Cartan decomposition of SO(2n) I am trying to understand the Cartan decomposition theory, on the following example : $G=SO(2n)$, and $K=U(n)$, and I'm interested in the manifold $G/K$ (an hermitian symmetric space). 
1) How do we see $U(n)$ as a subgroup of $SO(2n)$ ?
2) Do we have a Cartan decomposition to $G$ ? More precisely do we have for the Lie algebra of $G$ that
$\mathfrak{g} = \mathfrak{k} + \mathfrak{p}$, where $\mathfrak{k}$ is the Lie algebra of $K$, and also the +1 eigenspace of an involution $\theta$ on $\mathfrak{g}$, and $\mathfrak{p}$ is the -1 eigenspace of $\theta$ ? 
3) If so, what is $\theta$ is that case? And what is $\mathfrak{p}$ explicitly ?
 A: Let $h$ be an hermitian structure on $V=\bf C^n$.  The real part $q$ of $h$ is an euclidian structure on $\bf R^{2n}$$ =V_{\bf R}$, the real vector space underlying $V$. In coordinate, $q((z_1,...z_n)=h(z_1,...z_n)=z_1\bar z_1+...z_n\bar z_n$. 
If a $\bf C$-linear map preserve $h$ it is $\bf R$linear and preserves $q$. This proves that $U(n)\subset O(2n)$. 
Note that an element of $O(2n)$ is in $U(n)$ iff it is linear as a complex map, in other word  commutes with the complex multiplication $I$ (viewed as a $\bf R$-linear map).
Acting on $M(2n, R)$ the conjugation by $I$, denoted 
$i$ and defined by $i(X)= IXI^{-1}$ is linear,  is an involution ($I^2=-Id$ commute with everything), and a group automorphism. 
Therefore, the pair $(O(2n), U(n)) $ is a  symmetric pair as $U(n)\subset O(2n)$ is the fixed group of the involution $i$. 
To compute the Cartan decomposition, it is enough to compute the set of fixed point of $i$ acting on the Lie algebra $o(2n)$ of antysmmetric $2n$ matrices, or to compute the tangent space of the symmetric space at the point $I$. 
One finds that this is the set of antisymmetric $(2n,2n)$ matrice s.t. $ MI+IM=0$, or in block $\left( \begin{array}{cc} A & B \\
B & -A \end{array} \right)$, with $A$ antisymmetric, $B$ symmetric. Here we identiofy $\bf R^{2n}$ and $\bf C^{n}$ via the map $(x_1,...,x_n, y_1,...y_n)\to (x_1+iy_1,...,x_n+iy_n)$, and the matrix of $I$ is $\left( \begin{array}{cc} 0 & -Id \\
Id & 0 \end{array} \right)$.
