How does one construct a non-continuous representation $\rho:\operatorname{SL}_2(\mathbf{R})\rightarrow G$ for some connected (finite dimensional) Lie group $G$?
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asked Mar 23, 2012 at 10:26
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4$\begingroup$ You can embed $SL_2(\mathbf{R})$ into $SL_2(\mathbf{C})$ and twist by a wild automorphism of $\mathbf{C}$. $\endgroup$– YCorCommented Mar 23, 2012 at 12:37
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$\begingroup$ Right, good point! $\endgroup$– Hugo ChapdelaineCommented Mar 23, 2012 at 14:39
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$\begingroup$ 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. $\endgroup$– GuntramCommented Mar 24, 2012 at 10:00
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$\begingroup$ @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 mathoverflow.net/questions/64116/… (see pm's answer). $\endgroup$– MishaCommented Mar 24, 2012 at 20:33
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3 Answers
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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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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 Fréchet space $V$ of all continuous functions, and/or the Fréchet 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, $f(x)=\sin(x^2)$.
answered Mar 23, 2012 at 12:34
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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.
answered Mar 23, 2012 at 10:52