Define the map $$\pi : (\mathbb{R}^3-\{(0,0,0)\})\to S^2$$ by
$\pi(p)=\frac{p}{||p||}.$ Show that if $\Sigma_R$ is the sphere of
radius $R>0$, then the Gauss map of $\Sigma_R$ is $\pi|_{\Sigma_R}$ (which means the map $\pi$ restricted to the surface $\Sigma_R$.)
Compute the shape operator and the Gauss curvature of the sphere.
I know the Gauss maps a surface in $\mathbb{R}^3$ to the sphere $S^2,$ so $\pi(p)$ is a unit vector for all $p\in \sum$ such that $\pi(p)$ is orthogonal to the surface $\mathbb{R}^3$ at $p$. Also, we defined the Gauss curvature as: $ K(p) = \kappa_1 \kappa_2 .$
How can I do this problem?
Best Answer
By symmetry we are sufficient to consider a point $x=(0,0,R)$.
Eigenvalue of Gauss map : If $\pi$ is a Gauss map, we must calculate $d\pi_x\ v$ for any $v\in T_x \Sigma_R$.
By symmetry of $\Sigma_R$, we suffice to consider $v=R(0,1,0)$ (Note that $v\perp x$).
Then we must make a curve $c$ passing $c(0)=x$ on $\Sigma_R$ such that $c'(0)=v$.
For such $c$, we have easy candidate : Let $c(t) = R(0,\sin\ t,\ \cos\ t)$ (That is, $c(0)=x$).
Then $$d\pi\ c'(0) = \frac{d}{dt} \pi (c(t)) =(0,1,0) $$
Hence $d\pi$ has an eigenvalue $1/R$ wrt eigenvector $(0,1,0)$.
So by symmetry principal curvatre is $-1/R$ so that Gaussian curvature is $(-1/R)^2$.
Shape Operator : Here shape operator $S$ is $$ S_x(v) = -\frac{d}{dt} \pi(c(t)) $$ where $c(t)$ is any curve with $c(0)=x$ and $c'(0)=v$, and $v\in T_x \Sigma_R$.