Proof. Statements (i)–(vi) follow directly from the definitions. Statements (iv) and (v) follow from properties (i)–(iii) by induction.

Proof. Take generators $g_{1}^{\vec {v}_{1}}\in G^{P}_{j_1}$ and $g_{2}^{\vec {v}_{2}}\in G^{P}_{j_2}$ for some elements $g_1\in G_1$ and $g_2\in G_2$ as well as vectors $\vec {v}_{1} \in \mathcal {Q}_{i_1, j_1}$ and $\vec {v}_{2} \in \mathcal {Q}_{i_2, j_2}$. We want to show that their commutator is in $G^{P}_{j_1+j_2}$. If this is true for the generators, then by Lemma 7.3 of [Reference Green and Tao19] it holds for arbitrary two elements of $G_{j_1}$ and $G_{j_2}$, proving the lemma.

By (41), we have

\[ [g_{1}^{\vec{v}_{1}}, g_{2}^{\vec{v}_{2}}]=[g_1, g_2]^{\vec{v}_1 \cdot \vec{v}_2}\prod_\alpha g_\alpha^{Q_\alpha(\vec{v}_1, \vec{v}_2)} \]

where each $g_\alpha$ is a commutator of $k_1$ copies of $g_1$ and $k_2$ copies of $g_2$ for some ${k_1, k_2\geqslant 1}$ satisfying $k_1+k_2\geqslant 3$, and

\[ Q_\alpha(\vec{u}, \vec{w}):=(Q_\alpha(u_1,w_1), \ldots, Q_{\alpha}(u_t, w_t)) \]

for some polynomial $Q_\alpha (x_1, x_2)$ that has degree at most $k_1$ in $x_1$ and at most $k_2$ in $x_2$, and vanishes when $x_1 = 0$ or $x_2 = 0$.

By the filtration property of $G$, the commutator $[g_1, g_2]$ is in $G_{i_1+i_2}$. We moreover have that

\[ \vec{v}_1\cdot\vec{v}_2\in \mathcal{Q}_{i_1, j_1}\cdot\mathcal{Q}_{i_2, j_2}\subseteq\mathcal{Q}_{i_1+i_2,j_1+j_2} \]

by part (vi) of Lemma 3.1. Consequently, the element $[g_1, g_2]^{\vec {v}_1 \cdot \vec {v}_2}$ is contained in $G^{P}_{j_1+j_2}$.

We handle the terms $g_\alpha ^{Q_\alpha (\vec {v}_1, \vec {v}_2)}$ in a similar manner. If $g_\alpha$ is a commutator of $k_1$ copies of $g_1$ and $k_2$ copies of $g_2$, then $g_\alpha \in G_{k_1 i_1 + k_2 i_2}$ by the filtration property of $G$. The polynomial $Q_\alpha$ can then be written as $Q_\alpha (x_1, x_2) = \sum \nolimits _{ \substack {1\leqslant l_1\leqslant k_1,\\ 1\leqslant l_2\leqslant k_2}}\beta _{l_1, l_2}x_1^{l_1} x_2^{l_2}$, and so

\begin{align*} Q_\alpha(\vec{v}_1,\vec{v}_2) & = \sum_{ \substack{1\leqslant l_1\leqslant k_1,\\ 1\leqslant l_2\leqslant k_2}}\beta_{l_1, l_2} (\vec{v}_1)^{l_1} \cdot(\vec{v}_2)^{l_2}\in \sum_{ \substack{1\leqslant l_1\leqslant k_1,\\ 1\leqslant l_2\leqslant k_2}}\mathcal{Q}_{l_1 i_1 + l_2 i_2, l_1 j_1 + l_2 j_2}\\ & \subseteq \sum_{ \substack{1\leqslant l_1\leqslant k_1,\\ 1\leqslant l_2\leqslant k_2}}\mathcal{Q}_{k_1 i_1 + k_2 i_2, l_1 j_1 + l_2 j_2}\subseteq \mathcal{Q}_{k_1 i_1 + k_2 i_2, j_1+j_2} \end{align*}

by Lemma 3.1. Thus, $g_\alpha ^{Q_\alpha (\vec {v}_1, \vec {v}_2)}$ is contained in $G^{P}_{j_1+ j_2}$ for each $\alpha$ which implies that $[g_{1}^{\vec {v}_{1}}, g_{2}^{\vec {v}_{2}}]\in G^{P}_{j_1+j_2}$.

Proof. We first decompose

\[ {{\vec{P}(\textbf{x})}\choose{i}}=\sum_{\textbf{j}}\vec{v}_{i, \textbf{j}}{{\textbf{x}}\choose{\textbf{j}}}. \]

and note that $\vec {v}_{i,\textbf {j}}\in \mathcal {P}_{i, |\textbf {j}|}$. Therefore, $g_i^{\vec {v}_{i,\textbf {j}}}\in G^{P}_{|\textbf {j}|}$ by definition of $G_{|\textbf {j}|}^{P}$. Using (40) and (41), we regroup the terms of

\[ g^{P}(\textbf{x}) = \prod_{i=0}^{s} g_i^{{\vec{P}(\textbf{x})}\choose{i}} = \prod_{i=0}^{s} g_i^{\sum_{\textbf{j}}\vec{v}_{i, \textbf{j}}{{\textbf{x}}\choose{\textbf{j}}}} \]

to bring all the elements involving the same monomial ${{\textbf {x}}\choose {\textbf {j}}}$ together. Thus, the Taylor coefficient of ${{\textbf {x}}\choose {\textbf {j}}}$ is of the form

(14)\begin{equation} \prod_{i=0}^{s} g_i^{\sum_{\textbf{j}}\vec{v}_{i, \textbf{j}}} \prod_\alpha g_\alpha^{\vec{v}_{\alpha}}, \end{equation}

where the terms $g_\alpha ^{\vec {v}_{\alpha }}$ come from applying (40) and (41). For each label $\alpha$, the element $g_\alpha$ is a commutator consisting of $k_r$ copies of $g_{i_r}$ for $1\leqslant r\leqslant n$, and so $g_\alpha \in G_{i_1 k_1 + \cdots + i_n k_n}$ by the filtration property of $G$. The vector $\vec {v}_\alpha$ is a rational multiple of ${\vec {v}_{i_1, |\textbf {j}_1|}^{l_1}} \cdots {\vec {v}_{i_n, |\textbf {j}_n|}^{l_n}}$ for some $1\leqslant l_1\leqslant k_1$, …, $1\leqslant l_n\leqslant k_n$ and $|\textbf {j}_1|+ \cdots + |\textbf {j}_n|\geqslant |\textbf {j}|$. Therefore, $\vec {v}_\alpha \in \mathcal {Q}_{i_1 l_1 + \cdots + i_n l_n, |\textbf {j}_1|+ \cdots + |\textbf {j}_n|}$ by part (vi) of Lemma 3.1. It follows from parts (iv) and (v) of the same lemma that

\[ \mathcal{Q}_{i_1 l_1 + \cdots + i_n l_n, |\textbf{j}_1|+ \cdots + |\textbf{j}_n|}\subseteq\mathcal{Q}_{i_1 l_1 + \cdots + i_n l_n, |\textbf{j}|}\subseteq\mathcal{Q}_{i_1 k_1 + \cdots + i_n k_n, |\textbf{j}|}, \]

implying that $g_\alpha ^{\vec {v}_\alpha }\in G^{P}_{|\textbf {j}|}$ for each $\alpha$. Thus, the coefficient (14) is in $G^{P}_{|\textbf {j}|}$, as claimed.

Proof. Whenever $i=1$ or $j=1$, the lemma follows by definition. Other cases follow by induction on $(i,j)$.

Proof. Since $\vec {P}$ satisfies the filtration condition, we have by Lemma 3.5 that $\mathcal {P}_{i,j}=\mathcal {Q}_{i,j}$. By definition of $G^{P}_j$, the group is generated by elements of the form $g^{\vec {v}}$ for $i\in \mathbb {N}_+$, $g\in G_i$, $\vec {v}=\sum \nolimits _{r=1}^{l_i+k_i}a_r\vec {v}_r$ and $a_r\in \mathbb {R}$. Letting $\vec {w}=\sum \nolimits _{r=1}^{l_i}a_r\vec {v}_r$, we observe that $g^{\vec {w}}\in G^{P}_{j+1}$, and so

\[ g^{\vec{v}}=g^{\vec{v}-\vec{w}} = (g^{a_{l_i+1}})^{\vec{v}_{l_i + 1}}\cdots (g^{a_{l_i+k_i}})^{\vec{v}_{l_i + k_i}} \mod G^{P}_{j+1}. \]

Thus each generator $g^{\vec {v}}$ in $G^{P}_{j}$ is a product of an element from $G^{P}_{j+1}$ and $h_r^{\vec {v}_r}$ for $l_i+1 \leqslant r\leqslant l_i+k_i$ and $h_r=g^{a_r}\in G_i$.

Proof. The case $j>2i$ follows from the fact that the polynomial map ${{\vec {P}(x,y)}\choose {i}}$ has degree $2i$. For the case $j=2i$, note that the only monomial of ${{\vec {P}(x,y)}\choose {i}}$ of degree $2i$ is ${{y}\choose {2i}}$, which comes from the term $y^{2}$ in $\vec {P}(x,y)$, and one can verify directly that the coefficient of ${{y}\choose {2i}}$ in ${{\vec {P}(x,y)}\choose {i}}$ is $(0,0,{(2i)!}/{i!}, {(2i)!}/{i!})$. Consequently, $\mathcal {P}_{i,2i}={\textrm {Span}}\{\vec {v}_3\}$.

In the case $i+1\leqslant j\leqslant 2i-1$, we have

(18)\begin{equation} \mathcal{P}_{i,j}\subseteq\{{0}\}\times\{0\}\times\mathbb{R}\times\mathbb{R} \end{equation}

because the polynomials ${{x}\choose {i}}$ and ${{x+y}\choose {i}}$ both have degree $i$, and we claim that (18) is an equality. Since $\mathcal {P}_{i,2i}\subseteq \mathcal {P}_{i,j}$, we know that $\mathcal {P}_{i,j}$ contains $\vec {v}_3$, and it remains to show that $\mathcal {P}_{i,j}$ contains a vector of the form $(0,0,a,b)$ for some $a \neq b$. To this goal, we look at the coefficient of ${{y}\choose {2i-1}}$ in ${{\vec {P}(x,y)}\choose {i}}$. Note that

\[ {{y+y^{2}}\choose{i}}={{y^{2}}\choose{i}}+y{{y^{2}}\choose{i-1}}+R(y), \]

where $R(y)$ has degree $2i-2$. In particular, the polynomial $y{{y^{2}}\choose {i-1}}$ has degree $2i-1$, and so has a non-zero coefficient at ${{y}\choose {2i-1}}$. Therefore, the coefficient of ${{y}\choose {2i-1}}$ in ${{\vec {P}(x,y)}\choose {i}}$ is of the form $(0,0,a,b)$ for distinct integers $a\neq b$, implying that $\mathcal {P}_{i,j}=\{0\}\times \{0\}\times \mathbb {R}\times \mathbb {R}$, as claimed.

To handle the case $1\leqslant j\leqslant i$, we look at $\mathcal {P}_{i,i}$ and note that $\mathcal {P}_{i,j}\supseteq \mathcal {P}_{i,i}$ for these values of $j$ by Lemma 3.1. By the previous case, we already know that $\mathcal {P}_{i,j}\supseteq \{0\}\times \{0\}\times \mathbb {R} \times \mathbb {R}$. The space $\mathcal {P}_{i,i}$ moreover contains $\vec {v}_1$, which is the coefficient of ${{x}\choose {i}}$ in ${{\vec {P}(x,y)}\choose {i}}$, and $\vec {v}_2$, which is the coefficient of ${{x}\choose {i-1}}y$ by Lemma 3.7 and (17). Thus, $\mathcal {P}_{i,i}$ is all of $\mathbb {R}^{4}$.

Proof. The proof proceeds by verifying that $\mathcal {P}_{i_1,j_1}\cdot \mathcal {P}_{i_2,j_2}\subseteq \mathcal {P}_{i_1+i_2, j_1+j_2}$ for all ${i_1,i_2,j_1,j_2\geqslant 1}$ on a case-by-case basis using Lemma 4.2. The details are rather tedious and unsophisticated; therefore, we leave them to the reader.

Proof. The lemma follows from combining Lemmas 4.3 and 3.6 with the structural information on $\mathcal {P}_{i,j}$ that we obtain from Lemma 4.2.

Proof. By Lemma C.1, $g_i^{p^{i}}\in \varGamma _i$ mod $G_{i+1}$, and so $g_i^{\vec {v}} = (\gamma _i h)^{\vec {v}}$ for some $\gamma _i\in \varGamma _i$ and $h\in G_{i+1}$. Using (40), Lemmas 3.1 and 3.8, we note that $h^{\vec {v}}\in G_{j+1}^{P}$, and similarly for commutators emerging from applying (40). We, therefore, have $g_i^{p^{i}\vec {v}}\in \varGamma _j^{P}\mod G^{P}_{j+1}$, from which we deduce that $p^{i}\eta (g_i^{\vec {v}})\in \mathbb {Z}$. The assumption that ${\eta (g_i^{a\vec {v}})\in \mathbb {Z}}$ for some non-zero integer $a$ further implies that $\gcd (a, p^{i})\eta (g_i^{\vec {v}})\in \mathbb {Z}$. Taking $p$ sufficiently large guarantees that $\gcd (a, p^{i})=1$, and so $\eta (g_i^{\vec {v}})\in \mathbb {Z}$.

Proof. The case $j>id_2$ follows trivially from the fact that the polynomial ${{\vec {P}(x,y)}\choose {i}}$ has degree $id_2$.

In the process of deducing the other cases, we shall use the fact that

(22)\begin{equation} \vec{P}(x,y)=\vec{v}_1 x + \vec{P}(0,y)= \vec{v}_1 x + \vec{v}_2 Q(y) + \vec{v}_3 R(y) \end{equation}

and (16). Combining (22) and (16), we rewrite ${{\vec {P}(x,y)}\choose {i}}$ as

(23)\begin{align} {{\vec{P}(x,y)}\choose{i}} = \vec{v}_1{{x}\choose{i}}+\sum_{l=1}^{i-1}{{x}\choose{l}}{{\vec{P}(0,y)}\choose{i-1}}+\left(0, {{Q(y)}\choose{i}}, {{R(y)}\choose{i}}, {{Q(y)+R(y)}\choose{i}}\right). \end{align}

We can further rewrite the last vector in the sum as

(24)\begin{align} & \left(0, {{Q(y)}\choose{i}}, {{R(y)}\choose{i}}, {{Q(y)+R(y)}\choose{i}} \right)\nonumber\\ & \quad= \vec{v}_2 {{Q(y)}\choose{i}} + \sum_{l=1}^{i-1}\vec{v}_4{{Q(y)}\choose{l}}{{R(y)}\choose{i-l}}+\vec{v}_3{{R(y)}\choose{i}}. \end{align}

From (23) and (24), we see that the only monomials in ${{\vec {P}(x,y)}\choose {i}}$ of degree greater than $(i-1)d_2+d_1$ in ${{\vec {P}(x,y)}\choose {i}}$ have coefficients of the form $a\vec {v}_3$ for some $a\in \mathbb {Z}$. In particular, the value $a$ will be non-zero for the coefficient of ${{y}\choose {i d_2}}$, implying the case $(i-1)d_2+d_1+1 \leqslant j \leqslant id_2$ by Lemma 4.6.

To deduce the case $i d_1+1\leqslant j\leqslant (i-1)d_2+d_1$, we note that the projection of ${{\vec {P}(x,y)}\choose {i}}$ onto the first coordinate has degree $i$, while its projection onto the second coordinate has degree $id_1$, and so

\[ \mathcal{P}_{i,j}\subseteq 0\times 0 \times \mathbb{R} \times \mathbb{R} \]

for these values of $j$. We claim this is an equality. We already know that ${\vec {v}_3\in \mathcal {P}_{i,j}}$ since its multiple is the coefficient of ${{y}\choose {i d_2}}$. We claim that the coefficient of the monomial ${{y}\choose {(i-1)d_2+d_1}}$ is of the form $a\vec {v}_3+b\vec {v}_4$ for integers $a,b$ such that $b\neq 0$, which will imply this case. This follows from (24), the assumption $d_1< d_2$, and the observation that the polynomial ${Q(y)}{{R(y)}\choose {i-1}}$ has degree $(i-1)d_2+d_1$, thus contributing to the coefficient of ${{y}\choose {(i-1)d_2+d_1}}$.

The next case, $i+1\leqslant j\leqslant id_1$, only happens if $d_1>1$, and so we make this assumption. Since the projection of ${{\vec {P}(x,y)}\choose {i}}$ onto the first coordinate has degree $i$, we deduce that

\[ \mathcal{P}_{i,j}\subseteq 0\times \mathbb{R} \times \mathbb{R} \times \mathbb{R}. \]

To prove that this is an equality, it remains to show in the light of the previous cases that a vector of the form $a\vec {v}_2+b\vec {v}_3+c\vec {v}_3$ is in $\mathcal {P}_{i,j}$ for some $a\neq 0$. Since the projection of ${{\vec {P}(x,y)}\choose {i}}$ onto the second coordinate has degree $i d_1$, the coefficient of ${{y}\choose {i d_1}}$ is of this form, implying this case.

If $d_1 =1$, then the same argument implies that $\vec {v}_2\in \mathcal {P}_{i,i}$.

Finally, the case $1\leqslant j\leqslant i$ follows from combining the previous case, Lemma 3.1 and the observation that $\vec {v}_1$ is the coefficient of ${{x}\choose {i}}$.

Proof. Suppose that the sequence $g^{P}\in {\textrm {poly}}(\mathbb {Z}^{2},G^{P}_\bullet )$ is not $O_{M}(A^{-c_M})$- equidistributed. By Theorem 2.9, there exists a non-trivial horizontal character $\eta :G^{P}\to \mathbb {R}$ of complexity at most $cA$ for some $c>0$ to be chosen later, such that $\eta \circ g^{P}\in \mathbb {Z}$. Let $j$ be the largest natural number such that $\eta |_{G^{P}_j}\neq 0$. By assumption, $\eta$ annihilates ${G^{P}_{j+1}}$.

If $j=1$, we proceed exactly as in Theorem 4.5, and so we assume that $j>1$. For any $i\geqslant 1$, we define

\[ \xi_{i,k}(h)=\eta(h^{\vec{v}_k}) \]

for $h\in G_i$ and each $k\in \{1,2,3,4\}$ such that $\vec {v}_k\in \mathcal {P}_{i,j}$ but $\vec {v}_k\notin \mathcal {P}_{i,j+1}$. The maps $\xi _{i,k}$ define $i$th level characters on $G$ by Corollary 3.9. By Lemma 3.6, if all of $\xi _{i,k}$ were trivial, so would be $\eta$, implying that at least one of $\xi _{i,k}$ is non-trivial. The bound on the modulus of $\eta$ and the fact that the vectors $\vec {v}_k$ have entries of size $O(1)$ imply that $|\xi _{i,k}|\leqslant A$, provided that the constant $c$ is appropriately chosen.

Our goal is to show that for all pairs $(i,k)$ as above, we have $\xi _{i,k}(g_i)\in \mathbb {Z}$. Since at least one of these $\xi _{i,k}$ is non-trivial and of modulus at most $A$, we obtain a contradiction of the $A$-irrationality of $g$, implying that $g^{P}$ is $O_{M}(A^{-c_M})$-equidistributed. By enumerating all such pairs $(i,k)$, we observe that all we have to show is that $\eta$ sends the following elements to $\mathbb {Z}$:

1. (i) $g_j^{\vec {v}_1}$;

2. (ii) $g_{{j}/{d_1}}^{\vec {v}_2}$, if $d_1|j$;

3. (iii) $g_{{(j+d_2-d_1)}/{d_2}}^{\vec {v}_4}$, if $d_2| (j-d_1)$;

4. (iv) $g_{{j}/{d_2}}^{\vec {v}_3}$, if $d_2|j$.

That $\eta (g_j^{\vec {v}_1})\in \mathbb {Z}$ follows from observing that this is precisely the coefficient of ${{x}\choose {j}}$ is $\eta (g_j^{\vec {v}_1})$; showing that other elements are sent to $\mathbb {Z}$ is a bit more involved.

We assume $d_1|j$, and we claim that $\eta (g_{{j}/{d_1}}^{\vec {v}_2})\in \mathbb {Z}$. From Lemma 3.7 we know that the coefficient of ${{x}\choose {{j}/{d_1}-1}}{{y}\choose {d_1}}$ is of the form

\[ a\eta(g_{{j}/{d_1}}^{\vec{v}_2})+b\eta(g_{{j}/{d_1}}^{\vec{v}_3}) + \sum_{i={j}/{d_1}+1}^{s} \eta(g_i^{\vec{w}_i}) \]

for some integers $a,b$ such that $a\neq 0$, and some integer vectors $\vec {w}_i\in {\textrm {Span}}\{\vec {v}_2, \vec {v}_3, \vec {v}_4\}$. If $i\geqslant {j}/{d_1}+1$, then $j\leqslant id_1 - d_1$, and so $j+1\leqslant id_1$. It follows from this that $\vec {w}_i\in \mathcal {P}_{i,j+1}$, therefore $\eta (g_i^{\vec {w}_i})=0$. We moreover have that $\vec {v}_3\in \mathcal {P}_{{j}/{d_1},j+1}$, and so $\eta (g_{{j}/{d_1}}^{\vec {v}_3})=0$ as well. From this, Lemma 4.6 and the vanishing of the coefficient of ${{x}\choose {{j}/{d_1}-1}}{{y}\choose {d_1}}$ mod $\mathbb {Z}$, we conclude that $\eta (g_{{j}/{d_1}}^{\vec {v}_2})\in \mathbb {Z}$.

We are left with showing that the elements in (iii) and (iv) are sent to $\mathbb {Z}$ by $\eta$. Note that since $1\leqslant d_1< d_2$, the number $d_2$ cannot simultaneously divide $j-d_1$ and $j$; therefore, only one of these cases at a time is available. We assume that $d_2$ divides $j$, and the other case will follow similarly. To show that $\eta (g_\frac {j}{d_2}^{\vec {v}_3})\in \mathbb {Z}$, we look at the coefficient of ${{y}\choose {d_2}}$, which is of the form

(25)\begin{equation} a\eta(g_{{j}/{d_2}}^{\vec{v}_3}) + \sum_{i={j}/{d_2}+1}^{s} \eta(g_i^{\vec{w}_i}) \end{equation}

for some non-zero integer $a$ and vectors $w_i\in {\textrm {Span}}\{\vec {v}_2, \vec {v}_3, \vec {v}_4\}\cap \mathcal {P}_{i,j}\cap \mathbb {Z}^{4}$. By Lemma 5.1, we have $w_i\in \mathcal {P}_{i,j+1}$ unless $j = id_1$ or $j=(i-1)d_2+d_1$. If $d_1$ divides $j$, then we have already shown that $\eta (g_{{j}/{d_1}}^{\vec {v}_2})\in \mathbb {Z}$, and we moreover have $\eta (g_{{j}/{d_1}}^{\vec {v}_3})=0$ and $\eta (g_{{j}/{d_1}}^{\vec {v}_4})=0$ since $\vec {v}_3, \vec {v}_4\in \mathcal {P}_{{j}/{d_1}, j+1}$. Therefore, $\eta (g_{{j}/{d_1}}^{\vec {w}})\in \mathbb {Z}$ for any $w\in {\textrm {Span}}\{\vec {v}_2, \vec {v}_3, \vec {v}_4\}\cap \mathcal {P}_{i,j}\cap \mathbb {Z}^{4}$. The case $j=(i-1)d_2+d_1$ does not happen by our assumption that $d_2$ does not divide $j-d_1$. Thus the sum in (25) vanishes mod $\mathbb {Z}$, implying that $a\eta (g_{{j}/{d_2}}^{\vec {v}_3})\in \mathbb {Z}$. That $\eta (g_{{j}/{d_2}}^{\vec {v}_3})\in \mathbb {Z}$ follows by Lemma 4.6.

Proof. The statement is trivial for $j>id_2$. To obtain the expressions for $\mathcal {P}_{i,j}$ for other values of $j$, we make two observations. First, we note from Lemma 3.7 that for $0\leqslant k\leqslant i-1$, the coefficient of ${{x}\choose {k}}{{y}\choose {l}}$

\[ a_{k,l}\vec{v}_2^{i-k} + b_{k,l}\vec{v}_3^{i-k} \]

for some integers $a_{k,l}$ which satisfy $a_{k,l}=0$ if $l>(i-k)d_1$ and $b_{k,l}=0$ if $l>(i-k)d_2$. Moreover, the numbers $a_{k,(i-k)d_1}$ and $b_{k, (i-k)d_2}$ are non-zero since $Q$ and $R$ have degrees $d_1$, $d_2$, respectively. By substituting $k=0,1$ and using $d_1< d_2$, we deduce that

\[ \vec{v}_2^{i}\in\mathcal{P}_{i,i d_1}, \quad \vec{v}_2^{i-1}\in\mathcal{P}_{i,(i-1)d_1+1}, \quad \vec{v}_3^{i}\in\mathcal{P}_{i,i d_2}, \quad \vec{v}_3^{i-1}\in\mathcal{P}_{i,(i-1)d_2+1}, \]

but

\[ \vec{v}_2^{i}\notin\mathcal{P}_{i,i d_1+1}, \quad \vec{v}_2^{i-1}\notin\mathcal{P}_{i,(i-1)d_1+2}, \quad \vec{v}_3^{i}\notin\mathcal{P}_{i,i d_2 + 1}, \quad \vec{v}_3^{i-1}\notin\mathcal{P}_{i,(i-1)d_2+2}. \]

Second, we observe that for $k>1$, we have $\vec {v}_2^{i-k}\in {\textrm {Span}}\{\vec {v}_2^{i}, \vec {v}_2^{i-1}\}$ and $\vec {v}_3^{i-k}\in {\textrm {Span}}\{\vec {v}_3^{i}, \vec {v}_3^{i-1}\}$, and moreover $(i-k')d_r+k'\leqslant (i-k)d_r + k$ for any $k'< k$ and $r\in \{1, 2\}$. From this, we see that to specify a basis for $\mathcal {P}_{i,j}$, it is sufficient to look at whether $\vec {v}_2^{i-1}$, $\vec {v}_2^{i}$, $\vec {v}_3^{i-1}$ and $\vec {v}_3^{i}$ are in $\mathcal {P}_{i,j}$ or not. The statements for $j>i$ follow by combining these observations.

Lastly, we note that $\vec {v}_1$ is the coefficient of ${{x}\choose {i}}$, which together with $\mathcal {P}_{i,j}\supseteq \mathcal {P}_{i,j+1}$ implies the case $1\leqslant j\leqslant i$.

Proof. Suppose that the sequence $g^{P}\in {\textrm {poly}}(\mathbb {Z}^{2},G^{P}_\bullet )$ is not $O_{M}(A^{-c_M})$- equidistributed. By Theorem 2.9, there exists a non-trivial horizontal character $\eta :G^{P}\to \mathbb {R}$ of complexity at most $cA$ for an appropriately chosen $c>0$, such that $\eta \circ g^{P}\in \mathbb {Z}$. Let $j$ be the largest natural number such that $\eta |_{G^{P}_j}\neq 0$. By assumption, $\eta$ annihilates ${G^{P}_{j+1}}$.

Like in Theorems 4.5 and 5.3, we use the properties of $\eta$ and the structural information on $G^{P}$ contained in Lemma 6.1 to contradict the $A$-irrationality of $g$. We do this by inspecting the coefficients of $\eta \circ g^{P}$. For any $i\geqslant 1$, we define

\[ \xi_{i,k,l}(h)=\eta(h^{\vec{v}_k^{l}}) \]

for $h\in G_i$ and each $k\in \{1,2,3\}$, $l\in \{i-1,i\}$ such that $\vec {v}_k^{l}\in \mathcal {P}_{i,j}$ but $\vec {v}_k^{l}\notin \mathcal {P}_{i,j+1}$. The maps $\xi _{i,k,l}$ define $i$th level characters on $G$ by Corollary 3.9. By definition of $G^{P}_{j}$, the group is generated precisely by $G^{P}_{j+1}$ and the elements of the form $h^{\vec {v}_k^{l}}$ for $h\in G_i$, $\vec {v}_k^{l}\in \mathcal {P}_{i,j}\setminus {\mathcal {P}_{i,j+1}}$ and $i\geqslant 1$. Therefore, if all of $\xi _{i,k,l}$ were trivial, so would be $\eta$, implying that at least one of $\xi _{i,k,l}$ is non-trivial. The bound on the modulus of $\eta$ and the fact that the vectors $\vec {v}_k^{l}$ have entries of size $O(1)$ imply that $|\xi _{i,k,l}|\leqslant A$, provided that the constant $c$ is appropriately chosen.