Timeline for amalgamated product of groups and representation theory
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 3, 2013 at 17:48 | vote | accept | Andriy Regeta | ||
| Feb 14, 2013 at 16:57 | comment | added | Misha | Andriy: Google "fiber product", then Marc's answer will make sense. | |
| Feb 14, 2013 at 16:18 | comment | added | Andriy Regeta | Sorry, what I wrote here is a stupid remark, I think, I got an answer! | |
| Feb 14, 2013 at 13:28 | comment | added | Andriy Regeta | Now I convinced myself that I can forget about $C$, so I have just a free product of two groups (in fact, those groups are algebraic). But if representations of $G$ are just pairs of representations of $A$ and $B$, then representations of $Γ$ (in the answer below given by @HW) are just pairs of representations of $F_1$ and $F_2$ which seems to be impossible to me (following from the answer of @HW). How should one understand $Rep(A) \times_{Rep(C)} Rep(B)$ ? | |
| Feb 13, 2013 at 22:26 | comment | added | Amin | Your question is a bit unclear to me. As said by Marc Hoyois, the representations (finite or not) of the amalgamated product are just pairs of representations of A and B that coincide when restricted to C, by the very universal property of the amalgamated product. What else did you want ? | |
| Feb 13, 2013 at 20:57 | comment | added | Alain Valette | To paraphrase Mark: If you are interested in infinite dimensional representations, then there is no connection either: view the free group $\mathbb{F}_2$ as the free product of two copies of $\mathbb{Z}$. A unitary representation of $\mathbb{F}_2$ is a pair of unitaries on a Hilbert space, and that's about all you can say, while a unitary representation of $\mathbb{Z}$ is completely described by the spectral theorem (basically, it corresponds to a positive measure on the circle). | |
| Feb 13, 2013 at 17:52 | comment | added | Misha | @Marc: Maybe you should post your comment as an answer. This formula holds even if we think of $Rep$ as a scheme-theoretic functor (when the target is an algebraic group). | |
| Feb 13, 2013 at 17:11 | comment | added | Andriy Regeta | @Marc Hoyois, thanks! I am just on the way to understand your answer. | |
| Feb 13, 2013 at 16:16 | comment | added | Marc Hoyois | $Rep(G)$ is still formally related to $Rep(A)$, $Rep(B)$, and $Rep(C)$ even in the finite-dimensional case. In fact, we have $Rep(G)=Rep(A)\times_{Rep(C)}Rep(B)$ because a representation is just a functor out of $G$ and amalgamated product is the categorical pushout. | |
| Feb 13, 2013 at 13:55 | history | edited | HJRW | CC BY-SA 3.0 |
Added gr.group-theory tag
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| Feb 13, 2013 at 13:44 | answer | added | HJRW | timeline score: 6 | |
| Feb 13, 2013 at 13:26 | comment | added | Andriy Regeta | I see, Thanks! I am interested in general about not only finite-dimensional representations, but such information is, of course, very interesting! (and probably, it holds then also in infinite dimensional case). | |
| Feb 13, 2013 at 12:59 | comment | added | user6976 | If you are interested in finite dimensional representations, then there is no connection because the amalgamated product may have no non-trivial faithful finite dimensional representations while $A$ and $B$ have faithful finite dimensional representations. | |
| Feb 13, 2013 at 12:43 | history | asked | Andriy Regeta | CC BY-SA 3.0 |