Is $SO(N)$ a sphere? [closed]

Question

Is $SO(N)$ as a metric space with a Killing form metric, isometric to a sphere $S^M$ (for any $M$) with standard metric on the sphere?

Answer 1

Spheres are simply connected but $\operatorname{SO}(3)$ is not simply connected. Hence they cannot be homeomorphic, and a fortiori they cannot be isometric.

If $a$ is a unit quaternion, i.e. a quaternion of absolute value $1,$ then $a^{-1}$ is also a unit quaternion, and $x\mapsto a^{-1}xa$ is an isometry of the $3$-dimensional sphere of unit "pure" quaternions (i.e. quaternions whose real part is zero).

Thus we have a mapping that takes $a$ to a member of $\operatorname{SO}(3),$ and that mapping is a homomorphism (actually an isomorphism). But the images of $a$ and $-a$ under that mapping are equal: both correspond to the same rotation of the sphere of unit pure quaternions. Therefore a path along the sphere of unit quaternions that starts at $a$ and ends at $-a$ is mapped to a path in $\operatorname{SO}(3)$ that goes starts and ends at the same point: a loop. But that path cannot be contracted to a point since $a$ and $-a$ are two different points. Therefore the corresponding path from the rotation corresponding to $a$ back to the rotation corresponding to $a$ cannot be contracted to a point. So $\operatorname{SO}(3)$ is not simply connected.