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Let $G$ be a simple complex algebraic group. What are its complex irreducible finite-dimensional representations?

Before you start voting to close the question, I never said "rational". I am asking about abstract representations of $G$ as a group.

I can a make an obvious conjecture but I have no idea how to prove it, even for $SL_2$. Yet, it could be something well-known. Pick a rational irreducible representation $(V,\rho)$ and a field endomorphism $\tau: {\mathbb C} \rightarrow {\mathbb C}$. The field endomorphism extends to an endmorphism of $G$: just apply $\tau$ to each matrix entry of of $g\in G\subseteq GL_n(\mathbb C)$. Thus, we get a twisted rational representation $(V,\tau\circ \rho)$ that clearly remains irreducible.

STUPID WABBIT CONJECTURE: all irreducible representations are twisted rational representations.

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  • $\begingroup$ Whyever would one be interested in this question? $\endgroup$ – Kevin Buzzard Jan 13 '17 at 11:23
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    $\begingroup$ Your conjecture is I think trivially false because I think it doesn't include representations of the form $\tau\circ\rho$ where $\tau$ is just an injection of fields $\mathbb{C}\to\mathbb{C}$. So there's uncountably many more for you I think... $\endgroup$ – Kevin Buzzard Jan 13 '17 at 11:25
  • $\begingroup$ I guess Weil wrote down an abstract presentation of the group $SL_2(k)$ for any field $k$. That might be a place to start. $\endgroup$ – Kevin Buzzard Jan 13 '17 at 11:31
  • $\begingroup$ OK, I change to an endomorphism. Obviously, it does not have to be surjective. $\endgroup$ – Bugs Bunny Jan 13 '17 at 12:02
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    $\begingroup$ @KevinBuzzard: the question is more natural than it seems, since the structure of $G(k)$ is tightly controlled by $k$-subgroups of $G$. For example, a Tits system is a group-theoretic notion (relevant to Bruhat decomposition and uniform simplicity proofs), but for a connected reductive $G$ over an infinite field $k$ the only Tits systems in $G(k)$ satisfying certain natural conditions are the "standard" ones: see arxiv.org/pdf/1103.5970v7.pdf. Might you like it if any $k'$-linear representation of $G(k)$ is "algebraic" up to some $k\rightarrow k'$? $\endgroup$ – nfdc23 Jan 13 '17 at 14:11
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If $\rho:G(\mathbf{C}) \rightarrow {\rm{GL}}(V)(\mathbf{C})$ is such an abstract linear representation, by irreducibility (or mere semisimplicity) the Zariski closure $H \subset {\rm{GL}}(V)$ of the image has connected reductive identity component (due to Lie-Kolchin). But $G(\mathbf{C})$ is perfect with simple quotient modulo its finite center, so $H^0$ is semisimple, and then even simple in the sense of algebraic groups. The group $G(\mathbf{C})$ is generated by its "unipotent" subgroups that have no nontrivial finite quotient, so $H$ is even connected.

Thus, your question amounts to asking if $G(\mathbf{C}) \rightarrow H(\mathbf{C})$ arises from a $\mathbf{C}$-homomorphism $G \rightarrow H$ up to a field map $\mathbf{C} \rightarrow \mathbf{C}$. This is a (very) special case of the setting that was completely analyzed in an optimal way by Borel and Tits in their paper Homomorphismes "abstraits" de groupes algébriques simples in Annals of Math 97 (1973), pp. 499-571. There they work over general infinite fields, with connnected semisimple groups that are absolutely simple. The Introduction presents their main results (A) and (B) which address your question (in view of the initial part of this answer). See 8.18 for a certain standard counterexample over $\mathbf{R}$ and 8.19 for a conjecture in wider generality which requires some caveats in characteristic 2 (the necessity of which is explained in a Remark in the Introduction of Igor Rapinchuk's 2013 paper "On abstract representations of the groups of rational points of algebraic groups and their deformations" in Algebra and Number Theory, vol. 7).

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Just to complete the answer, every complex irreducible representation of $G(\mathbb{C})$ ($G$ is simply connected) is an $r$-fold tensor product of the form $\rho _1\circ \sigma _1\otimes \cdots \otimes \rho _r \circ \sigma _r$, where each $\rho _i$ is an algebraic irreducible representation of $G$ and $\sigma _i$ are $distinct$ embeddings of the field $\mathbb{C}$ into itself. This readily follows from the results of Borel-Tits (see Theorem (10.3) of the article of Borel-Tits mentioned by nfdc23).

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