# Broken symmetries realized nonlinearly

I'm trying to understand some concepts of spontaneous symmetry breaking, I'll write first the statement that I can't understand and later my questions.

STATEMENT

Consider a group $G$ and a subgroup $H \subset G$. In particular: If $V_a$ are the generators of $H$ and $A_a$ are the remaining generators of G, then we can choose a representation of the group to take the form:

$g(ξ, u) = e^{iξ·A} e^{iu·V}$

where $\xi^a$ are the Goldstone bosons, $e^{iu·V} ∈ H$ and $e^{iξ·A} ∈ G/H$.

For a general $g ∈ G$, we have from closure of G that

$g\, e^{iξ·A} = e^{iξ′·A} e^{iu′·V},$

where $ξ′ = ξ′(ξ, g)$ and $u′ = u'(ξ, g)$ are analytic functions due the Lie group structure.

In this manner, as required, the Goldstone fields linearly realize the symmetries of the preserved subgroup H and nonlinearly realize the remaining broken symmetries.

QUESTIONS

1.- How can I see that $e^{iξ·A} ∈ G/H$?

2.- What is the definition of linearly and nonlinearly realizing symmetries?

3.- How can I see from above that if $g∈ H$ then the relation between $\xi$ and $\xi'$ is linear and otherwise it's nonlinear?

I'm new with group theory so I hope you can explain this extensively but still as formal as possible.

1. The factors $e^{i\xi\cdot A}$ and $e^{iu\cdot V}$ are chosen this way so there is a unique label $(u,\xi)$ for each group element $g=e^{i\xi\cdot A}e^{iu\cdot V}$. To see that $\xi$ is a label for $G/H$, it's enough to check that $g$ and $gh$ are associated with the same value $\xi$ when $h\in H$. This is easy to see: set $h=e^{iu_h\cdot V}$, so that $gh=e^{i\xi\cdot A}e^{iu\cdot V}e^{iu_h\cdot V}=e^{i\xi\cdot A}e^{iu'\cdot V}$. Hence, $gh\mapsto (u',\xi)$.
Some remarks: (i) note that $G/H$ should be generally viewed as a topological quotient space (and not a quotient group), because $H$ is not always a normal subgroup. A simple example is the case $G=SO(3)$, $H=SO(2)\subset SO(3)$. The quotient space $SO(3)/SO(2)$ is homeomorphic to the 2-sphere, and spherical polar angular coordinates can be obtained naturally this way, but the 2-sphere is certainly not a group. (ii) there is some arbitrariness in the choice of representative $\xi(g)$. However, once a representative $\xi$ is chosen for $g$, it also represents $gh$ for all $h\in H$.
1. To see the difference between linearly and nonlinearly realized symmetries, it helps to think of symmetries as homomorphisms of the symmetry group into the space of functions of fields, with multiplication' mapping to composition. For example, let $\vec\phi\in\mathcal{F}$ be a label for fields, and $G$ a symmetry group. Then in general, $G$ can act on $\vec\phi$ by choosing functions $f_g(\vec\phi):\mathcal{F}\rightarrow\mathcal{F}$ in such a way that $f_g(f_{g'}(\phi))=f_{gg'}(\phi)$. A linearly realized symmetry corresponds to the special case where $f_g(\phi)$ is a linear map for every $g\in G$.
2. The relationship between $\xi$ and $\xi'$ when $g\in H$ is not generally linear, it depends on your convention for $\xi(g)$. When $\xi$ is small, however, we can define $\xi$ to be locally flat' coordinates for the tangent space of $G/H$ near some point $\phi_0$ (presumably the VEV of the field theory). In this case, the action of $H$ on $\xi$ is linear as long as $\xi$ is very small. This is easy to see in the example of spherical coordinates: here $\vec\phi_0$ is given by a unit vector $\hat n_0(\theta,\phi)$, and the rotation group $SO(2)$ is the set of rotations about the axis $\hat n_0$. This group acts linearly on the tangent space spanned by the angular tangent vectors $\hat\theta(\hat n_0)$, $\hat\phi(\hat n_0)$. In general, however, the multiplication $(u,\xi)(v,\zeta)=(u',\xi')$ involves solving a nonlinear equation for $u'$ and $\xi'$ in terms of $u$, $v$, $\xi$ and $\zeta$. This multiplication rule is generically nonlinear.
• Thanks for your answer it was really helpful. I'm still confused about something. I've read that the important thing of this is that the symmetries of $H$ are realized linearly and the broken ones nonlinearly. If The relationship between ξ and ξ′ when g∈H is not generally linear, how is this statement true? – marRrR Nov 29 '15 at 16:07
• The truth of the statement depends on what space $G$ and $H\subset G$ act on. If you consider the action of $G$ on the tangent space of $G/H$ near some point $p\in G/H$, then $H$ will act linearly, while a generic $g\in G$ will change $p$ itself. Hence, you can think of the nonlinearity as coming from a sort of parallel transport. – TotallyRhombus Nov 29 '15 at 17:06
• So if I consider $g \, e^{iξ⋅A}=e^{iξ'⋅A}\,e^{iu'⋅V}$ as in my question then if $g∈H$ then $ξ'=ξ'(ξ,g)$ and $u'=u′(g)$ where the relation for $ξ'$ is linear? If this is true, how can I see that? – marRrR Nov 29 '15 at 17:28
• In general $\xi'(\xi,g)$ is not linear in $\xi$, as can be seen from the Baker-Campbell-Hausdorff formula, unless there are special constraints on $[V,A]$. However, the power series expansion for $\xi'(\xi,g)$ starts at first order in $\xi$, so the transformation is approximately linear when $\xi$ is small (and is exactly linear on the tangent space). – TotallyRhombus Nov 29 '15 at 18:09