G-sets, G-spaces and Covering Spaces

These notes amplify pages 68–72 of Hatcher’s “Algebraic Topology”.

Here, G will always be a discrete group (though the definitions make sense and are useful for topological groups, also with spaces replaced by many other kinds of mathematical object).

G-spaces

Definition 1 A (left) G-space Y is a space Y equipped with an action map G×Y → Y, usually written (g, y) ↦ gy, that satisfies the axioms

(i) 1y = y for any y ∈ Y; (ii) g(g′y) = (gg′)y for any y ∈ Y, g, g′ ∈ G. (2)

The orbit of y ∈ Y is the subspace Gy = {gy ∈ Y : g ∈ G}. The stabilizer of y ∈ Y is the subgroup H_{y} = {g ∈ G : gy = y} of G.

If Z is another G-space, a map f: Y → Z is a G-map if f(gy) = g(f(y)) for all y ∈ Y and g ∈ G.

In view of (ii), we may omit many parentheses. For each g ∈ G, the map y ↦ gy is a homeomorphism of Y, usually written simply as g: Y → Y, with inverse homeomorphism g^{−1}: Y → Y.

The axioms imply that H_{y} is indeed a subgroup of G (proof omitted). We have the useful identity

H_{gy} = gH_{y}g^{−1}.  (3)

To see this, note that gH_{y}g^{−1} ⊂ H_{gy} because for any h ∈ H_{y}, (ghg^{−1})gy = ghy = gy. Similarly, we have g^{−1}H_{gy}g ⊂ H_{y}, which implies H_{gy} ⊂ gH_{y}g^{−1}.

If f: Y → Z is a G-map, it is clear that H_{y} ⊂ H_{f(y)} for any y ∈ Y.

Lemma 4 The orbits in a G-space Y form a partition of Y: y ∈ Gy for any y ∈ Y, and for any y, z ∈ Y, either Gy = Gz or Gy and Gz are disjoint.

Proof Clearly, y = 1y ∈ Gy. Suppose w ∈ Gy ∩ Gz, i. e. w = gy = hz for some g, h ∈ G. Then g^{−1}w = g^{−1}gy = 1y = y. Given any k ∈ G, ky = kg^{−1}w = kg^{−1}hz ∈ Gz shows that Gy ⊂ Gz; similarly, Gz ⊂ Gy.

Definition 5 Given a G-space Y, the orbit space Y/G of Y is the set of all orbits in Y, topologized as a quotient space of Y by means of the obvious quotient map q: Y → Y/G given by q(y) = Gy.

The G-space Y is free if for every y ∈ Y, gy ≠ y for all g ≠ 1 in G. This implies that as g varies, the points gy are all distinct. We consider a stronger condition:

Every point y ∈ Y has a neighborhood U such that U ∩ gU = ∅ for all g ≠ 1. (6)

Then as g varies, the sets gU are all disjoint and open. It follows that q: Y → Y/G is a covering map: since q^{−1}(q(U)) = ∐_{g} gU, the set GU = q(U) is open in Y/G and is evenly covered by the sets gU ⊂ Y.

G-sets A G-set is simply a discrete G-space. It is transitive if there is only one orbit.

Example Any disjoint union of G-sets is a G-set.

Example Given a subgroup H ⊂ G, denote by G/H the set of all cosets gH = {gh : h ∈ H} of H in G. It is a transitive G-set with action g′(gH) = (g′g)H. (This does not conflict with the notation of Definition 5: G/H is the orbit space of the action of H on G given by right multiplication, (h, g) ↦ gh^{−1}.)

The stabilizer of the coset H = 1H ∈ G/H is just H_{H} = H. Observe that the number of elements of G/H is the index of H in G, if finite.

G-sets are easily classified. We note that each orbit is itself a G-set.

Theorem 7 Let G be a discrete group.

(a) Any G-set Y is the disjoint union of its orbits;

(b) For any y ∈ Y, the orbit Gy is isomorphic to the G-set G/H_{y};

(c) The G-sets G/H and G/K are isomorphic if and only if the subgroups H and K of G are conjugate.

Proof Lemma 4 takes care of (a).

For (b), we map ϕ: G/H_{y} → Gy by ϕ(gH_{y}) = gy; since ghy = gy for any h ∈ H_{y}, this is well defined. It is visibly surjective. Suppose that ϕ(gH_{y}) = ϕ(kH_{y}), i. e. gy = ky. Then k^{−1}gy = y, k^{−1}g ∈ H_{y}, and kH_{y} = k(k^{−1}g)H_{y} = gH_{y}.

Equation (3) gives the necessity in (c). Conversely, we define the isomorphism of G-sets G/H ≅ G/gHg^{−1} by kH ↦ kHg^{−1} = (kg^{−1})(gHg^{−1}) for any k ∈ G.

We need to determine the automorphism group of G/H. With that in hand, we can write down the automorphism group of any G-set.

Theorem 8 Given a subgroup H of G, the automorphism group Aut(G/H) of the G-set G/H is isomorphic to the group N(H)/H, where N(H) = {g ∈ G : gHg^{−1} = H} is the normalizer of H in G, the largest subgroup of G in which H is a normal subgroup.

The action of Aut(G/H) on G/H is transitive if and only if H is normal in G.

Proof Take any G-map θ: G/H → G/H. Then θ(H) = k^{−1}H for some k ∈ G. (The reason to use k^{−1} here will appear later.) Because θ is a G-map, we have

θ(gH) = gθ(H) = gk^{−1}H  for any g ∈ G,  (9)

so that the element k ∈ G completely determines θ. Now take g = h ∈ H; since hH = H, we must have hk^{−1}H = k^{−1}H, which reduces to khk^{−1} ∈ H, hence kHk^{−1} ⊂ H. This is enough to ensure that equation (9) is well defined for all g, as

θ(ghH) = ghk^{−1}H = gk^{−1}(khk^{−1})H = gk^{−1}H = θ(gH).

We note that θ is automatically surjective, as θ(gkH) = gkk^{−1}H = gH for any g ∈ G. However, it is not automatically 1-1. Suppose θ(gH) = θ(H). Then gk^{−1}H = k^{−1}H, which reduces to kgk^{−1} ∈ H, then to g ∈ k^{−1}Hk. For θ to be 1-1, this must imply g ∈ H, i. e. k^{−1}Hk ⊂ H, or H ⊂ kHk^{−1}. This, together with kHk^{−1}H ⊂ H, shows that k ∈ N(H).

The action of Aut(G/H) on G/H is transitive if and only if θ(H) = k^{−1}H can be any element of G/H. Since k ∈ N(H) is arbitrary, this is equivalent to N(H) = G.

Thus given k ∈ N(H), we have the automorphism θ_{k} of G/H defined by (9). Now k ↦ θ_{k} is a homomorphism of groups from N(H) to Aut(G/H), because

θ_{m}(θ_{k}(gH)) = θ_{m}(gk^{−1}H) = gk^{−1}m^{−1}H = θ_{mk}(gH).

We saw that this homomorphism is surjective, and its kernel is clearly H.

Applications to covering spaces

Proposition 10 Let p: X̃ → X be a covering map, and put π = π_{1}(X, x_{0}). Then the fibre F = p^{−1}(x_{0}) is a π-set, and the stabilizer of any point x̃ ∈ F is Hx̃ = p_{∗}(π_{1}(X̃, x̃)).

Proof Given a point x̃ ∈ F and a loop γ in X at x_{0}, lift γ to a path γ̃ ending at x̃. We define [γ]x̃ = γ̃(0) ∈ F; it is well defined, because if γ ≃ δ, γ̃ ≃ δ̃. If [γ] ∈ Hx̃, γ̃ is a loop, and [γ] = p_{∗}[γ̃]; and conversely.

We must verify the axioms (2). If γ is the constant path c_{x_{0}} at x_{0}, γ̃ = cx̃, and we have 1x̃ = [c_{x_{0}}]x̃ = cx̃(0) = x̃. Let β be another loop at x_{0}. We lift it to a path β̃ ending at γ̃(0), so that [β]([γ]x̃) = [β]γ̃(0) = β̃(0). Then β · γ lifts to β̃ · γ̃, and ([β][γ])x̃ = [β · γ]x̃ = (β̃ · γ̃)(0) = β̃(0).

Remark Some writers avoid the contortions in this proof by making the action of π on F a right action.

Corollary 11 Assume that X is path-connected. Then each path component of X̃ intersects the π-set F in exactly one orbit. In particular, X̃ is path-connected if and only if F is a transitive π-set.

Proof First, each orbit in F lies in one path-component of X̃. If x̃_{1} and x̃_{2} belong to the same orbit, say x̃_{2} = [γ]x̃_{1}, the loop γ lifts to a path in X̃ from x̃_{2} to x̃_{1}.

Conversely, if x̃_{1} and x̃_{2} are two points of F that lie in the same path-component of X̃, there is a path β in X̃ from x̃_{2} to x̃_{1}, hence [p^{∘}β]x̃_{1} = x̃_{2}, and the two points are in the same orbit.

Finally, each path component of X̃ contains a point of F. Given x̃ ∈ X̃, choose a path β in X from p(x̃) to x_{0}, and lift it to a path β̃ in X̃ from x̃ to some point β̃(1) ∈ F.

Corollary 12 If X and X̃ are path-connected, the number of sheets in the covering is the index in π of the stabilizer Hx̃ of any point x̃ ∈ F.

Now suppose that p_{1}: X̃_{1} → X and p_{2}: X̃_{2} → X are two covering maps, with fibres F_{1} = p_{1}^{−1}(x_{0}) and F_{2} = p_{2}^{−1}(x_{0}). It is obvious that a map f: X̃_{1} → X̃_{2} over X (one that satisfies p_{2}^{∘}f = p_{1}) induces a π-map of π-sets f|F_{1}: F_{1} → F_{2}, where π = π_{1}(X, x_{0}). We have the converse, for well-behaved X.

Theorem 13 Assume that X is path-connected and locally path-connected. Then any π-map F_{1} → F_{2} extends uniquely to a map f: X̃_{1} → X̃_{2} of covering spaces over X.

Proof As X̃_{1} is a covering space of X, it too is locally path-connected, and its path components are therefore open. It suffices to construct f on each path component of X̃_{1}, which reduces the problem to the known case where X̃_{1} is path-connected.

A homeomorphism of X̃ over X is known as a deck transformation. These form the automorphism group Aut(X̃) of X̃. We combine Theorems 13 and 8.

Corollary 14 With X as in Theorem 13, suppose that X̃ is a path-connected covering space and x̃_{0} ∈ F = p^{−1}(x_{0}). Then the group Aut(X̃) of deck transformations of X̃ is isomorphic to N(H)/H, where H = p_{∗}(π_{1}(X̃, x̃_{0})) and N(H) denotes the normalizer of H in π.

The action of Aut(X̃) on F is transitive if and only if H is normal in π. If so, we have a regular or normal covering space, π/H is a group, and we can identify X with the orbit space X̃/(π/H).

Remark Even if H is normal in π, the π-action (via π/H) on X̃ given by Corollary 14 is unrelated to the π-action on the fibre F given by Proposition 10 (unless π/H happens to be abelian); in particular, the action on X̃ does not in general extend the action on F. Write q: π → π/H for the quotient homomorphism; then in terms of the proof of Theorem 8, the π-set action on F is given by g′q(g) = q(g′)q(g), using left multiplication, while by equation (9), the deck transformation action induces on F ≅ π/H the action θ_{k}q(g) = q(g)q(k^{−1}), using right multiplication.

Universal covering spaces A simply-connected covering space of X is called a universal covering space of X. Under the conditions of Theorem 13, it is unique up to homeomorphism. A further condition is needed to ensure existence.

Corollary 15 With X as in Theorem 13, suppose that X̃ is a universal covering space. Then X̃ has the group π of deck transformations, and given any subgroup H of π, the orbit space X̃/H is a covering space of X with π_{1}(X̃/H) ≅ H.

Proof As the subgroup {1} of π is normal, we have the group π of deck transformations of X̃. Then p: X̃ → X factors to give a covering map p′: X̃/H → X, and p^{′−1}(x_{0}) is the π-set π/H, with stabilizer group H ≅ π_{1}(X̃/H).