maximal compact subgroup as fixed points of some involution on p-adic group? - MathOverflow

maximal compact subgroup as fixed points of some involution on p-adic group?

2009-11-16T03:55:23Z

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?

Answer by Peter McNamara for maximal compact subgroup as fixed points of some involution on p-adic group?

2009-11-16T04:19:23Z

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.

Answer by Jeff Adler for maximal compact subgroup as fixed points of some involution on p-adic group?

2010-03-09T04:06:48Z

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.