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If $G$ is the real points of a semisimple algebraic group and $\rho:G\to GL(n,\mathbb R)$ is continuous representation. Is $\rho$ an algebraic morphism?

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    $\begingroup$ I do not understand why this was closed: it is well known and classical (but is not really discussed at length in many books) but definitely non-trivial. $\endgroup$ – Venkataramana Sep 30 '15 at 5:36
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    $\begingroup$ I agree with Venkataramana, this question seems very appropriate. If it gets closed, I will vote to re-open it. $\endgroup$ – GH from MO Sep 30 '15 at 5:57
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    $\begingroup$ The standard counterexample without the "simply connected" hypothesis in Venkataramana's answer is the inverse of the analytic isomorphism ${\rm{SL}}_n(\mathbf{R}) \rightarrow {\rm{PGL}}_n(\mathbf{R})$ for odd $n > 1$ (i.e., take the semisimple group of interest to be ${\rm{PGL}}_n$). Also, for an alternative treatment as a consequence of general procedures over any field of characteristic 0, see the answer by user27056 at mathoverflow.net/questions/114974/… $\endgroup$ – grghxy Sep 30 '15 at 7:25
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    $\begingroup$ since G is semi-simple, this analytic isomorphism exists only as a covering;(see the example of $SL_3(\mathbb R)$ in my answer). I think there are no other counterexamples, because of the "rigidity" of semisimple group representations $\endgroup$ – Venkataramana Sep 30 '15 at 7:28
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    $\begingroup$ One other point: probably you are tacitly assuming (as is everyone giving comments/answers) that the original semisimple group in question is connected (in the sense of algebraic groups), as without that Lie algebras lose a bit of their control on the situation. Even over $\mathbf{R}$ there are subtleties if the group of $\mathbf{R}$-points is disconnected for the analytic topology, but it is an important (and remarkable) theorem of Cartan that such disconnectedness never happens when the given (connected!) semisimple group is simply connected (in the sense of algebraic groups!). $\endgroup$ – grghxy Sep 30 '15 at 7:31
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If $G$ is a real simple algebraic group and $\rho$ is a finite dimensional continuous representation, and if $G$ is not the group of complex points and $G(\mathbb C)$ is simply connected, then $\rho$ is indeed algebraic. This follows, for example, by going to the Lie algebra level and using classification of lie algebra representations.

If $G=SL_2(\mathbb C)$, for example, you can have $\rho\otimes {\overline \rho}$ which is not quite algebraic, but can be viewed to be algebraic, if we view the complex group $SL_2(\mathbb C)$ as the group of real points of the Weil restriction of scalars $R_{\mathbb C/\mathbb R}(SL_2)$.

The same sort of thing holds for any simply connected complex simple algebraic group, viewed as the group of real points by restriction of scalars.

If $G(\mathbb C)$ is not simply connected, then there is some (small) ambiguity. Take $$G_0=SL_3(\mathbb R)\subset G_0(\mathbb C)= SL_3(\mathbb C)/\text{scalars}.$$ We can view $SL_3(\mathbb R)$ as an algebraic group in $G_0(\mathbb C)$; then the standard representation of $SL_3(\mathbb R)$ is not algebraic with this algebraic structure, but algebraic, if we view $SL_3(\mathbb R)$ as real points of $SL_3(\mathbb C)$.

[Edit] Since @jerry has asked for details, I give them; I am sure that all this is explained in some textbook, but not having one at hand, I cannot give references.

The first point is that given a complex semi-simple Lie algebra $\mathfrak g$ , there is a simply connected complex algebraic group $G$ with Lie Algebra $\mathfrak g$ (this is perhaps due to Hermann Weyl). So any representation of the Lie algebra $\mathfrak g$ integrates to a holomorphic representation of the complex group $G$. That holomorphic finite dimensional representations of $G$ are algebraic is also well known, perhaps by using the Borel Weil Theorem.

Secondly, if $G_0$ is an algebraic semisimple group defined over $\mathbb R$ with $G(\mathbb C)$ simply connected, and $\rho:G(\mathbb R) \rightarrow SL_n(\mathbb C)$ is a representation, it is known (as Dimitrov has remarked) that $\rho $ is smooth and hence yields a complex representation of the Lie algebra $\mathfrak g_0$, and therefore of its complexification $\mathfrak g$. Since $G$ is simply connected, this gives an algebraic representation of $G$. Since $G= G_0(\mathbb C)$ it follows that this complex representation is an algebraic representation of $G_0(\mathbb C)$.

Incidentally, these kinds of algebraicity results are proved in great generality by Borel and Tits in an Annals paper "abstract homomorphisms of algebraic groups" (it is in French).

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  • $\begingroup$ Sorry for being stupid. Could you please explain a bit why by going to the Lie algebra and coming back we know that the representation is indeed algebraic? Since this involves taking log and exp, which are not algebraic. Is there any extra problem when the group is not split? $\endgroup$ – Jerry Sep 30 '15 at 3:48
  • $\begingroup$ On "the other end of things", how do you get from knowing $\rho$ is continuous to knowing it's smooth? I know the story from there, but on the face of it you need smoothness to talk about tangent spaces and Lie algebras. Do you use something like smooth approximation...? $\endgroup$ – Kevin Casto Sep 30 '15 at 4:52
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    $\begingroup$ @KevinCasto: For homomorphisms of Lie groups, smoothness is automatic from continuity. This is proved in Proposition 1.2.22 of Terry Tao's book Hilbert's Fifth Problem and Related Topics. $\endgroup$ – Vesselin Dimitrov Sep 30 '15 at 5:07
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    $\begingroup$ @Jerry: The Lie algebra is semi-simple, and hence its representation theory is described in terms of highest weights, which essentially means that all representations are explicitly constructed, hence algebraic. $\endgroup$ – Venkataramana Sep 30 '15 at 5:38

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