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Forgive me, I have a (potentially) very trivial question about a special case of the following problem:

Let $e_1, \ldots{}, e_n$ be the usual basis of $\mathbf{R}^n$ and let $\varphi_1, \ldots{}, \varphi_n$ be the dual basis.

(a) Show that $\varphi_{i_1} \wedge \ldots{} \wedge \varphi_{i_k} (e_{i_1}, \ldots, e_{i_k}) = 1$.

I take it that $1 \leq i_1 \lt \ldots{} \lt i_k \leq n$ and that $\varphi_{i}(e_j) = \delta_{i, j}$ where $\delta_{i, j}$ is the Kronecker delta, by a previous theorem in the book.

My work is the following:

\begin{align} \varphi_{i_1} \wedge \ldots{} \wedge \varphi_{i_k} &= k! \operatorname{Alt}(\varphi_{i_1} \otimes \ldots{} \otimes \varphi_{i_k})\\ &= \sum_{\sigma \in S_k} \operatorname{sgn}\sigma \,(\varphi_{i_1} \otimes \ldots{} \otimes \varphi_{i_k})\,, \end{align}

which gives:

$$ \varphi_{i_1} \wedge \ldots{} \wedge \varphi_{i_k}(e_{i_1}, \ldots, e_{i_k}) = \sum_{\sigma \in S_k} \operatorname{sgn}\sigma \,(\varphi_{i_1}(e_{\sigma(i_1)})\cdot \ldots{} \cdot \varphi_{i_k}(e_{\sigma(i_k)})) = (\star)\,.$$

But consider the example $k = 2$ and $n = 4$. Then I can choose $i_1 = 3$ and $i_2 = 4$. Now, since $S_2 = \{ (), (12) \}$, we have:

$$(\star) = (1)(1 \cdot 1) + (-1)(1 \cdot 1) = 0\,.$$

And this contradicts what I have to show? What on earth am I not seeing?

Ideally, I would have every term in the sum equal to 0 except for the identity permutation...

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  • $\begingroup$ The indices of the tensor product should be permuted by $\sigma$ $\endgroup$ – Sou Nov 15 '18 at 2:39
  • $\begingroup$ @KelvinLois Do you mean $\varphi_{i_1} \otimes \ldots{} \otimes \varphi_{i_k}$? My book's definition does not require that. It defines $\operatorname{Alt}(T)(v_1, \ldots{}, v_k) = \frac{1}{k!} \sum_{\sigma \in S_k} \operatorname{sgn} \sigma \cdot T(v_{\sigma(1)}, \ldots{}, v_{\sigma(k)})$. $\endgroup$ – Richard Nov 15 '18 at 2:51
  • $\begingroup$ Ok then when you apply it to $e_i$’s you should get one because other indices that is permuted nontrivially would lead to $\delta_{ij}=0$ $\endgroup$ – Sou Nov 15 '18 at 2:57
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The second summand in your last equation (the one with $\star$ on the left hand side) should be $(-1)*(0 \cdot 0)$ since you are evaluating $\phi_4$ on $e_3$ and $\phi_3$ on $e_4$.

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  • $\begingroup$ But that term corresponds to $\sigma = (12)$ and hence $\sigma(3) = 3$ and $\sigma(4) = 4$... So am I not evaluating $\varphi_3$ on $e_3$ and $\varphi_4$ on $e_4$? $\endgroup$ – Richard Nov 15 '18 at 2:40
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    $\begingroup$ ah i see what's bugging you. The $\sigma$ should range over permutations of the indices $i_1, \ldots, i_k$. $\endgroup$ – hunter Nov 15 '18 at 2:54
  • $\begingroup$ Thank you so much. That must be it. $\endgroup$ – Richard Nov 15 '18 at 2:54

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