# Computing $n$-th external power of standard simplectic form

Miguel yesterday at 17:18. 1 answers, 0 views

I need some help:

Define a 2-form on $R^n$ by $\omega=dx_1\wedge dx_2+dx_3\wedge dx_4+...+dx_{2n-1}\wedge dx_{2n}$. How to compute $\omega^n:=\omega\wedge\omega\wedge\ldots\wedge\omega$?

Norbert 04/25/2013.

Note that $$\omega=\sum\limits_{k=1}^n x_{2k-1}\wedge x_{2k}$$ so $$\omega^n=\sum\limits_{k_1=1}^n\ldots\sum\limits_{k_n=1}^n x_{2k_1-1}\wedge x_{2k_1}\wedge\ldots\wedge x_{2k_n-1}\wedge x_{2k_n}$$ Summands here are non-zero iff $k_1,\ldots,k_n$ are all distinct numbers, so summands in bijective correspondence with permutations of numbers $1,\ldots, n$. Thus $$\omega^n=\sum\limits_{\sigma\in\mathfrak{S}_n}x_{2k_{\sigma(1)}-1}\wedge x_{2k_{\sigma(1)}}\ldots x_{2k_{\sigma(n)}-1}\wedge x_{2k_{\sigma(n)}}$$ Since we are dealing with external product we may say that \begin{align} \omega^n &=\sum\limits_{\sigma\in\mathfrak{S}_n}(-1)^{2\operatorname{sgn}(\sigma)}x_{1}\wedge x_{2}\ldots x_{2n-1}\wedge x_{2n}\\ &=\sum\limits_{\sigma\in\mathfrak{S}_n}x_{1}\wedge x_{2}\ldots x_{2n-1}\wedge x_{2n}\\ &=n!\; x_{1}\wedge x_{2}\ldots x_{2n-1}\wedge x_{2n} \end{align}