3

I am trying to show by induction that pullbacks and the exterior derivative commute. I know that this question has been discussed on this site, eg. here and here. However, none of these questions employ the approach that I'm interested in:

So if $f: X \rightarrow Y$ is a smooth map of manifolds and $\alpha \in \Omega^p(Y)$ is a differential p-form on $Y$, then I want to show that

$$f^*(d\alpha) = d(f^*(\alpha)) \textit{ in } \Omega^{p+1}(X)$$

So my approach is via induction on the $p$, exploiting that

  1. Pullbacks commute with wedge products $f^{*}(\alpha \wedge \beta) = f^{*}(\alpha)\wedge f^{*}(\beta)$
  2. And that the exterior derivative is unique under the conditions that (i) $d^2 = 0$, (ii) $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^k \alpha \wedge d\beta$, (iii) $dg$ for a smooth map $g \in C^{\infty}(X)$ is just the usual derivative

So the induction step is via the commutativity with wedge products, but I'm struggling to unwrap the definitions for the base case. Here I just need to verify that for a smooth map $a\in C^{\infty}(Y)$ the following holds:

$$f^* (da) = d(f^*(a))$$

I think this should just be a notation chase. Let $f(x) = y$, so the LHS is just

$$f^*(da)\rvert_x = (T^*f)(da\rvert_y) \in T_x^*X$$

Whereas the RHS is

$$d(f^*a)\rvert_x = d(a\circ f)\rvert_x \in T_x^{*} X$$

How can I show that these two expressions are equal? Did I get something wrong?

gen
  • 1,608
  • It's not just a notation chase. Use the definitions of the notation you're using. For example, what does $T^*f$ mean? My suggestion is that you should initially just do this in local coordinates and translate your proof back to abstract notation after that.. – Deane Oct 24 '21 at 18:03
  • Yes, I do think that it's the abstraction that made it hard for me to understand this. And based on the answer below I also understand that it's the chain rule that is required here, but to be honest I haven't been able to see how it fits into the above formulation. – gen Oct 24 '21 at 18:11
  • Could you explain what the definition of $T^*f$ is? I'm not familiar with that notation. – Deane Oct 24 '21 at 19:03

1 Answers1

1

The chain rule tells you that $d(a\circ f)_x=da_{f(x)}\circ df_x$, which is what you need to conclude.

Balloon
  • 8,624
  • Sorry for the late follow-up, could you rewrite your answer using the same formulae as the ones I used in my question? Looking back I'm a little confused how your expressions correspond to mine. – gen Oct 24 '21 at 17:14