Calculate Pullback of differential form

Question

Let $\varphi:\mathbb{R}^n \setminus \{0\} \to \mathbb{R}^n \setminus \{0\}$ be given by $$ \varphi(x)=\frac{x}{||x||} $$ where $x=(x_1,...,x_n)$, and $||x||=\sqrt{x_1^2+\cdots +x_n^2}$. For $n-1$ differential form $$ \omega = \sum_{k=1}^n (-1)^{k+1} x^k dx^1 \wedge \cdots \wedge dx^{k-1} \wedge dx^{k+1} \wedge \cdots \wedge dx^{n} $$ We need find its pullback $\varphi^*(\omega)$, and prove that $$ d(\varphi^*(\omega))=0 $$

I have proved that $d\left(\frac{\omega}{||x||^n} \right)=0$, and $d\omega =n dx^1 \wedge \cdots \wedge dx^n$, but I don't know here how to calculate $\varphi^*(\omega)$.

Answer 1

The pullback acts by substitution. I'm not sure how to interpret $x^k$, so I'll interpret it as $x_k$ for the sake of the demonstration:

$$\varphi^*(\omega)=\sum\limits_{k=1}^n (-1)^{k+1}\left(\frac{x_k}{||x||}\right) d\left(\frac{x_1}{||x||}\right)\wedge d\left(\frac{x_2}{||x||}\right)\wedge\cdots \wedge d\left(\frac{x_{k-1}}{||x||}\right) \wedge d\left(\frac{x_{k+1}}{||x||}\right)\wedge\cdots\wedge d\left(\frac{x_n}{||x||}\right).$$ I'll leave it to you to calculate $d\left(\frac{x_j}{||x||}\right).$

EDIT: A hint on the calculation. For a function $f$, we find its exterior derivative via $$df=\sum\limits_{i=1}^n\frac{\partial f}{\partial x^i} dx^i.$$ So, you need to calculate the partial derivatives of $f(x)=\frac{x_j}{||x||}.$ You should get that $$\frac{\partial}{\partial x^i}\left(\frac{x_j}{||x||}\right)=\frac{\delta_i^j}{||x||}-\frac{x_ix_j}{||x||^3}.$$

Answer 2

You only need to use these facts to derive the formula, which is easy to remember because it turns out to be a simple substitution:

$1).\ $ pullback commutes with the exterior derivative.

$2).\ $ pullback distributes over the wedge product.

$3).\ $ in local coordinates, $d\phi=\sum^n_{k=1}\partial_kdx_k.$

$4).\ \phi^*(f)=f\circ \phi$ on $0-$ forms.

Now, if $\omega(x) = \sum_{k=1}^n (-1)^{k+1} x^k dx^1 \wedge \cdots \wedge dx^{k-1} \wedge dx^{k+1} \wedge \cdots \wedge dx^{n},$

then

$(\phi^*\omega)(x)=\sum_{k=1}^nf_k(\phi(x)) \phi^*(dx^1 \wedge \cdots \wedge dx^{k-1} \wedge dx^{k+1} \wedge \cdots \wedge dx^{n}))$,

where $f_k:\mathbb R^n\setminus \{0\}\to \mathbb R:(x_1,\cdots,\cdots, x_n)\mapsto (-1)^{k-1}x_k$.

Using the facts outined above, we get

$(\phi^*\omega)(x)=\sum_{k=1}^nf_k(\phi(x)) (d\phi^*x^1 \wedge \cdots \wedge d\phi^*x^{k-1} \wedge d\phi^*x^{k+1} \wedge \cdots \wedge d\phi^*x^{n}))=\sum_{k=1}^nf_k(\phi(x)) (dx^1\circ \phi) \wedge \cdots \wedge d(x^{k-1}\circ \phi) \wedge d(x^{k+1}\circ \phi) \wedge \cdots) \wedge d(x^{n}\circ\phi))=\sum_{k=1}^nf_k(\phi(x)) d\phi^1 \wedge \cdots \wedge d\phi^{k-1} \wedge d\phi^{k+1} \wedge \cdots \wedge d\phi^{n}).$

This last item is the formula you can use to calculate the pullback, with $f_k(\phi(x))=(-1)^{k+1}\frac{x^k}{||x||},\ d\phi^k=\sum^n_{j=1}\partial_j\phi^kdx^j$ and noting that any repeated $1-$form $dx^k$ in the sum will force the corresponding term to $0$.