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 $H$.


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".


1 Answer 1


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 $(\pi_v^{n_v} \mathcal{O}_v)^{\perp} = \pi^{m-n_v}\mathcal{O}_v$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.