Proof. Fix $m\in \mathbb Z$ and let $S\subseteq \mathbb Z$ be a $(\delta ,2)$ -recurrent set. Let $(X,\mathcal B,\mu ,T)$ be an MPS, with $A\subseteq X$ having $\mu (A)>\delta $ . Consider the MPS $(X,\mathcal B,\mu ,T^m)$ . Since $\mu (A)>\delta $ , there exists $n\in S$ such that $\mu (A\cap (T^{m})^{-n}A \cap (T^{m})^{-2n}A)>0$ , meaning $\mu (A\cap T^{-mn}A\cap T^{-2(mn)}A)>0$ . Since $mn\in mS$ , this proves $mS$ is $(\delta ,2)$ -recurrent.

Proof of Theorem 1.1

Let $\delta <\delta '<\tfrac 12$ . We will construct an increasing sequence of finite sets $S_1\subseteq S_2\subseteq \cdots $ so that $S_n$ is $({1}/{n},2)$ -recurrent, and $S_n^{\wedge 2}$ is $\delta '$ -non-recurrent. Setting $S:=\bigcup _{n=1}^\infty S_n$ , we get that S is a set of $2$ -recurrence, while every finite subset of $S^{\wedge 2}$ is $\delta '$ -non-recurrent. Lemma 2.3 then implies S is $\delta $ -non-recurrent.

To define $S_1$ , we apply Lemma 2.4 to find an $S_1\subseteq \mathbb Z$ which is $(1,2)$ -recurrent, while $S_1^{\wedge 2}$ is $\delta _1$ -non-recurrent for some $\delta _1>\delta '$ . We define the remaining $S_n$ inductively: suppose $n\in \mathbb N$ and that $S_n$ has been chosen to be $({1}/{n},2)$ -recurrent, while $S_n^{\wedge 2}$ is $\delta _n$ -non-recurrent for some $\delta _n>\delta '$ . Let $\eta <\tfrac 12$ so that $2\eta \delta _n>\delta '$ . We will find $S_{n+1}\supset S_n$ so that $S_{n+1}$ is $({1}/({n+1}),2)$ -recurrent and $S_{n+1}^{\wedge 2}$ is $2\eta \delta _n$ -non-recurrent. To do so, apply Lemma 2.4 to find a finite $R \subseteq \mathbb Z$ which is $({1}/({n+1}),2)$ -recurrent such that $R^{\wedge 2}$ is $\eta $ -non-recurrent. By Lemma 2.1, choose $m\in \mathbb N$ so that $(S_n^{\wedge 2})\cup m^2(R^{\wedge 2})$ is $2\eta \delta _n$ -non-recurrent. Now $S_{n+1}:= S_n \cup mR$ is the desired set: $mR$ is $({1}/({n+1}),2)$ -recurrent, by Lemma 2.2, while $S_{n+1}^{\wedge 2}= (S_n^{\wedge 2})\cup m^2(R^{\wedge 2})$ . Since $2\eta \delta _n> \delta '$ , this completes the inductive step of the construction.

Proof. Let $\eta <\tfrac 12$ and choose r large enough to find the E and U provided by Lemma 3.3, with $m(E)>\eta $ . Let $(X,\mathcal B,\mu ,T) = (\mathbb T^r,\mathcal B,m,R_{\boldsymbol {\beta }})$ be the group rotation on $\mathbb T^r$ determined by $\boldsymbol {\beta }$ . For $n\in BH(\boldsymbol {\beta },\mathbf y;k,\eta )$ , we have $R_{\boldsymbol {\beta }}^n E \subseteq E+U$ , so $E\cap R_{\boldsymbol {\beta }}^n E=\varnothing .$ Since $R_{\boldsymbol {\beta }}$ is invertible, this means $E\cap R_{\boldsymbol {\beta }}^{-n}E =\varnothing $ , as well.

Proof of Lemma 2.4

Let $\delta>0$ and $\eta <\tfrac 12$ . Choose k large enough to satisfy the conclusion of Lemma 3.5. With this k, choose $r>k$ and $\varepsilon $ small enough to satisfy the conclusion of Lemma 3.4. Let $\boldsymbol {\beta }\in \mathbb T^r$ be generating and let $BH=BH(\boldsymbol {\beta },\mathbf y;k,\varepsilon )$ , where $\mathbf y=(\tfrac 12,\ldots ,\tfrac 12)\in \mathbb T^r$ , so that $BH$ is $\eta $ -non-recurrent. Finally, let $S=\sqrt {BH}$ , so that S is $(\delta ,2)$ -recurrent, by Lemma 3.5. Since $S^{\wedge 2}\subseteq BH$ , we get that $S^{\wedge 2}$ is $\eta $ -non-recurrent, as desired.

Proof. We first dispense with the case where $\beta $ is rational. In this case, the sequence $e(n^2\beta )$ is periodic, so we fix a period $p\in \mathbb N$ such that $e((pn+q)^2\beta )=e(q^2\beta )$ for every n and $q\in \mathbb N$ . For $0\leq r\leq p$ and $N\in \mathbb N$ , we can then write $B_{pN+r}(f_1,f_2)$ as

$$ \begin{align*} \frac{1}{pN+r}\bigg(\sum_{n=pN+1}^{pN+r} e(n^2\beta)\cdot T^nf_1\cdot T^{2n}f_2+\sum_{q=1}^{p}e(q^2\beta)\sum_{n=0}^{N-1} T^{pn+q}f_1\cdot T^{2(pn+q)}f_2\bigg). \end{align*} $$

For large N, the sum $\sum _{n=pN+1}^{pN+r} e(n^2\beta )\cdot T^nf_1\cdot T^{2n}f_2$ can be ignored, and we have

$$ \begin{align*} \lim_{N\to\infty} B_{N}(f_1,f_2)= \lim_{N\to\infty} \frac{1}{p}\sum_{q=1}^{p}e(q^2\beta) \frac{1}{N}\sum_{n=0}^{N-1} T^{pn+q}f_1\cdot T^{2(pn+q)}f_2. \end{align*} $$

Now, [Reference Frantzikinakis6, Theorem A] implies that the Kronecker factor of $\mathbf X$ (which is itself a factor of $\mathbf A_2$ ) is characteristic for the averages above. This proves the first assertion of the lemma when $\beta $ is rational.

We now assume $\beta \in (0,1)$ is irrational and consider two cases, based on whether equation (8.3) holds. When equation (8.3) fails we write the coefficient $e(n^2\beta )$ in terms of $\bar {g}T^{p(n)}g$ , where $g\in L^\infty (\mu )$ and p is a polynomial. When equation (8.3) holds, we write $e(n^2\beta )$ as $g_0S^ng_1S^{2n}g_2$ , where $\mathbf Y=(Y,\mathcal D,\nu ,S)$ is an ergodic $MPS$ such that $\mathbf X\times \mathbf Y$ is ergodic and $g_i\in L^\infty (\nu )$ . In each case, we write $B_N(f_1,f_2)$ as a familiar multiple ergodic average and apply known results.

For the first case, we assume equation (8.3) fails. We fix $k\in \mathbb N$ such that $1\neq e(k\beta )\in \mathcal E(\mathbf X)$ , meaning $e(k\beta )$ is a non-trivial eigenvalue of $\mathbf X$ . Let $g\in L^2(\mu )$ be a corresponding eigenfunction, so that $g\circ T= e(k\beta ) g$ and $|g|= 1\, \mu $ -almost everywhere. Then $e(k\beta )^m = \bar {g}\cdot T^{m}g \mu $ -almost everywhere. In particular,

(8.4) $$ \begin{align} e(k\beta)^{kn^2+2nj} = \bar{g}\cdot T^{kn^2+2jn}g \quad \text{ for all } j,n\in \mathbb Z \end{align} $$

in $L^2(\mu )$ . Then,

$$ \begin{align*} &\lim_{N\to\infty}B_N(f_1,f_2)\\&\quad= \lim_{N\to\infty}\frac{1}{k}\sum_{j=0}^{k-1} \frac{1}{N}\sum_{n=0}^{N-1} e((nk+j)^2\beta) T^{nk+j} f_1 \cdot T^{2nk+2j} f_2\\ &\quad= \lim_{N\to\infty} \frac{1}{k}\sum_{j=0}^{k-1} \frac{1}{N}\sum_{n=0}^{N-1} e(j^2\beta) e(k\beta)^{kn^2+2jn} T^{nk+j} f_1 \cdot T^{2nk+2j} f_2\\ &\quad= \lim_{N\to\infty} \frac{1}{k}\kern-2pt\sum_{j=0}^{k-1}\kern-2pt e(j^2\beta) \frac{1}{N }\!\sum_{n=0}^{N-1} \overline{g} \cdot T^{kn^2+2jn}g \cdot T^{nk+j}f_1 \cdot T^{2nk+2j}f_2 \quad \!\text{by equation}~({8.4})\kern-0.1pt. \end{align*} $$

The polynomial exponents $p_1(n)=kn^2+2jn$ , $p_2(n)=nk+j$ , $p_3(n)=2nk+2j$ are, in the terminology of [Reference Frantzikinakis6], essentially distinct and not type ( $e_1$ ). Therefore, [Reference Frantzikinakis6, Theorem B] asserts that $f_1$ and $f_2$ in equation (8.1) can be replaced with $P_{\mathbf A_2}f_1$ and $P_{\mathbf A_2}f_2$ , respectively, without changing the value of the limit. This proves the first assertion of the lemma in the case where $\mathcal E(\mathbf X)\cap \{e(n\beta )\}_{n\in \mathbb Z}\neq \{1\}$ .

Now we assume that equation (8.3) holds. We will prove that $\lim _{N\to \infty } B_N(f_1,f_2)=0$ for all $f_1, f_2\in L^\infty (\mu )$ . This implies equation (8.2), since $P_{\mathbf A_2} f_i \in L^\infty (\mu )$ .

Consider the system $\mathbf Y=(\mathbb T^2,\mathcal D,m,S)$ , where $S(x,y)=(x+\beta ,y+x)$ ; this $\mathbf Y$ is ergodic since $\beta $ is irrational. As discussed in §7.1, the eigenvalues of $\mathbf Y$ are $\{e(n\beta )\}_{n\in \mathbb Z}$ . Thus, $\mathbf Y$ has no non-trivial eigenvalues in common with $\mathbf X$ , by equation (8.3). The product system $(\mathbb T^2\times X, m \times \mu , S\times T)$ is therefore ergodic, by Lemma 6.2. We will write $B_N(f_1,f_2)$ as an element of $L^2(m\times \mu )$ . First observe that for all $(x,y)\in \mathbb T^2$ , we have

$$ \begin{align*} e(n^2\beta)&= e(y)e(-2y-2nx -n(n+1) \beta)e(y+2nx+n(2n+1)\beta) \\ &= g_0\cdot g_1(S^n(x,y)) \cdot g_2(S^{2n}(x,y)), \end{align*} $$

where $g_0(x,y):=e(y)$ , $g_1(x,y):=e(-2y)$ , $g_2(x,y):=e(y)$ . So

(8.5) $$ \begin{align} e(n^2\beta) \cdot T^n f_1 \cdot T^{2n}f_2= g_0\otimes 1_{X} \cdot (S\times T)^n g_1\otimes f_1 \cdot (S\times T)^{2n}g_2\otimes f_2 \in L^2(m\times \mu). \end{align} $$

When computing the limit of the averages of the right-hand side in equation (8.5), Theorem 7.1 allows us to replace each $g_i\otimes f_i$ with its projection $\phi _i:=P_{\mathbf Z}(g_i\otimes f_i)$ , where $\mathbf Z$ is the Kronecker factor of $\mathbf Y\times \mathbf X$ . By Lemma 6.3 and Observation 7.2, $g_i\otimes f_i$ is orthogonal to every eigenfunction of $\mathbf Y\times \mathbf X$ , so $\phi _i =0$ . Thus, the limit of the averages is $0$ in $L^2(m\times \mu )$ . Since $B_N(f_1,f_2)$ belongs to the natural embedding of $L^2(\mu )$ in $L^2(m\times \mu )$ , this proves $\lim _{N\to \infty } B_N(f_1,f_2)=0$ in $L^2(\mu )$ .

Proof. Uniformly approximating g by trigonometric polynomials, it suffices to prove the lemma in the case where g is a character of $\mathbb T^r$ . In this case, we can write $g(n^2\boldsymbol {\beta })$ as $e(n^2\alpha )$ for some $\alpha \in [0,1)$ and apply Lemma 8.2.

Observation 9.1. If $\alpha \in Z$ and $\beta \in W$ are generating elements, then the closed subgroup $\Gamma $ of $Z\times W$ generated by $(\alpha ,\beta )$ is a joining of Z with W: $\pi _1(\Gamma )$ is generated by $\alpha $ and $\pi _2(\Gamma )$ is generated by $\beta $ .

Notation 9.2. If $\Gamma $ is a joining of Z with W and $t\in \Gamma $ , we write $t_1$ for $\pi _1(t)$ and $t_2$ for $\pi _2(t)$ .

Observation 9.3. If $\Gamma $ is a joining of $\mathbb T^d$ with $\mathbb T^r$ , then its identity component is also a joining of these groups. To see this, note that since $\Gamma $ is a closed subgroup of a finite-dimensional torus, its identity component $\Gamma _0$ has finite index in $\Gamma $ . The images of $\pi _1$ and $\pi _2$ therefore have finite index in their respective codomains $\mathbb T^d$ and $\mathbb T^r$ . Since these codomains are connected, they have no proper closed finite index subgroups, so the images $\pi _1(\Gamma _0)$ , $\pi _2(\Gamma _0)$ must equal their respective codomains.

Proof. Write $\chi (n\alpha + n^2\beta )$ as $\chi (\alpha )^n\chi (\beta )^{n^2} = e(n\gamma _1 + n^2\gamma _2)$ , where at least one of $\gamma _1, \gamma _2\in [0,1)$ is irrational. Weyl’s theorem then implies the limit of the averages is $0$ .

Proof. (i) Approximating f by trigonometric polynomials, it suffices to prove the special case where f is a non-trivial character $\chi \in \widehat {Z}$ . Under this assumption, we will show that the limit of the averages in equation (10.1) is $0$ . In this case, $f(n\alpha +n^2\beta )=\chi (\alpha )^n\chi (\beta )^{n^2}$ . Connectedness of Z implies $\chi ^n\not \equiv 1$ for all $n\in \mathbb N$ . Since $\alpha $ and $\beta $ generate Z, Lemma 10.1 implies $\chi (\alpha )^n\neq 1$ or $\chi (\beta )^n\neq 1$ for all $n\in \mathbb N$ . Lemma 10.2 then implies $\lim _{N\to \infty } ({1}/{N})\sum _{n=1}^N \chi (n\alpha + n^2\beta )=0$ .

(ii) Assuming Z has finitely many connected components $Z_j$ and identity component $Z_0$ , let $A_j:=\{n\in \mathbb Z:n\alpha +n^2\beta \in Z_j\}$ , and let p be the index of $Z_0$ in Z. We claim that each $A_j$ is a union of infinite arithmetic progressions of the form $p\mathbb Z+q$ . To prove this, it suffices to prove $A_j+p = A_j$ . To do so, observe that $p\alpha , p\beta \in Z_0$ . We will show that if $n\kern1.2pt{\in}\kern1.2pt A_j$ , then $n\kern1.2pt{-}\kern1.2pt p\kern1.2pt{\in}\kern1.2pt A_j$ ; in other words, if $n\alpha \kern1.2pt{+}\kern1.2pt n^2\beta \kern1.2pt{\in}\kern1.2pt Z_j$ , then $(n\kern1.2pt{+}\kern1.2pt p)\alpha \kern1.2pt{+}\kern1.2pt (n\kern1.2pt{+}\kern1.2pt p)^2\beta \kern1.2pt{\in}\kern1.2pt Z_j$ . Now fix $n,j$ with $n\alpha +n^2\beta \in Z_j$ . Then

$$ \begin{align*}(n+p)\alpha + (n+p)^2\beta = n\alpha +n^2\beta + p\alpha + (2n+p)p\beta \in Z_j+Z_0=Z_j,\end{align*} $$

as desired. Similarly, we can show that if $n\in A_j$ , then $n+p\in A_j$ , so that $A_j+p=A_j$ .

Fix $q\in \mathbb Z$ . We claim $\alpha _0:=(1+2q)p\alpha $ and $\beta _0:=p^2\beta $ generate $Z_0$ . To see this, note that the closed subgroup they generate is contained in $Z_0$ , and has finite index in the subgroup generated by $\alpha $ and $\beta $ , while $Z_0$ has no proper finite index closed subgroup.

We decompose the limit in equation (10.1) as

(10.3) $$ \begin{align} \frac{1}{p}\sum_{q=0}^{p-1} \lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N f((pn+q)\alpha + (pn+q)^2\beta). \end{align} $$

Thinking of q as fixed, so that $(pn+q)\alpha + (pn+q)^2\beta \in Z_i$ for some i, it suffices to prove that the limit is $0$ when f is a character $\chi $ of Z which is not constant on $Z_i$ (and therefore not constant on $Z_0$ ). We fix such a $\chi $ and write

$$ \begin{align*} \chi((pn+q)\alpha + (pn+q)^2\beta) &= \chi(q\alpha+q^2\beta)\chi(n(1+2q)p\alpha+n^2p^2\beta)\\ &=\chi(q\alpha+q^2\beta)\chi(n\alpha_0+n^2\beta_0). \end{align*} $$

Since $\alpha _0$ and $\beta _0$ generate $Z_0$ , we have

$$ \begin{align*} \lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N \chi((pn+q)\alpha + (pn+q)^2\beta) = \chi(q\alpha+q^2\beta) \lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N \chi(n\alpha_0+n^2\beta_0) = 0, \end{align*} $$

by part (i). This shows that the averages in equation (10.3) converge to $0$ when f is a character which is not constant on $Z_i$ , completing the proof of part (ii).

To prove part (iii), fix $\alpha ,\alpha ', \beta \in Z$ and assume $H:=\overline {\langle \alpha \rangle }=\overline {\langle \alpha '\rangle }$ is a connected subgroup of Z. It suffices to prove that equation (10.2) holds when f is a character $\chi $ of Z. If $\chi (H)=\{1\}$ , then $\chi (n\alpha +n^2\beta )=\chi (n\alpha '+n^2\beta )=\chi (n^2\beta $ ), so the averages in equation (10.2) are equal. Now assume $\chi (H)\neq \{1\}$ . We will prove that both sides of equation (10.2) are $0$ . First note that $\chi (H)=\{z\in \mathbb C:|z|=1\}$ , since H is compact and connected, and its image under $\chi $ is a non-trivial compact connected subgroup of $\mathcal S^1$ . Since $\alpha $ and $\alpha '$ generate dense subgroups of H, $\chi (\alpha )$ and $\chi (\alpha ')$ generate dense subgroups of $\chi (H)$ , and hence they are both not roots of unity. Lemma 10.2 then implies both limits in part (iii) are $0$ .

Proof. Let $I_2= \int \int f_0(z)f_1(z+t)f_2'(z+2t)\, dz\, dt$ . We will prove that

(11.4) $$ \begin{align} |I-I_2|\leq \kappa\|f_0\|_{L^2(m)}\|f_1\|_{L^2(m)} \end{align} $$

and that $I_2 = I_W$ . We first prove the special case where each $f_i$ is a trigonometric polynomial. Expanding each $f_i$ as $\sum _{\chi \in \widehat {Z}} \hat {f}_i(\chi )\chi $ and simplifying, we get

$$ \begin{align*} I &= \sum_{\chi, \psi,\tau \in \widehat{Z}} \int \hat{f}_0(\chi)\chi(z) \hat{f}_1(\psi)\psi(z+t) \hat{f}_2(\tau)\tau(z+2t)\, dz \, dt\\[-2pt] &= \sum_{\chi, \psi, \tau \in \widehat{Z}} \int \hat{f}_0(\chi) \hat{f}_1(\psi) \hat{f}_2(\tau) \chi\psi\tau(z) \psi\tau^2(t)\, dz\, dt\\[-2pt] &= \sum_{\chi,\psi,\tau\in\widehat{Z}} \hat{f}_0(\chi) \hat{f}_1(\psi) \hat{f}_2(\tau) \int \chi\psi\tau(z)\, dz \int \psi \tau^2(t)\, dt. \end{align*} $$

At least one of $\int \psi \tau ^2(t)\, dt$ or $\int \chi \psi \tau (z)\, dz$ is zero unless $\psi \tau ^2$ and $\chi \psi \tau $ are both trivial; this triviality occurs exactly when $\psi = \tau ^{-2}$ and $\chi =\tau $ . The sum in the last line above may therefore be restricted to values of $\chi ,\psi ,$ and $\tau $ satisfying these identities, and we get

$$ \begin{align*} I=\sum_{\tau \in \widehat{Z}} \hat{f}_0(\tau)\hat{f}_1(\tau^{-2})\hat{f}_2(\tau). \end{align*} $$

As noted in equation (11.2), $\hat {f}_2'(\tau ) = \hat {f}_2(\tau )$ for $\tau \in \widehat {W}$ and $\widehat {f'}_2(\tau )=0$ for $\tau \notin \widehat {Z}\setminus \widehat {W},$ so

$$ \begin{align*} I_2 = \sum_{\tau \in \widehat{W}} \hat{f}_0(\tau)\hat{f}_1(\tau^{-2})\hat{f}_2(\tau). \end{align*} $$

Then,

$$ \begin{align*} |I - I_2| & = \bigg|\sum_{\tau \notin \widehat{W}} \hat{f}_0(\tau)\hat{f}_1(\tau^{-2})\hat{f}_2(\tau)\bigg|\\ &\leq \sum_{\tau\in \widehat{Z}} \kappa |\hat{f}_0(\tau) \hat{f}_1(\tau^{-2 })| \ \ \quad \text{ since } |\hat{f}_2(\tau)|<\kappa \text{ for } \tau\notin \widehat{W}\\ &\leq \kappa \|\hat{f}_0\|_{l^2}\|\hat{f}_1\|_{l^2} \qquad\qquad \ \ \text{Cauchy--Schwarz, assuming } \tau\mapsto \tau^2 \text{ is injective}\\ &= \kappa \|f_0\|_{L^2(m)}\|f_1\|_{L^2(m)}, \quad \text{Plancherel}, \end{align*} $$

where $\|\cdot \|_{l^2}$ denotes the $l^2$ norm for functions on $\widehat {Z}$ .

To prove $I_2=I_W$ , replace t with $t+s$ in the $dt$ integral in $I_2$ , then integrate s over K, using the fact that $f_2'(z+s)=f_2(z)$ for all $z\in Z$ , $s\in K$ :

$$ \begin{align*} I_2 &= \int \int f_0(z) f_1(z+t)f_2'(z+2t) \, dz \, dt\\[-2pt] &= \int_K \int \int f_0(z) f_1(z+t+s) f_2'(z+2t+2s) \, dz \, dt\, dm_K(s)\\[-2pt] &= \int \int f_0(z) \int_K f_1(z+t+s)\, dm_K(s)\, f_2'(z+2t) \, dz \, dt \\[-2pt] &= \int \int f_0(z) f_1'(z+t) f_2'(z+2t) \, dz\, dt. \end{align*} $$

A similar manipulation, replacing z with $z+s$ , lets us replace $f_0$ with $f_0'$ , completing the proof that $I_2=I_W$ , and hence $|I - I_W|\leq \kappa \|f_0\|_{L^2(m)}\|f_1\|_{L^2(m)}$ . This proves eqaution (11.4).

Proof. (i) Let $\chi _j \in \widehat {Z}$ for $j\leq k$ , and let K be the kernel of $\pi $ . We first consider those $\chi _j$ where $\chi _j|_K$ is constant. In this case, $\chi _j$ can be written as $\chi _j'\circ \pi $ , where $\chi _j'\in \widehat {\mathbb T}^r$ , and $\int (g_{s}\circ \pi )\, \overline {\chi }_j\, dm$ can be written as

$$ \begin{align*} \int g_{s}\, \overline{\chi}_j'\, dm_{\mathbb T^r}. \end{align*} $$

So choose g by Lemma 3.9 to make these integrals vanish for such $\chi _j$ . For those j where $\chi _j|_K$ is not constant, write $\int _Z f(z)\, dz$ as $\int _{Z} \int _{K} f(z+t) dm_K(t)\, dm(z).$ Then

$$ \begin{align*} \widehat{g_s\circ \pi}(\chi_j)=\int g_{s}(\pi(z)) \overline{\chi_j(z)}\, dz &= \int \int g_{s}(\pi(z+t)) \overline{\chi_j(z+t) }\, dm_K(t)\, dz \\[-2pt] &= \int g_{s}(\pi(z))\overline{\chi_j(z)} \, dz\, \int_{K} \overline{\chi_j(t)}\, dm_K(t) \\[-2pt] &=0, \end{align*} $$

where the last line follows from the fact that $\chi _j|_K$ is a non-trivial character of K.

(ii) Since $\Gamma $ is an affine joining of $\mathbb T^d$ with $\mathbb T^r$ , by definition, there is a joining $\Gamma _0$ so that the integral over $\Gamma $ is a convex combination of integrals over translates of $\Gamma _0$ . To prove part (ii), it therefore suffices to find a g subordinate to U so that $\int \chi (w_1)g(w_2)\, dm_{\Gamma +t}(w)=0$ for every $t\in \mathbb T^{d}\times \mathbb T^{r}$ . We will use the identity