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The wedge sum of $k$ symplectic 2-forms is given by ( if $\omega = \sum_i e_i^* \wedge f_i^*$)

$$ \omega^k = k! \sum_{1\le i_1 <...<i_k \le n} (e_{i_1}^* \wedge f_{i_1}^*) \wedge ... \wedge (e_{i_k}^* \wedge f_{i_k}^*).$$

But now I read that $\omega^n(e_1,f_1,...,e_n,f_n)=1.$ This does not make sense to me, afais the result is $\omega^n(e_1,f_1,...,e_n,f_n)= n!$

Edit: Sorry, it was kind of stupid to me to not include the reference, but I wanted to ask this question and forgot where I saw it. But it was actually an answer here on math.stackexchange click me.

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user167575
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    Different authors define the wedge product differently. What you are seeing is a symptom of this. –  Aug 27 '15 at 21:39
  • You mean the wedge product? In any case, I don't see any reason why that $k!$ prefactor should be there. – Qiaochu Yuan Aug 27 '15 at 21:49
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    @Qiaochu: He's summing over all permutations, and then by specifying the order of the indices, pulling out a $k!$, I think. –  Aug 27 '15 at 22:12
  • @MikeMiller yes that's exactly how I arrived at this equation for $\omega^k$ ( I used $i_1 < i_2 <...$ instead of summing over all indices independently) What do you mean by " what you are seeing is a symptom of this?" – user167575 Aug 27 '15 at 22:53
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    Some people stick a $p!q!/(p+q)!$ in. front of your formula for the wedge product. In any case, the exciting thing about $\omega^n$ isn't its specific value - it's that it's nonzero. –  Aug 27 '15 at 23:55
  • @Mike: no, that's only if we're talking about antisymmetric tensors; the OP is explicitly using the wedge product symbol, so there should be a unique expression giving the $k^{th}$ wedge power. – Qiaochu Yuan Aug 28 '15 at 00:37
  • it may be that I am wrong, but I am currently not sure that my question was answered in the comments here, but maybe I don't see it. – user167575 Aug 28 '15 at 02:29
  • Sorry to have made that mistake at the end of my answer! (and thanks to Ted for suggesting I correct it) – Jason DeVito - on hiatus Aug 28 '15 at 17:27

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There's a reason $\dfrac{\omega^n}{n!}$ shows up all over complex geometry as the induced volume form. You are correct and there is an error in whatever you're reading.

Ted Shifrin
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  • Thank you, I included the reference. If you think there is a reason that the author of the answer made this error, I would love to know, otherwise you have fully answered my question, I guess. – user167575 Aug 28 '15 at 10:31