Im stuck in this exercise on page 278 of Analysis II of Amann and Escher

Suppose $M:=\{(x,y,z)\in\Bbb R^3: x^2+y^2=1\}$ and $$\nu:M\to\mathrm S^2,\quad (x,y,z)\mapsto (x,y,0)$$ Show that $\nu\in C^\infty(M,\mathrm S^2)$ and that $T_p\nu$ is symmetric with eingenvalues $0$ and $1$.

(Here $T_p\nu$ is the tangential of $\nu$ around $p$, and $i_M$ is the canonical injection $i_M:M\to\Bbb R^n$ for some $M\subset\Bbb R^n$.)

My work so far: clearly $\nu$ is continuous, we want to show now that it is also smooth. Let

$$ \psi_\pm:M\setminus F_{\pm}\to(-\pi,\pi)\times\Bbb R,\quad (x,y,z)\mapsto(\arg(\mp x\mp iy),z) $$

That is $\{\psi_-,\psi_+\}$ is an atlas of $M$. Now define $$ \varphi_-:\mathrm S^2\setminus H_2\to (-\pi,\pi)\times (0,\pi),\quad (x,y,z)\mapsto \left(\arg\left(\frac{x+iy}{\sin(\arccos z)}\right),\arccos z\right) $$ where $H_2:=(-\infty,0]\times\{0\}\times \Bbb R$ and $\arg$ is the principal argument, defined in $(-\pi,\pi)$. Its easy to see that $\varphi_-$ is a chart and together with $\varphi^*:=\varphi_-\circ\sigma$, where $\sigma:=(123)$ is a permutation, they are an atlas of $\mathrm S^2$.

Now observe that, under the action of $\nu$, $\psi_-$ and $\varphi_-$ map the same territories and $$ \nu_{\psi_-,\varphi_-}(\theta,r)=(\varphi_-\circ\nu)(\cos\theta,\sin\theta,r)=\varphi_-(\cos\theta,\sin\theta,0)=(\theta,\pi/2)\\\implies[\partial(\nu_{\psi_-,\varphi_-})(\theta,r)]=\begin{bmatrix}1&0\\0&0\end{bmatrix} $$

A similar result can be shown for $\psi_+$ and $\varphi_+(x,y,z):=\varphi_-(-x,-y,z)$. Thus $\nu\in C^\infty(M,\mathrm S^2)$ as expected.

Now to define $T_p\nu$, for some $p\in M$, we need to use the functional equation $$ T_{\psi(p)}\nu_{\psi,\varphi}\circ T_p\psi=T_{\nu(p)}\varphi\circ T_p\nu $$

where $\psi$ is a chart around $p\in M$ and $\varphi$ is a chart around $\nu(p)\in\mathrm S^2$. Also we knows that $T_{x_0}g=(g(x_0),\partial g(x_0))$ for a parametrization $g$ and it must hold that $(T_{\psi(p)}g)(T_p\psi)=i_{T_pM}$ for $g=i_M\circ \psi^{-1}$. Hence the tangential part $A_\psi$ of the chart $\psi$ is defined by the equation $\partial \psi^{-1}(x_0)A_\psi=I_n$, for $M\subset\Bbb R^n$.

Observe that $$ \psi_\pm^{-1}(\theta,r)=(\mp \cos\theta,\mp \sin\theta,r)\implies[\partial\psi^{-1}_\pm(\theta,r)]=\begin{bmatrix}\pm\sin\theta&0\\\mp\cos\theta&0\\0&1\end{bmatrix}\implies A_\psi=\begin{bmatrix}???\end{bmatrix} $$ But Im unable to find an $A_\psi$ that holds my hypothesis when $\cos\theta=0$ or $\sin\theta=0$. Probably Im doing an overkill or something but I dont know how to continue. Some help will be appreciated, thank you.


Ok, I solved it. It was not as hard as I thought, I just needed a little more of thinking.

Define $g_\pm:=i_M\circ \psi_\pm^{-1}$ and $V_\pm:=\psi_\pm(M)$, then

$$ T_{\psi(p)}g:T_{\psi(p)}V\to T_p\Bbb R^2\quad\text{and}\quad T_p\psi:T_pM\to T_{\psi(p)}V\tag1 $$

for some $p:=(x,y,z)\in M$. This together with $T_{\psi(p)}g\circ T_p\psi=i_{T_pM}$ means that the tangential part of $T_p\psi$ is just the transpose of the tangential part of $T_{\psi(p)}g$, that is, $A_{\psi_\pm}=[\partial\psi^{-1}_\pm(\theta,r)]^T$. Also observe that

$$ T_{\psi(p)}\nu_{\psi,\varphi}=\big((\varphi\circ \nu)(p),\partial\nu_{\psi,\varphi}(\psi(p))\big)\tag2 $$ because $\nu_{\psi,\varphi}$ is a function between open sets in euclidean spaces. Also $$ \varphi^{-1}_\pm(\theta,\eta)=(\mp\cos\theta\sin\eta,\mp\sin\theta\sin\eta,\cos\eta)\\\implies[\partial\varphi^{-1}_\pm(\theta,\eta)]=\begin{bmatrix}\pm\sin\theta\sin\eta&\mp\cos\theta\cos\eta\\\mp\cos\theta\sin\eta&\mp\sin\theta\cos\eta\\0&-\sin\eta\end{bmatrix}\tag3 $$ and for such $p:=(x,y,z)$ $$ (\varphi_\pm\circ\nu)(p)=\varphi_\pm(x,y,0)=(\arg(\mp x\mp iy),\pi/2)\quad\text{and}\quad \psi_\pm(x,y,z)=(\arg(\mp x\mp iy),z)\tag4 $$ Thus the tangent part of $T_p\nu$ have the form $$ \begin{bmatrix}\sin\theta&0\\\mp\cos\theta&0\\0&-1\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}\sin\theta&\mp\cos\theta&0\\0&0&1\end{bmatrix}=\begin{bmatrix}\sin\theta&0\\\mp\cos\theta&0\\0&0\end{bmatrix}\begin{bmatrix}\sin\theta&\mp\cos\theta&0\\0&0&1\end{bmatrix}\\=\begin{bmatrix}\sin^2\theta& \mp\sin\theta\cos\theta&0\\\mp \sin\theta\cos\theta&\cos^2\theta&0\\0&0&0\end{bmatrix}\tag5 $$ And according to the expected, the tangential part of $T_p\nu$ is symmetric, with characteristic polynomial $$ q(x):=(x-\sin^2\theta)(x-\cos^2\theta)x-\sin^2\theta\cos^2\theta x=x(x^2-x)=x^2(x-1)\tag6 $$ so the eigenvalues of the tangential part of $T_p\nu$ are $0$ and $1$.


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.