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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?

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  • $\begingroup$ You mean $|\alpha|^2 + |\beta|^2 = 1$, which just says that $\det_{\mathbb{C}} = 1$. $\endgroup$ – Travis Oct 11 '14 at 13:07
  • $\begingroup$ Well, yes, it does. I used it while computing eigenvalues. I'll correct the missing $2$s. $\endgroup$ – Don Oct 11 '14 at 13:11
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    $\begingroup$ Note that the characteristic polynomial of $A \in SU(2)$ has real coefficients, so the eigenvalues are complex conjugates. Moreover, the product of the eigenvalues is $\det_{\mathbb C} A = 1$. What do these together imply about the eigenvalues? $\endgroup$ – Travis Oct 11 '14 at 13:16
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    $\begingroup$ First, note that if two matrices are in the same conjugacy class, they must have the same eigenvalues, so all of the matrices $\text{diag}(e^{i\theta}, e^{-i\theta}) \in SU(2)$, $\theta \in [0, \pi]$ are in distinct conjugacy classes. To prove that this exhausts them, you need something like a converse for this statement, and it is provided by the Spectral Theorem, see en.wikipedia.org/wiki/Normal_matrix . $\endgroup$ – Travis Oct 11 '14 at 14:23
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    $\begingroup$ I realized our comments here essentially contain a proof, which I've written up as an answer for posterity. $\endgroup$ – Travis Oct 11 '14 at 14:33
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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 $$\begin{pmatrix}\alpha & \beta \\ -\bar{\beta} & \alpha\end{pmatrix} \leftrightarrow (\alpha, \beta)$$ 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:

We can decompose any element $(\gamma, \delta)$ of $\mathbb{C}^2$ as $$(\gamma, \delta) \leftrightarrow \Re \gamma + (\Im \gamma + \delta) \in \mathbb{R} \oplus \mathbb{R}^3,$$ and $\Re \gamma$ corresponds to a real diagonal matrix.

Now, $SU(2)$ acts on $\mathbb{C}^2$ by conjugation, but notice that for any real diagonal matrix, say the one corresponding to $(\gamma, 0)$, we have $$\begin{pmatrix}\alpha & \beta \\ -\bar{\beta} & \alpha\end{pmatrix} \begin{pmatrix}\gamma & 0\\ 0 & \gamma\end{pmatrix} \begin{pmatrix}\alpha & \beta \\ -\bar{\beta} & \alpha\end{pmatrix}^{-1} = \begin{pmatrix}\gamma & 0\\ 0 & \gamma\end{pmatrix}.$$ In other words, the conjugation action of $SU(2)$ on $\mathbb{C}^2$ fixes every real diagonal matrix.

On the other hand, we can show that the conjugation action maps any element of $\mathbb{R}^3$ (that is, any element $(\lambda, \mu) \in \mathbb{C}^2$ with $\Re \lambda = 0$) to such an element, so $SU(2)$ acts on $\mathbb{R}^3$ by conjugation. In fact, it's easy to show that the conjugation action of $SU(2)$ on $\mathbb{C}^2$ preserves the quadratic form $(\lambda, \mu) \mapsto |\lambda|^2 + |\mu|^2$, so it preserves its restriction to $\mathbb{R}^3$, which is just the usual norm-square there. This defines a map $SU(2) \to SO(3)$, and in fact we can show it is a double cover; in particular $SU(2)$ acts transitively on $\mathbb{R}^3$.

In short, the conjugacy class of $$\begin{pmatrix}\alpha & \beta \\ -\bar{\beta} & \alpha\end{pmatrix} \in SU(2)$$ consists of exactly the elements $$\begin{pmatrix}\alpha' & \beta' \\ -\bar{\beta}' & \alpha'\end{pmatrix} \in SU(2)$$ 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.

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    $\begingroup$ Thank you. I've read a little about $SU_2$ and $S^3$. Could you tell me how we can identify the conjugacy classes with circles of latitude of $S^3$? $\endgroup$ – Don Oct 11 '14 at 14:37
  • $\begingroup$ You're welcome, and sure, the details are a little involved, so I simply appended it to my answer. $\endgroup$ – Travis Oct 11 '14 at 15:31

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