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.


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.


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 Answer 1

  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.

  • $\begingroup$ 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? $\endgroup$
    – marRrR
    Nov 29, 2015 at 16:07
  • $\begingroup$ 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. $\endgroup$
    – TLDR
    Nov 29, 2015 at 17:06
  • $\begingroup$ 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? $\endgroup$
    – marRrR
    Nov 29, 2015 at 17:28
  • $\begingroup$ 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). $\endgroup$
    – TLDR
    Nov 29, 2015 at 18:09

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