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)?
