Maximal compact subgroup of a real reductive group

Definition

A maximal compact subgroup of a real reductive Lie group $G$ is a compact subgroup $K\subseteq G$ not properly contained in any other compact subgroup. For a Cartan involution of $G$ , its fixed-point subgroup

K=G^\theta=\{g\in G:\theta(g)=g\}

is maximal compact, and its Lie algebra is the $+1$ -eigenspace $\mathfrak k$ in the Cartan decomposition $\mathfrak g=\mathfrak k\oplus\mathfrak p$ . Conversely, after conjugacy, maximal compact subgroups arise this way. Hence $K$ is not canonical as a subgroup, but its conjugacy class is canonical.

Conjugacy and topology

Every compact subgroup of $G$ is contained in a maximal compact subgroup, and any two maximal compact subgroups are conjugate in $G$ . Moreover, the inclusion $K\hookrightarrow G$ is a homotopy equivalence in the standard Cartan setting because $G$ is diffeomorphic to $K\times\mathfrak p$ . Thus many topological invariants of $G$ can be computed from $K$ Knapp, Chapter VI, §4.

Role in representation theory

Restricting a representation of $G$ to the compact group $K$ decomposes it into finite-dimensional $K$ -types. Their multiplicities formulate admissibility and provide the algebraic skeleton of Harish–Chandra modules. The subgroup also enters the decompositions $G=K\exp(\mathfrak p)$ and $G=KAN$ , so it connects representation theory to harmonic analysis and geometry.

Example and warning

For $G=\mathrm{GL}_n(\mathbb R)$ , one may take $K=\mathrm O(n)$ ; for $G=\mathrm{SL}_n(\mathbb R)$ , one takes $\mathrm{SO}(n)$ . A maximal compact subgroup is generally not a Cartan subgroup: these are different notions, despite the common use of the letter $K$ in Cartan theory.

References

A. W. Knapp, Lie Groups Beyond an Introduction, 2nd ed., Birkhäuser, 2002. Publisher record. Relevant: Chapter VI, §4 on maximal compact subgroups.
S. Helgason, Differential Geometry, Lie Groups, and Symmetric Spaces, American Mathematical Society, 2001. DOI record. Relevant: Chapter VI on the global structure associated with $K$ .