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One can show that the group $\text{Ext}^1(\mathbf Q, \mathbf Z)$ (calculated in $Ab$) identifies naturally with $\mathbf A_f/\mathbf Q$, where $\mathbf A_f$ is the additive group of finite adèles. More precisely, the long exact sequence of $\mathbf Z \to \mathbf Q \to \mathbf Q/\mathbf Z$ identifies it with $\mathbf A_f/\mathbf Q$. The group $\mathbf A_f/\mathbf Q$ has a nice topology coming from the fact that $\mathbf Q$ is a discrete subgroup of $\mathbf A_f$. It is a locally compact group.

I am wondering if there is any way to define the topology on $\text{Ext}^1(\mathbf Q, \mathbf Z)$ intrinsically, without identifying it with $\mathbf A_f/\mathbf Q$, or even without looking at any particular exact sequence...

Bruno Joyal
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  • From the long exact sequence $0\to 2\pi iZ\to C\xrightarrow{\exp} C^\times\to0$ and thr fact that $C$ is a divisible abelian group, we have an exact sequence $\hom(Q,C)\to\hom(Q,C^\times)\to Ext^1(Q,Z)\to0$. The two homs have natural topologies: maybe that's the one? – Mariano Suárez-Álvarez May 18 '14 at 05:40
  • If we do it with $0\to Z\to R\to S^1\to0$, again $R$ is injective, and we get $\hom(Q,R)\to\hom(Q,S^1)\to Ext^1(Q,Z)\to 0$; now $\hom(Q,S^1)$ is clearly locally compact, as it is the Pontriagin dual of $Q$. – Mariano Suárez-Álvarez May 18 '14 at 05:46
  • @Mariano: Interesting! I had thought that it would come from the Pontryagin topology on hom(Q, Q/Z), both factors considered as discrete groups. I'd be curious to know if it coincides with the topology you describe... But what I'd really like to know is whether the topology can be defined without looking at any particular exact sequence. Perhaps by calculating Ext's in the category of locally compact groups? – Bruno Joyal May 18 '14 at 05:46

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