I am studying lecture notes on Groups and Representations by Prof. Andre Lukas. I am unable to completely understand notesnd the examples worked out in the symmetry-breaking application section (pages 98-99 of the notes).
Say we want to find the symmetry breaking of $SU(5)$ to a subgroup $H_1 = SU(4)$ and $H_2 = SU(3)\times SU2)$ in the 24 (adjoint) representation of $SU(5)$. The branching is as follows:
$$ 24 \mapsto[15\oplus4\oplus\bar{4}\oplus1]_{H_1} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;24 \mapsto [(8,1)\oplus(1,3)\oplus(3,2)\oplus(\bar{3},2)\oplus(1,1)]_{H_2} . $$ The text then says
To understand the orbits we can diagonalise to $v = \langle \phi \rangle = diag(v_1,...v_5)$, where $v_i \in R$ and $\sum v_i = 0$. The values of the four independent $v_i$ (modulo overall scaling) classify the orbits. For generic choices neither of $H_1$ and $H_2$ are unbroken and only $U(1)^4$ survives (with 24-4=20 resulting in Goldstone modes). For the non-generic choice $v = diag(v, v, v, v, -4v)$, $H_1 \times U(1)$ is unbroken (with 24-15-1 = 8 Goldstone modes) and for $v = diag(2v, 2v, 2v,-3v, -3v)$ $H_2 \times U(1)$ is unbroken (with 24-8-3-1 = 12 Goldstone modes).
I think the condition that is being used to come up with the above non-generic choices of $v=\langle\phi\rangle$ is that $$ [v, T] = 0 $$ where $T\in\mathscr{L}(H)$. This is because of equation (5.114) in the notes and the fact that the action of the Lie algebra on the adjoint representation is the commutation. Is my reasoning valid? If not, how are these non-generic choices being found?
