MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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.

share|cite|improve this answer
That doesn't really answer the question: is GL_n of Z_p the fixed points of some natural involution? – Ben Webster Nov 16 '09 at 4:36
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. – Peter McNamara Nov 16 '09 at 4:42
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. – Ben Webster Nov 16 '09 at 16:26
@Ben, I like your interpretation of the question, but I disagree with your implicit claim that the question was even well-defined as stated. – S. Carnahan Nov 16 '09 at 18:40
I think that Peter's answer is an insightful one to the precise question asked: "what's the possible p-adic analog[ue]?" – Pete L. Clark Mar 9 '10 at 4:15

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.