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.

单词	EverySymplecticManifoldHasEvenDimension
释义	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.
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