Pulling back differential forms

We have seen that if $f \colon M \to N$ is a smooth map then it has a derivative or tangent map $T_x(f)$ that acts on tangent vectors in $T_xM$ by sending them to $T_{f(x)}N$. Moreover $T_x(f)$ is linear. Recall that if $X \colon V \to W$ is a linear map between vector spaces then it has an adjoint or dual $X ^* \colon W^* \to V^*$ defined by

$$X^*(\xi)(v) = \xi(X(v))$$

where $\xi \in W^*$ and $v \in V$. Notice that $X^*$ goes in the opposite direction to $X$. So we have a linear map called the cotangent map

$$T^*_x(f) \colon T_{f(x)} N \to T_x M$$

which is just the adjoint of the tangent map. It is defined by

$$T^*_x(f)(\omega)(X) =\omega(T_x(f)(X).$$

This action defines a map on differential forms called the pull-back by $f$ and denoted $f^*$. if $\omega \in \Omega^k(N)$ then we define $f^*(\omega) \in \Omega^k(M)$ by

$$f^*(\omega)(x)(X_1, \dots, X_k) = \omega(f(x))(T_x(f)(X_1), \dots, T_x(f)(X_k))$$

for any $X_1, \dots, X_k$ in $T_xM$.

Notice that if $\phi$ is a zero form or function on $N$ then $f^{-1}(\phi) = \phi \circ f$. The pull back map

$$f^{*} \colon \Omega^{q}(N) \to \Omega^{q}(M).$$

satisfies the following proposition.

Proposition 5.4   If $f \colon M \to N$ is a smooth map and $\omega$ and $\mu$ is a differential form on $N$ then:

1.

$df^{*}(\omega) = f^{*}(d\omega)$, and

2.

$f^{*}(\omega \wedge \mu) = f^{*}(\omega) \wedge f^{*}(\mu)$.

Proof. Exercise. $\qedsymbol$