Vacuum manifold of Glashow-Weinberg-Salam & lack of vortex solutions

Question

Update: may have figured it out (below)

I've been reading Preskill's notes on vortices, and he's included a couple of exercises at the end of section 1.3.

In the first, you must look at the breaking of $G = SU(2)_1\times SU(2)_2 \times U(1)_Y$ down to $H = U(1)_Q$ where $U(1)_Q$ is generated by

$Q = p T_3^{(1)} + q T_3^{(2)} + r Y$,

($p, q$ and $r$ being integers with no common factor) and show that $\pi_1 (G/H) \cong \mathbb{Z}_r$. I thought I had shown this, but the next exercise made me think I'm definitely not understanding the idea.

(I argued that, for a given $r$, the curve in $G$ generated by $Q$ will have winding number $r$. A curve in $G$ is categorised homotopically by its winding number $n$, and if composed with the $Q$ curve, the resultant curve will have winding number $n+r$ (and be homotopically equivalent to any curve with that winding number). Since modding out $H$ identifies any points lying along the $Q$ curve, this means that any two curves whose difference in winding number is $r$ differ from each other only by a curve which is homotopically equivalent to the $Q$ curve, which is now just a single point. Thus the group of homotopically distinct curves is $\mathbb{Z}_r$.)

In the next question however, he asks why the standard $SU(2) \times U(1)$ GWS model does not admit stable vortex solutions. From what I read this seems to be $SU(2) \times U(1)_Y \rightarrow U(1)_{EM}$, with the unbroken generator

$Q = T_3 + \frac{1}{2} Y$.

Reading online I do see a lot of people saying the vacuum manifold of the GWS is $S^3$, which is of course simply-connected. But I don't understand why the vacuum manifold is $S^3$. What is the deciding difference with question 1? Is my previous argument nonsense?

I'm also in the market for a systematic way to calculate the vacuum manifold of a quotient group (though as you may be able to tell my mathematical level is not high enough to tackle a pure-maths approach).

Answer 1

Possible solution to my own question

Here's what I think is going on:

Generators can be normalised however you want. The key is the relative normalisation between them. In the first exercise, because there were no common factors, it was necessary to exponentiate by at least $2\pi$ in order to get back to the same spot (i.e. a closed loop in $G$). This means that any closed loop would have to have as winding number a multiple of $r$, and the quotient space's fundamental group is $\mathbb{Z}_r$ as argued above.

In the second question, a full loop would have to go around the $SU(2)$ part twice in order to go around the $U(1)$ once and form a closed loop. So the quotient loops have winding number $1$ and the fundamental group of the quotient space is $\mathbb{Z}_1 = 1$, i.e. trivial!