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Let $G$ be either a finite group or a Lie group. Let $r_1$ and $r_2$ be two linear representations of $G$. $r_1$ is an ordinary representation but $r_2$ may be projective. Is there a criterion to decide the following question: given $G$, $r_1$, $r_2$, does there exist an ordinary linear representation $R$ such that $r_1 \otimes R$ is isomorphic to $r_2 \otimes R$? As a slight refinement of this question: does there exist an ordinary linear representation $R$ and two ordinary linear representations $T_1$ and $T_2$ where $T_1$ and $T_2$ are each the direct sum of several copies of the trivial representation (possibly a different number of copies in $T_1$ and $T_2$) such that $r_1 \otimes T_1 \otimes R$ is isomorphic to $r_2 \otimes T_2 \otimes R$?

One obvious case when the answer is "no" is that $r_2$ is not an ordinary representation. A second case is if the group is, for example, $SU(2)$ and $r_1$ is a spin $S_1$ representation and $r_2$ is a spin $S_2$ representation with $S_1 > S_2$. Then, $r_1 \otimes R$ will contain higher spin representations than $r_2 \otimes R$ will. Is there a general answer?

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If $r_1$ and $r_2$ are two ordinary linear representations of the same degree of a finite group $G$ and $R$ is the regular representation, then $r_1\otimes R\cong r_2\otimes R$. –  Richard Stanley Apr 25 '13 at 17:48
Does "degree" mean the same thing as dimension of a representation? In which case are you saying that if $r_2$ is an ordinary representation then the answer is always "yes" to the second question, as I can tensor in $T_1$ and $T_2$ to make the dimension of $r_1 \otimes T_1$ match that of $r_2 \otimes T_2$ and then pick $R$ to be the regular representation? –  Matt Hastings Apr 25 '13 at 17:56
Is $R$ allowed to be infinite-dimensional (in the Lie case)? If so, and we take $R$ to be left regular representation of $G$ on $L^2(G)$, then I believe Richard Stanley's observation would still work, with $r\otimes R$ being isomorphic to an amplification of $R$. (Fell's absorption principle.) –  Yemon Choi Apr 26 '13 at 1:24
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1 Answer

up vote 3 down vote accepted

If $G$ is a compact connected Lie group, then the ring of characters is an integral domain (it is the ring of Weyl group invariants in the co-ordinate ring of a maximal torus). Let $r_1,r_2, R, T_1, T_2$ be as in the question. Assume that $r_1,r_2$ are ordinary irreducible representations

If $t_1,t_2$ are the dimensions of $T_1, T_2$, and $\chi _1$ and $\chi _2$ are the characters of $r_1,r_2$, and since the character of $R$ may be divided out in the integral domain, it follows from the assumptions that $t_1\chi _1=t_2\chi _2$. Hence $\chi _1$ occurs in the $t_2$ fold direct sum of $\chi _2$ with itself; that is, $\chi _1=\chi _2$. This is supposing that $r_2$ is an ordinary representation.

If $G$ is finite, then the comment of Richard Stanley already gives a counter-example.

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Thanks! So it seems the question has a simple answer in both the finite and Lie cases, but the answer is very different in these two cases. –  Matt Hastings Apr 25 '13 at 18:41
I would say the difference comes from the connectedness or otherwise of the group. If $G$ is not conected, then the group of connected components may well be of the type given by Richard Stanley. –  Aakumadula Apr 26 '13 at 10:57
See my comment above. I think the difference is finiteness, but perhaps I have made an error –  Yemon Choi Apr 26 '13 at 18:29
I mean the following; suppose $F$ is a finite group for which Richard Stanley's example works. Suppose $G$ is a Lie group whose connected component group $G/G^0$ is $F$. Then for $G$ also, there exist two reps $r_1,r_2$ as above. –  Aakumadula Apr 27 '13 at 3:43
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