Proof. Since $T_a$ is measure-preserving, we have

$$ \begin{align*} &\int_{G/\Gamma} F(b\Gamma, a^{P_1(n)}b\Gamma, \ldots, a^{P_t(n)}b\Gamma)\,d\nu(b\Gamma)\\[3pt] &\quad= \int_{G/\Gamma} F(a^m b\Gamma, a^{m+P_1(n)}b\Gamma, \ldots, a^{m+P_t(n)}b\Gamma)\,d\nu(b\Gamma) \end{align*} $$

for any $m,n\in \mathbb {N}$ . Consequently,

$$ \begin{align*} &\int_{G/\Gamma} F(b\Gamma, a^{P_1(n)}b\Gamma, \ldots, a^{P_t(n)}b\Gamma)\,d\nu(b\Gamma)\\[3pt] &\quad=\mathop{\mathbb{E}}\limits_{m\in[N]} \int_{G/\Gamma} F(a^m b\Gamma, a^{m+P_1(n)}b\Gamma, \ldots, a^{m+P_t(n)}b\Gamma)\,d\nu(b\Gamma), \end{align*} $$

from which the lemma follows.

Proof. The connectedness of $G/\Gamma $ implies that $G=G^0\Gamma $ , and so there exist $\alpha \in G^0$ and $\gamma \in \Gamma $ such that $a = \alpha \gamma ^{-1}$ . Then

$$ \begin{align*} a^n b\Gamma = (\alpha\gamma^{-1})^n b\Gamma = (\alpha\gamma^{-1})^n b \gamma^{n}\Gamma. \end{align*} $$

It follows from normality of $G^0$ and the fact that $\alpha $ and b are elements of $G^0$ that the sequence $g_b(n) = (\alpha \gamma ^{-1})^n b \gamma ^{n}$ takes values in $G^0$ . Since the sequences $h_1(n) = a^n b$ and $h_2(n) = \gamma ^{n}$ are adapted to $G_{\bullet }$ , and the set ${{\mathrm{poly}}}(\mathbb {Z}, G_{\bullet })$ is a group, we deduce that $g_b = h_1 h_2$ is adapted to $G^0_{\bullet } = G_{\bullet }\cap G^0$ .

We want a more precise description of $g_b$ , and for this we shall use some results from [Reference LeibmanLei09, §§ 11–13]. Let $g=g_b$ for the identity $b=1$ ; that is, $g(n) = (\alpha \gamma ^{-1})^n \gamma ^n$ . Leibman showed in [Reference LeibmanLei09, § 11.2] that

(19) $$ \begin{align} g(n) &= \prod_{1\leqslant k_1 \leqslant s} (A^{k_1-1}\alpha)^{q_{k_1}(n)} \prod_{1\leqslant k_2 < k_1 < s}[A^{k_1-1}\alpha, A^{k_2-1}\alpha]^{q_{k_1, k_2}(n)}\nonumber\\[6pt] &\quad\prod_{1\leqslant k_3 < k_2 < k_1 < s}[[A^{k_1-1}\alpha, A^{k_2-1}\alpha], A^{k_3-1}\alpha]^{q_{k_1, k_2, k_3}(n)} \ldots, \end{align} $$

where $Ax = [x,\gamma ]$ and $q_{k_1, \ldots , k_r}$ are integral polynomials with $\deg q_{k_1, \ldots , k_r} \leqslant k_1 + \cdots +~k_r$ . More explicitly, we have

(20) $$ \begin{align} g(n) &= \alpha^n (A\alpha)^{{n}\choose{2}} (A^2\alpha)^{{n}\choose{3}} \cdots [A\alpha, \alpha]^{{n}\choose{3}}[A^2\alpha,\alpha]^{{n}\choose{4}} \cdots\nonumber\\ &\quad[A^2\alpha,A\alpha]^{4{{n+1}\choose{5}}}[A^3\alpha,A\alpha]^{5{{n+1}\choose{6}}}\cdots. \end{align} $$

The coefficients of g can be analysed using a family of subgroups of $G^0$ introduced in [Reference LeibmanLei09, § 12]. For $k_1, \ldots , k_l\in \mathbb {N}_+$ , we let $G^0_{(k_1,\ldots ,k_l)}$ be the subgroup of $G^0$ generated by all l-fold commutators of elements of the form $A^{k_1-1}h_1$ , …, $A^{k_l-1}h_l$ for $h_1, \ldots , h_l\in G^0$ . (A one-fold commutator is any element $h\in G$ . For $l>1$ , an l-fold commutator is an element of the form $[h_i, h_j]$ , where $h_i$ is an i-fold commutator, $h_j$ is an j-fold commutator and $i+j = l$ .) We then define

$$ \begin{align*} G^0_{k,l} = \langle G^0_{(k_1, \ldots, k_i)}: i\geqslant l,\; k_1+\cdots+k_l\geqslant k\rangle \end{align*} $$

for integers $1\leqslant l\leqslant k$ and set $G^0_{k,l} = G^0_{l,l}$ whenever $l>k$ .

The following lemma lists some basic properties of the groups $G^0_{k,l}$ that we shall use.

Proof. Properties (i)–(iv) are proved in [Reference LeibmanLei09, Lemma 12.2]. For $k\geqslant 2$ , the statement $G_k = G^0_k$ in (v) is true by definition, and the statement $G_k = G^0_{k,1}$ is proved in [Reference LeibmanLei09, Lemma 12.3]. To finish the proof of (v), it remains to show that $G^0_{k,1} = \langle A^{k-1}G^0, G^0_{k,2}\rangle = \langle A G^0_{k-1}, G^0_{k,2}\rangle $ for $k\geqslant 2$ . For $k = 2$ , this is true by definition of $G^0_{k,1}$ and the fact that $G^0_{k,2}\geqslant G^0_{k,3}\geqslant \cdots \,$ , which follows from part (iv). We assume that the statement is true for some $k\geqslant 2$ . That $G_{k+1}^0$ contains $\langle A G^0_k, G^0_{k+1,2}\rangle $ follows from the fact that both $A G^0_k$ and $G^0_{k+1,2}$ are contained in the $(k+1)$ th element of the lower central series of G, which is precisely $G^0_{k+1}$ . For the other direction, we observe that

$$ \begin{align*} G^0_{k+1} &= [G_k, G] = [G^0_k, \langle G^0, \gamma \rangle] \leqslant \langle [G^0_k, G^0], [G^0_k,\gamma] \rangle \\[3pt] &\leqslant \langle [A^{k-1} G^0, G^0], [G^0_{k,2}, G^0], AG^0_k\rangle \leqslant \langle G^0_{k+1,2}, AG_k^0\rangle. \end{align*} $$

A similar argument shows that $G^0_{k+1} = \langle A^{k}G^0, G^0_{k+1,2}\rangle $ .

Before we prove property (vi), we recall that ${(G^0)^\nabla _k = \langle G_{k+1}, [G_j, G_{k-j}]: 1\leqslant j < k\rangle }$ . That (vi) holds for $k=1$ can be verified by inspection. For $k\geqslant 2$ , we observe that $[A^{j-1} G^0, A^{k-j-1}G^0]\leqslant [G_j^0, G_{k-j}^0]$ , and so

$$ \begin{align*} G^0_{k,2} \leqslant \langle [G^0_j, G^0_{k-j}]: 1\leqslant j < k\rangle; \end{align*} $$

when coupled with property (v), this implies that $(G^0)^\nabla _k \geqslant \langle G^0_{k,2}, G^0_{k+1}\rangle $ . For the converse, we have

$$ \begin{align*} [G_j^0, G^0_{k-j}] = [\langle A^{j-1}G^0, G^0_{j,2}\rangle, \langle A^{k-j-1}G^0, G^0_{k-j,2}\rangle] \leqslant \langle G^0_{k,2}, G^0_{k,3}, G^0_{k,4}\rangle \leqslant G^0_{k,2}, \end{align*} $$

for each $1\leqslant j < k$ , from which it follows that $(G^0)^\nabla _k \leqslant \langle G^0_{k,2}, G^0_{k+1}\rangle $ .

Letting $g(n) = \prod _{i=1}^s g_{i}^{{n}\choose {i}}$ , we observe from (19), (20) as well as parts (v) and (vi) of Lemma 3.3 that

(21) $$ \begin{align} g_i = A^{i-1}\alpha \operatorname{mod} \; (G^0)_{i}^\nabla. \end{align} $$

For an arbitrary $b\in G^0$ , we have $g_b(n) = a^n b \gamma ^n = b (\alpha _b \gamma ^{-1})^n\gamma ^n$ , where $\alpha _b = \alpha [\alpha ,b] Ab$ , as observed in [Reference LeibmanLei09, § 11.3]. Letting $g_b(n) = \prod _{i=0}^s g_{b,i}^{{n}\choose {i}}$ , it is therefore true that

(22) $$ \begin{align} g_{b,i} = A^{i-1}\alpha_b = A^{i-1}\alpha \operatorname{mod} \; (G^0)^\nabla_i \end{align} $$

for all $i\in \mathbb {N}_+$ .

For $i=1$ , we have $g_{b,1} = \alpha $ mod $G^0_2$ , and we claim that $g_{b,i}$ is irrational. The ergodicity of a implies that for almost every b, the sequence $n\mapsto a^n b$ is equidistributed in $G/\Gamma $ , and so the same is true for the sequence $g_b$ in $G^0/\Gamma ^0$ . Consequently, the projection $\pi (g_b): \mathbb {Z}\to G^0/(G^0_2\Gamma ^0)$ is equidistributed as well. Since $\pi (g_b(n)) = \pi (b) + \pi (\alpha ) n$ , it follows that $\pi (\alpha )$ is an irrational element of $G^0/G^0_2$ , and so $g_{b,1}$ is an irrational element of $G^0$ .

Before proving that $g_{b,i}$ are irrational for $i>1$ , we discuss some properties of the map $A: G\to G$ . From the definition of the filtration $G^0_{\bullet }$ we observe that $AG^0_i\leqslant G^0_{i+1}$ for all $i\geqslant 1$ (this is also a consequence of parts (iv) and (v) of Lemma 3.3). Therefore the map $A_i := A\mid _{G^0_i}$ takes values in $G^0_{i+1}$ , and moreover $A_i(\Gamma _i)\leqslant \Gamma _{i+1}$ . We also observe that the projection $\overline {A}_i: G^0_i\to G^0_{i+1}/(G^0)^\nabla _{i+1}$ is a (continuous) group homomorphism because

$$ \begin{align*} A(xy) = [xy, \gamma] = [x,\gamma][[x,\gamma],y][y,\gamma] = Ax [Ax,y] Ay = Ax Ay \operatorname{mod} G^0_{2i+1, 2} \end{align*} $$

for any $x,y\in G^0_i$ and $G^0_{2i+1,2}\leqslant G^0_{i+1,2}\leqslant (G^0)^\nabla _{i+1}$ by parts (iv) and (vi) of Lemma 3.3. From part (v) of Lemma 3.3 it follows that $\overline {A}_i$ is surjective. Finally, we note, using parts (iii) and (v) of Lemma 3.3, that $A_i((G^0)^\nabla _i)\leqslant (G^0)^\nabla _{i+1}$ .

Suppose that $g_{b,i}$ is irrational but $g_{b,i+1}$ is not for some $1\leqslant i < s$ . Then there exists a non-trivial $(i+1)$ th-level character $\eta _{i+1}:G^0_{i+1}\to \mathbb {R}$ such that $\eta _{i+1}(g_{b,i+1})\in \mathbb {Z}$ . From (22) and the fact that $\eta _{i+1}$ vanishes on $(G^0)^\nabla _{i+1}$ , we deduce that $\eta _{i+1}(g_{b,i+1})=\eta _{i+1}(A^i \alpha )$ . We also let $\overline {\eta }_{i+1}:G^0_{i+1}/(G^0)^\nabla _{i+1}\to \mathbb {R}$ be the induced map.

Let $\eta _i := \eta _{i+1}\circ A_i: G^0_i\to \mathbb {R}$ . It is an ith-level character as a consequence of four facts: the vanishing of $\eta _{i+1}$ on $(G^0)_{i+1}^\nabla $ , the inclusion $(G^0_{i+1,2})\leqslant (G^0)_{i+1}^\nabla $ (both of which imply that $\eta _i = \overline {\eta }_{i+1}\circ \overline {A}_i$ is a continuous group homomorphism), the inclusion $A_i((G^0)^\nabla _i)\leqslant (G^0)^\nabla _{i+1}$ , and the fact that $\eta _i(\Gamma _i)\leqslant \mathbb {Z}$ . Moreover, it satisfies

$$ \begin{align*}\eta_i(g_{b,i}) = \eta_i(A^{i-1}\alpha) = \eta_{i+1}(A^i \alpha) = \eta_{i+1}(g_{b,i+1}),\end{align*} $$

implying that $\eta _i(g_{b,i})\in \mathbb {Z}$ . The non-triviality of $\eta _{i+1}$ implies that $\overline {\eta }_{i+1}$ and $\overline {A}_i$ are surjective maps onto non-trivial groups; hence $\eta _i$ is non-trivial. This contradicts the irrationality of $g_{b,i}$ . By induction, $g_{b,1}$ , …, $g_{b,s}$ are all irrational, implying that $g_b$ is irrational.

Proof. We take the cases $i=0$ and $i>0$ separately. For $i>0$ , we recall from (12) that $Z_i(G/\Gamma ) = G/G_{i+1}\Gamma $ . Since $G/\Gamma = G^0/\Gamma ^0$ by connectedness of $G/\Gamma $ , and $G_j = G^0_j$ for $j\geqslant 2$ , it follows that

$$ \begin{align*}Z_i(G^0/\Gamma^0) = Z_i(G/\Gamma) = G/G_{i+1}\Gamma = G^0/G^0_{i+1}\Gamma^0.\end{align*} $$

For $i=0$ , we have $Z_i(G/\Gamma ) = G/G^0\Gamma = 1 = G^0/G^0\Gamma ^0 = Z_i(G^0/\Gamma ^0)$ .

Proof. We first show the equivalence of (ii) and (iii), followed by the equivalence of (i) and (iii).

The implication (iii) $\implies $ (ii) follows from the fact that the polynomial ${{u}\choose {j}}$ is a sum of the polynomials $1, u, \ldots , u^j$ . For the converse, we similarly note that $u^j$ is a sum of the polynomials $1, u, \ldots , {{u}\choose {j}}$ .

To prove the equivalence of (i) and (iii), we set ${{u}\choose {k}} = \sum _{j=0}^k c_{jk} u^j$ for each $k\in \mathbb {N}$ , so that $Q_i(u) = \sum _{k=0}^d b_{ik}\sum _{j=0}^k c_{jk} u^j$ . Importantly, $c_{jk}\neq 0$ for any $0\leq j\leq k$ . This allows us to rewrite

$$ \begin{align*} 0 &= Q_0(x) + Q_1(x+P_1(y)) + \cdots + Q_t(x+P_t(y))\\ &= \sum_{i=0}^t \sum_{k=0}^d b_{ik}\sum_{j=0}^k c_{jk} (x+P_i(y))^j\\ &= \sum_{j=0}^k \sum_{i=0}^t \sum_{k=j}^d b_{ik} c_{jk} (x+P_i(y))^j. \end{align*} $$

The relation $(Q_0, \ldots , Q_t)$ is a sum of homogeneous algebraic relations if and only if for every $0\leq j\leq d$ , we have

(27) $$ \begin{align} \sum_{k=j}^d c_{jk}\sum_{i=0}^t b_{ik} (x+P_i(y))^j = 0, \end{align} $$

and it is immediate from this expression that (iii) implies (i). To prove the implication (i) $\implies $ (iii), we assume that the relation $(Q_0, \ldots , Q_t)$ is indeed a sum of homogeneous algebraic relations, and so (27) holds for $0\leq j\leq d$ . Taking $j = d$ implies (26) for $j=d$ , $k=d$ , that is,

$$ \begin{align*} b_{0d} x^d + b_{1d} (x+P_1(y))^d + \cdots + b_{td} (x+P_t(y))^d = 0. \end{align*} $$

Taking partial derivatives with respect to x of the expression above $d-j$ times implies (26) for $k = d$ and $0\leq j \leq d$ .

Thus, (27) equals

(28) $$ \begin{align} \sum_{k=j}^{d-1} c_{jk}\sum_{i=0}^t b_{ik} (x+P_i(y))^j = 0. \end{align} $$

Running the same argument as above, we prove (26) for $k = d-1$ and $0\leq j\leq d-1$ . Downward induction on k proves (iii) for all required values of k and j.

Proof. The equivalence of (ii) and (iii) follows trivially from the definition of $W^{\prime }_k$ , and we focus on showing the equivalence of (i) and (ii) instead. The inhomogeneity of $\vec {P}$ implies the existence of a non-trivial algebraic relation

$$ \begin{align*} (Q_0(u), \ldots, Q_t(u)) = \bigg(\sum_k b_{0k} {{u}\choose{k}}, \ldots, \sum_k b_{tk} {{u}\choose{k}} \bigg) \end{align*} $$

that is not a sum of homogeneous algebraic relations. By Lemma 4.1, this means that there exist $k\in \mathbb {N}_+$ and $0\leq j \leq k$ for which

(31) $$ \begin{align} b_{0k} {{x}\choose{j}} + b_{1k} {{x+P_1(y)}\choose{j}} + \cdots + b_{tk} {{x+P_t(y)}\choose{j}} \neq 0. \end{align} $$

We claim that in fact we can take $j=k$ . We define the discrete derivative of $Q\in \mathbb {R}[u]$ to be $\partial Q(u) = Q(u+1) - Q(u)$ , and the partial discrete derivative of $R\in \mathbb {R}[x,y]$ with respect to x to be $\partial _x R(x,y) = R(x+1, y) - R(x,y)$ . Observing that $\partial {{u}\choose {k}} = {{u+1}\choose {k}}-{{u}\choose {k}} = {{u}\choose {k-1}}$ , we deduce that $\partial _x {{x+P_i(y)}\choose {k}} = {{x+P_i(y)}\choose {k-1}}$ . It follows that if

$$ \begin{align*} b_{0k} {{x}\choose{j}} + b_{1k} {{x+P_1(y)}\choose{j}} + \cdots + b_{tk} {{x+P_t(y)}\choose{j}} = 0, \end{align*} $$

then applying the partial discrete derivative with respect to x to the expression above $k-j$ times would imply

$$ \begin{align*} b_{0k} {{x}\choose{j}} + b_{1k} {{x+P_1(y)}\choose{j}} + \cdots + b_{tk} {{x+P_t(y)}\choose{j}} = 0, \end{align*} $$

contradicting (31). We can thus assume that $j=k$ in (31).

Since

$$ \begin{align*} Q_0(x) + Q_1(x+P_1(y)) + \cdots + Q_t(x+P_t(y)) = 0, \end{align*} $$

we have

$$ \begin{align*} R(x,y) = -\sum_{j\neq k} \sum_{i=0}^t b_{ij} {{x+P_i(y))}\choose{j}} \in \sum_{j\neq k} W_j, \end{align*} $$

and so $W^c_k = W_k\cap \sum _{j\neq k} W_j$ is non-trivial. Thus (ii) implies (i) by contrapositive. The argument can be reversed, and so (i) and (ii) are in fact equivalent.

Proof. Suppose that $\mathcal {A}_i(\vec {P}) \geq t$ , and let

$$ \begin{align*} Q_0(x) + Q_1(x+P_1(y)) + \cdots + Q_t(x+P_t(y)) = 0 \end{align*} $$

be an algebraic relation of degree $d\geq t$ . Taking the partial derivative with respect to x of the expression above $d-t$ times, we can assume that the relation has degree t. The homogeneity of $\vec {P}$ and Lemma 4.1 imply the existence of non-trivial algebraic relations of the form

(32) $$ \begin{align} a_0 {{x}\choose{t}} + a_1 {{x+P_i(y)}\choose{t}} + \cdots + a_t {{x+P_t(y)}\choose{t}} = 0. \end{align} $$

Relation (32) and the formula

$$ \begin{align*} {{x+P_i(y)}\choose{t}}={{x}\choose{t}}+{{x}\choose{t-1}}P_i(y) + {{x}\choose{t-2}}{{P_i(y)}\choose{2}}+ \cdots +{{P_i(y)}\choose{t}}, \end{align*} $$

imply

$$ \begin{align*} a_1 {{P_i(y)}\choose{k}} + \cdots + a_t {{P_t(y)}\choose{k}} = 0 \end{align*} $$

for $1\leq k\leq t$ . This gives us t equations

$$\begin{align*} a_1 P_1(y) + \cdots + &a_t P_t(y) = 0, \nonumber \\[3pt] a_1 P_1(y)^2 + \cdots + &a_t P_t(y)^2 = 0, \nonumber \\[3pt] \vdots \qquad\qquad\qquad\quad &\vdots \nonumber \\[3pt] a_1 P_1(y)^t + \cdots + &a_t P_t(y)^t = 0. \end{align*}$$

The invertibility of the Vandermonde matrix and the distinctness of the polynomials $P_1, \ldots , P_t$ imply that these t equations can only be satisfied if $a_1 = \cdots = a_t = 0$ , which also implies $a_0 = 0$ . This contradicts the non-triviality of (32).

Proof. We fix $k\in \mathbb {N}_+$ and denote

$$ \begin{align*} A_1 = {{\mathrm{Span}}}_{\mathbb{R}}\{\vec{P}^k(x,y): x, y\in\mathbb{R}\},\ \ &A_2 = {{\mathrm{Span}}}_{\mathbb{R}}\{\vec{P}^j(x,y): x, y\in\mathbb{R},\; 1\leq j\leq k\},\\[3pt] A_3 = {{\mathrm{Span}}}_{\mathbb{R}}\bigg\{{{\vec{P}(x,y)}\choose{k}}: x, y\in\mathbb{R}\bigg\},\ \ &A_4 = {{\mathrm{Span}}}_{\mathbb{R}}\bigg\{{{\vec{P}(x,y)}\choose{j}}: x, y\in\mathbb{R},\; 1\leq j\leq k\bigg\} \end{align*} $$

for the four spaces mentioned in the statement of the lemma. It is clear that $A_1\subseteq A_2$ and $A_3 \subseteq A_4$ . To prove the converse inclusions, we note that $({\partial }/{\partial _x})\vec {P}(x,y)^k = k \vec {P}(x,y)^{k-1}$ and $\partial _x{{\vec {P}(x,y)}\choose {k}} = {{\vec {P}(x,y)}\choose {k-1}}$ , where ${\partial }/{\partial _x}$ is the usual partial derivative with respect to x and $\partial _x$ is the partial discrete derivative with respect to x defined in the proof of Lemma 4.1. For every $R(x,y)\in A_1$ and $h\neq 0$ , the expression

$$ \begin{align*}\cfrac{R(x+h, y)-R(x, y)}{h}\end{align*} $$

is still in $A_1$ , and so $({\partial }/{\partial _x})R(x,y)\in A_1$ by the closeness of $A_1$ . Applying ${\partial }/{\partial _x}$ to $\vec {P}(x,y)^k$ exactly $k-j$ times with the observations above, we deduce that $\vec {P}(x,y)^j$ is still in $A_1$ . Hence $A_2\subseteq A_1$ . Analogously, applying $\partial _x$ exactly $k-j$ times to ${{\vec {P}(x,y)}\choose {k}}$ , we deduce that ${{\vec {P}(x,y)}\choose {j}}\in A_3$ , and so $A_4\subseteq A_3$ .

It remains to show that $A_2 = A_4$ . For this, we note that $\vec {P}(x,y)^k$ is a linear combination of ${{\vec {P}(x,y)}\choose {1}}, \ldots , {{\vec {P}(x,y)}\choose {k}}$ , and conversely ${{\vec {P}(x,y)}\choose {k}}$ is a linear combination of $\vec {P}(x,y), \ldots , \vec {P}(x,y)^k$ , from which the equality $A_2 = A_4$ follows.

Proof. The assumption $\mathcal {A}_i(\vec {P}) = s'$ implies that $(x+P_i(y))^{s'+1}$ is linearly independent of $(x+P_k(y))^{s'+1}$ for $k\neq i$ , hence $\mathcal {P}_{s'+1}$ contains $\vec {e}_i$ . The lemma then follows by the definition of $G^P$ .

Proof of Corollary 5.4 using Theorem 5.3

Let $\mathcal {A}_i(\vec {P}) = s$ . Let $(X,\mathcal {X},\mu ,T)$ be a totally ergodic system, $f_0, \ldots , f_t\in L^\infty (\mu )$ , and suppose that $\operatorname {\mathrm {\mathbb {E}}}(f_i\mid \mathcal {Z}_s)=0$ . By Theorem 1.4, the expression

(35) $$ \begin{align} \lim_{N\to\infty}\mathop{\mathbb{E}}\limits_{n\in [N]}\int_X f_0 \cdot T^{P_1(n)} f_1 \cdots T^{P_{t}(n)} f_{t} \,d\mu \end{align} $$

remains unchanged if we project the functions $f_0, \ldots , f_t$ onto the factor $\mathcal {Z}_{s_0}$ for some $s_0\in \mathbb {N}$ . If $s_0<s$ , then $\operatorname {\mathrm {\mathbb {E}}}(f_i\mid \mathcal {Z}_{s_0}) = 0$ and the limit (35) is 0, so we can assume that $s_0\geqslant s$ . Since the factor $\mathcal {Z}_{s_0}$ is an inverse limit of $s_0$ -step nilsystems, we can approximate X by totally ergodic nilsystems.

Let $(G/\Gamma , {\mathcal {G}}/\Gamma , \nu , T_a)$ be a totally ergodic nilsystem, and $G_{\bullet }$ be the lower central series filtration on G. Using (12), it suffices to show that if $f_0, \ldots , f_t\in L^\infty (\nu )$ and $f_i$ vanishes on each coset of $G_{s+1}\Gamma $ , then

$$ \begin{align*} \lim_{N\to\infty}\mathop{\mathbb{E}}\limits_{n\in[N]} \int_{G/\Gamma} f_0(b\Gamma)\cdot f_1(a^{P_1(n)}b\Gamma) \cdots f_t(a^{P_t(n)} b\Gamma) \,d\nu(b\Gamma) = 0. \end{align*} $$

Let $G^0_{\bullet }$ be the filtration on $G^0$ given by $G^0_{\bullet } = G_{\bullet }\cap G^0$ , and let $g_b\in {{\mathrm{poly}}}(\mathbb {Z}, G^0_{\bullet })$ be the irrational sequence defined in Proposition 3.2 for which $a^n b\Gamma = g_b(n)\Gamma $ . The irrationality of $g_b$ , Lemma 3.1 and Theorem 5.3 imply that

$$ \begin{align*} &\lim_{N\to\infty}\mathop{\mathbb{E}}\limits_{n\in[N]} \int_{G/\Gamma} f_0(b\Gamma)\cdot f_1(a^{P_1(n)}b\Gamma) \cdots f_t(a^{P_t(n)} b\Gamma) \,d\nu(b\Gamma)\\[6pt] &\quad= \int_{G^0/\Gamma^0} \lim_{N\to\infty}\mathop{\mathbb{E}}\limits_{m,n\in[N]} f_0(g_b(m)\Gamma^0)\\[6pt] &\qquad\cdot f_1(g_b(m+P_1(n))\Gamma^0) \cdots f_t(g_b(m+P_t(n))\Gamma^0) \,d\nu(b\Gamma^0)\\[6pt] &\quad = \int_{(G^0)^P/(\Gamma^0)^P} f_0 \otimes \cdots \otimes f_t \,d\nu^P, \end{align*} $$

where $(G^0)^P$ is the Leibman group for $\vec {P}$ and $\nu ^P$ is the Haar measure on $(G^0)^P/(\Gamma ^0)^P$ .

The assumption that $f_i$ vanishes on each coset of $G_{s+1}\Gamma $ in $G/\Gamma $ and Lemma 3.4 together imply that $f_i$ vanishes on each coset of $G^0_{s+1}\Gamma ^0$ inside $G^0/\Gamma ^0$ . By Lemma 5.2, the group $(G^0)^P$ contains $H=1^{i}\times G^0_{s+1} \times 1^{t-i}$ ; therefore

$$ \begin{align*} &\bigg|\int_{(G^0)^P/(\Gamma^0)^P} f_0 \otimes \cdots \otimes f_t \bigg| \leqslant \int_{(G^0)^P/H(\Gamma^0)^P} \bigg|\int_{x H(\Gamma^0)^P} f_0 \otimes \cdots \otimes f_t \bigg|\\[6pt] &\quad\leqslant\bigg(\prod_{j\neq i}\|f_j\|_\infty\bigg)\int_{(G^0)^P/H(\Gamma^0)^P} \bigg|\int_{x_i G^0_{s+1}\Gamma^0} f_i\bigg| = 0, \end{align*} $$

implying that $\mathcal {Z}_s$ is characteristic for the weak convergence of $\vec {P}$ at i.

Proof of Corollary 1.13

Let $\vec {P}\in \mathbb {R}[x,y]^{t+1}$ be an eligible homogeneous progression with $\mathcal {A}_i(\vec {P}) = s$ and let $(X,\mathcal {X},\mu ,T)$ be ergodic. By Theorem 1.4, there exists a Host–Kra factor that is characteristic for the weak and $L^2$ convergence of $\vec {P}$ . Since each Host–Kra factor is an inverse limit of nilsequences, we can approximate X by an ergodic nilsystem $(G/\Gamma , {\mathcal {G}}/\Gamma , \nu , T_a)$ . The compactness of $G/\Gamma $ and the assumption that G is generated by the connected component $G^o$ and a imply that $a^r \in G^o$ for some $r\in \mathbb {N}_+$ , and hence

(36) $$ \begin{align} \mathop{\mathbb{E}}\limits_{n\in[rN]}\prod_{i=1}^t T_a^{P_i(n)}f_i &= \mathop{\mathbb{E}}\limits_{j\in[r]} \mathop{\mathbb{E}}\limits_{n\in[N]} \prod_{i=1}^t T_a^{P_i(r(n-1)+j)} f_i \nonumber\\[3pt] &= \mathop{\mathbb{E}}\limits_{j\in[r]} \mathop{\mathbb{E}}\limits_{n\in[N]} \prod_{i=1}^t (T_a^r)^{\tilde{P}_{i,j}(n)} (T_a^{P_i(j)} f_i), \end{align} $$

where $\tilde {P}_{i,j}(n) = ({P_i(r(n-1)+j) - P_i(j)})/{r}$ . This is where we use the fact that $\vec {P}$ is eligible. The definition of eligibility implies that for any $0\leqslant j < r$ , the progression

$$ \begin{align*} \vec{\tilde{P}}_j(x,y) = (x,\; x+\tilde{P}_{1,j}(y),\ldots,\; x+\tilde{P}_{t,j}(y)) \end{align*} $$

is homogeneous and that $\mathcal {A}_i(\vec {\tilde {P}}_j) = \mathcal {A}_i(\vec {P})$ for every $0\leq i < r$ .

If $s>0$ , suppose that $\operatorname {\mathrm {\mathbb {E}}}(f_i\mid \mathcal {Z}_s(T_a)) = 0$ . Then the equality $\mathcal {Z}_s(T_a) = \mathcal {Z}_s(T_a^r)$ and the $T_a$ -invariance of $\mathcal {Z}_s$ imply that $\operatorname {\mathrm {\mathbb {E}}}(T_a^{P_i(j)} f_i\mid \mathcal {Z}_s(T_a^r)) = 0$ . We deduce from Corollary 5.4 and the total ergodicity of $T_a^r$ on each connected component of $G/\Gamma $ that the expression in (36) converges to 0 as $N\to \infty $ .

If $s=0$ , suppose that $\operatorname {\mathrm {\mathbb {E}}}(f_i\mid \mathcal {K}_{\mathrm {rat}}(T_a)) = 0$ . The total ergodicity of $T_a^r$ implies that $\mathcal {K}_{\mathrm {rat}}(T_a) = \mathcal {Z}_0(T_a^r)$ , and so $\operatorname {\mathrm {\mathbb {E}}}(T_a^{P_i(j)} f_i\mid \mathcal {Z}_0(T_a^r)) = 0$ . Again, it follows from Corollary 5.4 and the total ergodicity of $T_a^r$ on each connected component of $G/\Gamma $ that the expression in (36) converges to 0 as $N\to \infty $ .

Proof. From the definition of inhomogeneous relations it follows that each inhomogeneous relation must have degree at least $2$ . Thus, the fact that $\vec {P}$ has algebraic complexity at most $1$ immediately implies that it is homogeneous.

To prove that $\vec {P}$ is eligible, we fix $r\in \mathbb {N}_+$ and $0\leq j < r$ . We show that the progression

$$ \begin{align*} \vec{\tilde{P}}(x,y) = (x,\; x+\tilde{P}_{1,j}(y),\ldots,\; x+\tilde{P}_{t,j}(y)), \end{align*} $$

where $\tilde {P}_{i,j}(y) = ({P_i(r(y-1)-j) - P_i(y)})/{r}$ , also has algebraic complexity at most $1$ , from which the eligibility of $\vec {P}$ will follow easily. Indeed, suppose first that $\vec {\tilde {P}}$ satisfies an algebraic relation of degree $2$ :

$$ \begin{align*} \sum_{i=0}^t a_{i2}\bigg(x+\frac{P_i(r(y-1) - j)-P_i(j)}{r}\bigg)^2 + a_{i1}\bigg(x+\frac{P_i(r(y-1) - j)-P_i(j)}{r}\bigg) = 0. \end{align*} $$

Setting $y' = r(y-1) - j,\; x' = x r,\; a_{i2}' = a_{i2}/r^2,\; a_{i1}' = a_{i1}/r$ for brevity and rearranging, we deduce that

$$ \begin{align*} \sum_{i=0}^t(a_{i2}'(x'+P_i(y'))^2 + (a^{\prime}_{i1} - 2a^{\prime}_{i2}P_i(j)) (x'+P_i(y')) + a^{\prime}_{i2}P_i(j)^2 - a^{\prime}_{i1} P_i(j)) = 0. \end{align*} $$

The homogeneity of $\vec {P}$ implies that

$$ \begin{align*} \sum_{i=0}^t a^{\prime}_{i2}(x'+P_i(y'))^2 = 0, \end{align*} $$

and the fact that $\vec {P}$ has algebraic complexity at most $1$ further implies that $a^{\prime }_{02} = \cdots = a_{t2}' = 0$ . The claim $a_{02} = \cdots = a_{t2} = 0$ follows by rescaling. Thus, $\vec {\tilde {P}}$ satisfies no algebraic relation of degree 2. It follows by induction that $\vec {\tilde {P}}$ satisfies no algebraic relation of degree $d> 2$ since each such relation

(37) $$ \begin{align} Q_0(x) + Q_1(x+\tilde{P}_{1,j}(y)) + \cdots + Q_t(x+\tilde{P}_{t,j}(y)) = 0 \end{align} $$

would induce an algebraic relation of degree $d-1$ by partially differentiating (37) with respect to x. This establishes the claim that $\vec {\tilde {P}}$ has algebraic complexity at most $1$ . Thus, every algebraic relation satisfied by $\vec {\tilde {P}}$ is of the form

$$ \begin{align*} a_0 x &+ a_1 \bigg(x+\frac{P_1(r(y-1)+j) - P_1(j)}{r}\bigg) \\[3pt] & + \cdots + a_t \bigg(x+\frac{P_t(r(y-1)+j) - P_t(j)}{r}\bigg) = 0 \end{align*} $$

and corresponds to an algebraic relation

$$ \begin{align*} a_0 x + a_1 (x+P_1(y)) + \cdots + a_t (x+P_t(y)) = 0 \end{align*} $$

satisfied by $\vec {P}$ . This one-to-one correspondence between the algebraic relations satisfied by $\vec {\tilde {P}}$ and $\vec {P}$ implies the eligibility of $\vec {P}$ .

Proof of Corollary 1.15(ii)

Let $(X, \mathcal {X}, \mu , T)$ be a totally ergodic system, and suppose that $\vec {P}\in \mathbb {R}[x,y]^{t+1}$ is an integral progression with algebraic complexity at most $1$ . This implies that $\vec {P}$ is homogeneous since each inhomogeneous algebraic relation must have degree at least 2. For each $0\leqslant i\leqslant t$ , let $P_i(y) = \sum _{j=1}^d a_{i,j} Q_j(y)$ and $L_i(y_1, \ldots y_d) = \sum _{j=1}^d a_{i,j} y_j$ for some $a_{i,j}\in \mathbb {Z}$ and integral polynomials $Q_1, \ldots , Q_d$ . Letting

$$ \begin{align*} \vec{L}(x,y_1, \ldots, y_d) = (x,\; x+L_1(y_1, \ldots, y_d),\ldots, x+L_t(y_1, \ldots, y_d)), \end{align*} $$

we observe that $\vec {P}(x,y) = \vec {L}(x, Q_1(y), \ldots , Q_d(y))$ . It follows that $\vec {L}$ also has algebraic complexity at most $1$ , since each algebraic relation of degree $(j_0, \ldots , j_t)$ between terms of $\vec {L}$ would immediately imply an algebraic relation of the same degree between terms of $\vec {P}$ after substituting $y_i = Q_i(y)$ .

Using the same argument as in the proof of Corollary 5.4, we reduce the question of understanding

(38) $$ \begin{align} \lim_{N\to\infty}\mathop{\mathbb{E}}\limits_{n\in [N]} \int_X \prod_{i=0}^t T^{P_i(n)} f_i \,d\mu \end{align} $$

to understanding

(39) $$ \begin{align} \lim_{N\to\infty}\mathop{\mathbb{E}}\limits_{x,y\in[N]} F(g^P(x,y)) \end{align} $$

for each essentially bounded function $F: (G/\Gamma )^{t+1}\to \mathbb {C}$ and an irrational sequence $g\in {{\mathrm{poly}}}(\mathbb {Z}, G_{\bullet })$ for some filtration $G_{\bullet }$ on G. Following the same method to analyse

(40) $$ \begin{align} \lim_{N\to\infty}\mathop{\mathbb{E}}\limits_{y_1, \ldots, y_d\in [N]} \int_X \prod_{i=0}^t T^{L_i(y_1, \ldots, y_d)} f_i \,d\mu, \end{align} $$

we deduce that understanding (40) comes down to estimating

(41) $$ \begin{align} \lim_{N\to\infty}\mathop{\mathbb{E}}\limits_{x,y_1, \ldots, y_d\in[N]} F(g^L(x,y_1, \ldots, y_d)), \end{align} $$

where

$$ \begin{align*}g^L(x,y_1, \ldots, y_d) = (g(x), g(x+L_1(y_1, \ldots, y_d)), \ldots, g(x+L_t(y_1, \ldots, y_d))).\end{align*} $$

By Theorem 5.3, the limit in (39) equals $\int _{G^P/\Gamma ^P} F$ ; by [Reference Green, Tao, Bárány, Solymosi and SágiGT10, Theorem 11] (while Theorem 11 of [Reference Green, Tao, Bárány, Solymosi and SágiGT10] has been shown to fail in general, its corrected version, to be found in the arXiv version of the paper at https://arxiv.org/abs/1002.2028, still holds in this case, as the system under consideration is translation-invariant), the limit in (41) is $\int _{G^L/\Gamma ^L} F$ for some subgroup $G^L\leqslant G^{t+1}$ . From the fact that $\max _i \mathcal {A}_i(\vec {P})\leqslant ~1$ we deduce that $G^P = \langle h_1^{\vec {v}_1}, G_2^{t+1}: h_1\in G_1, \vec {v}_1\in \mathcal {P}_1 \rangle $ ; similarly, the construction of the group $G^L$ in [Reference Green, Tao, Bárány, Solymosi and SágiGT10] and the fact that $\vec {L}$ has algebraic complexity at most $1$ reveal that $G^L = \langle h_1^{\vec {v}_1}, G_2^{t+1}: h_1\in G_1, \vec {v}_1\in \mathcal {L}_1 \rangle $ , where

$$ \begin{align*} \mathcal{L}_1 = {{\mathrm{Span}}}_{\mathbb{R}}\{(x, x+L_1(y_1, \ldots, y_d), \ldots, x+L_t(y_1, \ldots, y_d)): x,y_1, \ldots, y_d\in\mathbb{R}\}. \end{align*} $$

We observe that $\mathcal {P}_1 = \mathcal {L}_1$ ; from this it follows that $G^P = G^L$ , and so the limits in (39) and (41) are equal. This implies that (38) and (40) are equal as well.

Proof of Theorem 5.3 using Theorem 6.7

Let $F:(G/\Gamma )^{t+1}\to \mathbb {R}$ be a continuous function. By the Stone–Weierstrass theorem, Lipschitz functions on a compact set form a dense subset of the algebra of continuous functions. Approximating F by a sequence of Lipschitz functions if necessary, we can assume without loss of generality that F is Lipschitz. We let M be the maximum of the complexity of $G/\Gamma $ and the Lipschitz norm of F.

Let $g\in {{\mathrm{poly}}}(\mathbb {Z}, G_{\bullet })$ be an irrational sequence. For each $N\in \mathbb {N}_+$ , we let $A_N$ be the maximal real number A for which g is $(A_N, N)$ -irrational. We claim that $A_N\to \infty $ as $N\to \infty $ . If not, then there exist some number $A>0$ and an index $i\in \mathbb {N}_+$ with the property that $g_i$ is not $(A,N)$ -irrational for all $N\in \mathbb {N}_+$ . We fix this i. It follows that there exists a sequence of non-trivial ith-level characters $\eta _{N}: G_i\to \mathbb {R}$ of modulus at most A such that $\|\eta _N(g_i)\|_{\mathbb {R}/\mathbb {Z}}<A/N^i$ . Since there are only finitely many ith-level characters of modulus bounded by A, we conclude that there exists a non-trivial ith-level character $\eta $ of modulus at most A such that $\|\eta (g_i)\|_{\mathbb {R}/\mathbb {Z}} < A/N^i$ for all $N\in \mathbb {N}_+$ . Taking $N\to \infty $ , we see that $\eta (g_i)~\in ~\mathbb {Z}$ , contradicting the irrationality of $g_i$ .

It therefore follows from Theorem 6.7 that

$$ \begin{align*} \mathop{\mathbb{E}}\limits_{x,y\in[N]} F(g^P(x,y)\Gamma^{t+1}) = \int_{G^P/\Gamma^P} F + O_M(A_N^{-c_M}). \end{align*} $$

Since M is constant, letting $N\to \infty $ sends the error term to 0, implying that $g^P$ is equidistributed on $G^P/\Gamma ^P$ as claimed.

Proof of Proposition 7.5 using Theorem 6.7

Let $g\in {{\mathrm{poly}}}(\mathbb {Z}, G_{\bullet })$ be A-irrational and N-periodic. We claim that g is $(A,Nk)$ -irrational for all sufficiently large $k\in \mathbb {N}_+$ . If not, then there exists $1\leqslant i\leqslant s$ such that for each $k\in \mathbb {N}_+$ there exists an ith-level character $\eta _{i,k}:G_i\to \mathbb {R}$ of complexity at most A satisfying $\|\eta _{i,k}(g_i)\|_{\mathbb {R}/\mathbb {Z}}< A/(Nk)^i$ . The N-periodicity of $g_i$ implies that $g_i^{N^i}\in \Gamma _i$ mod $G_{i+1}^\nabla $ [Reference Candela and SisaskCS12, Lemma 5.3]; hence $\eta _{i,k}(g_i)\in ({1}/{N^i})\mathbb {Z}$ . Thus, $\eta _{i,k}(g_i)\in \mathbb {Z}$ whenever $k^i>A$ . In particular, since we can take k arbitrarily large, there exists a non-trivial ith-level character $\eta _{i,k}$ of complexity at most A for which $\eta _{i,k}(g_i)\in \mathbb {Z}$ , contradicting the A-irrationality of g. Hence g is $(A,Nk)$ -irrational for all sufficiently large $k\in \mathbb {N}_+$ .

Applying Theorem 6.7, we deduce that

$$ \begin{align*} \mathop{\mathbb{E}}\limits_{x,y\in\mathbb{Z}/N\mathbb{Z}} F(g^P(x,y)\Gamma^{t+1}) & = \mathop{\mathbb{E}}\limits_{x,y\in[Nk]} F(g^P(x,y)\Gamma^{t+1}) + O(1/k)\\ & = \int_{G^P/\Gamma^P} F + O_M(A^{-c_M}) + O(1/k) \end{align*} $$

for all sufficiently large $k\in \mathbb {N}_+$ . Taking $k\to \infty $ finishes the proof.