Timeline for Can the free module in the representation ring be characterised this way?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 11, 2023 at 10:55 | vote | accept | HenrikRüping | ||
| Oct 11, 2023 at 10:12 | comment | added | Jeremy Rickard | Alternatively, and this works in any characteristic, for $F$ the trivial module $\dim\operatorname{Hom}(F,FG)=1$, so if $[V]=\alpha[FG]$ is a rational multiple of $[FG]$ in the representation ring, then $\dim\operatorname{Hom}(F,V)=\alpha$ is an integer. | |
| Oct 11, 2023 at 9:57 | comment | added | Will Sawin | @HenrikRüping For the $K$-theory version of the question, the $K$-theory group is freely generated by irreducibles or indecomposables (depending on the definition) so in particular is torsion-free, and this argument shows that the class of $U$ is a scalar multiple of the class of $FG$. If $F$ has characteristic zero then the trivial representation occurs in $FG$ with multiplicity one, showing that the class of $U$ is an integer multiple of the class of $FG$. I guess in characteristic $p$ there is the problem of making the tensor product in $K$-theory wel-defined. | |
| Oct 11, 2023 at 8:42 | comment | added | HenrikRüping | I dont understand the last step. For a general ring $R$ it could happen that $V\oplus V$ is free, although $V$ is not. For example, for the ring $M_2(F_2)$ and $V = F_2^2$ this is the case. Krull-Schmidt (e.g. any module can be written uniquely as a sum of indecomposables) should still hold for $M_2(F_2)$. Maybe there is a reason that this cannot happen for group rings, and I just dont see it yet. | |
| Oct 11, 2023 at 8:15 | history | answered | Jeremy Rickard | CC BY-SA 4.0 |