M}\omega$, which converts a condition on the derivative of $J$ into a symplectic orthogonality condition against infinitesimal orbit directions. The sign in the displayed formula depends on the convention used for $\iota_{\xi_M}\omega$; changing the convention changes both sides of the moment-map identity consistently, so the final orthogonal subspace is the same. Regularity is what permits the equality $T_xJ^{-1}(\mu)=\ker DJ_x$; without it, the theorem gives no smooth constraint manifold and no uniform tangent-space model for the fibre. For example, the rotation action of $S^1$ on $\mathbb C$ has moment map $J(z)=|z|^2/2$, and the level $J^{-1}(0)=\{0\}$ is not a regular level because $DJ_0=0$. The theorem also does not assert that $\omega|_{J^{-1}(\mu)}$ is nondegenerate. In the same circle example with $r>0$, the level $J^{-1}(r^2/2)$ is a circle, and the restriction of any two-form to a one-dimensional manifold is zero. Thus the level set is a constraint manifold, not yet the reduced symplectic manifold. The next issue is to identify exactly which group directions produce the degeneracy of the restricted form. [definition: Isotropy Group of a Coadjoint Value] Let $G$ be a Lie group and let $\mu\in\mathfrak g^*$. The isotropy group of $\mu$ for the coadjoint action is \begin{align*} G_\mu=\{g\in G: \operatorname{Ad}^*_{g^{-1}}\mu=\mu\}. \end{align*} Its Lie algebra is denoted $\mathfrak g_\mu$. [/definition] The use of $G_\mu$ rather than all of $G$ is forced by equivariance: if $J$ is equivariant, the element $g$ sends $J^{-1}(\mu)$ to $J^{-1}(\operatorname{Ad}^*_{g^{-1}}\mu)$. Thus only the stabiliser of $\mu$ acts on the fixed level set $J^{-1}(\mu)$. To build a quotient that removes exactly the null directions of the restricted form, we need to identify those null directions with the $G_\mu$-orbits. [quotetheorem:10069] [citeproof:10069] This identifies the directions that must be collapsed, under the smooth-level-set hypotheses. The reverse inclusion is a finite-dimensional symplectic linear algebra statement: after the tangent space to the level is written as the symplectic orthogonal of $T_x(G\cdot x)$, taking the symplectic orthogonal again recovers the orbit tangent space. The coadjoint stabiliser then selects the part of that orbit tangent space that actually stays inside the fixed level $J^{-1}(\mu)$, since infinitesimally this is the condition $\operatorname{ad}^*_\xi\mu=0$. Equivariance is essential here: without it, $G_\mu$ need not preserve the level set, so its orbits would not define directions inside $J^{-1}(\mu)$ at all. A concrete way this can happen is to start with an equivariant moment map $J:M\to\mathfrak g^*$ and replace it by $J_c=J+c$, where $c\in\mathfrak g^*$ is not fixed by the coadjoint action. The infinitesimal Hamiltonian equations are unchanged, because the components have only been shifted by constants, but equivariance can fail. For $x\in J_c^{-1}(\mu)$ and $g\in G_\mu$, equivariance of $J$ gives \begin{align*} J_c(g\cdot x)=\operatorname{Ad}^*_{g^{-1}}(\mu-c)+c, \end{align*} which need not equal $\mu$. Thus even elements stabilising $\mu$ may move points out of the chosen level for the shifted, non-equivariant moment map. Regularity is also essential, because at singular fibres the restricted form can have different rank on different pieces, and the phrase "the kernel at $T_xJ^{-1}(\mu)$" no longer captures the full local geometry. In the rotation action of $S^1$ on $\mathbb C$ with $J(z)=|z|^2/2$, the singular fibre $J^{-1}(0)=\{0\}$ has no circle orbit to quotient and $DJ_0=0$, so the expected hypersurface tangent space $\ker DJ_0=T_0\mathbb C$ is not the tangent space of a regular level. The kernel computation would predict directions coming from symplectic orthogonals in $T_0\mathbb C$, while the actual fibre is a single point; this is the local reduced-space failure at the named singular level. The theorem also does not say that the quotient is already a manifold. If the $G_\mu$-action has non-identity stabilisers, different orbit types can occur in the same level set; if the action is not proper, the orbit space may fail to be Hausdorff. Thus the kernel computation gives the infinitesimal reason for quotienting, but separate global hypotheses are needed before the quotient can carry a smooth symplectic form. This motivates the following definition of the quotient on which the restricted two-form can become nondegenerate. [definition: Reduced Space at a Regular Value] Let $(M,\omega)$ carry a Hamiltonian $G$-action with equivariant moment map $J:M\to\mathfrak g^*$. For $\mu\in\mathfrak g^*$, the reduced space at $\mu$ is \begin{align*} M_\mu=J^{-1}(\mu)/G_\mu. \end{align*} [/definition] This definition is topological until hypotheses are imposed. The next section gives the standard assumptions under which the quotient is a smooth manifold and the reduced two-form exists. [example: Circle Level Sets in Complex Space] Let $M=\mathbb C^n$ with $z_j=q_j+ip_j$ and \begin{align*} \omega_0=\sum_{j=1}^n dq_j\wedge dp_j. \end{align*} For the diagonal action $e^{i\theta}\cdot z=e^{i\theta}z$, the infinitesimal vector field for the generator $1\in\mathfrak s^1\cong\mathbb R$ is \begin{align*} \xi_M(z)=\sum_{j=1}^n\left(-p_j\frac{\partial}{\partial q_j}+q_j\frac{\partial}{\partial p_j}\right). \end{align*} Since $|z_j|^2=q_j^2+p_j^2$, the function \begin{align*} J(z)=\frac{1}{2}\sum_{j=1}^n(q_j^2+p_j^2) \end{align*} has differential \begin{align*} dJ=\sum_{j=1}^n(q_j\,dq_j+p_j\,dp_j). \end{align*} Also, \begin{align*} \iota_{\xi_M}\omega_0=\sum_{j=1}^n\left(dq_j(\xi_M)\,dp_j-dp_j(\xi_M)\,dq_j\right)=\sum_{j=1}^n(-p_j\,dp_j-q_j\,dq_j)=-dJ. \end{align*} Thus $J$ is a moment map for the diagonal circle action, with the sign convention $\iota_{\xi_M}\omega_0=dJ^\xi$. For $r>0$, \begin{align*} z\in J^{-1}(r^2/2)\quad\text{means}\quad \frac{1}{2}\sum_{j=1}^n |z_j|^2=\frac{r^2}{2}. \end{align*} Equivalently, \begin{align*} \sum_{j=1}^n |z_j|^2=r^2, \end{align*} so \begin{align*} J^{-1}(r^2/2)=S^{2n-1}_r\subset\mathbb C^n. \end{align*} The derivative at $z$ is \begin{align*} DJ_z(v)=\sum_{j=1}^n(q_j a_j+p_j b_j) \end{align*} for $v=(a_1+ib_1,\dots,a_n+ib_n)$. If $z\in S^{2n-1}_r$, then choosing $v=z$ gives \begin{align*} DJ_z(z)=\sum_{j=1}^n(q_j^2+p_j^2)=r^2\ne 0, \end{align*} so $r^2/2$ is a regular value. The quotient map is the Hopf map \begin{align*} S^{2n-1}_r\to\mathbb CP^{n-1},\qquad z\mapsto [z_1:\cdots:z_n]. \end{align*} Two points $z,w\in S^{2n-1}_r$ have the same image exactly when they span the same complex line, so $w=\lambda z$ for some $\lambda\in\mathbb C^\times$. Since both have norm $r$, \begin{align*} r^2=\|w\|^2=\|\lambda z\|^2=|\lambda|^2\|z\|^2=|\lambda|^2r^2, \end{align*} hence $|\lambda|=1$ and $\lambda\in S^1$. Therefore the fibres are precisely the diagonal $S^1$-orbits, and \begin{align*} S^{2n-1}_r/S^1\cong\mathbb CP^{n-1}. \end{align*} The reduced two-form is the unique form whose pullback to $S^{2n-1}_r$ is the restriction of $\omega_0$; under the above identification, it is proportional to the Fubini-Study form, with the proportionality depending on the radius parameter $r$. [/example] The example already shows the pattern of reduction: an odd-dimensional contact-type level set loses the circle direction and becomes an even-dimensional symplectic quotient. In general, the same mechanism is controlled by freeness and properness. ## Marsden-Weinstein-Meyer Reduction in the Free and Proper Case The main question is now whether the quotient $J^{-1}(\mu)/G_\mu$ carries a natural symplectic form. A quotient form must be characterised by pullback, since forms on the quotient are detected by pulling them back to the level set. [definition: Basic Differential Form] Let $P$ be a smooth manifold with a smooth action map \begin{align*} a:H\times P\to P,\qquad a(h,p)=h\cdot p, \end{align*} of a Lie group $H$. A differential form $\alpha\in\Omega^k(P)$ is basic if \begin{align*} g^*\alpha=\alpha\quad\text{for all }g\in H, \end{align*} and \begin{align*} \iota_{\xi_P}\alpha=0\quad\text{for all }\xi\in\mathfrak h. \end{align*} [/definition] Basic forms are exactly the forms that descend through a principal bundle quotient. For reduction, the candidate is the restricted form $i^*\omega$ on $J^{-1}(\mu)$, where $i:J^{-1}(\mu)\hookrightarrow M$ is inclusion. The reduction theorem is therefore the statement that this candidate is basic, descends, and becomes nondegenerate after the vertical directions have been removed. [quotetheorem:6847] [citeproof:6847] The theorem is the central construction of the chapter. Each hypothesis prevents a specific failure. The regular value condition makes $J^{-1}(\mu)$ a smooth constraint manifold; at a singular value, as in the zero level of $J(z)=|z|^2/2$ for the circle action on $\mathbb C$, the fibre can collapse to a singular stratum rather than a regular hypersurface. Freeness prevents residual stabilisers: if a group element fixes a point of the level, the quotient is typically an orbifold or stratified space rather than a manifold. Properness prevents pathological orbit spaces; without it, quotient spaces can fail to be Hausdorff, so smooth manifold charts cannot be expected. The theorem also has a precise scope. It proves the smooth free-and-proper form of reduction, not singular reduction, orbifold reduction, or reduction at nonregular values. Those variants are important, but they require additional language about orbit types and strata. Within the present hypotheses, the dimension count records how much has been removed. [remark: Dimension of the Reduced Space] If $G_\mu$ acts freely and properly on $J^{-1}(\mu)$, then \begin{align*} \dim M_\mu=\dim M-\dim G-\dim G_\mu. \end{align*} When $G$ is abelian, $G_\mu=G$, so the formula becomes $\dim M_\mu=\dim M-2\dim G$. [/remark] The dimension formula explains why circle reduction of $\mathbb C^n$ lowers dimension from $2n$ to $2n-2$. The same theorem also produces families of compact symplectic manifolds from products of spheres. [example: Polygon Spaces as Reduced Products of Spheres] Fix positive numbers $\ell_1,\dots,\ell_m$, and identify $S^2_{\ell_j}\subset\mathbb R^3\cong\mathfrak{so}(3)^*$ with the coadjoint orbit of radius $\ell_j$, equipped with its Kirillov-Kostant-Souriau symplectic form. On the product \begin{align*} M=S^2_{\ell_1}\times\cdots\times S^2_{\ell_m}, \end{align*} the diagonal $SO(3)$-action rotates every vector at once: \begin{align*} A\cdot(v_1,\dots,v_m)=(Av_1,\dots,Av_m). \end{align*} The moment map is the sum of the orbit moment maps, so \begin{align*} J(v_1,\dots,v_m)=v_1+\cdots+v_m. \end{align*} A point of $J^{-1}(0)$ is an ordered list of vectors with fixed lengths and zero total sum: \begin{align*} |v_j|=\ell_j\text{ for each }j,\quad v_1+\cdots+v_m=0. \end{align*} From such a list, define vertices by $x_0=0$ and \begin{align*} x_k=v_1+\cdots+v_k\quad\text{for }1\leq k\leq m. \end{align*} Then the $k$th edge is \begin{align*} x_k-x_{k-1}=v_k, \end{align*} so its length is \begin{align*} |x_k-x_{k-1}|=|v_k|=\ell_k. \end{align*} The polygon closes because \begin{align*} x_m=v_1+\cdots+v_m=J(v_1,\dots,v_m)=0=x_0. \end{align*} Thus $J^{-1}(0)$ parametrises closed ordered spatial polygons with side lengths $\ell_1,\dots,\ell_m$ and a chosen initial vertex at the origin. The quotient by $SO(3)$ removes rigid rotations. If $A\in SO(3)$, the vertices associated to $(Av_1,\dots,Av_m)$ satisfy \begin{align*} Av_1+\cdots+Av_k=A(v_1+\cdots+v_k)=Ax_k. \end{align*} Thus the new polygon is obtained from the old one by rotating every vertex about the origin. Conversely, rotating a polygon about the origin sends each edge vector $v_k=x_k-x_{k-1}$ to $Av_k$, so it gives the corresponding $SO(3)$-orbit in $J^{-1}(0)$. Therefore \begin{align*} J^{-1}(0)/SO(3) \end{align*} is the moduli space of closed spatial polygons with the prescribed side lengths, modulo rigid rotations. The stabiliser records where the quotient is smooth. If a polygon is non-collinear, then at least two edge vectors $v_a$ and $v_b$ are not parallel. Any rotation fixing $(v_1,\dots,v_m)$ fixes both $v_a$ and $v_b$. A rotation in $SO(3)$ that fixes two nonparallel vectors fixes the plane they span and also fixes its oriented normal vector, so it is the identity rotation. Hence non-collinear polygons have identity stabiliser. If all edges are collinear along a line $L$, then every rotation about $L$ fixes each vector $v_j$, so the stabiliser contains a copy of $SO(2)$ and is positive-dimensional. On the free non-collinear locus, symplectic reduction gives the natural symplectic structure on the smooth space of polygon shapes. [/example] This example also indicates what can fail outside the free and proper setting: stabilisers can create orbifold or singular strata. The present course stays with the smooth version, because it contains the main mechanism without the additional language of singular reduction. ## Reduced Dynamics and Reconstruction of Hamiltonian Trajectories Reduction is most useful when it respects dynamics. The problem is to start with a Hamiltonian $H:M\to\mathbb R$ that has symmetry, restrict to a conserved level of $J$, and obtain an ordinary Hamiltonian system on $M_\mu$. [definition: Invariant Hamiltonian] Let $(M,\omega)$ be a symplectic manifold with a smooth action of a Lie group $G$. A Hamiltonian $H:M\to\mathbb R$ is $G$-invariant if \begin{align*} H(g\cdot x)=H(x) \end{align*} for all $g\in G$ and all $x\in M$. [/definition] Invariant Hamiltonians are constant on symmetry orbits, so their values are compatible with quotienting. After restriction to $J^{-1}(\mu)$, the next task is to name the function on the reduced space whose pullback is the restricted Hamiltonian. This function is the energy that the reduced Hamiltonian vector field will use. [definition: Reduced Hamiltonian] Let $H:M\to\mathbb R$ be a $G_\mu$-invariant Hamiltonian and let $M_\mu=J^{-1}(\mu)/G_\mu$. The reduced Hamiltonian $H_\mu:M_\mu\to\mathbb R$ is defined by \begin{align*} H_\mu([x])=H(x) \end{align*} for $x\in J^{-1}(\mu)$. [/definition] The definition is well posed because $H$ is constant on the $G_\mu$-orbit of $x$. This turns the quotient map into an energy-preserving projection: the value assigned to a reduced point does not depend on which representative of the orbit is chosen. The next question is dynamical rather than set-theoretic. If a trajectory starts on the moment level, the original Hamiltonian vector field should remain tangent to that level, and after quotienting its motion should be governed by the Hamiltonian vector field of $H_\mu$ rather than by an unrelated vector field. [quotetheorem:10070] [citeproof:10070] The theorem separates the analysis of a symmetric Hamiltonian system into a horizontal part and a vertical part. The horizontal part is the lower-dimensional trajectory on $M_\mu$; the vertical part is the motion along the $G_\mu$-orbit that was collapsed by the quotient. The hypotheses again have concrete roles. $G$-invariance of $H$ is what makes $H$ constant on orbits and makes $X_H$ preserve the moment map; if $H$ is perturbed by a non-invariant term, Noether conservation can fail and a trajectory can leave $J^{-1}(\mu)$. Completeness on the time interval is a domain condition: without it, the vector field may be defined but its trajectories may stop before the interval on which one wants to compare original and reduced motion. The reduction hypotheses are also still needed for dynamics. If the quotient is singular or non-Hausdorff, the phrase "Hamiltonian vector field on $M_\mu$" is no longer an ordinary smooth-manifold statement. Even in the free and proper case, the theorem deliberately stops at the projected curve: it determines which reduced point evolves in $M_\mu$, but not which representative of the corresponding $G_\mu$-orbit is occupied upstairs. The missing data is dynamical rather than cosmetic, because different lifts of the same reduced curve can have different vertical velocities along the symmetry orbit. To complete the original initial-value problem, the course therefore isolates the problem of lifting a reduced trajectory while recovering exactly the group motion that makes the lift solve Hamilton's equation. The next definition names that lifting task. It is needed because reduction replaces a point of $J^{-1}(\mu)$ by its orbit, so solving the reduced equations gives the evolution of an equivalence class rather than a physical phase-space curve. Reconstruction asks for the additional group variable that turns this reduced motion back into a trajectory upstairs. [definition: Reconstruction Problem] Let $\bar\gamma:I\to M_\mu$ be a trajectory of the reduced Hamiltonian system. The reconstruction problem is to find a curve $\gamma:I\to J^{-1}(\mu)$ such that \begin{align*} \pi\circ\gamma=\bar\gamma \end{align*} and $\gamma$ is a trajectory of the original Hamiltonian vector field $X_H$. [/definition] Reconstruction is not merely choosing any lift of $\bar\gamma$. A lift must also have the correct vertical component, which is the group motion along the symmetry directions. [remark: Connection Form View of Reconstruction] When the principal bundle $\pi:J^{-1}(\mu)\to M_\mu$ is equipped with a connection form, a reduced trajectory has a horizontal lift, and the missing vertical part is encoded by a curve in $G_\mu$. The resulting group equation is the reconstruction equation. Its exact form depends on the chosen connection and on the mechanical system, so the course uses it as a structural principle rather than a single universal formula. [/remark] The physical meaning is most familiar in mechanics with rotational symmetry. Fixing angular momentum and quotienting rotations leaves shape variables; reconstruction then restores the orientation variable. [example: Angular Momentum Reduction for a Central Force] Consider a particle of mass $m$ moving in $\mathbb R^3_0:=\mathbb R^3\setminus\{0\}$ under a central potential $V(|q|)$, with phase space $T^*\mathbb R^3_0$ and Hamiltonian \begin{align*} H(q,p)=\frac{|p|^2}{2m}+V(|q|). \end{align*} Identify $\mathfrak{so}(3)^*$ with $\mathbb R^3$ using the Euclidean dot product. For the rotation action of $SO(3)$, the infinitesimal generator determined by $\Omega\in\mathbb R^3\cong\mathfrak{so}(3)$ is \begin{align*} \Omega_{T^*\mathbb R^3_0}(q,p)=(\Omega\times q,\Omega\times p). \end{align*} The angular momentum map is \begin{align*} J(q,p)=q\times p, \end{align*} because its $\Omega$-component is \begin{align*} J^\Omega(q,p)=\Omega\cdot(q\times p). \end{align*} Fix $\mu\ne 0$ and impose $J(q,p)=\mu$. Then \begin{align*} q\cdot\mu=q\cdot(q\times p)=0 \end{align*} and \begin{align*} p\cdot\mu=p\cdot(q\times p)=0, \end{align*} so both $q$ and $p$ lie in the plane perpendicular to $\mu$. Thus motion with this angular momentum is represented in the fixed oriented plane $\mu^\perp$. The coadjoint stabiliser $SO(3)_\mu$ consists of rotations preserving the vector $\mu$, hence rotations about the axis $\mathbb R\mu$, so \begin{align*} SO(3)_\mu\cong SO(2). \end{align*} Quotienting by $SO(3)_\mu$ removes the residual rotation angle around that axis. Choose polar coordinates in $\mu^\perp$ with orientation chosen so that positive planar angular momentum points in the direction of $\mu$. Write \begin{align*} q=r e_r \end{align*} with $r>0$, and use Hamilton's equation $\dot q=p/m$, so $p=m\dot q$. Since \begin{align*} \dot q=\dot r\,e_r+r\dot\theta\,e_\theta, \end{align*} we get \begin{align*} p=m\dot r\,e_r+mr\dot\theta\,e_\theta. \end{align*} The scalar angular momentum in this oriented plane is \begin{align*} p_\theta=(q\times p)\cdot\frac{\mu}{|\mu|}. \end{align*} Substituting $q=r e_r$ and $p=m\dot r\,e_r+mr\dot\theta\,e_\theta$ gives \begin{align*} p_\theta=\left(r e_r\times(m\dot r\,e_r+mr\dot\theta\,e_\theta)\right)\cdot\frac{\mu}{|\mu|}. \end{align*} The term $e_r\times e_r$ vanishes, and $e_r\times e_\theta=\mu/|\mu|$ by the chosen orientation, so \begin{align*} p_\theta=mr^2\dot\theta. \end{align*} Since $q\times p=\mu$, this also gives \begin{align*} p_\theta=|\mu|. \end{align*} The kinetic energy splits visibly from the orthonormal frame $(e_r,e_\theta)$: \begin{align*} |\dot q|^2=|\dot r\,e_r+r\dot\theta\,e_\theta|^2=\dot r^2+r^2\dot\theta^2. \end{align*} Therefore \begin{align*} \frac{|p|^2}{2m}=\frac{m}{2}|\dot q|^2=\frac{m}{2}\dot r^2+\frac{m}{2}r^2\dot\theta^2. \end{align*} Using $p_\theta=mr^2\dot\theta$, the angular term becomes \begin{align*} \frac{m}{2}r^2\dot\theta^2=\frac{(mr^2\dot\theta)^2}{2mr^2}=\frac{p_\theta^2}{2mr^2}. \end{align*} On the level $J=\mu$, this is \begin{align*} \frac{p_\theta^2}{2mr^2}=\frac{|\mu|^2}{2mr^2}. \end{align*} Thus the radial reduced Hamiltonian is \begin{align*} H_\mu(r,p_r)=\frac{p_r^2}{2m}+V(r)+\frac{|\mu|^2}{2mr^2}, \end{align*} where $p_r=m\dot r$, and the effective potential is \begin{align*} V_{\mathrm{eff}}(r)=V(r)+\frac{|\mu|^2}{2mr^2}. \end{align*} The angle $\theta$ is not a coordinate on the fully reduced space; it is the representative of the collapsed $SO(3)_\mu$-orbit. After solving the radial equation, reconstruction recovers this angle from \begin{align*} \dot\theta=\frac{p_\theta}{mr^2}=\frac{|\mu|}{mr^2}. \end{align*} At $\mu=0$, the coadjoint stabiliser is all of $SO(3)$, because every rotation fixes the zero vector, and there is no distinguished angular momentum axis. [/example] This example is the model for the chapter's philosophy. Conserved quantities select a constraint manifold, symmetry orbits describe the redundant directions, and the quotient carries the symplectic form and Hamiltonian flow needed to study the true degrees of freedom. The same construction also connects this course to neighbouring subjects: topologically, reduction builds new manifolds and stratified spaces from group actions; in geometric invariant theory, moment-map quotients are the symplectic counterpart of algebraic quotients; and in coadjoint orbit geometry, reduction explains why orbit method examples naturally carry symplectic forms. Reduction produces concrete quotients and stratified spaces from Hamiltonian actions, making symmetry a source of examples rather than just structure. The next chapter develops this idea in the especially computable setting of toric geometry and related global models. # 10. Toric and Elementary Global Examples This chapter turns the Hamiltonian formalism from the previous lectures into a source of global examples. The guiding question is: how much of a compact symplectic manifold can be seen from a torus action and its moment map? In the basic toric cases the answer is remarkably concrete: projective spaces, products, and blow-ups are encoded by convex polygons or polytopes. The point is not to prove the full [Delzant classification theorem](/theorems/10072), but to use it as a map of the landscape. Moment images give a visual calculus for symplectic size, fixed points, products, and elementary surgeries. This chapter closes the foundational part of the course by connecting Hamiltonian actions, reduction, and global constructions in a family of examples where the geometry can be computed by hand. ## Hamiltonian Torus Actions and Moment Images The basic problem is to understand symmetries generated by several commuting Hamiltonians at once. A single Hamiltonian $H:M\to \mathbb R$ gives an $S^1$-action when its flow is periodic; a torus action packages $k$ commuting periodic Hamiltonian flows and asks for a single map recording all conserved quantities. [definition: Hamiltonian Torus Action] Let $(M,\omega)$ be a symplectic manifold, and let $T$ be a torus with Lie algebra $\mathfrak t$. A smooth action $T\curvearrowright M$ is Hamiltonian if there exists a smooth map \begin{align*} \mu:M\longrightarrow \mathfrak t^* \end{align*} such that, for every $\xi\in\mathfrak t$, the component function $\mu^\xi:M\to\mathbb R$ defined by $\mu^\xi(x)=\mu(x)(\xi)$ satisfies \begin{align*} d\mu^\xi = \iota_{X_\xi}\omega, \end{align*} where $X_\xi$ is the vector field generated by $\xi$. [/definition] The definition says that every infinitesimal torus symmetry is generated by a Hamiltonian, and the moment map records these Hamiltonians in a coordinate-free way. Because $T$ is abelian, no coadjoint correction appears; after choosing a basis of $\mathfrak t$, the moment map is just a tuple of commuting Hamiltonians. [example: Rotating The Two-Sphere] Identify $\mathbb{CP}^1$ with $S^2$ and scale the Fubini-Study form so that the total symplectic area is $\lambda>0$. For the circle action \begin{align*} e^{i\theta}\cdot [z_0:z_1]=[z_0:e^{i\theta}z_1], \end{align*} use the normalization \begin{align*} \mu([z_0:z_1])=\lambda\frac{|z_1|^2}{|z_0|^2+|z_1|^2}. \end{align*} The denominator is positive on projective space, so $0\le |z_1|^2/(|z_0|^2+|z_1|^2)\le 1$, and therefore $0\le \mu([z_0:z_1])\le \lambda$. Conversely, for every $t\in[0,\lambda]$, the point $[\sqrt{\lambda-t}:\sqrt t]$ has \begin{align*} \mu([\sqrt{\lambda-t}:\sqrt t])=\lambda\frac{t}{(\lambda-t)+t}=t. \end{align*} Thus $\mu(\mathbb{CP}^1)=[0,\lambda]$. The fixed points are exactly the coordinate points. Indeed, if $[z_0:e^{i\theta}z_1]=[z_0:z_1]$ for every $\theta$, then for each $\theta$ there is a nonzero scalar $c_\theta$ with $z_0=c_\theta z_0$ and $e^{i\theta}z_1=c_\theta z_1$. If both $z_0$ and $z_1$ were nonzero, the first equation would give $c_\theta=1$, and then the second would give $e^{i\theta}=1$ for every $\theta$, impossible. Hence one coordinate is zero, giving $[1:0]$ and $[0:1]$. Their moment-map values are \begin{align*} \mu([1:0])=\lambda\frac{0}{1+0}=0 \end{align*} and \begin{align*} \mu([0:1])=\lambda\frac{1}{0+1}=\lambda. \end{align*} If $z_0z_1\ne 0$ and $e^{i\theta}$ fixes $[z_0:z_1]$, the same two equations force $c_\theta=1$ and then $e^{i\theta}=1$, so the stabilizer is trivial. The interval therefore records the orbit stratification: endpoints are fixed orbits, and interior points come from free circle orbits. [/example] The two-sphere example shows the smallest possible match between half the symplectic dimension and the torus dimension. This motivates isolating the maximal effective case, because only then should the moment image have enough coordinates to encode the full family of invariant directions. [definition: Symplectic Toric Manifold] A symplectic toric manifold is a connected symplectic manifold $(M^{2n},\omega)$ equipped with an effective Hamiltonian action of an $n$-torus $T^n$. [/definition] The word effective excludes redundant torus directions, so the definition sets up the strongest finite-dimensional form of Hamiltonian abelian symmetry. The next question is then global rather than local: once a compact toric or Hamiltonian torus manifold has a moment map, what shape can its image have? [quotetheorem:10071] This theorem is stated here as a structural result rather than proved in this course. Its proof studies the components $\mu^\xi$ as Morse-Bott functions, uses their critical sets to control the images of level sets, and then combines the resulting interval-convexity statements for different $\xi\in\mathfrak t$. This mechanism is a central bridge from Hamiltonian dynamics to [convex geometry](/page/Convex%20Geometry). The hypotheses are substantial: compactness prevents non-closed or unbounded moment images, as in the standard rotation of $\mathbb C$ whose moment image is a ray, while connectedness prevents the image from being a disconnected union of polytopes. Hamiltonianity is also essential, since a symplectic torus action need not admit a global moment map at all, for example a translation action on a symplectic torus. The theorem says that the image is controlled by fixed points, but it does not classify the manifold, determine stabilisers away from the fixed set, or say that every compact convex polytope can occur as a smooth toric moment image. [example: The Moment Triangle of Projective Plane] Let $\mathbb{CP}^2$ carry the Fubini-Study form scaled by $\lambda>0$, and let $T^2$ act by \begin{align*} (e^{i\theta_1},e^{i\theta_2})\cdot [z_0:z_1:z_2]=[z_0:e^{i\theta_1}z_1:e^{i\theta_2}z_2]. \end{align*} With the standard normalization, a moment map is \begin{align*} \mu([z_0:z_1:z_2])=\lambda\left(\frac{|z_1|^2}{|z_0|^2+|z_1|^2+|z_2|^2},\frac{|z_2|^2}{|z_0|^2+|z_1|^2+|z_2|^2}\right). \end{align*} Write $\mu([z_0:z_1:z_2])=(u_1,u_2)$ and set $D=|z_0|^2+|z_1|^2+|z_2|^2$. Since $D>0$ and each $|z_j|^2\ge 0$, we have \begin{align*} u_1=\lambda\frac{|z_1|^2}{D}\ge 0 \end{align*} and \begin{align*} u_2=\lambda\frac{|z_2|^2}{D}\ge 0. \end{align*} Also, \begin{align*} u_1+u_2=\lambda\frac{|z_1|^2+|z_2|^2}{D}. \end{align*} Because $|z_1|^2+|z_2|^2\le |z_0|^2+|z_1|^2+|z_2|^2=D$, this gives $u_1+u_2\le \lambda$. Hence $\mu(\mathbb{CP}^2)$ is contained in \begin{align*} \{(u_1,u_2)\in\mathbb R^2:u_1\ge 0,\ u_2\ge 0,\ u_1+u_2\le \lambda\}. \end{align*} Conversely, take any $(u_1,u_2)$ satisfying $u_1\ge 0$, $u_2\ge 0$, and $u_1+u_2\le\lambda$. The point $[\sqrt{\lambda-u_1-u_2}:\sqrt{u_1}:\sqrt{u_2}]$ is defined because the three entries are nonnegative and not all zero. Its denominator is \begin{align*} (\lambda-u_1-u_2)+u_1+u_2=\lambda. \end{align*} Therefore \begin{align*} \mu([\sqrt{\lambda-u_1-u_2}:\sqrt{u_1}:\sqrt{u_2}])=\lambda\left(\frac{u_1}{\lambda},\frac{u_2}{\lambda}\right)=(u_1,u_2). \end{align*} Thus the moment image is exactly the closed triangle \begin{align*} \mu(\mathbb{CP}^2)=\{(u_1,u_2)\in\mathbb R^2:u_1\ge 0,\ u_2\ge 0,\ u_1+u_2\le \lambda\}. \end{align*} The coordinate points map to the three vertices: \begin{align*} \mu([1:0:0])=(0,0), \end{align*} \begin{align*} \mu([0:1:0])=(\lambda,0), \end{align*} and \begin{align*} \mu([0:0:1])=(0,\lambda). \end{align*} They are fixed because multiplying a single nonzero homogeneous coordinate by a phase does not change the projective point. Conversely, if $[z_0:z_1:z_2]$ is fixed by every $(e^{i\theta_1},e^{i\theta_2})$, then for every pair $(\theta_1,\theta_2)$ there is a nonzero scalar $c_{\theta_1,\theta_2}$ such that \begin{align*} z_0=c_{\theta_1,\theta_2}z_0,\quad e^{i\theta_1}z_1=c_{\theta_1,\theta_2}z_1,\quad e^{i\theta_2}z_2=c_{\theta_1,\theta_2}z_2. \end{align*} If two coordinates were nonzero, varying the corresponding phase would force contradictory values of $c_{\theta_1,\theta_2}$. Hence exactly one coordinate is nonzero, so the fixed points are precisely $[1:0:0]$, $[0:1:0]$, and $[0:0:1]$. The open edge $u_1=0$ has $z_1=0$ and $z_0,z_2\ne 0$, the open edge $u_2=0$ has $z_2=0$ and $z_0,z_1\ne 0$, and the open edge $u_1+u_2=\lambda$ has $z_0=0$ and $z_1,z_2\ne 0$. The interior consists exactly of points with $z_0z_1z_2\ne 0$. Thus the triangle records the orbit stratification: vertices are fixed orbits, open edges have one circle of stabilizer, and interior points have trivial stabilizer. [/example] The triangle computation also explains why moment polytopes are more than pictures: their faces record isotropy type, while their affine lengths record symplectic areas of invariant spheres. Products behave in the expected way and give the next basic family. [example: Product of Two Projective Lines] Write a point of $\mathbb{CP}^1\times\mathbb{CP}^1$ as $([z_0:z_1],[w_0:w_1])$, and let the two circle factors rotate $z_1$ and $w_1$ respectively. With the same normalization as for $\mathbb{CP}^1$, the product moment map is \begin{align*} \mu([z_0:z_1],[w_0:w_1])=\left(\lambda_1\frac{|z_1|^2}{|z_0|^2+|z_1|^2},\lambda_2\frac{|w_1|^2}{|w_0|^2+|w_1|^2}\right). \end{align*} Since $|z_0|^2+|z_1|^2>0$ and $|w_0|^2+|w_1|^2>0$, the two components satisfy \begin{align*} 0\le \lambda_1\frac{|z_1|^2}{|z_0|^2+|z_1|^2}\le \lambda_1 \end{align*} and \begin{align*} 0\le \lambda_2\frac{|w_1|^2}{|w_0|^2+|w_1|^2}\le \lambda_2. \end{align*} Therefore $\mu(\mathbb{CP}^1\times\mathbb{CP}^1)\subseteq [0,\lambda_1]\times[0,\lambda_2]$. Conversely, choose any $(s,t)\in[0,\lambda_1]\times[0,\lambda_2]$. The point \begin{align*} ([\sqrt{\lambda_1-s}:\sqrt{s}],[\sqrt{\lambda_2-t}:\sqrt{t}]) \end{align*} is well-defined, and its first denominator is $(\lambda_1-s)+s=\lambda_1$, while its second denominator is $(\lambda_2-t)+t=\lambda_2$. Hence \begin{align*} \mu([\sqrt{\lambda_1-s}:\sqrt{s}],[\sqrt{\lambda_2-t}:\sqrt{t}])=\left(\lambda_1\frac{s}{\lambda_1},\lambda_2\frac{t}{\lambda_2}\right)=(s,t). \end{align*} Thus \begin{align*} \mu(\mathbb{CP}^1\times\mathbb{CP}^1)=[0,\lambda_1]\times[0,\lambda_2]. \end{align*} The moment polygon is therefore a rectangle. On the ruling $\mathbb{CP}^1\times\{p\}$, the second factor is constant, so the product form restricts to $\lambda_1\omega_{FS}$ and its symplectic area is $\lambda_1$. On the ruling $\{q\}\times\mathbb{CP}^1$, the first factor is constant, so the product form restricts to $\lambda_2\omega_{FS}$ and its symplectic area is $\lambda_2$. Thus changing the product form changes the side lengths of the rectangle while leaving the underlying smooth product $\mathbb{CP}^1\times\mathbb{CP}^1$ unchanged. [/example] These examples suggest a dictionary: compact toric manifolds should correspond to certain integral convex polytopes. The next section states the exact local combinatorial conditions needed for that dictionary. ## Delzant Polytopes and the Classification Statement The classification problem asks which convex polytopes occur as moment images of compact symplectic toric manifolds, and whether the polytope determines the manifold. Not every convex polytope is permitted, because the local model near a fixed point remembers the integral lattice of circle subgroups in the torus. [definition: Delzant Polytope] Let $\Delta\subset \mathfrak t^*$ be a compact convex polytope for a torus $T$ with integral lattice $\Lambda\subset\mathfrak t$. The polytope $\Delta$ is Delzant if it satisfies the following three conditions: 1. $\Delta$ is simple: exactly $n$ facets meet at each vertex. 2. $\Delta$ is rational: each facet has an inward normal vector lying in $\Lambda$. 3. $\Delta$ is smooth: at each vertex, the primitive inward normals to the $n$ facets meeting there form a $\mathbb Z$-basis of $\Lambda$. [/definition] Simplicity is a local manifold condition, rationality records compatibility with the torus lattice, and smoothness rules out orbifold singularities. These three conditions are exactly what is visible in the standard linear model for a torus action near a fixed point. [example: Triangle and Rectangle Are Delzant] For the triangle \begin{align*} \Delta_\lambda=\{(u_1,u_2):u_1\ge 0,\ u_2\ge 0,\ u_1+u_2\le \lambda\}, \end{align*} the facet $u_1=0$ has primitive inward normal $(1,0)$, the facet $u_2=0$ has primitive inward normal $(0,1)$, and the facet $u_1+u_2=\lambda$ has primitive inward normal $(-1,-1)$. At $(0,0)$ the meeting normals are $(1,0)$ and $(0,1)$, and \begin{align*} \det((1,0),(0,1))=1\cdot 1-0\cdot 0=1. \end{align*} At $(\lambda,0)$ the meeting normals are $(0,1)$ and $(-1,-1)$, and \begin{align*} \det((0,1),(-1,-1))=0\cdot(-1)-(-1)\cdot 1=1. \end{align*} At $(0,\lambda)$ the meeting normals are $(1,0)$ and $(-1,-1)$, and \begin{align*} \det((1,0),(-1,-1))=1\cdot(-1)-(-1)\cdot 0=-1. \end{align*} Since a pair of vectors in $\mathbb Z^2$ with determinant $\pm 1$ is a $\mathbb Z$-basis of $\mathbb Z^2$, the triangle satisfies the smoothness condition at every vertex. For the rectangle \begin{align*} [0,\lambda_1]\times[0,\lambda_2], \end{align*} the four facets have primitive inward normals $(1,0)$, $(-1,0)$, $(0,1)$, and $(0,-1)$. The vertex $(0,0)$ uses $(1,0)$ and $(0,1)$, with determinant $1$. The vertex $(\lambda_1,0)$ uses $(-1,0)$ and $(0,1)$, with \begin{align*} \det((-1,0),(0,1))=(-1)\cdot 1-0\cdot 0=-1. \end{align*} The vertex $(0,\lambda_2)$ uses $(1,0)$ and $(0,-1)$, with \begin{align*} \det((1,0),(0,-1))=1\cdot(-1)-0\cdot 0=-1. \end{align*} The vertex $(\lambda_1,\lambda_2)$ uses $(-1,0)$ and $(0,-1)$, with \begin{align*} \det((-1,0),(0,-1))=(-1)\cdot(-1)-0\cdot 0=1. \end{align*} Thus exactly two facets meet at each vertex, all facet normals are primitive lattice vectors, and the normals at every vertex form a $\mathbb Z$-basis. The triangle and rectangle are therefore Delzant polygons, which is the combinatorial reason they correspond to smooth compact toric surfaces. [/example] The basis condition is the point at which smoothness enters the dictionary. If the primitive inward normals at a vertex span a sublattice of index greater than one, the local quotient model has a finite stabiliser and gives an orbifold point rather than a smooth point. [example: A Non-Delzant Corner] Consider a polygon in $\mathbb R^2$ with two adjacent inward primitive normals $v_1=(1,0)$ and $v_2=(1,2)$ at a vertex. The vector $v_1$ is primitive because $\gcd(1,0)=1$, and $v_2$ is primitive because $\gcd(1,2)=1$, so the corner is rational. The two normals are linearly independent: if $a(1,0)+b(1,2)=(0,0)$, then \begin{align*} (a+b,2b)=(0,0). \end{align*} The second coordinate gives $2b=0$, hence $b=0$, and then the first coordinate gives $a=0$. Thus the two adjacent facets have independent normals, so this two-dimensional corner is simple. The determinant of the ordered pair $v_1,v_2$ is \begin{align*} \det(v_1,v_2)=1\cdot 2-1\cdot 0=2. \end{align*} A pair of integer vectors forms a $\mathbb Z$-basis of $\mathbb Z^2$ exactly when this determinant is $\pm 1$. Since the determinant is $2$, the subgroup generated by $(1,0)$ and $(1,2)$ has index $2$ in $\mathbb Z^2$, so these normals do not form a $\mathbb Z$-basis. Therefore the corner satisfies the simple and rational conditions but fails the smoothness condition in the definition of a Delzant polytope. Geometrically, the corresponding local toric model has a $\mathbb Z/2\mathbb Z$ quotient singularity rather than a smooth fixed point chart. [/example] This failure example shows why the Delzant conditions are not cosmetic restrictions. The classification theorem is the conceptual endpoint of the toric story, but the proof is not part of this foundations course. We use it as context for recognizing examples and for interpreting operations on polytopes as symplectic operations on manifolds. [quotetheorem:10072] The construction behind the classification starts with one copy of $\mathbb C$ for each facet, chooses a subtorus determined by the primitive facet normals, and takes a symplectic quotient at the level fixed by the support constants. Uniqueness is then obtained by comparing equivariant local normal forms over corresponding faces and checking that the transition data agree. This material belongs to the next stage of the subject, so here the theorem functions as a guide for examples rather than as a tool we prove. Compactness is reflected by the compactness of the polytope; without it, affine toric examples such as $\mathbb C^n$ have noncompact moment images and require a different classification statement. Connectedness rules out disjoint unions, and effectiveness removes invisible torus factors that would not change the polytope. The Delzant hypotheses are exactly the smooth local fixed-point conditions: dropping smoothness leads to toric orbifolds, dropping rationality loses compatibility with the torus lattice, and dropping simplicity takes us outside the smooth toric-manifold model. The theorem also does not classify arbitrary Hamiltonian torus spaces with smaller torus dimension, non-Hamiltonian symplectic torus actions, or Hamiltonian spaces with singular reduced pieces. [remark: Translations of Moment Maps] A moment map is determined only up to addition of a constant element of $\mathfrak t^*$. Translating a moment polytope therefore does not change the Hamiltonian torus action. The intrinsic information is the integral affine shape, including facet normals and relative support numbers. [/remark] The classification statement gives a compact way to remember the examples already computed. It also predicts that changing a polygon by a legitimate Delzant operation should correspond to a symplectic construction, and the most important elementary operation is blow-up. ## Symplectic Blow-Up and Cutting Corners The local question behind blow-up is how to replace a point in a symplectic manifold by the space of complex directions through that point while preserving a symplectic form. In toric dimension four, this operation has a visible moment-map signature: it cuts off a corner of the moment polygon. [definition: Symplectic Blow-Up at a Point] Let $(M^{2n},\omega)$ be a symplectic manifold and let $p\in M$. A symplectic blow-up of size $\varepsilon>0$ at $p$ is a symplectic manifold $(\widetilde M,\widetilde\omega)$ obtained from a symplectic embedding of the standard ball of capacity $\varepsilon$ centred at $p$ by deleting the ball interior and applying symplectic reduction to the boundary Hopf circle action. [/definition] The collapsed boundary is called the exceptional divisor. This definition is local, but in toric examples it becomes combinatorial. A torus fixed point maps to a vertex of the moment polytope, and removing a standard ball centred at that fixed point removes a simplex-shaped neighbourhood of that vertex. [quotetheorem:10073] [citeproof:10073] The new edge represents the exceptional divisor. Its affine length is the blow-up size, and its normal vector is the sum of the two primitive inward normals adjacent to the original vertex. The fixed-point hypothesis is what makes the operation toric: a non-fixed point does not map to a vertex, and blowing it up would generally destroy the given torus action rather than produce a new polygon by cutting a corner. The smallness hypothesis is also necessary, because a cut that reaches another vertex or passes through another face changes the combinatorics and may fail to describe an embedded local blow-up. The theorem is local in nature: it identifies the moment polygon of the toric blow-up but does not classify all symplectic forms on the resulting smooth blow-up or all possible non-toric blow-ups. This corner-cutting model leads naturally to symplectic cutting, where the new face is produced by reducing a Hamiltonian level set. [example: One-Point Blow-Up of Projective Plane] Start with the moment triangle for $(\mathbb{CP}^2,\lambda\omega_{FS})$: \begin{align*} \Delta_\lambda=\{(u_1,u_2)\in\mathbb R^2:u_1\ge 0,\ u_2\ge 0,\ u_1+u_2\le \lambda\}. \end{align*} The fixed point $[1:0:0]$ maps to $(0,0)$. A toric blow-up of size $\varepsilon$, with $0<\varepsilon<\lambda$, cuts the Delzant corner at affine distance $\varepsilon$ from $(0,0)$ along the two adjacent edges, by *Toric Blow-Up Cuts a Delzant Corner*. The two original edges adjacent to $(0,0)$ are $u_2=0$ and $u_1=0$. The point at distance $\varepsilon$ along $u_2=0$ is $(\varepsilon,0)$, and the point at distance $\varepsilon$ along $u_1=0$ is $(0,\varepsilon)$. The new edge is the line through $(\varepsilon,0)$ and $(0,\varepsilon)$. A point $(u_1,u_2)$ lies on this line exactly when \begin{align*} u_1+u_2=\varepsilon. \end{align*} Since the corner removed is the small triangle containing $(0,0)$, namely the region with $u_1\ge 0$, $u_2\ge 0$, and $u_1+u_2<\varepsilon$, the remaining moment polygon is \begin{align*} \widetilde\Delta_{\lambda,\varepsilon}=\{(u_1,u_2)\in\mathbb R^2:u_1\ge 0,\ u_2\ge 0,\ u_1+u_2\le\lambda,\ u_1+u_2\ge\varepsilon\}. \end{align*} Its vertices are obtained by solving pairs of boundary equations: \begin{align*} u_2=0,\ u_1+u_2=\varepsilon \quad\Longrightarrow\quad (u_1,u_2)=(\varepsilon,0), \end{align*} \begin{align*} u_2=0,\ u_1+u_2=\lambda \quad\Longrightarrow\quad (u_1,u_2)=(\lambda,0), \end{align*} \begin{align*} u_1=0,\ u_1+u_2=\lambda \quad\Longrightarrow\quad (u_1,u_2)=(0,\lambda), \end{align*} and \begin{align*} u_1=0,\ u_1+u_2=\varepsilon \quad\Longrightarrow\quad (u_1,u_2)=(0,\varepsilon). \end{align*} Thus the blow-up changes the triangle into a quadrilateral. The new edge runs from $(\varepsilon,0)$ to $(0,\varepsilon)$, so its affine length is $\varepsilon$ in the primitive edge direction $(1,-1)$. This edge represents the exceptional sphere, hence the exceptional sphere has symplectic area $\varepsilon$. The coordinate line corresponding to the bottom edge originally had affine length $\lambda$ from $(0,0)$ to $(\lambda,0)$; after the cut, its proper transform has length \begin{align*} \lambda-\varepsilon. \end{align*} The same calculation holds for the left edge, whose remaining segment runs from $(0,\varepsilon)$ to $(0,\lambda)$. Thus the blow-up introduces an exceptional sphere of area $\varepsilon$ and reduces by $\varepsilon$ the areas of the two coordinate lines passing through the blown-up fixed point. [/example] The blow-up example raises a more general construction problem: how can a Hamiltonian level set be collapsed to create a new symplectic boundary face? Lerman's symplectic cut answers this question and supplies the mechanism behind the corner-cutting picture. [quotetheorem:10074] The course uses the cut construction as a stated model operation; its proof applies symplectic reduction to $M\times\mathbb C$ with the diagonal circle action and moment map $H(x)-|z|^2$. The quotient at level $a$ keeps the region $H<a$ where $z\ne 0$ and replaces the boundary level $H=a$ by the reduced space $H^{-1}(a)/S^1$. The regular value assumption ensures that $H^{-1}(a)$ is a smooth hypersurface, so the collapsed boundary has the expected codimension-two reduced piece. Freeness ensures that the quotient $H^{-1}(a)/S^1$ is a smooth manifold; if the action has finite stabilisers, the same construction naturally produces an orbifold, and if the level is singular then the quotient may be stratified rather than smooth. Thus the theorem describes the clean smooth case, while more general cuts require singular or orbifold reduction. In toric geometry, the cutting Hamiltonian is an affine function on the moment polytope, so cutting at a regular level intersects the polytope with a half-space. [example: Cutting the Projective Plane Triangle] For the blow-up of $\mathbb{CP}^2$ at the vertex $(0,0)$, assume $0<\varepsilon<\lambda$ and start from the moment triangle \begin{align*} \Delta_\lambda=\{(u_1,u_2)\in\mathbb R^2:u_1\ge 0,\ u_2\ge 0,\ u_1+u_2\le\lambda\}. \end{align*} Take the affine Hamiltonian $H(u_1,u_2)=u_1+u_2$. The sublevel region being removed is \begin{align*} \Delta_\lambda\cap\{H<\varepsilon\}=\{(u_1,u_2):u_1\ge 0,\ u_2\ge 0,\ u_1+u_2<\varepsilon\}. \end{align*} Therefore the part of the original triangle that remains after the cut is obtained by keeping exactly those points of $\Delta_\lambda$ with $H\ge\varepsilon$, namely \begin{align*} \Delta_\lambda\cap\{H\ge\varepsilon\}=\{(u_1,u_2):u_1\ge 0,\ u_2\ge 0,\ u_1+u_2\le\lambda,\ u_1+u_2\ge\varepsilon\}. \end{align*} The new boundary face comes from the level set $H=\varepsilon$, so inside the triangle it is the segment \begin{align*} \{(u_1,u_2):u_1\ge 0,\ u_2\ge 0,\ u_1+u_2=\varepsilon\}. \end{align*} Its endpoints are found by setting one coordinate equal to zero. If $u_2=0$ and $u_1+u_2=\varepsilon$, then $u_1=\varepsilon$, giving $(\varepsilon,0)$. If $u_1=0$ and $u_1+u_2=\varepsilon$, then $u_2=\varepsilon$, giving $(0,\varepsilon)$. Thus the cut replaces the vertex $(0,0)$ by the new edge from $(\varepsilon,0)$ to $(0,\varepsilon)$, while the remaining old vertices are $(\lambda,0)$ and $(0,\lambda)$. By the *[Symplectic Cut Construction](/theorems/10074)*, the new facet is the reduced space at level $\varepsilon$, and by *Toric Blow-Up Cuts a Delzant Corner* this is exactly the polygon obtained from the toric one-point blow-up at $(0,0)$. The calculation shows that reduction, cutting, and blowing up give the same local modification of the moment triangle. [/example] ## What the Examples Teach The final problem is to extract the geometric lessons without relying on the full toric classification proof. The examples show that Hamiltonian torus actions convert symplectic geometry into integral affine convex geometry, but only in the presence of enough symmetry. [explanation: Reading Geometry from the Polytope] Vertices correspond to fixed points of the torus action, edges in a toric surface correspond to invariant symplectic spheres, and facet normals record circle subgroups that stabilise points [lying over](/theorems/2876) the corresponding facets. Affine lengths of edges encode symplectic areas, while the total Euclidean-looking shape must be interpreted in the integral affine coordinates determined by the torus lattice. For $\mathbb{CP}^1$, the interval remembers the area of the sphere. For $\mathbb{CP}^2$, the triangle remembers the three coordinate lines and their common area scale. For $\mathbb{CP}^1\times\mathbb{CP}^1$, the rectangle remembers the two independent ruling areas. For the blow-up of $\mathbb{CP}^2$, the cut triangle remembers both the exceptional divisor and the reduced areas of curves passing through the blown-up point. [/explanation] This chapter also prepares the transition to more flexible and more rigid phenomena. Darboux's theorem says that all symplectic manifolds look the same infinitesimally, while moment polytopes show that global Hamiltonian symmetry can make large-scale invariants computable. The next topics move beyond toric symmetry and ask which global features remain visible when no large torus action is present. Toric examples show how a large symmetry group can make global symplectic geometry almost combinatorial. The next chapter moves beyond that highly structured setting and asks which phenomena remain rigid even when no large torus action is available. # 11. Rigidity Beyond Darboux Chapter 4 proved Darboux's theorem: locally every symplectic manifold looks like the standard model. After the global toric examples of Chapter 10, this chapter explains why that local flexibility still does not extend to global embeddings. The central question is whether a symplectic embedding is constrained only by volume, or whether the symplectic form remembers two-dimensional information that volume cannot detect. The prerequisites are Darboux's theorem, the standard symplectic form on $\mathbb R^{2n}$, symplectic embeddings, and the basic relation between $\omega^n$ and symplectic volume. The answer is supplied by symplectic capacities and by Gromov's non-squeezing theorem. Capacities package embedding obstructions into numerical invariants, while non-squeezing gives the first striking obstruction: a ball cannot be symplectically squeezed through a cylinder whose symplectic radius is too small, even when there is abundant volume. ## Measuring Symplectic Size The problem is to assign a number to a symplectic manifold that is invariant under symplectomorphism, behaves predictably under scaling, and turns embeddings into inequalities. Volume has some of these features, but it scales like a $2n$-dimensional quantity. Non-squeezing shows that symplectic geometry also has a two-dimensional scale. [definition: Symplectic Embedding] Let $(M,\omega_M)$ and $(N,\omega_N)$ be symplectic manifolds. A smooth embedding $\varphi:M\to N$ is a symplectic embedding if \begin{align*} \varphi^*\omega_N=\omega_M. \end{align*} [/definition] A symplectic embedding preserves the full two-form, not just the induced volume form. To compare possible sources and targets without solving every embedding problem separately, we want a numerical invariant that decreases under symplectic embeddings and has the same scaling as symplectic area. [definition: Symplectic Capacity] Fix $n\ge 1$, and let $\mathrm{Symp}_{2n}$ denote the class of $2n$-dimensional symplectic manifolds. A symplectic capacity is a function \begin{align*} c:\mathrm{Symp}_{2n}\to [0,\infty] \end{align*} assigning to each symplectic manifold $(M,\omega)$ a value $c(M,\omega)$ such that: 1. If there is a symplectic embedding $(M,\omega_M)\hookrightarrow (N,\omega_N)$, then $c(M,\omega_M)\le c(N,\omega_N)$. 2. For every nonzero $\lambda\in\mathbb R$, $c(M,\lambda\omega)=|\lambda|c(M,\omega)$. 3. In standard symplectic space $(\mathbb R^{2n},\omega_0)$, the ball $B^{2n}(r)$ and cylinder $Z^{2n}(r)$ have the same capacity: \begin{align*} c(B^{2n}(r),\omega_0)=c(Z^{2n}(r),\omega_0)=\pi r^2. \end{align*} [/definition] The monotonicity axiom turns any capacity into an obstruction machine: if $c(M)>c(N)$, no symplectic embedding $M\hookrightarrow N$ exists. The conformality axiom records the fact that multiplying the symplectic form by $\lambda$ multiplies symplectic area by $|\lambda|$, not by $|\lambda|^n$. [example: Capacity Obstruction From Axioms] Consider the standard ball $B^{2n}(r)\subset \mathbb R^{2n}$ and the standard cylinder $Z^{2n}(R)=B^2(R)\times \mathbb R^{2n-2}$, with $r,R\ge 0$. Suppose $c$ is a symplectic capacity and that there is a symplectic embedding $B^{2n}(r)\hookrightarrow Z^{2n}(R)$. By monotonicity of $c$ under symplectic embeddings, \begin{align*} c(B^{2n}(r))\le c(Z^{2n}(R)). \end{align*} By the ball-cylinder normalisation axiom for capacities, \begin{align*} c(B^{2n}(r))=\pi r^2\quad\text{and}\quad c(Z^{2n}(R))=\pi R^2. \end{align*} Substituting these two normalised values into the monotonicity inequality gives \begin{align*} \pi r^2\le \pi R^2. \end{align*} Since $\pi>0$, division by $\pi$ gives $r^2\le R^2$. Because $r$ and $R$ are radii, they are nonnegative, so $r^2\le R^2$ is equivalent to $r\le R$. Thus the capacity axioms alone rule out a symplectic embedding of a ball into a cylinder with smaller symplectic radius. The obstruction depends on the two-dimensional area scale $\pi r^2$, not on the total $2n$-dimensional volume of the target. [/example] The normalisation axiom is not merely a convention: it says that balls and cylinders measure the same symplectic width. Gromov's theorem is the deep result ensuring that this normalisation is compatible with actual symplectic embeddings. ## Non-Squeezing Darboux's theorem might suggest that symplectic embeddings are locally flexible enough to behave like volume-preserving embeddings. The test case is a ball mapped into an infinite cylinder: volume gives no obstruction because the cylinder has infinite volume, but the first symplectic coordinate plane has a finite area scale. The problem is whether a symplectic map can squeeze the ball through a smaller two-dimensional symplectic window while spreading it in the remaining directions. [quotetheorem:10075] [citeproof:10075] The theorem is often called the symplectic camel theorem: the ball cannot pass through a smaller symplectic eye. Each hypothesis carries information. The restriction $n\ge 2$ separates the theorem from the two-dimensional case, where the obstruction is ordinary area rather than a new higher-dimensional rigidity phenomenon. The target must be a symplectic cylinder over one coordinate plane; the result does not say that every domain with first-coordinate area $\pi R^2$ has the same embedding behaviour. The embedding must preserve $\omega_0$, since smooth or volume-preserving embeddings can exploit the infinite length of the cylinder in the remaining coordinates. This is the first place where Darboux's local normal form is visibly insufficient, and it motivates capacities as global invariants designed to remember this two-dimensional obstruction. [remark: Dimension Two] When $n=1$, the statement reduces to the area obstruction for domains in $\mathbb R^2$. The new phenomenon begins in dimensions at least four, where volume and symplectic width are different kinds of data. [/remark] The theorem should be read as a rigidity theorem beyond Darboux. Darboux removes local invariants of the form, but it does not say that a large domain can be symplectically rearranged through an arbitrarily narrow symplectic window. [example: Volume Does Not Decide Embeddings] Let $n\ge 2$ and choose $r>R$. Define a linear map on $\mathbb R^{2n}$ by \begin{align*} F(q_1,p_1,\dots,q_n,p_n)=(a q_1,a p_1,b q_2,b p_2,\dots,b q_n,b p_n), \end{align*} where $0<a<R/r$ and $b=a^{-1/(n-1)}$. Then the first symplectic coordinate satisfies \begin{align*} (aq_1)^2+(ap_1)^2=a^2(q_1^2+p_1^2)\le a^2|x|^2<a^2r^2<R^2 \end{align*} for every $x\in B^{2n}(r)$, so $F(B^{2n}(r))\subset Z^{2n}(R)$. The Jacobian determinant of $F$ is \begin{align*} \det DF=a^2b^{2n-2}. \end{align*} Using $b=a^{-1/(n-1)}$, this becomes \begin{align*} a^2b^{2n-2}=a^2\left(a^{-1/(n-1)}\right)^{2n-2}=a^2a^{-2}=1. \end{align*} Thus $F$ preserves the standard $2n$-dimensional volume form. However, \begin{align*} F^*\omega_0=a^2\,dq_1\wedge dp_1+b^2\sum_{i=2}^n dq_i\wedge dp_i, \end{align*} which is not equal to $\omega_0$ because $a<1$ and hence $b>1$. So this volume-preserving embedding is not symplectic. By *Gromov Non-Squeezing Theorem*, a symplectic embedding $B^{2n}(r)\hookrightarrow Z^{2n}(R)$ would force $r\le R$, contradicting the choice $r>R$. The point is that preserving volume only controls the top exterior power, while preserving $\omega_0$ also preserves the two-dimensional symplectic area scale detected by the cylinder. [/example] ## Capacities From Non-Squeezing Once non-squeezing is known, the ball-cylinder comparison can be promoted into a systematic invariant. The guiding question is: how large a standard ball can fit symplectically into a given manifold? [definition: Gromov Width] Let $(M,\omega)$ be a $2n$-dimensional symplectic manifold. Its Gromov width is \begin{align*} c_G(M,\omega)=\sup\{\pi r^2: B^{2n}(r)\text{ symplectically embeds into }(M,\omega)\}. \end{align*} [/definition] This definition converts the existence problem for balls into a number. The supremum may be infinite, and it need not be achieved by an embedding of a ball of maximal radius. [quotetheorem:10076] [citeproof:10076] For this result, the deep input is concentrated in the cylinder upper bound. The fixed dimension hypothesis matters because the definition tests embeddings of standard $2n$-balls into $2n$-manifolds, so monotonicity is obtained by composition inside the same symplectic category. The conformality statement also uses that multiplying a symplectic form by a nonzero scalar rescales symplectic areas linearly; it is not a statement about Euclidean radius alone. The boundary cases are the standard ball and the standard cylinder: inclusion gives the lower bounds, while volume controls the ball upper bound and non-squeezing controls the cylinder upper bound. The theorem does not classify all symplectic embeddings into $M$; it records only the largest standard ball detectable by this particular capacity, which is why later capacities refine the Gromov width rather than replace embedding theory by a single number. [example: Gromov Width Of A Standard Cylinder] Let $Z^{2n}(R)=B^2(R)\times\mathbb R^{2n-2}$ with $R\ge 0$. If $x=(q_1,p_1,\dots,q_n,p_n)\in B^{2n}(R)$, then $|x|^2<R^2$, and hence \begin{align*} q_1^2+p_1^2\le q_1^2+p_1^2+\cdots+q_n^2+p_n^2=|x|^2<R^2. \end{align*} Thus $x\in Z^{2n}(R)$, so the standard inclusion gives a symplectic embedding $B^{2n}(R)\hookrightarrow Z^{2n}(R)$. By the definition of Gromov width, the number $\pi R^2$ belongs to the set whose supremum defines $c_G(Z^{2n}(R))$, and therefore \begin{align*} c_G(Z^{2n}(R))\ge \pi R^2. \end{align*} For the reverse inequality, let $\rho\ge 0$ be any radius such that there is a symplectic embedding $B^{2n}(\rho)\hookrightarrow Z^{2n}(R)$. By *Gromov Non-Squeezing Theorem*, this forces $\rho\le R$. Since both radii are nonnegative, squaring preserves the inequality: \begin{align*} \rho^2\le R^2. \end{align*} Multiplying by the positive number $\pi$ gives \begin{align*} \pi\rho^2\le \pi R^2. \end{align*} Every element $\pi\rho^2$ in the defining set for $c_G(Z^{2n}(R))$ is therefore at most $\pi R^2$, so \begin{align*} c_G(Z^{2n}(R))\le \pi R^2. \end{align*} Combining the two inequalities gives \begin{align*} c_G(Z^{2n}(R))=\pi R^2. \end{align*} The cylinder has infinite ordinary volume, but its Gromov width is exactly the two-dimensional symplectic area of its circular cross-section. [/example] ## Balls, Cylinders, And Ellipsoids Capacities become useful when they give fast tests for proposed embeddings. The simplest targets are balls, cylinders, and ellipsoids, because their defining inequalities expose the different roles of volume and symplectic radii. [definition: Symplectic Ellipsoid] For positive numbers $a_1,\dots,a_n$, the symplectic ellipsoid $E(a_1,\dots,a_n)\subset\mathbb C^n\cong\mathbb R^{2n}$ is \begin{align*} E(a_1,\dots,a_n)=\left\{z\in\mathbb C^n:\sum_{i=1}^n \frac{\pi |z_i|^2}{a_i}<1\right\}, \end{align*} equipped with the restriction of the standard symplectic form. [/definition] The parameter $a_i$ is the symplectic area of the $i$th coordinate disc at the corresponding intercept. Because ellipsoids are defined by explicit axis parameters, the first question is which inclusions and exclusions can already be read from those parameters. The next theorem records the elementary tests used before invoking sharper four-dimensional capacity sequences. [quotetheorem:10077] [citeproof:10077] This theorem gives only the first layer of ellipsoid geometry. The positivity of the parameters is necessary because a zero axis would collapse the defining domain and would no longer give a genuine symplectic ellipsoid. The coordinatewise hypothesis is a sufficient inclusion test, not a necessary [embedding criterion](/theorems/2195): for example, permuting the axes gives a symplectomorphic ellipsoid even though the displayed coordinatewise inequalities may fail before relabelling. The cylinder obstruction is also only necessary, since satisfying $\min_i a_i\le \pi R^2$ does not by itself construct an embedding into the cylinder. In dimension four, complete ellipsoid embedding criteria involve subtler capacity sequences, but these elementary tests already show how symplectic geometry sees coordinate-plane areas and why sharper invariants are needed. [example: A Simple Ellipsoid Test] In $\mathbb C^2$, write $A=\pi R^2$. The ball $B^4(R)$ is the ellipsoid $E(A,A)$, because \begin{align*} E(A,A)=\{(z_1,z_2):\pi |z_1|^2/A+\pi |z_2|^2/A<1\}. \end{align*} Substituting $A=\pi R^2$ gives \begin{align*} \pi |z_1|^2/A+\pi |z_2|^2/A=(|z_1|^2+|z_2|^2)/R^2, \end{align*} so the inequality is exactly $|z_1|^2+|z_2|^2<R^2$. Assume now that $a\le A$ and $b\le A$. If $(z_1,z_2)\in E(a,b)$, then \begin{align*} \pi |z_1|^2/a+\pi |z_2|^2/b<1. \end{align*} Since $a,b,A$ are positive and $a\le A$, $b\le A$, we have $1/A\le 1/a$ and $1/A\le 1/b$. Therefore \begin{align*} \pi |z_1|^2/A+\pi |z_2|^2/A\le \pi |z_1|^2/a+\pi |z_2|^2/b<1. \end{align*} Thus $(z_1,z_2)\in E(A,A)=B^4(R)$, proving $E(a,b)\subset B^4(R)$. If instead $a>A$ while $b$ is chosen small, for example $b<A^2/a$, then the standard four-dimensional ellipsoid volume formula gives \begin{align*} \operatorname{Vol}(E(a,b))=ab/2<A^2/2=\operatorname{Vol}(B^4(R)). \end{align*} So volume can be smaller than the ball's volume even though the coordinatewise inclusion test fails because the first axis satisfies $a>A$. In that situation, volume alone does not justify expecting a symplectic embedding; one must check capacity-type obstructions. [/example] The previous example tests when an entire ellipsoid sits inside a ball by comparing each axis. The complementary question asks how large a ball can be placed inside an ellipsoid, which is the local model for computing lower bounds on Gromov width. [example: Ball Into Ellipsoid] For $r>0$, the ball $B^{2n}(r)$ is the ellipsoid $E(\pi r^2,\dots,\pi r^2)$, because \begin{align*} E(\pi r^2,\dots,\pi r^2)=\left\{z\in\mathbb C^n:\sum_{i=1}^n \frac{\pi |z_i|^2}{\pi r^2}<1\right\}. \end{align*} For each $i$, \begin{align*} \frac{\pi |z_i|^2}{\pi r^2}=\frac{|z_i|^2}{r^2}. \end{align*} Hence the defining inequality becomes \begin{align*} \sum_{i=1}^n \frac{|z_i|^2}{r^2}<1. \end{align*} Multiplying by $r^2>0$ gives \begin{align*} \sum_{i=1}^n |z_i|^2<r^2, \end{align*} which is exactly the standard ball condition. Now assume $\pi r^2\le \min_i a_i$. Then $\pi r^2\le a_i$ for every $i$, so, since all quantities are positive, \begin{align*} \frac{1}{a_i}\le \frac{1}{\pi r^2}. \end{align*} If $z\in B^{2n}(r)$, then \begin{align*} \sum_{i=1}^n |z_i|^2<r^2. \end{align*} Using the reciprocal inequality term by term gives \begin{align*} \sum_{i=1}^n \frac{\pi |z_i|^2}{a_i}\le \sum_{i=1}^n \frac{\pi |z_i|^2}{\pi r^2}. \end{align*} The right-hand side is \begin{align*} \sum_{i=1}^n \frac{|z_i|^2}{r^2}=\frac{1}{r^2}\sum_{i=1}^n |z_i|^2<1. \end{align*} Therefore $z\in E(a_1,\dots,a_n)$, and so \begin{align*} B^{2n}(r)\subset E(a_1,\dots,a_n). \end{align*} Thus every $r$ with $\pi r^2\le \min_i a_i$ contributes the value $\pi r^2$ to the defining set for the Gromov width of $E(a_1,\dots,a_n)$. This gives the lower bound $c_G(E(a_1,\dots,a_n))\ge \min_i a_i$, and it is exactly the bound detected by comparing the coordinate axes; sharper capacities are needed for more delicate four-dimensional embedding questions. [/example] ## Rigidity After Darboux The chapter's lesson is that symplectic geometry has no local invariants but has strong global embedding obstructions. Darboux says every point has coordinates in which $\omega$ is standard; non-squeezing says that no coordinate trick can force a ball through a cylinder of smaller symplectic radius. [explanation: The Two-Dimensional Nature Of Symplectic Rigidity] The expression $\omega_0=\sum_i dq_i\wedge dp_i$ pairs coordinates into symplectic planes. A symplectic map may mix all coordinates, but it cannot erase the area information detected by pseudo-holomorphic curves and encoded by capacities. This is why the obstruction in non-squeezing is $\pi r^2\le \pi R^2$, not a comparison of $2n$-dimensional volumes. This distinction is the starting point for modern symplectic rigidity. Later constructions refine the Gromov width into families of capacities, especially in dimension four, and connect embedding questions with Reeb dynamics, holomorphic curves, and Floer-theoretic invariants. [/explanation] Darboux's theorem says the local model is always standard, but the global embedding problem shows that symplectic geometry still has strong constraints. Those rigidity questions lead naturally to the more special situation of integrable systems, where enough conserved quantities allow the geometry to be described by action-angle coordinates. # 12. Integrable Systems and Action-Angle Coordinates ## Complete Involutive Families and Liouville Integrability This chapter is the course's transition from general Hamiltonian dynamics to the special systems that can be solved by geometric coordinates. It assumes the preceding material on symplectic manifolds from Chapter 2, Hamiltonian vector fields and Poisson brackets from Chapter 5, Lagrangian submanifolds from Chapters 1 and 3, and Darboux-type local normal forms from Chapter 4. The main goal is to explain why $n$ independent conserved quantities in involution on a $2n$-dimensional phase space produce invariant Lagrangian tori, and how action-angle coordinates turn the dynamics on those tori into linear motion. A Hamiltonian $H:M\to\mathbb R$ on a $2n$-dimensional symplectic manifold $(M,\omega)$ gives a single vector field $X_H$. The problem of integrability asks when the motion of $X_H$ can be reconstructed from conserved quantities rather than from solving a general system of differential equations. Having many conserved quantities is not enough: if their Hamiltonian flows interfere with each other, the common level sets need not carry a coherent torus action. The symplectic form turns functions into vector fields, so the right compatibility condition between conserved quantities is expressed by the Poisson bracket. [definition: Poisson Commuting Functions] Let $(M,\omega)$ be a symplectic manifold. Smooth functions $f,g:M\to\mathbb R$ Poisson commute if \begin{align*} \{f,g\} := \omega(X_f,X_g)=0. \end{align*} [/definition] Poisson commuting functions are conserved along each other's Hamiltonian flows, since \begin{align*} X_f(g)=dg(X_f)=\{g,f\}. \end{align*} This statement is weaker than the geometric compatibility needed for integrability: a first integral may be constant along one flow without producing a commuting multi-parameter family of flows. This motivates the next structural question: does conservation in both directions also force the two Hamiltonian flows to commute as flows on phase space? [quotetheorem:10078] [citeproof:10078] This theorem explains why involution is stronger than ordinary conservation: it gives a commuting family of vector fields, so the flows can be assembled into local $\mathbb R^k$-actions rather than treated separately. The hypothesis $\{f,g\}=0$ is essential; if the bracket is nonzero, the commutator of the Hamiltonian vector fields records precisely that obstruction. The theorem does not say that the flows are complete or that their orbits are compact, so additional hypotheses are still needed before invariant tori appear. The natural next definition packages the exact amount of data needed for a $2n$-dimensional phase space, namely $n$ independent commuting Hamiltonians. [definition: Completely Integrable System] Let $(M^{2n},\omega)$ be a symplectic manifold. A completely integrable system on $M$ is a smooth map \begin{align*} F=(f_1,\dots,f_n):M\to\mathbb R^n \end{align*} such that the functions Poisson commute pairwise and $dF_x:T_xM\to\mathbb R^n$ has rank $n$ on a dense open set. [/definition] The rank condition says that the $n$ conserved quantities are independent at regular points. Without this condition, repeated or functionally dependent integrals can make the common level set too large, so the system has not really reduced half of the phase-space directions. Since the manifold has dimension $2n$, their common regular level sets have dimension $n$; this motivates asking whether these level sets are the Lagrangian leaves predicted by symplectic linear algebra. [quotetheorem:10079] [citeproof:10079] The statement turns an analytic condition, the existence of conserved quantities, into a geometric conclusion: the regular phase space is foliated by Lagrangian leaves. The regular value hypothesis is essential; at singular fibers the dimension can jump and the fiber can fail to be a smooth submanifold. The conclusion is also local over the regular value set and does not imply compactness, so a regular fiber may be $\mathbb R^n$ rather than a torus. The next examples show the two main sources of such systems in the course, separable Hamiltonians and moment maps. [example: Harmonic Oscillator] On $(\mathbb R^{2n},\omega_0=\sum_{k=1}^n dq_k\wedge dp_k)$, let \begin{align*} H_i(q,p)=\frac{1}{2}(p_i^2+q_i^2),\qquad F=(H_1,\dots,H_n). \end{align*} With the convention $\iota_{X_f}\omega_0=df$, the coordinate formula for the Poisson bracket is \begin{align*} \{f,g\}=\sum_{k=1}^n\left(\frac{\partial f}{\partial q_k}\frac{\partial g}{\partial p_k}-\frac{\partial f}{\partial p_k}\frac{\partial g}{\partial q_k}\right). \end{align*} For $H_i$, the only nonzero partial derivatives are $\partial H_i/\partial q_i=q_i$ and $\partial H_i/\partial p_i=p_i$. Hence, if $i\neq j$, every summand in $\{H_i,H_j\}$ contains a derivative of $H_i$ or $H_j$ in the wrong coordinate pair, so \begin{align*} \{H_i,H_j\}=0. \end{align*} Also $\{H_i,H_i\}=q_i p_i-p_i q_i=0$, so the functions Poisson commute pairwise. For $c_i>0$, the equation $H_i=c_i$ is exactly \begin{align*} \frac{1}{2}(p_i^2+q_i^2)=c_i, \end{align*} equivalently \begin{align*} p_i^2+q_i^2=2c_i. \end{align*} Thus the $i$th coordinate pair lies on the circle of radius $\sqrt{2c_i}$ in the $(q_i,p_i)$-plane, and \begin{align*} F^{-1}(c)=S^1_{\sqrt{2c_1}}\times\cdots\times S^1_{\sqrt{2c_n}}\cong T^n. \end{align*} For $H=H_1+\cdots+H_n=\frac{1}{2}\sum_i(p_i^2+q_i^2)$, the Hamiltonian vector field satisfies \begin{align*} X_H=\sum_{i=1}^n\left(p_i\frac{\partial}{\partial q_i}-q_i\frac{\partial}{\partial p_i}\right), \end{align*} because $\iota_{X_H}\omega_0=dH=\sum_i(q_i\,dq_i+p_i\,dp_i)$. Therefore \begin{align*} \dot q_i=p_i,\qquad \dot p_i=-q_i. \end{align*} Writing $q_i=r_i\cos\theta_i$ and $p_i=-r_i\sin\theta_i$ gives \begin{align*} \dot q_i=-r_i\sin\theta_i\,\dot\theta_i=p_i=-r_i\sin\theta_i \end{align*} and \begin{align*} \dot p_i=-r_i\cos\theta_i\,\dot\theta_i=-q_i=-r_i\cos\theta_i, \end{align*} so $\dot\theta_i=1$ wherever the displayed factors are nonzero, and hence everywhere by continuity along the circle. The flow rotates each circle at unit speed, so on $F^{-1}(c)\cong T^n$ the motion is linear with frequency vector $(1,\dots,1)$. [/example] The harmonic oscillator is the model in which action-angle coordinates can be written by inspection. The general theorem says that compact regular fibers of any completely integrable system have the same local structure, although the coordinates may not extend globally. ## Lagrangian Torus Fibrations and Action Variables Once the regular fibers are known to be Lagrangian, the next problem is to find coordinates adapted to the fibration. Ordinary Darboux coordinates flatten the symplectic form near a point, but they do not remember the torus fibration. Action variables are the special coordinates that record the symplectic areas swept out by cycles on the fibers. [definition: Lagrangian Torus Fibration] A Lagrangian torus fibration is a surjective submersion \begin{align*} \pi:(M^{2n},\omega)\to B^n \end{align*} such that each fiber $\pi^{-1}(b)$ is a compact connected Lagrangian submanifold diffeomorphic to $T^n$. [/definition] The submersion condition supplies a smooth family of tori, while the Lagrangian condition is what makes the fibers compatible with the symplectic form. This motivates the search for base coordinates defined by periods around cycles on the fibers, since those periods are invariant under Hamiltonian motion along each torus. [definition: Action Coordinates] Let $\pi:(M^{2n},\omega)\to B^n$ be a Lagrangian torus fibration over a contractible open set $U\subset B$, and suppose a $1$-form $\lambda\in\Omega^1(\pi^{-1}(U))$ is chosen with $d\lambda=\omega|_{\pi^{-1}(U)}$. For a smoothly varying basis $\gamma_1(b),\dots,\gamma_n(b)$ of $H_1(\pi^{-1}(b);\mathbb Z)$, the action functions are the maps $I_i:U\to\mathbb R$ defined by \begin{align*} I_i(b)=\frac{1}{2\pi}\int_{\gamma_i(b)}\lambda, \qquad i=1,\dots,n. \end{align*} [/definition] Changing the homology basis changes the action variables by an integral linear transformation. On a sufficiently small contractible neighbourhood where the closed $1$-form ambiguity in the primitive has fixed fiberwise periods, changing the primitive changes the actions by constants determined by those periods. This introduces the next problem: after choosing action variables on the base, can they be completed by periodic angle variables along the fibers so that the symplectic form is standard? [example: Action for the One-Dimensional Harmonic Oscillator] For $(\mathbb R^2,dq\wedge dp)$ with $H(q,p)=\frac{1}{2}(p^2+q^2)$, take $\lambda=p\,dq$. Fix $E>0$ and set $r=\sqrt{2E}$. The energy level $H=E$ is the circle $p^2+q^2=2E=r^2$, and the Hamiltonian orientation is given by the parametrization $q=r\cos\theta$ and $p=-r\sin\theta$ for $0\leq \theta\leq 2\pi$. Along this parametrized circle, \begin{align*} dq=-r\sin\theta\,d\theta. \end{align*} Therefore \begin{align*} p\,dq=(-r\sin\theta)(-r\sin\theta\,d\theta)=r^2\sin^2\theta\,d\theta. \end{align*} Integrating over one full circuit gives \begin{align*} \int_{H=E}p\,dq=\int_0^{2\pi}r^2\sin^2\theta\,d\theta. \end{align*} Since $\int_0^{2\pi}\sin^2\theta\,d\theta=\pi$, this becomes \begin{align*} \int_{H=E}p\,dq=r^2\pi=(2E)\pi=2\pi E. \end{align*} Thus the action coordinate is \begin{align*} I=\frac{1}{2\pi}\int_{H=E}\lambda=\frac{1}{2\pi}(2\pi E)=E. \end{align*} With this orientation, the conjugate angle is the oscillator phase $\theta$, so the action records the energy and the angle records position along the invariant circle. [/example] This computation shows the intended normal form: the action is constant on the invariant circle, and the angle parametrises motion along it. For a general completely integrable system, the obstruction is that regular common level sets need not come with preferred coordinates, periods, or cycles a priori. Near a compact regular fiber, one must show that the [commuting Hamiltonian flows](/theorems/10078) organize the fiber as a torus and that the surrounding symplectic form can be written in action-angle coordinates. [quotetheorem:10080] [proofunderconstruction:10080] In these coordinates the geometry has been simplified as far as possible near $L$. Compactness and regularity are doing real work here: without compactness the period lattice need not be a full lattice, and without regularity the torus fibration can break at singular leaves. The theorem is local near one fiber, so it does not choose compatible actions over a large base containing nontrivial loops. The remaining information in the Hamiltonian is the function $H(I)$, and this motivates writing Hamilton's equations in the new coordinates. [quotetheorem:6999] [citeproof:6999/symplectic-geometry-i-foundations] The theorem gives the precise meaning of solving the system by quadratures in this setting: after the symplectic change of coordinates, the differential equation is integrated by affine functions of time on a torus. Its assumption $H=H(I)$ is essential; angle dependence would reintroduce forces in the $I$-directions and destroy invariance of the coordinate tori. The theorem also describes the motion on a fixed torus, not whether an orbit is periodic or dense; that depends on rational relations among the frequency components. This prepares the global question of whether these local straight-line descriptions can be patched across different tori. ## Arnold-Liouville Theorem and Monodromy as a First Global Warning The local action-angle theorem is powerful, but it is not a global classification theorem. The next question is what prevents the local action variables from being chosen consistently over all regular values. The answer is already visible in examples: the homology basis of the tori can return transformed after continuation around a loop in the base. [quotetheorem:1353] [citeproof:1353] This theorem gathers the structural conclusions into one statement: compact regular common level sets are not arbitrary Lagrangian manifolds, but invariant tori with translation dynamics. Each hypothesis excludes a real failure mode: noncompact regular fibers can be cylinders or copies of $\mathbb R^n$, and singular fibers may have pinches or lower-dimensional strata. The theorem does not classify how the neighbouring tori fit together over a large region of the base. Its hypotheses matter because singular fibers are where much of the interesting global topology of integrable systems is stored, and they are also where obstructions to global action-angle coordinates first appear. [example: Toric Moment Maps As Integrable Systems] Let $e_1,\dots,e_n$ be the standard basis of the Lie algebra $\mathfrak t^n$, and write \begin{align*} \mu_i=\langle \mu,e_i\rangle . \end{align*} By the defining property of the moment map, the fundamental vector field $(e_i)_M$ satisfies \begin{align*} \iota_{(e_i)_M}\omega=d\mu_i. \end{align*} With the chapter's convention $\iota_{X_f}\omega=df$, this means \begin{align*} X_{\mu_i}=(e_i)_M. \end{align*} Because $T^n$ is abelian, its coadjoint action is trivial, so an equivariant moment map is invariant under the $T^n$-action. Therefore, for all $i,j$, \begin{align*} X_{\mu_j}(\mu_i)=(e_j)_M(\mu_i)=0. \end{align*} Using $d\mu_i(X_{\mu_j})=\omega(X_{\mu_i},X_{\mu_j})$, we get \begin{align*} \{\mu_i,\mu_j\}=\omega(X_{\mu_i},X_{\mu_j})=d\mu_i(X_{\mu_j})=X_{\mu_j}(\mu_i)=0. \end{align*} Thus the components of $\mu$ Poisson commute pairwise. On the open set where the $T^n$-action is free, the fundamental vector fields $(e_1)_M,\dots,(e_n)_M$ are linearly independent. To see that $d\mu_1,\dots,d\mu_n$ are independent there, suppose \begin{align*} a_1d\mu_1+\cdots+a_nd\mu_n=0. \end{align*} For every tangent vector $Y$, \begin{align*} 0=\sum_{i=1}^n a_i d\mu_i(Y)=\sum_{i=1}^n a_i\omega((e_i)_M,Y)=\omega\left(\sum_{i=1}^n a_i(e_i)_M,Y\right). \end{align*} Nondegeneracy of $\omega$ gives \begin{align*} \sum_{i=1}^n a_i(e_i)_M=0. \end{align*} Since the action is free, the fundamental vector fields are independent, so $a_1=\cdots=a_n=0$. Hence $d\mu$ has rank $n$ on the free locus, and $\mu$ is a completely integrable system there. At a point in the free locus, the vectors $X_{\mu_i}=(e_i)_M$ span the tangent space to the $T^n$-orbit. Since each $\mu_i$ is constant along every torus orbit, each orbit lies inside one fiber of $\mu$; both have dimension $n$, so the connected regular fibers are exactly the $T^n$-orbits. The moment polytope records what happens away from the free locus: at boundary faces some circle subgroups acquire fixed directions, so the torus fibers lose dimensions and degenerate. [/example] Toric systems are the most rigid and combinatorial examples: the base has a polytope structure, and singular fibers occur in a controlled way. The spherical pendulum shows a different phenomenon, where a single singular fiber affects the topology of the regular fibration around it. [example: Spherical Pendulum] Represent $T^*S^2$ as pairs $(q,p)\in\mathbb R^3\times\mathbb R^3$ with $\lVert q\rVert=1$ and $q\cdot p=0$. Take the vertical unit vector to be $e_3=(0,0,1)$, and use the Hamiltonian \begin{align*} H(q,p)=\frac{1}{2}(p_1^2+p_2^2+p_3^2)+q_3. \end{align*} Rotation about the vertical axis has momentum \begin{align*} J(q,p)=q_1p_2-q_2p_1. \end{align*} Using the canonical bracket in the ambient cotangent coordinates, the derivatives of $H$ are $\partial H/\partial q=(0,0,1)$ and $\partial H/\partial p=(p_1,p_2,p_3)$. The derivatives of $J$ are $\partial J/\partial q=(p_2,-p_1,0)$ and $\partial J/\partial p=(-q_2,q_1,0)$. Therefore \begin{align*} \{H,J\}=0(-q_2)+0(q_1)+1\cdot 0-p_1p_2-p_2(-p_1)-p_3\cdot 0. \end{align*} The only remaining terms cancel: \begin{align*} \{H,J\}=-p_1p_2+p_1p_2=0. \end{align*} Thus energy and vertical angular momentum Poisson commute, so $(H,J)$ is an integrable system on the regular locus. At a regular value $(E,j)$, the differentials $dH$ and $dJ$ are independent. Since $T^*S^2$ has dimension $4$, the common level set \begin{align*} (H,J)^{-1}(E,j) \end{align*} has dimension $4-2=2$. When a regular component is compact and connected, the *Liouville-Arnold Theorem* identifies it with a two-torus $T^2$, and the flows of $H$ and $J$ become translations in action-angle coordinates. The upright equilibrium is $q=e_3$, $p=0$. At this point, \begin{align*} H(e_3,0)=1. \end{align*} Also, \begin{align*} J(e_3,0)=0. \end{align*} Hence $(1,0)$ is a singular value of the energy-momentum map. A loop in the regular value set going once around this singular value transports a homology basis $(\alpha,\beta)$ of $H_1(T^2;\mathbb Z)$ to the basis with $\alpha$ unchanged and $\beta$ replaced by $\beta+\alpha$. Relative to the ordered basis $(\alpha,\beta)$, this is the nontrivial element of $GL(2,\mathbb Z)$ whose first column is $(1,0)$ and whose second column is $(1,1)$. The spherical pendulum therefore has local invariant tori on regular levels, but the singular upright fiber prevents one global choice of action-angle coordinates around it. [/example] This basis change is the simplest obstruction to global action-angle coordinates. Locally the period lattice can be chosen smoothly, but globally it may have nontrivial monodromy. [definition: Monodromy Of A Lagrangian Torus Fibration] Let $\pi:M\to B$ be a Lagrangian torus fibration and fix $b\in B$. The monodromy representation is the homomorphism \begin{align*} \rho:\pi_1(B,b)\to \operatorname{Aut}(H_1(\pi^{-1}(b);\mathbb Z)) \end{align*} obtained from the Gauss-Manin local system of the homology groups of the fibers. [/definition] If the monodromy is nontrivial, there is no single global choice of action variables on the whole base. Thus the Liouville-Arnold theorem should be read as a local normal form theorem near a regular compact fiber, not as a promise that all regular fibers can be simultaneously straightened. [remark: Global Meaning Of Action-Angle Coordinates] Global action-angle coordinates require more than local integrability. They require the torus bundle to be topologically simple enough and the integral affine structure on the base to have no monodromy obstruction. This is the first warning that symplectic geometry has no local invariants near points but still has strong global invariants for fibrations, embeddings, and Hamiltonian dynamics. [/remark] The course now ends with a typical symplectic pattern. Darboux's theorem removed local pointwise structure, moment maps and reduction organised symmetric systems, and non-squeezing revealed global rigidity. Integrable systems combine these themes: locally the dynamics becomes linear on tori, while globally the arrangement of those tori carries topology that cannot be erased by symplectic coordinates. ## Beyond This Course This course has built a pathway from linear symplectic algebra to global rigidity. A useful next step is to revisit each major object from the boundary outward: Liouville domains lead to contact manifolds, Reeb flows, and filling questions; Hamiltonian actions lead to equivariant geometry, representation theory, and geometric invariant theory; symplectic capacities lead to embedding obstructions, displacement energy, and Floer-theoretic invariants; integrable systems lead to Poisson geometry and singular Lagrangian torus fibrations. For an internal reading path, continue with [Cambridge III Differential Geometry](/page/Cambridge%20III%20Differential%20Geometry) for the manifold background behind forms and vector fields, [Cambridge II Dynamical Systems](/page/Cambridge%20II%20Dynamical%20Systems) for flows and recurrence, and [Cambridge III Riemannian Geometry](/page/Cambridge%20III%20Riemannian%20Geometry) for a contrasting geometric structure governed by metrics rather than closed nondegenerate two-forms. The next specialist topics are contact geometry, Hamiltonian group actions, symplectic reduction, and Floer homology. These topics preserve the central lesson of the page: local normal forms are powerful, but the global arrangement of symplectic data carries information that coordinates alone cannot remove. ## References and Further Reading A standard first reference is Ana Cannas da Silva, *Lectures on Symplectic Geometry*, which develops Darboux's theorem, Hamiltonian actions, moment maps, and reduction at a level close to this course. For a broader and more detailed treatment, see Dusa McDuff and Dietmar Salamon, *Introduction to Symplectic Topology*. For contact geometry and the boundary viewpoint on Liouville domains, see Hansjorg Geiges, *An Introduction to Contact Topology*. For integrable systems and torus fibrations, see Michèle Audin, *Torus Actions on Symplectic Manifolds*. For Floer-theoretic directions beyond this course, McDuff and Salamon's *J-holomorphic Curves and Symplectic Topology* is a standard continuation.