Cartan decomposition of a unitary group?

Question

For local fields $F$, we consider two case 1) $E$=quadratic extension of $F$ , 2) $E = F \times F$.

Let V be a 2-dim hermition space over E.

In 1) case, by Cartan decompostion $U(2)$ can be decomposed as $KMK$. (here, $K$ is a compact subgroup of $U(2)$ and M={$x \in E^\times | \left\vert x \right\vert \le 1 $}

In 2) case, $U(2)=GL_2 (F)$ the cartan decomposition of $U(2)$ is $KMK$ where $K$ is compact subgroup and $M$={ $\begin{pmatrix} x & 0 \\\ 0 & y \end{pmatrix} \in GL_{2}(F)$ | $|x|\le|y|$ }.

My question arises here.

If we consider $U(1)$, then, how can we think the Cartan decomposition of it in the above two cases? I don't know what the above $M$ should be in these case.

Answer 1

Let me formulate the Cartan decomposition for a reductive group over a local field (for more information see e.g. http://www.math.tau.ac.il/~bernstei/Unpublished_texts/unpublished_texts/Bernstein93new-harv.lect.from-chic.pdf section 2.1):

Let $G$ be a reductive group defined over over a local field $F$ of characteristic $0$. Let $A$ be the maximal split torus. That is, the maximal algebraic subgroup defined over $F$ of $G$ which is isomorphic to $(F^{\times})^k$ for some $k$. Note that it is unique up to a conjugation. Let $K$ be the maximal compact subgroup of $G$. Note that in the Archemedian case it is an algebraic subgroup and it is unique up to conjugation. In the non-Archemedian it is not algebraic and often it is not unique up to a conjugation. If your group is defined over the ring of integers $O$ of $F$, you can take $K=G(O)$.

The week version of Cartan decomposition says: $$G=KAK.$$

For the stronger version, let $\Lambda=Mor(F^\times, A)$ be the group of co-characters of $A$. Let $\Lambda^{++}$ be the Wile chamber of $\Lambda$. ruffly spicing it is the fundamental domain for $W:=N_G(A)/Z_G(A)$. We have an obvious map $\Lambda \times F^\times \to A$. Let $A^+$ be:\ 1. In the Archemedian case the image of $\Lambda^{++} \times \mathbb{R}_{>0}$. 2. In the non-Archemedian case the image of $\Lambda^{++} \times \pi$, where $\pi$ is the uniformizer of $F$.

Than $$G=KA^+K.$$ Moreover, any $K\times K$ double co-set have a unique element from $A^+$.

The case of $U(1)$, this is not that interesting. First of all the group is commutative, so we do not need to write $K$ twice. In the non-split case, the group is compact, then $A$ is trivial and the Cartan decomposition says $G=K$. In the split case, $A$ is one dimensional and $A^+$ is either $\mathbb{R}_{>0}$ or $\pi^\mathbb{Z}$.