Dimension of the pseudo unitary group is $(p+q)^2$ where $U_{p,q} = M ∈ M_{p+q} (\Bbb C) : ^tMJ_{p,q}M = J_{p,q}.$ where $J_{p,q}=Diag(1...1,-1...-1)$ with $1\ p$ times and $-1\ q$ times.
How can we see that the dimension of this group is $(p+q)^2$? I can't see any reference or proof of this?
What about the dimension of $SU_{p,q}$?
Thank you for your help
Let $U(p,q) = \lbrace M \in M_{p+q}(\mathbb{C}) | ^t\overline{M}JM = J \rbrace$. One can compute the Lie algebra to be
$\mathfrak{u}(p,q) = \lbrace M \in M(p,q) | ^t\overline{M}J + JM = 0 \rbrace$. Now we will compute the explicit form as block matrix as below.
$M = \begin{bmatrix} A & B \\ C & D\end{bmatrix}$.
One can compute with above equations to get that the $M \in \mathfrak{u}(p,q)$ if and only if the matrix $M$ satisfies
$A + (^t\overline{A}) = 0 = D + (^t\overline{D})$ and $B = (^t\overline{C})$. That is $A \in \mathfrak{u}(p)$, $D \in \mathfrak{u}(q)$ and $B$ is any $p \times q$ matrix with complex entries. Now the only thing left is to add up the dimension of the spaces of matrices $A,B$ and $D$ which are $p^2, 2pq$ and $q^2$ respectively. Thus we have the answer.
One can caluculate the dimension of $SU(p,q)$ from here. Since the condition of determinant one imposes a linear equation in the Lie algebra of matrices having trace zero.