Question

Q: How does one construct a non continuous representation $\rho:SL_2(\mathbf{R})\rightarrow G$ for some connected (finite dimensional) Lie group $G$?

Answer 1

Example can be found, for instance, in Boris Weisfeiler's paper "Abstract homomorphisms of big subgroups of algebraic groups", pages 149-150, see

http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.ndml/1175197662

His example of a discontinuous representation $\rho$ of $SO(n, {\mathbb R})$ to a semidirect product $H$ of $SO(n, {\mathbb R})$ with the abelian group ${\mathbb R}^N$ (the Lie algebra of $SO(n)$), works for $SL(2, {\mathbb R})$ as well. Actually, Weisfeiler's example is even more dramatic: The image of the compact group $SO(n)$ under $\rho$ is dense in the noncompact Lie group $H$. Weisfeiler's paper also lists many positive results on rigidity of abstract homomorphisms of Lie groups.

Answer 2

There are natural function spaces on Lie groups that are nevertheless not continuous (and, thus, are not representations in any usual, useful sense). For example, already on $G=\mathbb R$, the Frechet space $V$ of all continuous functions, and/or the Frechet space of bounded continuous functions, with the translation action of $G$, are not repn spaces, in the sense that $G\times V\rightarrow V$ is not continuous. The reason is the existence of not-uniformly-continuous continuous functions. For example, $f(x)=\sin(x^2)$.

Answer 3

A partial answer:

No measurable constructions are possible. Any measurable group homomorphism between locally compact groups is automatically continuous, in fact $C^\infty$ for Lie groups. You can have a look at the answers in an old question of mine:

http://mathoverflow.net/questions/64116/are-measurable-automorphism-of-a-locally-compact-group-topological-automorphisms

If I would like to find something non continuous, I personally would start with finding some non measurable automorphism of the circle first.