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Let $K$ be a non-Archimedean local field. Its additive group $K^+$ is a locally compact Hausdorff abelian group. My question is the following:

Is $K^+$ isomorphic to its Pontryagin dual $\widehat{K^+}$ as a topological group?

Remarks:

1) Fix a non-trivial unitary character $\psi:K^+ \rightarrow \mathbb{C}^{\times}$. For each $a \in K$, the map $a\psi: K^+ \rightarrow \mathbb{C}^{\times}, \; x \mapsto \psi(ax)$ is a unitary character of $K^+$, and the map $a \mapsto a\psi$ gives an abstract group isomorphism from $K^+$ onto $\widehat{K^+}$ (see e.g. [1, sect. 1.17 Prop.]). Unfortunately, the argument given in [1, sect. 1.17 Prop.] does not prove (as far as I can see) that the previous map is an homeomorphism.

2) Let $p \in \mathbb{N}$ be the characteristic of the residue field of $K$. It is known that as a topological field, $K$ is isomorphic either to a finite extension of $\mathbb{Q}_p$ or to some field of formal Laurent series $\mathbb{F}_q((t))$ where $q$ is a power of $p$. Since the additive group of $\mathbb{Q}_p$ is known to be isomorphic to its Pontryagin dual (see e.g. the introductory paragraphs in [2]), the question can be reduced to the case where $K=\mathbb{F}_q((t))$. But I can't find any references or arguments which hint to a positive answer.

Thank you in advance.

References:

[1] C. J. BUSHNELL AND G. HENNIART, The Local Langlands Conjecture for GL(2), Springer, 2006.

[2] L. CORWIN, Some remarks on self-dual locally compact abelian groups, Trans. Amer. Math. Soc. 148 (1970), 613-622.

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1 Answer 1

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A reference is the paper "Locally compact modules" by Martin D. Levin.

In Levin's terminology, the field $\mathbb{F}_q((t))$ can be seen as the completion of $\mathbb{F}_q(t)$ at an infinite valuation with respect to the ring $\mathbb{F}_q[t^{-1}]$. It then follows from §3, Lemma 1.(i) that $\mathbb{F}_q((t))$ is self-dual. Levin states this without proof, for the proof he refers to Weil's "Basic Number Theory", page 41.

I think the following works as a proof:

If $K$ is a non-archimedean local field, then we can find a compact discrete valuation ring $R \subseteq K$, with uniformizing parameter $\pi$, such that $K$ can be written as a topological union $K = \bigcup_{k} \pi^{-k} R$. The Pontryagin dual of $R = \varprojlim R/\pi^kR$ is given by $\varinjlim\; (\pi^{-k}R)/R = K/R$.

Consider the short exact sequence of locally compact $R$-modules $0 \to R \to K \to K/R \to 0$. Applying Pontryagin duality gives a short exact sequence of $R$-modules $0 \to R \to K^\vee \to K/R \to 0$, where $K^\vee$ is the Pontryagin dual of $K$. So we can see $R$ as a compact $R$-submodule of $K^\vee$. Because $K/R$ is discrete, continuity of the projection $K^\vee \to K/R$ shows that $R \subseteq K^\vee$ is a compact open $R$-submodule.

The inclusion $R \subseteq K^\vee$ extends to a $K$-linear map $f: K \to K^\vee$, which is the identity on $R$. Because $K$ and $K^\vee$ are both 1-dimensional $K$-vector spaces (by the argument you gave), we know that $f$ is a vector space isomorphism. We claim that it is a homeomorphism. For each $\lambda \in K$, $f$ restricts to a homeomorphism from $\lambda + R$ to $f(\lambda)+R$, because it is a bijection with compact domain. Further, $K$ is the disjoint union of the cosets $\lambda + R$, where $\lambda$ ranges over a set of representatives of $K/R$. Because each $f(\lambda)+R$ is open, $f$ is an homeomorphism from $K$ to an open subset of $K^\vee$. But because $f$ is bijective, $f$ must be an homeomorphism.

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