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!

  • 3
    $\begingroup$ You don't want $\langle \operatorname{Frob}_k \rangle$ to have the suspace topology from $G_k$, but instead the discrete topology. $\endgroup$ – Lukas Heger Jan 6 '19 at 21:32

Not saying anything about the Weil group at this point. Trying to settle the first question instead with the following construction that smacks of cheating.

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.


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