Conjugacy classes in $SU_2$

Question

I'm trying to find all conjugacy classes in $SU_2$.

Matrices in $SU_2$ are of the form:

$M = \begin{bmatrix} \alpha & \beta \\ - \bar{\beta} & \bar{\alpha} \end{bmatrix}$ and $|\alpha|^2+|\beta|^2 =1$

I thought I could use Jordan decomposition to do this and here's what I've got so far:

$W(M) = X^2 - trM \cdot X + \det M = X^2 - (\alpha + \bar{\alpha})X + 1$

so the eigenvalues are: $\lambda_{1} = \frac{\alpha + \bar{\alpha} + \sqrt{(\alpha + \bar{\alpha})^2 +4}}{2}, \ \ \lambda_{2} = \frac{\alpha + \bar{\alpha} - \sqrt{(\alpha + \bar{\alpha})^2 +4}}{2}$.

1) $\lambda_1 \neq \lambda_2$ if $(\alpha + \bar{\alpha})^2 \neq 4, \ \ \alpha = a+bi, \ \ a \neq \sqrt{2}, - \sqrt{2} $

and then $M$ is diagonizable and $M = S\begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} S^{-1}$

2) a) if $\beta=0, (\alpha + \bar{\alpha})^2 = 4$, then $M$ is diagonal

b) if $\beta=0, (\alpha + \bar{\alpha})^2 \neq 4$, then $M$ is not diagonizable and $M = S\begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix} S^{-1}$

But in order to find conjugacy classes I need those $S \in SU_2$ and I don't know what to do next. Maybe there's a better way to approach this problem?

I'll be grateful for all your help.

Update:

If we let $S = \begin{bmatrix} x & y \\ - \bar{y} & \bar{x} \end{bmatrix} $

then in case 1) $SMS^{-1} = \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix}$ and after multiplying those matrices I got these four equations:

$\bar{xy}(\bar{\alpha} - \alpha) + \bar{\beta} (\bar{x})^2 + \beta (\bar{y})^2=0$

$xy(\bar{\alpha} - \alpha) + \beta x^2 + \beta y^2=0$

$\bar{\alpha} y \bar{y} + \alpha x \bar{x} + \beta \bar{y} x - \bar{\beta} y \bar{x} = \lambda_1$

$\bar{\alpha} y \bar{y} + \alpha x \bar{x} - \beta \bar{y} x + \bar{\beta} y \bar{x} = \lambda_2$.

Could those equations be helpful?

Answer 1

Since the characteristic polynomial of any matrix $A \in SU(2)$ has real coefficients, its eigenvalues are complex conjugates $\lambda, \bar{\lambda}$; since $\lambda \bar{\lambda} = \det A = 1$, the eigenvalues are $e^{i \theta}, e^{-i \theta}$ for some unique $\theta \in [0, \pi]$.

If two matrices are in the same conjugacy class, they are similar and so have the same eigenvalues. Thus, all of the diagonal matrices $$\begin{pmatrix} e^{i \theta} & 0 \\ 0 & e^{-i\theta} \end{pmatrix} \in SU(2), \qquad \theta \in [0, \pi],$$ are in distinct classes.

Conversely, since any $A \in SU(2)$ is normal, by the Spectral Theorem it is unitarily equivalent (conjugate via an element of $U(2)$) to a diagonal matrix, which must be of the above form. In fact, we can easily show that it is conjugate via an element of $SU(2)$, so the classes containing the above diagonal matrices actually exhaust the conjugacy classes.

Edit Via the identification $$\begin{pmatrix}\alpha & \beta \\ -\bar{\beta} & \bar\alpha\end{pmatrix} \leftrightarrow (\alpha, \beta)$$ we can regard $SU(2)$ as the subset of $\mathbb{C}^2$ for which $|\alpha|^2 + |\beta|^2 = 1$, namely as the unit sphere $\mathbb{S}^3 \subset \mathbb{R}^4 \cong \mathbb{C}^2$.

As @Don pointed out in his comment, the conjugacy classes are latitude $2$-spheres in $\mathbb{S}^3$ (except for the north and south pole, which are singleton classes). There are several ways to see this, and here's one:

The conjugation action of $SU(2)$ on $\Bbb C^2$ decomposes into actions on the trace and tracefree parts of a matrix, via $$A \mapsto \left(\frac{1}{2} \operatorname{tr} A\right) I + \left[A - \left(\frac{1}{2} \operatorname{tr} A\right) I \right].$$

On the other hand, conjugation fixes the identity matrix, so the action on the tracefree part is trivial, and one can show that the action on the set of tracefree matrices preserves a definite quadratic form $Q$, inducing a map $SU(2) \to SO(3)$; this map is a double cover and in particular acts transitively on each of the spheres $\{Q(x) = r\}$ of the quadratic form.

In short, the conjugacy class of $$\begin{pmatrix}\alpha & \beta \\ -\bar{\beta} & \bar\alpha\end{pmatrix} \in SU(2)$$ consists of exactly the elements $$\begin{pmatrix}\alpha' & \beta' \\ -\bar{\beta}' & \bar\alpha'\end{pmatrix} \in SU(2)$$ such that $$\Re \alpha = \Re \alpha',$$ but this is just the intersection of $SU(2) \cong \mathbb{S}^3$ with a hyperplane with coordinate $\Re \alpha$.

Remark Much of the above can be said more cleanly in the language of the quaternions $\mathbb{H}$, which I avoided above because people often encounter $SU(2)$ before they do quaternions: In summary, the group multiplication rule for $SU(2)$ can be identified with multiplication of (unit) quaternions, the splitting $\mathbb{C}^2 \cong \mathbb{R} \oplus \mathbb{R}^3$ is just the decomposition of $\mathbb{H}$ into real and imaginary parts, and these are the irreducible subrepresentations of $\mathbb{H}$ regarded as an $SU(2)$-module under the conjugation action. Then, the conjugacy classes (the orbits under the conjugation action) are just the intersections of the unit sphere $SU(2) \cong \mathbb{S}^3$ with the hyperplanes with given (quaternionic) real part.