1
$\begingroup$

Suppose that $\phi : M \rightarrow N$ is a smooth map between differential manifolds. Let $\omega, \eta$ be forms on $N$. Is there an easy proof for the fact that $$\phi^*(\omega \wedge \eta) = \phi^* \omega \wedge \phi^* \eta?$$ In other words, $\phi^* : \Omega(N) \rightarrow \Omega(M)$ is a graded algebra homomorphism. (By easy, I mean does not require expanding both sides using the definition.)

$\endgroup$
1
$\begingroup$

I've read your question wrong, here is a more general answer you want, I guess.

We will use this following proprieties:

$$\phi^*(\omega_1+\omega_2)= \phi^*(\omega_1)+\phi^*(\omega_2)$$

$$\omega \wedge \eta = \frac {(m+l)!}{m!l!}Alt(\omega \otimes \eta) $$

$$Alt(\omega \otimes \eta)=\frac {1}{(m+l)!} \sum_{\sigma \in S_ {m+l }} (\omega \otimes \eta)\circ\sigma $$

By that, $$\phi^*(\omega \wedge \eta)= \frac {(m+l)!}{m!l!}\phi^*(Alt(\omega \otimes \eta)) = \frac {(m+l)!}{m!l!}\phi^*(\frac {1}{(m+l)!} \sum_{\sigma \in S_ {m+l }} (\omega \otimes \eta)\circ\sigma) =\\ =\frac {1}{m!l!} \sum_{\sigma \in S_ {m+l }}\phi^* (\omega \otimes \eta)\circ\sigma $$

Now if we prove that $\phi^*(\omega \otimes \eta)\circ\sigma = (\phi^*\omega \otimes \phi^*\eta)\circ\sigma $, we have that

$$\frac {1}{m!l!} \sum_{\sigma \in S_ {m+l }}\phi^* (\omega \otimes \eta)\circ\sigma = \frac {1}{m!l!} \sum_{\sigma \in S_ {m+l }} (\phi^*\omega \otimes\phi^*\eta)\circ\sigma = \phi^*(\omega)\wedge \phi^*(\eta)$$

Proving that $\phi^*(\omega \otimes \eta)=(\phi^*\omega \otimes \phi^*\eta)$ is easy.

$\endgroup$
2
$\begingroup$

Let $p$ be a point $M$, $v,w \in T_pM$ and $\omega_1, \omega_2, \omega$ and $\eta $ be 1-forms of $M$ . We have to use that: $$(\omega_1 \wedge \omega_2)_p(v,w) = \begin{vmatrix} {\omega_1}_p(v) & {\omega_1}_p(w) \\ {\omega_2}_p(v) & {\omega_2}_p(w) \end{vmatrix}$$

$\phi^*(\omega \wedge\eta)_p(v,w) = (\omega \wedge \eta)_{\phi (p)}(d \phi_p(v),d \phi_p(w)) = \begin{vmatrix} {(\omega)}_{\phi(p)}(d \phi_p(v)) & {(\omega)}_{\phi(p)}(d \phi_p(w)) \\ {(\eta)}_{\phi(p)}(d \phi_p(v) & {(\eta)}_{\phi(p)}(d \phi_p(w)) \end{vmatrix} = \begin{vmatrix} {(\phi^*\omega)}_p(v) & {(\phi^*\omega)}_p(w) \\ {(\phi^*\eta)}_p(v) & {(\phi^*\eta)}_p(w) \end{vmatrix} = (\phi^*\omega) \wedge (\phi^*\eta)_p(v,w) $.

So we have $\phi^*( \omega \wedge \eta) = \phi^*(\omega) \wedge \phi^*(\eta).$

$\endgroup$
  • 1
    $\begingroup$ I see what you are trying to do, I like you aprroach. But what if $\omega=\sum f_{i_1,\ldots, i_k} dx_{i_1}\wedge\cdots\wedge dx_{i_k}\in \Gamma^\infty \Big(\Lambda^k\big( T^* M \big)\Big) $ and $\eta=\sum g_{j_1,\ldots, j_\ell} dx_{j_1}\wedge\cdots\wedge dx_{j_\ell}\in \Gamma^\infty \Big(\Lambda^\ell\big( T^* M \big)\Big)$ are arbitrary forms ( of degree $k$ and $\ell$ respectively) ? $\endgroup$ – EternalBlood Mar 19 '17 at 3:05
  • $\begingroup$ Maybe using this first case, linearity and induction. $\endgroup$ – Danuso Rocha Mar 19 '17 at 3:18
  • $\begingroup$ Thanks for alert me, @EternalBlood. I gave the general answer. :) $\endgroup$ – Danuso Rocha Mar 19 '17 at 5:16

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.