# Classification of Continuous Group Homs from Circle to $SL_2(\mathbb{R})$

Problem. Classify all continuous group homomorphisms $$f:\mathbb{S}^1\to SL_2(\mathbb{R})$$.

Attempt 1. My initial thought was to look at the induced Lie algebra map $$df:\mathbb{R}\to \mathfrak{sl}_2$$. Since we have a basis $$E,F,H$$ (denoted $$e,f,h$$ in the link), I thought we can classify $$f$$ according to what $$df(1)=:v$$ is. However, since $$\mathbb{S}^1$$ is not simply connected, Lie's Second Theorem does not hold, so we do not know if there exists an $$f$$ given $$v$$.

Attempt 2. My other idea was to look at the rational points $$x$$ on $$\mathbb{S}^1$$ which have order $$q$$ ($$f$$ is determined by these points by continuity). Since $$f$$ is a homomorphism, $$f(x)$$ must have order dividing $$q$$. But then I do not know which elements in $$SL_2(\mathbb{R})$$ has finite order. I thought of using Iwasawa decomposition, but I could not make it work.

I guess other things one could try is to try looking at the universal cover $$\widetilde{SL_2(\mathbb{R})}$$, but that restricts us to a map $$f$$ that lifts.

Question 1. How does one approach the above problem? Can we salvage either of my approaches?

Question 2. More generally, how does one approach these classification problems when Lie's second theorem is not available to us?

• See also the “Representations” section at en.m.wikipedia.org/wiki/Circle_group Nov 1 '20 at 9:14
• Okay, sure. We can decompose into irreps, and we have a classification of irreps for the circle. But suppose we did not know about this result. Then how would we approach the problem directly? Nov 1 '20 at 9:34

Looking at the induced Lie algebra map will work fine. $$df(1)$$ is some $$X \in \mathfrak{sl}_2(\mathbb{R})$$ which exponentiates to a one-parameter subgroup $$\varphi : \mathbb{R} \to SL_2(\mathbb{R})$$ and next you want to figure out which of these one-parameter subgroups are periodic with period $$2\pi$$ (or $$1$$ depending on your preferred conventions). Thinking about the eigenvalues of $$X$$ it's not hard to see that this happens iff the eigenvalues of $$X$$ are purely imaginary, so they are necessarily two conjugate pairs $$is, -is$$ (because $$\text{tr}(X) = 0$$), and moreover $$s$$ is an integer (or an integer multiple of $$2\pi$$, again depending on your preferred conventions). Can you finish it from here?