Timeline for Inequivalence of group representations preserved under tensor product?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 26, 2013 at 1:24 | comment | added | Yemon Choi | 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.) | |
| Apr 25, 2013 at 18:41 | vote | accept | Matt Hastings | ||
| Apr 25, 2013 at 18:23 | answer | added | Venkataramana | timeline score: 3 | |
| Apr 25, 2013 at 17:56 | comment | added | Matt Hastings | 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? | |
| Apr 25, 2013 at 17:48 | comment | added | Richard Stanley | 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 $r_1\otimes R\cong r_2\otimes R$. | |
| Apr 25, 2013 at 17:34 | history | asked | Matt Hastings | CC BY-SA 3.0 |