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…
Best Answer
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 $e_4$.