# Compute the exterior product of $n$ copies of $\omega=dx_1 \wedge dx_2+\cdots+dx_{2n-1}\wedge dx_{2n}$

Let $$\omega$$ be the $$2$$-form in $$\mathbb{R}^{2n}$$ given by $$\omega=dx_1 \wedge dx_2+dx_3\wedge dx_4+\cdots dx_{2n-1}\wedge dx_{2n}$$ Compute the exterior product of $$n$$ copies of $$\omega$$.

(Chapter 1, Exercise 7 in Differential Forms and Applications by Manfredo P. do Carmo)

The wording of the problem is kind of difficult for me to understand, but I suppose that the exercise asks us to calculate $$\omega \wedge \cdots \wedge \omega$$.

So, I think the only way one can get a non-zero coefficient is that if we permute the parentheses $$(dx_1 \wedge dx_2)$$ up to $$(dx_{2n-1}\wedge dx_{2n})$$. There are $$n!$$ such permutations. Since each permutation can be arranged to $$dx_1\wedge dx_2 \wedge dx_3 \wedge dx_4 \wedge \cdots \wedge dx_{2n-1}\wedge dx_{2n}$$ with an even number of transpositions, we always get $$+1$$ as the coefficient of the differential form. So, the answer should be

$$\omega\wedge\cdots\wedge\omega=\color{green}{n!}\,dx_1\wedge dx_2 \wedge dx_3\wedge dx_4 \wedge \cdots \wedge dx_{2n-1}\wedge dx_{2n}$$

Is that right? If I write my argument exactly like here, would it be considered a complete calculation and receive full points in an exam of differential manifolds?

• It is correct, except for the missing argument showing that all of these permutations are equal to $dx_1 \wedge dx_2 \wedge \cdots \wedge dx_{2n}$ with a plus sign. Can you add that in? – darij grinberg May 26 '19 at 15:52
• @darijgrinberg Thanks for taking the time to read it. Doesn't saying that all of them can be rearranged to $dx_1 \wedge dx_2 \wedge \cdots \wedge dx_{2n}$ with an even number of transpositions suffice? – stressed out May 26 '19 at 15:54
• Yeah, but why is the number of transpositions even? – darij grinberg May 26 '19 at 15:54
• @darijgrinberg Because if I want to move $dx_1$ to the first position, I will have to jump over, let's say $2m$, elements (i.e. $2m$ transpositions), then if I want to do the same to $dx_2$, again I have to jump over some even number of elements. I know that it's kind of a handwavy argument, but it's right. Right? can I say it in a less handwavy way? – stressed out May 26 '19 at 15:57
• @stressedout: That proof can be done slickly by inducting on $k$ and then on $s$. – darij grinberg May 26 '19 at 16:07

Your argument is correct, except for the fact that you need to justify why each of your permuted pairs of $$dx_{i}\wedge dx_{i+1}$$ when wedged together form $$dx_1\wedge\cdots\wedge dx_{2n}$$ without a sign. This is because: given forms $$\alpha,\beta$$ of degrees $$k,\ell$$ respectively, $$\alpha\wedge\beta=(-1)^{k\ell} \beta\wedge \alpha.$$ In particular, here $$\deg(\alpha)=\deg(\beta)=2$$ so that $$(dx_{i}\wedge dx_{i+1})\wedge (dx_j\wedge dx_{j+1})=(-1)^4 (dx_j\wedge dx_{j+1})\wedge (dx_i\wedge dx_{i+1})=(dx_j\wedge dx_{j+1})\wedge (dx_i\wedge dx_{i+1})$$ for any choice of $$i\ne j$$ (both even).