Set of zeroes as coset space

Question

I am currently studying Chapter 6 of Coleman S. - Aspects of Symmetry.

We study a spontaneously broken gauge theory in two spatial dimensions where the Lagrangian reads:

$$ \mathcal{L} = -\frac{1}{4}F_{\mu\nu}^aF^{\mu\nu a} + D_\mu\phi^\dagger\cdot D^\mu\phi - U(\phi) $$

The pattern of symmetry breaking is $G\mapsto H$ and $U$ is $G$-invariant. We assume that the zeros of the potential $U$ is the coset space $G/H$, also $U$ is always greater than or equal to zero and equals zero for the groundstates.

It is argued that in order to find non-singular finite energy solutions of the classical equations of motion where $\phi(\infty,\theta)$ depends non-trivially on $\theta$, we can choose the gauge fields so that $\mathbf{e}_\theta\cdot\mathbf{D}\phi$ goes to zero sufficiently fast, since

$$ \mathbf{e}_\theta\cdot\mathbf{D}\phi = \frac{1}{r}\frac{d\phi}{d\theta}+ieA_0^aT^a\phi. \tag{3.44} $$

What confuses me however, is the remark that follows:

Note that our simplifying assumption - that the set of zeros of $U$ is the coset space $G/H$ - is critical for this argument, for it is only this assumption that allows us to assert that we can cancel an infinitesimal change in $\phi$ (the first term in the covariant derivative) with an infinitesimal group transformation (the second term).

I just don't see how this assumption is necessary. What is the intuition behind the coset space as the set of zeros?

Answer 1

I thought a little about this question and I'm still not a 100% sure how to answer it. However, I think the crucial point about this assumption is that (with the assumption) the angular derivative of $\phi$ at "infinity" is always proportional to the action of a Lie algebra element of $\mathfrak{g}$, i.e. $$\partial_\theta \phi= i \lambda^a T_a \phi$$ is satisfied asympotically at $r\rightarrow\infty$ with $\alpha^a$ depends on $\theta$. This is because $G$ has a natural transitive action on the quotient space $G/H$, i.e. given points $p,q\in G/H$ there is always a $g\in G$ such that $g\cdot p=q$. Thus, we can choose a fixed point in $G/H$, call it $\phi_0$, and write the asymptotic behaviour of $\phi$ as $$\phi(\infty,\theta)= g(\theta)\cdot\phi_0 = \exp\left(i\alpha^a(\theta) T_a\right)\cdot \phi_0.$$ Only if this is the case, the partial derivative can be compensated as a gauge of $A$ and, thus, the covariant derivative vanishes fast enought.

Now, what happens if we do not assume the space of minima to be $G/H$? Let's denote the space of minima by $Z$ and as above take a $\phi_0\in Z$. We will seperate this in two cases:

1. The connected component of $\phi_0$ in $Z$ is $G/H$: In this case nothing changes and the argument is still valid. Note, that given a minimum $\phi_0\in Z$ always gives rise to a whole $G/H$ subspace in $Z$ since $U$ is supposed to be $G$-invariant and thus $G\cdot\phi_0 \cong G/H \subseteq Z$.

2. The other case: If the connected component, call it $Z_{\phi_0}$ of $\phi_0$ is not $G/H$ we have to be more careful. In this case, the field $\phi$ has additional asymptotic degrees of freedom. It can oscillate between different $G/H$ subspaces. To illustrate this, imagine a symmetry breaking pattern $SO(3)\rightarrow SO(2)$, then the quotient is the 2-sphere $SO(3)/SO(2)=S_2$. The space $Z$ could for example be equal to $I\times S_2$, with $I$ some interval. For a finite action/energy we assume $\phi(x)\in Z$ assymptotically. The movement of $\phi$ in the $S_2$ subspaces generated by the $G$-action can be compensated by the gauge field. However, this is not the case if $\phi$ moves in $I$ as $\theta$ varies. Thus, any change of $\phi$ in the $I$-component results in an infinte contribution to the kinetic energy.

Anyway, all the standard symmetry breaking potentials I have seen so far had the property that $Z=G/H$. We could in principle allow for potentials as in case 1 above. But I don't think it will be useful for anything.

As a conclution, I think the assumption is done to ensure that the change of $\phi$ "at infinity" can be compensated in the gauge field $A$. Without the assumption, this is not true in general. However, I think there are still solutions with finite energy, but we propably get trouble when trying to quantize the theory. But this is just speculation from my side.

Although it is not a full answer to your question, I hope these thoughts might help you!