Hello. I have some question on Cartan decomposition of unitary group, especially $U(2)$.


I am interested in local situation, that is p-adic or archimedian.


Let $F$ be a local field and $E$ be its quadratic extension. Let $V$ be a hyperbolic hermitian vector space of dimension $2$ and we consider $U(V)$.


Then by Cartan decomposition, (see [Cartan decomposition of a unitary group?](https://mathoverflow.net/questions/114209/cartan-decomposition-of-a-unitary-group)) we can decompose $U(V)=KM^{+}K$ where $M^+=${$x \in E^{\times}||x|\le1 $}.


Here, I am just curious whether the center of $U(V)$, that is just $U(1)$ by diagonal embedding into $U(V)$, is contained into $K$. Since $U(1)$ is compact, it seems that it is possible to take $K$ from the beginning to contain the center. Am I right?(As I don't know the exact shape of $K$, I am not sure it)


Any words or comments will be greatly appreciated.



On the Cartan decomposition of unitary group
*Theorem*: Let $G$ be a reductive algebraic group over a local field $F$, let $K$ be any maximal compact subgroup of $G(F)$, and let $Z = Z(G)$. Then $K \cap Z(F)$ is the unique maximal compact subgroup of $Z(F)$.


*Proof*: Let $W$ be the maximal compact subgroup of $Z(F)$. Then the natural multiplication map $K \times W \to G(F)$ has image a compact subgroup of $G(F)$ containing $K$; since $K$ is maximal by assumption, the image is equal to $K$ and thus $W \subseteq K$. QED.


In particular, in your case $Z$ is $U(1)$ and hence $Z(F)$ is compact, and thus $Z(F)$ is contained in $K$ for every maximal compact $K$.


