Show that if image of a connected $2$ surface under Gauss map is a single point on unit sphere then $S$ is a part of a plane.

Question

Let us take $p\in S$ and consider the plane $\{x\in\Bbb{R^3}|(x-p)\cdot N(p)=0\}$ where $N$ is the gauss map on $S$. Basically, it is the tangent space of $S$ at $p$, $S_p$.
By the question, $\exists$ unit vector $ v\in \Bbb{R^3}$, $N(S)=\{v\}$.
So, $S_p=\{x\in\Bbb{R^3}|(x-p)\cdot v=0\}$.
Now, take another point $q\in S$, then $S_q=\{x\in\Bbb{R^3}|(x-q)\cdot v=0\}$
So, $\forall q\in S$, the tangent space $S_q$ is parallel to $S_p$.
Now, I want to show all these tangent planes are same with $S_p$.
But from here I can't proceed further. Is there any alternative method? Can anybody explain it? Thanks for assistance in advance.

Answer 1

You are very close to a correct answer. Try and use the other givens. As you said, all of the tangent planes to $S$ are parallel. We want to consider $p,q\in S$ and show that $S_p=S_q$. Let $P$ be the unique plane going through the origin, which is parallel to all of these $S_p$ (this is usually called the tangent space). For any $r\in S,$ we then have $S_r=r+P$. Since $S$ is connected, we can find a differentiable path $\gamma:I\rightarrow S$ from $p$ to $q$. For any $t\in I$, we have $\gamma(t)+\gamma'(t)\in S_{\gamma(t)}=P+\gamma(t)$ or equivalently $\gamma'(t)\in P$. By the fundamental thm of calc, $$q=\gamma(1)=\gamma(0)+\int_0^1\gamma'(t)dt.$$ Since $\gamma'(t)\in P$ for all $t\in I$, we have $\int_0^1\gamma'(t)dt\in P.$ Thus $q=p+\int_0^1\gamma'(t)dt\in p+P.$ Because $q-p\in P$, we see that $S_p=p+P=q+P=S_q,$ as desired.

Answer 2

Even easier, show that every point in $S$ satisfies the equation $$(x-p)\cdot v = 0.$$ If you are given the surface parametrically, say $x(s,t)$, then note that $x_s\cdot v = 0$ and $x_t\cdot v = 0$ for all $s,t$ (why?), and so the function $(x-p)\cdot v$ is constant on $S$. (Here you use connectedness. How?)