Let $\{e_{1},\ldots,e_{n}\}$ be the usual basis for $\mathbb{R}^{n}$ and let $\{\varphi_{i},\ldots,\varphi_{n}\}$ be the dual basis.

Show that $$\varphi_{i_{1}}\wedge\cdots\wedge \varphi_{i_{k}}(e_{i_{1}},\ldots,e_{i_{k}})=1$$

The wedge product is defined by $$\omega\wedge \eta=\frac{(k+l)!}{k!l!}\text{Alt}(\omega \otimes \eta)$$ where, say $\omega\in \Lambda^{k}(V)$, $\eta\in \Lambda^{l}(V)$ and $\omega\wedge \eta\in\Lambda ^{k+l}(V)$

and alternation

$\text{Alt}(T)(v_{1},\ldots,v_{n}):=\dfrac{1}{k!}\displaystyle\sum_{\sigma\in S_{k}}\text{sgn}\;\cdot T(v_{\sigma(1)},\ldots,v_{\sigma(k)})$ with $S_{k}$ being the set of all permutations of the numbers $1$ to $k$.

there are two steps that I don't get in the following solution:

$$ \begin{align} & \varphi_{i_{1}}\wedge\cdots\wedge \varphi_{i_{k}}(e_{1},\ldots,e_{n}) \\[6pt] = & \frac{k!}{1!\cdots 1!}\text{Alt}(\varphi_{i_{1}}\otimes\cdots\otimes\varphi_{i_{k}})(e_{1},\ldots,e_{k}) \\[6pt] = & \sum_{\sigma\in S_{k}}\text{sgn}(\sigma)\varphi_{i_{1}}(e_{\sigma(1)})\cdots \varphi_{i_{k}}(e_{\sigma(k)})=1 \end{align} $$

I am really new to multilinear algebra and will appreciate if someone can explain why in the second line the coefficient becomes $k!$ and why in the last line we get 1? Which elements do we exactly permute?

Thank you


The coefficient $\frac{k!}{1! \cdots 1!}$ comes from the coefficient $\frac{(k + l)!}{k!l!}$ in the wedge product formula. They're generalizing that formula to a wedge product of arbitrarily many alternating tensors:

$$\omega_1 \wedge \omega_2 \wedge \cdots \wedge \omega_r = \frac{(k_1 + k_2 + \cdots + k_r)!}{k_1!k_2!\cdots k_r!}Alt(\omega_1 \otimes \omega_2 \cdots \otimes \omega_r)$$

Where $\omega_i \in \Lambda^{k_i}(V)$. In your case, we have:

$$\varphi_{i_1} \wedge \cdots \wedge \varphi_{i_n} = \frac{(1 + \cdots + 1)!}{1! \cdots 1!}Alt(\varphi_{i_1} \otimes \cdots \otimes \varphi_{i_k})$$

For the last line, let's look at an example where $k = 3$:

$$\sum_{\sigma \in S_3} sgn(\sigma)\varphi_{i_1}(e_{i_{\sigma(1)}})\varphi_{i_2}(e_{i_{\sigma(2)}})\varphi_{i_3}(e_{i_{\sigma(3)}}) $$

$$=\varphi_{i_1}(e_{i_1})\varphi_{i_2}(e_{i_2})\varphi_{i_3}(e_{i_3}) - \varphi_{i_1}(e_{i_2})\varphi_{i_2}(e_{i_1})\varphi_{i_3}(e_{i_3}) - \varphi_{i_1}(e_{i_1})\varphi_{i_2}(e_{i_3})\varphi_{i_3}(e_{i_2}) - \varphi_{i_1}(e_{i_3})\varphi_{i_2}(e_{i_2})\varphi_{i_3}(e_{i_1}) + \varphi_{i_1}(e_{i_2})\varphi_{i_2}(e_{i_3})\varphi_{i_3}(e_{i_1}) + \varphi_{i_1}(e_{i_3})\varphi_{i_2}(e_{i_1})\varphi_{i_3}(e_{i_2}) $$

What you're doing is permuting all the possible indices of the $e's$. But you can see that only one of these terms is going to be nonzero, and that's the first one where $\sigma$ is the identity, because $\phi_i(e_j)$ is nonzero only when $i = j$.

  • $\begingroup$ Thank you very much Alex. Really liked your anwer. Just on thing that bothers me. In the equation $$\varphi_{i_1} \wedge \cdots \wedge \varphi_{i_n} = \frac{(1 + \cdots + 1)!}{1! \cdots 1!}Alt(\varphi_{i_1} \otimes \cdots \otimes \varphi_{i_k})$$ where all the $1$'s come from? Is it because each $\varphi\in\Lambda^{1}(V)$ is a linear functional on $V$ i.e belongs to the dual space. $\endgroup$ – user124471 May 5 '14 at 18:27
  • 1
    $\begingroup$ Yep, the 1's come from the fact that each of the $\varphi_i$ are one-forms. $\endgroup$ – Alex Zorn May 5 '14 at 19:02
  • $\begingroup$ Brilliant! many thanks $\endgroup$ – user124471 May 6 '14 at 14:56

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