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Dec 1, 2020 at 3:35 comment added YCor @QiaochuYuan thanks, I was not careful. I should have said, for $T$ a profinite abelian group.
Dec 1, 2020 at 3:31 comment added Qiaochu Yuan @YCor: really? That doesn't seem right. If, say, $T = S^1$ then what I'd expect (using that Pontryagin duality is an equivalence of categories) is $\text{Hom}(\prod G_p, S^1) \cong \oplus_p \text{Hom}(G_p, S^1)$.
Nov 30, 2020 at 23:13 comment added curious math guy @YCor What is the argument in that case?
Nov 30, 2020 at 23:03 comment added YCor If $T$ is a compact abelian group and Hom refers to continuous group homomorphisms, this is an isomorphism.
Nov 30, 2020 at 22:46 history asked curious math guy CC BY-SA 4.0