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 $$f^*\sigma=\frac{1}{|p|^n}\sigma$$

  • $\begingroup$ 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. $\endgroup$ – Nastassja Sep 12 '14 at 8:36
  • $\begingroup$ Let $\gamma(t)$ be any curve whose velocity is $v$ at $p$, that is $\dot\gamma(0)=v$, then $df_p(v)=\frac{d}{dt}|_{t=0}(f\circ \gamma)$ $\endgroup$ – Xipan Xiao Sep 12 '14 at 14:16
  • 1
    $\begingroup$ For example, let $\gamma(t)=p+tv$, we have $df_p(v)=\frac{d}{dt}(\frac{p+tv}{|p+tv|})=\frac{d}{dt}\frac{p+tv}{\sqrt{\langle p+tv, p+tv \rangle}}$=$\frac{v}{|p+tv|}-(p+tv)\frac{\langle v, p+tv \rangle}{|p+tv|^3}|_{t=0}$ $=\frac{v}{|p|}-p\frac{\langle v, p \rangle}{|p|^3}$ $\endgroup$ – Xipan Xiao Sep 12 '14 at 15:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.