A connected closed surface ,invariant under orthogonal transformation, is sphere In the method of moving planes, there is a explain why the connected closed surface with a plane of symmetry in every direction is sphere. If a connected closed surface is invariant under orthoganal tansformation , it must be sphere. 
I really don't know how to start to prove it . So I prove it by my way. If assume it is convex, I can prove it is  rotational symmetry about the center of mass. But I fail to prove it is convex (the domain bounded by surface is convex). 
I want to know :
First, why a connected closed surface which is invariant under orthoganal tansformation is sphere? If this is a trivial question which can find answer in some book, just tell me which book I should read is enough.
Second, how to prove it is convex ? 
 A: 
If a connected closed surface is invariant under orthogonal transformation, it must be sphere.

$\newcommand{\Reals}{\mathbf{R}}$Lemma: If $S \subseteq \Sigma$ are closed, connected, non-empty surfaces in $\Reals^{3}$, then $\Sigma = S$.
Proof: Since $S$  and $\Sigma$ are surfaces, $S$ is open in $\Sigma$ (a local chart for $S$ is a local chart for $\Sigma$). Since $S$ is closed in $\Reals^{3}$, $S$ is closed in $\Sigma$. Since $\Sigma$ is connected, the closed-and-open subset $S$ is all of $\Sigma$.

Let $\Sigma \subset \Reals^{3}$ be a closed, connected, non-empty surface invariant under the action of the special orthogonal group $SO(3)$ acting by rotations, let $p$ be a point of $\Sigma$, and let $S$ be the sphere centered at the origin and containing $p$. (We cannot have $p = 0$, since no rotation-invariant surface contains the origin.)
Note that $S \subset \Sigma$ because $\Sigma$ is $SO(3)$-invariant. Since $S$ is a closed, connected, non-empty surface, the lemma guarantees $\Sigma = S$.
