Stabilizer subgroup of adjoint action

Question

Given $b \in \mathfrak{su}(n)$, how can I find the stabilizer $\text{stab}(b)$ for the adjoint action of $SU(n)$ on $\mathfrak{su}(n)$ given by $Ad_U(b) = UbU^{\dagger}$ without using coordinates? The $SU(4)$ case is particularly important in my application.

Answer 1

I have no idea what you mean "without using coordinates". The best you can hope to do without a lot of work is to determine the conjugacy class of the stabiliser. For that, all you need to know is the spectrum.

Clearly, if $b \in \mathfrak{su}(n)$, then $ib$ is traceless and hermitian. It can therefore be diagonalised via (special) unitary transformations.

For example, if $b \in \mathfrak{su}(4)$, let $\lambda_1\leq\lambda_2\leq\lambda_3\leq\lambda_4$, with $\sum_i \lambda_i = 0$, be the eigenvalues of $ib$. Up to conjugation we have five possibilities:

1. $\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$, with stabiliser a maximal torus: $S(U(1)\times U(1) \times U(1) \times U(1))$;

2. $\lambda_1 = \lambda_2 < \lambda_3 < \lambda_4$, with stabiliser $S(U(2)\times U(1)\times U(1))$;

3. $\lambda_1 = \lambda_2 < \lambda_3 = \lambda_4$, with stabiliser $S(U(2) \times U(2))$;

4. $\lambda_1 = \lambda_2 = \lambda_3 < \lambda_4$, with stabiliser $S(U(3) \times U(1))$; and

5. $\lambda_1 = \lambda_2 = \lambda_3 = \lambda_4 = 0$, with stabiliser $SU(4)$.