CP(2) = SU(3)/U(2)? In my understanding the complex projective line $CP^1 = \mathbb{C}^2/\mathbb{C^*}$ where $\mathbb{C^*}$ is $\mathbb{C}$ without $0$, i.e. just 2 complex coordinates and a homogeneous factor.
And the complex projective plane  $CP^2 = \mathbb{C}^3/\mathbb{C^*}$: 3 complex coordinates and a homogeneous factor.
I read in an article that the complex projective line $CP^1 = SU(2)/U(1)$ and
the complex projective plane $CP^2 = SU(3)/U(2)$.
I wonder how this is meant.
How is it related to the idea of coordinates and homogeneous factors?  
 A: Let $A \in SU(3)$, $A=\matrix{(x_1& x_2&x_3)}$ where $x_1 \in \mathbb{C}^3$, $x_2 \in \mathbb{C}^3$, $x_3 \in \mathbb{C}^3$, with $\|x_1\|=\|x_2\|=\|x_3\|=1$ and $\langle x_1,x_2\rangle=\langle x_2,x_3 \rangle =\langle x_1,x_3 \rangle=0$.
$U(2)$ acts on the first two vectors of $A$. If $B \in U(2)$, $\left(\matrix{x_1 & x_2}\right)B$ describes all the orthonormal basis of the plane $(x_1,x_2)$, when $B$ describes $U(2)$. So, an element in $SU(3)/U(2)$ is the same thing that a plane $P$ in $\mathbb{C}^3$ and a unit vector $x_3$ orthogonal to the plane $P$. So it is the same that a unit vector $x_3$ in $\mathbb{C}^3$.
More generally, in other words, let $A_1,A_2 \in SU(n)$, let $c(A_1)$ and $c(A_2)$ the last columns of $A_1$ and $A_2$. $c(A_1)$ and $c(A_2)$ are unit vectors. We have: $c(A_1)=e^{i\theta}c(A_2)$ for some $\theta \in \mathbb{R}$ if and only if $A_1=A_2\left(\matrix{B&0 \\0 & \det(B)^{-1}}\right)$ for some $B \in U(n-1)$.
If we define an action of $U(n-1)$ on $SU(n)$ by $A*B=A\left(\matrix{B&0 \\0 & \det(B)^{-1}}\right)$ for $A\in SU(n)$ and $B \in U(n-1)$, we have $\exists B \in U(n-1),A_1=A_2*B$ if and only if $\exists \theta, c(A_1)=e^{i\theta}c(A_2)$. 
Let the quotient map that maps $x \in \mathbb{C}^n-\{0\}$ to $\overline{x} \in CP^{n-1}:=(\mathbb{C}^n-\{0\})/\mathbb{C}^*$.
Let $\equiv $ the equivalence relation $A_1\equiv A_2$ for $A_1,A_2 \in SU(n)$ if and only if $\exists B\in U(n-1),A_1=A_2*B$.
(1) We have $A_1\equiv A_2$ if and only if $\overline{c(A_1)}=\overline{c(A_2)}$. 
So there is a map on $SU(n)/U(n-1)$ to $CP^{n-1}$. It is monic because of (1). It's onto because every column $x\in \mathbb{C}^n-\{0\}$ such that $\|x\|=1$ may be completed to a matrix in $SU(n)$.
