# Make a short exact sequence of abstract groups into a short exact sequence of topological groups (motivated by the Weil Group)

Let $$1 \to H_1 \to G \to H_2 \to 1$$ be a short exact sequence of abstract groups.

Question: If $$H_1$$, $$H_2$$ have fixed topologies, can we endow $$G$$ with a topology such that the sequence above becomes a short exact sequence of topological groups? If yes, is this topology then uniquely determined?

Motivation: I want to understand the topology of the Weil group $$W_K$$ which is a dense subgroup of the absolute Galois group $$G_K$$ (which has this weird profinite topology I find hard to get). However, we have a short exact sequence of abstract groups

$$1 \to I_K \to W_K \to \langle \operatorname{Frob}_k \rangle \to 1$$ where $$I_K$$ denotes the inertia subgroup of $$G_K$$ and $$\operatorname{Frob}_k : x \mapsto x^{|k|}$$ is the Frobenius element of the absolute Galois group $$G_k$$ of the residue field $$k$$ of $$K$$.

If I understand it correctly, this sequence is not a short exact sequence of topological groups if we give $$W_K$$ the subspace topology of $$G_K$$. However, if we give $$I_K$$ and $$\langle \operatorname{Frob}_k \rangle$$ the subspace topologies of $$G_K$$ and $$G_k$$ respectively, we can give $$W_K$$ a unique topology such that this sequence becomes a short exact sequence of topological groups.

I heard that this causes $$W_K$$ to have a finer topology than the usual subspace topology, $$I_K$$ to be open and the maximal compact subgroup of $$W_K$$ etc. but I do not really understand that.

Could you please explain this to me? Thank you!

• You don't want $\langle \operatorname{Frob}_k \rangle$ to have the suspace topology from $G_k$, but instead the discrete topology. – Lukas Heger Jan 6 '19 at 21:32

Let $$i:H_1\to G$$ be the monomorphism, and $$p:G\to H_2$$ be the epimorphism in this short exact sequence. Exactness means that $$N:=i(H_1)=ker(p)\unlhd G$$. If I got it right $$H_1$$ and $$H_2$$ are topological groups, and you want to give a topology to $$G$$ so that $$G$$ becomes a topological group, and the mappings $$i,p$$ are continuous.
Unless I made a silly mistake this can be achieved by giving $$G$$ the topology defined by letting the cosets of $$N$$ to form a basis. In other words, the open sets of $$G$$ are the arbitrary unions of cosets of $$N$$. This is trivially a topology $$\tau$$. Furthermore:
1. $$p:(G,\tau)\to H_2$$ is continuous for much the same reason that any mapping from a discrete space to any topological space is continuous. The preimage of any subset of $$H_2$$ is a union of cosets of $$N$$.
2. $$i:H_1\to(G,\tau)$$ is continuous for much the same reason that any map to a space with a trivial topology is continuous. The preimage of any open subset $$\in\tau$$ is either all of $$H_1$$ or the empty set.
3. $$(G,\tau)$$ is a topological group. For if $$x,y\in G$$ are arbitrary, then the product map $$m:G\times G\to G, m(x,y)=xy$$ satisfies $$m(xN,yN)\subseteq xyN$$. Any open set containing $$xy$$ contains the coset $$xyN$$, so continuity of $$m$$ follows. Continuity of the inverse mapping is verified in the same way.
Seems to me that you really wanted to ask a slightly different question. I'm afraid I cannot suggest one. Anwyway, it seems morally certain to me that the topology $$G$$ needs to have to satisfy items 1 to 3 is not unique at all.