## Topology: Topological Groups

This article assumes you know some basic group theory. The motivation here is to consider groups whose underlying operations are continuous with respect to its topology.

Definition. A topological group G is a group with an underlying topology such that:

• the product map m: G × G → G is continuous;
• the inverse map i: G → G is continuous.

Examples

1. Any group can be a topological group by endowing it with the discrete topology.
2. The complex plane C, real line R, and their subspaces ZQ are all topological groups under addition.
3. C* = C – {0} and R* = R – {0} are also topological groups under multiplication.
4. More interesting topological groups include GLn(R) and SLn(R), the group of non-singular n × n real matrices, and the group of n × n real matrices with determinant 1 (respectively). The topology of these groups are given by the subspace topology upon embedding them in $\mathbf{R}^{n^2}.$

The first property we want to prove is:

Proposition. On a topological group G, the following maps are homeomorphisms.

• Left-multiplication: for g in G, $l_g:G \to G, x\mapsto gx.$
• Right-multiplication: for g in G, $r_g:G \to G, x\mapsto xg.$
• Inverse: $i:G\to G$ above.
• Conjugation: for g in G, $c_g(x) = gxg^{-1}.$

Proof.

Since multiplication and inverse maps are continuous, the above maps $l_g, r_g, i, c_g$ are all continuous too. Plus, they’re bijective and their inverses are given by $l_{g^{-1}}, r_{g^{-1}}, i, c_{g^{-1}}$ respectively, which are all continuous. So the maps are homeomorphisms. ♦

In particular, for any $x, y\in G$ there’s a homeomorphism of G which maps x to y; indeed, $l_{yx^{-1}}(x) = y.$

On an intuitive level, this means the topology of G is completely uniform: e.g. to check what happens near any point g, it suffices to check near the identity e. [ This should not be confused with uniform topological spaces, which is an entirely different thing. ]

Next, the following result is useful.

Proposition 2. Let A, B be subsets of a topological group G. If A is open in G, then so are AB and BA, where $AB=\{xy : x\in A, y\in B\}.$

Proof.

This follows from $AB = \cup_{b\in B} Ab$ which is a union of open subsets of G since right-multiplication is a homeomorphism. Same goes for BA. ♦

The case for closed subsets is not so nice, but at least we have:

Proposition 3. Let C be a closed subset of G and K be a compact subspace of G. Then CK and KC are closed in G.

Proof.

We’ll show that GCK is open in G; the case for KC is similar.

Fix x in G-CK and take the map $f:G\times G\to G$ which takes $(a,b)\mapsto ab^{-1}.$ Clearly f is continuous, so $f^{-1}(G-C)\subseteq G\times G$ is open. Now for any $y\in K,$ the ordered pair $(x,y)\in f^{-1}(G-C)$ because if $f(x,y) = xy^{-1} \in C,$ we would have $x = (xy^{-1})y \in CK.$

So $(x,y)\in U_y\times V_y$ for some open subsets $x\in U_y\subseteq G$ and $y\in V_y\subseteq G$ such that $f(U_y, V_y)\subseteq G-C.$ By compactness of K, it can be covered by finitely many $V_{y_1}, \ldots, V_{y_n}:$

let $V := V_{y_1} \cup \ldots \cup V_{y_n} \supseteq K\$ and $\ U:= U_{y_1} \cap \ldots \cap U_{y_n}\ni x.$

Then $f(U, K) \subseteq f(U,V) \subseteq \cup_k f(U_{y_k}, V_{y_k}) \subseteq G-C.$ Thus $x\in U\subseteq G-CK$ so indeed GCK is open. ♦

## Subgroups

First, we look at the separation axioms on a topological group.

Proposition 4. The following are equivalent for a topological group G.

1. G satisfies T3.
2. G satisfies T2.
3. G satisfies T1.
4. The trivial subgroup {e} is closed.

Proof

The only non-trivial direction is (4→1). In fact, we’ll prove any topological group G is regular, from which (3→1) follows. The final case of (4→3) follows by left-multiplying G with g, thus every {g} is closed.

Consider the continuous map $f:G\times G\to G$ which takes $(a,b)\mapsto ab^{-1}.$ If x is contained in an open subset U of G, then $f^{-1}(U)$ contains (x, e) so we have $(x,e)\in V\times W\subseteq f^{-1}(U)$ for some open subsets VW of G containing xe respectively.

Now $x\in V$. Furthermore, $V\cap (X-U)W =\emptyset$ since any element in the intersection corresponds to $a\in V, b\in W$ such that $ab^{-1}\not\in U$ which is a contradiction. By proposition 2, (XU)W is an open subset containing XU, so any topological group is regular. ♦

It’s ok if {e} is not closed, for we can take the closure which will turn out to be a subgroup, in fact, a normal subgroup of G. More generally:

Proposition 5. If H is a subgroup of G, then so is cl(H). If N is a normal subgroup of G, then so is cl(N).

Note.

The term “normal” unfortunately has a slight ambiguity here. It can refer to normal subgroups, or normal topological spaces (as in, any two disjoint closed subsets can be separated by open subsets). We’ll only refer to the former in this article.

Proof.

Take xy in cl(H). Now the continuous map $f:G\times G\to G$ which takes $(a,b)\mapsto ab^{-1}$ gives us $f(H, H)\subseteq H.$ Since cl(H) is a closed subset of G containing H, $f^{-1}(\text{cl}(H))$ is a closed subset of G × G containing H × H. Thus $f^{-1}(\text{cl}(H))$ contains cl(H × H) = cl(H) × cl(H) and we have $f(\text{cl}(H), \text{cl}(H))\subseteq \text{cl}(H)$ which, together with $e\in \text{cl}(H)$, proves that cl(H) is a subgroup of G.

For the second statement, each conjugancy map $c_g:G\to G, x\mapsto gxg^{-1}$ is a homeomorphism and maps N to N. Thus it must map cl(N) to cl(N) also. ♦

The closure can be nicely classified as follows:

Proposition 6. If H is a subgroup of G, then cl(H) is the intersection of all HU, where U is an open subset containing e.

Proof.

Now $x\in \text{cl}(H)$ iff every open subset V containing x intersects H, which holds iff every open subset U containing e intersects $x^{-1}H.$ And this holds iff $x\in HU^{-1}.$ Now recall that U is open iff $U^{-1}$ is open. ♦

Proposition 7. The connected component Y containing e is a closed normal subgroup of G.

Proof.

Y is closed in G since every connected component is closed.

Now suppose $x,y\in Y.$ Since left-multiplication by x is a homeomorphism of G, it must map the connected component of e to that of x. But x and e both have Y as their connected component, so xYY and thus $xy\in Y.$

Next, inverse is a homeomorphism so it must take the connected component of e to the connected component of i(e)=e, i.e. i(Y) = Y. This proves Y is a subgroup. Normality follows from the fact that the conjugation map is a homeomorphism taking e to itself. ♦

Note.

Proposition 7 still holds if we replace “connected components” with “path-connected components“.

So far we’ve been looking at closed subgroups. What about open ones?

Proposition 8. An open subgroup H of G is also closed. Furthermore, a closed subgroup H of finite index in G is open.

Proof.

The complement GH is a union of (left) cosets of H. For the first statement, each coset is open since left-multiplication is a homeomorphism. Thus GH is open and H must be closed. For the second statement, each coset is closed and GH is a union of finitely many closed subsets, so it’s closed too, and H is open. ♦

Corollary. If G is a connected group, then any neighbourhood U containing e generates the whole group (in the algebraic sense).

Proof.

Replace U by U ∩ i(U) and we may assume i(U) = U. Now consider $V:=\cup_{n=1}^\infty U^n$ where Un is the collection of products of n elements from U. It’s easy to see that V is an open subgroup of G. By the above proposition V is a clopen subset so V=G. It’s clear that the group <U> generated by U must contain V; thus <U>=G. ♦

## Continuous Homomorphisms

Between topological groups, naturally we’ll be looking at homomorphisms which preserve the topology, i.e. continuous homomorphisms. For example: the following is obvious.

Property. If $f:G\to H$ is a continuous map of topological groups, then for any closed (resp. open) subgroup K of H, $f^{-1}(K)$ is a closed (resp. open) subgroup of G.

If K is normal in H, then $f^{-1}(K)$ is normal in G.

In particular, if H is Hausdorff, then the kernel of f is a closed normal subgroup of G.

The corresponding result doesn’t hold for the im(f), though we do know it’s a subgroup of H. E.g. consider $f:\mathbf{Z}^2 \to \mathbf{R}$ which takes (ab) to a+b√2. The homomorphism is clearly continuous if Z2 is given the discrete topology. However, the image of f is not closed.

But since the continuous image of a connected (resp. compact) set is also connected (resp. compact), at least we have:

Property. If $f:G\to H$ is a continuous homomorphism of topological groups and G is connected (resp. compact), then im(f) is a connected (resp. compact) group.

For example, take the above map $f:\mathbf{Z}^2\to \mathbf{R}$ which takes (ab) to a+b√2. This induces a group isomorphism from Z2 to im(f) which is continuous but not a homeomorphism.