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Is it true that all continuous finite dimensional $p$-adic representations of $Gal(\bar{K}/K)$ are semisimple, where $K$ is a number field, i.e. if $\rho:Gal(\bar{K}/K) \mapsto GL_n(\bar{\mathbb{Q}}_p)$ is a continuous representation, where $K$ is a number field (even $\mathbb{Q}$ if you want), is it semisimple?

Replacing $\mathbb{Q}_p$ with $\mathbb{C}$, the result is indeed true, using the existence of an invariant Haar measure over the complex numbers. So the natural question (related) is if it exists a $\mathbb{Q}_p$-valued Haar measure in a compact group? Of course some conditions must be removed from the classical formulation (for example it does not make sence to ask that if $f$ is a positive continuous function, then its integral is positive), but the existence of a non-trivial invariant measure such that the volume of the group is 1 should be enough. Any reference is welcome as well

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Actually, you don't need Haar measure to prove semisimplicity of Galois representations over $\mathbb{C}$. It isn't hard to show that any continuous $\mathbb{C}$-representation of a profinite group must have finite image, so this already follows from the representation theory of finite groups. – Kevin Ventullo May 12 '11 at 21:38
See also notes of Colmez, . Around page 10 he comments on p-adic distributions (and non-existence of "p-adic haar measures"). – mnr Jun 27 '11 at 6:26
up vote 17 down vote accepted

This can't possibly work. Many p-adic Galois representations are not semisimple. For instance, $\mathbb{Z}_p$ occurs as a quotient of the Galois group of $\mathbb{Q}$ (as the Galois group of the cyclotomic $\mathbb{Z}_p$-extension) and there are non-semisimple $\mathbb{Q}_p$-representations of $\mathbb{Z}_p$, e.g. the map sending $x \in \mathbb{Z}_p$ to $\begin{pmatrix} 1 & x \\ 0 & 1\end{pmatrix}$. (This is pretty fortunate, since otherwise all Galois cohomology groups would be forced to be trivial.)

As for Haar measure: what would be a $\mathbb{Q}_p$-valued Haar measure on $\mathbb{Z}_p$? It would have to give $p^n \mathbb{Z}_p$ measure $1/p^n$, but these are getting large rather than small p-adically; so you couldn't sensibly integrate continuous $\mathbb{Q}_p$-valued functions against such a measure - the "Riemann sums" would not converge. The lack of anything resembling Haar measure is one of the reasons why p-adic analysis has a very different flavour from real analysis.

EDIT: As some other posters have pointed out, it's possible to define a nice $\mathbb{Q}_p$-valued, and even $\mathbb{Q}$-valued, measure on a suitably nice class of sets of $\mathbb{Z}_p$; e.g. you can clearly do this for open compact subsets. This then gives you a linear functional on locally constant functions (analogous to the step functions used in measure theory over $\mathbb{R}$). But this won't extend to a linear functional on continuous $\mathbb{Q}_p$-valued functions, for the reasons I sketched above, and so it doesn't imply anything about semisimplicity of continuous representations.

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You might encounter the question: Why would it be unfortunate if Galois cohomology were trivial? :) – Minhyong Kim May 12 '11 at 23:48
In which case, I think I'd respond: Because nobody would pay me to do research on it. :-) – David Loeffler May 13 '11 at 7:50

A remark on the (lack of) $p$-adic valued Haar measure: there can exist topologically irreducible continuous representations of pro-$p$ groups (e.g. $GL_n(\mathbb Z_p)$ on infinite-dimensional $p$-adic Banach spaces. This is pretty strong evidence that there is no simple analogue of Haar measure in the $p$-adic situation. (Remember that the existence of Haar measure means that this is impossible for representations of compact groups on real or complex spaces: topologically irreducible representations are necessarily finite-dimensional.)

Another remark, on semi-simplicity: It is conjectured that the representations of the absolute Galois group $G_K$ of $K$ occuring on the $p$-adic etale cohomology of a smooth proper variety over a number field $K$ is semi-simple. (This is a part of the so-called Tate conjecture.) As far as I know, this is very wide open other than in the abelian variety case (and hence in the case of $H^1$ in general, since $H^1$ can always be thought of as the $H^1$ of the Albanese). As far as I know, there is no general representation-theoretic argument of the type you envisage which would prove it; it is something deep and special about the particular representations appearing in etale cohomology.

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As has been pointed out, the conclusion is false, so that no analogue of Haar measure can hope to prove it; but there is some analogue of Haar measure on $\operatorname{GL}_n(\mathbb Q_p)$, introduced (I think) by Julia Gordon: – L Spice May 13 '11 at 3:37
Err, sorry, I meant $\operatorname{GL}_n(\overline{\mathbb Q_p})$. Of course there is a very literal analogue of Haar measure on $\operatorname{GL}_n(\mathbb Q_p)$. :-) – L Spice May 13 '11 at 14:18
Thanks Matthew for the answer. I was wondering why one cannot generalize the complex case to prove that the p-adic rep's comming from geometry are semisimple. A natural questios was if it happens that all p-adic Galois representations turn out to be semisimple, but we just don't know how to prove it, but by your comment (and David's example) it is clear that this is not the case and there is something deep about etale cohomology. – A. Pacetti May 14 '11 at 23:58

Well, actually, the "motivic Haar measure" LSpice refers to is an analogue of Haar measure that lives on ${\operatorname{GL}}_n({\mathbb C}((t)))$, not on ${\operatorname{GL}}_n({\mathbb Q}_p)$, and takes values in the Grothendieck ring of varieties, so I think it's not quite relevant here. What is closer to this discussion though is the fact that the usual Haar mesaure on ${\operatorname{GL}}_n({\mathbb Q}_p)$ if it's reasonably normalized (e.g. so that the volume of ${\operatorname{GL}}_n({\mathbb Z}_p)$
is $1$), actually takes values in $\mathbb Q$ on all reasonable sets that you ever want to consider. More precisely, one can define a sigma-algebra of the so-called "definable sets", and the volumes of definable sets just are in $\mathbb Q$. In this sense they are in $\mathbb Q_p$ already, so the trick is that for these sets you do not need any completion of $\mathbb Q$ in order to define their volumes, and so you do not need to worry about using the $p$-adic metric... Most sets one works with turn out to be automatically definable, so this fact may be handy in some other situation.

Added some hours later: It was my first post on mathoverflow, and I am still not allowed to add comments to others' posts :) -- so this should be a comment to the comment by LSpice that appears in Emerton's post. Talking about measure on $\operatorname{GL}_n(\overline{{\mathbb Q}_p})$, there is a paper by E.Hrushovski and D. Kazhdan that talks about integration in algebraically closed valued fields (using logic). As a first approximation, as far as I understand, the values of this measure are something like equivalence classes of definable sets over the residue field (I am certainly being imprecise here). There are several papers by Yimu Yin (the ones to start with are, and aimed at clarifying this fundamental work of Hrushovski and Kazhdan in a slightly simplified setting. Unfortunately, I do not know of any non-technical introductory paper about this. There is a short note by Moshe Kamenski -- maybe this is the best place to start. Also, I hope someone corrects me here if I made any errors in this description.

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Could you please give some reference on this result? – Anatoly Kochubei May 13 '11 at 5:21
I should have included some references: the fact I mentioned follows from the theory of motivic integration by R. Cluckers and F. Loeser; the paper is the most relevant overview. – JGordon May 13 '11 at 5:28
Dear Anatoly, thanks for asking for references - your question appeared just as I was typing them up. Yoav Yaffe and I have another introductory note on this:; the original article by R. Cluckers and F. Loeser appeared in Inventiones, for the arxiv version see – JGordon May 13 '11 at 5:32
Thank you very much. – Anatoly Kochubei May 13 '11 at 5:47

Regarding the natural related question: looking for a $p$-adic valued Haar measure on $p$-adic stuff is generally doomed to failure. The problem is that small sets tend to have large $p$-power index and hence large $p$-adic measure. To see this explicitly in the simplest case, look at the description of the ring of distributions on $\mathbb{Z}_p$ given by Amice in terms of power series. Using this description, it's easy to verify that there is no nontrivial translation-invariant $p$-adic valued distribution on $\mathbb{Z}_p$.

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