Spontaneous symmetry breaking of $SU(2)$ in real and complex scalar fields 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?
 A: You've made a bit of a hash of the setup, and hence the questions you assume are begged.
You must first observe your first action is SO(3)-invariant, so, as you correctly note, SO(3) is broken down to SO(2), your unbroken U(1).
However, your second, complex action is actually SO(6)-invariant, if only you wrote it as a theory of 6-vectors, the real and imaginary parts of your complex scalars. The potential breaks that theory's SO(6) to SO(5), which has 10 unbroken generators, not just one. Your  suboptimal spherical basis for the SO(3) subgroup of SO(6) you are considering makes that hard to see. So not only do you have a surviving U(1), you have a surviving SO(5).
A few technical details.

*

*The first basis you write is hermitian imaginary matrices, so their exponentials are only real orthogonal matrices when multiplied by i. You might as well chuck the i s to get to SO(3) and real structure constants.


*The second, spherical, basis you use is strictly equivalent to the first, the similarity transform which achieves this provided in Wikipedia. So you are not achieving anything by using it, beyond making the key-point opaque beyond access. If you used the same basis, you'd see that the SO(3) subgroup they span, a direct sum of two 3×3 SO(3) blocks embedded in 6×6  SO(6) matrices, does not overlap the unbroken SO(5) matrices, and so does not help you to explore its breaking structure, notably the surviving generators: the SO(3) subgroup your wrote is completely broken. You may write the 15 generators of SO(6) in the generalized form of your first (rotation) basis, stick your r at the bottom component, and observe the 10 SO(5) matrices rotating the 0-components to each other.


*You might translate all this  to complex notation, of course, but why bother?


*In any case, the answer to your nominal question is: Yes, the reality or complexity of the representation does very much affect the breaking of the group in question. Recall the full breaking in the Standard Model, with a complex doublet representation;  sharply contrasted to that of the Georgi-Glashow model, with a real triplet representation.  And the custodial SO(4) ~ SU(2)×SU(2)  symmetry of the Higgs potential.
