Topology: Quotients of Topological Groups

Topology for Coset Space

This is really a continuation from the previous article. Let G be a topological group and H a subgroup of G. The collection of left cosets G/H is then given the quotient topology. This quotient space, however, satisfies an additional property.

Definition. A map f : X → Y between two topological spaces is said to be open if, for any open subset U of X, f(U) is open in Y.

Proposition 1. The map p : G → G/H is open.

Proof. Let U be an open subset of G. From proposition 2 hereUH is open in G. Since p^{-1}(p(U)) = UH is open, so is p(U) by definition of quotient topology. ♦

warning In general, not every quotient map pX → X/~ is open. For example, glue the endpoints of I = [0, 1] together and form the quotient map p:I\to S^1. Then U = (1/2, 1] is open in I but p(U) is not open in S1.  

We should say something about open maps since this is our first encounter with them. Don’t worry too much for now – they don’t appear all that often.

Properties of Open Maps

  • The identity map on any X is open.
  • If fX → Y and gY → Z are open, then so is gfX → Z.
  • If f : X → Y is open and f(X)\subseteq Z\subseteq Y, then restricting to f : X → Z still gives an open map since if f(U) is open in Y, then f(U) ∩ Zf(U) is open in Z.
  • Restricting the domain doesn’t give an open map in general: e.g. if we restrict id: R → R to the closed interval [0, 1], the image of [0, 1] is not open. However, restricting f : X → Y to an open subset U of X gives an open map U → Y.
  • The projection map \prod_i X_i \to X_i is open whether we use the product or box topology. To prove that, use the fact that for any map f:A\to B, we have f(\cup_i A_i) = \cup_i f(A_i).
  • If each f_i : X_i \to Y_i is open, then the resulting f:\prod_i X_i \to \prod_i Y_i which takes (x_i) \mapsto (f_i(x_i)) is also open. Again, its proof requires f(\cup_i A_i) = \cup_i f(A_i).

Exercises

Generally, open continuous maps preserve local properties. For example, prove the following for an open continuous fX → Y.

  • If X is locally connected, then so is f(X).
  • If X is locally path-connected, then so is f(X).
  • Let X and Y be Hausdorff. If X is locally compact, then so is f(X).

Answers (Highlight to Read)

  • If y is in f(X), pick x in X such that f(x)=y. Then x is contained in some connected open subset U of X. So f(x) lies in f(U) which is an open and connected subset of Y and hence f(X).
  • Same as connected.
  • If y is in f(X), pick x in X such that f(x)=y. Then x is contained in some open subset U of X whose closure cl(U) is compact. So f(x) is contained in f(U), which is open in Y and hence f(X). Also f(cl(U)) is compact and hence closed in f(X), so the closure of f(U) in f(X) is contained in f(cl(U)) and thus compact also. 

blue-lin

Properties of Coset Space

Let H be a subgroup of the topological group G. The first result is straightforward.

Theorem 2. G/H is discrete if and only if H is open in G.

Proof.

G/H is discrete iff every singleton subset is open; by the definition of quotient topology, this holds iff every left coset gH is open in G, which holds iff H is open in G. ♦

The next result is trickier.

Theorem 3. The following are equivalent.

  1. G/H is T2.
  2. G/H is T1.
  3. H is closed in G.

Proof.

Let pG → G/H be the projection map, which is open by proposition 1. Now (1→2) is obvious. (2→3) follows from H = p^{-1}(\{e\}); if G/H is T1, then {e} is closed in G/H, hence H is closed in G.

Finally, suppose H is closed in G; to show that G/H is Hausdorff, recall that it suffices to show that the image of the diagonal map \Delta :G/H \to G/H \times G/H is closed in G/H × G/H. Now its complement is:

U := (G/H \times G/H) - \text{im}(\Delta) = \{(p(g), p(g')) : g^{-1}g' \not\in H\}.

Let q = (p, p) : G × G → G/H × G/H, which is open because p is. Then q^{-1}(U) = \{(g, g')\in G\times G : g^{-1}g'\not \in H\} is open since the continuous map f: G\times G\to G, (x, y)\mapsto x^{-1}y gives us q^{-1}(U) = f^{-1}(G-H).

Since q is surjective, we have U = q(q^{-1}(U)) which is open since q is open. ♦

blue-linGroup Quotient

Now suppose N is a normal subgroup of the topological group G. It turns out the group operations for G/N are continuous with respect to the quotient topology, i.e. G/N is also a topological group.

Theorem 4. The product map m':G/N \times G/N \to G/N and inverse map i':G/N \to G/N are continuous.

Proof.

First, we prove a general result:

Algebraic Quotient Lemma. Suppose p_i : X_i\to Y_i is a collection of quotient maps which are open. Let q = (p_i) : \prod_i X_i \to \prod_i Y_i be the projection map which takes (x_i) \mapsto (p_i(x_i)).

Then q is a quotient map. [ Compare this result with the warning after example 2 here. ]

Proof of AQL.

We need to show that V\subseteq \prod_i Y_i is open if and only if q^{-1}(V) is open. (→) is obvious since q is continuous. For (←), suppose q^{-1}(V) is open; we have q(q^{-1}(V)) = V since q is surjective. Since each pi is open, so is q. And since q is surjective q(q^{-1}(V))=V, which must be open. ♦

We resume our proof of theorem 4 to show m’ is continuous (i’ is easy). Applying it to G → G/N, we see that the projection qG × G → G/N × G/N is a quotient map too. Thus, showing m’ is continuous is equivalent to showing m’qG × G → G/N is continuous. But m’q is obtained via composing the product map G × G → G with projection G → G/N. Thus, we’re done. ♦

Theorem 5. The quotient group G/N is discrete if and only if N is open in G. Also, G/N is Hausdorff if and only if N is closed in G.

Proof.

Follows immediately from theorem 3 above. ♦

Corresponding to the first isomorphism theorem for groups, we have:

Theorem 6. If f:G\to H is a continuous homomorphism of topological groups, this induces a continuous injective map g:G/\text{ker}(f)\to H.

Proof. The only non-trivial part is continuity. If pG → G/ker(f) is the projection map, then gpf is continuous, so by the universal property of quotient topologyg is continuous. ♦

warningIn general, G/ker(f) doesn’t have the subspace topology from H. For example, f can be an injective continuous map which does not provide G with the subspace topology from H.

E.g. take \mathbf{Z}\to S^1, where m \mapsto (\cos(m\sqrt 2), \sin(m\sqrt 2)).

Next, recall (proposition 7 here) that the connected component of e\in G is a closed normal subgroup N of G.

Theorem 7. If N is the connected component of e, then G/N is totally disconnected.

Proof.

For each g in G, left-multiplication l_g:G\to G, x \mapsto gx is a homeomorphism. Thus, the connected components of G are the cosets gN, for various g. More generally, we’ll prove:

Lemma. In a topological space X, denote x ~ y if they belong to the same connected component. If the projection map p : X → X/~ is open, then X/~ is totally disconnected.

Proof of Lemma.

Suppose there’s a connected subset Y of X/~ comprising of more than a point. Let Z := p^{-1}(Y) which contains points from more than one connected component, so it is disconnected, i.e. Z=U\cup U' for some non-empty disjoint open subsets UU’ of Z. totally_disconnected_proof Now, if Z contains x, then it contains the entire connected component <x> of x. We claim this holds for U as well. Indeed, we have \left<x\right> = (U\cap \left<x\right>) \cup (U'\cap \left<x\right>) as a disjoint union; since <x> is connected and U\cap \left<x\right>\ne\emptyset we must have U\cap \left<x\right> = \left<x\right> \implies U\supseteq \left<x\right>. Thus U and U’ are both unions of connected components of X.

Write UZ ∩ V and U’Z ∩ V’ for open subsets VV’ of X. Since p is open, p(V) and p(V’) are open subsets of X/~ satisfying:

  • p(V) \cup p(V') = p(V\cup V') \supseteq p(U\cup U') = p(p^{-1}(Y)) = Y;
  • p(V) \cap p(V')\cap Y=\emptyset: indeed, if it contains y, then pick x\in V, x'\in V' such that p(x) = p(x’) = y. Then x, x'\in p^{-1}(Y) = Z so they’re in UU’ respectively. Since p(x) = p(x’), x and x’ must belong to the same connected component, which contradicts x\in V, x'\in V'.

Hence, p(V) ∩ Y and p(V’) ∩ Y are disjoint non-empty open subsets of Y with union Y, which contradicts our assumption that Y is connected. ♦

Examples of Group Quotients

  1. Consider Q as a topological subgroup of R. The quotient R/Q has the coarsest topology.
  2. Take \mathbf{Z}\subset \mathbf{R}. The map \exp: \mathbf{R}\to S^1 which takes t\mapsto (\cos(2\pi t), \sin(2\pi t)) has kernel equal to Z. By theorem 5 above, this gives a bijective continuous homomorphism R/Z → S1. Since the image is Hausdorff, if we could prove R/Z is compact, then we’d have shown that R/Z and S1 are isomorphic topological groups. But then R/Z is the continuous image of the composition [0, 1] → R → R/Z. Case closed.
  3. If G and H are topological groups, then so is G × H (with the product topology). The continuous bijective group homomorphism (G × H)/H → G is a homeomorphism since it’s open. Thus (G × H)/H and G are isomorphic topological groups.
  4. Take \mathbf{Z}^2\subset \mathbf{R}^2. The same reasoning holds as before and one can show that R2/Zis isomorphic to S1 × S1.
  5. Finally, take \text{SL}_n(\mathbf{R}) \subset \text{GL}_n(\mathbf{R}). The determinant map GLn(R) → R* induces a bijective continuous homomorphism GLn(R)/SLn(R) → R*. Now we can’t use the same trick since the groups aren’t compact. Instead, we compose: R* → GLn(R) → GLn(R)/SLn(R) → R*, where the first map takes c to the diagonal matrix with entries (c, 1, …, 1). This gives the identity map on R*; since the first two maps are continuous, the last map is open.
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