As is well known, maximal compact subgroup of real Lie group is just the fixed points of Cartan involution.
Now the question is what's the possible p-adic analog?
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2 Answers
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There are some complications in the p-adic case. For GL_n, every maximal compact subgroup is conjugate to GL_n of the ring of integers, but for more general groups (eg PGL_2, SU_3) there can be more than one conjugacy class of maximal compact subgroups. One thing that you can say is that every maximal compact subgroup is the stabaliser of a point on the Bruhat-Tits building.
answered Nov 16, 2009 at 4:19
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$\begingroup$ That doesn't really answer the question: is GL_n of Z_p the fixed points of some natural involution? $\endgroup$– Ben Webster ♦Nov 16, 2009 at 4:36
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1$\begingroup$ I realise it doesn't answer the question, but provides some information on maximal compact subgroups that might be useful. I'll see if I can go away and give any sort of a better answer. $\endgroup$ Nov 16, 2009 at 4:42
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$\begingroup$ Fair enough. It's wise in a situation like that to say something like "I realize this isn't quite your question, but..." just so people don't think you misread. $\endgroup$– Ben Webster ♦Nov 16, 2009 at 16:26
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$\begingroup$ @Ben, I like your interpretation of the question, but I disagree with your implicit claim that the question was even well-defined as stated. $\endgroup$– S. Carnahan ♦Nov 16, 2009 at 18:40
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$\begingroup$ I think that Peter's answer is an insightful one to the precise question asked: "what's the possible p-adic analog[ue]?" $\endgroup$ Mar 9, 2010 at 4:15
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Here is how the real and p-adic situations are the same.
Let $G$ be a connected reductive algebraic group defined over a field $F$ not of characteristic two. Let $\theta$ be an involution of $G$ defined over $F$. Then the group $G^\theta$ of fixed points is a reductive algebraic subgroup of $G$.
Here are two ways in which they are different.
In the real case, one can always choose $\theta$ so that the group of rational points of $G^\theta$ is compact. In the p-adic case, compact reductive groups are quite rare, and so in most cases there is no analogous way to choose $\theta$.
Second, compact subgroups do not play the same roles in the real and p-adic cases. Think of the fields themselves. In the p-adic case, the maximal compact subring is the ring of integers. In the real case, there are no nontrivial compact subrings. There is a ring of integers, but it is not compact. Moreover, since $G^\theta$ has smaller dimension than $G$, it cannot be an open subgroup, and maximal compact subgroups are always open in the p-adic case. Thus, even in the rare cases where $G^\theta$ is compact, it is not maximal.
