Timeline for Sets M,N with G action such that C[M] = C[N] as G modules, how are they related ?
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| Sep 11, 2012 at 20:27 | comment | added | Will Sawin | The number of orbits of $M \times M \times M \times ... \times M$ for all lengths of the product do not characterize the representation. For instance consider to actions of $G \times G$ on a set, one factoring through the first $G$ and one being the same action factoring through the second $G$. | |
| Sep 11, 2012 at 20:24 | comment | added | Qiaochu Yuan | @Alexander: no. $M^k$ and $N^k$ have the same number of orbits if and only if $\sum_{g \in G} \text{Fix}_M(g)^k = \sum_{g \in G} \text{Fix}_N(g)^k$ for all $k$ if and only if the multiset $\{ \text{Fix}_M(g) \}$ is the same as the multiset $\{ \text{Fix}_N(g) \}$, but this is strictly weaker than the claim that $\text{Fix}_N(g) = \text{Fix}_M(g)$ for all $g$ since, for example, $M$ and $N$ may be related by an outer automorphism of $G$ which exchanges two elements $g_1, g_2$ such that $\text{Fix}_N(g_1) \neq \text{Fix}_N(g_2)$. | |
| Sep 11, 2012 at 20:02 | comment | added | Alexander Chervov | Ahh... about MxMxM...xM it is also obviously yes. May be it is characterization ? | |
| Sep 11, 2012 at 19:30 | comment | added | Alexander Chervov | Might add that \sum d_i^2 = number of orbits in MxM, where d_i are dims(irreps) in C[M]. | |
| Sep 11, 2012 at 19:28 | comment | added | Alexander Chervov | Wow, so fast :) Agree, thank you ! By the way what about MxMxM..xM ? more than two times ? | |
| Sep 11, 2012 at 19:26 | history | answered | Benjamin Steinberg | CC BY-SA 3.0 |