I saw that "Over an algebraically closed field of characteristic 0, semisimple representations are isomorphic if and only if they have the same character" in the Wikipedia page , which does not mention the condition that the group is finite. However, I can only find reference for the result about representations of finite group. Does anyone know any reference for infinite group or could someone give a proof or disproof of this argument, please?
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2$\begingroup$ The same result applies to compact groups, on which there is a huge literature. If you do a Google scholar search on compact groups, you will find dozens of references. For non-compact groups, the situation is far more complicated. $\endgroup$– David HandelmanCommented Aug 30, 2018 at 13:38
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2$\begingroup$ I think you are reading too much into that Wikipedia page, and the intention of the page is to be discussing the case of finite groups from the start (even if some properties listed carry over to other groups). For example, it says $\chi(g)$ is a sum of roots of unity related to the order of $g$: it is just built in that elements of $G$ have finite order, so reasonably enough it is intended that $G$ be finite. The last section of the page mentions Lie groups as a generalization, but up to this point the page (to me) clearly was meant to be focusing on the case of finite groups. $\endgroup$– KConradCommented Aug 30, 2018 at 14:40
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2 Answers
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Yes, it is true, and one doesn't even need to assume the field $k$ is algebraically closed. Section 7 of Lam's "A First Course in Noncommutative Rings" is a good reference for character theory for $k$-algebras. In fact, Theorem 7.19 says exactly what you want: If $M$ and $M'$ are finite-dimensional semisimple representations of a $k$-algebra $R$ with the same character, then they are isomorphic. The only assumption on $k$ is that it has characteristic 0.
Additionally, one sees there that in positive characteristic, over an algebraically closed field, the same result is true for simple representations (Corollary 7.21), although not for semisimple representations in general.
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$\begingroup$ Thanks a lot! I was out of Internet for ten days, sorry. $\endgroup$– LongmaCommented Sep 11, 2018 at 8:23
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$\begingroup$ Theorem 7.19 in Lam's book actually only talks about a finite-dimensional $k$-algebra $R$ as becomes clear from the preceding paragraph. I think that, at the very least, you need that $R$ is Artin, because otherwise $R/\mathrm{rad}(R)$ may not be semisimple. $\endgroup$– ClaudiusCommented Mar 4 at 8:31
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I think it is true. First note it is well known if $A$ is a finite dimensional algebra over an algebraically closed field $K$ of characteristic 0 then two semisimple modules are isomorphic if and only if they have the same character by Wedderburn theory. Basically if $e$ is the primitive idempotent corresponding to the projective cover of a simple its trace will tell you the number of copies of the simple in a semisimple module.
Now let $V,W$ be finite dimensional semisimple $KG$-modules with the same character and let $I$ be the intersection of the annihilator ideals of these modules. Then $A=KG/I$ is finite dimensional and $V,W$ are semisimple $A$-modules with the same character so isomorphic.
answered Aug 30, 2018 at 16:56
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$\begingroup$ As @Alex, points out you don't need algebraically closed since if you divide the trace of the primitive idempotent by the dimension of the endomorphism algebra of the simple you get the multiplicity of the simple as a constituent. $\endgroup$ Commented Aug 30, 2018 at 18:41