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?
Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)
|
2
|
|
|
|
|
8
|
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. |
|||||||||||||||
|
You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.
|
5
|
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. |
||
|
|
