# Basic Symplectic Geometry

Definition 6.1

Let ${\displaystyle M}$ be a smooth manifold. A 2-form ${\displaystyle \omega \in \Omega ^{2}(M)}$ is called a "symplectic form" if ${\displaystyle d\omega =0}$ and ${\displaystyle \omega }$ is nondegenerate. In this case, we call the pair ${\displaystyle (M,\omega )}$ a "symplectic manifold".

One can think of the closed condition as follows. By Stokes' Theorem, for any 3-dimensional submanifold with boundary ${\displaystyle V\subseteq M}$, we have

${\displaystyle \int _{\partial V}\omega =\int _{V}d\omega =0.}$
In particular, if ${\displaystyle \Sigma _{1}}$ and ${\displaystyle \Sigma _{2}}$ are surfaces inside ${\displaystyle M}$ such that ${\displaystyle \Sigma _{1}\cup \Sigma _{2}=\partial V}$ for some ${\displaystyle V}$, then (provided we orient things correctly),
${\displaystyle \int _{\Sigma _{1}}\omega =\int _{\Sigma _{2}}\omega .}$

What about non-degeneracy? Since ${\displaystyle \omega }$ is a 2-form, for each ${\displaystyle p\in M}$, we get an alternating map

${\displaystyle \omega :T_{p}M\times T_{p}M\to \mathbb {R} .}$
This defines a map
{\displaystyle {\begin{aligned}T_{p}M&\to T_{p}^{*}M\\v&\mapsto \iota _{v}\omega ,\end{aligned}}}
where ${\displaystyle \iota _{v}\omega }$ is defined by
${\displaystyle \langle \iota _{v}\omega ,w\rangle =\omega (v,w)\quad {\text{for all}}\quad w\in T_{p}M.}$
We say that ${\displaystyle \omega }$ is "nondegenerate" if the map ${\displaystyle T_{p}M\to T_{p}^{*}M}$ is an isomorphism for all ${\displaystyle p\in M}$.

Exercise 6.1

Let ${\displaystyle V}$ be a vector space with ${\displaystyle \mathrm {dim} _{\mathbb {R} }V=2n}$. Show that ${\displaystyle \omega \in \Lambda ^{2}V^{*}}$ satisfies ${\displaystyle \omega ^{n}\neq 0}$ if and only if the induced map ${\displaystyle V\to V^{*}}$ as described above is an isomorphism.

Suppose we also have an inner product ${\displaystyle \langle \,,\,\rangle }$, and so an isomorphism ${\displaystyle V\to V^{*}}$. Show that, if the composition ${\displaystyle J:V{\overset {\omega }{\longrightarrow }}V^{*}{\overset {\langle ,\rangle }{\longrightarrow }}V}$ is orthogonal with respect to ${\displaystyle \langle \,,\,\rangle }$, then ${\displaystyle J^{2}=-1}$, so ${\displaystyle J}$ defines a complex structure on ${\displaystyle V}$.

Conversely, suppose we have a complex structure ${\displaystyle J}$ and inner product ${\displaystyle \langle \,,\,\rangle }$ on ${\displaystyle V}$, which is compatible with ${\displaystyle J}$ in the sense that ${\displaystyle \langle Jv,Jw\rangle =\langle v,w\rangle }$. Prove that ${\displaystyle \omega (u,v):=\langle u,Jv\rangle }$ is a skew non-degenerate 2-form.

Example 6.1

The following are examples of symplectic manifolds.

1. Kähler manifolds. If ${\displaystyle X}$ is an almost Hermitian manifold with Hermitian form ${\displaystyle \omega }$, then ${\displaystyle \omega }$ is a nondegenerate 2-form by Exercise on local Kahler. So ${\displaystyle \omega }$ is a symplectic form if and only if ${\displaystyle d\omega =0}$, i.e., if and only if ${\displaystyle X}$ is an almost Kähler manifold.
2. Cotangent bundles. Consider ${\displaystyle \pi :T^{*}N\to N}$. This has a canonical 1-form on it, defined by ${\displaystyle \theta _{(n,\xi )}=\pi ^{*}\xi }$ for ${\displaystyle n\in N}$ and ${\displaystyle \xi \in T_{n}^{*}N}$. Then (claim) the 2-form ${\displaystyle \omega =d\theta }$ is actually a symplectic 2-form. So ${\displaystyle (T^{*}N,\omega )}$ is naturally a symplectic manifold (and is a good model for the phase space of classical physics---think about the case ${\displaystyle N=\mathbb {R} ^{n}}$)

Exercise 6.2

Take local coordinates ${\displaystyle x_{i}}$ on ${\displaystyle U\cong \mathbb {R} ^{n}\subset N}$. Let ${\displaystyle y_{i}}$ be coordinates on the fibres of ${\displaystyle T^{*}U\cong \mathbb {R} ^{n}\times (\mathbb {R} ^{n})^{*}\subset T^{*}N}$.

Check that in these coordinates ${\displaystyle \theta =\Sigma _{i=1}^{n}y_{i}dx_{i}}$ and ${\displaystyle \omega :=d\theta =\Sigma _{i=1}^{n}dy_{i}\wedge dx_{i}}$ and that this is indeed non-degenerate.

The construction of the isomorphism ${\displaystyle T_{p}M\to T_{p}^{*}M}$ above globalises to give an isomorphism between vector fields and 1-forms: given a vector field ${\displaystyle v}$ on a symplectic manifold ${\displaystyle (M,\omega )}$, we get a 1-form ${\displaystyle \iota _{v}\omega }$ satisfying

${\displaystyle \langle \iota _{v}\omega ,w\rangle =\omega (v,w)}$
for all vector fields ${\displaystyle w}$ on ${\displaystyle M}$.

Definition 6.2

We say that a vector field ${\displaystyle v}$ on ${\displaystyle (M,\omega )}$ is "symplectic" if the 1-form ${\displaystyle \iota _{v}\omega }$ is closed, i.e., ${\displaystyle d\iota _{v}\omega =0}$. We say that ${\displaystyle v}$ is "Hamiltonian" if ${\displaystyle \iota _{v}\omega }$ is exact, i.e., ${\displaystyle \iota _{v}\omega =-dH}$ for some smooth function ${\displaystyle H}$ on ${\displaystyle M}$. In this case, we call the function ${\displaystyle H}$ a "Hamiltonian" for the vector field ${\displaystyle v}$. We write ${\displaystyle \mathrm {symp} (M,\omega )}$ and ${\displaystyle \mathrm {ham} (M,\omega )}$ respectively for the spaces of symplectic and Hamiltonian vector fields on ${\displaystyle M}$.

Remark 6.1

Note that the Hamiltonian ${\displaystyle H}$ is only defined up to constant translation. Indeed ${\displaystyle C^{\infty }(M,\mathbb {R} )\to \mathrm {ham} (M,\omega )}$ given by ${\displaystyle H\mapsto v_{H}}$ (where ${\displaystyle v_{H}}$ is the unique vector field satisfying ${\displaystyle \iota _{v_{H}}\omega =-dH}$) has kernel precisely the constants ${\displaystyle \mathbb {R} }$.

Example 6.2

Consider ${\displaystyle M=\mathbb {R} ^{2}}$ with coordinates ${\displaystyle (p,q)}$, and the symplectic form ${\displaystyle \omega =dp\wedge dq}$. Consider the Hamiltonian ${\displaystyle H={\frac {1}{2}}(p^{2}+q^{2})}$. We have

${\displaystyle dH=pdp+qdq}$
For any vector field ${\displaystyle v}$
${\displaystyle {\text{if}}\quad v=a{\frac {\partial }{\partial p}}+b{\frac {\partial }{\partial q}}\quad {\text{then}}\quad \iota _{v}\omega =adq-bdp.}$
So taking ${\displaystyle b=p}$ and ${\displaystyle a=-q}$, we get
${\displaystyle -dH=\iota _{v}\omega \quad {\text{for}}\quad v=-q{\frac {\partial }{\partial p}}+p{\frac {\partial }{\partial q}}.}$
So ${\displaystyle v}$ above is a Hamiltonian vector field.

Remark 6.2

On ${\displaystyle \mathbb {R} ^{2}}$, every symplectic vector field is Hamiltonian, since every closed ${\displaystyle 1}$-form is exact.

Example 6.3

Consider the the torus ${\displaystyle M=\mathbb {R} ^{2}/\mathbb {Z} ^{2}}$. If ${\displaystyle (\theta _{1},\theta _{2})}$ are coordinates on ${\displaystyle \mathbb {R} ^{2}}$, then the symplectic form ${\displaystyle d\theta _{1}\wedge d\theta _{2}}$ descends to a symplectic form ${\displaystyle \omega }$ on ${\displaystyle M}$. The vector field ${\displaystyle v={\frac {\partial }{\partial \theta _{2}}}}$ has ${\displaystyle \iota _{v}\omega =-d\theta _{1}}$, which is closed but not exact (as ${\displaystyle \theta _{1}}$ is not a function on ${\displaystyle M}$). Hence ${\displaystyle v}$ is symplectic but not Hamiltonian.

Remark 6.3

Let ${\displaystyle (X,\omega )}$ be a symplectic manifold and assume that ${\displaystyle \phi _{t}:X\to X}$ is a smoothly varying family of diffeomorphisms. Then we have

${\displaystyle {\frac {d}{dt}}\phi _{t}^{*}\omega =d\iota _{v_{t}}\omega +\iota _{v_{t}}d\omega ,}$
where ${\displaystyle v_{t}={\frac {d\phi _{t}}{dt}}}$ is the vector field defined by ${\displaystyle \phi _{t}}$ and ${\displaystyle \iota _{v_{t}}d\omega }$ is the 2-form given by ${\displaystyle (\iota _{v_{t}}d\omega )(u,w)=(d\omega )(v_{t},u,w)}$. (The left hand side of this equation is called the "Lie derivative" of ${\displaystyle \omega }$ in the direction ${\displaystyle v_{t}}$.) In particular, since ${\displaystyle \omega }$ is symplectic, ${\displaystyle {\frac {d}{dt}}\phi _{t}^{*}\omega =d\iota _{v_{t}}\omega }$ is ${\displaystyle 0}$ if and only if ${\displaystyle v_{t}}$ is a symplectic vector field. In particular, specialising to the case where ${\displaystyle v_{t}=v}$ is constant in ${\displaystyle t}$, we see that ${\displaystyle v}$ is a symplectic vector field if and only if flowing along ${\displaystyle v}$ preserves the symplectic form ${\displaystyle \omega }$.

Example 6.4

Consider ${\displaystyle M=\mathbb {R} ^{2}}$ with coordinates ${\displaystyle (p,q)}$ and symplectic form ${\displaystyle dp\wedge dq}$. Consider the Hamiltonian

${\displaystyle H={\frac {p^{2}}{2m}}+V(q),}$
where ${\displaystyle V}$ is some smooth real-valued function. The associated Hamiltonian vector field is
${\displaystyle v_{H}=-V'(q){\frac {\partial }{\partial p}}+{\frac {p}{m}}{\frac {\partial }{\partial q}}.}$
The equations for flow along ${\displaystyle v_{H}}$ are therefore
{\displaystyle {\begin{aligned}{\dot {p}}&=-V'(q)\\{\dot {q}}&={\frac {p}{m}}.\end{aligned}}}
Interpreting ${\displaystyle q}$ as position, ${\displaystyle p}$ as momentum, ${\displaystyle m}$ as mass and ${\displaystyle V}$ as potential enegery, these are precisely the equations of motion in classical mechanics.

Remark 6.4

The vector field ${\displaystyle X_{H}=-v_{H}}$ is called the "symplectic gradient of ${\displaystyle H}$". It is the vector field dual to ${\displaystyle dH}$ under ${\displaystyle \omega :TM\to T^{*}M}$ and, once we have picked a compatible almost complex structure and metric, is precisely ${\displaystyle -J\nabla H}$. Since this is orthogonal to ${\displaystyle \nabla H}$, flowing under ${\displaystyle X_{H}}$ preserves ${\displaystyle H}$. Formally this follows from ${\displaystyle X_{H}(H)=dH(X_{H})=\omega (X_{H},X_{H})=0}$.

Exercise 6.3

Let ${\displaystyle S^{1}}$ act on ${\displaystyle \mathbb {C} ^{n}}$ diagonally in the usual way. Show, by differentiating, that the infinitesimal vector field defined by ${\displaystyle X_{\xi }(z)={\frac {d}{dt}}|_{t=0}({\text{exp}}(t\xi ).z)}$, is ${\displaystyle X_{\xi }(z_{1},\ldots ,z_{n})=i\xi (z_{1},\ldots ,z_{n})}$ for ${\displaystyle \xi \in {\text{Lie}}(S^{1})\cong \mathbb {R} }$. Putting ${\displaystyle \xi =1}$, show that this is really ${\displaystyle \Sigma _{i=1}^{n}r_{i}{\frac {\partial }{\partial \theta _{i}}}}$ where ${\displaystyle z_{j}=r_{j}e^{i\theta _{j}}}$. Furthermore, verify that this is a Hamiltonian vector field with Hamiltonian ${\displaystyle {\frac {-1}{2}}\Sigma _{i=1}^{n}r_{i}^{2}+c}$ for any constant ${\displaystyle c}$.

Exercise 6.4

Let ${\displaystyle X}$ be a symplectic vector field on ${\displaystyle (M,\omega )}$ and ${\displaystyle \gamma }$ a path in ${\displaystyle M}$. Denote by ${\displaystyle T_{\epsilon }}$ the surface swept out after time ${\displaystyle \epsilon }$ by the flow along the vector field ${\displaystyle X}$ starting at ${\displaystyle \gamma }$. Then define ${\displaystyle {\text{Flux}}_{\gamma }(X):=\lim _{\epsilon \to 0}{\frac {\int _{T_{\epsilon }}\omega }{\epsilon }}}$. Prove that ${\displaystyle X}$ is Hamiltonian ${\displaystyle \Leftrightarrow }$${\displaystyle {\text{Flux}}_{\gamma }(X)=0}$ for every ${\displaystyle \gamma \in H_{1}(M)}$.