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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?) 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.

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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 $K$.

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  • $\begingroup$ Thanks David. The proof is really simple. In fact, I tried to find out the exact form of K. I am really appreciaited for shedding a light. $\endgroup$
    – Jude
    Commented May 7, 2013 at 1:18
  • $\begingroup$ The "exact form of K" depends rather on whether $F$ is archimedean or nonarchimedean. For $F = \mathbb{R}$, you can use the fact that the indefinite special unitary group in 2 variables, $SU(1,1)$, is isomorphic to $SL(2)$ and the max compact subgroup of the latter is well known. $\endgroup$
    – David Loeffler
    Commented May 7, 2013 at 6:42
  • $\begingroup$ Thanks for letting me know the archimedean case. It helps to broaden my understanding. $\endgroup$
    – Jude
    Commented May 7, 2013 at 16:01

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