Suppose we're looking at an $SU(N)$ symmetry in the adjoint representation. The kinetic part of the Lagrangian yields a term

$$ \Delta \mathcal{L} = -g^2 Tr\left[[T^a, \Phi][T^b, \Phi]\right]A^a_{\mu}A^{b\mu} $$

When $\Phi$ acquires a VEV (call it $\Phi_0$) the broken generators are those whose associated gauge bosons acquire mass, via the above term of the Lagrangian. This amounts to saying that the unbroken generators are those whose commutator with the VEV vanishes:

$$ [T^a, \Phi_0] = 0 $$

Let's say generically the VEV has the form

$$ \Phi_0 = diag(a, a, ..., a, b, b, ... b) $$

where there are $n_a$ of $a$, $n_b$ of b, and $Tr[\Phi_0] = 0$. My understanding is that there are effectively three cases for generators that will commute with the VEV:

1) $$ T^a = \begin {bmatrix} A & 0 \\ 0 & 0 \\ \end {bmatrix} $$

where A is an $n_a \times n_a$ block

2) $$ T^a = \begin {bmatrix} 0 & 0 \\ 0 & B \\ \end {bmatrix} $$

where B is an $n_b \times n_b$ block

3) $T^a$ is a diagonal matrix.

I understand why generators from case 1) would produce generators of SU($n_a$), and why generators from case 2) would produce generators of SU($n_b$). What I can't figure out is why diagonal generators (often proportional to the VEV) are generators of U(1). My understanding was that the only generator of U(1) was the identity. Why is it that any traceless diagonal matrix can produce U(1)?

  • $\begingroup$ ... because they also commute with the VEV.... $\endgroup$
    – user178876
    Commented Feb 18, 2018 at 15:39
  • $\begingroup$ Yes, but why do they generate U(1) and not some other symmetry? How can we show that they are specifically U(1) generators? $\endgroup$ Commented Feb 18, 2018 at 15:41
  • $\begingroup$ ... because their commutation relations with all other surviving generators are trivial... $\endgroup$
    – user178876
    Commented Feb 18, 2018 at 15:42
  • $\begingroup$ My group theory background isn't that strong. Why is that sufficient to show they're U(1) generators? $\endgroup$ Commented Feb 18, 2018 at 15:56
  • $\begingroup$ ... well, if a generator commutes with all other generators, it generates a U(1)... I guess you either want to read a very basic group theory intro, or convince yourself by other means... $\endgroup$
    – user178876
    Commented Feb 18, 2018 at 15:59