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Let $V$ be a real vector space and $\operatorname{Alt}^k(V)$ the set of alternating $k$-forms, i.e. $\alpha \in \operatorname{Alt}^k(V)$ is a map $\alpha: V^k \to \mathbb{R}$ such that

  • $\alpha$ is linear in every of its $k$ components and

  • $\alpha(v_1,\dots,v_k) = 0$ if $v_1,\dots,v_k \in V$ are linearly dependent.

One can also show that the second criterion is equivalent to each of the following statements:

  • $\alpha(v_1,\dots,v_k) = 0$ if $v_i = v_j$ for some $i \neq j$,

  • $\alpha(v_1,\dots,v_i,\dots,v_j,\dots,v_k) = - \alpha(v_1,\dots,v_j,\dots,v_i,\dots,v_k) $, i.e. interchanging two variables results in a change of sign,

  • $\alpha(v_{\sigma(1)},\dots,v_{\sigma(k)}) = \operatorname{sgn}(\sigma) \cdot \alpha(v_1,\dots,v_k)$.

For $\alpha \in \operatorname{Alt}^k(V), \beta \in \operatorname{Alt}^k(V)$ we can define the exterior product

$(\alpha \wedge \beta)(v_1,\dots,v_{r+s}) := \frac{1}{r! s!} \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \cdot \alpha(v_{\sigma(1)},\dots,v_{\sigma(r)}) \beta(v_{\sigma(r+1)},\dots,v_{\sigma(r+s)})$.

and one can show that $\alpha \wedge \beta \in \operatorname{Alt}^{r+s}(V)$.

Now I want to show that $\alpha \wedge \beta = (-1)^{rs} \cdot (\beta \wedge \alpha)$.

I wrote both definitions out

$(\alpha \wedge \beta)(v_1,\dots,v_{r+s}) := \frac{1}{r! s!} \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \cdot \alpha(v_{\sigma(1)},\dots,v_{\sigma(r)}) \beta(v_{\sigma(r+1)},\dots,v_{\sigma(r+s)})$ and

$(\beta \wedge \alpha)(v_1,\dots,v_{r+s}) := \frac{1}{r! s!} \sum_{\mu \in S_n} \operatorname{sgn}(\mu) \cdot \beta(v_{\mu(1)},\dots,v_{\mu(s)}) \alpha(v_{\mu(s+1)},\dots,v_{\mu(s+r)})$.

For me, both thinks looks the same: I mean since we consider all possible permutations the addends in both wedge products should be the same, isn't it? But this it obviously not true (otherwise we would have equality) but I can't see it. Where is my fallacy?

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  • $\begingroup$ In the equation for $\beta \wedge \alpha$, shouldn't it be $\beta(v_{\mu(1)}, \ldots, v_{\mu(s)}) \alpha(v_{\mu(s+1)}, \ldots, v_{\mu(r+s)}$? $\endgroup$ – Daniel Schepler Oct 25 '17 at 22:57
  • $\begingroup$ Yes, you're right. Thank you! $\endgroup$ – Diglett Oct 26 '17 at 6:26
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The point you're missing is: to get the same term in the second term as for $\sigma$ in the first term, you need $\mu(1) = \sigma(r+1)$, $\ldots$, $\mu(s) = \sigma(r+s)$, $\mu(s+1) = \sigma(1)$, $\ldots$, $\mu(s+r) = \sigma(r)$. So now, since $\mu$ and $\sigma$ are different permutations, you need to find a relation between $\operatorname{sgn}(\mu)$ and $\operatorname{sgn}(\sigma)$.

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