Complex projective space as a $U(1)$ quotient

Question

$\DeclareMathOperator\U{U}\DeclareMathOperator\SU{SU}$As is well known, one can view $\mathbb{CP}^n$ as a quotient of the unit $(2n + 1)$-sphere in $\mathbb{C}^{n+1}$ under the action of $\U(1)$, since every line in $\mathbb{C}^{n+1}$ intersects the unit sphere in a circle.

Moreover, we have $S^{2n + 1} = \SU(n+1)/\SU(n)$, where $\SU(n)$ embeds into the bottom right-hand corner (say).

My question is: Is there an embedding $j$ of $\U(1)$ into $\SU(n+1)/\SU(n)$ that gives the $\U(1)$ action as a left (right) multiplication, i.e. such that $A.e^{i \theta} = Aj(e^{i \theta})$, for all $A \in \SU(n+1)/\SU(n)$?

For $n=1$, it's easy: $\SU(2) = S^3$, and we embed $e^{i\theta}$ as
$\left( \begin{array}{cc} e^{i \theta} & 0 \\\\ 0 & e^{-i \theta} \end{array} \right)$.

Answer 1

The $U(1)$ group is the torus in $SU(N+1)$ which commutes with $SU(N)$. In the example given in the question where $SU(N)$ is chosen as the bottom $N$-dimensional block. $U(1)$ consists of the diagonal matrices

$$\operatorname{diag}\{\exp(Ni\theta), \underbrace{\exp(-i\theta), \dots, \exp(-i\theta)}_\text{$N$ times}\}.$$

Please observe that the restriction to the bottom $N$-dimensional block is proportional to the unit matrix thus it commutes with the whole of $SU(N)$, also it belongs to $SU(N+1)$, since it has a unit determinant.

The reason that the $U(1)$ and the $SU(N)$ factors commute is due to a theorem by A. Borel which states that the denominator subgroup of homogeneous Kaehlerian spaces must be the centralizer of a torus. In our case the torus is the $U(1)$ subgroup and the certralizer is $SU(N) \times U(1)$.

Answer 2

My original answer was unsalvegable so I've deleted it and am posting a new "answer". As with the first one, I don't rate this as particularly an answer but more just trying to understand what's going on.

I was initially having trouble understanding Scott's answer, but now I think I do and I think it gives the matrix representation wanted which isn't quite what David wrote.

We have $SU(n+1)$ and inside this we have $SU(n)$ and quotient out to get $S^{2n+1}$. We also have a slightly larger subgroup which is $S(U(n) \times U(1))$, which contains $SU(n)$, such that the quotient is $\mathbb{CP}^n$.

Now, $S(U(n) \times U(1))$ is $U(n)$ via $A \mapsto (\det A^{-1},A)$ and the inclusion $SU(n) \to S(U(n) \times U(1))$ goes over to the standard inclusion. Here, $SU(n)$ is a normal subgroup and $U(n)$ is the semi-direct product of $SU(n)$ and $U(1)$ with the map $U(1) \to U(n)$ given by $\lambda \mapsto (\lambda, 1,\dots,1)$ (diagonal matrix). When taken over to $S(U(n) \times U(1))$ this becomes $\lambda \mapsto (\lambda^{-1},\lambda,1,\dots,1)$.

So then $SU(n+1)/S(U(n) \times U(1)) \cong (SU(n+1)/SU(n))/U(1)$ where $U(1) \to SU(n+1)$ is the map $\lambda \to (\lambda^{-1},\lambda,1,\dots,1)$.

This isn't the same as David's, I know, so it may not be what you want (since that answer's been accepted). Presumably only one satisfies the condition that you want and presumably it's David's since that answer's been accepted. Still, I was confused and I think I've straightened myself out now.

Answer 3

From wikipedia: $\mathbb{CP}^n$ is a Symmetric space of type AIII for $p=n$, $q=1$. There are embeddings of both $U(1)$ and $SU(n)$ into $S(U(n) \times U(1)) \subset SU(n+1)$ which give you your quotient by right multiplication.