# Number of Goldstone modes when the symmetry is broken?

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?

I'm not sure what your trouble is, but here is what the language means: The "generic case" is $$v=\operatorname{diag}(v_1,v_2,v_3,v_4, -v_1-v_2-v_3 -v_4),$$ traceless alright.

"Generic" means no two of these 5 parameters are equal, since, if two are, then there are 3 su(2) generators T commuting with v; if three are, then there are 8 su(3) such generators; if four, then 15 su(4) generators; if two and three, then 11 $$su(3)\oplus su(2)$$ ones.

All leftover unequal diagonal entries are only corresponding to U(1)s acting on their respective 1d subspaces.

So the goldston counting works as detailed, and for the generic case we started with, you have 4 surviving u(1)s, hence 24-4=20 goldstons.

• Well, my trouble was quite trivial; I just wanted to confirm whether my understanding was correct, and it turns out it is. Thanks! Dec 23, 2022 at 23:10