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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 $\omega^n$.

PtF
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    I don't know how this is in the context of sympletic manifolds, but as far as I know the wedge product satisfies $\omega \wedge \omega = 0$. – Gold Jun 11 '13 at 12:08
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    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
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    @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 - on hiatus Jun 11 '13 at 14:38

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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.

Jason DeVito - on hiatus
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    Let me know if you need further help - I can always fill in more details! – Jason DeVito - on hiatus 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 - on hiatus 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 (http://math.stackexchange.com/questions/1411971/symplectic-form-and-wedge-sum) – Jason DeVito - on hiatus Aug 28 '15 at 17:29
  • @JasonDeVito, sir I need help understanding lemma 2 and 3. Can you help me ? – creative Oct 31 '17 at 05:30
  • @Khichudi: You could ask a separate question on the site, if you want. In case it helps, $e_i^\ast$ is dual do the basis ${e_1,..,e_n, f_1,..., f_n}$ meaning that $e_1^\ast(e_1) = 0$ and $e_1^\ast (x) = 0$ for any other $x$ in the basis. – Jason DeVito - on hiatus Oct 31 '17 at 15:41
  • @JasonDeVito Forgive me for reviving an older thread, but don't both lemmas 2 & 3 continue to hold - with the proviso that $\omega$ is non-zero - even when $\omega$ is allowed to be degenerate? I may be missing something rather obvious here, but I believe $\omega^n \ne 0$ holds regardless of whether $\omega$ is symplectic or not. As an aside, there is a typo in your previous comment, it should read $e^*_1(e_1) = 1$. – Matthew Buck Jan 05 '21 at 17:22
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    @MatthewBuck: I agree my comment has a typo. Lemma 2 amd 3 are not valid for general nonzero forms. For example, in $\mathbb{R}^4$, the form $\omega= e_1\wedge e_2$ satisfies $\omega^2=0$. – Jason DeVito - on hiatus Jan 06 '21 at 02:45
  • @JasonDeVito Thanks for the feedback. I see where my confusion stemmed from: what the integer $n$ is referring to. I stand by my claim that Lemma 2 holds in general, provided we don't insist on $\dim V = 2n$. That is, if $\omega \ne 0$ is an alternating bilinear form on $V$, then there exists a basis ${u_1,\ldots,u_k,e_1,\ldots,e_n,f_1,\ldots,f_n}$ satisfying the properties listed above. Then we have $\omega = \sum_{i=1}^n e_i^* \wedge f_i^*$ and $\omega^n \ne 0$. I appreciate that this is nothing more than applying your result to $\omega$ restricted to the subspace $\mathrm{span}{e_i,f_i}$. – Matthew Buck Jan 06 '21 at 09:56
  • @Matthew: I should have been more explicit about what $n$ was - sorry about that. – Jason DeVito - on hiatus Jan 06 '21 at 13:56