Groups, cosets, and classifying spaces

The topology of a symmetry group $G$, an orbit $G/H$, and a classifying space $BG$ answers three different questions. A loop in $G$ probes the global group. A sphere in a vacuum orbit can label an order-parameter sector. A map from spacetime into $BG$ specifies the topology of a principal $G$-bundle. The spaces are related by fibrations, but their homotopy groups are not interchangeable.

The example to keep in mind is

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

Both orbits have $\pi_2=\mathbb Z$, while $\pi_1(SU(2))=0$ and $\pi_1(SO(3))=\mathbb Z_2$. The boundary homomorphism to the circle subgroup is multiplication by $1$ in the first fibration and by $2$ in the second, with compatible generators. This difference matters when translating an orbit winding into a subgroup bundle.

Prerequisites are loops, homotopy, and the covering-group distinction. All homotopy groups below are based at the identity or its coset; $\pi_0$ denotes path components. The groups are finite-dimensional Lie groups, and subgroups carry their embedded topology.

Choose the space first

Space	Construction	Typical question
$G$	The group itself as a manifold	Does a loop of group transformations contract?
$G/H$	Left cosets $gH$ for a specified closed subgroup $H\subseteq G$	What topology does the orbit of an order parameter have?
$BG$	Base of a universal principal $G$-bundle $EG\to BG$	Which principal $G$-bundles exist over a given base?

If $H$ is the stabilizer of a vector $v$ in a continuous finite-dimensional representation, the map $gH\mapsto gv$ identifies the homogeneous space with its orbit in the usual compact-group setting. A vacuum manifold equals this orbit only when all relevant minima form that orbit. Several disconnected orbits or extra flat directions require the actual vacuum space. Also, $G/H$ need not be a group: multiplication of cosets is well-defined as a group operation only when $H$ is normal.

For instance, $SO(3)/SO(2)$ is the sphere of directions obtained by rotating a fixed unit vector. It has no noncontractible loops, even though the rotation group itself does. Replacing $\pi_1(G/H)$ by $\pi_1(G)$ would already give the wrong answer here.

The subgroup fibration and its maps

For a Lie group $G$ and a closed subgroup inclusion $i:H\hookrightarrow G$, the projection $p(g)=gH$ is a locally trivial principal $H$-bundle:

$$H\xrightarrow{i}G\xrightarrow{p}G/H.$$

The closed-subgroup hypothesis is essential to this general statement; a dense immersed subgroup is not a substitute. See Mitchell, §3, p. 5.

For connected $G$, its low-degree homotopy sequence is

$$\begin{aligned} \pi_2(G)&\xrightarrow{p_*}\pi_2(G/H) \xrightarrow{\partial}\pi_1(H) \xrightarrow{i_*}\pi_1(G)\\ &\xrightarrow{p_*}\pi_1(G/H) \xrightarrow{\partial}\pi_0(H)\longrightarrow 0. \end{aligned}$$

Exactness says that the image at a term is the kernel of the next map. At component terms, interpret this as exactness of pointed sets; do not assume that every entry is an abelian group. This is the fibration sequence of Hatcher, Theorem 4.41, p. 376.

The boundary map has a concrete meaning. Pull the bundle back along a based sphere $f:S^2\to G/H$. Choose lifts over its two hemispheres. Along the equator, the lifts differ by a transition function $S^1\to H$. Its homotopy class is $\partial[f]$, with sign set by the equator and transition-function conventions. Thus $\partial$ turns an orbit sphere into an $H$-bundle clutching loop.

If $\pi_2(G)=0$, exactness identifies

$$\pi_2(G/H)\xrightarrow[\ \partial\ ]{\cong} \ker\!\left[\pi_1(H)\xrightarrow{i_*}\pi_1(G)\right].$$

If in addition $H$ is connected, then

$$\pi_1(G/H)\cong \pi_1(G)/\operatorname{im}i_*.$$

These conclusions require the stated hypotheses. In the examples below $\pi_2(G)=0$ follows directly from $SU(2)\cong S^3$ and invariance of higher homotopy groups under coverings. If $H$ is disconnected, retain $\pi_0(H)$: for $G=SU(2)$ and $H=\{\pm I\}$ it supplies the $\mathbb Z_2$ fundamental group of $G/H=SO(3)$.

Two sphere orbits, two circle windings

Fix the two inclusions explicitly:

$$i_{\mathrm{SU}}(e^{i\theta})= \begin{pmatrix}e^{i\theta}&0\\0&e^{-i\theta}\end{pmatrix}, \qquad 0\leq\theta\leq2\pi,$$ $$i_{\mathrm{SO}}(e^{i\varphi})= \begin{pmatrix} \cos\varphi&-\sin\varphi&0\\ \sin\varphi&\cos\varphi&0\\ 0&0&1 \end{pmatrix}, \qquad 0\leq\varphi\leq2\pi.$$

Each parameter runs once around its own circle subgroup. Choose the spin covering and axis conventions so that the first torus maps to the second by $\varphi=2\theta$. The induced map between the circle fibers is then $z\mapsto z^2$, while the induced map between their sphere cosets is the identity after identifying the orbits. These maps are recorded in the atlas: the circle in SU(2), the circle in SO(3), and the spin covering.

For $SU(2)$, the fibration is the Hopf bundle. Both $\pi_2(S^3)$ and $\pi_1(S^3)$ vanish, so

$$0\longrightarrow\mathbb Z \xrightarrow{\partial_{\mathrm{SU}}}\mathbb Z \longrightarrow0$$

forces the boundary to be an isomorphism. Orient the sphere generator so that $\partial_{\mathrm{SU}}(1)=1$. This agrees with the Hopf calculation in Hatcher, p. 375 and Example 4.45, pp. 377–378.

For $SO(3)$, a $2\pi$ rotation lifts to a path from $I$ to $-I$ in $SU(2)$; a $4\pi$ rotation lifts to a closed loop. Hence the subgroup inclusion induces reduction modulo $2$:

$$0\longrightarrow\mathbb Z \xrightarrow{\partial_{\mathrm{SO}}}\mathbb Z \xrightarrow{\bmod 2}\mathbb Z_2 \longrightarrow0.$$

Exactness gives $\operatorname{im}\partial_{\mathrm{SO}}=2\mathbb Z$. The sign and the factor follow together from naturality of the boundary map under the covering of fibrations:

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

Reversing the chosen sphere generator changes both signs, not the relative factor. A unit-degree sphere therefore has subgroup winding $1$ for the first circle and $2$ for the second. It does not follow that a physically normalized magnetic charge doubles: the electric charge convention and Lie-algebra generator must also be compared.

Compare the SU(2) sphere orbit and SO(3) sphere orbit, then reproduce the exact sequences in the topology workbench.

Classifying spaces and bundles over a sphere

Choose a universal principal bundle $G\to EG\to BG$, with $EG$ contractible. For a CW complex $X$, pullback gives a bijection

$$[X,BG]\ \longleftrightarrow\ \{\text{principal }G\text{-bundles over }X\}/\cong.$$

Here $[X,BG]$ means unbased homotopy classes, and isomorphisms cover the identity of $X$. This classifies the bundle, without specifying a connection or a solution of field equations. See Mitchell, Theorems 7.4 and 7.6, p. 15.

Apply the homotopy sequence to the contractible total space. It gives

$$\pi_n(BG)\cong\pi_{n-1}(G)\quad(n\geq2), \qquad \pi_1(BG)\cong\pi_0(G).$$

In particular, $BG$ is simply connected for connected $G$. Then bundles over the oriented sphere $S^2$ are classified by $\pi_2(BG)\cong\pi_1(G)$. Equivalently, trivialize the bundle on the two hemispheres and use a loop in $G$ as the equatorial transition function. The shift is Mitchell, Corollary 11.2, p. 23.

Structure group	$\pi_1(G)$	Principal $G$-bundles on $S^2$
$SU(2)$	$0$	One isomorphism class
$SO(3)$	$\mathbb Z_2$	Two isomorphism classes
$U(1)$	$\mathbb Z$	Integer clutching winding

The first two entries differ even though the two sphere orbits above agree. Neither entry is a calculation of $\pi_2(G/H)$.

What a defect calculation does and does not say

Suppose an order parameter approaches a chosen vacuum space $M$ on a small linking sphere or at spatial infinity. In three spatial dimensions, component choices, loops, and two-spheres in $M$ are the usual topological inputs for domain walls, strings, and point defects, respectively. This is the setting of Kibble, §4, pp. 1391–1393. Fixed-basepoint sectors use homotopy groups; without fixed boundary data, one must also account for the appropriate changes of basepoint and identifications.

In a gauge theory, the Higgs field on the sphere describes a reduction of the gauge bundle to its stabilizer $H$. If the $G$-bundle extends smoothly over a ball, it is trivial there. Its $H$ clutching class $m$ must therefore obey $i_*(m)=0$. The two calculations above allow every integer $m$ for the stated $SU(2)$ circle, and only even $m$ for the stated $SO(3)$ circle. An odd $SO(2)$ winding instead induces the nontrivial $SO(3)$ bundle on the sphere, which cannot be the restriction of a smooth bundle over that ball.

This is an extension obstruction. It does not prove that a particular action admits a finite-energy monopole, determine its mass, or establish dynamical stability. Singular defects and different boundary conditions pose different extension questions. For the physical setup, continue to topological defects; for a compact map reference, use the topology formulas.

As a check, explain why $\pi_1(SO(3))=\mathbb Z_2$ does not imply strings for the vacuum orbit $SO(3)/SO(2)$. Then try the global-form exercises, which combine quotient descent with these topology distinctions.