Let $F:V \rightarrow W$ be a linear map. Show that $F^{\ast}(\omega \wedge \eta)=(F^{\ast}\omega) \wedge (F^{\ast}\eta)$ for all $\omega \in \Lambda^{p}(W) , \eta \in \Lambda^{q}(W)$.

Where $F^{\ast}\omega$ denotes the pullback of $\omega$, $\wedge$ is the wedge product and $\Lambda^{p}(W)$ is set of all alternating $p$-tensors on W.

So far I have $F^{\ast}(\omega \wedge \eta)(v_{1},\dotsc,v_{p+q})=\omega \wedge \eta(F(v_{1}),\dotsc, F(v_{p+q}))$

$=\frac{(p+q)!}{p!q!}A(\omega \otimes \eta)(F(v_{1}),\dotsc, F(v_{p+q}))$

$=\frac{(p+q)!}{p!q!}\frac{1}{(p+q)!}\displaystyle\sum_{\sigma \in S_{p+q}}\operatorname{sgn}(\sigma)(\omega \otimes \eta)(w_{\sigma(1)},\dotsc,w_{\sigma(p+q)})$

Where $w_{i}$ denotes $F(v_{i})$. I am unsure on how to proceed from here, so any help would be greatly appreciated!


Try starting from the other direction and use

$(F^*\omega) \otimes (F^*\eta)(v_{\sigma(1)},\ldots, v_{\sigma(p+q)}) = (F^*\omega)(v_{\sigma(1)},\ldots,v_{\sigma(p)})\cdot (F^*\eta)(v_{\sigma(p+1)},\ldots, v_{\sigma(p+q)}) = \omega(F(v_{\sigma(1)}),\ldots,F(v_{\sigma(p)}))\cdot\eta(F(v_{\sigma(p+1)}),\ldots,F(v_{\sigma(p+q)})) = \omega \otimes \eta (F(v_{\sigma(1)}),\ldots,F(v_{\sigma(p+q)}))$

and then compare to what you already have.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.