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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.

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1 Answer 1

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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.

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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

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