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An important result in Harmonic analysis states that the Pontryagin dual $\widehat{A}$ of an locally compact abelian group $A$, defined as $\hom_{\mathbb{Z}}(A,T)$ where the hom is taken over continuous group homomorphisms and $T=\mathbb{R}/\mathbb{Z}$, is again locally compact.

Here and always I will always say "(locally) compact" when I mean "(locally) compact Hausdoff".

Recently I have been thinking a lot about how this sort of sitatuion generalizes when we examine morphisms of modules over rings other than $\mathbb{Z}$. In particular, I have been curious about the following:

If $R$ is a compact Discrete Valuation Ring does that imply that the dual $\widehat{M}$ of a locally compact module $M$ will be locally compact, where I define $\widehat{M}=\hom_R(M,T)$ with hom taken over continuous module homomorphisms and $T=\mathrm{Frac}(R)/R=K/R$?

The standard proofs for abelian groups use the theory of integration heavily and I'm not figuring out how to generalize that theory to the case of modules over other rings.

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  • $\begingroup$ Just to check that I'm not missing anything, compactness for a DVR is equivalent to completeness and the residue field being finite, right? $\endgroup$ Aug 2, 2022 at 3:18
  • $\begingroup$ @EricWofsey Yes, that is exactly correct $\endgroup$
    – Milo Moses
    Aug 2, 2022 at 4:08
  • $\begingroup$ Is it true that a locally compact module module over a compact DVR must actually have a neighborhood base of compact sets that are submodules? This seems plausible, and I can prove the dual is always locally compact if it is true. $\endgroup$ Aug 2, 2022 at 19:17
  • $\begingroup$ @EricWofsey that’s an interesting question. I found a link to a computer algebra system which seems to claim that topological modules over non-archemedian rings (by which I assume they mean local complete) have a basis of open sub modules near the identity: leanprover-community.github.io/mathlib_docs/topology/algebra/…. Taking the closures of these sub modules will again yield sub modules, which will eventually be compact. $\endgroup$
    – Milo Moses
    Aug 3, 2022 at 0:48
  • $\begingroup$ Also, I found the following paper which discusses the subject: link.springer.com/content/pdf/10.1007%2FBF02018165. I am trying to decipher it, and I should be able to say whether it gives you the result you want soon. $\endgroup$
    – Milo Moses
    Aug 3, 2022 at 0:59

2 Answers 2

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Here is a partial answer. Suppose that $M$ is not just locally compact, but has a neighborhood base of $0$ consisting of compact submodules. Then I claim $\widehat{M}$ is locally compact. To prove this, let $f\in\widehat{M}$. Since $T$ is discrete, $\ker(f)$ is open, so it contains some compact open submodule $K\subseteq M$. The set $U$ of continuous homomorphisms $M\to T$ that vanish on $K$ is then open in the compact-open topology. I claim $U$ is compact, so it is a compact neighborhood of $f$ in $\widehat{M}$.

To prove this, note first that $U$ can naturally be identified with $\widehat{M/K}$. Since $K$ is open, $M/K$ is discrete, so the compact-open topology on $\widehat{M/K}$ is just the product topology. Also, $M/K$ must be a torsion $R$-module (if $x\in M/K$ were a non-torsion element then continuity of scalar multiplication would imply $Rx$ is not discrete). So $\widehat{M/K}$ is actually a subspace of the product $\prod_{x\in M/K}\pi^{-n_x}R/R$, where $\pi$ is a uniformizer and $n_x$ is such that $\pi^{n_x}x=0$. Since $R$ is compact, so is $\pi^{-n_x}R/R\cong R/(\pi^{n_x})$, and thus so is the product $\prod_{x\in M/K}\pi^{-n_x}R/R$. The subset of this product consisting of the homomorphisms $M/K\to T$ is closed, and thus $\widehat{M/K}$ is compact.

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  • $\begingroup$ It seems like you are asserting the existence of a submodule $K$ that is simultaneously compact and open; I cannot see how that is justified from your givens. To me, what follows from the definitions is an inclusion of submodules $U\leq K\leq \ker(f)$ with $U$ open and $K$ compact. At this point, the definition of the compact open topology gives us that $\widehat{M/K}$ is open and your argument in the second paragraph gives us that $\widehat{M/U}$ is compact, yielding the desired conclusion of local compactness. $\endgroup$
    – Milo Moses
    Aug 6, 2022 at 18:16
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    $\begingroup$ A submodule that contains a neighborhood of $0$ is automatically open, since it then contains a neighborhood of each of its points. $\endgroup$ Aug 6, 2022 at 23:43
  • $\begingroup$ Of course! I totally forgot about that fact. Using this I can deduce a full answer; I posted it below. $\endgroup$
    – Milo Moses
    Aug 7, 2022 at 20:28
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Here is a solution to the problem, based of Eric Wofsey's partial response. Given any $f\in \widehat{M}$, $\ker(f)$ is clopen and so by local compactness there is an inclusion of subsets $U\leq K \leq \ker(f)$ with $U$ open and $K$ compact. Letting $V(A,B)$ denote the set consisting of maps in $\widehat{M}$ mapping $A$ into $B$, we obtain an inclusion of subsets

$$V(\ker(f),0)\leq V(K,0)\leq V(U,0).$$

Since $f\in V(\ker(f),0)$, we have that $V(K,0)$ is an open neighborhood of $f$. It remains to show that $V(U,0)$ is compact. To do this, we note that a morphism acts by $0$ on $U$ if and only if it acts by $0$ on the (open) submodule $\tilde{U}$ generated by $U$. Hence, $V(U,0)=V(\tilde{U},0)=\widehat{M/\tilde{U}}$ so by the second paragraph of Eric Wofsey's answer we are done.

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