Wedge product: $\,\omega \wedge \eta= (-1)^{kl} \eta \wedge \omega$

Question

Prove $\omega \wedge \eta = (-1)^{kl} \eta \wedge w$ where $\omega \in \Lambda^{k}(V)$ and $\eta \in \Lambda^{l}(V)$.

This is from page 79, in M. Spivak's Calculus on manifolds.

My progress:

$\omega \wedge \eta = \dfrac{(k+l)!}{k!l!} \text{Alt}(w \otimes \eta) $

so I am stuck calculating $\text{Alt}(w \otimes \eta) $:

$$\text{Alt}(w \otimes \eta)(v_1,...,v_{k+l}) = \dfrac{1}{(k+l)!}\sum_{\sigma \in S_{k+l}} \text{sgn}\sigma (\omega \otimes \eta) (v_{\sigma (1)},...,v_{\sigma (k+l)}) $$

$$= \sum_{\sigma \in S_{k+l}} \text{sgn}\sigma\, \omega (v_{\sigma (1)},...,v_{\sigma (k)}) \eta (v_{\sigma (k+1)},...,v_{\sigma (k+l)}).$$

My assumption is I must do something with $\sigma$ however, I am not sure what exactly. Any help please.

Answer 1

Hint. It suffices to show that the sign of the permutation $$ \tau=\left(\begin{array}{ccccccccccccccccccc} 1&2&3&\cdots& k&k+1&k+2&\cdots&k+\ell \\ \ell+1 & \ell+2 & \ell+3 & \cdots & \ell+k & 1 & 2 & \cdots & k \end{array} \right) $$ is equal to $(-1)^{k\ell}$.

Next, observe that \begin{align} \text{sgn}\,\sigma\,(\omega \otimes \eta) (v_{\sigma (1)},...,v_{\sigma (k+\ell)}) &=\text{sgn}\,\sigma\,\omega(v_{\sigma (1)},...,v_{\sigma (k)})\, \eta(v_{\sigma (k+1)},...,v_{\sigma (k+\ell)})\\ &=\text{sgn}\,\sigma\,\eta(v_{\sigma (k+1)},...,v_{\sigma (k+\ell)}) \omega(v_{\sigma (1)},...,v_{\sigma (k)})\\ &=\text{sgn}\,\sigma\,(\eta\otimes\omega ) (v_{\sigma\tau (1)},...,v_{\sigma \tau(\ell+k)}) \\ &=\text{sgn}\,\tau^{-1}\, \text{sgn}\,\sigma\tau\,(\eta\otimes\omega ) (v_{\sigma\tau (1)},...,v_{\sigma\tau (\ell+k)})\\ &=(-1)^{k\ell}\, \text{sgn}\,\tau\sigma\,(\eta\otimes\omega ) (v_{\sigma\tau (1)},...,v_{\sigma\tau (k+l)}). \end{align}