## Topological Groups

We now begin the fascinating look at the vastly fruitful subject of “topological groups”. The basic idea is that while someone who has only done basic point-set topology (openness, compactness, connectedness, etc.) may think that topology and algebra are as far apart as one can get, often the two are inseparable. For example, think about the most easily understood and most well liked topological space: with the usual metric. But this isn’t just a topological space it is a group under the usual addition. And in fact, there is a fascinating and unavoidable interplay between these two structures. For one, the metric (and thus the topology) is based entirely on the group theoretic structure of . You then begin to realize that a lot of the most common spaces that one deals with in mathematics: , , , etc. have a perfectly reasonable group theoretic structure already “built in”. And, it is not too hard to see that just as with the interaction between the topological and group structures is an interesting and vastly important aspect of the theory of those spaces. Thus, it makes sense to ask in general what can one say about a set which has both a topological and group theoretic structure?

So, as usual we start with a vague notion of what we want to study “something that is a group and a topological space” and we start to ask ourselves “Is that really what I want?” The answer is all likelihood is no. Just like manifolds where our intuitive idea was “something that looked like was not good enough (we figured that to tame the idea we needed second countability and Hausdorfness) saying that a topological group is any set with both a group theoretic and topological structure is just too broad. We want a stronger, more intimate, connection between the two, something that will also (hopefully) smooth out some of the wrinkles that overwhelming generality often harbors. With that said we are now left with the task of creating this “intimate connection”.

If someone came up to you and asked “what is the most important thing in topology” you may (as I would) reply “continuous functions of course!”. Now, if someone were to ask you the same question but with group theory in lieu of topology you might (after considerably more time) reply “Hmm…well everything has to do with how the elements interact with each other and their inverse. Maybe the functions and ?”. With those well-placed (who knows what someone would actually say) answers one has the key to the definition of a topological group. We have two important functions and the idea that it is important for functions to be continuous. Is there nothing more obvious then to combine these two in the beautiful matrimony known as:

**Topological Group: **Let be such that is a group and is a topological space. Then, if the maps and are continuous then is called a *topological group.*

*Remark:* As is customary we shall denote by the concatenation . Also, the above definitions of shall be used uniformly throughout any discussion of topological groups.

While the above may give you vague idea of why we should define a topological group to be this way I understand that it in no way shows that this is a moral imperative. I will (if I remember) try to point out along the way things that “should be true” in topological groups that are only guaranteed with these conditions.

So, motivation aside, let’s begin:

Our first theorem is nothing too deep, just a statement about continuous maps but it will turn out to be quite useful in the future.

**Theorem:** Let have both a topological and group structure, then is a topological group if and only if the map is continuous.

**Proof:** Suppose that is a topological group. Note, that by assumption is continuous and so by a problem from a long time ago so is . Thus,

is continuous. But, .

Conversely, suppose that is continuous. Then, . Thus, since it follows that is continuous.

So, using this we see that

is continuous. But, . The conclusion follows.

**Corollary:** It follows that since given continuous maps that is continuous and so by the above so must

Thus, if is closed then so is:

where . It follows that if and are continuous functions from a topological group with closed identity into itself which agree on a dense subsets, then they must be the same function. This is startling since we have proved this theorem before and we needed the codomain to be Hausdorff! Don’t worry, it will get *much* better.

We now show that some other common maps are continuous, and in fact, are homeomorphisms.

**Theorem:** Define and . Then, and are homeomorphisms.

**Proof:** It is clear that and are bijective (this is easy and you must have done it while discussing Cayley’s theorem). Also, each are continuous since they are the restriction of to and respectively. For the same reason they’re inverses and are continuous. The conclusion readily follows.

It turns out (unsurprisingly) that is also a homeomorphism.

**Theorem:** Let be defined as usual, then is a homeomorphism.

**Proof:** Clearly it is injective since

.

Also, it’s surjective since

and so .

It’s assumed that is continuous and since it’s idempotent the conclusion follows.

**Corollary:** Using the above we can see that if we define , , and then if is open (or respectively closed) then and are open (closed respectively).

We can also prove a similar fact that will come to be important when we talk about spaces of cosets.

**Theorem:** Let and define and then is open implies that is open.

**Proof:** This follows since and and each is open. .

*Remark:* The analogous theorem does not apply if one replaces open by closed (think and )

We now begin discussing a series of seemingly useless theorems that will save our asses when we discuss a very neat theorem (which I shall not allude to…it’s a fun surprise!)

**Symmetric Neighborhood:** Let be open. Then, it is called *symmetric* if .

**Theorem:** Let be a neighborhood of then there exists some symmetric neighborhood of such that .

**Proof:** This follows quite easily from the two following lemmas.

**Lemma:** Let be arbitrary, then

**Proof: **Obvious. .

**Theorem:** If is as above then .

**Proof:** Also obvious. .

So, using the above and noticing that and so is a neighorhood of contained within . Noting that finishes the argument.

The next theorem baffled me for a while until one realizes a neat little trick (try to look for it in the proof!)

**Theorem:** Let be a neighborhood of . Then, there exists some neighborhood of such that (where )

**Proof:** Notice that the map as define above is surjective and thus . Furthermore, we see that where each are a pair of open subspaces of . Thus,

And thus is the union of the product of two open sets. But, notice that since that and so for some . Thus, putting this together we see that which is open and

from where the conclusion follows.

To prove the next theorem we need a tedious to prove (yet intuitive) little theorem.

**Theorem: **Define . Then, is continuous.

**Proof:** We prove the case for and let the reader make the obvious generalization (it can be done by induction). Note that since we must only find a continuous map from to and the conclusion will follow.

To do this note that

is continuous since it is the product of continuous maps. And thus

given by

is continuous. But,

The conclusion follows by previous comment.

From this we can prove a nice little theorem

**Theorem:** Let be a neighborhood of and . Then, there exists some symmetric neighborhood of such that .

**Proof:** Using the exact same idea as in the proof when by the continuity and surjectivity of we have that

Where are open. Also, we are able to find some such that . Thus, noticing that

we have the desired neighborhood of such that . Now, since we proved earlier that every neighborhood of contains a symmetric neighborhood of the conclusion follows.

I know that those seemed awfully unimportant, and while they are no immediate value upon themselves they will be an indispensable tool in a matter of paragraphs.

Until then, we prove a quick theorem which shall come into play later as an interesting tidbit

**Theorem:** Let be a symmetric neighborhood of . Then, (I use to denote subgroup).

**Proof: **Clearly .

Now, if then for some and . So, and since is symmetric we see that and thus .

Now, if and then .

The conclusion follows.

Now comes to the real culmination of this post, a startling theorem relating topological groups to the separation axioms. We begin with a surprising yet easy to prove theorem.

**Theorem:** Let be a topological group with closed. Then, is .

**Proof:** This follows readily from the fact that is a closed mapping (since it’s a homeomorphism) for any and . Thus, all the singletons in are closed. .

Even more surprising is the following result.

**Theorem:** If is a topological group with closed then is Hausdorff.

**Proof:** One need merely note that

where was defined in the first theorem. Thus, since is continuous and closed it follows that is closed and thus from a previous theorem is Hausdorff. .

Don’t have a heart-attack but it even goes one step further.

**Theorem:** Let be a topological group such that is closed, then is regular.

**Proof:**

**Lemma:** Let be as above and let be a closed set not containing . Then, there exists disjoint neighborhoods such that and .

**Proof:** Clearly is a neighborhood of and so by previous comment we may find a symmetric neighborhood of such that . We claim that . To see this we prove that if then . To see this merely note that is a neighborhood of and for if not then for some and so . But, since is symmetric it follows that and thus which is a contradiction.

Thus, and are the disjoint neighborhoods containing and respectively.

Using this we note that if is closed and then and so we may find open sets containing and respectively. Thus, and respectively. They are both open and disjoint since .

That just blows my mind. I mean, to some that may have been obvious or mundane but when I first proved that (upon directive of my book of course) I was stunned. What’s shocking about the above is that you may ask “Where did you use the fact that is closed?” I didn’t you can separated a closed set and a point not in that set in *any* topological group. But, a regular space is and so we needed the closedness of .

From this we have the astonishing corollary

**Corollary: **Let be a topological group, then:

There are many, many more interesting theorems about topological groups which could have been put here if it weren’t for the finiteness of time. For example, all that a topological group requires to be embeddable in is that it is separable and has a countable neighborhood base at the identity element.

[…] a topological ring is a ring with a topology such that the underlying group structure forms a topological group and the product map is continuous. The first thing we note is […]

Pingback by Rings of Functions (Pt. I) « Abstract Nonsense | August 1, 2011 |

what is D(V) in the last proof?

Comment by ravik | October 12, 2012 |