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Introduction:

Let $K$ be a number field and let's denote with $\mathbb A_K$ the ring of adeles which is also a locally compact group (with respect to the addition). Remember that the topology is the restricted product topology with respect to the rings $\mathcal O_v$ for any nonarchimedean place $v$.

By taking the standard character $\psi_v$ on each local fied $K_v$ (here $v$ ranges on all places of $K$, also the archimedean ones), we can obtain tha standard character $$\psi=\prod \psi_v:\mathbb A_K\to S^1\,.$$

Tate in his famous thesis proved that $\psi$ composed with the multiplication induces an isomorphism of topological groups $\mathbb A_K\cong\widehat{\mathbb A_K}$ (here $\widehat{\mathbb A_K}$ is the Pontryagin dual).


Furthermore we have the following symmetric pairing: $$\left<,\right>:\mathbb A_K\times\mathbb A_K\to S^1$$ $$(a,b)\mapsto\psi(ab)$$

and for any subset $H\subseteq\mathbb A_K$ we can define the annihilator (or the orthogonal complement):

$$H^\perp:=\{b\in\mathbb A_K\colon\left<b,H\right>=1\}$$

which is always a closed subgroup of $\mathbb A_K$.

Another important result in Tate's thesis is the following: If we consider the embedding $K\subset \mathbb A_K$, then $K^\perp=K$. In this post I'd like to know if it is possible to calculate explicitly $H^\perp$ for a particular $H$.

Question:

Suppose that $H=\prod H_v\subset \mathbb A_K$ is an additive subgroup constructed in the following way:

  1. For all but finitely many nonarchimedean $v$, we have $H_v=\mathcal O_v$.
  2. For all nonarchimedean $v$ there is $n_v\in\mathbb Z$ such that $H_v=\mathfrak p_v^{n_v}\mathcal O_v$. Here $\mathfrak p_v$ is the maximal ideal of $\mathcal O_v$. Note that due to property $1$, all but finitely many $n_v$ are equal to $0$.
  3. If $v$ is archimedean then $H_v=K_v$ (namely $H_v$ is equal to $\mathbb C$ or to $\mathbb R$ depending on the nature of $v$).

What is the explicit expression of $H^\perp$? Is there a way to calculate it?

Is it true in general that $H^\perp=\prod H_v^\perp$? Here by $H_v^\perp$ I mean the annihilator of $H_v$ in $K_v$ with respect to the "local pairing".

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Yes, the formula is true.

The inclusion $\prod_{v}{H_v^{\perp}} \subset H^{\perp}$ is immediate: Let $a = (a_v)$ be an adele such that $a_v \in H_v^{\perp}$ for all places $v$. Let $b = (b_v) \in H$, i.e. $b_v \in H_v$ for all places $v$. Then $\psi_v(a_v b_v) = 1$ for all $v$ and hence $$ \psi(ab) = \prod_{v}{\psi_v(a_vb_v)} = 1\,. $$

The inclusion $H^{\perp} \subset \prod_{v}{H_v^{\perp}}$ is not much harder: Let $a = (a_v) \in H^{\perp}$ that is to say $\psi(a b) = 1$ for all $b \in H$. We need to show that $a_v \in H_v^{\perp}$ for all $v$. So let us fix a place $v$ of $K$. Let $i_v : K_v \rightarrow \Bbb{A}_K$ be the inclusion $x \mapsto (0, \dots, 0, x,0, \dots, 0)$ at $v$; it is a continous morphism of topological groups respecting also the multiplication on $\Bbb{A}$. Let $x \in H_v$. We have $\psi_v = \psi \circ i_v$ and $i_v(x) \in H$. It follows that $$ 1 = \psi(ai_v(x)) = \psi(i_v(a_v)i_v(x)) = \psi(i_v(a_vx)) = \psi_v(a_vx)\,, $$ which implies $a_v \in H_v^{\perp}$.

One more property of $H^{\perp}$:

In genreal, if $B$ is a subgroup of a LCAG $A$, which is open, then $B^{\text{an}}:= \{\chi \in \hat{A}\, :\, \chi(B) = 1\} \subset \hat{A}$ is compact and if $B$ compact, then $B^{\text{an}}$ is open.

Since the isomorphism $\Bbb{A} \cong \widehat{\Bbb{A}}$ is also topological, the same holds for $H^{\perp} \subset \mathbb{A}$. You took a manifestly open subgroup $H$ of $\Bbb{A}$ and so its annihilator should be compact, which is true, since for the archimedean places you get $K_v^{\perp} = \{0\}$, while for all non-archimedean places you get at least a compact "side" $H_v^{\perp}$; and since moreover $\mathcal{O}_v^{\perp} = \mathcal{O}_v$ (with respect to the local pairing) for all but finitely many non-archimedean $v$, you get a compact rectangle for $H^{\perp}$. That $\mathcal{O}_v^{\perp} = \mathcal{O}_v$ holds for all but finitely many non-archimedean $v$ follows from the fact that the local extensions $K_v/\Bbb{Q}_p$ are unramified for all but finitely many places $v$, which implies that the conductor of $\psi_v$ (i.e., the largest fractional ideal on which $\psi_v$ is tirival) is equal to $\mathcal{O}_v$. In general, the conductor of $\psi_v$ is the inverse different of the extension $K_v/\Bbb{Q}_p$.

So your calculation reduces to calculating the anhilators of subgroups $\pi_v^{n_v}\mathcal{O}_v \subset K_v$, where $\pi_v \in \mathcal{O}_v$ is a uniformizer. If $\psi_v : K_v \rightarrow S^{1}$ has conductor $\pi_v^{m}\mathcal{O}_v$, you can check that $(\pi_v^{n_v} \mathcal{O}_v)^{\perp} = \pi^{m-n_v}\mathcal{O}_v$.

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