Symmetry breaking, stabilizers, and vacuum orbits

The unbroken subgroup is the stabilizer of a specified order parameter under a specified group action:

$$H_v=\{g\in G:\rho(g)v=v\}.$$

Solving only the infinitesimal equation $d\rho(X)v=0$ finds its Lie algebra and hence the identity component, but can miss disconnected transformations. Those components can change the topology of $G/H_v$ even when the unbroken Lie algebra stays the same.

We work with compact internal groups acting continuously on finite-dimensional real order-parameter spaces. The first example is a real scalar triplet in four-dimensional Lorentzian spacetime with canonical kinetic normalization. The second is a real symmetric traceless tensor. They have closely related continuous stabilizers but different orbits: a sphere for a vector and a projective plane for an unoriented axis.

From an action to an orbit

The orbit is $\mathcal O_v=\{\rho(g)v:g\in G\}$. The map $gH_v\mapsto\rho(g)v$ identifies it with $G/H_v$; the stabilizer is closed, and for compact $G$ this is an embedded homogeneous orbit. Replacing $v$ by $\rho(g_0)v$ conjugates the stabilizer to $g_0H_vg_0^{-1}$. See Zuber, chapter 1, exercise A, equation (1.49), p. 62.

For a matrix representation, first solve the infinitesimal linear system. Then solve the full group equation, retaining elements away from the identity. Every element of $\ker\rho$ automatically lies in every stabilizer. Finally, determine which orbit is selected by the potential or effective action. A stationary vector does not by itself establish a stable minimum, and a group-invariant potential can have several inequivalent minimizing orbits.

A vector order parameter

Take $G=SO(3)$ acting on $\phi\in\mathbb R^3$ by ordinary rotations and choose $v_0=v e_3$, $v>0$. A rotation that fixes $v_0$ fixes the third basis vector. Orthogonality then leaves a rotation in its perpendicular plane:

$$H_{v_0}= \left\{\begin{pmatrix}R(\theta)&0\\0&1\end{pmatrix} :0\leq\theta<2\pi\right\}\cong SO(2), \qquad \mathcal O_{v_0}\cong SO(3)/SO(2)\cong S^2.$$

This stabilizer is connected. A rotation taking $e_3$ to $-e_3$ does not fix the vector, even though it preserves its axis.

To make this orbit the actual classical minimum set, specify

$$\mathcal L=\frac12\partial_\mu\phi\cdot\partial^\mu\phi -\frac\lambda4(\phi\cdot\phi-v^2)^2, \qquad \lambda>0,$$

with metric $(+---)$. The potential is nonnegative and vanishes precisely when $|\phi|=v$, so its full minimum set is that sphere. Write $\phi=(\pi_1,\pi_2,v+h)$. The potential Hessian at $v_0$ is

$$\left.\frac{\partial^2V}{\partial\phi_a\partial\phi_b}\right|_{v_0} =2\lambda v^2\delta_{a3}\delta_{b3}.$$

With the stated kinetic term, the radial fluctuation has $m_h^2=2\lambda v^2$ and the two tangential fluctuations are massless at tree level. Their number equals $\dim SO(3)-\dim SO(2)=2$. This is the relativistic global-symmetry setting; the same counting is not a general rule for nonrelativistic systems or for gauge redundancies. The Hessian argument is the one developed in Zuber, §4.1.1, equations (4.3)–(4.5), pp. 140–142.

The displayed scalar Lagrangian also has the reflection symmetries of $O(3)$. If the full internal group is taken to be $O(3)$, its stabilizer is $O(2)$ and $O(3)/O(2)$ is still $S^2$. Our $SO(3)$ calculation explicitly uses the connected rotation subgroup; it does not claim this potential has only that symmetry.

The same triplet with gauge fields

Gauge the rotation action and use

$$D_\mu\phi=\partial_\mu\phi+g A_\mu\times\phi, \qquad \mathcal L=-\frac14 F^a_{\mu\nu}F^{a\mu\nu} +\frac12 D_\mu\phi\cdot D^\mu\phi-V(\phi).$$

Expanding around the constant representative $v_0$ gives

$$\frac12g^2|A_\mu\times v_0|^2 =\frac12g^2v^2\left(A_\mu^1A^{1\mu}+A_\mu^2A^{2\mu}\right).$$

Thus two vector fields have tree-level mass $gv$ and the third has no mass term. The tangential scalar coordinates supply the longitudinal polarizations of the massive vectors. The invariant scalar length, bundle reduction, and spectrum carry the physical information; the choice of direction $v_0$ is a gauge choice. A gauge-fixed nonzero scalar expectation is not a gauge-invariant order parameter for spontaneous breaking of local redundancy. Compare Zuber, §5.2.2, equations (5.30)–(5.32), p. 167 for the corresponding kinetic-term mechanism in the abelian example.

The same local triplet action can be defined with global gauge group $SU(2)$: use its real adjoint representation, with the axis convention of the spin-cover map. Its stabilizer is the full diagonal torus

$$\widetilde H= \{\operatorname{diag}(e^{i\theta},e^{-i\theta})\}\cong U(1).$$

It includes $-I$, since the center acts trivially on the entire adjoint representation. Therefore the two global choices give

$$SU(2)/U(1)\cong S^2, \qquad SO(3)/SO(2)\cong S^2,$$

but the group and subgroup maps differ. With each circle given period $2\pi$ in its own parameter, the induced boundary maps are

$$\partial_{SU(2)}(n)=n, \qquad \partial_{SO(3)}(n)=2n.$$

The topology lesson derives these signs and normalizations from the spin cover. The circle in SU(2) covers the circle in SO(3) twice. Whether fundamental doublet matter is present distinguishes the permitted global groups even though this triplet cannot. The global-form lesson explains this descent test.

A tensor order parameter retains an axis flip

Now take $G=SO(3)$ acting on real symmetric traceless $3\times3$ matrices by

$$Q\longmapsto RQR^T.$$

This is a five-dimensional representation. Choose the uniaxial candidate

$$Q_0=q\left(e_3e_3^T-\frac13 I_3\right) =\frac q3\operatorname{diag}(-1,-1,2), \qquad q\ne0.$$

The one-dimensional eigenspace along $e_3$ and the two-dimensional perpendicular eigenspace have distinct eigenvalues. A stabilizing rotation must preserve these eigenspaces but may send $e_3$ to either sign. The full answer is

$$H_{Q_0}= \left\{\operatorname{diag}(A,\det A):A\in O(2)\right\} \subset SO(3).$$

The determinant of this $3\times3$ matrix is $(\det A)^2=1$, so both components belong to the ambient rotation group. The element $\operatorname{diag}(1,-1,-1)$ fixes $Q_0$ but sends the vector $v_0$ to $-v_0$. It is exactly a component missed by solving only the Lie-algebra stabilizer. We have $H_{Q_0}\cong O(2)$, with identity component $SO(2)$.

The orbit consists of

$$Q(n)=q\left(nn^T-\frac13I_3\right), \qquad |n|=1, \qquad Q(n)=Q(-n).$$

Conversely, equality of two such matrices identifies their one-dimensional eigenspaces, so their unit vectors differ only by sign. Hence

$$\mathcal O_{Q_0}\cong SO(3)/H_{Q_0}\cong\mathbb{RP}^2.$$

This explicit construction establishes the orbit of the candidate, without assuming a potential selects it. For example, a potential depending only on $\operatorname{tr}Q^2$ fixes a norm in a five-dimensional space and generally leaves a larger minimum set containing several rotation orbits. Additional invariant dependence can change that conclusion.

The double cover $S^2\to\mathbb{RP}^2$ gives $\pi_1(\mathbb{RP}^2)=\mathbb Z_2$ and $\pi_2(\mathbb{RP}^2)=\mathbb Z$; see Hatcher, Example 1.43, p. 74, and Proposition 4.1, p. 342. The component term in the subgroup exact sequence explains the first result directly: $\pi_1(SO(2))\to\pi_1(SO(3))$ is surjective, and the remaining $\pi_0(H_{Q_0})=\mathbb Z_2$ supplies the orbit fundamental group. Thus an axis and a vector have the same continuous stabilizer algebra but different loop sectors. The integer $\pi_2$ statement is based; changing the lift through the nontrivial loop reverses the sphere generator because the deck transformation is antipodal, with degree $-1$ on $S^2$ (Hatcher, §2.2, property (f), p. 134).

Interpret the result with its physical assumptions

For a global order parameter whose full vacuum manifold is the stated orbit, the vector example has no nontrivial loop sector, whereas the director example has a $\mathbb Z_2$ loop sector. Turning these sectors into defects requires the action, boundary conditions, and allowed field configurations. In a gauge theory, the same question also involves the bundle and its reduction. Kibble, §2, p. 1388 and §4, pp. 1391–1393 supplies the orbit and homotopy setup.

The topology workbench covers the two sphere fibrations, not the tensor/director example. Use its results only after choosing the correct inclusion. The space atlas records the curated orbits, and topological defects explains why a homotopy sector is separate from a finite-energy or dynamically stable solution.