Is it true that $S^n$ and $SO(n+1,\mathbb R)$ are homeomorphic ?

Question

Is it true that $S^n$ and $SO(n+1,\mathbb R)$ are homeomorphic ?

(Where $S^n:=\{x \in \mathbb R^{n+1} : ||x||=1\}$ and $SO(n+1,\mathbb R):=\{ A \in M_{n+1}(\mathbb R) : AA^t=I , \det(A)=1 \}$ )

Answer 1

It is only true for $n=1$. Indeed, $\pi_1(SO(n)) = \mathbb Z/2\mathbb Z$ for $n\ge 3$, thus they cannot be spheres.

(It can be proved that $SO(3) \cong \mathbb{RP}^3$, see here. Then use the fibration

$$0\to SO(n) \to SO(n+1) \to \mathbb S^{n+1} \to 0$$

one can show that $\pi_1(SO(n)) = \mathbb Z/2\mathbb Z$ for $n\ge 3$)

More directly: The dimension of $SO(n)$ is $\frac 12 n(n+1)$.