Definition 5.30.1. A *topological group* is a group $G$ endowed with a topology such that multiplication $G \times G \to G$, $(x, y) \mapsto xy$ and inverse $G \to G$, $x \mapsto x^{-1}$ are continuous. A *homomorphism of topological groups* is a homomorphism of groups which is continuous.

## 5.30 Topological groups, rings, modules

This is just a short section with definitions and elementary properties.

If $G$ is a topological group and $H \subset G$ is a subgroup, then $H$ with the induced topology is a topological group. If $G$ is a topological group and $G \to H$ is a surjection of groups, then $H$ endowed with the quotient topology is a topological group.

Example 5.30.2. Let $E$ be a set. We can endow the set of self maps $\text{Map}(E, E)$ with the compact open topology, i.e., the topology such that given $f : E \to E$ a fundamental system of neighbourhoods of $f$ is given by the sets $U_ S(f) = \{ f' : E \to E \mid f'|_ S = f|_ S\} $ where $S \subset E$ is finite. With this topology the action

is continuous when $E$ is given the discrete topology. If $X$ is a topological space and $X \times E \to E$ is a continuous map, then the map $X \to \text{Map}(E, E)$ is continuous. In other words, the compact open topology is the coarsest topology such that the “action” map displayed above is continuous. The composition

is continuous as well (as is easily verified using the description of neighbourhoods above). Finally, if $\text{Aut}(E) \subset \text{Map}(E, E)$ is the subset of invertible maps, then the inverse $i : \text{Aut}(E) \to \text{Aut}(E)$, $f \mapsto f^{-1}$ is continuous too. Namely, say $S \subset E$ is finite, then $i^{-1}(U_ S(f^{-1})) = U_{f^{-1}(S)}(f)$. Hence $\text{Aut}(E)$ is a topological group as in Definition 5.30.1.

Lemma 5.30.3. The category of topological groups has limits and limits commute with the forgetful functors to (a) the category of topological spaces and (b) the category of groups.

**Proof.**
It is enough to prove the existence and commutation for products and equalizers, see Categories, Lemma 4.14.11. Let $G_ i$, $i \in I$ be a collection of topological groups. Take the usual product $G = \prod G_ i$ with the product topology. Since $G \times G = \prod (G_ i \times G_ i)$ as a topological space (because products commutes with products in any category), we see that multiplication on $G$ is continuous. Similarly for the inverse map. Let $a, b : G \to H$ be two homomorphisms of topological groups. Then as the equalizer we can simply take the equalizer of $a$ and $b$ as maps of topological spaces, which is the same thing as the equalizer as maps of groups endowed with the induced topology.
$\square$

Lemma 5.30.4. Let $G$ be a topological group. The following are equivalent

$G$ as a topological space is profinite,

$G$ is a limit of a diagram of finite discrete topological groups,

$G$ is a cofiltered limit of finite discrete topological groups.

**Proof.**
We have the corresponding result for topological spaces, see Lemma 5.22.2. Combined with Lemma 5.30.3 we see that it suffices to prove that (1) implies (3).

We first prove that every neighbourhood $E$ of the neutral element $e$ contains an open subgroup. Namely, since $G$ is the cofiltered limit of finite discrete topological spaces (Lemma 5.22.2), we can choose a continuous map $f : G \to T$ to a finite discrete space $T$ such that $f^{-1}(f(\{ e\} )) \subset E$. Consider

This is a subgroup of $G$ and contained in $E$. Thus it suffices to show that $H$ is open. Pick $t \in T$ and set $W = f^{-1}(\{ t\} )$. Observe that $W \subset G$ is open and closed, in particular quasi-compact. For each $w \in W$ there exist open neighbourhoods $e \in U_ w \subset G$ and $w \in U'_ w \subset W$ such that $U_ wU'_ w \subset W$. By quasi-compactness we can find $w_1, \ldots , w_ n$ such that $W = \bigcup U'_{w_ i}$. Then $U_ t = U_{w_1} \cap \ldots \cap U_{w_ n}$ is an open neighbourhood of $e$ such that $f(gw) = t$ for all $w \in W$. Since $T$ is finite we see that $\bigcap _{t \in T} U_ t \subset H$ is an open neighbourhood of $e$. Since $H \subset G$ is a subgroup it follows that $H$ is open.

Suppose that $H \subset G$ is an open subgroup. Since $G$ is quasi-compact we see that the index of $H$ in $G$ is finite. Say $G = Hg_1 \cup \ldots \cup Hg_ n$. Then $N = \bigcap _{i = 1, \ldots , n} g_ iHg_ i^{-1}$ is an open normal subgroup contained in $H$. Since $N$ also has finite index we see that $G \to G/N$ is a surjection to a finite discrete topological group.

Consider the map

We claim that this map is an isomorphism of topological groups. This finishes the proof of the lemma as the limit on the right is cofiltered (the intersection of two open normal subgroups is open and normal). The map is continuous as each $G \to G/N$ is continuous. The map is injective as $G$ is Hausdorff and every neighbourhood of $e$ contains an $N$ by the arguments above. The map is surjective by Lemma 5.12.6. By Lemma 5.17.8 the map is a homeomorphism. $\square$

Definition 5.30.5. A topological group is called a *profinite group* if it satisfies the equivalent conditions of Lemma 5.30.4.

If $G_1 \to G_2 \to G_3 \to \ldots $ is a system of topological groups then the colimit $G = \mathop{\mathrm{colim}}\nolimits G_ n$ as a topological group (Lemma 5.30.6) is in general different from the colimit as a topological space (Lemma 5.29.1) even though these have the same underlying set. See Examples, Section 109.77.

Lemma 5.30.6. The category of topological groups has colimits and colimits commute with the forgetful functor to the category of groups.

**Proof.**
We will use the argument of Categories, Remark 4.25.2 to prove existence of colimits. Namely, suppose that $\mathcal{I} \to \textit{Top}$, $i \mapsto G_ i$ is a functor into the category $\textit{TopGroup}$ of topological groups. Then we can consider

This functor commutes with limits. Moreover, given any topological group $H$ and an element $(\varphi _ i : G_ i \to H)$ of $F(H)$, there is a subgroup $H' \subset H$ of cardinality at most $|\coprod G_ i|$ (coproduct in the category of groups, i.e., the free product on the $G_ i$) such that the morphisms $\varphi _ i$ map into $H'$. Namely, we can take the induced topology on the subgroup generated by the images of the $\varphi _ i$. Thus it is clear that the hypotheses of Categories, Lemma 4.25.1 are satisfied and we find a topological group $G$ representing the functor $F$, which precisely means that $G$ is the colimit of the diagram $i \mapsto G_ i$.

To see the statement on commutation with the forgetful functor to groups we will use Categories, Lemma 4.24.5. Indeed, the forgetful functor has a right adjoint, namely the functor which assigns to a group the corresponding chaotic (or indiscrete) topological group. $\square$

Definition 5.30.7. A *topological ring* is a ring $R$ endowed with a topology such that addition $R \times R \to R$, $(x, y) \mapsto x + y$ and multiplication $R \times R \to R$, $(x, y) \mapsto xy$ are continuous. A *homomorphism of topological rings* is a homomorphism of rings which is continuous.

In the Stacks project rings are commutative with $1$. If $R$ is a topological ring, then $(R, +)$ is a topological group since $x \mapsto -x$ is continuous. If $R$ is a topological ring and $R' \subset R$ is a subring, then $R'$ with the induced topology is a topological ring. If $R$ is a topological ring and $R \to R'$ is a surjection of rings, then $R'$ endowed with the quotient topology is a topological ring.

Lemma 5.30.8. The category of topological rings has limits and limits commute with the forgetful functors to (a) the category of topological spaces and (b) the category of rings.

**Proof.**
It is enough to prove the existence and commutation for products and equalizers, see Categories, Lemma 4.14.11. Let $R_ i$, $i \in I$ be a collection of topological rings. Take the usual product $R = \prod R_ i$ with the product topology. Since $R \times R = \prod (R_ i \times R_ i)$ as a topological space (because products commutes with products in any category), we see that addition and multiplication on $R$ are continuous. Let $a, b : R \to R'$ be two homomorphisms of topological rings. Then as the equalizer we can simply take the equalizer of $a$ and $b$ as maps of topological spaces, which is the same thing as the equalizer as maps of rings endowed with the induced topology.
$\square$

Lemma 5.30.9. The category of topological rings has colimits and colimits commute with the forgetful functor to the category of rings.

**Proof.**
The exact same argument as used in the proof of Lemma 5.30.6 shows existence of colimits. To see the statement on commutation with the forgetful functor to rings we will use Categories, Lemma 4.24.5. Indeed, the forgetful functor has a right adjoint, namely the functor which assigns to a ring the corresponding chaotic (or indiscrete) topological ring.
$\square$

Definition 5.30.10. Let $R$ be a topological ring. A *topological module* is an $R$-module $M$ endowed with a topology such that addition $M \times M \to M$ and scalar multiplication $R \times M \to M$ are continuous. A *homomorphism of topological modules* is a homomorphism of modules which is continuous.

If $R$ is a topological ring and $M$ is a topological module, then $(M, +)$ is a topological group since $x \mapsto -x$ is continuous. If $R$ is a topological ring, $M$ is a topological module and $M' \subset M$ is a submodule, then $M'$ with the induced topology is a topological module. If $R$ is a topological ring, $M$ is a topological module, and $M \to M'$ is a surjection of modules, then $M'$ endowed with the quotient topology is a topological module.

Lemma 5.30.11. Let $R$ be a topological ring. The category of topological modules over $R$ has limits and limits commute with the forgetful functors to (a) the category of topological spaces and (b) the category of $R$-modules.

**Proof.**
It is enough to prove the existence and commutation for products and equalizers, see Categories, Lemma 4.14.11. Let $M_ i$, $i \in I$ be a collection of topological modules over $R$. Take the usual product $M = \prod M_ i$ with the product topology. Since $M \times M = \prod (M_ i \times M_ i)$ as a topological space (because products commutes with products in any category), we see that addition on $M$ is continuous. Similarly for multiplication $R \times M \to M$. Let $a, b : M \to M'$ be two homomorphisms of topological modules over $R$. Then as the equalizer we can simply take the equalizer of $a$ and $b$ as maps of topological spaces, which is the same thing as the equalizer as maps of modules endowed with the induced topology.
$\square$

Lemma 5.30.12. Let $R$ be a topological ring. The category of topological modules over $R$ has colimits and colimits commute with the forgetful functor to the category of modules over $R$.

**Proof.**
The exact same argument as used in the proof of Lemma 5.30.6 shows existence of colimits. To see the statement on commutation with the forgetful functor to $R$-modules we will use Categories, Lemma 4.24.5. Indeed, the forgetful functor has a right adjoint, namely the functor which assigns to a module the corresponding chaotic (or indiscrete) topological module.
$\square$

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