MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a (countable) discrete abelian group and denote by $\hat{G}$ its Pontryagin dual, i.e. the compact abelian group of group homomorphisms $\chi:G \longrightarrow \mathbb{T}$. Recall that, for a subgroup $H \subset G$, the annihilator is given by $H^\perp = \{\chi \in \hat{G} \mid \chi(g) = 1~\forall g \in H\}$.

If $H_1$ and $H_2$ are subgroups of $G$, then $H_1^\perp H_2^\perp$, the subgroup of $\hat{G}$ generated by $H_1^\perp$ and $H_2^\perp$, is contained in $(H_1 \cap H_2)^\perp$.

$\textbf{Question 1:}$ Do we actually have $H_1^\perp H_2^\perp = (H_1 \cap H_2)^\perp$?

More specifially, let $\alpha$ and $\beta$ be commuting, injective group endomorphisms of $G$. These correspond to commuting, surjective group endomorphisms $\hat{\alpha},\hat{\beta}$ of $\hat{G}$, e.g. given by $\chi \mapsto \chi \circ \alpha$. Note that we have $\alpha(G)^\perp = \ker \hat{\alpha}$. $\textbf{Question 1}$ now transforms into:

$\textbf{Question 2:}$ Does $\ker\hat{\alpha}\ker\hat{\beta} = \ker\widehat{\alpha\beta}$ hold?

Comments: We may assume $G$ to be countable or $\alpha(G) \cap \beta(G) = \alpha\beta(G)$ if that helps. Note that a similar statement does hold: $$\alpha(G)\beta(G) = G \Longleftrightarrow \ker\hat{\alpha} \cap \ker\hat{\beta} = \{1_\hat{G}\}$$ Here $\alpha(G)\beta(G)$ denotes the subgroup of $G$ generated by $\alpha(G)$ and $\beta(G)$. From this equivalence, one can deduce that the two questions from above have a positive answer in the case where $\alpha$ or $\beta$ has finite cokernel.

share|cite|improve this question
up vote 1 down vote accepted

The answer to both questions is yes. Indeed, by Pontryagin duality the inclusion $(H_1\cap H_2)^\perp \subset H_1^\perp H_2^\perp$ you want is equivalent to $$ (H_1^\perp H_2^\perp)^\perp\subset H_1\cap H_2 \tag{$*$} $$ where for $\Sigma\subset\hat G$ we write $\Sigma^\perp=\{g\in G:\chi(g)=1 \text{ for all }\chi\in\Sigma\}$. Now suppose $a\notin H_1\cap H_2$. Without loss of generality, say $a\notin H_1$. Then there is a character $\eta\in H_1^\perp$ such that $\eta(a)\neq 1$ (just lift a character of $G/H_1$ that is nontrivial at $aH_1$). So $a$ is not in $H_1^{\perp\perp}$, hence a fortiori not in the smaller set $(H_1^\perp H_2^\perp)^\perp$. So we have shown that $a\notin H_1\cap H_2 \Rightarrow a\notin (H_1^\perp H_2^\perp)^\perp$, whence $(*)$ by contraposition.

share|cite|improve this answer
Thanks for the quick reply which apparently does the job. The part about lifting a suitable character from $G/H_1$ to $G$ can even be omitted since $G$ is discrete, so $H_1$ is closed and hence $H_1^{\perp\perp} = H_1$. – Nico Stammeier May 20 '14 at 12:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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