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
$\left( \begin{array}{cc} e^{i \theta} & 0 \\\\ 0 & e^{-i \theta} \end{array} \right)$.

$\DeclareMathOperator\U{U}\DeclareMathOperator\SU{SU}$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
$\left( \begin{array}{cc} e^{i \theta} & 0 \\\\ 0 & e^{-i \theta} \end{array} \right)$.

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
$\left( \begin{array}{cc} e^{i \theta} & 0 // 0 & e^{-i \theta} \\ \end{array} \right)$.

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
$\left( \begin{array}{cc} e^{i \theta} & 0 \\\\ 0 & e^{-i \theta} \end{array} \right)$.

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 $A \in SU(n+1)/SU(n)$?

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
$\left( \begin{array}{cc} e^{i \theta} & 0 // 0 & e^{-i \theta} \\ \end{array} \right)$.