Let $n \geqslant 2$ be an integer and consider the vector space $V=F^{2n}$ with the standard basis $e_1,\ldots,e_{2n}$.

Now the second exterior power $\Lambda^2(V)$ contains the element $$\omega=e_1 \wedge e_2 + e_3 \wedge e_4 + \cdots + e_{2n-1} \wedge e_{2n}.$$

Since $\Lambda^{2n}(V)$ is spanned by $x := e_1 \wedge \cdots \wedge e_{2n}$, we know that the $n$-fold wedge product $\omega \wedge \cdots \wedge \omega$ is a scalar multiple of $x$.

Question: How do we compute this scalar?

  • $\begingroup$ That's only for a $1$-form right? By definition, $v \wedge v = 0$ for every $v \in \Lambda^{1}(V)$. I don't know if it holds for higher powers. Correct me if I'm wrong. $\endgroup$ – user13451345 Mar 2 '16 at 23:03
  • $\begingroup$ Yeah you're right, sorry. $\endgroup$ – shalop Mar 2 '16 at 23:04
  • $\begingroup$ No worries, there was no harm done. $\endgroup$ – user13451345 Mar 2 '16 at 23:04
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    $\begingroup$ The scalar is equal to $n!$, which can be seen by multiplying out and distributing all of the terms above, then noticing that all of the terms with repeated $x_j$ vanish and so you get a sum over permutations in $S_n$. Reordering two at a time doesn't change the sign of the form. $\endgroup$ – shalop Mar 2 '16 at 23:24
  • $\begingroup$ @Shalop Thanks for the help! If you'd like, feel free to leave your comment as an answer so that this question doesn't remain in the unanswered section. $\endgroup$ – user13451345 Mar 2 '16 at 23:36

First distribute all terms in the product defining $\omega \wedge \cdots \wedge \omega$ (there are $n$ products). Notice that all of the terms with repeated $x_j$ will vanish by antisymmetricity of the wedge product. Thus you get $$ \omega^{\wedge n} = \sum_{\sigma \in S^*_n} e_{\sigma(1)}\wedge e_{\sigma(1)+1} \wedge \cdots \wedge e_{\sigma(n)}\wedge e_{\sigma(n)+1}$$

where $S_n^*$ denotes the group of all bijections from $\{1,3,5,...,2n-1\}$ unto itself. Next notice that if $\sigma \in S_n^*$, then $e_{\sigma(1)}\wedge e_{\sigma(1)+1} \wedge \cdots \wedge e_{\sigma(n)}\wedge e_{\sigma(n)+1} = e_1 \wedge \cdots \wedge e_n$, because reordering the terms two at a time does not change the sign of the $2n$-form. Consequently, $$\omega^{\wedge n} = \big| S_n^* \big| \cdot e_1 \wedge \cdots \wedge e_n = n! \cdot e_1 \wedge \cdots \wedge e_n$$


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