Prove that $SO(3)$ acts transitively on the unit sphere $S^2$ of $\Bbb R^3$

Question

Prove that $SO(3)$ acts transitively on the unit sphere $S^2$ of $\Bbb R^3$.

I think $S^2$ means a 2-D sphere and $SO(3)$ is the usual $SO(3)$ group. I'm unsure how to prove that $SO(3)$ acts transitively.

My guess is to show that like $SO(2)$, $SO(3)$ preserves distance, but I'm unsure how that can guarantee that it will act transitively on the sphere. Any ideas if I'm thinking about this right?

Answer 1

It suffices to show all of $S^2$ is the orbit of a single point, say $e_1$. The equation $Ae_1=x$ says that the first column of $A$ is $x\in S^2$. To create the rest of the matrix $A$, simply take any two vectors on our sphere that are orthogonal to each other and $x$. To do this, compute $e_1\times x$ and then normalize to obtain $y$ for example, then compute $x\times y$ and normalize to obtain $z$. Observe $\det[x~y~z]=\pm1$, so we can permute $y$ and $z$ as necessary to get either $A=[x~y~z]$ or $[x~z~y]$.

Answer 2

Let $e_1=(1,0,0)\in S^2$. Let $v=(a,b,c)\in S^2$, so that $a^2+b^2+c^2=1$. Suppose first that $b$ and $c$ are not both $0$. Consider the matrix $M_v$ given by

$$ M_v=\left(\begin{matrix} a & \sqrt{b^2+c^2} & 0\\ b & -\frac{ab}{\sqrt{b^2+c^2}} & \frac{c}{\sqrt{b^2+c^2}} \\ c & -\frac{ac}{\sqrt{b^2+c^2}} & \frac{-b}{\sqrt{b^2+c^2}} \\ \end{matrix}\right) $$

Then $M_v\in SO_3$ and $M_v(e_1)=v$. If $b$ and $c$ are both zero, then $v=\pm e_1$. If $v=e_1$, then $M_{e_1}\in SO_3$ sends $e_1$ to $v$, and if $v=(-e_1)$, then the matrix $M_{-e_1}={\sf diag}(-1,-1,1)\in SO_3$ sends $e_1$ to $v$.

So we have shown : for any $v$ there is a $M_v\in SO_3$ sending $e_1$ to $v$.

Then, for any $x,y\in SO_3$, the map $f=M_y(M_x)^{-1}\in SO_3$ sends $x$ to $y$.

Answer 3

Here is a geometric argument:

All rotations of $S^2\subset{\mathbb R}^3$ when written as linear maps $T:\ {\mathbb R}^3\to{\mathbb R}^3$ are elements of $SO(3)$. Therefore it is enough to show that for any two points $p$, $q\in S^2$ there is a rotation $T$ with $T(p)=q$.

When $p=q$ the identity will do. When $p$ and $q$ are antipodal points of $S^2$ then a $180^\circ$-rotation about any axis orthogonal to $p\vee q$ will do.

When $p$ and $q$ are linearly independent the line spanned by $p\times q$ is the axis of a rotation $T$ with $T(p)=q$.