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 $\omega^n:=\omega\wedge\omega\wedge\ldots\wedge\omega$?
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1Do you mean on $\mathbb{R}^{2n}$? What have you done so far? – 23rd Apr 25 '13 at 19:38
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yes actually is called the standar symplectic form for $R^{2n}$ n=2, $\omega^2$=$2dx_1$∧$dx_{2}$∧$dx_{3}$∧$dx_{4}$ – Miguel Apr 25 '13 at 19:41
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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 $$ \begin{align} \omega^n &=\sum\limits_{\sigma\in\mathfrak{S}_n}(-1)^{2\operatorname{sgn}(\sigma)}x_{1}\wedge x_{2}\ldots x_{2n-1}\wedge x_{2n}\\ &=\sum\limits_{\sigma\in\mathfrak{S}_n}x_{1}\wedge x_{2}\ldots x_{2n-1}\wedge x_{2n}\\ &=n!\; x_{1}\wedge x_{2}\ldots x_{2n-1}\wedge x_{2n} \end{align} $$
Norbert
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what do you mean with $sgn(σ)$, $sgn(σ)={1 if x>0, -1 if x<0, 0 if x=0$ – Miguel Apr 25 '13 at 20:29
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