Is there an easy way to prove that every Lie subgroup of $SU(2)$ is closed? That is, is it true that any $K=\exp_{SU(2)}(k)$ is closed for every subalgebra $k \subset \mathfrak{su}(2)$? If there isn't, is there a reference for this fact?
I thought that this was left as an exercise on the book on compact Lie groups by Sepanski, but I cannot find the reference. Thanks.
Yes, this is true. I'm going to use $Sp(1)$ instead of $SU(2)$ for convenience, but as they are Lie-isomrphic, this doesn't matter.
The Lie algebra of $Sp(1)$ is $\operatorname{Im}\mathbb{H}$ with commutator as Lie bracket.
Now, let $k\subseteq \mathfrak{sp}(1)$. We divide into cases depending on $\dim k$.
If $\dim k = 0$, the $K = \{1\}$ and if $\dim k = 3$, then $K = Sp(1)$, which are closed.
If $\dim k = 1$, then $k$ is spanned by some nonzero imaginary quaternion $r$. Then $\exp(k) = \{\cos\theta + r\sin\theta:\theta \in [0,2\pi)\}$ is closed.
If $\dim k = 2$, then $k$ cannot be a subalgebra. To see this, pick an orthonormal basis $\{r_1,r_2\}$ for $k$. Then $r_1 r_2 = -r_2 r_1$ is perpendicular to both $r_1$ and $r_2$. Thus, $[r_1,r_2] = r_1 r_2 - r_2r_1 = 2r_1r_2$ is perpendicular to both $r_1$ and $r_2$, so cannot be an element of $k$.
(As an aside, the same one can get the same result for $SO(3)$ by lifting to $SU(2)$. The result for $S^1$ is trivial, so the result is that every (connected) Lie group of rank $1$ has the property that every exponential of a subalgebra is closed. On the other hand, for any compact Lie group $G$ of rank $2$ or bigger, there are non-closed Lie subgroups. Specifically, one has $T^2\subseteq G$, and one can look at the exponential of $\operatorname{span}\{1, \pi\}$.)
