Extras

Topological Groups (Munkres Chapter 2, §22 Quotient Topology, Supplementary Exercises Section)

The crossover between the studies of topology and group theory gives rise to the concept of topological groups.

Definition. A topological group $G$ is a group that is also a topological space satisfying the $T_1$ axiom, such that the binary operation map $\cdot : G \times G \to G$ given by $(x,y) \mapsto x \cdot y$ and inverse map $^{-1} : G \to G$ given by $x \mapsto x^{-1}$ are continuous maps.

1. Let $H$ denote a group that is also a topological space satisfying the $T_1$ axiom. Show that $H$ is a topological group if and only if the map of $H \times H$ into $H$ sending $(x,y)$ into $x \cdot y^{-1}$ is continuous.

We denote the map of $H \times H$ into $H$ sending $(x,y)$ into $x \cdot y^{-1}$ by $h$.

Suppose that $H$ is a topological group. Since $H$ satisfies the $T_1$ axiom, each of its one-point subsets is closed. This implies that $\{x\} \times H$, where $x \in H$, is closed, since its complement with respect to $H \times H$, which is $(H \backslash \{x\}) \times H$, is open as it is a product of open sets of $H$.

Fix some $x_0 \in H$. Note that since the binary operation map $\cdot$ is continuous, its restriction $\cdot|_{x_0}$ to the subspace $\{x_0\} \times H$ is also continuous. Since the inverse map $^{-1}$ is continuous, the following composition of functions from $\{x_0\} \times H$ to $H$ given by

$$h|_{x_0} := \cdot|_{x_0} \circ i|_{x_0} \circ \phantom{}^{-1} \circ \pi_2$$

where $\pi_2$ is the projection to the second coordinate and $i|_{x_0}$ is the map from $H$ to $\{x_0\} \times H$ given by $y \mapsto (x_0,y)$, is continuous. We have that $i|_{x_0}$ is continuous because an open set in $\{x_0\} \times H$ can be expressed as $\bigcup_\alpha(U_\alpha \times V_\alpha)$, where $U_\alpha \subseteq \{x_0\}$ and $V_\alpha \subseteq H$ are respectively open, and $i|_{x_0}^{-1}\left(\bigcup_{\alpha}(U_\alpha \times V_\alpha)\right)=\bigcup_{\beta}V_\beta$, which is open, where $\beta$ comes from $\alpha$ such that $x_0 \in U_\alpha$.

Now, observe that $H \times H$ is a disjoint union of the sets $\{h\} \times H$ where each $h \in H$ is distinct. It is also vacuously true that $h|_{x_1}(x,y)=h|_{x_2}(x,y)$ for each $(x,y) \in (\{x_1\} \times H) \cap (\{x_2\} \times H)$ where $x_1 \neq x_2$ since $(\{x_1\} \times H) \cap (\{x_2\} \times H)=\emptyset$. This means that we can apply Pasting Lemma to construct a continuous function from $H \times H$ to $H$ given by $(x,y) \mapsto h|_{x_0}(x,y)$ if $(x,y) \in \{x_0\} \times H$, which is precisely $h$ itself.

Alternatively, one can view $h$ as the composition $\cdot \circ \phi$, where $\phi : H \times H \to H \times H$ is defined by $\phi(x,y)=(x,y^{-1})$, which is continuous due to the result in Exercise 18.10, as we know that the identity map $x \mapsto x$ and inverse map $y \mapsto y^{-1}$ are continuous.

Conversely, if the map $h$ of $H \times H$ into $H$ sending $(x,y)$ into $x \cdot y^{-1}$ is continuous, then $h$ is continuous in each variable separately, so if we fix $x$ to be the identity element $e$, then the map $h_e$ defined by $h_e(y)= h(e,y)=e \cdot y^{-1}=y^{-1}$ is continuous, and that is precisely the inverse map. Note as well that the map from $H \times H$ to $H$ defined by $(x,y) \mapsto x \cdot y$ is equivalent to $h(x,h(e,y))$, so it is continuous. Thus, $h$ is a topological group.

2. Show that the following are topological groups:

   (a) $(\mathbb{Z},+)$

   (b) $(\mathbb{R},+)$

   (c) $(\mathbb{R}_+,\cdot)$

   (d) $(S^1,\cdot)$, where we take $S^1$ to be the space of all complex numbers $z$ for which $|z|=1$.

   (e) The general linear group $\mathrm{GL}(n)$, under the operation of matrix multiplication. ($\mathrm{GL}(n)$ is the set of all nonsingular $n \times n$ matrices, topologised by considering it as a subset of Euclidean space of dimension $n^2$ in the natural way.)

(a) The binary operation map in $(\mathbb{Z},+)$ is $+ : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ defined by $(x,y) \mapsto x+y$. The inverse map $^{-1} : \mathbb{Z} \to \mathbb{Z}$ is defined by $x \mapsto -x$.

We assume that $\mathbb{Z}$ equips the subspace topology inherited from the standard topology of $\mathbb{R}$, which turns out to be equivalent to the discrete topology for $\mathbb{Z}$ as every subset of $\mathbb{Z}$ is the intersection of $\mathbb{Z}$ with some open sets of $\mathbb{R}$: if $A \subseteq \mathbb{Z}$, then $A = \mathbb{Z} \cap \bigcup_{a \in A}(a-\varepsilon,a+\varepsilon)$ where $0 < \varepsilon < 1$. This implies that $\mathbb{Z} \times \mathbb{Z}$ is also equipped with the discrete topology.

Therefore, $+$ and $^{-1}$ are both continuous as the inverse image of every open subset of $\mathbb{Z}$ under both of the maps are open, since every subset of $\mathbb{Z}$ and $\mathbb{Z} \times \mathbb{Z}$ is open.

(b) The binary operation map in $(\mathbb{R},+)$ is $+ : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ defined by $(x,y) \mapsto x+y$. The inverse map $^{-1} : \mathbb{R} \to \mathbb{R}$ is defined by $x \mapsto -x$.

If $(a,b)$ and $(c,d)$ are two open intervals in $\mathbb{R}$, then $(a,b)+(c,d) := \{x \in \mathbb{R} : x=y+z \text{ for some } y \in (a,b),\; z \in (c,d)\}$ is also an open interval; in fact, it is $(a+c,b+d)$.

Since any open set in $\mathbb{R}$ is a union of open intervals in $\mathbb{R}$, due to the argument above, it follows that its inverse image under $+$ is a union of products of open intervals in $\mathbb{R} \times \mathbb{R}$, and thus open. Hence, $+$ is continuous.

Now, for an open interval $(a,b)$ in $\mathbb{R}$, its inverse image under $^{-1}$ is $(-b,-a)$, which is also open. Since any open set in $\mathbb{R}$ is a union of open intervals, it follows that its inverse image under $^{-1}$ is open. Thus, $^{-1}$ is continuous.

3. Let $H$ be a subspace of $G$. Show that if $H$ is also a subgroup of $G$, then both $H$ and $\overline{H}$ are topological groups.

4. Let $\alpha$ be an element of $G$. Show that the maps $f_\alpha , g_\alpha : G \to G$ defined by

   $$f_\alpha(x)=\alpha \cdot x \quad \text{and} \quad g_\alpha(x)=x \cdot \alpha$$

   are homeomorphisms of $G$. Conclude that $G$ is a homogeneous space. (This means that for every pair $x, y$ of points of $G$, there exists a homeomorphism of $G$ onto itself that carries $x$ to $y$.)

   (In fact, the maps $f_\alpha$ and $g_\alpha$ are respectively a left and right group action. $G$ being a homogeneous space means that $G$ has a transitive group action.)

5. Let $H$ be a subgroup of $G$. If $x \in G$, define $xH = \{x \cdot h : h \in H\}$; this set is called a left coset of $H$ in $G$. Let $G/H$ denote the collection of left cosets of $H$ in $G$; it is a partition of $G$. Give $G/H$ the quotient topology.

   (a) Show that if $\alpha \in G$, the map $f_\alpha$ of the preceding exercise induces a homeomorphism of $G/H$ carrying $xH$ to $(\alpha \cdot x)H$. Conclude that $G/H$ is a homogeneous space.

   (b) Show that if $H$ is a closed set in the topology of $G$, then one-point sets are clsoed in $G/H$.

   (c) Show that the quotient map $p : G \to G/H$ is open.

   (d) Show that if $H$ is closed in the topology of $G$ and is a normal subgroup of $G$, then $G/H$ is a topological group.

6. The integer $\mathbb{Z}$ are a normal subgroup of $(\mathbb{R},+)$. The quotient $\mathbb{R}/\mathbb{Z}$ is a familiar topological group; what is it?

Galois Theory for Infinite Extensions

note

This section is based on a lecture of an advanced undergraduate abstract algebra course conducted by Prof. Ming-Lun Hsieh in National Taiwan University.

An application of topological groups is the study of Galois theory for infinite extensions.

Let $E/k$ be a Galois extension (in general), for which its degree of extension, $[E : k]$, may be infinite.

It follows that its Galois group, denoted as $\mathrm{Gal}(E/k)$, which is defined by the set of automorphisms of $E/k$, denoted as $\mathrm{Aut}_k(E)$ here, has infinite order.

Thanks to the fundamental theorem of Galois theory, if a Galois extension is finite, then there exists a one-to-one correspondence, which is also named Galois correspondence, between its intermediate field extensions and the subgroups of its Galois group, described as follows.

$$\begin{gather*} \{ K : \text{intermediate extensions in } E/K/k \} & \to & \{ H : \text{ subgroups of } \mathrm{Gal}(E/k) \} \\ K & \mapsto & \mathrm{Gal}(E/K) \\ E^H & \leftarrowtail & H \end{gather*}$$

Here, $E^H := \{ x \in E : \sigma(x)=x \text{ for all } \sigma \in H \}$ represents the $H$-invariant of $E$ (note that $H$ here denotes a subgroup of $\mathrm{Gal}(E/k)$).

Note that under this correspondence, the field $E$ itself necessarily corresponds to the trivial group $\{e\}$, whereas $k$ corresponds to the whole group $\mathrm{Gal}(E/k)$ itself.

However, if the extension is finite, i.e. $\mathrm{Gal}(E/k)$ becomes an infinite group, then in general this one-to-one correspondence may cease to exist. Still, not all hope is lost. We can still obtain a one-to-one correspondence when we restrict the condition of the subgroups a bit to be closed subgroups, but "closed" in what sense? This is where topology comes into play.

Krull Topology: A Topology on Galois Groups

Let $E/k$ be a Galois extension. To understand what is going on with "closed subgroups" just now, we will define a topology on $G=\mathrm{Gal}(E/k)$ such that the open neighbourhood of the identity element $e \in G$ constains $\mathrm{Gal}(E/F)$ where $F/k$ is a finite extension.

Definition (Krull topology). A subset of $U$ of $G$ is open, i.e. $U$ is in the Krull topology, if for all $\sigma \in U$, there exists a finite extension $F/k$ such that $\sigma \cdot \mathrm{Gal}(E/F)$ $\subset U$.

Here, $F/k$ is regarded as a "distance" and $\mathrm{Gal}(E/F)$ is regarded as a "distance-neighbourhood" in the sense that such "distance" is zero if and only if there is no intermediate fields between $F$ and $E$.

Under this topology, we can then say that a subgroup $H$ of $G$ is open (resp. closed) if $H$ is open (resp. closed, i.e. $G \backslash H$ is open).

Just like any other topological space, it is possible here for a subgroup to be both closed and open. A notable example is $\mathrm{Gal}(E/F) \subset G$, where $F/k$ is a finite extension.

Why is $\mathrm{Gal}(E/F)$ both open and closed?

Let $H = \mathrm{Gal}(E/F)$. Since $F$ is a finite field extension of $k$, it follows that $[G : H]$ is finite and $[G : H] = [F : k]$, so $G$ is equal to a finite disjoint union of its (left) cosets, i.e. $G = \bigsqcup_{i=1}^{n} x_i H$ for some $x_i, \ldots, x_n \in G$, so $H = G \backslash \bigsqcup_{i=1}^{n} x_i H$ is closed, since $x_i H$ is open for each $i = 1,\ldots,n$ (why?).

Fundamental Theorem of Galois Theory for Infinite Extensions

Theorem. Let $E/k$ be an infinite Galois extension, then there is a one-to-one correspondence as follows:

$$\begin{gather*} \{ K : \text{intermediate extensions in } E/K/k \} & \to & \{ H : \text{ \textbf{closed} subgroups of } \mathrm{Gal}(E/k) \} \\ K & \mapsto & \mathrm{Gal}(E/K) \\ E^H & \leftarrowtail & H \end{gather*}$$

Proof. Here, we denote $E/K/k$ as a tower of field extensions.

Let $K = \bigcup_i{K_i}$, where $K_i$ are finite extensions of $k$, then $\mathrm{Gal}(E/K)=\bigcap_{i \in I} \mathrm{Gal}(E/K_i)$.

Note that $\mathrm{Gal}(E/K_i)$ are closed and open, as previously proven, thus $\mathrm{Gal}(E/K)$ is closed.

Let $H = \mathrm{Gal}(E/K)$. We want to show that $K = E^H$. By definition, $K \subset E^H$. If $K \neq E^H$, then there exists $\alpha \in E^H \backslash K$ and there exists $\sigma \in \mathrm{Gal}(E/K)$ such that $\sigma(\alpha) \neq \alpha$. Since the set of homomorphisms $\mathrm{Hom}_{k}(E,$ $\overline{k}$$)$ can be mapped surjectively to $\mathrm{Hom}_k(k(\alpha),\overline{k})$, we then have $\sigma \in H$ but $\sigma(\alpha) \neq \alpha$, which leads to a contradiction.

Conversely, let $H$ be a closed subgroup of $G$ and $K=E^H$. We want to show that $H \simeq \mathrm{Gal}(E/K)$. By definition, $H \subset \mathrm{Gal}(E/K)$. Suppose that $H \subsetneq \mathrm{Gal}(E/K)$, then there exists $\sigma \in \mathrm{Gal}(E/k)$ such that $\sigma \notin H$. Since $H$ is closed in $G$, there exists a finite extension $F/k$ such that $\sigma \cdot \mathrm{Gal}(E/F) \cap H = \emptyset$.

By extending $F$ where required, we can assume that $F/k$ is Galois. Note that $\sigma \cdot \mathrm{Gal}(E/F) \cap H = \emptyset$ if and only if $\sigma \notin$ $H \cdot \mathrm{Gal}(E/F)$, if and only if $\sigma|_F$ $\notin H|_F$. Here, $H|_F$ represents the image of $H$ in $\mathrm{Gal}(F/k)$, i.e. $\mathrm{Im}\; H = \{\tau|_F \in \mathrm{Gal}(F/k) : \tau \in H \}$.

Since $\sigma|_F \notin \mathrm{Im}\; H$, there exists some $\alpha \in F^H$ such that $\sigma(\alpha) \neq \alpha$ by Galois Theory for Finite Extensions, but $\alpha \in F^H = F \cap E^H \subset K$, a contradiction.

Further reading

• Marks, S. (2020). Galois representations. (A note on the Tutorial on modular forms course conducted in 2020)

Topological Transcendental Fields

https://www.mdpi.com/2075-1680/11/3/118