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Every book on modular representation theory of finite groups introduces p-modular systems and describes how to reduce an ordinary representation $U$ to obtain one in characteristic p (call it $\overline{U}$), by choosing a lattice. They all prove that the resulting $kG$- module is not unique, but is well defined in the Grothendieck group.

Are there any general results about how rigid this process is. For example for which $U$ is it the case that $\overline{U}$ is uniquely determined up to isomorphism? Can any composition factor of $\overline{U}$ be forced to the socle by choosing an appropriate lattice? Can the number of indecomposable summands of $\overline{U}$ change depending on the choice of lattice? I haven't been able to find any results like these.

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  • $\begingroup$ As Geoff points out, this kind of question has been raised previously: mathoverflow.net/questions/130022, though I was too lazy to search for it (and will remove my redundant answer here). $\endgroup$ Apr 4 '14 at 16:01
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This question, or variants of it, appear here from time to time. With enough ramification, the reduction of an irreducible lattice can be made completely reducible ( see Feit's book on Representation theory). On the other hand, J.G. Thompson showed (in his "Vertices and Sources" that the reduction of an irreducible can be made to have a simple socle, and the socle can be made to correspond to any irreducible Brauer character which occurs in the (restriction to $p$-regular elements of) the character afforded by the lattice.

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  • $\begingroup$ Thanks, I had looked in Curtis & Reiner, Nagao-Tsushima, Benson but not in Feit! $\endgroup$
    – daveh
    Apr 4 '14 at 16:18
  • $\begingroup$ I've just looked at the Thompson paper and I don't see that result. The only general result is Theorem 1, and that seems to show that we choose any subcharacter of a PIM and find a lattice so it appears at the top when we reduce, but nothing about Brauer characters. $\endgroup$
    – daveh
    Apr 4 '14 at 17:04
  • $\begingroup$ Yes, that is the relevant result (or its dual). The reduced lattice has the same head as the PIMdoes, and that head is simple. A module with simple head must be indecomposable. The way to dualise it is to take the lattice affording the irreducible character as a pure submodule of the "lift" of the PIM. Then its reduction has the same socle as the PIM, which is simple. $\endgroup$ Apr 4 '14 at 17:18
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It is generally not rigid at all. Note that if $U$ lifts an absolutely irreducible $kG$ module, then $\overline{U}$ is certainly uniquely defined upto isomorphism (though absolutely irreducible $kG$-modules may not themselves lift uniquely!). Feit (Lemma 18.2 in "The Representation Theory of Finite Groups", p.70) shows that if the coefficient ring is suitably ramified, then for any filtration of $\overline{U}$, there is a lattice $U'$ such that $\overline{U'}$ is the direct sum of the successive quotients of the filtration (he constructs such a lattice). In particular, there one can find lattices $V$ that are completely reducible modulo the radical of the coefficient ring. I think that such lattices answer all your questions. Feit's original paper with the result mentioned is "On finite linear groups", J. Algebra 5, 378-400 (1967).

Edit: I have a shameless plug. For the following jargon and proofs, see here: http://arxiv.org/pdf/1312.4475.pdf Using the notation there, by Corollary 5.7 and Propositions 4.4 and 5.4, it follows that if $\mathcal{O}$ is ramified (at all, not just 'suitably'), then if $U$ has exponent $\pi$, it is determined upto isomorphism by $\overline{U}$. to be more precise, $\overline{U}=M\oplus \Omega_{kG} M$, and $U\cong \Omega_{\mathcal{O}G} M$ for some $kG$-module $M$ (everything is projective-free). I think that counts as a (non-trivial) example of when this is "rigid".

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