# basis free volume form for a symplectic vector space

It's easy to show, using a symplectic basis, that if $\omega$ is a symplectic form on a $2n$-dimensional vector space $V$, then $\omega^n \neq 0$ (see for example Jason DeVito's excellent answer to this question). I'd like to be able to prove it without choosing a basis (the symplectic analogue to a question I asked recently about inner products), and I think it should be possible, but so far I've been unable.

Let $\omega\in \Lambda^2 V^*$, where $V$ is an inner product space of dim $2n$.

I want to show that $\omega^n=0 \Leftrightarrow\omega$ is degenerate.

$\Leftarrow:$ $\omega$ is degenerate, so there is some element $u\in V$ with $\omega(u,v)=0, \forall v\in V.$ Then let $\{e^i\}$ be a basis containing $u$. Therefore $\omega^n(e_1\wedge\dotsb\wedge e_{2n})=0,$ and any other $2n$-vector is collinear with $e_1\wedge\dotsb\wedge e_{2n},$ so $\omega^n=0.$

$\Rightarrow:$ Given the canonical isomorphism $V\cong V^*$, we have an associated element $\omega^\sharp=\sum_i q^i\wedge p^i.$ Let $W$ be the span of the $q^i,p^i.$ If there are fewer than $2n$ distinct elements in $\{q^i,p^i\}_i$, then $\dim W<2n$ so there is a vector $u\in W^\perp$ which is in the kernel of $\omega.$ If there are $2n$ distinct $q^i,p^i$, hence exactly $n$ terms in the sum then $\omega^n=q^1\wedge p^1\wedge\dotsb \wedge p^n\wedge q^n$, but since $\omega^n=0$, there is some linear dependence among the $q^i,p^i$, so the dimension of their span is again less than $2n$, so we have an element in the kernel.

If there are more than $n$ terms, then $\omega^n$ will be a sum of volume elements, which we stipulate vanishes. Can I find an element of the kernel here? (One way maybe is to say $\omega^n=\operatorname{Pf}(\omega)\text{vol}$ which vanishes iff $\det\omega$ does, but this seems like cheating)

(Previous apparently nonsensical argument below:because $\omega$ is nondegenerate, we have an isomorphism $\tilde{\omega}\colon V\to V^*$ given by $v \mapsto (u\mapsto \omega(v,u))$. This induces an isomorphism $\det(\tilde{\omega})\colon \Lambda^{2n} V\to \Lambda^{2n} V^*$. Then let $u = u_1\wedge \dotsb \wedge u_{2n}$ and $v = v_1\wedge \dotsb \wedge v_{2n}$ be rank $2n$ multivectors in $\Lambda^{2n} V$ ... next step?? This doesn't seem to work right.)

• Here's a suggestion of something to try: what does it mean for a two form if $\omega^n = 0$? I have to admit that your question made me do a double-take: I am used to defining a symplectic form by the fact that $\omega^n \ne 0$. (Of course, the condition that $\tilde \omega$ is an isomorphism is an equivalent definition.) Dec 22, 2011 at 12:55