Here is a toy problem whose aim is to help me understand breaking global symmetries into subgroups and under which occasions that is possible. My question is: does a field being real or complex affect the ease with which global symmetry groups break into subgroups?
Consider a real scalar field $\phi$ in the $\mathbb{3}$ representation of $SU(2)$:
$$ \mathscr{L} = \frac{1}{2}\partial_{\mu}\phi^a\partial^{\mu}\phi_a -\frac{1}{2}m^2\phi^a\phi_a -(\phi^a \phi_a - v^2)^2 $$
Here, the generators are those corresponding to $SO(3)$, so their exponentiation gives real rotation matrices, $$ T^1 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & i \\ 0 & -i & 0 \\ \end{pmatrix}, \ \ \ T^2 = \begin{pmatrix} 0 & 0 & -i \\ 0 & 0 & 0 \\ i & 0 & 0 \\ \end{pmatrix}, \ \ \ T^3 = \begin{pmatrix} 0 & i & 0 \\ -i & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} $$
Under spontaneous symmetry-breaking, the field $\phi^a = \tilde{\phi^a} + s^a$, where $s^a$ is a real vector whose magnitude is $v$. We will have that the global symmetry breaks down to a non-trivial subgroup acting on $\tilde{\phi^a}$ if $s^a$ is invariant under the subgroup.
Note that $(r^a) = \begin{pmatrix} 0 \\ 0 \\ v \end{pmatrix}$ is invariant under the exponentiation of $T^3$; that is, if we have that $s^a = r^a$, then $\tilde{\phi^a}$ is invariant under the $U(1)$ subgroup of $SU(2)$ generated by $T^3$.
Because $SU(2)$ can rotate any real vector with magnitude $v$ onto $r^a$, we can always take $s^a = r^a$, and so we will always have $\tilde{\phi^a}$ invariant under a $U(1)$ subgroup of $SU(2)$ .
Compare this to a complex scalar field $\psi$ in the $\mathbb{3}$ representation:
$$ \mathscr{L} = \partial_{\mu}{\psi^{\dagger a}}\partial^{\mu}\psi_a -m^2{\psi^{\dagger a}}\psi_a -({\psi^{\dagger a}} \psi_a - v^2)^2 $$
Here, I will take the generators to exponentiate to group actions that mix the real and imaginary parts: $$ T^1 = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \\ \end{pmatrix}, \ \ \ T^2 = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \\ \end{pmatrix}, \ \ \ T^3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix} $$
Under spontaneous symmetry-breaking, the field $\psi^a = \tilde{\psi^a} + s^a$, where $s^a$ is a complex vector whose magnitude is $v$. We will have that the global symmetry breaks down to a non-trivial subgroup acting on $\tilde{\psi^a}$ if $s^a$ is invariant under the subgroup.
Note that $(r_\alpha^a) = \begin{pmatrix} 0 \\ ve^{i\alpha} \\ 0 \end{pmatrix}$ is invariant under the exponentiation of $T^3$ for arbitrary $\alpha$; that is, if we have that $s^a = r_\alpha^a$, then $\tilde{\phi^a}$ is invariant under the $U(1)$ subgroup of $SU(2)$ generated by $T^3$.
Because $SU(2)$ cannot rotate every complex vector with magnitude $v$ onto $r_{\alpha}^a$ (I believe for any $\alpha$, though please correct me if I am incorrect), we cannot always take $s^a = r^a$, and so we will only sometimes have $\tilde{\psi^a}$ invariant under a $U(1)$ subgroup of $SU(2)$ .
Is this breakdown correct? That is, are we always guaranteed to have a $U(1)$ subgroup preserved in the former case, while in the latter case we are not guaranteed to have a $U(1)$ subgroup preserved?
