Prove that this action admits three orbits I have the following problem:

Let $G=\operatorname{SU}(2)$ and let $T=\left \{\begin{pmatrix}e^{i\theta} & 0 \\
0 & e^{-i\theta}\end{pmatrix},\theta \in \mathbb{R}\right \}\cong \operatorname{U}(1)$ be a maximal torus in $G$. We denote by  $\mathcal{F}$ the flag manifold $G/T$ of $G$.
Let$$N(T)=\left \{\begin{pmatrix}e^{i\theta} & 0 \\
0 & e^{-i\theta}\end{pmatrix},\theta \in \mathbb{R}\right \}\cup \left \{\begin{pmatrix}0 & e^{i\theta} \\
e^{-i\theta} & 0\end{pmatrix},\theta \in \mathbb{R}\right \},$$ be the normalizer of $T$. Consider the matrix $A=\begin{pmatrix}i & 0 \\
0 & -i\end{pmatrix}$.
Prove that the set $C=\left \{[g]\in \mathcal{F},g^{-1}AgA^{-1}\in N(T)\right \}\subset S^2$ is composed of the north pole $N$, the south pole $S$ and the equator $E$. Conclude that the $T$-action on $C$ admits three orbits.

I have tried the following:
First, we have $\operatorname{SU}(2)=\left \{\begin{pmatrix}a & -\overline b \\
b & \overline a\end{pmatrix},\lvert a\rvert^2+\lvert b\rvert^2=1\right \}$. Pick a matrix $g=\begin{pmatrix}a & -\overline b \\
b & \overline a\end{pmatrix}\in \operatorname{SU}(2)$ such that $[g]\in C$, then $g^{-1}AgA^{-1}=\begin{pmatrix}\lvert a\rvert^2-\lvert b\rvert^2 & 2\overline{ab} \\
-2ab & \lvert a\rvert^2-\lvert b\rvert^2\end{pmatrix}\in N(T)$, which implies that either $\lvert a\rvert^2-\lvert b\rvert^2=0$ or $ab=0$. If $\lvert a\rvert^2-\lvert b\rvert^2=0$, then $g=\begin{pmatrix}a & -\overline b \\
b & \overline a\end{pmatrix}$, such that $\lvert a\rvert^2=\lvert b\rvert^2=\dfrac{1}{2}$.
If $ab=0$, then $a=0$ or $b=0$. The case where $a=0$ implies $g=\begin{pmatrix}0 & -\overline b \\
b & 0\end{pmatrix}$,  such that $\lvert b\rvert^2=1$, whereas the case where $b=0$, we obtain $g=\begin{pmatrix}a & 0 \\
0 & \overline a\end{pmatrix}$, such that $\lvert a\rvert^2=1$. Hence $$C=\left \{\left [\begin{pmatrix}0 & -1 \\
1 & 0\end{pmatrix}\right ]\right \}\cup \left \{\left [\begin{pmatrix}i & 0 \\
0 & -i\end{pmatrix}\right ]\right \}\cup  \left \{\left [\begin{pmatrix}a & -\overline b \\
b & \overline a\end{pmatrix}\right ],\lvert a\rvert^2=\lvert b\rvert^2=\frac{1}{2}\right \}.$$  Could you please help in identifying the poles and the equator with what I have obtained and explain how to show that the $T$-action on $C$ has three orbits?
 A: You have already identified the 3 orbits of $T$ on $C$ by the way you have written $C$.  Namely, the two points of $N(T)/T$ are both $T$-fixed, and so constitute 2 orbits.  For the 3rd, note that
$$
\operatorname{diag}(z, z^{-1})\begin{pmatrix}
1/\sqrt2 & 1/\sqrt2 \\
-1/\sqrt2 & 1/\sqrt2
\end{pmatrix}\operatorname{diag}(w, w^{-1}) = \begin{pmatrix}
z w/\sqrt2 & z w^{-1}/\sqrt2 \\
-z^{-1}w/\sqrt2 & z^{-1}w^{-1}/\sqrt2
\end{pmatrix}
$$
for all $z, w \in S^1$.  These matrices are all of the form $\begin{pmatrix} a & b \\ -\overline b & \overline a \end{pmatrix}$ with $\lvert a\rvert^2 = \lvert b\rvert^2 = 1/2$, and every such matrix arises in this way (just put $z = \sqrt{2a b}$ and $w = \sqrt{a/b}$, with compatibly chosen square roots).
As to identifying these with points of $S^2$, you first have to choose an identification of $\mathcal F$ with $S^2$.  One way to do this is to note that the map that sends $\begin{pmatrix} a & b \\ -\overline b & \overline a \end{pmatrix} \in G$ to the line through $\begin{pmatrix} a \\ -\overline b \end{pmatrix}$ is a surjection from $G$ to the set of lines in $\mathbb C^2$ for which the stabiliser of the line through $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ is $T$, so that we may $G$-equivariantly identify $\mathcal F = G/T$ with the space of lines in $\mathbb C^2$.  (A line in $\mathbb C^2$, which can be uniquely completed to a nested structure of a $0$-dimensional subspace of a $1$-dimensional subspace of a $2$-dimensional subspace of $\mathbb C^2$, is a particularly simple case of a flag; this is why we call $\mathcal F$ a flag manifold.)  Using the Hopf fibration allows us to map further to $S^2$, identified with the norm-$1$ imaginary quaternions, by sending the line through $\begin{pmatrix} a \\ -\overline b \end{pmatrix}$, where $\lvert a\rvert^2 + \lvert b\rvert^2 = 1$, to $(a - j\overline b)i(\overline a + b j)$, which, if you multiply it out, sends $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ (the image of every element of $T$) to $i$, which we may take to be the north pole; $\begin{pmatrix} 0 \\ 1 \end{pmatrix}$ (the image of every element of $N(T) \setminus T$) to $-i$, which is then the south pole; and $\begin{pmatrix} a \\ -\overline b \end{pmatrix}$ with $\lvert a\rvert^2 = \lvert b\rvert^2 = 1/2$ (the image of a point of $C$ that does not lift to $N(T)$) to the norm-$1$ quaternions of the form $x j + y k$ with $x, y \in \mathbb R$, which is then the equator.  Geometrically, the map $G \to S^2$ that I have specified, given by $\begin{pmatrix} a & b \\ -\overline b & a \end{pmatrix} \mapsto (a - j\overline b)i(\overline a + b j)$, is invariant under right translation by $T$ (which is why we may view it as a map $\mathcal F \to S^2$), and identifies left translation by $\operatorname{diag}(e^{i\theta}, e^{-i\theta})$ with rotation about the $i$-axis through an angle of $2\theta$, from $j$ towards $k$—which explains why the north pole and the south pole are singleton orbits.
