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

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You can embed $SL_2(\mathbf{R})$ into $SL_2(\mathbf{C})$ and twist by a wild automorphism of $\mathbf{C}$. –  YCor Mar 23 '12 at 12:37
Right, good point! –  Hugo Chapdelaine Mar 23 '12 at 14:39
For the existence of wild automorphisms of $\mathbf{C}$ and for the construction Misha quoted (due to Borel--Tits) you need the axiom of choice, however. –  Guntram Mar 24 '12 at 10:00
@Guntram: You are right of course, but if you eliminate the (uncountable) Axiom of Choice, then, at least in Solovay's model, all subsets of ${\mathbb R}$ are measurable and we are in the situation discussed in… (see pm's answer). –  Misha Mar 24 '12 at 20:33

3 Answers 3

up vote 7 down vote accepted

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

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.

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Wow ! I would have bet that no such things could exist ! –  BS. Mar 23 '12 at 12:31
Thanks a lot Misha for the reference. –  Hugo Chapdelaine Mar 23 '12 at 14:40

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, $f(x)=\sin(x^2)$.

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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:

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.

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