Timeline for Do doubly-transitive actions give rise to indecomposable representations for infinite groups?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 16, 2023 at 1:50 | history | became hot network question | |||
| Jun 15, 2023 at 19:46 | vote | accept | Kenta Suzuki | ||
| Jun 15, 2023 at 19:42 | answer | added | Will Sawin | timeline score: 8 | |
| Jun 15, 2023 at 18:54 | comment | added | Benjamin Steinberg | I guess if $G$ is triply transitive, then since $V_X$ looks like the permutation module for $G_y$, in that case it is indecomposable with only scalars as intertwiners by the above. But likely one doesn't need that. | |
| Jun 15, 2023 at 18:41 | comment | added | Benjamin Steinberg | The point is whenever T is an intertwiner on all finitely supported functions then the coefficients of y in Tx must be be the same as the coefficient of gy in Tgx. So if T is not diagonal, then using double transitivity you can use double transitivity to get each element to appear in the support of Tx, which contradicts finite support when X is infinite. | |
| Jun 15, 2023 at 18:36 | comment | added | Benjamin Steinberg | I’m not sure about this subspace. | |
| Jun 15, 2023 at 18:35 | comment | added | Benjamin Steinberg | If the set X is infinite the space of all finitely supported functions is an indecomposable module. This is because the Centralizer algebra is C | |
| Jun 15, 2023 at 17:49 | history | asked | Kenta Suzuki | CC BY-SA 4.0 |