Is there a comlete discription of algebraic groups having only finitely many irreducible representations? ( giving some references would be very kind )
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$\begingroup$ Is there some motivation for this question (i.e., an application in mind)? $\endgroup$– user76758Commented Feb 8, 2014 at 18:11
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$\begingroup$ @Aleksa: It would help to specify whether you include only finite dimensional representations, and also whether the algebraic group is taken over an arbitrary algebraically closed field. $\endgroup$– Jim HumphreysCommented Feb 8, 2014 at 21:05
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$\begingroup$ @Jim Humphreys: I am interested in finite dimensional representations. Firstly, what is known in characteristic zero and what happens in characteristic $p>0$? $\endgroup$– AleksaCommented Feb 9, 2014 at 6:19
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$\begingroup$ ...and of course over algebraically closed field. $\endgroup$– AleksaCommented Feb 9, 2014 at 6:27
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I am not sure about references, and I have only a partial answer to your question. If your group is complex reductive, then it has only finitely many isomorphism classes of irreducible representations if and only if it is itself finite. To see this, we use the Theorem of the Highest Weight to enumerate the irreps by dominant integral weights. There are finitely many of these if and only if the group is finite. For one of these directions, suppose that $G$ is complex reductive and has only finitely many dominant integral weights. Then, its complex dimension must be $0$, so $G$ is discrete. Now, consider a compact real form $K$ of $G$. Note that $K$ is discrete, and hence finite. Since $G$ is the complexification of $K$, $G$ must be finite.
answered Feb 8, 2014 at 18:09
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$\begingroup$ Ok, for reductive or linearly reductive only the finite groups hav that property. But what about algebraic groups that are not reductive? $\endgroup$– AleksaCommented Feb 8, 2014 at 18:39
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2$\begingroup$ I guess, for algebraic groups over ${\mathbb C}$, your answer is complete: a group has finitely many representations iff the connected component of unity is unipotent. $\endgroup$ Commented Feb 8, 2014 at 18:42
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2$\begingroup$ @Aleksa: If $\rho:G \rightarrow {\rm{GL}}(V)$ is a finite-dimensional irreducible representation over an algebraically closed field then the unipotent radical acts trivially, so it factors through the maximal reductive quotient. So really the case with reductive identity component is the essential one (for smooth affine groups over an alg. closed field). $\endgroup$ Commented Feb 9, 2014 at 0:09
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1$\begingroup$ Using the facts that the group in question is generated by a finite subgroup $F$ and by a connected normal unipotent subgroup $U$ (see, for instance, mathoverflow.net/questions/150949 ) and that the latter always possesses a $1$-dimensional trivial submodule, you can easily reduce the problem to the case of $F$. $\endgroup$ Commented Feb 9, 2014 at 10:00