Normalizer of SU$(2)$ in SU$(6)$

Question

Consider the $\mathfrak{su}(2)$ subalgebra of $\mathfrak{su}(6)$ embedded as $$\mathfrak{su(2)}=\text{Span}\{\mathbb{1}_3 \times \sigma^i\}, \quad i=1,2,3$$ with $\sigma^i$ the Pauli matrices and $\times$ the Kronecker product. What is the normalizer $$N=\left\{ g\in {\rm SU}(6) \; | \; gxg^{-1} \in \mathfrak{su}(2) \quad \forall \; x \in \mathfrak{su}(2) \right\}.$$ My guess is $S(U(3)\times U(2))$ but I'm not sure, and possibly missing something discrete.

Answer 1

This is correct. The complex span of $\mathfrak{su}(2)$ extended by the identity matrix (which commutes with everything) is the full matrix algebra $M_2(\mathbb{C})$, so the centralizer, $C$ say, of the matrix subalgebra $\langle I_3 \rangle \otimes \mathfrak{su}(2)$ of $M_6(\mathbb{C})$ is the centralizer of $\langle I_3 \rangle \otimes M_2(\mathbb{C})$ in $M_6(\mathbb{C})$, namely $M_3(\mathbb{C}) \otimes \langle I_2 \rangle$.

[This follows from the double centralizer theorem in the simple algebra $M_6(\mathbb{C})$ and the subalgebra $\langle I_3 \rangle \otimes M_2(\mathbb{C})$.]

Suppose that $g \in M_6(\mathbb{C})$ is an invertible matrix in the normalizer. Using a block matrix decomposition of $I_3 \otimes X$ for $X \in \mathfrak{su}(2)$ we have $$ g \left( \begin{matrix} X & 0 & 0 \\ 0 & X & 0 \\ 0 & 0 & X \end{matrix} \right) g^{-1} = \left( \begin{matrix} Y & 0 & 0 \\ 0 & Y & 0 \\ 0 & 0 & Y \end{matrix} \right) $$ for some $Y \in \mathfrak{su}(2)$ (depending on $X$). Thus $g$ induces an automorphism $\phi$ of $M_2(\mathbb{C})$ (namely $X \mapsto Y$ in the notation above) and so there exists an invertible matrix $S$ such that $\phi(X) = SXS^{-1}$ for all $X \in M_2(\mathbb{C})$. Hence $ (I_3 \otimes S)^{-1}g$ is in the centralizer $C$. Multiplying on the left by $I_3 \otimes S$ we get

$$ g \in \{ I_3 \otimes S : S \in \mathrm{GL}_2(\mathbb{C}) \} C = \{ I_3 \otimes S : S \in \mathrm{GL}_2(\mathbb{C}) \} \bigl(M_3(\mathbb{C}) \otimes \langle I_2 \rangle \bigr) $$

Therefore $g = T \otimes S$ for some invertible $S \in \mathrm{GL}_2(\mathbb{C})$ and $T \in \mathrm{GL}_3(\mathbb{C})$. We require that $T \otimes S$ is a special unitary matrix. Since $T \otimes S$ is unitary and $(T \otimes S)^\dagger = T^\dagger \otimes S^\dagger$ this implies that $S$ and $T$ are unitary. Hence $g \in S(U(3) \times U(2))$, embedding in $M_6(\mathbb{C})$ via $(T, S) \mapsto T \otimes S$.