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.)


2 Answers 2


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.


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

  • 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) ? $\endgroup$ Commented Mar 19, 2017 at 3:05
  • $\begingroup$ Maybe using this first case, linearity and induction. $\endgroup$ Commented Mar 19, 2017 at 3:18
  • $\begingroup$ Thanks for alert me, @EternalBlood. I gave the general answer. :) $\endgroup$ Commented Mar 19, 2017 at 5:16

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