approximate vanishing in Pontryagin dual

Question

Let $\{n_k\}\subseteq \mathbb{Z}$ to be any given sequence of integers, and suppose it satisfies the following property:

(*) For any $\lambda\in A\subseteq \mathbb{T}$(the unit circle), $|\lambda^{n_k}-1|\to 0$ as $k\to\infty$.

Question 1: Is it true that there exists a subsequence of $\{n_k\}$, say $\{n'_{l}\}$, such that $n'_{l}=0,\forall l\geq 1$ under the following cases:

Case1: $A=\mathbb{T}$ in (*).

Case2: $A$ is a dense subset (or dense subgroup) of $\mathbb{T}$.

Case 3: $A$ is a measurable subset of $\mathbb{T}$ with postive Lebesgue measure.

RK: Clearly if $\{n_k\}$ contains a bounded subsequence, then the answer is yes.

For case2, when $A$ is "good" enough, then the answer is no.

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A general version of the above question is the following:

Suppose $X$ is a compact metrizable abelian group, and $\{\phi_n\}\subseteq \widehat{X}$(the Pontryagin dual of $X$). And suppose it satisfies the following property:

(*) For any $x\in A\subseteq X$, $|\phi_n(x)-1|\to 0$ as $n\to \infty$.

Question 2: Can we prove that there exists a subsequence of $\phi_n$, say $\phi'_l$, such that $\phi'_l(x)=1 \forall x\in X, l\geq 1$ under the assumption that $A\subseteq X$ is a dense subset(or subgroup)?

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Answer 1

It seems the following.

First of all we have to formulate good questions.

– The positive answers mean that the sequence of powers is eventually zero (otherwise we can apply the answer to the subsequence of all non-zero members of the sequence).

– Since the characters are homomorphisms, if the set $A$ has a positive answer, then a positive answer has any set $B$ such that $A\subset B\subset \langle A\rangle$, where $\langle A\rangle$ is a subgroup generated by the set $A$.

– if $A$ consists of periodic elements of the group $\Bbb T$, then the sequence $\{n_k\}=\{k!\}$ yields a negative answer.

– If $A$ is a countable subset of the group $\Bbb T$, then $\Bbb T^A$ is a compact (and metrizable) topological group, so it is topologically periodic, that is for any element $x\in \Bbb T^A$ and any neighborhood $U$ of the unit there exists a positive integer $n$ such that $x^n\in U$. Let $x$ be an element of the group $\Bbb T^A$ such that $x(a)=a$ for each $a\in A$, $\{a_k: k\in\Bbb N\}$ be an enumeration of the set $A$ and $U_k=\{t\in \Bbb T^A: |t(a_i)-1|<1/k$ for all $1\le i\le k\}$. For each $k$ there exists a positive integer $n_k$ such that $x^{n_k}\in U_k$. The construction of the sequence $\{n_k\}$ implies that a seqence $\{a^{n_k}\}$ converges to the unit for each element $a\in A$, so in this case the answer is negative.

– There is a notion of an annulator $(\widehat{X}, A)$ of the set $A$, that is the subgroup of the group $\widehat X$ consisting of characters $\phi$ such that $\phi(A)=1$. The continuity of characters implies that $H(A)=H(\overline A)$. But if $\overline{A}$ is a (proper) subgroup of the group $X$, then by Th. 40 from [Pon] $\overline{A}=(X, (\widehat{X},\overline A))$, so for each element $a\in X\setminus\overline{A}$ there exists a character $\phi\in (\widehat{X}, A)$ such that $\phi(a)\ne 0$.

– Let $X=\Bbb T$, $A$ is Baire and each non-empty open subset of the set $A$ is dense in a non-empty open arc. Let $U=\{z\in\Bbb T: \operatorname{Re} z\ge 0\}$ be a neighborhood of the unit of the group $\Bbb T$. For each element $a\in A$ we can choose a number $k_a$ such that $a^{n_k}\in U$ for each $k>k_a$. For each natural number $k$ put $A_k=\{a\in A:k_a=k\}$. The continuity of power on the group $X$ implies that the set $A_k$ is closed in $A$ for each natural number $k$. Since $A=\bigcup_{k\in\Bbb N} A_k$, Baire Theorem implies that there exists a number $k$ such that a set $A_k$ has non-empty interior in $A$. The set $\overline{A_k}$ contains a non-empty open arc $V\subset$ of the circle $\Bbb T$. If the sequence $\{n_k\}$ is unbounded, then there exists a number $m>k$ such that $|n_m|>1/\mu(V)$, where $\mu$ is the standard measure on $\Bbb T$ such that $\mu(\Bbb T)=1$. But then $U\supset n_m\overline{A_k}\supset n_m V=\Bbb T$, a contradiction. As a corollary we have a positive answer for $A=\Bbb T$.

[Pon] Lev S. Pontrjagin, Continuous groups, 2nd ed., M., (1954) (in Russian).

PS. If your questions are of research interest, I can pose them at the seminar at our topological department and we, as specialists in theory of topological groups, can try to solve them further.

Update 1. A subset $A$ of a locally compact abelian topological group $G$ we shall call characteristic, if there is no sequence $\{\phi_k\}$ of non-trivial characters of the group $G$ such that $\{\phi_k(a)\}$ converges to the unit for each element $x\in A$. A subset $A$ of a topological group $G$ we shall call group dense, if $\overline{\langle A\rangle}=G$. Below $G$ is a locally compact abelian topological group, $G^*$ is the group of characters of the group $G$, and $A$ is a subset of the group $G$.

Proposition 1. Let $A\subset B\subset G$. Then if the subset $A$ is characteristic then the subset $B$ is characteristic too.$\square$

Proposition 2. Let $A, B$ be subsets of the group $G$ such that $A\subset B\subset \langle A \rangle$. Then the subset $A$ is characteristic iff the subset $B$ is characteristic.

Proof. It holds because the characters of the group $G$ are homomorphisms. $\square$

Proposition 3. If the group $G$ is non-periodic and $A$ is a characteristic subset of the group $G$, then the set $A$ contains a non-periodic element.

Proof. Let $x_0$ be a non-periodic element of the group $G$. By [Pon, $\S$ 40.A], for each natural number $k$ there exists a character $\psi_k:G\to\Bbb T$ such that $\psi_k(x_0^{k!})\ne 1$. Define a character $\phi_k$ of the group $G$ by putting $\phi_k(x)=\psi_k(x^{k!})$ for each element $x$ of the group $G$. If a subset $A$ consists of periodic elements of the group $G$, then members of the sequence $\{\phi_k(a)\}$ eventually equals to the unit for each element $a\in A$, but all characters $\phi_k$ are non-trivial.$\square$

Proposition 4. If the periodic group $G$ has a characteristic subset then the group $G$ has bounded exponent, that is there exists a number $n$ such that $x^n=e$ for each element $x\in G$.

Proof. Assume the converse. Then for each natural number $k$ there exists an element $x_k\in G$ such that $x_k^{k!}\ne e$. By [Pon, $\S$ 40.A], for each natural number $k$ there exists a character $\psi_k:G\to\Bbb T$ such that $\psi_k(x_k^{k!})\ne 1$. Define a character $\phi_k$ of the group $G$ by putting $\phi_k(x)=\psi_k(x^{k!})$ for each element $x$ of the group $G$. Since the group $G$ is periodic, the members of the sequence $\{\phi_k(x)\}$ eventually equals to the unit for each element $x\in G$, but all characters $\phi_k$ are non-trivial.$\square$

Proposition 5. Each characteristic subset of the group $G$ is group dense.

Proof. Let $A$ be a non-group dense subset of the group $G$. Put $H=\overline{\langle A\rangle}$. Let $(G^*, H)$ be the annulator of the set $H$, that is the subgroup of the dual group $G^*$ consisting of characters $\phi$ such that $\phi(H)=1$. By Theorem 53 from [Pon], $H=(G, (G^*, H))$. So for each element $x\in G\setminus H$ there exists a character $\phi\in (G^*, H)\subset (G^*, A)$ such that $\phi(x)\ne 1$.

For a subset $A$ of a periodic group $G$ as $\mbox{exp } A$ we denote the smallest number $n$ such that $a^n=e$ for each element $a\in A$.

Corollary 6. If $A$ is a characteristic subset $A$ of the periodic group $G$, then $\mbox{exp } A=\mbox{exp } G$.

Proof. It holds because $\mbox{exp } \overline{\langle A\rangle}=\mbox{exp } A$.

Answer 2

This is not an answer to the above problem, but the motivation to ask the above question.

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I want to prove the following:

(Statement) Given a countable discrete infinite group $G$ with Kazhdan's property (T), take $f\in\mathbb{Z}G$ which is not a right zero divisor in $\mathbb{Z}G$ and suppose the canonical action $G\curvearrowright \widehat{\mathbb{Z}G/\mathbb{Z}Gf}:=X$ is weak mixing, then the canonical map $\alpha: H^2(G, \mathbb{Z}Gf)\to H^2(G, \mathbb{Z}G)$ is injective.

(Note: $\widehat{\mathbb{Z}G/\mathbb{Z}Gf}$ denotes the Pontryagin dual of the countable discrete module $\mathbb{Z}G/\mathbb{Z}Gf$. And $\alpha$ is defined by treating any 2-cocycle $c(-,-): G\times G\to \mathbb{Z}Gf$ as a cocycle into $\mathbb{Z}G$.

(Failed Approach) We have the following fact: suppose the Homoclinic group $\Delta(X):=\{x\in X: sx\to e_X, \forall s\in G ~\text{with}~ s\to\infty\}\subseteq X$ is dense ($e_X$ denotes the identity of $X$ as a group), then action $G\curvearrowright \widehat{\mathbb{Z}G/\mathbb{Z}Gf}:=X$ is mixing, hence weak mixing.

So let us see whether we can prove the above statement under this further assumption.

Suppose $\alpha([c])=[0]$, i.e.,

$\exists$ a map $b: G\to \mathbb{Z}G$ s.t., $c(s,t)f=b(s)+sb(t)-b(st), \forall s,t\in G$. ------(1)

we want to show $c$ is actually a 2-coboundary, it suffices to show that $Image(b)\subset \mathbb{Z}Gf$ by the assumption that $f$ is not a right zero divisor in $\mathbb{Z}G$.

To do this, take any $x\in \Delta(X)$, apply it on both sides of (1), we get

$0=x(b(s)-b(st))+(s^{-1}x)(b(t))\in \mathbb{R}/\mathbb{Z}$.

For any fixed $t$, let $s\to\infty$, then we get $x(b(s)-b(st))\to 0, \forall x\in \Delta(X)$.

If we denote $\phi_s=b(s)-b(st)\in \widehat{X}$(follow the notation as in the above question),

Suppose we can show that for some s large enough, $\phi_s=0\in \widehat{X}$, i.e., $b(s)-b(st)\in \mathbb{Z}Gf$, then from (1) we get $sb(t)\in \mathbb{Z}Gf$, so $b(t)\in\mathbb{Z}Gf,\forall t\in G$.

This is the motivation to ask the above question; unfortunately, $\Delta(X)$ is a countable group in $X$, although we assume it is dense, it seems we could not get what we want...

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RKs: 1, in the above proof, I do not use (and I do not know how to use) the assumption that $G$ has property (T)

2, In fact the statement should hold for a more broader class of groups here.

3, The reason why the above statement should be true is too long to be mentioned here, but let me mention that for $G=SL_n(\mathbb{Z}), n>2$, then $H^2(G,\mathbb{Z}G)=0$ by Borel-Serre and the following conjecture due to Hopf: If $G$ is a finitely presented group, then $H^2(G,\mathbb{Z}G)=0,\mathbb{Z}$ or free abelian of infinity rank.