# Exterior differentiation under pullback question

In Loring Tu's book, An Introduction to manifolds, he uses the following proposition to prove that the pullback of a smooth $$k$$ form is a smooth $$k$$ form:

Proposition: Let $$F:N\rightarrow M$$ be a smooth map. If $$\omega$$ is a smooth k-form then $$\mathrm dF^{*}\omega=F^{*}\mathrm d\omega.$$

However, my question is: Isn't the exterior derivative, $$\mathrm d$$ is a map $$\Omega^{k}(M)\rightarrow \Omega^{k+1}(M)$$ (space of smooth $$k$$, $$k+1$$ forms), so isn't he assuming that $$F^{*}\omega$$ is smooth?

• There's no need to assume. If $F$ and $\omega$ are smooth, then $F^*\omega$ is automatically smooth as well. Commented Aug 10, 2020 at 16:44
• Yes. One can show that the pullback of a smooth form by a smooth function is smooth. Commented Aug 10, 2020 at 16:54
• Maybe what you've thinking of is that you can use the fact that the pullback of any smooth form by $F$ is a smooth form and the fact that $dF^*\omega=F^*d\omega$ to construct an explicit formula for the pullback of the smooth $k$-form? Commented Aug 10, 2020 at 18:14
• @RachidAtmai: I do not know how to define $d\phi$ unless $\phi$ is at least a $C^1$ differential form. We're not going to talk about currents and distributional derivatives here. Smoothness of $F^*\omega$ is a separate question from computing exterior derivatives, as I already said. When we take exterior derivative, only certain combinations of partial derivatives appear, and that does not establish smoothness. Commented Aug 10, 2020 at 22:08
• I do not own the book, so I can't verify whether OP is misinterpreting. But I surmise he is. The proof that the pullback of a smooth form by a smooth function is an entirely self-contained, separate argument. Start by showing $f^*dx^i$ is smooth. Commented Aug 10, 2020 at 22:11

The way it stands, the placement of Proposition 19.5 is a mistake, because $$F^*\omega$$ needs to be $$C^{\infty}$$ before one can take its exterior derivative. To fix this, in Proposition 19.7, replace the justification "(Proposition 19.5)" by "(Proposition 17.10)," and then move Proposition 19.7 to before Proposition 19.5.

I see that Arctic Char has proposed the same solution a while ago. I give it my ringing endorsement.

• Hi @Loring! :) :) Commented Aug 15, 2020 at 6:13

After reading the corresponding section I agree that the author is wrong in claiming that $$\mathrm d F^*\omega = F^* \mathrm d\omega$$ is used to show that $$F^* \omega$$ is smooth. This is not necessary.

First recall (I am using the second edition):

Proposition 19.7: If $$F : N \to M$$ is a $$C^\infty$$ map of manifolds and $$\omega$$ is a $$C^\infty$$ $$k$$-form on $$M$$, then $$F^*\omega$$ is a $$C^\infty$$ $$k$$-form on $$N$$.

Proof (Sketch) given in the book: In a local coordinates, $$\omega = \sum_I a_I \mathrm dy^{i_1} \wedge \cdots \wedge \mathrm dy^{i_k}.$$ for some local smooth functions $$a_I$$. Then

\begin{align} F^*\omega &= \sum (F^* a_I) F^* \mathrm dy^{i_1} \wedge \cdots \wedge F^* \mathrm dy^{i_k} \\ &= \sum (F^* a_I) \mathrm d(F^* y^{i_1}) \wedge \cdots \wedge \mathrm d (F^* y^{i_k}) \ \ \ \ \ (\text{Proposition }19.5)\\ &= \cdots \\ &= \sum (a_I \circ F)\frac{\partial (F^{i_1}, \cdots, F^{i_k})}{\partial (x^{j_1} \cdots x^{j_k})} \mathrm dx^J. \end{align}

Since

$$(a_I \circ F)\frac{\partial (F^{i_1}, \cdots, F^{i_k})}{\partial (x^{j_1} \cdots x^{j_k})}$$ are smooth, the author concludes that $$F^*\omega$$ is smooth.

Proposition 19.5 says that for any smooth $$k$$-form $$\omega$$ we have $$F^* \mathrm d \omega = \mathrm d F^* \omega$$.

As already pointed out by TedShifrin in the comment, only $$F^* dy^{i_l} = d (F^* y^{i_l})$$ is needed to show Proposition 19.7, and the proof in the book is exactly using just that. This fact is proved in the previous section (Proposition 17.10).

So I think it might be a typo to use Proposition 19.5 to prove Proposition 19.7. Indeed he needs only to use 17.10. Also it is confusing to put Proposition 19.5 before Proposition 19.7, that is, showing $$F^* \mathrm d\omega = \mathrm d F^*\omega$$ without first showing $$F^*\omega$$ is smooth. I did not check the whole book, but I guess the concept of $$C^1$$-differential form is not introduced. So it does not really make sense to talk about $$\mathrm d F^*\omega$$ without first showing that $$F^*\omega$$ is $$C^\infty$$, at least in the context of this book.

• This clears everything up. Thank you.
– user643073
Commented Aug 11, 2020 at 6:18
• Arctic Char, nice job getting Loring Tu's 'ringing endorsement'
– BCLC
Commented May 1, 2021 at 3:46