# every symplectic manifold has even dimension

All we need to prove is that every finite dimensional vector space $V$ with an anti-symmetric non-degenerate 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 non-degenerate $\omega(v_{0},\cdot)$ 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 anti-symmetry).

Let $V_{0}=\operatorname{span}\{v_{0},u_{0}\}$. Consider a space $V_{1}$ of ”orthogonal” elements to $V_{0}$ under $\omega$. That is:

$V_{1}=\left\{v_{1}\in V:\text{ for all v\in V_{0}},\>\omega(v,v_{1})=0\right\}$

We now prove $V=V_{0}\bigoplus V_{1}$:

• $V_{0}\bigcap V_{1}=\{0\}$

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:

$\omega(v_{0},w)=0\implies\beta\omega(v_{0},u_{0})=0\implies\beta=0$ similarly

$\omega(u_{0},w)=0\implies\alpha\omega(u_{0},v_{0})=0\implies\alpha=0$

So $w$ must be null.

• $V=V_{0}\bigoplus V_{1}$

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 $V=V_{0}\bigoplus V_{1}$

So the matrix representation of $\omega$ is block-diagonal in $V_{0}\bigoplus V_{1}$ and a restriction anti-symmetric bilinear for of $\omega$ to $V_{1}$ exists.

If $V_{1}$ is not null we can repeat the procedure with the restriction. Since $\operatorname{dim}(V)=\operatorname{dim}(V_{0})+\operatorname{dim}(V_{1})$ and $V$ is finite dimensional the procedure must stop at a finite step.

At the end we get a decomposition $V=\bigoplus_{i=0}^{k-1}V_{i}$, where $\operatorname{dim}(V_{i})=2$ and $\operatorname{dim}(V)=2k$ is even.

Title every symplectic manifold has even dimension EverySymplecticManifoldHasEvenDimension 2013-03-22 15:44:05 2013-03-22 15:44:05 cvalente (11260) cvalente (11260) 18 cvalente (11260) Theorem msc 53D05 AlternatingForm