Extras

Topological Groups (Munkres Chapter 2, Supplementary Exercises Section)

The crossover between the studies of topology and abstract algebra 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.

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.)

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.

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