Pull-back of a differential : I get confused with the variables

Question

I edited again my message with the remarks done in the comments.

I have a 2-form :

$$\alpha=\alpha_{\mu \nu} dx^\mu \wedge dx^\nu$$

I want to compute the pull back $F^{*}(d \alpha)$ to show that : $F^{*}(d \alpha)=dF^{*}( \alpha)$

But I make a mistake somewhere because I can't prove the equality.

$$ F : y \mapsto x $$

So when I write $x^\mu$ I have in fact a dependance $x^\mu(y^\nu)$.

$$d \alpha=d \alpha_{\mu \nu} \wedge dx^\mu \wedge dx^\nu \\= \frac{\partial \alpha_{\mu \nu}}{\partial x^\epsilon} dx^\epsilon \wedge dx^\mu \wedge dx^\nu$$

$$F^{*}(d \alpha)=F^{*}(\frac{\partial \alpha_{\mu \nu}}{\partial x^\epsilon}) F^{*}(dx^\epsilon) \wedge F^{*}(dx^\mu) \wedge F^{*}(dx^\nu)$$

I have :

$$ F^{*}(dx^\mu) = \frac{\partial x^\mu}{\partial y^i} dy^i$$

and

$$F^{*}(\frac{\partial \alpha_{\mu \nu}(x)}{\partial x^\epsilon})=\frac{\partial \alpha_{\mu \nu}(x(y))}{\partial x^\epsilon(y)}$$

And finally, I get :

$$F^{*}(d \alpha)=\frac{\partial \alpha_{\mu \nu}(x(y))}{\partial x^\epsilon(y)} \frac{\partial x^\epsilon}{\partial y^i}\frac{\partial x^\mu}{\partial y^j}\frac{\partial x^\nu}{\partial y^k} dy^i \wedge dy^j \wedge dy^k$$

On the other hand, I have :

$$F^{*}(\alpha)=F^{*}(\alpha_{\mu \nu}(x)) F^{*}(dx^\mu) \wedge F^{*}(dx^\nu)\\ =\alpha_{\mu \nu}(x(y)) \frac{\partial x^\mu}{\partial y^i} \frac{\partial x^\nu}{\partial y^j} dy^i \wedge dy^j$$

But here there is a problem when I differentiate :

$$dF^{*}(\alpha)=d(\alpha_{\mu \nu}(x(y)) \frac{\partial x^\mu}{\partial y^i} \frac{\partial x^\nu}{\partial y^j}) \wedge dy^i \wedge dy^j$$

Indeed I will have extra derivative term in $\frac{\partial^2 x^\mu}{\partial y^i \partial y^l} $ when I differentiate. And I don't have these terms in $F^{*}(d \alpha)$.

So where is my mistake ??

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[edit] According to the answer below, I see that my misunderstanding is in the fact that :

$$dF^{*}(\alpha)=d(\alpha_{\mu \nu}(x(y))) \wedge (\frac{\partial x^\mu}{\partial y^i}) dy^i \wedge ( \frac{\partial x^\nu}{\partial y^j} ) dy^j$$

We don't differentiate the terms $\frac{\partial x^\nu}{\partial y^j}$. But I don't understand why as the definition of the exterior derivative is the following :

With :

$$\alpha=\alpha_\mu dx^{\mu}$$

We have by definition :

$$ d\alpha=d\alpha_\mu \wedge dx^{\mu}$$

Thus in my example it should be :

$$dF^{*}(\alpha)=d(\alpha_{\mu \nu}(x(y)) \frac{\partial x^\mu}{\partial y^i} \frac{\partial x^\nu}{\partial y^j} ) \wedge dy^i \wedge dy^j$$

And not :

$$dF^{*}(\alpha)=d(\alpha_{\mu \nu}(x(y))) \frac{\partial x^\mu}{\partial y^i} \frac{\partial x^\nu}{\partial y^j} \wedge dy^i \wedge dy^j$$

ie the differential applies to all the terms including the chain derivative and not only on $\alpha_\mu$. Could someone clarify this for me (or at least give me an exact definition of the exterior derivative ?)

Answer 1

From $d \alpha=d \alpha_{\mu \nu} \wedge dx^\mu \wedge dx^\nu$, we have $$F^*(d\alpha) = d(\alpha_{\mu \nu} \circ F) \wedge d(x^\mu \circ F) \wedge d(x^\nu \circ F).$$ On the other hand, $$dF^*(\alpha) = d[(\alpha_{\mu \nu} \circ F) \wedge d(x^\mu \circ F) \wedge d(x^\nu \circ F)] = d(\alpha_{\mu \nu} \circ F) \wedge d(x^\mu \circ F) \wedge d(x^\nu \circ F). $$

As you can see, writing down the partial derivatives explicitly is of no particular use.

Answer 2

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Your derivation is mostly correct. If $F$ is not a change of coordinate, we usually write the Jacobian as $[\frac{\partial F^i}{\partial y^j}]$ instead of $[\pa{x^i}{y^j}]$. To get the desired result, we just need to expand the terms and use Poincare lemma as follows, \begin{align*} &d(\ai_{ab}\circ F\pa{F^a}{y^i}\pa{F^b}{y^j})dy^i\w dy^j\\ =&\pd_{y^h}(\ai_{ab}\circ F)dy^h\w\pa{F^a}{y^i}dy^i\w\pa{F^b}{y^j}dy^j+\ai_{ab}\circ F\frac{\pd F^a}{\pd y^h\pd x^i}\pa{F^b}{y^j}dy^h\w dy^i\w dy^j\\&+\ai_{ab}\circ F\frac{\pd F^b}{\pd y^h\pd y^j}\pa{F^a}{y^i}dy^h\w dy^i\w dy^j\\ =&(\ai_{ab})\pf\circ F\pf (\pd_{y^h})dy^h\w\pa{F^a}{y^i}dy^i\w\pa{F^b}{y^j}dy^j+\ai_{ab}\circ F\pa{F^b}{y^j}(d^2F^a)\w dy^j\\&-\ai_{ab}\circ F\pa{F^a}{y^i}(d^2F^b)\w dy^i\\ =&(\ai_{ab})\pf(\pd_{x^m})\pa{F^m}{y^h}dy^h\w\pa{F^a}{y^i}dy^i\w\pa{F^b}{y^j}dy^j\\ =&F\du(d\ai). \end{align*}