# Spivak Calculus on Manifolds - Problem 4-1

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...

• The indices of the tensor product should be permuted by $\sigma$ – Sou Nov 15 '18 at 2:39
• @KelvinLois Do you mean $\varphi_{i_1} \otimes \ldots{} \otimes \varphi_{i_k}$? My book's definition does not require that. It defines $\operatorname{Alt}(T)(v_1, \ldots{}, v_k) = \frac{1}{k!} \sum_{\sigma \in S_k} \operatorname{sgn} \sigma \cdot T(v_{\sigma(1)}, \ldots{}, v_{\sigma(k)})$. – Richard Nov 15 '18 at 2:51
• Ok then when you apply it to $e_i$’s you should get one because other indices that is permuted nontrivially would lead to $\delta_{ij}=0$ – Sou Nov 15 '18 at 2:57

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 $$e_4$$.
• 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 $e_4$? – Richard Nov 15 '18 at 2:40
• ah i see what's bugging you. The $\sigma$ should range over permutations of the indices $i_1, \ldots, i_k$. – hunter Nov 15 '18 at 2:54