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Since $\varphi_{i_j} \in \mathcal{T}(\mathbb{R}^n)$, for every $j = 1, \dots, k$, we have

\begin{align*} \varphi_{i_1}\wedge\dots\wedge\varphi_{i_k}(e_{i_1}, \dots, e_{i_k}) &= \frac{k!}{1!\dots 1!}\operatorname{Alt}(\varphi_{i_1}\otimes\dots\otimes\varphi_{i_k})(e_{i_1}, \dots, e_{i_k})\\ &= \sum_{\sigma \in S_k}\operatorname{sgn}(\sigma)\varphi_{i_1}(e_{\sigma(i_1)})\dots\varphi_{i_k}(e_{\sigma(i_k)})\\ &= 1. \end{align*}

I don't understand the final line. Why is $\phi_{i_1}(e_{\sigma(i_1)}) = 1$ for example? I thought that $\phi_{i_1}(e_{\sigma(i_1)}) = 1$ if $\sigma(i_1) = i_1$.

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  • $\begingroup$ Because you're summing over every permutation. $\endgroup$ – user137731 Dec 1 '15 at 21:57
  • $\begingroup$ @Bye_World OK I see that makes sense, however say we had instead $\phi_{i_1} \wedge \dots \wedge \phi_{i_k} (e_{j_1}, \dots , e_{j_k})$ instead. My book says that this is equal to $\text{sgn} \sigma$ where $\sigma$ is such that $i_{l} = j_{\sigma (l)}$ but I have it is $\text{sgn} \sigma$ such that $\sigma(j_l) = i_l$ I have no idea how they got their evaluation. $\endgroup$ – FACEIT Dec 1 '15 at 22:00
  • $\begingroup$ $\varphi_{i_1}(e_{\sigma(i_1)})\cdots\varphi_{i_k}(e_{\sigma(i_k)})=0$ unless $\sigma$ is the identity no? So the sum is just $1\pm0\pm0\pm0\pm\cdots$ $\endgroup$ – whacka Dec 1 '15 at 22:06
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As you point out, $\varphi_{i_1}(e_{\sigma(i_1)}) = 1$ if and only if $\sigma(i_1) = i_1$, and it is zero otherwise. Likewise, for every $j = 1, \dots, k$, $\varphi_{i_j}(e_{\sigma(i_j)}) = 1$ if and only if $\sigma(i_j) = i_j$, and it is zero otherwise.

Now note that the product $\varphi_{i_1}(e_{\sigma(i_1)})\dots\varphi_{i_k}(e_{\sigma(i_k)})$ is zero if any of the factors are zero. The only way that the product is not zero is if all of the factors are equal to $1$, which only happens if $\sigma(i_1) = 1, \dots, \sigma(i_k) = i_k$ (i.e. $\sigma$ is the identity permutation, $\operatorname{id}$). So the sum in the penultimate line has only one non-zero summand which is the one associated to the identity permutation. As the signature of the identity permutation is $1$, we see that

$$\sum_{\sigma \in S_k}\operatorname{sgn}(\sigma)\varphi_{i_1}(e_{\sigma(i_1)})\dots\varphi_{i_k}(e_{\sigma(i_k)}) = \operatorname{sgn}(\operatorname{id})\varphi_{i_1}(e_{i_1})\dots\varphi_{i_k}(e_{i_k}) = 1.$$

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  • $\begingroup$ Thank you for this answer - could you please see my comment on the original post to @Bye_World? $\endgroup$ – FACEIT Dec 1 '15 at 22:02
  • $\begingroup$ If what you wrote is what is in the book, then you are correct and what they've written is probably just a typo. $\endgroup$ – Michael Albanese Dec 1 '15 at 22:22

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