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Annihilators reverse inclusions and the double annihilator closes the subgroup

Statement

Assume the Axiom of Choice (The Axiom of Choice) and Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain). Let $G$ be a locally compact Hausdorff abelian group with dual $\hat{G}$ and bidual identification $G ≅ \hat{\hat{G}}$ of Pontryagin biduality: the evaluation map is a topological isomorphism.

(1) If $H \leq K \leq G$ then $K^{⊥} \leq H^{⊥}$ .

(2) For every subgroup $H \leq G$ one has $(H^{⊥})^{⊥} = \overline{H}$ ; in particular $H \subseteq (H^{⊥})^{⊥}$ and the double annihilator is closed.

(3) For a closed subgroup $H \leq G$ this reads $H^{⊥ ⊥} = H$ .

The same statements hold with the roles of $G$ and $\hat{G}$ exchanged, annihilators of subgroups of $\hat{G}$ being computed in $\hat{\hat{G}} ≅ G$ .

Facts & Assumptions

Given: A locally compact Hausdorff abelian group $G$ with dual $\hat{G}$ , subgroups $H \leq K \leq G$ , and the annihilator conventions of The annihilator of a subgroup.

[F1]

$H^{⊥} = {γ \in \hat{G} : γ (h) = 1 for all h \in H}$ is a subgroup of $\hat{G}$ , closed when $H$ is closed, and $(H)^{⊥} = (\overline{H})^{⊥}$ : a continuous character is trivial on $H$ exactly when it is trivial on the closure. For $L \leq \hat{G}$ the annihilator is $L^{⊥} = {x \in G : λ (x) = 1 for all λ \in L}$ . (The annihilator of a subgroup)

[F2]

For a closed subgroup $H$ of the locally compact Hausdorff abelian group $G$ , the quotient $G / H$ is a locally compact Hausdorff abelian group and the pullback $\hat{q}$ of the quotient map $q$ is a topological group isomorphism of $\hat{G / H}$ onto $H^{⊥}$ ; a composition of continuous homomorphisms is a continuous homomorphism. (The quotient of an LCA group by a closed subgroup is LCA, The dual of a quotient is the annihilator, The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ , The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection, Dual homomorphisms: continuity, and the annihilator of a closed subgroup, The Pontryagin dual with the compact-open topology)

[F3]

If $x \neq 0$ in a locally compact Hausdorff abelian group then some continuous character takes a value different from $1$ at $x$ . (Continuous characters separate points of an LCA group)

[F4]

The evaluation map $Φ_{G} : G \to \hat{\hat{G}}$ is an isomorphism of topological groups, so the roles of $G$ and $\hat{G}$ may be exchanged in the annihilator calculus. (Pontryagin biduality: the evaluation map is a topological isomorphism)

[F5]

$H \subseteq \overline{H}$ for every subgroup $H \leq G$ , and every open set containing a point of the closure meets the set. Moreover $\overline{H}$ is a subgroup: if $a, b \in \overline{H}$ and $U$ is any open neighbourhood of $a - b$ , continuity of subtraction supplies neighbourhoods $A ∋ a$ , $B ∋ b$ with $A - B \subseteq U$ ; choose $h \in H \cap A$ , $k \in H \cap B$ , so $h - k \in H \cap U$ . Hence $a - b \in \overline{H}$ , and $0 \in \overline{H}$ . (The annihilator of a subgroup, For $A \subseteq S \subseteq X$ the closure of $A$ in $S$ is ${\overline{A}}^{X} \cap S$ , while the interior only contains ${int}^{X} (A) \cap S$ , with equality when $S$ is open; and a dense subset of $X$ traces to a dense subset of every open $S$ , Topological group: multiplication and inversion are continuous)

Proof

1.1F1

Part (1): let $γ \in K^{⊥}$ and $h \in H \leq K$ ; then $γ (h) = 1$ , so $γ \in H^{⊥}$ . Hence $K^{⊥} \leq H^{⊥}$ .

1.2F1

The inclusion $H \subseteq (H^{⊥})^{⊥}$ always holds: if $h \in H$ then $γ (h) = 1$ for every $γ \in H^{⊥}$ , and this is exactly the defining condition for $h \in (H^{⊥})^{⊥}$ .

1.3F2F3

Let $H$ be closed and let $x \notin H$ . Then $x + H \neq 0$ in the quotient $G / H$ , which is a locally compact Hausdorff abelian group by [F2]; so by [F3] there is a character $χ$ of $G / H$ with $χ (x + H) \neq 1$ . Then $γ : = χ \circ q$ is a continuous homomorphism $G \to T$ , that is $γ \in \hat{G}$ ; it satisfies $γ (h) = χ (0 + H) = 1$ for every $h \in H$ , so $γ \in H^{⊥}$ , and $γ (x) = χ (x + H) \neq 1$ , so $x \notin (H^{⊥})^{⊥}$ .

2.1step 1.2step 1.3

For closed $H$ , step 1.3 shows $(H^{⊥})^{⊥} \subseteq H$ , and step 1.2 gives $H \subseteq (H^{⊥})^{⊥}$ ; hence $(H^{⊥})^{⊥} = H$ , which is (3).

3.1F1F5step 2.1

For an arbitrary subgroup $H \leq G$ , $H^{⊥} = (\overline{H})^{⊥}$ by [F1] and $\overline{H}$ is closed, so step 2.1 applied to $\overline{H}$ gives $(H^{⊥})^{⊥} = ((\overline{H})^{⊥})^{⊥} = \overline{H}$ ; in particular $H \subseteq (H^{⊥})^{⊥}$ and the double annihilator is closed. This is (2).

4.1F4step 1.1step 2.1step 3.1∎

Statements (1), (2) and (3) are proved in steps 1.1, 3.1 and 2.1. The exchange of roles is legitimate because $\hat{G}$ is again a locally compact Hausdorff abelian group and $Φ_{\hat{G}}$ identifies it with the bidual of $\hat{G}$ by [F4], so the same three arguments apply with $G$ replaced by $\hat{G}$ .

Depends on

Used by

Dependency tree · two levels

83 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

T. W. Koerner, Topological Groups (author lecture notes) (standard reference, not scraped)
Manfred Einsiedler and Thomas Ward, Ergodic Theory with a View Towards Number Theory, Appendix C (course-hosted full text) (standard reference, not scraped)