Lie group Isomorphism Between $SO(2)$ and $S^1$

Question

I'm trying to prove that $SO(2)$ and $S^1$ are isomorphic Lie groups. This isn't that hard, as we can take the map $$\phi : a + bi \mapsto \begin{pmatrix}a & -b \\ b & a\end{pmatrix}.$$ That $\phi$ is smooth, a group homomorphism, and injective are immediate. Surjectivity is harder to prove, and it can be done algebraically by going back to the definition of $SO(2)$, but just for fun I want to see if I can prove it from a totally topological perspective. Here's my stab at it:

Since $\phi$ is a Lie group homomorphism, it has constant rank, and since it's injective, it's an immersion. Since the source is compact, it's actually an embedding, and so $\phi(S^1)$ is an embedded 1-dimensional submanifold of $SO(2)$. All embedded codimension-0 submanifolds are open submaifolds, so $\phi(S^1)$ is open in $SO(2)$. It's also a Lie subgroup of of $SO(2)$, so it must be a union of connected components. But $SO(2)$ is connected, and this forces $\phi(S^1) = SO(2)$.

Is that argument valid? My main concern with it is that I'm not totally sure how to show that $SO(2)$ is connected without going back to the algebraic argument I'm trying to avoid.

Answer 1

We have that ${\rm SO}(2,\Bbb R)$ is compact, because it is closed and bounded in ${\rm Mat}(2,\Bbb R)$. It also acts transitively in $\Bbb S^1$. And since the stabilizer of $e_1 = (1,0)$ is ${\rm Stab}(e_1) = \{{\rm Id}_2\}$, the orbit map $f\colon {\rm SO}(2,\Bbb R) \to \Bbb S^1$ given by $f(A) = Ae_1$ is bijective. It is also continuous. Any continuous bijection from a compact space onto a Hausdorff space is a homeomorphism. This means that since $\Bbb S^1$ is connected, so is ${\rm SO}(2,\Bbb R)$.

Answer 2

You could (if you really wanted stay in topology-land) try and prove that $\mathbb{R}$ is the universal cover of $SO(2)$ via the covering map $t\mapsto \begin{pmatrix} \cos(t)&-\sin(t)\\ \sin(t)&\cos(t) \end{pmatrix}$ . Since $\mathbb{R}$ is (at least) connected, $\pi_0(SO(2))=0$.