What is $PU(n,m)$? I have seen this notation in several places. I tried to read about it in Wikipedia,

In terms of matrices, elements of U(n) are complex n×n unitary matrices, and elements of the centre are diagonal matrices equal to eiθ multiplied by the identity matrix. Thus, elements of PU(n) correspond to equivalence classes of unitary matrices under multiplication by a constant phase θ

I'm not sure what "correspond to equivalence classes of unitary matrices under multiplication by a constant phase" means.
If I want to write this group explicitly, what would be the description of this set?
$ PU\left(n,m\right)=\left\{ \begin{pmatrix}A_{n\times n} & B_{n\times m}\\
C_{m\times n} & D_{m\times m}
\end{pmatrix}\in U\left(n,m\right):\thinspace\thinspace\thinspace such\thinspace\thinspace\thinspace that...\right\}  $
Note: I am using the definition of $U(n,m)$ as
$ U\left(n,m\right)=\left\{ \begin{pmatrix}A_{n\times n} & B_{n\times m}\\
C_{m\times n} & D_{m\times m}
\end{pmatrix}\in GL_{n+m}\left(\mathbb{C}\right):AA^{*}-BB^{*}=I_{n},\thinspace DD^{*}-CC^{*}=I_{m},AC^{*}=BD^{*}\right\}  $
If someone can describe the elements of $PU$ explicitly it would be very helpful.
Thanks in advance.
 A: Here is a general definition. Suppose that $G$ is a subgroup of $GL(n,{\mathbb C})$. Then the group $PG$ is defined as the projection $\pi(G)$ of $G$ to the group $PGL(n,{\mathbb C})$ under the quotient map
$$
\pi: GL(n,{\mathbb C})\to PGL(n,{\mathbb C}). 
$$
The group $PGL(n,{\mathbb C})$ is the quotient of $GL(n,{\mathbb C})$ by its center, the subgroup of scalar complex matrices ${\mathbb C}^\times$, i.e. matrices of the form
$$
\lambda I_n, \lambda\in {\mathbb C}^\times,
$$
where $I_n\in GL(n,{\mathbb C})$ is the identity matrix.
It is very seldom true that the subgroup $PG$ can be embedded in $GL(n,{\mathbb C})$. For instance, the group $PU(p,q)$ cannot be embedded (as a subgroup) in $GL(p+q, {\mathbb C})$, unless $p+q=1$, which is, of course, utterly uninteresting. I am not going to prove this.
Nevertheless, the group $PGL(n,{\mathbb C})$ does embed in
$GL(N,{\mathbb C})$ for $N=n^2-1$. Hence, $PU(p,q)$ does embed in
$GL((p+q)^2-1,{\mathbb C})$ as a subgroup of invariants of some  polynomials (but I find it painful to write down these invariants explicitly).
For a subgroup $G< GL(n,{\mathbb C})$ the projection $\pi(G)$ is naturally isomorphic to the quotient $G/C_G$, where $C_G= G\cap {\mathbb C}^\times$.
Consider now the case when $G=U(p,q)$, $p+q=n$. My favorite definition of this group (same as Wikipedia's) is as the subgroup of $GL(n,{\mathbb C})$ consisting of matrices which preserve the pseudo-hermitian product on ${\mathbb C}^n$:
$$
\langle v, w\rangle = \sum_{k=1}^p v_k \bar{w}_k - \sum_{k=p+1}^n v_k \bar{w}_k,
$$
where $v=(v_1,...,v_n), w=(w_1,...,w_n)$. I find your definition to be nonstandard. I think, you are using transposes of the standard matrix representation, i.e. you are considering the action on the row-vectors rather than column-vectors. There are situations when this is useful, I just prefer the standard notation. My suggestion to you would be to use the standard notation as well. In particular, I prefer not to answer your question about $U(1,1)$ since our notation are different. However, for the rest of the answer, the difference is irrelevant.
Then the intersection of $G=U(p,q)$ with the subgroup of scalar matrices consists of scalar matrices of the form
$$
\lambda I_n, \lambda\in {\mathbb C}^\times, |\lambda|=1. 
$$
I will use the notation $H$ for this intersection (I do not think there is a standard notation here).
Answering your question in the comments: When do two matrices $a, b\in U(p,q)$ have the same projection to the $PU(p,q)$? It is when the two $H$-cosets are equal: $aH=bH$. Equivalently, when $b\in aH$. Equivalently, when there is $\lambda\in {\mathbb C}^\times, |\lambda|=1$ such that $b=\lambda a$.
