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Let me ask a question which could be quite stupid, but still:

let $G$ be a group which is an amalgamated product of subgroups $A$ and $B$ over $C$:$\; \;$ $G = A \ast_{C} B$ (subgroups are infinite!).

Question: How representation theory of $G$ is connected to representation theory of $A$ and $B$ (and $C$), in another words, how $Rep(G)$ is connected to $Rep(A)$ and $Rep(B)$?

Any comments are welcome!

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    $\begingroup$ 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. $\endgroup$
    – user6976
    Commented Feb 13, 2013 at 12:59
  • $\begingroup$ 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). $\endgroup$
    – Andriy Regeta
    Commented Feb 13, 2013 at 13:26
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    $\begingroup$ $Rep(G)$ 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. $\endgroup$
    – Marc Hoyois
    Commented Feb 13, 2013 at 16:16
  • $\begingroup$ @Marc Hoyois, thanks! I am just on the way to understand your answer. $\endgroup$
    – Andriy Regeta
    Commented Feb 13, 2013 at 17:11
  • $\begingroup$ @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). $\endgroup$
    – Misha
    Commented Feb 13, 2013 at 17:52

1 Answer 1

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Let me expand on Mark Sapir's comment with a concrete example. Burger and Mozes famously constructed a group of the form

$\Gamma = F_1*_H F_2$

where $F_1,F_2$ are finitely generated free groups, $H$ is of finite index on either side and $\Gamma$ is simple! Because finitely generated linear groups are residually finite (by a theorem of Mal'cev, say), it follows that $\Gamma$ has no non-trivial finite-dimensional representations. On the other hand, of course, the free groups $F_i$ have impossibly rich representation theories.

There are some positive results that hold in special cases, such as when amalgamating over cyclic subgroups. See, for instance, this recent preprint of Jack Button and the references therein.

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    $\begingroup$ There a much older (and easier) instance of a simple amalgam of 2 free groups (over an infinitely generated subgroup, hence infinitely presented). I don't have the reference available right now. It may be due to M. Hall or H. Neumann, I'm not sure. $\endgroup$
    – YCor
    Commented Feb 13, 2013 at 15:34
  • $\begingroup$ This recent preprint: Button - Groups possessing only indiscrete embeddings in $\operatorname{SL}(2,\mathbb C)$. $\endgroup$
    – LSpice
    Commented Sep 26, 2020 at 15:58
  • $\begingroup$ There's also Bhattacharjee, Meenaxi, Constructing finitely presented infinite nearly simple groups. Comm. Algebra 22 (1994), which gives an amalgam of this form without finite quotients. A more sophisticated (CAT(0)) example was given in Wise, Complete square complexes. Comment. Math. Helv. 82 (2007). $\endgroup$
    – HJRW
    Commented Sep 27, 2020 at 14:29

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