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Let $V$ be a vector space, $\mathcal B=\{w_1,\dots,w_k\}$ a orthonormal basis for $V$, and let $T=w_1^\star\otimes\dots\otimes w_k^\star\in\mathcal L^k(V)$.

I don't understand why is this equivalence true: $$w_1^\star\wedge\dots\wedge w_k^\star(w_1,\dots,w_k)=\frac 1 {k!}\sum_{\sigma\in S^k}(w_1^\star\otimes\dots\otimes w_k^\star)^\sigma(w_1,\dots,w_k)$$

Actually, I was sure that the equivalence was $$w_1^\star\wedge\dots\wedge w_k^\star(w_1,\dots,w_k)=\sum_{\sigma\in S^k}(w_1^\star\otimes\dots\otimes w_k^\star)^\sigma(w_1,\dots,w_k)$$ and this made sense to me since $w_1^\star\wedge\dots\wedge w_k^\star(w_1,\dots,w_k)=\mathrm{Det}|w_1,\dots,w_k|=1$, and $(w_1^\star\otimes\dots\otimes w_k^\star)^\sigma(w_1,\dots,w_k)=1$ if and only if $\sigma$ is the identity, otherwise is equal to zero (so the sum is equal to $1$).

Can you tell me where I'm wrong? Thank you in advance

Dr. Scotti
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    should the $S^d$ be $S^k$ instead? Also, I think the answer to this question depends on how you defined the wedge product. Some authors insert the factorials into the definition, while others do not. So i suggest going through the definitions again and seeing how things match up – peek-a-boo Jul 13 '19 at 23:00
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    There are different conventions. I personally prefer the second definition. You will find this discussed (and references given) numerous times on MSE. – Ted Shifrin Jul 14 '19 at 01:29
  • Ok, thank you. However if I integrate a form, e.g. $dx\wedge dy$, why doesn't the result depend on the definition I'm using? – Dr. Scotti Jul 14 '19 at 08:10

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