# Is $\hom_{R}(M,T)$ locally compact whenever $M$ is, $R$ being a compact DVR?

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.

• Just to check that I'm not missing anything, compactness for a DVR is equivalent to completeness and the residue field being finite, right? Aug 2, 2022 at 3:18
• @EricWofsey Yes, that is exactly correct Aug 2, 2022 at 4:08
• 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. Aug 2, 2022 at 19:17
• @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. Aug 3, 2022 at 0:48
• 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. Aug 3, 2022 at 0:59

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.

• 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. Aug 6, 2022 at 18:16
• A submodule that contains a neighborhood of $0$ is automatically open, since it then contains a neighborhood of each of its points. Aug 6, 2022 at 23:43
• Of course! I totally forgot about that fact. Using this I can deduce a full answer; I posted it below. Aug 7, 2022 at 20:28

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.