From a Group Action to a Quotient
A group action organizes a space into sets of points that can be moved to one another by the group. Identifying each such set to a single point produces a quotient space. The topology on that space is not chosen arbitrarily: it is defined so that a subset is open exactly when its preimage under the quotient map is open. For group actions by homeomorphisms, the quotient map has an additional useful property: it is always open.
Let \(G\) be a group and let \((X,\mathcal{T})\) be a topological space. An action of \(G\) on \(X\) assigns to each \(g\in G\) a map \(x\mapsto gx\) such that \(ex=x\) and \(g(hx)=(gh)x\) for all \(g,h\in G\) and \(x\in X\), where \(e\) is the identity element of \(G\). In this tutorial, assume each map \(x\mapsto gx\) is a homeomorphism of \(X\). A continuous action of a topological group has this property, but continuity of the action as a function of both variables is not needed for the openness proof below.
The relation \(\sim\) is an equivalence relation, so its classes really do partition \(X\). In particular, points in the same orbit are precisely the points identified by the quotient.
Proof. For reflexivity, \(x=ex\), so \(x\sim x\). If \(y=gx\), then \(x=g^{-1}y\), so \(y\sim x\); this proves symmetry. If \(y=gx\) and \(z=hy\), then \(z=h(gx)=(hg)x\), so \(x\sim z\); this proves transitivity. The class of \(x\) consists, by the definition of \(\sim\), of exactly the points \(gx\) as \(g\) ranges over \(G\), which is \(Gx\). \(\square\)
Why the Orbit Map Is Open
Proof. Let \(U\subseteq X\) be open. A point \(x\) belongs to \(p^{-1}(p(U))\) exactly when its orbit meets \(U\). This happens exactly when \(x=gu\) for some \(u\in U\) and \(g\in G\). Therefore $$ p^{-1}(p(U))=\bigcup_{g\in G}gU. $$ Each \(gU\) is open because the action by \(g\) is a homeomorphism. An arbitrary union of open sets is open, so \(p^{-1}(p(U))\) is open. By the definition of the quotient topology, \(p(U)\) is open in \(X/G\). Thus \(p\) is open. It is surjective because every orbit is \(p(x)\) for any one of its points. A surjective open map is a quotient map: if \(p^{-1}(W)\) is open, then \(W=p(p^{-1}(W))\) is open by openness of \(p\); the reverse implication follows from the definition of the quotient topology. \(\square\)
The formula in the proof is important: the preimage of the image of \(U\) is the union of all its translates, not usually just \(U\). The argument needs each group element to act by a homeomorphism so that every translate remains open. The orbit map is surjective, but it need not be one-to-one; it identifies every orbit to one point.
Worked Example: A Finite Rotation Group on the Circle
Let \(m\in\mathbb{N}\). Take \(X=S^1=\{z\in\mathbb{C}:|z|=1\}\) with its subspace topology from \(\mathbb{C}\), and let \(G=C_m=\{\zeta^j:0\le j<m\}\), where \(\zeta=e^{2\pi i/m}\). The action is multiplication: \(\zeta^jz\). It is an action by homeomorphisms because multiplication by a fixed unit complex number has the continuous inverse given by multiplication by its reciprocal.
For \(z=e^{it}\), its orbit is $$ C_mz=\{e^{i(t+2\pi j/m)}:0\le j<m\}. $$ Consider \(f_m:S^1\to S^1\) given by \(f_m(z)=z^m\). It is continuous, since it is the restriction of a polynomial map on \(\mathbb{C}\), and it is surjective: if \(w=e^{is}\), then \(f_m(e^{is/m})=e^{is}=w\).
The fibers of \(f_m\) are exactly the orbits. Indeed, for \(z=e^{it}\) and \(w=e^{is}\), $$ f_m(z)=f_m(w) \quad\Longleftrightarrow\quad e^{imt}=e^{ims} \quad\Longleftrightarrow\quad t-s=\frac{2\pi k}{m}\text{ for some }k\in\mathbb{Z}. $$ The last condition says \(z=\zeta^k w\), so \(z\) and \(w\) lie in the same orbit. The converse follows by substituting \(\zeta^k w\) into the formula: \((\zeta^k w)^m=\zeta^{km}w^m=w^m\).
To conclude that the quotient is homeomorphic to \(S^1\), we check that \(f_m\) is a quotient map. It is open: for any \(z=e^{it}\) in an open set \(U\subseteq S^1\), there is \(\delta>0\) such that the arc \(\{e^{is}:t-\delta<s<t+\delta\}\) lies in \(U\). Choose \(\delta\) smaller if necessary so that \(m\delta<\pi\). Its image under \(f_m\) is an open arc around \(e^{imt}\), so every point of \(f_m(U)\) has an open neighborhood contained in \(f_m(U)\). Thus \(f_m(U)\) is open. The map is a continuous, surjective, open map, hence a quotient map. Since its fibers are exactly the orbits, it induces a bijection \(X/G\to S^1\); the quotient definitions for \(p\) and \(f_m\) show that a subset of \(S^1\) is open exactly when its preimage under this bijection is open in \(X/G\). Consequently \(S^1/C_m\cong S^1\).
Check: when \(m=3\), the orbit of \(1\) is \(\{1,e^{2\pi i/3},e^{4\pi i/3}\}\). All three points have the same image under \(f_3\), namely \(1\), as \((e^{2\pi i/3})^3=e^{2\pi i}=1\) and \((e^{4\pi i/3})^3=e^{4\pi i}=1\).
Worked Example: The Antipodal Quotient of a Sphere
Let \(n\ge 0\), let \(S^n=\{x\in\mathbb{R}^{n+1}:\|x\|=1\}\) have its subspace topology from Euclidean space, and let \(G=C_2=\{e,a\}\) act by \(a(x)=-x\). The map \(x\mapsto -x\) is a homeomorphism, since it is continuous and is its own inverse. For every \(x\in S^n\), its orbit is \(\{x,-x\}\); these are two distinct points because \(x\ne0\).
A line through the origin in \(\mathbb{R}^{n+1}\) meets \(S^n\) in exactly two points, \(x\) and \(-x\). Thus sending the orbit \(\{x,-x\}\) to the line spanned by \(x\) gives a bijection between \(S^n/C_2\) and the set of lines through the origin, denoted \(\mathbb{RP}^n\). Give this set of lines the quotient topology from the map sending each unit vector to its line. Under that definition, the bijection is a homeomorphism: both spaces have the same points and the same criterion for a set of lines to be open, namely that its preimage in \(S^n\) is open. Hence the antipodal quotient is \(\mathbb{RP}^n\). For \(n=0\), \(S^0=\{-1,1\}\) and the quotient has one point, agreeing with the fact that \(\mathbb{R}^1\) has just one line through the origin.
Check: for \(n=2\), each point \(x=(x_1,x_2,x_3)\in S^2\) is identified with \(-x=(-x_1,-x_2,-x_3)\). These are exactly the two unit vectors on the line through \(x\); no other unit vector lies on that line.
Worked Example: All Rotations of the Circle
Let \(X=S^1\) with its usual topology and let \(G=S^1\) act on \(X\) by multiplication. Here \(G\) is the group of unit complex numbers. Given \(z,w\in S^1\), the group element \(g=wz^{-1}\) satisfies \(gz=w\). Thus the action is transitive: every point has the same orbit, namely \(S^1\). The quotient set has one point.
The quotient topology on a one-point set consists of \(\varnothing\) and the whole set. Indeed, their preimages under the orbit map are \(\varnothing\) and \(S^1\), both open, and these are all the subsets of a one-point set. Therefore the quotient is a one-point topological space. This differs from the finite cyclic action: a finite rotation group leaves many distinct orbits, whereas all rotations can move any point to any other.
A Common Pitfall: Openness Does Not Mean Injectivity
It is incorrect to infer that an open orbit map is a homeomorphism. A homeomorphism must be bijective, whereas an orbit map is generally many-to-one. For a concrete calculation, take \(C_3\) acting on \(S^1\) by rotations and let \(\zeta=e^{2\pi i/3}\). Since \(\zeta\ne1\), the points \(1\) and \(\zeta\) are distinct. But \(\zeta\cdot1=\zeta\), so \(1\) and \(\zeta\) belong to the same orbit and \(p(1)=p(\zeta)\). The orbit map is not injective and therefore is not a homeomorphism.
The correction is to distinguish the orbit map from a map defined on the quotient. In this example the orbit map sends \(z\) to its orbit, while the formula \(z\mapsto z^3\) sends each orbit to a point of \(S^1\). The latter descends to a homeomorphism from the quotient, as shown above; it is not the orbit map itself.
Check Your Understanding
Use the definitions and proofs above to answer the following questions.
- For a group action, explain why \(x\sim y\) and \(y\sim z\) imply \(x\sim z\).
- For an open set \(U\subseteq X\), write \(p^{-1}(p(U))\) as a union of translates and explain why each translate is open.
- For the action of \(C_4\) on \(S^1\), what is the orbit of \(1\), and what is its image under \(z\mapsto z^4\)?
- Why do the points \(x\) and \(-x\) determine the same point of the antipodal quotient of \(S^n\)?
- Can the orbit map be a homeomorphism if some orbit contains two distinct points? Explain.