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Let $\omega=\sum_j \omega_jdy^j$. We want to show that $d(F^{*}\omega)=F^{*}(d\omega)$, where $F: M \to N$ is a map from manifold $M$ to manifold $N$. Local coordinate systems of $M, N$ are $(y^1, \ldots, y^m), (x^1, \ldots, x^n)$ respectively.

Now

$$F^{*}\omega=\sum_{j,k}\omega_j\frac{\partial y^j}{\partial x^{k}}dx^k,$$

$$d(F^{*}\omega)=\sum_{j,k,i}\frac{\partial ( \omega_j\frac{\partial y^j}{\partial x^k} )}{\partial x^i}dx^i \wedge dx^k = \sum_{j,k,i}\frac{\partial \omega_j}{\partial x^i} \frac{\partial y^j}{\partial x^k} dx^i \wedge dx^k + \sum_{j,k,i} \omega_j \frac{\partial^2 y^j}{\partial x^i \partial x^k} dx^i \wedge dx^k.$$

$$d\omega=\sum_{j,l}\frac{\partial \omega_j}{\partial y^i} dy^l \wedge dy^j,\qquad F^{*}d\omega = \sum_{i,j,k,l} \frac{\partial \omega_j}{\partial y^l}\frac{\partial y^l}{\partial x^i} \frac{\partial y^l}{\partial x^i} dx^i \wedge dx^k = \sum_{i,j,k} \frac{\partial \omega_j}{\partial x^i} \frac{\partial y^l}{\partial x^i} dx^i \wedge dx^k.$$

My question is why $\displaystyle\sum_{j,k,i} \omega_j \frac{\partial^2 y^j}{\partial x^i\; \partial x^k} dx^i \wedge dx^k = 0$? Thank you very much.

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2 Answers 2

up vote 3 down vote accepted

Hint: $$\frac{\partial^2 y^j}{\partial x^i\partial x^k} = \frac{\partial^2 y^j}{\partial x^k\partial x^i}$$ but $$dx^i \wedge dx^k = - dx^k \wedge dx^i.$$

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Forget about $j$ for a moment.
The point is that for a twice continuously differentiable function $f$, you have
$$\sum_{k,i} \frac{\partial^2 f}{\partial x^i \partial x^k} dx^{i} \wedge dx^{k} = 0 \quad (*)$$ This results from:
a) Schwarz's lemma $\frac{\partial^2 f}{\partial x^i \partial x^k} =\frac{\partial^2 f}{\partial x^k \partial x^i} $
b) $dx^{i} \wedge dx^{k}=-dx^{k} \wedge dx^{i}$ (for $i\neq k)$
c) $dx^{i} \wedge dx^{i}=0$
and a regrouping of the sum $(*)$ of $n^2$ terms into three partial sums that I'll leave to you.

The sum you are interested in consists in multiplying the zero expression $(*)$ by $\omega_j$ on the left and adding for all $j$: obviously you will still obtain zero.

Warning: Schwarz $\neq$ Schwartz

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