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Let $G$ be a compact group and $K$ a finite extension of $Q_{p}$. If $\rho$ is a continuous representation of $G$ on a finite dimensional vector space over $K$, then it is well known that the semi-simplification of the reduction modulo $p$ of an integral model of $\rho$ only depends upon the isomorphism class over $K$ of $\rho$. Are there results or strategies that, given a $K$-representation $\rho$ as above, allow one to determine properties of all the possible integral models of $\rho$? For example: are there some conditions under which all the integral models of $\rho$ are semi-simple? Where can I find references? Thanks!

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Presumably over $K$ you mean "reduction modulo $\pi$" (for a uniformizer $\pi$ of $O_K$) rather than "reduction modulo $p$". Anyway, "semi-simple" for modules $M$ over a ring $R$ admitting a "nontrivial" ideal theory is not a good notion: there are $R$-submodules $I M$ for an ideal $I$ of $R$ (preserved by all $R$-linear automorphisms of $M$). One exercise-level result is that if the reduction is irreducible then integral model is unique up to $K^{\times}$-scaling. Do you have specific "properties of all integral models" in mind? That would clarify the motivation. – BCnrd Oct 11 '10 at 0:44

Let be focus on the two-dimensional case first. Finding an integral model for $\rho$ is equivalent to choosing a lattice in $K^2$ that is invariant under $\rho(G)$. So in this case you are asking for invariant points in the tree for $GL_2(K)$ under a compact subgroup $\rho(G)$ of $GL_2(K)$. There are various possibilities, and there is a nice paper of Bellaiche and Chenevier Sous-groupes de $GL_2$ et arbres discussing the possibilites.

When you say that you would like all integral models to be semi-simple, I'm not sure what you mean; but it say the represenation over $K$ is irreducible, then it will never happen that all the integral models have semi-simple reduction, if they are not themselves irreducible. (This is a proposition of Ribet, and is easily seen in the tree-based picture.)

Just to illustrate how one argues: if two points in the tree are $\rho(G)$-invariant, then so will be all the points on the line segment joining them. If $\rho(G)$ fixes an infinite half-line, this corresponds to an invariant one-dimensional space over $K$, and so means that $\rho$ is reducible over $K$. So if $\rho(G)$ is irreducible over $K$, then the fixed set of $\rho(G)$ will be bounded and convex, which already imposes some restrictions. (One an prove Ribet's result by these sort of considerations: basically, any extremal point of the fixed set of $\rho(G)$ will have non-simple reduction.)

Once you understand the $GL_2$ case you can imagine how to try and generalize things to the $GL_n$ case, although the combinatorics becomes more complictated, and (as Bellaiche and Chenevier note) it is harder to say anything nice, because when $n > 2$, $GL_n$ is the full automorphism group of the building.

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Just a nitpick: the second paragraph should end, "... unless the reductions are themselves irreducible." – Kevin Ventullo Oct 11 '10 at 1:33
Dear Kevin, Thanks --- this is now fixed. – Emerton Oct 11 '10 at 2:28
The Ribet-Bellaïche lemma is discussed by Mazur in his expository talk How can we construct abelian Galois extensions of basic number fields? at the Ribet Conference. See – Chandan Singh Dalawat Oct 11 '10 at 3:19
Dear Chandan, Thanks very much for pointing this out. – Emerton Oct 11 '10 at 3:51

For general representations, there are no such strategies and this is a very hard open problem, even for finite groups. Since your usual compact group will have various finite quotients, your problem is at least as hard. This rather older survey by Reiner is a good place to start. Particularly pages 171-173 should be of interest to you.

As for semi-simplicity, you need to be clear about what you mean. In the integral world, the classical notion of semi-simplicity makes little sense, because your representations always have subrepresentations. E.g. if $\Gamma$ is an integral model, then $p\Gamma$ is a subrepresentation. But at least the Krull-Schmidt theorem holds in your setting, even though the indecomposable summands of an integral model might not sit in irreducible $K$-representations. In fact, if the semi-simplification of the reduction of $\rho$ contains two non-isomorphic irreducible components, then I believe that there is almost no hope for all integral models to only have indecomposable summands that sit in irreducible $K$-representations. (Clearly, if the reduction itself is not semi-simple then your integral models will be at least as badly behaved.)

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