Vacuum Manifold of an SU(2) Theory
I am reading <PERSON>'s book "Aspects of Symmetry", specifically chapter 6 "Classical Lumps and their Quantum Descendants". He gives an example for the topological solutions for an $SU(2)$ theory, I copy the complete example below. He considers a theory of $SU(2)$ gauge fields and scalar fields.
"The [gauge] group is $SU(2)$, and the scalar fields transform like a complex isospinor, $K$. The zeros of $U$ [the potential] are all fileds such that $\bar{K}K$ is some positive number, $a^2$. (This is the Weinberg-Salam model, without leptons and without the U(1) gauge field.)
If we write the two-component complex $K$ field in terms of four real fields, we see that $G/H$ [the vacuum manifold] is a hypersphere, $S^3$. All mappings from $S^2$ to $S^3$ are homotopic to the trivial mapping ... Thus there are no non-trivial topological conservation laws."
I think I undertsand this, but there are some nagging doubts in my mind. The gauge group is $G=SU(2)$. 364
Now we want $H$, the group that leaves vacua invariant. I think that this is just the identity element. My reasoning is as follows: $SU(2)$ is the group of rotations of two component complex vectors. There is no element except the identity that can leave such a vector invariant, since the only such rotation there could be would be a rotation about the axis of the vector, which would require rotating into a higher dimesion than $SU(2)$ can access (I imagine the usual two-dimensional vector arrows in a plane, we cannot rotate around the axis of such a vector without including three dimesnional rotations, hence for $SO(2)$ we have a similar situation). This reasoning seems shaky to me at best. First, the vectors we act on with $SU(2)$ are complex, not real, so my imagining them as simple arrows in a plane seems simplisitc. Second, $SU(2)$ is a double cover of $SO(3)$ which certainly has elements apart from the identity that leave vectors invariant, they are the elements that rotate around the axis of the vector. If $H$ is just the identity, then we have the required $G/H=SU(2)/{I}\cong SU(2)\cong S^3$ as required.
Furthermore, in the next example he states that if the gauge group $G$ is any compact connected Lie group, and $H$ is any discrete subgroup, all mappings of a sphere into $G/H$ are homotopic to the trivial mapping (if anyone can direct me to a proof of this as well that would be fantastic) and states that the previous example (the one I copied above) is a specific case of this. The point I take away from this is that $H$ must be a discrete subgroup, which of course ${I}$ by itself is.
Thanks for any advice you can give! 364