# How to show $\omega^n$ is a volume form in a symplectic manifold $(M, \omega)$?

I have the following problem: Let $(M, \omega)$ be a symplectic manifold. How can I show $$\omega^n=\underbrace{\omega\wedge \ldots\wedge \omega}_{n-times},$$ satisfies $\omega^n(p)\neq 0$ for all $p\in M$. I believe that is not too dificult but I'm not used with exterior product.. Any help will be valuable...

This is important because the nondegeneracy condition on $ω$ is equivalent to the condition that M has an even dimension $2n$ and the top wedge product $\omega^n$ is nowhere vanishing on M, i.e., $ω^n$ is a volume form. In particular, $M$ must be orientable and is canonically oriented by $\omega^n$.

• I don't know how this is in the context of sympletic manifolds, but as far as I know the wedge product satisfies $\omega \wedge \omega = 0$. – user1620696 Jun 11 '13 at 12:08
• The nondegeneracy condition on $\omega$ is equivalent to the condition that $M$ has an even dimension $2n$ and the top wedge product $\omega^n$ is nowhere vanishing on $M$, i.e., $\omega^n$ is a volume form. In particular, $M$ must be orientable and is canonically oriented by $\omega^n$. – PtF Jun 11 '13 at 12:13
• @user1620696: That equations holds only for odd degree forms. In general, we have $\alpha \wedge \beta = (-1)^{|\alpha||\beta|}\beta \wedge \alpha$ where $|\alpha|$ denotes the degree of $\alpha$. – Jason DeVito Jun 11 '13 at 14:38

This is really a linear algebra problem - one can work one tangent space at a time. (In particular, all this works even if $\omega$ isn't closed.)

So, let $\omega$ be a non-degenerate antisymmetric bilinear form on a vector space $V$. I'd prove $\omega^n\neq 0$ via a sequence of lemmas.

Lemma 1: There is a basis $\{e_i, f_j\}$ for $V$ for which $\omega(e_i,e_j) = \omega(f_i,f_j) = 0$ and $$\omega(e_i,f_j) = \begin{cases} 1 & i = j\\ 0 & i\neq j\end{cases}.$$

This is proved by induction. At some point, you'll probably use the following fact: Given any proper subspace $W\subsetneq V$, the subspace $$W^\bot = \{v\in V: \omega(v,w) = 0\text{ for all } w\in W\}$$ has positive dimension - $\{0\} \subsetneq W^\bot$. This follows since $W^\bot$ is an intersection of $\dim W$ hyperplanes (codimension 1 planes), corresponding to $\ker(\omega(w_i,\cdot))$ where $\{w_i\}\subseteq W$ is a basis of $W$.

Such basis is called a symplectic basis. If $\omega$ is allowed to be degenerate, the statement is modified to have a basis of the form $\{e_i, f_j, u_k\}$ where $\omega(u_k,\cdot ) = 0$.

Lemma 2: With respect to a symplectic basis, $\omega$ has the form $\sum_i e_i^\ast\wedge f_i^\ast$

So, proving $\omega^n\neq 0$ for this particular $\omega$ gives the result.

Lemma 3: $$\omega^k = k! \sum_{1\leq i_1 < i_2 < \ldots < i_k \leq n } e^\ast_{i_1}\wedge f^\ast_{i_1} \wedge \ldots \wedge e^\ast_{i_k} \wedge f^\ast_{i_k}$$ so, in particular, $\omega^n = n! e^\ast_{1} \wedge f^\ast_{1} \wedge \ldots \wedge e^\ast_n \wedge f^\ast_n$, and hence $\omega^n(e_1,f_1,\ldots e_n,f_n) = n!\neq 0$, so $\omega^n$ is not $0$.

Lemma 3 is proved by induction.

• Let me know if you need further help - I can always fill in more details! – Jason DeVito Jun 11 '13 at 15:23
• Thanks, that clarifies a lot =D.. – PtF Jun 11 '13 at 15:32
• Jason, perhaps you should fix the typo in the penultimate line, as it's caused some confusion. $\omega^n(e_1,f_1,\dots,e_n,f_n) = n! \ne 0$ :) – Ted Shifrin Aug 28 '15 at 15:18
• @Ted: Done. To whomever I caused confusion - sorry! – Jason DeVito Aug 28 '15 at 17:25
• For those wondering why I edited this old post, apparently a typo at the end caused confusion to at least one person (math.stackexchange.com/questions/1411971/…) – Jason DeVito Aug 28 '15 at 17:29