# Pullback of a form under the retraction $r\colon \mathbb{R}^n\setminus\{0\}\to S^{n-1}$.

The following is from Spivak's DG Lemma 7 in Chapter 8, but I'm muddled in a computation.

Define two $(n-1)$-forms on $\mathbb{R}^n\setminus\{0\}$ by $$\sigma=\sum_{i=1}^n(-1)^{i-1}x^idx^1\wedge\cdots\wedge\widehat{dx^i}\wedge\cdots\wedge dx^n$$ $$\omega=\frac{1}{|x|^n}\sigma.$$

If $r(x)=\frac{x}{|x|}$ is the retraction, and $\iota\colon S^{n-1}\to\mathbb{R}^n\setminus\{0\}$ the inclusion, then $\omega=r^*\iota^*\sigma$.

Pick $p\in\mathbb{R}^n\setminus\{0\}$, and let $v_1,\dots,v_{n-1}$ be tangent vectors, then \begin{align*} r^*\iota^*\sigma_p(v_1,\dots,v_{n-1})&=(\iota\circ r)^*\sigma_p(v_1,\dots,v_{n-1})\\ &=\sigma_{\iota r(p)}(d(\iota\circ r)_p(v_1),\dots,d(\iota\circ r)_p(v_{n-1}))\\ &=\sigma_{p/|p|}(d(\iota\circ r)_p(v_1),\dots,d(\iota\circ r)_p(v_{n-1}))\\ &=\left(\sum_{i=1}^n(-1)^{i-1}x^i(p/|p|)dx^1|_{p/|p|}\wedge\cdots\wedge\widehat{dx^i|_{p/|p|}}\wedge\cdots\wedge dx^n|_{p/|p|}\right)(d(i\circ r)_p v_1,\dots,d(\iota\circ r)_p v_{n-1}) \end{align*}

On the other hand, \begin{align*} \omega_p(v_1,\dots,v_{n-1})&=\frac{1}{|p|^n}\sigma_p(v_1,\dots,v_{n-1})\\ &=\frac{1}{|p|^n}\left(\sum_{i=1}^n(-1)^{i-1}x^i(p)dx^1|_p\wedge\cdots\wedge\widehat{dx^i|_p}\wedge\cdots\wedge dx^n|_p\right)(v_1,\dots,v_{n-1})\\ \end{align*}

How can I see that these two expressions are equal to each other? Thanks.

To simplify, denote $$X=\sum_{i=1}^nx^i\frac{\partial}{\partial x^i}$$ and $\Omega=dx^1\wedge dx^2 \cdots \wedge dx^n$, then $\sigma=i(X)\Omega$, where $i$ is the Interior Product operator, that is, $\sigma(\cdots)=\Omega(X,\cdots)$.
$\forall p\in \mathbb{R}^n$, let $q=r(p)$. Note that $X$ has the following property: $$X_q=\sum_{i=1}^n\frac{x^i(p)}{|p|}\frac{\partial}{\partial x^i}=\frac{X_p}{|p|}$$ Now let $f=\iota\circ r$, to compute $f^*\sigma$ at $p$, let's choose an orthonormal basis at $p$ as $$\{e_1=\frac{p}{|p|}, e_2, \cdots, e_n\}$$ It's easy to verify the following for the differential map $f_*$: $$f_*(e_1)=0,\ f_*(e_i)=\frac{e_i}{|p|}, \forall i=2,\cdots,n$$ So we need only to compute $f^*\sigma$ against $\{e_2,\cdots, e_n\}$, and $$(f^*\sigma)_p(e_2,\cdots,e_n)=\sigma_q(f_*(e_2),\cdots,f_*(e_n))$$ $$=\Omega_q(\frac{X_p}{|p|},\frac{e_2}{|p|},\cdots,\frac{e_n}{|p|})$$ $$=\frac{1}{|p|^n}\Omega_p(X_p,e_2,\cdots,e_n)$$ $$=\frac{1}{|p|^n}\sigma_p(e_2,\cdots,e_n)$$ That is $$f^*\sigma=\frac{1}{|p|^n}\sigma$$

• Thanks Xipan, can you explain how you see that $f_*(e_1)=0$, and $f_*(e_i)=\frac{e_i}{|p|}$? I got stuck there in my own work trying to find $d(\iota\circ r)_p(v_i)$, because the formula I know for the differential is $df_p=\sum_i\frac{\partial f}{\partial x^i}(p)dx^i|_p$, but I'm having difficulty applying it. – Nastassja Sep 12 '14 at 8:36
• Let $\gamma(t)$ be any curve whose velocity is $v$ at $p$, that is $\dot\gamma(0)=v$, then $df_p(v)=\frac{d}{dt}|_{t=0}(f\circ \gamma)$ – Xipan Xiao Sep 12 '14 at 14:16
• For example, let $\gamma(t)=p+tv$, we have $df_p(v)=\frac{d}{dt}(\frac{p+tv}{|p+tv|})=\frac{d}{dt}\frac{p+tv}{\sqrt{\langle p+tv, p+tv \rangle}}$=$\frac{v}{|p+tv|}-(p+tv)\frac{\langle v, p+tv \rangle}{|p+tv|^3}|_{t=0}$ $=\frac{v}{|p|}-p\frac{\langle v, p \rangle}{|p|^3}$ – Xipan Xiao Sep 12 '14 at 15:33