[proofplan] Parts (1)–(3) are direct matrix computations: writing $t = \mathrm{diag}(e^{i\theta}, e^{-i\theta})$ and computing $sts^{-1}$ inverts the diagonal entries; computing $s^2$ gives $-I$; analysing $gtg^{-1} \in T$ for general $g$ shows $g$ either preserves the diagonal basis or swaps it. Part (4) uses spectral theory of unitary matrices: every $X \in \operatorname{SU}(2)$ has orthonormal eigenbasis, giving a unitary diagonaliser $Q$, which can be rescaled to lie in $\operatorname{SU}(2)$. Part (5) follows from (1) and (4): the eigenvalues are $\lambda$ and $\lambda^{-1}$, and the only freedom is which goes on the diagonal first, swapped by $s$. Part (6) translates conjugation invariants to the trace: $g \in G$ has characteristic polynomial $x^2 - (\operatorname{tr} g)x + 1$, so trace classifies conjugacy classes; the parameterisation $\lambda = e^{i\theta}$ and $\frac{1}{2}\operatorname{tr}\, g = \cos\theta$ exhibits the bijection with $[-1, 1]$. [/proofplan]

[step:Set up notation: parameterise the maximal torus $T$ and verify the relations $sts^{-1} = t^{-1}$ and $s^2 = -I$] Recall $G = \operatorname{SU}(2) = \{A \in \operatorname{Mat}_2(\mathbb{C}) : A^* A = I, \det A = 1\}$. The maximal torus is \begin{align*} T = \left\{ t_\theta := \begin{pmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{pmatrix} : \theta \in \mathbb{R} \right\} \subseteq G, \end{align*} isomorphic to $S^1$ via $t_\theta \mapsto e^{i\theta}$. Set \begin{align*} s := \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \in G. \end{align*} *Verification of (1): $sts^{-1} = t^{-1}$.* Since $s^* s = I$ and $\det s = 1$, $s \in G$. Compute $s^{-1}$: a direct check gives $s^{-1} = s^* = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. For $t = t_\theta \in T$, \begin{align*} s t_\theta s^{-1} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} e^{-i\theta} & 0 \\ 0 & e^{i\theta} \end{pmatrix} = t_{-\theta} = t_\theta^{-1}. \end{align*} *Verification of (2): $s^2 = -I$.* Direct computation: \begin{align*} s^2 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I. \end{align*} Since $-I$ commutes with every matrix, $-I \in Z(G)$. [/step]

[step:Determine the normaliser $N_G(T)$] We show $N_G(T) = T \cup sT$. *Inclusion $\supseteq$.* The torus $T$ is abelian, so $T \subseteq N_G(T)$. For $sT$: any element $st_0 \in sT$ acts on $T$ by conjugation as $(st_0) t (st_0)^{-1} = s (t_0 t t_0^{-1}) s^{-1} = s t s^{-1} = t^{-1} \in T$ (using abelianness of $T$ and Step 1), so $sT \subseteq N_G(T)$. *Inclusion $\subseteq$.* Let $g = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in N_G(T)$. The condition $g T g^{-1} = T$ means: for every $\theta \in \mathbb{R}$, there exists $\theta'(\theta) \in \mathbb{R}$ with $g t_\theta g^{-1} = t_{\theta'}$. Choose $\theta$ such that $e^{i\theta}$ and $e^{-i\theta}$ are distinct (any $\theta \notin \pi\mathbb{Z}$ works; fix $\theta = \pi/2$ for concreteness, so $t_{\pi/2} = \mathrm{diag}(i, -i)$). The eigenvalues of $g t_\theta g^{-1}$ equal those of $t_\theta$, namely $e^{i\theta}, e^{-i\theta}$. The eigenvectors of $t_\theta$ are $\mathbf{e}_1 = (1, 0)^\top$ (eigenvalue $e^{i\theta}$) and $\mathbf{e}_2 = (0, 1)^\top$ (eigenvalue $e^{-i\theta}$). Conjugation maps eigenspaces to eigenspaces, so $g$ permutes the lines $\mathbb{C}\mathbf{e}_1$ and $\mathbb{C}\mathbf{e}_2$. Two cases. *Case A: $g$ preserves both lines.* Then $g$ is diagonal, $g = \mathrm{diag}(a, d)$. The conditions $g \in G$ — namely $|a| = |d| = 1$ and $ad = 1$ — give $d = \bar{a} = a^{-1}$, so $g \in T$. *Case B: $g$ swaps the lines.* Then $g \mathbf{e}_1 \in \mathbb{C}\mathbf{e}_2$ and $g \mathbf{e}_2 \in \mathbb{C}\mathbf{e}_1$, i.e., $g = \begin{pmatrix} 0 & b \\ c & 0 \end{pmatrix}$ for some $b, c \in \mathbb{C}$. The conditions $|c|^2 = 1 = |b|^2$ (from $g^* g = I$) and $-bc = 1$ (from $\det g = 1$) give $c = -b^{-1} = -\bar{b}$. Compute $s^{-1} g$: \begin{align*} s^{-1} g = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & b \\ c & 0 \end{pmatrix} = \begin{pmatrix} -c & 0 \\ 0 & b \end{pmatrix}. \end{align*} Using $c = -\bar{b}$, this is $\mathrm{diag}(\bar{b}, b)$. Setting $b = e^{i\psi}$ (since $|b| = 1$), $s^{-1} g = \mathrm{diag}(e^{-i\psi}, e^{i\psi}) = t_{-\psi} \in T$. Hence $g = s \cdot t_{-\psi} \in sT$. Combining both cases, $N_G(T) \subseteq T \cup sT$. This proves $N_G(T) = T \cup sT$, establishing (3). [/step]

[step:Show every conjugacy class meets $T$ via spectral diagonalisation]We prove (4): for every $X \in G$, there exists $g \in G$ with $g X g^{-1} \in T$. *Spectral structure of $X \in G$.* The matrix $X \in \operatorname{SU}(2)$ is unitary ($X^* X = I$). The spectral theorem for unitary matrices on $\mathbb{C}^2$ provides an orthonormal basis $\mathbf{f}_1, \mathbf{f}_2$ of $\mathbb{C}^2$ consisting of eigenvectors of $X$, with eigenvalues $\lambda_1, \lambda_2 \in S^1$ (eigenvalues of unitary matrices have modulus $1$). Since $\det X = \lambda_1 \lambda_2 = 1$, we have $\lambda_2 = \lambda_1^{-1}$. Write $\lambda := \lambda_1$, so the eigenvalues are $\lambda, \lambda^{-1}$. *The unitary diagonaliser.* Form the matrix $Q := (\mathbf{f}_1 \mid \mathbf{f}_2) \in \operatorname{Mat}_2(\mathbb{C})$ with columns $\mathbf{f}_1, \mathbf{f}_2$. Orthonormality of $\{\mathbf{f}_1, \mathbf{f}_2\}$ is equivalent to $Q^* Q = I$, i.e., $Q \in \operatorname{U}(2)$. By construction, $X \mathbf{f}_j = \lambda_j \mathbf{f}_j$, which translates to $Q^{-1} X Q = \mathrm{diag}(\lambda, \lambda^{-1})$. *Adjusting to $\operatorname{SU}(2)$.* The matrix $Q \in \operatorname{U}(2)$ has $|\det Q| = 1$, so $\det Q \in S^1$. Choose $\varepsilon \in \mathbb{C}$ with $\varepsilon^2 = \det Q$ (any square root in $S^1$ works; explicitly, if $\det Q = e^{i\beta}$, take $\varepsilon = e^{i\beta/2}$). Define $Q' := \varepsilon^{-1} Q \in \operatorname{Mat}_2(\mathbb{C})$. Then \begin{align*} \det Q' = \varepsilon^{-2} \det Q = (\det Q)^{-1} \det Q = 1, \end{align*} and $(Q')^* Q' = |\varepsilon|^{-2} Q^* Q = 1 \cdot I = I$ (since $|\varepsilon| = 1$). Hence $Q' \in \operatorname{SU}(2) = G$. Moreover $Q'^{-1} X Q' = (\varepsilon^{-1} Q)^{-1} X (\varepsilon^{-1} Q) = \varepsilon Q^{-1} X Q \varepsilon^{-1} = Q^{-1} X Q = \mathrm{diag}(\lambda, \lambda^{-1}) \in T$ (the scalars $\varepsilon, \varepsilon^{-1}$ commute with everything). Setting $g := (Q')^{-1} \in G$ gives $g X g^{-1} = \mathrm{diag}(\lambda, \lambda^{-1}) \in T$, proving (4).[/step]

[guided]The argument has two distinct ideas. First, spectral theory: every unitary matrix on $\mathbb{C}^2$ is diagonalisable with an *orthonormal* eigenbasis. Why orthonormal? Because for a unitary $X$, eigenvectors with distinct eigenvalues are automatically orthogonal: if $X \mathbf{v} = \lambda \mathbf{v}$ and $X \mathbf{w} = \mu \mathbf{w}$ with $\lambda \neq \mu$, then $\langle \mathbf{v}, \mathbf{w} \rangle = \langle X^* X \mathbf{v}, \mathbf{w} \rangle = \langle X \mathbf{v}, X \mathbf{w} \rangle = \lambda \bar{\mu} \langle \mathbf{v}, \mathbf{w} \rangle$, so $(1 - \lambda \bar{\mu}) \langle \mathbf{v}, \mathbf{w} \rangle = 0$. If $\lambda \neq \mu$ and $|\lambda| = |\mu| = 1$, then $\lambda \bar{\mu} \neq 1$ (else $\lambda = 1/\bar{\mu} = \mu$), forcing $\langle \mathbf{v}, \mathbf{w} \rangle = 0$. For repeated eigenvalues, Gram-Schmidt within the eigenspace produces an orthonormal basis. Second, the determinant adjustment. The diagonaliser $Q$ found above lies in $\operatorname{U}(2)$ with $\det Q \in S^1$ but possibly $\det Q \neq 1$. We rescale by an arbitrary scalar $\varepsilon^{-1}$ with $\varepsilon^2 = \det Q$. The rescaling does not affect conjugation, since scalar matrices commute with everything: $(\varepsilon^{-1} Q)^{-1} X (\varepsilon^{-1} Q) = Q^{-1} X Q$. And the determinant becomes $\det(\varepsilon^{-1} Q) = \varepsilon^{-2} \det Q = 1$. The unitarity is preserved because $|\varepsilon| = 1$. This rescaling trick is a $\mathbb{Z}/2$-cover phenomenon: $\operatorname{SU}(2)$ is the connected double cover of $\operatorname{SO}(3)$, and the relationship $\operatorname{U}(2) = \operatorname{SU}(2) \cdot Z$ where $Z = \{\lambda I : |\lambda| = 1\} \cong S^1$ is the centre of $\operatorname{U}(2)$ shows that we can always normalise within $\operatorname{SU}(2)$.[/guided]

[step:Determine $\mathcal{C} \cap T$ for non-central conjugacy classes] We prove (5): if $g \in G \setminus \{\pm I\}$ has conjugacy class $\mathcal{C}$, then $\mathcal{C} \cap T = \{t, t^{-1}\}$ for some $t \neq \pm I$, and these two elements are distinct. *Existence of $t \in \mathcal{C} \cap T$.* By Step 3 (part (4)), $\mathcal{C} \cap T$ is non-empty: choose $t \in \mathcal{C} \cap T$. Write $t = \mathrm{diag}(\lambda, \lambda^{-1})$ with $\lambda \in S^1$. Since $g \notin \{\pm I\}$, $t \notin \{\pm I\}$ (conjugates of $\pm I$ are $\pm I$, since $\pm I$ are central), so $\lambda \neq \pm 1$, i.e., $\lambda \neq \lambda^{-1}$. *$t$ and $t^{-1}$ are conjugate in $G$.* By Step 1 (part (1)), $sts^{-1} = t^{-1}$ with $s \in G$, so $t^{-1} \in \mathcal{C}$. Thus $\{t, t^{-1}\} \subseteq \mathcal{C} \cap T$. Distinctness: $t = t^{-1}$ would force $\lambda = \lambda^{-1}$, so $\lambda = \pm 1$, hence $t \in \{\pm I\}$, contradiction. *Reverse inclusion $\mathcal{C} \cap T \subseteq \{t, t^{-1}\}$.* Let $t' \in \mathcal{C} \cap T$. Then $t' = h g h^{-1}$ for some $h \in G$. The eigenvalues of $t'$ equal those of $g$, which equal those of $t$, namely $\{\lambda, \lambda^{-1}\}$. Writing $t' = \mathrm{diag}(\mu, \mu^{-1})$ with $\mu \in S^1$, the multi-set $\{\mu, \mu^{-1}\}$ equals $\{\lambda, \lambda^{-1}\}$. Hence either $\mu = \lambda$ (giving $t' = t$) or $\mu = \lambda^{-1}$ (giving $t' = t^{-1}$). In both cases $t' \in \{t, t^{-1}\}$. Combining the inclusions, $\mathcal{C} \cap T = \{t, t^{-1}\}$ with $|\mathcal{C} \cap T| = 2$. [/step]

[step:Establish the bijection from conjugacy classes to $[-1, 1]$ via $\frac{1}{2}\operatorname{tr}$] We prove (6): the map $A \mapsto \frac{1}{2}\operatorname{tr}\, A$ induces a bijection from the set of conjugacy classes of $G$ to $[-1, 1]$. *Well-definedness.* Trace is a class function: $\operatorname{tr}(gAg^{-1}) = \operatorname{tr}(A)$ by cyclicity of the trace. So the map factors through conjugacy classes. *Image lies in $[-1, 1]$.* Every $A \in G$ has eigenvalues $\lambda, \lambda^{-1}$ with $\lambda \in S^1$ (Step 3). Writing $\lambda = e^{i\theta}$ for some $\theta \in \mathbb{R}$, \begin{align*} \frac{1}{2}\operatorname{tr}\, A = \frac{1}{2}(\lambda + \lambda^{-1}) = \frac{1}{2}(e^{i\theta} + e^{-i\theta}) = \cos\theta \in [-1, 1]. \end{align*} *Surjectivity.* For $r \in [-1, 1]$, choose $\theta \in \mathbb{R}$ with $\cos\theta = r$ (such $\theta$ exists: e.g., $\theta = \arccos r \in [0, \pi]$). Then $t_\theta = \mathrm{diag}(e^{i\theta}, e^{-i\theta}) \in T \subseteq G$ has $\frac{1}{2}\operatorname{tr}\, t_\theta = \cos\theta = r$. *Injectivity.* Suppose $A, B \in G$ have $\operatorname{tr}\, A = \operatorname{tr}\, B$. We show $A$ and $B$ are conjugate. [claim:Two elements $A, B \in G$ are conjugate iff they have the same characteristic polynomial] [proof] *($\Rightarrow$)* Conjugate matrices have equal characteristic polynomials (a standard linear-algebra fact: $\det(xI - hAh^{-1}) = \det(h(xI - A)h^{-1}) = \det(xI - A)$). *($\Leftarrow$)* Suppose $A$ and $B$ have the same characteristic polynomial. By Step 3, $A$ is conjugate to some $t_A = \mathrm{diag}(\lambda, \lambda^{-1}) \in T$ and $B$ is conjugate to some $t_B = \mathrm{diag}(\mu, \mu^{-1}) \in T$. The eigenvalues of $A$ are the roots of its characteristic polynomial, equal to those of $B$; so $\{\lambda, \lambda^{-1}\} = \{\mu, \mu^{-1}\}$ as multi-sets. Either $\mu = \lambda$ (so $t_A = t_B$, hence $A$ and $B$ are conjugate to a common element, hence to each other) or $\mu = \lambda^{-1}$ (so $t_B = t_A^{-1}$, but by Step 1 (part 1), $t_A$ and $t_A^{-1}$ are conjugate via $s$, so $A$ and $B$ are again conjugate). [/proof] [/claim] The characteristic polynomial of $A \in G$ is \begin{align*} \det(xI - A) = x^2 - (\operatorname{tr}\, A) x + \det A = x^2 - (\operatorname{tr}\, A) x + 1. \end{align*} So the characteristic polynomial is determined by $\operatorname{tr}\, A$. If $\operatorname{tr}\, A = \operatorname{tr}\, B$, the characteristic polynomials agree, so by the claim, $A$ and $B$ are conjugate. This completes injectivity. *Conclusion.* The map $A \mapsto \frac{1}{2}\operatorname{tr}\, A$ descends to a well-defined map from conjugacy classes to $[-1, 1]$ that is surjective (by the parameterisation $\lambda = e^{i\theta}$) and injective (by the claim). It is therefore a bijection.[/step]

[guided]The trace classifies conjugacy classes because, for $2 \times 2$ matrices of determinant $1$, the characteristic polynomial $x^2 - (\operatorname{tr}\, A)x + 1$ is determined by the trace. Knowing the characteristic polynomial determines the multi-set of eigenvalues, and over $\mathbb{C}$ a $2 \times 2$ matrix with two distinct eigenvalues is determined up to conjugation by its eigenvalues. The remaining edge cases — repeated eigenvalues at $\pm 1$, where the $2 \times 2$ matrix could conceivably be the diagonal $\pm I$ or a non-diagonal Jordan block — collapse because $\operatorname{SU}(2)$ matrices with eigenvalue $\lambda$ of multiplicity $2$ must be normal (unitary matrices are normal) and hence diagonal in some unitary basis, so the only possibilities are $\pm I$ themselves. This is why the bijection extends cleanly to the boundary $r = \pm 1$. The image is $[-1, 1]$ because the eigenvalues $\lambda, \lambda^{-1}$ live on the unit circle, so $\lambda + \lambda^{-1} = 2 \cos \theta$ for $\theta = \arg \lambda$, ranging over $[-2, 2]$; halving gives $[-1, 1]$. Geometrically, the map $A \mapsto \frac{1}{2}\operatorname{tr}\, A$ identifies the space of conjugacy classes of $\operatorname{SU}(2)$ with the closed interval $[-1, 1]$. The interior $(-1, 1)$ corresponds to the *generic* (non-central) classes, each consisting of a $2$-sphere's worth of matrices (in fact $\operatorname{SU}(2) / T \cong S^2$). The endpoints correspond to the central elements: $\frac{1}{2}\operatorname{tr}\, A = 1$ to $\{I\}$ and $\frac{1}{2}\operatorname{tr}\, A = -1$ to $\{-I\}$.[/guided]