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I'm reading Tate's thesis, and I'm stuck on one of his proofs. Let $k$ denote the completion of an algebraic number field at a prime divisor $p$. Tate claims that the additive group $k^+$ of $k$ is isomorphic, both topologically and algebraically, to its character group. Let $\xi \mapsto \chi(\xi)$ be a non-trivial character of $k^+$ and define a map from $k^+$ to $\widehat{k^+}$ by $\eta \mapsto \chi(\eta \xi)$. He proves that the map is injective and bicontinuous, but I don't understand why the map is surjective. In the proof he first shows that the image of the map is dense in $\widehat{k^+}$. By bicontinuity, the image is locally compact. Then he concludes, "Local compactness implies completeness and therefore closure..." (p. 309 of Cassels & Frohlich).

Could someone explain that sentence? How does local compactness imply completeness, and what does completeness even mean in the case of a character group? Is there a metric on the character group? I've looked in several books (e.g. Lang's Algebraic Number Theory, Ramakrishnan & Valenza) and could not find an explanation. I'd appreciate any help.

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    $\begingroup$ Not an answer to your question, but for the case of just $\mathbb{Q}_p$ here is a nice alternative, elementary explanation: jstor.org/stable/2319000 Best :) $\endgroup$ – Alex Youcis Sep 15 '13 at 1:58
  • $\begingroup$ Thanks. I'm using that as a back-up reference. I like that it doesn't use Pontryagin duality. $\endgroup$ – Not a grad student Sep 15 '13 at 13:09
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Any topological abelian group naturally carries a uniform space structure, and there is a notion of completion for uniform spaces. It is not too hard to show that a locally compact (Hausdorff) abelian group is complete in this uniformity; this was discussed in another MSE post in somewhat broken English, but the idea should be clear: if $x_\alpha$ is a Cauchy net, then the differences $x_\alpha - x_\beta$ lie within a compact neighborhood of the identity for all $\alpha, \beta$ sufficiently large. Hold such $\beta$ fixed and let such $\alpha$ vary; by compactness, the limit

$$\lim_\alpha x_\alpha - x_\beta$$

exists. If this limit is $x$, then $x+x_\beta$ is the limit of the original Cauchy net.

This applies in particular to the Pontryagin dual $\widehat{k^+}$: it is locally compact, hence complete as a uniform space. So is the image of $k$ under the map $i: k^+ \to \widehat{k^+}$ induced by the pairing on $k^+$, according to Tate. In other words, $i(k^+)$ is complete as a subspace of $\widehat{k^+}$, and since complete subspaces of complete Hausdorff spaces are closed, this completes Tate's argument.

I actually hadn't known this argument; it's nice. But I think the surjectivity you were asking about can also be deduced by other means. I'll be sketchy here. Let $\mathcal{O} \hookrightarrow k^+$ be the ring of integers in the completion, viewed as an additive subgroup. We have an exact sequence

$$0 \to \mathcal{O} \to k^+ \to k^+/\mathcal{O} \to 0$$

where the quotient is a discrete group (for example, in the paradigmatic case where $k^+ = \mathbb{Q}_p$ and $\mathcal{O} = \mathbb{Z}_p$, the quotient will be a Prüfer group $\mathbb{Z}[1/p]/\mathbb{Z}$). Now apply the Pontryagin dual functor to this exact sequence. It turns out that the Pontryagin dual of the discrete group $k^+/\mathcal{O}$ is isomorphic to the compact group $\mathcal{O}$ (and so the dual of $\mathcal{O}$ will be $k^+/\mathcal{O}$ again). Thus the dual sequence takes the form

$$0 \to \mathcal{O} \to \widehat{k^+} \to k^+/\mathcal{O} \to 0.$$

There is a map from the first short exact sequences to the second exact sequence that is induced from the pairing on $k^+$, that restricts to the identity maps on the ends. One concludes that the map in the middle $k^+ \to \widehat{k^+}$ is an isomorphism of topological abelian groups, by a topological version of the short five lemma (see remark 4 here).

To answer another question in the post: if the underlying space is Hausdorff and the uniformity admits a countable sub-base -- which turns out to be the case here -- then the uniform space is in fact metrizable, i.e., the uniformity is indeed induced from a metric.

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  • $\begingroup$ Hi! I was about to ask another very similar question, but I guess you could just explain it in a comment here since this answer was already so good. Why is it so, as Tate argued, that we get that the map $\phi: k^+ \hookrightarrow \widehat {k^+}$ has dense image just because $X(\eta \xi)=1 \quad \forall \eta \implies \xi = 0$? My feeling is that we can reduce to showing that $\phi$ has dense image near $1$ but I haven't been able to work out all the details. $\endgroup$ – Rodrigo Mar 16 '14 at 22:40
  • $\begingroup$ Well, I actually did write up a question. I'd be really glad if you could answer it (either here or there) math.stackexchange.com/questions/715038/…. $\endgroup$ – Rodrigo Mar 17 '14 at 2:17
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There is a proof of this fact in Bushnell, Henniart "The local Langlands conjecture for GL(2)", see 1.7, Proposition. In fact, it works for any nonarchimedean local field (not necessarily a $p$-adic field). Maybe the explanation there would help.

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  • $\begingroup$ I took a look at that, but unfortunately it doesn't address completeness in $\hat{k}$. $\endgroup$ – Not a grad student Sep 14 '13 at 16:36

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