Why is $(A^\perp)^\perp = A$?

Question

On page 52 of this paper, Iwasawa considered the bilinear symmetric non-degenerate pairing $\Phi_n \times \Phi_n \rightarrow \mathbb{Q}_p/\mathbb{Z}_p$ defined by $$\langle \alpha, \beta \rangle_n := \text{the class of } T_n(\alpha\beta) \text{ in } \mathbb{Q}_p / \mathbb{Z}_p$$ where $\Phi_n = \mathbb{Q}_p(\zeta_n)$, $\zeta_n$ being $p^{n+1}$-th root of unity and $T_n$ is the trace of the field extension $\Phi_n/\mathbb{Q}_p$. He then wrote

For any closed subgroup $A$ of [the additive group] $\Phi_n$, we denote by $A^\perp$ the annihilator of $A$ in $\Phi_n$ relative to this pairing. Then $A^\perp$ is a closed subgroup of $\Phi_n$ such that $(A^\perp)^\perp = A$, ...

Unfortunately, I do not see the reasons for his statement that $(A^\perp)^\perp = A$. I know that it is true in the case of non-degenerate bilinear form and $A$ is a subspace but the pairing given is not a bilinear form. So how can I prove that?

Answer 1

The trace form is a perfect pairing in the sense of harmonic analysis, i.e. it identifies $\Phi_n$ with its Pontryagin dual, see Weil's "Basic Number Theory" or Tate's thesis. Therefore, your statement follows from the Pontryagin duality, for example Proposition 3.6.1 in the book "Principles of Harmonic Analysis" by Deitmar/Echterhoff.