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?



Inequivalence of group representations preserved under tensor product?
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.


