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The question to follow has already been asked by the OP at https://math.stackexchange.com/questions/3454735/on-self-duality-of-non-archimedean-local-fields. Due to a lack of feedback, the OP felt compelled to post the same question here in hopes of a better feedback.

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 isomorphism of abstract groups 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 (as a topological group) 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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    $\begingroup$ Yes, at the level of isomorphism types, Pontryagin exchanges restricted direct sums and unrestricted direct products of finite abelian group. Since $\mathbf{F}_q(\!(t)\!)$ ($q$ power of the prime $p$) is isomorphic to $\mathbf{F}_p^{(\omega)}\times\mathbf{F}_p^{\omega}$, it is then isomorphic to its Pontryagin dual. $\endgroup$
    – YCor
    Dec 2, 2019 at 16:40
  • $\begingroup$ To get something more canonical, one should check that there is for a finite-dimensional vector space $V$ over $K$, a $\mathrm{GL}(V)$-equivariant isomorphism from the linear dual $V^*$ to $\hat{V}$. Again it's easier to do in characteristic zero. $\endgroup$
    – YCor
    Dec 2, 2019 at 16:42

1 Answer 1

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Here is a general simple and self-contained proof for the self-duality of local fields, which in particular should answer the questions above.

Let $k$ be any local field. Fix a non-trivial character $\psi\in \hat{k}$ and consider the map $$\rho:k\to \hat{k},\qquad a\mapsto a.\psi$$ precisely as described in your first remark. It is straightforward that $\rho$ is a continuous monomorphism. Thus the dual map $$\rho^*:\hat{\hat{k}}\to \hat {k}$$ is a continuous epimorphism. But we also have the evaluation map $$\mathrm{ev}:k \to \hat{\hat{k}}$$ which is an isomorphism of topological groups by Pontryagin's theorem. A direct computation verifies that $\rho^*\circ \mathrm{ev}=\rho$, which shows in turn that $\rho$ is an isomorphism.

The fact that $\rho$ is an homeomorphism follows from the open mapping theorem.

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