Pullbacks: interaction with wedge product and exterior derivative.

A reference for this material is Chapters 12 and 14 of John M. Lee. Introduction to smooth manifolds. Second edition. Grad. Texts in Math., 218. Springer, New York, 2013. xvi+708pp. ISBN: 978-1-4419-9981-8.

Note that if \(g \in \Omega ^0 (N)\) is just a function, then \(f^{\ast } g = g \circ f\).

Solution:

For vector fields \(X_1, \ldots , X_{k + \ell } \in \mathfrak {X}(M)\), we compute

\begin{align*} f^{\ast } (\omega & \wedge \eta ) (X_1, \ldots , X_{k + \ell } ) = \omega \wedge \eta (f_{\ast } X_1 , \ldots , f_{\ast } X_{k + \ell } ) \\ & = \frac {1}{k! \, \ell ! }\sum _{\sigma \in S_{k + \ell }} \text {sign} (\sigma ) \omega ( f_{\ast } X_{\sigma (1) }, \ldots , f_{\ast } X_{\sigma (k)} ) \eta ( f_{\ast } X_{\sigma (k + 1 )}, \ldots , f_{\ast } X_{\sigma (k + \ell )} ) \\ & = \frac {1}{k! \, \ell ! }\sum _{\sigma \in S_{k + \ell }} \text {sign} (\sigma ) f^{\ast } \omega ( X_{\sigma (1) }, \ldots , X_{\sigma (k)} ) f^{\ast } \eta ( X_{\sigma (k + 1 )}, \ldots , X_{\sigma (k + \ell )} ) \\ & = ( f^{\ast } \omega ) \wedge (f^{\ast } \eta ) (X_1, \ldots , X_{k + \ell }). \end{align*}

To prove the second identity, we first prove it for functions. That is, if \(\omega \in \Omega ^0 (M)\), then

\begin{align*} d f^{\ast } \omega (X) & = d (\omega \circ f) (X)\\ & = X (\omega \circ f) \\ & = f_{\ast } X (\omega ) \\ & = d\omega (f_{\ast } X ) \\ & = f^{\ast } d \omega (X) \end{align*}

Now we work in local coordinates. Let \((x_1, \ldots , x_m)\) be coordinates on \(M\) and \((y_1, \ldots , y_n)\) coordinates on \(N\). Recall that

\[ f_{\ast } \frac {\partial }{\partial x ^i } = \sum _{j = 1 } ^n \frac {\partial y ^j}{\partial x ^i} \frac {\partial }{\partial y ^j}. \]
We now do another elementary case:
\[ \omega = dy^a . \]
Then
\[ f^{\ast } \omega = f^{\ast } dy^a = \sum _{i = 1 } ^m \frac {\partial y ^a}{\partial x ^i} d x ^i. \]
We compute
\[ df ^{\ast } \omega = \sum _{i,j = 1}^m \frac {\partial ^2 y ^a}{\partial x^i \partial x ^j } dx^i \wedge dx^j . \]
For each pair \(i < j\), we know
\[ \frac {\partial ^2 y ^a}{\partial x^i \partial x ^j } dx^i \wedge dx^j + \frac {\partial ^2 y ^a}{\partial x^j \partial x ^i } dx^j \wedge dx^i = 0 , \]
so
\[ df ^{\ast } d y ^a = 0 . \]
Now consider
\[ \omega = g \, dy ^{i_1} \wedge \cdots \wedge dy ^{i_k} . \]
Then, using the first property and the Leibniz rule,
\begin{align*} d f^{\ast } \omega & = d [ ( f^{\ast }g ) \, f^{\ast }dy^{i_1} \wedge \cdots f^{\ast } dy ^{i_k}] \\ & = [ d( f^{\ast }g ) ] \wedge f^{\ast }dy^{i_1} \wedge \cdots f^{\ast } dy ^{i_k} \\ & \,\,\,\,\,\, + \sum _{a = 1 }^k (-1 ) ^{a-1} (f^{\ast }g) \, f^{\ast } dy^{i_1 } \wedge \cdots \wedge (df ^{\ast } dy ^{i_a} ) \wedge \cdots \wedge f^{\ast } dy^{i_k} \\ & = [ f^{\ast } dg ] \wedge f^{\ast }dy^{i_1} \wedge \cdots f^{\ast } dy ^{i_k} \\ & = f^{\ast } [ d g \wedge dy^{i_1} \wedge \cdots \wedge dy ^{i_k} ] \\ & = f ^{\ast } d \omega . \end{align*}