# Abstract Nonsense

## Topological Groups (Space of Cosets)

In this post we discuss the canonical way of combining the quotient topology and the quotient group together in one unified idea. But, we start more generally.

Remark: If you remember from algebra there were both left and right cosets and they weren’t (in non-abelian cases) generally equal. Thus, to speak of cosets without mention of left or right is technically incorrect. That said, for ease in this post coset will always refer to left coset. Everything is word-for-word the same for right cosets.

Recall from algebra that given a group $G$ and some $H\leqslant G$ the relation $a\sim b\Leftrightarrow a=bh,\text{ }h\in H$ was in fact an equivalence relation. Then, remember that the equivalence classes of this relation were all of the form $gH,\text{ }g\in G$ and that the set of all these equivalence classes were called the cosets of $H$ (denoted $G/H$) with each individual $gH$ being a coset. Remember then that if it happened that $H\unlhd G$ there was a canonical way to make $G/H$ into a group by defining

$aH\star bH=(ab)H$

With this group structure we called $G/H$ the quotient group of $G$ by $H$ and we were then presented with the canonical homomorphism

$\eta:G\to G/H:g\mapsto gH$

In fact though, normality was absolutely crucial to define the group structure in that way thus one rarely talked about $G/H$ outside of the context when it was the quotient group. We don’t take that view though, for regardless of whether $H$ is normal or not the map $\eta$ as above is still a surjective map and thus let’s us imbue upon it the quotient topology determined by $\eta$. If that’s true we will (keeping with previous notation) write

$\eta:G\leadsto G/H:g\mapsto gH$

with $\eta$ (for now) reserved for the map sending each element to it’s coset.

We first show that the canonical quotient map ($\eta$) is, in fact, open.

Theorem: Let $\eta:G\leadsto G/H$ be the canonical quotient map. Then, $\eta$ is open.

Proof: Recall that $\eta(U)$ is open if and only if $\eta^{-1}(\eta(U))$ is open. Notice that firstly that

$\displaystyle \eta(U)=\eta\left(\bigcup_{u\in U}\{u\}\right)=\bigcup_{u\in U}\eta(\{u\})=\bigcup_{u\in U}\left\{uH\right\}$

We claim that:

Lemma: $\eta^{-1}(\eta(U))=UH$

Proof: Let $g\in \eta^{-1}(\eta(U))$ then $\eta(g)=gH\in\eta(U)$ which implies that $gH=uH$ for some $u\in U$. So, $gH\in \eta(U)$ and so by the above this implies that $gH=uH$ for some $u\in U$. So, since $g\in gH$ and $gH=uH$ it follows that $g\in uH$ and so $g=uh$ for some $h\in H$. Thus, $g\in UH$.

Conversely, let $uh\in UH$ then $\eta(uh)=uhH$ but it is not to hard to see that since $H$ is a group and $h\in H$ that $uhH=uH$. But, $uH\in\eta(U)$ and so $uh\in \eta^{-1}(\eta(U))$. The conclusion follows. $\blacksquare$

From this, we see that if $U\subseteq G$ is open then $\eta(U)$ is open since $\eta^{-1}(\eta(U))=UH$ and we proved earlier that the product of an open set with any other set is open. The conclusion follows. $\blacksquare$

Now the obvious question is if $H$ happens to be normal is the space of cosets $G/H$ a topological group? The answer is yes as the following shows:

Theorem: Let $N\unlhd G$ then $\left(G/N,\star,\mathfrak{J}\right)$ is a topological group with $aN\star bN=abN$ and $\mathfrak{J}$ being the quotient topology determined by the canonical homomorphism.

Proof: It is an elementary fact that $G/N$ with this operation is, in fact, a group with $\left(gN\right)^{-1}=g^{-1}N$. Thus, it remains to show that the maps

$\vartheta:G/N\times G/N\to G/N:\left(gN,hN\right)\mapsto ghN$

and

$\varphi:G/N\to G/N:g\mapsto g^{-1}N$

are continuous. But, we proved in our first post pertaining to topological groups that this is equivalent to showing the map

$\overset{\sim}{\alpha}:G/N\to G/N:\left(gN,hN\right)\mapsto gh^{-1}N$

is continuous, thus we do this.

To do this we first note that since $\eta:G\leadsto G/N$ is open then so is (as was proven way back in the day) $\eta\times\eta:G\times G\leadsto G/N\times G/N$ and thus as the $\leadsto$ indicates $\eta\times \eta$  is a quotient map. Thus, to show that

$\overset{\sim}{\alpha}:G/N\times G/N\to G/N$

is continuous we need only show (by the characteristic property of quotient maps) that

$\overset{\sim}{\alpha}\circ(\eta\times\eta):G\times G\to G/N$

is continuous. But, we claim that the following diagram commutes



From where the result will follow immediately since $\eta\circ\alpha:G\times G\to G/N$ is the composition of continuous maps. So, to do this we note that

$\overset{\sim}{\alpha}(\eta\times\eta)(g,h)=\overset{\sim}{\alpha}\left(gN,hN\right)=gh^{-1}N$

and

$\eta\left(\alpha\left(g,h\right)\right)=\eta\left(gh^{-1}\right)=gh^{-1}N$

from where the conclusion follows. $\blacksquare$

We are now prepared to prove the main theorem of this post, one which I like to call the first topological group isomorphism theorem. But, before get into it we need to discuss some terminology.

You will undoubtedly remember from your algebra days that if $\pi:G\to G'$ is an algebraic epimorphism then $\ker\pi\unlhd G$ and $G/\ker\pi\cong G'$. Well, we also know from the above lemma that if we give $G/\ker\pi$ the quotient topology formed by the canonical homomorphism then it’s a topological group. So, the question is if $G,G'$ are topological groups and $\pi:G\to G'$ is a topological group epimorphism (a continuous algebraic epimorphism… abbreviated T.G. epimorphism) then are $G/\ker\pi\overset{\text{T.G}}{\cong}G'$ where $\overset{\text{T.G}}{\cong}$ means that there is a T.G. isomorphism (both an algebraic isomorphism and a homeomorphism) between $G\ker\pi$ and $G'$ (if this is true it is said that $G\ker\pi$ and $G'$ are T.G. isomorphic)? The answer is yes if we also stipulate that $\pi$ is a quotient map.

Theorem: Let $\pi:G\to G'$ be a  T.G. epimorphism which is also a quotient map. Then, $G/\ker\pi\overset{\text{T.G}}{\cong}G'$

Proof: Let

$\varphi:G/\ker\pi\to G':g\ker\pi\mapsto\pi(g)$

This map is well-defined since if $g_1\ker\pi=g_2\ker\pi$ then $g_1h_1=g_2h_2$ for some $h_1,h_2\in\ker\pi$ and so

$\varphi\left(g_2\ker\pi\right)=\varphi\left(g_1h_2h_1^{-1}\ker\pi\right)=\pi(g_1h_2h_1^{-1})=\pi(g_1)=\varphi\left(g_1\ker\pi\right)$

It is also clearly a bijection (this is exactly the same as in the first isomorphism theorem) and it’s a homomorphism because

$\varphi\left(g_1g_2\ker\pi\right)=\pi(g_1g_2)=\pi(g_1)\pi(g_2)=\varphi(g_1\ker\pi)\varphi(g_2\ker\pi)$

Thus, it remains to show that it’s bicontinuous. We first claim that the diagram



commutes. (the $\iota$ is just the identity map, as embarrassing as this is to say I couldn’t figure out how to make a triangular diagram look nice. This diagram would mean the same thing as if you collapsed the upper right vertex into the upper left and were left with a triangle. This will most likely happen every time I need to use a triangular diagram) To see this we merely note that

$\varphi(\eta(g))=\varphi\left(g\ker\pi\right)=\pi(g)$

Thus by the characteristic property of quotient maps it follows that $\varphi$ is continuous. Now, since $\pi:G\to G'$ is is a quotient map the next step is to realize that



commutes since

$\varphi^{-1}(\pi(g))=g\ker\pi=\eta(g)$

We may conclude that $\varphi^{-1}$ is continuous. The theorem follows. $\blacksquare$

Remark: Notice that if $G/N$ is compact and $G'$ Hausdorff then we don’t need the openess of $\pi$.

Now that we’ve done that we can prove some more trivial theorems.

Theorem: Let $N\unlhd G$ then, $G/N$ is regular if and only if $N$ is closed.

Proof: As proven earlier it suffices to show that it’s $T_1$. To do this merely need to show that $\eta^{-1}(gN)$ is closed. But, as we’ve proven earlier this is merely $gN$ (since it’s $\eta^{-1}(\eta(g))$) and thus a translation of a closed set and thus closed.

Conversely, if $G/N$ is regular than it’s $T_1$ and so $N$ is closed in $G/N$. Thus, so is $\eta^{-1}(N)=N$.

The conclusion follows. $\blacksquare$

This is actually a weaker version of a theorem which says that if $H\leqslant G$ is closed then $G/H$ is closed, but we omit the proof of this here.

Theorem: Let $G$ be first or second countable and $N\unlhd G$. Then, $G/N$ is first or second countable.

Proof: This follows since $G/N$ is the open map image of $G$ under $\eta$. $\blacksquare$

With this we can prove a cool equivalence

Theorem: $\mathbb{R}/\mathbb{Z}\overset{\text{T.G.}}{\cong}\mathbb{S}^1$ where $\mathbb{R}$ has the usual topology and addition and $\mathbb{S}^1$ has the usual topology and complex multiplication.

Proof: It is trivial to check that $\gamma:\mathbb{R}\to\mathbb{S}^1:\theta\mapsto e^{2\pi i\theta}$ is an open T.G. epimorphism and so from the above $\mathbb{R}/\ker\gamma\overset{\text{T.G}}{\cong}\mathbb{S}^1$. But, $\ker\gamma=\left\{x\in\mathbb{R}:\gamma(x)=e_{\mathbb{S}^1}=1\right\}$ and thus solving $e^{2\pi i x}=1$ we see this happens precisely when $x\in\mathbb{Z}$. Thus, $\ker\gamma=\mathbb{Z}$ and so the conclusion follows. $\blacksquare$

Now, from this we can prove something even more surprising. But first, we need to define a certain topological group. Note, I spend enough time on these posts and I don’t have that much more to spare. That said, what I am about to discuss is a crude and degrading introduction to a very important topological group which I will come back to without doubt. Consequently, I omit some of the laborious details for any interested reader to prove.

Theorem: Let $\text{SO}(2)\subseteq\mathfrak{M}^{2\times 2}(\mathbb{R})$ be the set of all two-by-two rotation matrices under matrix multiplication. In other words,

$\displaystyle \text{SO}(2)=\left\{\begin{bmatrix} \cos(\theta)& -\sin(\theta) \\ \sin(\theta)& \cos(\theta) \end{bmatrix}:\theta\in\left[0,2\pi\right)\right\}$

We can topologize $\text{SO}(2)$ by considering it as a subspace of $\mathbb{R}^4$ (this is a routine proof). Thus, with this topology and group structure $\text{SO}(2)$ is a topological group.

Proof: It remains to show that

$\alpha:\text{SO}(2)\times\text{SO}(2)\to\text{SO}(2)$

given by

$\displaystyle \alpha:\left(\begin{bmatrix} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{bmatrix},\begin{bmatrix} \cos(y) & -\sin(y) \\ \sin(y) & \cos(y)\end{bmatrix}\right)\mapsto \begin{bmatrix}\cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{bmatrix}\begin{bmatrix}\cos(y) & -\sin(y) \\ \sin(y) & \cos(y) \end{bmatrix}^{-1}$

is continuous. But, it is easy to verify that

$\begin{bmatrix}\cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{bmatrix}\begin{bmatrix}\cos(y) & -\sin(y) \\ \sin(y) & \cos(y) \end{bmatrix}=\begin{bmatrix}\cos(x+y) & -\sin(x+y) \\ \sin(x+y) & \cos(x+y)\end{bmatrix}$

And that these matrices are orthogonal (i.e. $A^T=A^{-1}$) and so

$\displaystyle \alpha:\left(\begin{bmatrix} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{bmatrix},\begin{bmatrix} \cos(y) & -\sin(y) \\ \sin(y) & \cos(y)\end{bmatrix}\right)\mapsto \begin{bmatrix}\cos(x+y) & \sin(x+y) \\ -\sin(x+y) & \cos(x+y)\end{bmatrix}$

But, remembering that $\text{SO}(2)$ is homeomorphic to a subspace of $\mathbb{R}^4$ and the corresponding map

$\gamma:\mathbb{R}^4\times\mathbb{R}^4\to\mathbb{R}^4$

is trivially continuous. The conclusion follows. $\blacksquare$

From this  we can construct the following T.G. isomorphism

Theorem: $\text{SO}(2)\overset{\text{T.G}}{\cong}\mathbb{S}^1$.

Proof: Each $z\in\mathbb{S}^1$ is of the form $e^{ix}$ for some $x\in[0,2\pi)$. Then, define

$\gamma:S^1\to\text{SO}(2):e^{ix}\mapsto\begin{bmatrix}\cos(x) & -\sin(x) \\ \sin(x) & \cos(x)\end{bmatrix}$

This is trivially continuous since the mapping

$\overset{\sim}{\gamma}:K\to\mathbb{R}^4:(\cos(x),\sin(x))\mapsto (\cos(x),-\sin(x),\sin(x),\cos(x))$

is continuous (and the fact that the LHS of the mapping is homeomorphic to $\mathbb{S}^1$ and $\text{SO}(2)$ is homeomorphic to a subspace of the RHS) where $K=\left\{(\cos(x),\sin(x)):x\in[0,2\pi)\right\}$. It’s injective because $e^{ix}$  injective on $x\in [0,2\pi)$ and surjectivity is obvious. Thus, $\gamma$ is a continuous bijective mapping of a compact (since $\mathbb{S}^1\subseteq\mathbb{R}^2$ is closed and bounded) space into a Hausdorff space, thus by previous theorem automatically a homeomorphism.

So, it remains to show that it’s a homomorphism but this follows from an earlier observation

$\gamma\left(e^{ix}e^{iy}\right)=\gamma\left(e^{i(x+y)}\right)=R_{x+y}=R_xR_y=\gamma\left(e^{ix}\right)\gamma\left(e^{iy}\right)$

Where

$R_x=\begin{bmatrix}\cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{bmatrix}$

and the others are defined similarly.

By previous comment the conclusion follows. $\blacksquare$

Corollary: $\mathbb{R}/\mathbb{Z}\overset{\text{T.G.}}{\cong}\text{SO}(2)$

That will wrap it up for this post, there is much more that can be done with these subgroups though. I will probably take a post or two to just do problems when I’m finished with what I want to say.