# Some clarifications regarding the pullback of differential forms

I recently started studying pullbacks and differential forms. As far as I understand, one can define, for a smooth map $f:M\rightarrow N$, the pullback of $f$ to be the map $$f^{*}:\Omega(N)\rightarrow\Omega(M)$$ such that, for all $\omega\in\Omega^{k}(N)$, $p\in M$ and $(v_{1},...,v_{k})\in T_{p}M\times...\times T_{p}M$, $$f^{*}\omega(p)(v_{1},...,v_{k})=\omega(f(p))\left(f_{*}(v_{1}),...,f_{*}(v_{k})\right)$$ (Notice that $f^{*}\omega(p)\in\bigwedge^{k}\left(T_{p}^{*}M\right)$).

I am also familiar with the following properties of the pull-back:

$\bullet$ $f^{*}(\omega\wedge\alpha)=f^{*}(\omega)\wedge f^{*}(\alpha)$
$\bullet$ $(f\circ g)^{*}=g^{*}\circ f^{*}$
$\bullet$ $f$ is $\mathbb{R}$-linear.

Now, I have seen this done: (assume $\phi:M\supseteq U\rightarrow \phi(U)\subseteq\mathbb{R}^m$ and $\omega'\in\Omega^{k}(\phi(U))$) $$\phi^{*}(\omega')=\phi^{*}(\underset{I}{\sum}a_{I}\text{dx}_{I})=\underset{I}{\sum}\phi^{*}(a_{I})\cdot\phi^{*}(\text{dx}_{I})=\underset{I}{\sum}(a_{I}\circ\phi)\cdot\text{d}\phi_{I}$$

My problems with this:

1) I think I understand the first equality: $\{dx_{i_1}\wedge...\wedge dx_{i_k}\,:\,1\le i_1 <...< i_k \le m\}$ is a basis for $\Omega^k(\phi(U))$ (and $dx_I:=dx_{i_1}\wedge...\wedge dx_{i_k}$). The second one, however, I do not: why is $\phi^*(a_I dx_I)=\phi^*(a_I)\cdot \phi^*(dx_I)$? This does not seem to come from the properties of the pullback above. Also, how can $\phi(a_I)$ make sense? $a_I\in\mathbb{R}$, and only the pullback of differential forms is defined. Finally: what operation is the $\cdot$ ?
2) About the third equality: how is $\phi(a_I)=a_I\circ \phi$?

One can think of a smooth function $g: N \rightarrow \mathbb{R}$ as an element of $\Omega^0(N)$. In this case, the pullback of $f^* g$ is well-defined, and tracing through the definition, $f^*g(p) = g(f(p)) = (g \circ f)(p)$. Moreover, given $g \in \Omega^0$ and $\omega \in \Omega^k$, $f^* (g \omega) = f^*g \wedge f^*\omega$ by the first property, and since $f^* g$ is a function, $f^*g \wedge f^*\omega = (g \circ f ) \cdot f^* \omega$.
This is what they are doing in that example; think of $dx_I$ as a $k$-form, and $a_I$ as a 0-form.
• You said: "since $f^* g$ is a function, $f^*g \wedge f^*\omega = (g \circ f ) \cdot f^* \omega$." Can you please tell me why that follows? – Soap Mar 16 '17 at 16:48
• Sure, but I'm not sure which part is confusing you. Is it clear why $f^* g = g \circ f$? Otherwise, is it clear why for example $2 \wedge dx = 2 dx$? It is the same thing. – asdjfksj Mar 16 '17 at 17:40
• Just looking at the definition and properties that I have for the pullback, it is not clear why $f^*g=g\circ f$. Is it defined to be like this in the particular case of $g$ being a zero-form? – Soap Mar 16 '17 at 17:54
• The general definition is: $f^*\omega(p)( v_1 \cdots v_k ) = \omega(f(p))(f_* v_1, \cdots f_* v_k)$. If $\omega$ is a zero form, it takes no arguments, so that the definition reduces to $f^* \omega(p)() = \omega(f(p)) () = \omega(f(p)) = (\omega \circ f) (p)$ – asdjfksj Mar 16 '17 at 17:56