Proof. See Proposition 9.3 of Koymans and Pagano [Reference Koymans and Pagano13].

Proof. Hoeffding’s inequality shows that the proportion of F that are $\epsilon $ -bad is at most

(3.1) $$ \begin{align} \frac{|\{F \in V : F \ \mathrm{is} \ \epsilon\mathrm{-bad}\}|}{|V|} \leq 2\exp(-2\epsilon^2|X|). \end{align} $$

Define

$$\begin{align*}a := |X| - \prod_{i = 1}^m \left(|X_i| - 1\right). \end{align*}$$

By Lemma 3.1, we see that the kernel of d is an a-dimensional vector space. Combining this with equation (3.1), we infer that

$$\begin{align*}\frac{|\{g \in \mathcal{A}(X) : g \ \mathrm{is} \ \epsilon\mathrm{-bad}\}|}{|\mathcal{A}(X)|} \leq \frac{|\{F \in V : F \ \mathrm{is} \ \epsilon\mathrm{-bad}\}|}{|\mathcal{A}(X)|} \leq 2^{a + 1} \cdot \exp(-2\epsilon^2|X|), \end{align*}$$

which is the theorem.

Proof. This follows from the Chebotarev density theorem; see [Reference Smith18, Proposition 6.5].

Proof. This is an easy consequence of the large sieve inequality stated in the work of Jutila [Reference Jutila12, Lemma 3]; see Proposition 6.6 in Smith [Reference Smith18].

Proof. Let $\kappa :=c_4+\delta $ . Since $r<t_1^{c_4}$ , it suffices to show that

$$\begin{align*}\left| |X(a)|-2^{-|\mathcal{M}|}|X|\right|\leq r t_1^{-\kappa}\cdot 2^{-|\mathcal{M}|}|X|. \end{align*}$$

We proceed by induction on r. Define

$$\begin{align*}X_j(a,x_1):=\left\{x_j\in X_j: \left(\frac{x_1}{x_j}\right)=a(1,j)\text{ if }(1,j)\in\mathcal{M}\right\}. \end{align*}$$

First consider $(1,j)\in \mathcal {M}$ , where $j> k$ . Apply Proposition 5.1 to

$$\begin{align*}K=\mathbb{Q}(\sqrt{-1},\sqrt{p}:p\in P), \quad L=\mathbb{Q}(\sqrt{x_1}) \end{align*}$$

and

$$\begin{align*}F:\sigma\mapsto \begin{cases} 1-2^{-|P|-2} &\text{if } \sigma=\left(\frac{KL/\mathbb{Q}}{x_j}\right)\text{ for some }x_j\in X_j(a,x_1),\\ \hfil -2^{-|P|-2} &\text{otherwise.} \end{cases} \end{align*}$$

Notice that $\left (\frac {KL/\mathbb {Q}}{x_j}\right )$ is independent of the choice $x_j\in X_j(a,x_1)$ . By Siegel’s theorem, for $D_1$ sufficiently large, we have $1-\beta>D_1^{-c_1/6}$ if $\beta $ is an exceptional real zero of $L(s,\chi _d)$ with $|d|<D_1$ . Then

$$\begin{align*}t_j^{\beta} < t_j\exp\left(-(\log t_j)^{5/6}\right). \end{align*}$$

We have the bounds

$$\begin{align*}\log |\Delta_{K_0}\Delta_L|\ll |P|\log t_1 \leq (\log t_1)^2\ll(\log t_j)^{\frac{2}{c_6}} \end{align*}$$

and

$$\begin{align*}|\operatorname{\mathrm{Gal}}(KL/\mathbb{Q})| = 2^{|P| + 2} \leq 2^{1+\log t_1}<t_1<\exp\left((\log t_j)^{\frac{1}{c_6}}\right). \end{align*}$$

Since $c_6>3$ , we have by a double application of Proposition 5.1 and partial summation

$$\begin{align*}\left|\sum_{s_j < p < t_j}F\left(\left(\frac{KL/\mathbb{Q}}{p}\right)\right)\right| \leq t_j\exp\left(-(\log t_j)^{1/3}\right) \end{align*}$$

for sufficiently large $D_1$ . Repeating this for the field $K/\mathbb {Q}$ , we get after possibly enlarging $D_1$

$$\begin{align*}\left||X_j(a,x_1)|-\frac{1}{2}|X_j|\right|\leq 2t_j\exp\left(-(\log t_j)^{1/3}\right)<t_1^{-1}|X_j|. \end{align*}$$

Next consider $(1,j)\in \mathcal {M}$ , where $j\leq k$ . Note that $\frac {1}{4}-c_2c_3-c_5-2\kappa> 0$ by assumption, so we can fix a positive constant $\epsilon $ such that $2\epsilon <\frac {1}{4}-c_2c_3-c_5-2\kappa $ . The large sieve in Proposition 5.2 gives

$$\begin{align*}\sum_{x_1\in X_1}\left|\sum_{x_j\in X_j}\left(\frac{x_1}{x_j}\right)\right|\ll_{\epsilon} t_jt_1^{3/4+\epsilon}. \end{align*}$$

From the identity

$$\begin{align*}|X_j(a,x_1)|=\frac{1}{2}\sum_{x_j\in X_j}\left(a(1,j)\left(\frac{x_1}{x_j}\right)+1\right) =\frac{a(1,j)}{2}\sum_{x_j\in X_j}\left(\frac{x_1}{x_j}\right)+\frac{1}{2}|X_j|, \end{align*}$$

we deduce that for sufficiently large $D_1$ ,

$$\begin{align*}\sum_{x_1\in X_1}\left| |X_j(a,x_1)| -\frac{1}{2}|X_j| \right| =\frac{1}{2}\sum_{x_1\in X_1}\left|\sum_{x_j\in X_j}\left(\frac{x_1}{x_j}\right)\right| \leq t_1^{-\frac{1}{4}+c_2c_3+\epsilon}|X_1||X_j|.\end{align*}$$

Let $B_1:=c_5+\kappa $ and $B_2:=\kappa +\epsilon $ ; then $B_1+B_2< \frac {1}{4}-c_2c_3-\epsilon $ . We conclude that

$$\begin{align*}\left| |X_j(a,x_1)|-\frac{1}{2}|X_j| \right|<t_1^{-B_2}|X_j| \text{ for all }(1,j)\in\mathcal{M}\text{ and }j\leq k \end{align*}$$

holds with at most $kt_1^{-B_1}|X_1|$ exceptions $x_1 \in X$ . Call the set of exceptions $X_1^{\text {bad}}(a)$ .

We bound the size of the set of exceptions $X^{\text {bad}}(a)=X(a)\cap \pi _1^{-1}(X_1^{\text {bad}}(a))$ in X. First, fix some $x_1\in X_1$ and move $x_1$ to P. Apply the induction hypothesis to

$$\begin{align*}X_2\times X_3\times \dots \times X_k\times X_{k+1}(a,x_1)\times\dots\times X_{r}(a,x_1). \end{align*}$$

Let us briefly explain the value of k to which we apply the induction hypothesis. Let $k_{\text {old}}$ be the current value of k, and let $k_{\text {new}}$ be the value of k to which we apply the induction hypothesis. We choose $k_{\text {new}}$ as the smallest integer satisfying

$$\begin{align*}k_{\text{new}} \geq k_{\text{old}} - 1, \quad t_{k_{\text{new}} + 2}> \max(e^{(\log t_2)^{c_6}}, e^{D_1^{c_1}}), \end{align*}$$

and we choose $k_{\text {new}} = r - 1$ if there is no such integer.

Since $t_1> r^{c_4}$ with $c_4 < \frac {1}{8}$ , we get

$$\begin{align*}|X(a)\cap \pi_1^{-1}(x_1)| \leq \left(1 + \frac{2}{t_1}\right)^r \cdot 2^{-|\mathcal{M}|+k}\frac{|X|}{|X_1|} \leq 2^{-|\mathcal{M}|+k+1}\frac{|X|}{|X_1|} \end{align*}$$

and hence

(5.1) $$ \begin{align} |X^{\text{bad}}(a)| \leq 2^{k+1}k t_1^{-B_1 }\cdot 2^{-|\mathcal{M}|}|X| <2^{j_1+1}j_1 t_1^{-B_1 }\cdot 2^{-|\mathcal{M}|}|X| <t_1^{-\kappa} \cdot \frac{|X|}{2^{|\mathcal{M}| + 1}}. \end{align} $$

For $x_1\notin X^{\text {bad}}(a)$ , we look at

$$\begin{align*}X_2(a,x_1)\times\dots\times X_r(a,x_1). \end{align*}$$

Then we obtain

$$ \begin{align*} |X(a)\setminus X^{\text{bad}}(a)| &=\sum_{x_1\in X_1\setminus X_1^{\text{bad}}(a)}|X(a)\cap \pi_1^{-1}(x_1)| \\ &=\sum_{x_1\in X_1\setminus X_1^{\text{bad}}(a)}|(X_2(a,x_1)\times\dots\times X_r(a,x_1))(a)|, \end{align*} $$

which lies between

$$\begin{align*}\left(1\pm \frac{2}{t_1}\right)^r\left(1\pm \frac{2}{t_1^{B_2}}\right)^k \left(1 \pm(r-1)t_1^{-\kappa}\right)\cdot 2^{-|\mathcal{M}|}|X| \end{align*}$$

by the induction hypothesis. Since $r<t_1^{c_4}<t_1^{\kappa }$ , $1-c_4>\kappa $ and $B_2>\kappa $ , we have

$$ \begin{align*} \left(1+ \frac{2}{t_1}\right)^r &\left(1+ \frac{2}{t_1^{B_2}}\right)^k \frac{1+(r-1)t_1^{-\kappa}}{1 +(r-\frac{1}{2})t_1^{-\kappa}} = \left(1+ \frac{2}{t_1}\right)^r\left(1+ \frac{2}{t_1^{B_2}}\right)^k \left(1- \frac{ \frac{1}{2}t_1^{-\kappa}}{1 + (r - \frac{1}{2})t_1^{-\kappa}}\right)\\ &\qquad\qquad <\exp\left(2rt_1^{-1}+2kt_1^{-B_2}-\frac{1}{4}t_1^{-\kappa}\right)< \exp\left(2t_1^{-(1-c_4)} +2c_5t_1^{-B_2}\log t_1 -\frac{1}{4}t_1^{-\kappa}\right) <1 \end{align*} $$

and similarly

$$\begin{align*}\left(1- \frac{2}{t_1}\right)^r\left(1- \frac{2}{t_1^{B_2}}\right)^k \frac{1 -(r-1)t_1^{-\kappa}}{1 - (r-\frac{1}{2})t_1^{-\kappa}} = \left(1- \frac{2}{t_1}\right)^r \hspace{-0.1cm} \left(1- \frac{2}{t_1^{B_2}}\right)^k \hspace{-0.1cm} \left(1+ \frac{ \frac{1}{2}t_1^{-\kappa}}{1 - (r - \frac{1}{2})t_1^{-\kappa}}\right)>1. \end{align*}$$

We conclude that the sum lies between

(5.2) $$ \begin{align} \left(1 \pm\left(r-\frac{1}{2}\right)t_1^{-\kappa}\right)\cdot 2^{-|\mathcal{M}|}|X|. \end{align} $$

Adding the contributions from equations (5.1) and (5.2) completes the inductive step.

Proof. Fix some $x\in X$ , and write $W(a):=W(x, a)$ . We will show that

$$\begin{align*}\sum_{a\in\mathbb{F}_2^{\mathcal{M}\sqcup\mathcal{N}}}\left(|W(a)|-2^{-m}\cdot k_2!\right)^2 \leq \frac{2^{|P|+k_0+1} k_1^2}{k_2}\cdot 2^{-2m+|\mathcal{M}\sqcup\mathcal{N}|}(k_2!)^2 .\end{align*}$$

Then the proposition follows from the Cauchy-Schwarz inequality.

The average of $|W(a)|$ over a is $2^{-m}\cdot k_2!$ , since $|\mathcal {P}(k_2)|=k_2!$ and there are m Legendre symbol conditions to satisfy. Now

$$\begin{align*}|W(a)|^2=|\{(\sigma_1,\sigma_2)\in\mathcal{P}(k_2)\times\mathcal{P}(k_2): M(\sigma_1(x))\hspace{-0.05cm}\restriction_{S(\sigma_1)}=a\hspace{-0.05cm}\restriction_{S(\sigma_1)},\ M(\sigma_2(x))\hspace{-0.05cm}\restriction_{S(\sigma_2)}=a\hspace{-0.05cm}\restriction_{S(\sigma_2)}\}|. \end{align*}$$

We have $\sum _{a\in \mathbb {F}_2^{\mathcal {M}\sqcup \mathcal {N}}}|W(a)|^2=\sum _{\sigma _1,\sigma _2\in \mathcal {P}(k_2)}|W(\sigma _1,\sigma _2)|$ , where

$$\begin{align*}W(\sigma_1,\sigma_2):=\{a\in\mathbb{F}_2^{\mathcal{M}\sqcup\mathcal{N}}: M(\sigma_1(x))\restriction_{S(\sigma_1)}=a\restriction_{S(\sigma_1)},\ M(\sigma_2(x))\restriction_{S(\sigma_2)}=a\restriction_{S(\sigma_2)}\}. \end{align*}$$

We fix some $\sigma _1,\sigma _2\in \mathcal {P}(k_2)$ and bound $|W(\sigma _1,\sigma _2)|$ . Let $ d:=|\{i\in [k_2]:\sigma _1(i)\leq k_1,\ \sigma _2(i)\leq k_1\}|$ . We have

$$ \begin{align*} |S(\sigma_1)\cap S(\sigma_2)| &=|\left\{(i,j)\in\mathcal{M}: (\sigma_1(i),\sigma_1(j)),\ (\sigma_2(i),\sigma_2(j))\in ([k_0]\times[k_1])\cup ([k_1]\times[k_0])\right\}| \\ &\qquad\qquad\qquad\qquad\qquad +|\left\{(p,j)\in\mathcal{N}:\sigma_1(j),\ \sigma_2(j)\in [k_1] \right\}| \leq d(|P|+k_0). \end{align*} $$

Therefore the conditions fix at least $2m-d(|P|+k_0)$ arguments of $a\in W(\sigma _1,\sigma _2)$ . Then

$$\begin{align*}|W(\sigma_1,\sigma_2)|\leq 2^{-2m+d(|P|+k_0)+|\mathcal{M}\sqcup\mathcal{N}|}.\end{align*}$$

Given some $d\leq k_1$ , we bound the number of $(\sigma _1,\sigma _2)\in \mathcal {P}(k_2)\times \mathcal {P}(k_2)$ that gives the same d. There are $\binom {k_2}{d}$ ways to pick the indices that map to $[k_1]$ under $\sigma _1$ and $\sigma _2$ . Then there are at most $(\frac {k_1!}{(k_1-d)!}(k_2-d)!)^2$ ways to pick a pair of $(\sigma _1,\sigma _2)$ in such a way. Hence the total number is bounded by

$$\begin{align*}\binom{k_2}{d}\left(\frac{k_1!}{(k_1-d)!}(k_2-d)!\right)^2\leq (k_2!)^2\left(\frac{k_1!}{(k_1-d)!}\right)^2\frac{ (k_2-d)!}{k_2!} \leq (k_2!)^2\left(\frac{k_1^2}{k_2}\right)^d .\end{align*}$$

The average of $|W(a)|^2$ is bounded by

$$\begin{align*}(k_2!)^2\sum_{d\geq 0}\left(\frac{k_1^2}{k_2}\right)^d \cdot 2^{-2m+d(|P|+k_0)} =\frac{k_2}{k_2-2^{|P|+k_0}k_1^2}\cdot 2^{-2m}(k_2!)^2. \end{align*}$$

Then the variance of $|W(a)|$ is bounded by

$$ \begin{align*} \frac{k_2}{k_2-2^{|P|+k_0}k_1^2}\cdot 2^{-2m}(k_2!)^2-(2^{-m}\cdot k_2!)^2 &=\frac{2^{|P|+k_0}k_1^2}{k_2-2^{|P|+k_0}k_1^2}\cdot 2^{-2m}(k_2!)^2 \\ &\leq \frac{2^{|P|+k_0+1}k_1^2}{k_2}\cdot 2^{-2m}(k_2!)^2 .\end{align*} $$

Multiplying by $2^{|\mathcal {M}\sqcup \mathcal {N}|}$ gives the required estimate.

Proof. Without loss of generality, assume that $\mathcal {M}= \{(i,j):1\leq i < j\leq r\}$ and $\mathcal {N}= P\times [r]$ , $X_i=\{x_i\}$ for $i\leq k_0$ . Let $m:=k_1|P|+\frac {1}{2}k_0(k_0-1)+k_0(k_1-k_0)$ as in Proposition 5.8. Apply the triangle inequality to the sum we wish to bound,

(5.3) $$ \begin{align} 2^{-|\mathcal{M}\sqcup\mathcal{N}|}&\sum_{a\in\mathbb{F}_2^{\mathcal{M}\sqcup\mathcal{N}}} \left|k_2!\cdot |X|- 2^m\sum_{\sigma\in\mathcal{P}(k_2)}|X_S(\sigma, a)|\right| \notag\\ &\qquad\qquad\qquad\quad \ + 2^{-|\mathcal{M}\sqcup\mathcal{N}|+m}\sum_{\sigma\in\mathcal{P}(k_2)}\sum_{a\in\mathbb{F}_2^{\mathcal{M}\sqcup\mathcal{N}}} \left||X_S(\sigma, a)|-2^{|\mathcal{M}\sqcup\mathcal{N}|-m}|X(\sigma, a)|\right|. \end{align} $$

For the first sum in equation (5.3), noting that

$$\begin{align*}\sum_{x\in X}|W(x, a)|=\sum_{\sigma\in\mathcal{P}(k_2)}|X_S(\sigma, a)|, \end{align*}$$

we obtain by Proposition 5.8 an upper bound

$$\begin{align*}\left(\frac{2^{|P|+k_0+1}}{k_2}\right)^{1/2} k_1\cdot k_2!\cdot |X|<k_2^{-\delta_1}\cdot k_2!\cdot |X|.\end{align*}$$

Now consider the second sum of equation (5.3). For each $\sigma \in \mathcal {P}(k_2)$ , we can partition X into $2^m$ sets according to $\tilde {a}:S(\sigma )\rightarrow \{\pm 1\}$ as follows:

$$\begin{align*}X_S(\sigma,\tilde{a})= \{x_1\}\times\dots\times\{x_{k_0}\}\times X_{k_0+1}(\tilde{a},\tilde{P})\times \dots \times X_{k_1}(\tilde{a},\tilde{P})\times X_{k_1+1}\times\dots\times X_r, \end{align*}$$

where $\tilde {P}=\{x_1\}\cup \dots \cup \{x_{k_0}\}\cup P$ and $X_i(\tilde {a},\tilde {P})$ is the subset of those $q \in X_i$ satisfying

$$\begin{align*}\left(\frac{q}{x_j}\right) = \widetilde{a}(\sigma^{-1}(i), \sigma^{-1}(j)) \text{ for } j \in [k_0] \text{ and } \left(\frac{q}{p}\right) = \widetilde{a}(p, \sigma^{-1}(i)) \text{ for } p \in P. \end{align*}$$

We first bound the contribution of $\sigma \in \mathcal {P}(k_2)$ with $|X_i(\tilde {a},\tilde {P})|< 2^{-|\tilde {P}|}k_2^{-c_7}|X_i|$ for some $\tilde {a}:S(\sigma )\rightarrow \{\pm 1\}$ and some $k_0< i\leq k_1$ in the sum. For each $\sigma \in \mathcal {P}(k_2)$ and $k_0<i\leq k_1$ , we have the upper bound

$$\begin{align*}\sum_{\tilde{a}:|X_i(\tilde{a},\tilde{P})|\leq 2^{-|\tilde{P}|}k_2^{-c_7} |X_i|}\left|X_S(\sigma,\tilde{a})\right| \leq k_2^{-c_7}|X|. \end{align*}$$

For each $\tilde {a}$ , there are $2^{|\mathcal {M}\sqcup \mathcal {N}|-m}$ many a satisfying $a\restriction _{S(\sigma )}=\tilde {a}$ , so the contribution of such a is bounded by

$$\begin{align*}\sum_{\sigma\in\mathcal{P}(k_2)}\sum_{k_0<i\leq k_1}\sum_{\tilde{a}:|X_i(\tilde{a},\tilde{P})|\leq 2^{-|\tilde{P}|}\cdot k_2^{-c_7}\cdot |X_i| } |X_S(\sigma,\tilde{a})| \leq k_1 k_2^{-c_7}\cdot k_2!\cdot |X| < k_2^{-\delta_1}\cdot k_2! \cdot |X|. \end{align*}$$

For the remaining terms, we have $|X_i(\tilde {a},\tilde {P})| \geq 2^{-|\tilde {P}|} k_2^{-c_7} |X_i|$ for all $k_0< i\leq r$ . Bound each summand by Proposition 5.7

$$\begin{align*}\left||X_S(\sigma, a)|-2^{|\mathcal{M}\sqcup\mathcal{N}|-m} |X(\sigma, a)|\right| \leq t_{k_0 + 1}^{-\delta_2}|X_S(\sigma, a)|; \end{align*}$$

then summing over $\sigma $ and a gives the required estimate.

Proof. We proceed by induction on $|V|$ . The case $|V| = 0$ is trivial. Let v be the smallest element in V. Define $\tilde {k}$ to be the k from the proposition if $v = 1$ , and define $\tilde {k}$ to be r if $v \neq 1$ . Fix some $x \in X_v$ . Put $B_1 := c_4 + c_5 + \delta $ and $B_2 := c_4 + \delta $ . Following the proof of Proposition 5.7, we get that

$$\begin{align*}\left||X_j(a, x)| - \frac{1}{2} |X_j| \right| < t_1^{-B_2} |X_j| \text{ for all } 1 \leq j \leq \tilde{k} \text{ with } j \neq v \end{align*}$$

holds for $x \in X_v$ with at most $\tilde {k} t_1^{-B_1} |X_v|$ exceptions, while for $j> \tilde {k}$ , we always get

$$\begin{align*}\left||X_j(a, x)| - \frac{1}{2} |X_j| \right| < t_1^{-1} |X_j|. \end{align*}$$

Just as in the proof of Proposition 5.7, define $X_v^{\text {bad}}(a)$ to be the set of exceptions. We split the sum in the proposition as

(5.4) $$ \begin{align} \sum_{Q \in \pi_V(X) \ \mathrm{ poor }} |X(a, Q)| &= \sum_{\substack{Q \in \pi_V(X) \ \mathrm{ poor } \\ \pi_v(Q) \not \in X_v^{\text{bad}}(a)}} |X(a, Q)| + \sum_{\substack{Q \in \pi_V(X) \ \mathrm{ poor } \\ \pi_v(Q) \in X_v^{\text{bad}}(a)}} |X(a, Q)| \nonumber \\ &\leq \sum_{\substack{Q \in \pi_V(X) \ \mathrm{ poor } \\ \pi_v(Q) \not \in X_v^{\text{bad}}(a)}} |X(a, Q)| + \sum_{\substack{Q \in \pi_V(X) \\ \pi_v(Q) \in X_v^{\text{bad}}(a)}} |X(a, Q)|. \end{align} $$

We first treat the latter sum in equation (5.4). In the case $v = 1$ , we apply Proposition 5.7 to the prebox

$$\begin{align*}(X_2 \times \dots \times X_k \times X_{k + 1}(a, x) \times \dots \times X_r(a, x), P \cup \{x\}) \end{align*}$$

for $x \in X_1^{\text {bad}}(a)$ and the natural restrictions of a, U, V, $\mathcal {M}$ and $\mathcal {N}$ . Then the latter sum is bounded by

$$\begin{align*}\sum_{\substack{Q \in \pi_V(X) \\ \pi_1(Q) \in X_1^{\text{bad}}(a)}} |X(a, Q)| = \sum_{x \in X_1^{\text{bad}}(a)} |X(a) \cap \pi_1^{-1}(x)| \leq |X_1^{\text{bad}}(a)| \cdot 2^{-|\mathcal{M}| + k + 1} \frac{|X|}{|X_1|}. \end{align*}$$

A small computation shows that this is at most

$$\begin{align*}t_1^{-c_4 - \delta} \cdot \frac{|X|}{2^{|\mathcal{M}| + 1}} \end{align*}$$

for sufficiently large $D_1$ . Now suppose that $v \neq 1$ so that $1 \in U$ . Then apply Proposition 5.7 with $k = r - 1$ , the prebox

$$\begin{align*}(X_1 \times \dots \times X_{v - 1} \times X_{v + 1} \times \dots \times X_r, \varnothing) \end{align*}$$

and the natural restrictions of a, U, V, $\mathcal {M}$ and $\mathcal {N}$ . Crucially, we have that this choice of k satisfies the requirements of Proposition 5.7 for sufficiently large $D_1$ due to our assumption $c_5 \log t_u> r + 10$ for all $u \in U$ . Then a similar computation shows that the latter sum is again at most

$$\begin{align*}t_1^{-c_4 - \delta} \cdot \frac{|X|}{2^{|\mathcal{M}| + 1}}. \end{align*}$$

It remains to bound the former sum in equation (5.4). We first treat the case $v = 1$ . Take a poor $Q \in \pi _V(X)$ with $x := \pi _1(Q) \not \in X_1^{\text {bad}}(a)$ . Then we claim that $\pi _{V - \{1\}}(Q)$ is poor for the prebox

$$\begin{align*}(X_2(a, x) \times \dots \times X_r(a, x), P \cup \{x\}). \end{align*}$$

Suppose that $\pi _{V - \{1\}}(Q)$ is not poor. Then we get for all $u \in U$ that

$$\begin{align*}\left||X_u(a, Q)| - \frac{|X_u(a, x)|}{2^{|V| - 1}}\right| \leq t_2^{-B_2} (|V| - 1) |X_u|. \end{align*}$$

But from this, we deduce that for all $u \in U$ ,

$$ \begin{align*} \left||X_u(a, Q)| - \frac{|X_u|}{2^{|V|}}\right| &\leq \left||X_u(a, Q)| - \frac{|X_u(a, x)|}{2^{|V| - 1}}\right| + \left|\frac{|X_u(a, x)|}{2^{|V| - 1}} - \frac{|X_u|}{2^{|V|}}\right| \\ &\leq t_2^{-B_2} (|V| - 1) |X_u| + t_1^{-B_2} |X_u| \leq t_1^{-B_2} |V| |X_u|, \end{align*} $$

establishing the claim. Now we can easily bound the former sum in equation (5.4) using the induction hypothesis. Finally, we deal with the case that $v \neq 1$ so that $1 \in U$ . In this case, we apply the induction hypothesis to the prebox

$$\begin{align*}(X_1(a, x) \times \dots \times X_{v - 1}(a, x) \times X_{v + 1}(a, x) \times \dots \times X_r(a, x), P \cup \{x\}) \end{align*}$$

with $k = r - 1$ .

Proof. Define $\mathcal {T}_k=\{\mathbf {t}:X(\mathbf {t}) \subseteq S_r(N) \text { and } X(\mathbf {t})\cap W\neq \varnothing \}$ . Our aim is to estimate $|V|$ in terms of

$$\begin{align*}\int_{\mathcal{T}_k}|V\cap X(\mathbf{t})|\frac{dp_1\cdots dp_kds_{k+1}\cdots ds_r}{s_{k+1}\cdots s_r},\end{align*}$$

where $dp_i$ is $1$ if $p_i\equiv 1\bmod 4$ is prime and $0$ otherwise.

Consider $n=(q_1,\dots ,q_r)\in S_r(N)$ with exactly k prime factors less than $D_1$ . Then $n\in X(\mathbf {t})$ if and only if $q_i=p_i$ for $1\leq i\leq k$ and

$$\begin{align*}s_i<q_i<\left(1+\frac{1}{e^{i-k}\log D_1}\right)s_i \text{ for }k<i\leq r.\end{align*}$$

If $n\in W$ and

(5.5) $$ \begin{align} n\prod_{i=k+1}^r \left(1+\frac{1}{e^{i-k}\log D_1}\right)<N, \end{align} $$

then

$$ \begin{align*} \begin{aligned} \int_{\substack{\mathbf{t}\in\mathcal{T}_k:\\n\in X(\mathbf{t})}}\frac{dp_1\cdots dp_kds_{k+1}\cdots ds_r}{s_{k+1}\cdots s_r} &=\int_{q_{k+1}\left(1+\frac{1}{e\log D_1}\right)^{-1}}^{q_{k+1}}\cdots\int_{q_r\left(1+\frac{1}{e^{r-k}\log D_1}\right)^{-1}}^{q_r}\frac{ds_{k+1}\cdots ds_r}{s_{k+1}\dots s_r} \\ &=\prod_{i=k+1}^r \log\left(1+\frac{1}{e^{i-k}\log D_1}\right) .\end{aligned} \end{align*} $$

If equation (5.5) does not hold or $n\notin W$ , then

$$\begin{align*}\int_{\substack{\mathbf{t}\in\mathcal{T}_k:\\n\in X(\mathbf{t})}}\frac{dp_1\cdots dp_kds_{k+1}\cdots ds_r}{s_{k+1}\cdots s_r} \leq\prod_{i=k+1}^r \log\left(1+\frac{1}{e^{i-k}\log D_1}\right). \end{align*}$$

There exists some constant $C>0$ such that any n that does not satisfy equation (5.5) lies in

$$\begin{align*}N\left(1-\frac{C}{\log D_1}\right) \leq N\prod_{i=k+1}^r\left(1+\frac{1}{e^{i-k}\log D_1}\right)^{-1}\leq n\leq N, \end{align*}$$

which we bound by

$$\begin{align*}\Phi_r(N)-\Phi_r\left(N\left(1-\frac{C}{\log D_1}\right)\right). \end{align*}$$

Then

$$\begin{align*}\sum_{k=0}^{\infty}\prod_{i=k+1}^r \log\left(1+\frac{1}{e^{i-k}\log D_1}\right)^{-1}\int_{\mathcal{T}_k}|V\cap X(\mathbf{t})|\frac{dp_1\cdots dp_kds_{k+1}\cdots ds_r}{s_{k+1}\cdots s_r} \end{align*}$$

is bounded above by $|V|$ and below by

$$ \begin{align*} & |V\cap W| + O\left(\Phi_r(N) - \Phi_r\left(N\left(1-\frac{C}{\log D_1}\right)\right)\right)= \\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad |V| + O\left(\Phi_r(N) - \Phi_r\left(N\left(1-\frac{C}{\log D_1}\right)\right)\right) + O\left(\epsilon \Phi_r(N)\right). \end{align*} $$

Similarly,

$$ \begin{align*} \sum_{k=0}^{\infty} \prod_{i=k+1}^r \log\left(1+\frac{1}{e^{i-k}\log D_1}\right)^{-1} &\int_{\mathcal{T}_k}|X(\mathbf{t})|\frac{dp_1\cdots dp_kds_{k+1}\cdots ds_r} {s_{k+1}\cdots s_r}\\ & =\Phi_r(N) + O\left(\Phi_r(N) - \Phi_r\left(N\left(1-\frac{C}{\log D_1}\right)\right)\right) + O\left(\epsilon \Phi_r(N)\right). \end{align*} $$

The result follows from the estimate $|V\cap X(\mathbf {t})|=(\delta +O(\epsilon ))|X(\mathbf {t})|$ for $\mathbf {t} \in \mathcal {T}_k$ .

Proof. Suppose we have some box $X\subseteq V_i$ and $d_i=p_1\cdots p_m$ . For any element $x\in X$ , there are prime factors $q_1,\dots ,q_m$ of x such that

$$\begin{align*}q_i=p_i \text{ if } p_i<D_1 \quad \text{ and }\quad \frac{1}{2}p_i<q_i<2p_i\text{ if } p_i\geq D_1. \end{align*}$$

If $d_i<N^{2/3}$ , we deduce from equation (4.1) that there exists some constant $C>0$ with

$$\begin{align*}\begin{aligned} |V_i|& \leq \Phi_{r-m}\left(\frac{2^m N}{d_i}\right) \cdot\prod_{p_i\geq D_1}\left|\left\{q_i\text{ prime}:\frac{1}{2}p_i<q_i<2p_i,\ q_i\equiv 1\bmod 4\right\}\right|\\ &\leq \Phi_r(N)\cdot \frac{C^m }{d_i}\prod_{p_i\geq D_1}\frac{p_i}{\log p_i} \ll\frac{\Phi_r(N)}{\log d_i}. \end{aligned} \end{align*}$$

Notice that $d_i>D_1^{2^{i-1}}$ . Then

$$\begin{align*}\left|\bigcup_{d_i<N^{2/3}}V_i\right| \ll\Phi_r(N)\sum_{i\geq 1}\frac{1}{2^{i-1}\log D_1} \ll\frac{\Phi_r(N)}{\log D_1}. \end{align*}$$

If $d_i\geq N^{2/3}$ , then $d_{i+1}\geq N^{4/3}>N$ . Therefore there is at most one i such that $d_i\geq N^{2/3}$ and $V_i$ is not empty. Then for sufficiently large $D_1$ ,

$$ \begin{align*} |V_i|&\leq \left|\left\{x\in S_r(N):d_i\mid x\right\}\right|\cdot\prod_{p_i\geq D_1}\left|\left\{q_i\text{ prime}:\frac{1}{2}p_i<q_i<2p_i,\ q_i\equiv 1\bmod 4\right\}\right|\\ &\leq \frac{N}{d_i}\prod_{p_i\geq D_1}\frac{2p_i}{\log p_i} \ll \frac{N}{\log d_i} \ll \frac{N}{\log N}, \end{align*} $$

which fits into the error bound.

Proof. By our Erdős–Kac result – that is, equation (4.2) – it suffices to show that

$$\begin{align*}\left|\frac{|\{d\in S_r(N) : \operatorname{\mathrm{rk}}_4 \operatorname{\mathrm{Cl}}^+(d)=k\}|}{|S_r(N)|}-\lim_{n\rightarrow\infty} P(n|k)\right|\ll (\log\log N)^{-c} \end{align*}$$

for any r satisfying equation (4.3). We can find some $W\subseteq S_r(N)$ that is comfortably spaced above $D_1$ and $(c_{10}\log \log \log N)$ -regular by Theorem 4.1 and Siegel-less above $D_1$ by Proposition 5.14 so that

$$\begin{align*}|W|\geq (1-\epsilon)\Phi_r(N)\text{ with } \epsilon \ll (\log\log N)^{-c} \end{align*}$$

for some absolute constant $c> 0$ . Then, applying Theorem 5.13, we see that we can restrict to acceptable boxes by introducing an error $\ll (\log \log N)^{-c}$ . In other words, it suffices to show that we have for any acceptable box $X \subseteq S_r(N)$

(5.6) $$ \begin{align} \left|\frac{|\{x\in X :\operatorname{\mathrm{rk}}_4 \operatorname{\mathrm{Cl}}^+(x)=k\}|}{|X|}-\lim_{n\rightarrow\infty} P(n|k)\right|\ll(\log\log N)^{-c}. \end{align} $$

Take X to be an acceptable box. Then one can check that there exist constants that satisfy the requirements in Theorem 5.9 applied to the prebox $(X_1\times \dots \times X_r, \varnothing )$ . Then Theorem 5.9 shows that the Rédei matrices of $x\in X$ are equidistributed amongst all $r \times r$ symmetric matrices over $\mathbb {F}_2$ with column sum zero, up to reordering some columns and rows, with an error within the statement. Since reordering columns and rows does not change the rank, we can assume that the Rédei matrix is a random $r \times r$ symmetric matrix over $\mathbb {F}_2$ with column sum zero.

Let A be the matrix obtained from the Rédei matrix after removing a column and a row. It is a classical fact that $\operatorname {\mathrm {rk}}_4 \operatorname {\mathrm {Cl}}^+(x)$ is equal to the corank of A. But A is a random $(r - 1) \times (r - 1)$ symmetric matrix. By [Reference MacWilliams15, Theorem 2], we have for all $n \geq k \geq 0$

$$ \begin{align*} P(n|k) &= \frac{N(n, n - k)}{2^{\frac{n(n + 1)}{2}}} = \frac{1}{2^{\frac{n(n + 1)}{2}}} \prod_{i = 1}^{\frac{n - k}{2}} \frac{2^{2i}}{2^{2i} - 1} \cdot \prod_{i = 0}^{n - k - 1} (2^{n - i} - 1) \\ &= \frac{1}{2^{\frac{k(k + 1)}{2}}} \prod_{i = 1}^{\frac{n - k}{2}} \frac{2^{2i}}{2^{2i} - 1} \cdot \prod_{i = 0}^{n - k - 1} (1 - 2^{i - n}) \end{align*} $$

if $n - k \equiv 0 \bmod 2$ and

$$ \begin{align*} P(n|k) &= \frac{N(n, n - k)}{2^{\frac{n(n + 1)}{2}}} = \frac{1}{2^{\frac{n(n + 1)}{2}}} \prod_{i = 1}^{\frac{n - k - 1}{2}} \frac{2^{2i}}{2^{2i} - 1} \cdot \prod_{i = 0}^{n - k - 1} (2^{n - i} - 1) \\ &= \frac{1}{2^{\frac{k(k + 1)}{2}}} \prod_{i = 1}^{\frac{n - k - 1}{2}} \frac{2^{2i}}{2^{2i} - 1} \cdot \prod_{i = 0}^{n - k - 1} (1 - 2^{i - n}) \end{align*} $$

if $n - k \equiv 1 \bmod 2$ , where $N(n, k)$ denotes the number of symmetric $n \times n$ -matrices with coefficients in $\mathbb {F}_2$ and rank k. Using this, one directly bounds the difference $P(r - 1|k) - \lim _{n\rightarrow \infty } P(n|k)$ , completing the proof.

Proof that Proposition 6.3 implies Theorem 6.1

From Erdős—Kac (see equation (4.2)), it follows that we only need to consider r satisfying equation (4.3). For each such r, we apply Proposition 5.13 with W as in Definition 6.2; the required lower bound for $|W|$ follows from the material in Section 4 and Proposition 5.14.

Proof that Proposition 6.6 implies Proposition 6.3

Take N to be a large integer, and take X to be a nice box for N. If N is sufficiently large and $n_2> n_{\text {max}}$ , we have

$$\begin{align*}\lim_{k \rightarrow \infty} P(k|n_2) = O\left(\frac{1}{\log \log \log \log N}\right). \end{align*}$$

Then it follows from equation (5.6) that

$$\begin{align*}\left|\left|X \cap \mathcal{D}_{n_2, n_3}(N)\right| - Q(n_2|n_3) \cdot \left|X \cap \mathcal{D}_{n_2}(N)\right|\right| \leq 2 \left|X \cap \mathcal{D}_{n_2}(N)\right| \leq \frac{A|X|}{\log \log \log \log N} \end{align*}$$

for a sufficiently large constant $A> 0$ . From now on, suppose that $n_2 \leq n_{\text {max}}$ . We deduce from Hoeffding’s inequality that the proportion of S in $\mathbb {F}_2^r$ failing equation (6.1) or equation (6.2) is bounded by

$$\begin{align*}O\left(\exp\left(-2^{-20n_{\text{max}}^2} \cdot k_{\text{gap}}\right)\right). \end{align*}$$

Given $S \not \in \langle R \rangle $ , the proportion of $a:\mathcal {M} \rightarrow \{\pm 1\}$ with $S \in \mathcal {V}_{a,2}$ is $O(0.5^r)$ . Taking the union over all S in $\mathbb {F}_2^r$ failing equation (6.1) or equation (6.2) proves that the proportion of nongeneric a is at most

$$\begin{align*}O\left(\exp\left(-2^{-20n_{\text{max}}^2} \cdot k_{\text{gap}}\right)\right). \end{align*}$$

Put $k_2 := \lfloor 0.25k_{\text {gap}}\rfloor $ . Then we have for all $\sigma \in \mathcal {P}(k_2)$ that $a:\mathcal {M} \rightarrow \{\pm 1\}$ is generic if and only if $\sigma (a)$ is generic, where $\sigma (a)$ is defined in the natural way. Theorem 5.9 implies that

$$\begin{align*}\sum_{a:\mathcal{M} \rightarrow \{\pm 1\}} \left|2^{-|\mathcal{M}|} \cdot |X| - \frac{1}{k_2!} \sum_{\sigma \in \mathcal{P}(k_2)} |X(\sigma(a))|\right| \leq (k_2^{-\delta_1} + t_{k + 1}^{\prime -\delta_2}) \cdot |X|, \end{align*}$$

where $\delta _1$ and $\delta _2$ are small, positive absolute constants. Restricting this sum to the nongeneric a shows that the union of $X(a)$ over all nongeneric a is within the error term of Proposition 6.3. We now deal with the $a:\mathcal {M} \rightarrow \{\pm 1\}$ that fail equation (6.3). Let j be an integer satisfying $k < j \leq r$ . We say that $a, a': \mathcal {M} \rightarrow \{\pm 1\}$ are equivalent at j, which we write as $a \sim _j a'$ , if $a(i, j) = a'(i, j)$ for all $1 \leq i \leq k$ . Since our box is $\eta $ -regular, we see that k is roughly equal to $\log \log D_1$ . In particular, if N is sufficiently large, we get

$$\begin{align*}k \leq 2 \log \log D_1 = \frac{1}{5} \log \log \log N. \end{align*}$$

Then there are at most $2^{\frac {1}{5} \log \log \log N}$ equivalence classes. Furthermore, if $a:\mathcal {M} \rightarrow \{\pm 1\}$ is such that equation (6.3) fails for some fixed j, we have that

$$\begin{align*}\left|\bigcup_{a': a \sim_j a'} X(a')\right| \leq \frac{1}{(\log t_{k + 1})^{100}} \cdot |X|, \end{align*}$$

where the union is over all $a':\mathcal {M} \rightarrow \{\pm 1\}$ equivalent to $a:\mathcal {M} \rightarrow \{\pm 1\}$ at j. Summing this over all choices of j and all equivalence classes, we stay within the error term of Proposition 6.3. So far, we have shown

$$ \begin{align*} &\left|\left|X \cap \mathcal{D}_{n_2, n_3}(N)\right| - Q(n_2|n_3) \cdot \left|X \cap \mathcal{D}_{n_2}(N)\right|\right| \\ &\qquad\qquad\qquad\quad \leq \sum_{\substack{a \text{ generic} \\ a \text{ sat. eq. } (6.3) \\ n_2(a) = n_2}} \sum_{\substack{\text{rk}(\text{Art}_2) = n_2 - n_3 \\ \text{Art}_2 \text{ valid}}} \left||X(a, \operatorname{\mathrm{Art}}_2)| - 2^{-n_2(n_2 + 1)}|X(a)|\right| + \frac{A|X|}{\log \log \log \log N}. \end{align*} $$

Note that we could have further restricted the sum over Artin pairings to only those with bottom row identically $0$ . However, the displayed inequality suffices for our purposes. We now apply Proposition 6.6 for every generic $a:\mathcal {M} \rightarrow \{\pm 1\}$ for X such that it satisfies equation (6.3) and $n_2(a) = n_2$ , and all valid Artin pairings $\text {Art}_2$ with $\text {rk}(\text {Art}_2) = n_2 - n_3$ . Since there are at most

$$\begin{align*}2^{n_2(n_2 + 1)} \leq 2^{n_{\text{max}}(n_{\text{max}} + 1)} \end{align*}$$

valid Artin pairings, we get

$$\begin{align*}\sum_{\substack{a \text{ generic} \\ a \text{ sat. eq. } (6.3) \\ n_2(a) = n_2}} \hspace{-0.1cm} \sum_{\substack{\text{rk}(\text{Art}_2) = n_2 - n_3 \\ \text{Art}_2 \text{ valid}}} \hspace{-0.1cm} \left||X(a, \operatorname{\mathrm{Art}}_2)| - 2^{-n_2(n_2 + 1)}|X(a)|\right| \leq 2^{n_{\text{max}}(n_{\text{max}} + 1)} \frac{A|X|}{(\log \log \log \log N)^4} \end{align*}$$

as desired.