Abstract Nonsense

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: $\mathbb{R}$ 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 $\mathbb{R}$. You then begin to realize that a lot of the most common spaces that one deals with in mathematics: $\mathbb{R}^n$, $\mathbb{C}[X,\mathbb{R}]$, $\ell_2$, etc. have a perfectly reasonable group theoretic structure already “built in”. And, it is not too hard to see that just as with $\mathbb{R}$ 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 $\mathbb{R}^n$ 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 $g_1*g_2$ and $g^{-1}$?”. 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 $(G,*,\mathfrak{J})$ be such that $(G,*)$ is a group and $(G,\mathfrak{J})$ is a topological space. Then, if the maps $\theta:G\times G\to G:(g_1,g_2)\mapsto g_1*g_2$ and $\phi:G\to G:g\mapsto g^{-1}$ are continuous then $G$ is called a topological group.

Remark: As is customary we shall denote $g_1*g_2$ by the concatenation $g_1g_2$. Also, the above definitions of $\theta,\phi$ 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 $G$ have both a topological and group structure, then $G$ is a topological group if and only if the map $\alpha:G\times G\to G:(g_1,g_2)\mapsto g_1g_2^{-1}$ is continuous.

Proof: Suppose that $G$ is a topological group. Note, that by assumption $\phi:G\to G$ is continuous and so by a problem from a long time ago so is $\iota\times\phi:G\times G\to G\times G:(g_1,g_2)\mapsto(g_1,g_2^{-1})$. Thus,

$\theta\circ(\iota\times \phi):G\times G\to G:(g_1,g_2)\overset{\iota\times\phi}{\longmapsto}(g_1,g_2^{-1})\overset{\theta}{\longmapsto}g_1g_2^{-1}$

is continuous. But, $\alpha=\theta\circ(\iota\times\phi)$.

Conversely, suppose that $\alpha:G\times G\to G$ is continuous. Then, $\alpha\mid_{\{e\}\times G}:\{e\}\times G\to G:(e,g)\mapsto eg^{-1}=g^{-1}$. Thus, since $G\approx \{e\}\times G$ it follows that $\phi:G\to G:g\mapsto g^{-1}$ is continuous.

So, using this we see that

$\alpha\circ(\iota\oplus\phi):G\times G\to G:(g_1,g_2)\overset{\iota\oplus\phi}{\longmapsto}(g_1,g_2^{-1})\overset{\alpha}{\longmapsto}g_1(g_2^{-1})^{-1}=g_1g_2$

is continuous. But, $\alpha\circ(\iota\oplus\phi)=\theta$. The conclusion follows. $\blacksquare$

Corollary: It follows that since given continuous maps $\varphi,\psi:G\to G$ that $\varphi\oplus\psi:G\to G\times G:g\mapsto(\varphi(g),\psi(g))$ is continuous and so by the above so must

$\alpha\circ(\varphi\oplus\psi):G\to G:g\overset{\varphi\oplus\psi}{\longmapsto}(\varphi(g),\psi(g))\overset{\alpha}{\longmapsto}\varphi(g)(\psi(g))^{-1}$

Thus, if $\{e\}$ is closed then so is:

$A(\varphi,\psi)=\left(\alpha\circ(\varphi\oplus\psi)\right)^{-1}(\{e\})$

where $A(\varphi,\psi)=\left\{g\in G:\varphi(g)=\psi(g)\right\}$. It follows that if $\psi$ and $\varphi$ 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 $L_k:G\to G:g\mapsto kg$ and $R_k:G\to G:g\mapsto gk$. Then, $L_k$ and $R_k$ are homeomorphisms.

Proof: It is clear that $L_k$ and $R_k$ 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 $\theta$ to $\{k\}\times G$ and $G\times\{k\}$ respectively. For the same reason they’re inverses $L_{k^{-1}}$ and $R_{k^{-1}}$ are continuous. The conclusion readily follows. $\blacksquare$

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

Theorem: Let $\phi:G\to G$ be defined as usual, then $\phi$ is a homeomorphism.

Proof: Clearly it is injective since

$\phi(g_1)=\phi(g_2)\implies g_1^{-1}=g_2^{-1}\implies g_2=g_1$.

Also, it’s surjective since

$g\in G\implies g^{-1}\in G$ and so $\phi(g^{-1})=(g^{-1})^{-1}=g$.

It’s assumed that $\phi$ is continuous and since it’s idempotent the conclusion follows. $\blacksquare$

Corollary: Using the above we can see that if we define $xU=\left\{xu:u\in U\right\}$, $Ux=\left\{ux:u\in U\right\}$, and $U^{-1}=\left\{u^{-1}:u\in U\right\}$ then if $U$ is open (or respectively closed) then $xU,Ux,$ and $U^{-1}$ 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 $U,V\subseteq G$ and define $UV=\left\{uv:u\in U\text{ and }v\in V\right\}$ and $VU\left\{vu:v\in V\text{ and }u\in U\right\}$ then $U$ is open implies that $UV$ is open.

Proof: This follows since $\displaystyle UV=\bigcup_{v\in V}Uv$ and $VU=\bigcup_{v\in V}vU$ and each $vU,Uv$ is open. $\blacksquare$.

Remark: The analogous theorem does not apply if one replaces open by closed (think $\{0\}$and $(0,1)$)

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 $U\subseteq G$ be open. Then, it is called symmetric if $U^{-1}=U$.

Theorem: Let $V$ be a neighborhood of $e$ then there exists some symmetric neighborhood $U$ of $e$ such that $U\subseteq V$.

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

Lemma: Let $E,K\subseteq G$ be arbitrary, then $(E\cap K)^{-1}=E^{-1}\cap K^{-1}$

Proof: Obvious. $\blacksquare$.

Theorem: If $E$ is as above then $E=(E^{-1})^{-1}$.

Proof: Also obvious. $\blacksquare$.

So, using the above and noticing that $e\in V\implies e^{-1}=e\in V^{-1}$ and so $V\cap V^{-1}$ is a neighorhood of $e$ contained within $V$. Noting that $(V\cap V^{-1})^{-1}=V^{-1}\cap(V^{-1})^{-1}=V\cap V^{-1}$ finishes the argument. $\blacksquare$

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 $U$ be a neighborhood of $e$. Then, there exists some neighborhood $V$ of $e$ such that $V^2\subseteq U$ (where $V^2=VV$)

Proof: Notice that the map $\theta:G\times G\to G$ as define above is surjective and thus $\theta(\theta^{-1}(U))=U$. Furthermore, we see that $\displaystyle \vartheta^{-1}(U)=\bigcup_{j\in\mathcal{J}}\left(V_j\times O_j\right)$ where each $V_j,O_j$ are a pair of open subspaces of $G$. Thus,

$\displaystyle U=\theta(\theta^{-1}(U))=\theta\left(\bigcup_{j\in\mathcal{J}}(V_j\times O_j)\right)=\bigcup_{j\in\mathcal{J}}\theta\left(V_j\times O_j\right)=\bigcup_{j\in\mathcal{J}}V_jO_j$

And thus $U$ is the union of the product of two open sets. But, notice that since $(e,e)\overset{\theta}{\longmapsto}ee=e$ that $(e,e)\in\theta^{-1}(U)$ and so $(e,e)\in V_{j_0}\times O_{j_0}\implies e\in V_{j_0}\cap O_{j_0}$ for some $j_0\in \mathcal{J}$. Thus, putting this together we see that $e\in V_{j_0}\cap O_{j_0}$ which is open and

$\displaystyle (V_{j_0}\cap O_{j_0})^2\subseteq V_{j_0}O_{j_0}\subseteq\bigcup_{j\in\mathcal{J}}V_jO_j=U$

from where the conclusion follows. $\blacksquare$

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

Theorem: Define $\theta_n:G^n\to G:(g_1,\cdots,g_n)\mapsto g_1\cdots g_n$. Then, $\theta_n$ is continuous.

Proof: We prove the case for $\theta_3$ and let the reader make the obvious generalization (it can be done by induction). Note that since $G^3\approx G\times (G\times G)$ we must only find a continuous map from $G\times (G\times G)$ to $G$ and the conclusion will follow.

To do this note that

$\iota\times\theta:G\times (G\times G)\to G\times G:\left(g_1,(g_2,g_3\right)\mapsto \left(g_1,\theta(g_2,g_3)\right)$

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

$\theta\circ(\iota\times\theta):G\times(G\times G)\to G$

given by

$\left(g_1,(g_2,g_2)\right)\overset{\iota\times\theta}{\longmapsto}\left(g_1,\theta(g_1,g_3)\right)\overset{\theta}{\longmapsto}\theta(g_1,\theta(g_2,g_3))$

is continuous. But,

$\theta(g_1,\theta(g_2,g_3))=\theta(g_1,g_2g_3)=g_1(g_2g_3)=g_1g_2g_3$

The conclusion follows by previous comment. $\blacksquare$

From this we can prove a nice little theorem

Theorem: Let $U$ be a neighborhood of $e$ and $n\in\mathbb{N}$. Then, there exists some symmetric neighborhood $V$ of $e$ such that $V^n\subseteq U$.

Proof: Using the exact same idea as in the proof when $n=2$ by the continuity and surjectivity of $\theta_n$ we have that

$\displaystyle U=\bigcup_{j\in\mathcal{J}}\left(\left(V_1\right)_j\cdots\left(V_n\right)_j\right)$

Where $(V_1)_j,\cdots,(V_n)_j$ are open. Also, we are able to find some $(V_1)_{j_0},\cdots,(V_n)_{j_0}$ such that $e\in(V_1)_{j_0}\cap\cdots\cap(V_n)_{j_0}$. Thus, noticing that

$\displaystyle \left((V_1)_{j_0}\cap\cdots\cap(V_n)_{j_0}\right)^n\subseteq (V_1)_{j_0}\cdots(V_n)_{j_0}\subseteq\bigcup_{j\in\mathcal{J}}\left((V_1)_j\cdots(V_n)_j\right)=U$

we have the desired neighborhood $V$ of $e$ such that $V^n\subseteq U$. Now, since we proved earlier that every neighborhood of $e$ contains a symmetric neighborhood of $e$ the conclusion follows. $\blacksquare$

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 $V$ be a symmetric neighborhood of $e$. Then, $\displaystyle \bigcup_{n=1}^{\infty}V^n=\Omega\leqslant G$ (I use $\leqslant$ to denote subgroup).

Proof: Clearly $e\in\Omega$.

Now, if $v\in\Omega$ then $v=v_1\cdots v_m$ for some $m\in\mathbb{N}$ and $v_1,\cdots,v_m\in V$. So, $v^{-1}=v_m^{-1}\cdots v_1^{-1}$ and since $V$ is symmetric we see that $v_m^{-1},\cdots,v_1^{-1}\in V$ and thus $v\in V^m\subseteq\Omega$.

Now, if $v=v_1\cdots v_m$ and $v'=v'_1\cdots v'_\ell$ then $vv'=v_1\cdots v_m v'_1\cdots v'_\ell\in V^{m+\ell}\subseteq\Omega$.

The conclusion follows. $\blacksquare$

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 $G$ be a topological group with $\{e\}$ closed. Then, $G$ is $T_1$.

Proof: This follows readily from the fact that $L_g$ is a closed mapping (since it’s a homeomorphism)  for any $g\in G$ and $L_g(\{e\})=\{g\}$. Thus, all the singletons in $G$ are closed. $\blacksquare$.

Even more surprising is the following result.

Theorem: If $G$ is a topological group with $\{e\}$ closed then $G$ is Hausdorff.

Proof: One need merely note that

$G\times G\supseteq \Delta=\left\{(x,y):x=y\right\}=\left\{(x,y):xy^{-1}=e\right\}=\alpha^{-1}(\{e\})$

where $\alpha:G\times G\to G$ was defined in the first theorem. Thus, since $\alpha$ is continuous and $\{e\}$ closed it follows that $\Delta$ is closed and thus from a previous theorem $G$ is Hausdorff. $\blacksquare$.

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

Theorem: Let $G$ be a topological group such that $\{e\}$ is closed, then $G$ is regular.

Proof:

Lemma: Let $G$ be as above and let $C\subseteq G$ be a closed set not containing $e$. Then, there exists disjoint neighborhoods $U,V$ such that $e\in U$ and $C\subseteq V$.

Proof: Clearly $G-C$ is a neighborhood of $e$ and so by previous comment we may find a symmetric neighborhood $V$ of $e$ such that $V^2\subseteq G-C$. We claim that $\overline{V}\subseteq G-C$. To see this we prove that if $c\in C$ then $c\notin D(V)$. To see this merely note that $gV$ is a neighborhood of $g$ and $gV\cap V=\varnothing$ for if not then $v_1=gv_2$ for some $v_1,v_2\in V$ and so $g=v_2v_1^{-1}$. But, since $V$ is symmetric it follows that $v_2,v_1^{-1}\in V$ and thus $g=v_2v_1^{-1}\in V^2\subseteq C-G$ which is a contradiction.

Thus, $V$ and $G-\overline{V}$ are the disjoint neighborhoods containing $e$ and $C$ respectively. $\blacksquare$

Using this we note that if $C\subseteq G$ is closed and $g\notin C$ then $e\notin g^{-1}C$ and so we may find open sets $U,V$ containing $e$ and $g^{-1}C$ respectively. Thus, $g\in gU$ and $C\subseteq gV$ respectively. They are both open and disjoint since $gU\cap gV=L_g(U\cap V)=L_g(\varnothing)=\varnothing$.

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 $\{e\}$ 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 $T_1$ and so we needed the closedness of $\{e\}$.

From this we have the astonishing corollary

Corollary: Let $G$ be a topological group, then:

$\{e\}\text{ is closed }\Leftrightarrow G\text{ is } T_1\Leftrightarrow G\text{ is } T_2\Leftrightarrow G\text{ is }T_3$

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 $\mathbb{R}^\infty$ is that it is separable and has a countable neighborhood base at the identity element.