Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Maximal compact subgroup
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
===Proofs=== For a real semisimple Lie group, Cartan's proof of the existence and uniqueness of a maximal compact subgroup can be found in {{harvtxt|Borel|1950}} and {{harvtxt|Helgason|1978}}. {{harvtxt|Cartier|1955}} and {{harvtxt|Hochschild|1965}} discuss the extension to connected Lie groups and connected locally compact groups. For semisimple groups, existence is a consequence of the existence of a compact [[Complexification (Lie group)|real form]] of the noncompact semisimple Lie group and the corresponding [[Cartan decomposition]]. The proof of uniqueness relies on the fact that the corresponding [[Riemannian symmetric space]] ''G''/''K'' has [[negative curvature]] and Cartan's fixed point theorem. {{harvtxt|Mostow|1955}} showed that the derivative of the exponential map at any point of ''G''/''K'' satisfies |d exp ''X''| β₯ |X|. This implies that ''G''/''K'' is a [[Hadamard space]], i.e. a [[complete metric space]] satisfying a weakened form of the parallelogram rule in a Euclidean space. Uniqueness can then be deduced from the [[Bruhat-Tits fixed point theorem]]. Indeed, any bounded closed set in a Hadamard space is contained in a unique smallest closed ball, the center of which is called its [[circumcenter]]. In particular a compact group acting by isometries must fix the circumcenter of each of its orbits.
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)