every symplectic manifold has even dimension
All we need to prove is that every finite dimensional vector space^{} $V$ with an antisymmetric nondegenerate linear form^{} $\omega $ has an even dimension^{} $2k$. This is only a linear algebra^{} result. In the case of a symplectic manifold $V$ is just the tangent space at a point, and thus its dimension equals the manifold’s dimension.
Pick any not null vector ${v}_{0}\in V$. Since $\omega $ is nondegenerate $\omega ({v}_{0},\cdot )$ is a not null linear form. Therefore there exists a not null vector ${u}_{0}$ such that $\omega ({v}_{0},{u}_{0})=1$
Now ${v}_{0}$ and ${u}_{0}$ are linearly independent^{} because if ${v}_{0}=\lambda {u}_{0}$ then $\omega ({v}_{0},{u}_{0})=\omega (\lambda {u}_{0},{u}_{0})=\lambda \omega ({u}_{0},{u}_{0})=0$ (by antisymmetry).
Let ${V}_{0}=\mathrm{span}\{{v}_{0},{u}_{0}\}$. Consider a space ${V}_{1}$ of ”orthogonal^{}” elements to ${V}_{0}$ under $\omega $. That is:
${V}_{1}=\{{v}_{1}\in V:\text{for all}v\in {V}_{0},\omega (v,{v}_{1})=0\}$
We now prove $V={V}_{0}\oplus {V}_{1}$:

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${V}_{0}\bigcap {V}_{1}=\{0\}$
Suppose $w\in {V}_{0}\bigcap {V}_{1}$ is not null, then it can be written $w=\alpha {v}_{0}+\beta {u}_{0}$ because it belongs to ${V}_{0}$. Since it also belongs to ${V}_{1}$ is is ”orthogonal” to both ${v}_{0}$ and ${u}_{0}$. That is:
$\omega ({v}_{0},w)=0\u27f9\beta \omega ({v}_{0},{u}_{0})=0\u27f9\beta =0$ similarly
$\omega ({u}_{0},w)=0\u27f9\alpha \omega ({u}_{0},{v}_{0})=0\u27f9\alpha =0$
So $w$ must be null.

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$V={V}_{0}\oplus {V}_{1}$
Suppose $w\in V$. Let $\alpha =\omega ({v}_{0},w)$, $\beta =\omega ({u}_{0},w)$, ${w}_{0}=\alpha {u}_{0}\beta {v}_{0}$.
Then $\omega ({v}_{0},{w}_{0})=\alpha =\omega ({v}_{0},w)$ and $\omega ({u}_{0},{w}_{0})=\beta =\omega ({u}_{0},w)$.
Considering ${w}_{1}=w{w}_{0}$ we have $w={w}_{0}+{w}_{1}$ (by construction) and $\omega ({v}_{0},{w}_{1})=\omega ({v}_{0},w{w}_{0})=\omega ({v}_{0},w)\omega ({v}_{0},{w}_{0})=\omega ({v}_{0},w)\omega ({v}_{0},w)=0$ and similarly for $\omega ({u}_{0},{w}_{1})$
So ${w}_{1}\in {V}_{1}$, ${w}_{0}\in {V}_{0}$ and $w={w}_{0}+{w}_{1}$ and thus $V={V}_{0}\oplus {V}_{1}$
So the matrix representation^{} of $\omega $ is blockdiagonal in ${V}_{0}\oplus {V}_{1}$ and a restriction antisymmetric bilinear^{} for of $\omega $ to ${V}_{1}$ exists.
If ${V}_{1}$ is not null we can repeat the procedure with the restriction. Since $\mathrm{dim}(V)=\mathrm{dim}({V}_{0})+\mathrm{dim}({V}_{1})$ and $V$ is finite dimensional the procedure must stop at a finite step.
At the end we get a decomposition $V={\oplus}_{i=0}^{k1}{V}_{i}$, where $\mathrm{dim}({V}_{i})=2$ and $\mathrm{dim}(V)=2k$ is even.
Title  every symplectic manifold has even dimension 

Canonical name  EverySymplecticManifoldHasEvenDimension 
Date of creation  20130322 15:44:05 
Last modified on  20130322 15:44:05 
Owner  cvalente (11260) 
Last modified by  cvalente (11260) 
Numerical id  18 
Author  cvalente (11260) 
Entry type  Theorem 
Classification  msc 53D05 
Related topic  AlternatingForm 