A tensor budget and a contracting kernel

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

Lessons 3–9 showed that the normalized transforms $g_p$ of the supports are flat: they correlate with no product of a multiplicative and an additive character. Flatness alone allows $g_p$ to be very sparse, concentrated on a few frequencies. This lesson and the next show that coverage of every large prime excludes this on a large set of primes. For an absolute $\delta_0>0$, the balanced primes whose transform has $L^1$ norm $\gamma_p=\frac1p\sum_v|g_p(v)|\ge\delta_0$ carry harmonic mass at least $0.15L$ in the band $0.05L\le\log\log p\le0.9L$ (Theorem 1.1).

The proof is by contradiction. A sparse spectrum at a prime $p$ defines a small kernel $b_p$ on $\mathbb F_p$, and the product over many such primes defines a nonnegative weight $W(n)$. Two estimates of the average of $W(a-d)$ over pairs $a\in A$, $d\in D$ conflict. In this lesson, an operator-norm bound for a tensor product of $2\times2$ block matrices shows that this average is small (Proposition 4.2): each prime contributes a contraction factor $1-1.6/p$. Lesson 11 evaluates the average of $W$ over primes, where it is about $1$, and uses coverage to transfer this to the pairs $(a,d)$.

Prerequisites. The inverse Goldbach problem, Section 2; Square-root bounds and collision stability, Theorem 3.1 and Lemma 4.1; Anchor codes and decorrelation of higher characters, Section 6 (the transforms $F_p,g_p$).

1. The statement

Recall from (6.1) of Lesson 9 the functions $F_p$ and $g_p$, with $\begin{gathered} \sum_{x\in\mathbb F_p}|F_p(x)|^2=p,\\ \frac1p\sum_{v\in\mathbb F_p}|g_p(v)|^2=1,\\ g_p(0)=0,\\ g_p(-v)=\overline{g_p(v)}, \end{gathered}$ and $F_p(x)=p^{-1/2}\sum_vg_p(v)e_p(vx)$. Write $\gamma_p=\frac1p\sum_{v\in\mathbb F_p}|g_p(v)|$ for the probability $L^1$ norm of $g_p$. By Cauchy–Schwarz, $\gamma_p\le1$.

Theorem 1.1 (a supply of nonsparse transforms [OpenAI-013, Proposition 5.1]). There is an absolute constant $\delta_0>0$ such that, for all sufficiently large $L$, $\begin{aligned} &\sum_{\substack{0.05L\le\log\log p\le0.9L\\1/3\le\sigma_p\le2/3\\\gamma_p\ge\delta_0}}\frac1p\ \\ &\quad \ge\ 0.15L .\quad(1.1) \end{aligned}$

This theorem uses only the sieve estimates and coverage; it does not use the flatness of Lesson 9. The proof occupies this lesson and Lesson 11.

2. Parameters

Throughout this lesson and the next, $\log X=e^L,\qquad R=\sqrt X,$ $\begin{gathered} A_0=A\cap(R+N_*,X],\\ D_0=D\cap[-X,-R-N_*),\\ m_A=|A_0|,\\ m_D=|D_0|. \end{gathered}$ For every prime $p\le R$, the reductions of $A_0$ lie in $S_p$ and those of $D_0$ in $T_p:=S_p^c$, by Lemma 6.1 of Lesson 1. Put $\begin{gathered} \kappa_p=\frac{|T_p|}{|S_p|}=\frac{1-\sigma_p}{\sigma_p},\\ \text{so that}\\ \frac1{\sigma_p}+\frac1{1-\sigma_p}-4=\kappa_p+\kappa_p^{-1}-2 . \end{gathered}$

Lemma 2.1. For large $L$, $\begin{gathered} \sum_{p\le R}\bigl(\kappa_p+\kappa_p^{-1}-2\bigr)\frac{\log p}p\ll L,\\ \sum_{p\le R}\frac{|\log\kappa_p|}p\ll\sqrt L .\quad(2.1) \end{gathered}$

Proof. Apply Lemma 4.1 of Lesson 2 at $Y=X(\log X)^{40}$, with the uniform measures on $A\cap(\sqrt Y+N_*,Y]$ and $D\cap[-Y,-\sqrt Y-N_*)$. By Theorem 3.1 of Lesson 2 their point masses are at most $(\log Y)^4/\sqrt Y$, so $b=4$, and the prime cutoff $\sqrt Y/(\log Y)^5$ exceeds $R$. The first bound follows since $\log\log Y\ll L$. For the second, $(\log t)^2\le t+t^{-1}-2$ for $t>0$ (with $t=e^y$ this is $y^2\le2\cosh y-2$), and Cauchy–Schwarz gives $\begin{aligned} \sum_{p\le R}\frac{|\log\kappa_p|}p&\le\Bigl(\sum_{p\le R}\frac{(\log\kappa_p)^2\log p}p\Bigr)^{1/2}\\ &\quad \Bigl(\sum_p\frac1{p\log p}\Bigr)^{1/2}, \end{aligned}$ and the last series converges. $\square$

3. A budget for centred tensor coordinates

Let $\mathbf e_x$ be the standard basis vector of $x$ in $\mathbb C^{\mathbb F_p}$, with the counting inner product. Let $Q_{S_p}$ be the orthogonal projection onto $\begin{aligned} \mathcal H_{S_p}&=\Bigl\{f\in\mathbb C^{\mathbb F_p}:\operatorname{supp}f\subseteq S_p,\\ &\quad \ \sum_xf(x)=0\Bigr\}, \end{aligned}$ and define $Q_{T_p}$, $\mathcal H_{T_p}$ likewise. Explicitly $Q_{S_p}f=\mathbf 1_{S_p}\bigl(f-\mathbb E_{S_p}f\bigr)$. For a squarefree $t$ whose prime factors are at most $R$, define $\begin{gathered} \begin{aligned} &B_A(t)\\ &\quad=\Bigl\|\frac1{m_A}\sum_{a\in A_0}\bigotimes_{p\mid t}Q_{S_p}\mathbf e_{a\bmod p}\Bigr\|^2, \end{aligned}\\ B_A(1)=1, \end{gathered}$ and $B_D(t)$ with $D_0$ and $T_p$.

Lemma 3.1 (the primitive energy identity). For squarefree $q\le R$, $\begin{aligned} &\sum_{\substack{h\bmod q\\ (h,q)=1}}\Bigl|\frac1{m_A}\sum_{a\in A_0}e_q(ha)\Bigr|^2\\ &\quad =\sum_{t\mid q}t\,B_A(t)\prod_{p\mid q/t}\kappa_p,\quad(3.1) \end{aligned}$ and the same holds for $D_0$ with $B_D$ and $\kappa_p^{-1}$. All terms on the right are nonnegative.

Proof. By the Chinese remainder theorem, $\sum_{(h,q)=1}e_q(hn)=\prod_{p\mid q}\bigl(p\mathbf 1_{p\mid n}-1\bigr)$. For $x,y\in S_p$, $Q_{S_p}\mathbf e_x=\mathbf e_x-\mathbf 1_{S_p}/|S_p|$, so $\langle Q_{S_p}\mathbf e_x,Q_{S_p}\mathbf e_y\rangle=\mathbf 1_{x=y}-1/|S_p|$, and since $p/|S_p|=1+\kappa_p$, $p\mathbf 1_{x=y}-1=\kappa_p+p\,\langle Q_{S_p}\mathbf e_x,Q_{S_p}\mathbf e_y\rangle .$ Expand the left side of (3.1) over two independent uniform elements $a,a'$ of $A_0$, multiply out the product of these identities over $p\mid q$, and collect the terms with a centred factor exactly at the primes of $t$: their average is $t\,\|\frac1{m_A}\sum_a\bigotimes_{p\mid t}Q_{S_p}\mathbf e_a\|^2\prod_{p\mid q/t}\kappa_p$. For $D_0$, $p/|T_p|=1+\kappa_p^{-1}$. $\square$

Let $\mathcal B_L=\{p : 0.05L\le\log\log p\le0.9L\}$, and for fixed $C_*>0$ let $\mathcal T_{C_*}$ be the set of products of at most $C_*L$ distinct primes of $\mathcal B_L$, including $1$.

Proposition 3.2 (the tensor budget [OpenAI-013, (5.4)]). For each fixed $C_*$, $\begin{gathered} \sum_{t\in\mathcal T_{C_*}}B_A(t)\le e^{o(L)}\frac R{m_A},\\ \sum_{t\in\mathcal T_{C_*}}B_D(t)\le e^{o(L)}\frac R{m_D}.\quad(3.2) \end{gathered}$ In particular, from the term $t=1$, $m_A,\ m_D\le Re^{o(L)}. \tag{3.3}$

Proof. For $t\in\mathcal T_{C_*}$, $\log t\le C_*Le^{0.9L}=o(\log X)$, so $t<R$ and $U=R/t=X^{1/2-o(1)}$, uniformly. Among the positive integers $u\le U$: the proportion that are not squarefree is at most $\sum_pp^{-2}<\frac12$; the proportion not coprime to $t$ is at most $\sum_{p\mid t}1/p\le C_*L\exp(-e^{0.05L})=o(1)$; and, by (2.1), $\begin{aligned} &\frac1{\lfloor U\rfloor}\sum_{u\le U}\sum_{p\mid u}|\log\kappa_p|\\ &\quad \le\sum_{p\le U}\frac{|\log\kappa_p|}p\ll\sqrt L, \end{aligned}$ so by Markov's inequality only a proportion $O(L^{-1/4})$ has inner sum above $L^{3/4}$. Hence at least $cU$ integers $u\le U$, with an absolute $c>0$, are squarefree, coprime to $t$, and satisfy $\prod_{p\mid u}\kappa_p\ge e^{-L^{3/4}},\qquad\prod_{p\mid u}\kappa_p^{-1}\ge e^{-L^{3/4}}.$ The large sieve (2.1) of Lesson 1, with $a_n=\mathbf 1_{A_0}(n)/m_A$ on at most $X+1$ integers and $Q=R$, bounds the sum of the left side of (3.1) over squarefree $q\le R$ by $2X/m_A=2R^2/m_A$. Writing $q=tu$ and keeping, on the right of (3.1), only the term with the given $t\in\mathcal T_{C_*}$ and $u$ as above, $\begin{aligned} \frac{2R^2}{m_A}&\ge\sum_{t\in\mathcal T_{C_*}}tB_A(t)\\ &\quad \sum_{\substack{u\le R/t\\ u\ \text{squarefree},\ (u,t)=1}}\prod_{p\mid u}\kappa_p\\ &\ge cRe^{-L^{3/4}}\sum_{t\in\mathcal T_{C_*}}B_A(t). \end{aligned}$ The pairs $(t,u)$ are distinct for distinct terms, and all terms are nonnegative. The bound for $D_0$ is the same with $\kappa_p^{-1}$. $\square$

4. A contracting local kernel

Fix a sufficiently small absolute $\epsilon_0>0$, and let $0<\delta_0\le\epsilon_0^2$. Suppose, for a sequence of arbitrarily large $L$, that (1.1) fails. Let $\begin{gathered} \begin{aligned} \mathcal S&=\Bigl\{p\in\mathcal B_L:\tfrac13\le\sigma_p\le\tfrac23,\ \gamma_p\\ &<\delta_0\Bigr\}, \end{aligned}\\ H=\sum_{p\in\mathcal S}\frac1p . \end{gathered}$ Mertens' estimate gives $\sum_{p\in\mathcal B_L}1/p=0.85L+o(1)$. Outside the balanced range $\kappa_p+\kappa_p^{-1}-2\ge\frac12$, and $\log p\ge e^{0.05L}$ on $\mathcal B_L$, so by (2.1) the unbalanced primes of $\mathcal B_L$ have harmonic mass $O(Le^{-0.05L})=o(1)$. Since (1.1) fails, $H\ge0.70L-o(L)$, and in particular $H\ge0.68L, \tag{4.1}$ also after the deletion of any one prime from $\mathcal S$.

The sparse spectrum. For $p\in\mathcal S$ put $E_p=\{v\in\mathbb F_p:|g_p(v)|>1\}$. It is symmetric and avoids $0$, and $\begin{gathered} |E_p|\le\epsilon_0p,\\ \sum_{v\in E_p}|g_p(v)|^2\ge(1-\epsilon_0^2)p .\quad(4.2) \end{gathered}$ Indeed $|E_p|\le\sum_v|g_p(v)|=p\gamma_p<\delta_0p$, and off $E_p$, $|g_p(v)|^2\le|g_p(v)|$, so $\sum_{v\notin E_p}|g_p|^2<\delta_0p\le\epsilon_0^2p$.

Put $t_0=0.85$ and $b_p(x)=\frac{t_0}p\sum_{h\in E_p}e_p(hx),$ a real function because $E_p$ is symmetric. Let $\Pi_p$ be the orthogonal projection of $\mathbb C^{\mathbb F_p}$ onto the span of the additive characters $e_p(h\,\cdot)$, $h\in E_p$. The matrix with entries $b_p(x-y)$ is $t_0\Pi_p$. Since $0\notin E_p$ and $F_p=\sum_v(g_p(v)/\sqrt p)\,e_p(v\,\cdot)$, Parseval and (4.2) give $\begin{gathered} \Pi_p\mathbf 1=0,\\ \|\Pi_pF_p-F_p\|\le\epsilon_0\sqrt p,\\ \|\Pi_pF_p\|^2\ge(1-\epsilon_0^2)p .\quad(4.3) \end{gathered}$ Let $B_p^{(2)}$ be the matrix with entries $b_p(x-y)^2$. Its eigenvalue at the additive character $s$ is $\frac{t_0^2}p\#\{(h,h')\in E_p^2:h+h'=s\}$$\le t_0^2|E_p|/p$, so $\begin{gathered} \|B_p^{(2)}\|\le t_0^2\epsilon_0,\\ |b_p(x)|\le t_0\epsilon_0 .\quad(4.4) \end{gathered}$ With $e_p^*=|E_p|/p$, direct summation gives $\begin{gathered} \frac1{p-1}\sum_{x\ne0}b_p(x)=-\frac{t_0e_p^*}{p-1},\\ \begin{aligned} &\frac1{p-1}\sum_{x\ne0}b_p(x)^2\\ &\quad =\frac{t_0^2\bigl(e_p^*-(e_p^*)^2\bigr)}{p-1}, \end{aligned}\quad(4.5) \end{gathered}$ because $\sum_xb_p(x)=0$, $b_p(0)=t_0e_p^*$, and $\sum_xb_p(x)^2=t_0^2|E_p|/p$. Both means are $O(\epsilon_0/p)$.

Block coordinates. Put $u_S=\mathbf 1_{S_p}/|S_p|$ and $u_T=\mathbf 1_{T_p}/|T_p|$. For $x\in S_p$, $\mathbf e_x=u_S+Q_{S_p}\mathbf e_x$, and for $y\in T_p$, $\mathbf e_y=u_T+Q_{T_p}\mathbf e_y$. So for a kernel $h(x-y)$, with matrix also denoted $h$, the map $\begin{aligned} L_p[h]&= \begin{pmatrix} u_S^*hu_T&u_S^*hQ_{T_p}\\ Q_{S_p}hu_T&Q_{S_p}hQ_{T_p} \end{pmatrix}\\ &\quad \colon\ \mathbb C\oplus\mathcal H_{T_p}\to\mathbb C\oplus\mathcal H_{S_p} \end{aligned}$ represents $h$ in the coordinates $(1,Q_{S_p}\mathbf e_x)$ and $(1,Q_{T_p}\mathbf e_y)$: $h(x-y)=(1,Q_{S_p}\mathbf e_x)^*L_p[h](1,Q_{T_p}\mathbf e_y)$. For complex $u,v$ define $\begin{gathered} \begin{aligned} L_p(u,v)&=L_p\bigl[(1+ub_p)(1+vb_p)\bigr], \end{aligned}\\ \begin{aligned} U_p(u,v)&=\frac1{p-1}\sum_{x\ne0}\bigl(1\\ &\quad +ub_p(x)\bigr)\bigl(1\\ &\quad +vb_p(x)\bigr). \end{aligned} \end{gathered}$

Lemma 4.1 (local contraction [OpenAI-013, Lemma 5.2]). For $\epsilon_0$ small enough, uniformly for $p\in\mathcal S$ and $|u|,|v|\le1.03$, $\begin{gathered} \|L_p(u,v)\|\le1+\frac Cp,\\ |U_p(u,v)|\le1+\frac Cp, \end{gathered}$ and for all large $p\in\mathcal S$, $\begin{gathered} \begin{aligned} \|L_p(1,1)\|&\le U_p(1,1)\Bigl(1-\frac{1.6}p\Bigr), \end{aligned}\\ U_p(1,1)=1+O\Bigl(\frac{\epsilon_0}p\Bigr).\quad(4.6) \end{gathered}$ All constants are absolute.

Proof. The constant kernel $1$ has $L_p[1]=\begin{pmatrix}1&0\\0&0\end{pmatrix}$, because $u_S$ and $u_T$ are probability vectors and vectors of $\mathcal H_{T_p}$ have sum zero. In $\mathbb C^{\mathbb F_p}$, $\begin{gathered} u_S=\frac{\mathbf 1+\sqrt{\kappa_p}F_p}p,\\ u_T=\frac{\mathbf 1-F_p/\sqrt{\kappa_p}}p, \end{gathered}$ as one checks on $S_p$ and $T_p$. Since $\Pi_p$ is self-adjoint and kills constants, $u_S^*\Pi_pu_T=-p^{-2}\|\Pi_pF_p\|^2$. Since $F_p$ is constant on $S_p$ and on $T_p$, $Q_{S_p}F_p=Q_{T_p}F_p=0$, so for example $Q_{S_p}\Pi_pu_T=-p^{-1}\kappa_p^{-1/2}Q_{S_p}(\Pi_pF_p-F_p)$, which has norm $O(\epsilon_0/\sqrt p)$ by (4.3); the balance bounds $\kappa_p^{\pm1/2}$ absolutely, and the other side block is the same. The supports are disjoint, so $Q_{S_p}Q_{T_p}=0$ and $\begin{aligned} \|Q_{S_p}\Pi_pQ_{T_p}\|&=\|Q_{S_p}(\Pi_p-\tfrac12I)Q_{T_p}\|\\ &\le\tfrac12 . \end{aligned}$ By (4.4) and $\|u_S\|,\|u_T\|=O(p^{-1/2})$, the square kernel contributes $O(\epsilon_0/p)$ to the scalar block, $O(\epsilon_0/\sqrt p)$ to the side blocks, and $O(\epsilon_0)$ to the lower-right block.

Since $(1+ub_p)(1+vb_p)$$=1+(u+v)b_p+uvb_p^2$, the scalar block of $L_p(u,v)$ is $s_p(u,v)=1+O(1/p)$, the side blocks are $O(\epsilon_0/\sqrt p)$, and the lower-right block has norm at most $2\cdot1.03\cdot t_0\cdot\frac12+O(\epsilon_0)<0.95$. At $u=v=1$ the scalar block is real and, by (4.3), $\begin{aligned} s_p(1,1)&=1-\frac{2t_0}{p^2}\|\Pi_pF_p\|^2+O\Bigl(\frac{\epsilon_0}p\Bigr)\\ &\le1-\frac{1.7-O(\epsilon_0)}p . \end{aligned}$ The norm of a $2\times2$ block operator is at most the norm of the $2\times2$ matrix of the block norms, here at most the largest eigenvalue of $\begin{pmatrix}|s_p|&C\epsilon_0/\sqrt p\\ C\epsilon_0/\sqrt p&0.95\end{pmatrix}$. Since $|s_p|-0.95\ge0.04$ for large $p$, this eigenvalue is at most $|s_p|+O(\epsilon_0^2/p)$. This gives the first bound and $\|L_p(1,1)\|\le1-(1.7-O(\epsilon_0))/p$. By (4.5), $U_p(u,v)=1+O(\epsilon_0/p)$ on the polydisc. Decreasing $\epsilon_0$, $1-(1.7-O(\epsilon_0))/p$$\le U_p(1,1)(1-1.6/p)$. $\square$

Tensoring. Choose a large fixed $C_K$ and put $K=\lceil C_KL\rceil$. Define the nonnegative weight $\begin{aligned} &W(n)\\ &\quad=\Bigl(\sum_{\substack{I\subseteq\mathcal S\\|I|\le K}}\prod_{p\in I}b_p(n)\Bigr)^2 .\quad(4.7) \end{aligned}$ For a polynomial $\mathsf P(u,v)$, scalar or operator valued, let $\mathcal R_K\mathsf P$ be the sum of its Taylor coefficients of degree at most $K$ in each variable, evaluated at $u=v=1$. Put $\begin{gathered} \mathcal L(u,v)=\bigotimes_{p\in\mathcal S}L_p(u,v),\\ \mathcal U(u,v)=\prod_{p\in\mathcal S}U_p(u,v),\\ U=\mathcal U(1,1). \end{gathered}$ Expanding $\prod_p(1+ub_p(n))(1+vb_p(n))$$=\sum_{I,J}u^{|I|}v^{|J|}\prod_{p\in I}b_p(n)$$\prod_{p\in J}b_p(n)$ shows that $\mathcal R_K\mathcal L$ represents the kernel $W(x-y)$ in the full tensor coordinates, and that $\mathcal R_K\mathcal U$ is the mean of $W$ over independent uniform unit residues at the primes of $\mathcal S$.

Proposition 4.2 (the pair average is small [OpenAI-013, (5.13)]). For large $L$ in the sequence, $\begin{aligned} &\frac1{m_Am_D}\sum_{a\in A_0}\sum_{d\in D_0}W(a-d)\\ &\quad \le U\exp\bigl(-1.6H\\ &\qquad +o(L)\bigr)\Bigl(\frac{R^2}{m_Am_D}\Bigr)^{1/2},\quad(4.8) \end{aligned}$ and $U=e^{O(\epsilon_0L)}$. The same holds after deleting one prime from $\mathcal S$.

Proof. Truncation. Let $r_0=1.03$. By Lemma 4.1 and $\sum_{p\in\mathcal S}1/p\le0.85L+o(L)$, both $\mathcal L$ and $\mathcal U$ have norm at most $e^{C_0L}$ on $|u|,|v|\le r_0$, with an absolute $C_0$. Cauchy's coefficient formula applies to polynomials with values in finite-dimensional operator spaces: the coefficient of $u^iv^j$ has norm at most $e^{C_0L}r_0^{-i-j}$. Summing the geometric series over $i>K$ or $j>K$, $\begin{aligned} &\|\mathcal R_K\mathcal L-\mathcal L(1,1)\|+|\mathcal R_K\mathcal U-U|\\ &\quad \ll e^{C_0L}r_0^{-K}. \end{aligned}$ By Lemma 4.1, $U=e^{O(\epsilon_0L)}$ and $\|\mathcal L(1,1)\|\le U\prod_{p\in\mathcal S}(1-1.6/p)$$\le Ue^{-1.6H}$. Choose $C_K$ so large that $e^{C_0L}r_0^{-K}\le e^{-5L}$, which is $o(Ue^{-1.6H})$ since $H\le0.85L+o(L)$ and $\epsilon_0$ is small.

The coordinates actually used. Let $\begin{gathered} \begin{aligned} V_A&=\frac1{m_A}\sum_{a\in A_0}\bigotimes_{p\in\mathcal S}\bigl(1,\\ &\quad Q_{S_p}\mathbf e_{a\bmod p}\bigr), \end{aligned}\\ \begin{aligned} V_D&=\frac1{m_D}\sum_{d\in D_0}\bigotimes_{p\in\mathcal S}\bigl(1,\\ &\quad Q_{T_p}\mathbf e_{d\bmod p}\bigr). \end{aligned} \end{gathered}$ The tensor space is the orthogonal sum of the subset modes $\bigotimes_{p\in I}\mathcal H_{S_p}$, and the squared norm of the component of $V_A$ in the mode $I$ is $B_A(\prod_{p\in I}p)$. Every monomial kept in $\mathcal R_K\mathcal L$ involves at most $2K$ primes, and at every other prime the local factor is $L_p[1]$, which has only a scalar block. So $\mathcal R_K\mathcal L=P_A(\mathcal R_K\mathcal L)P_D$, where $P_A,P_D$ keep only the modes with at most $2K$ centred factors. Proposition 3.2 with any fixed $C_*>2C_K$ gives $\|P_AV_A\|^2\le e^{o(L)}R/m_A$ and $\|P_DV_D\|^2\le e^{o(L)}R/m_D$.

The dimension of the tensor space plays no role: only the modes with few centred factors are seen by the truncated kernel, and their total weight is controlled by the large sieve through Proposition 3.2.

5. Exercises

5.1. Show that $(\log t)^2\le t+t^{-1}-2$ for all $t>0$.

5.2. Let $\Pi$ be an orthogonal projection on a Hilbert space and $Q_1,Q_2$ orthogonal projections with $Q_1Q_2=0$. Show that $\|Q_1\Pi Q_2\|\le\frac12$, and give an example with equality.

5.3. Show that the norm of a block operator $\begin{pmatrix}A&B\\C&D\end{pmatrix}$ is at most the norm of the real $2\times2$ matrix $\begin{pmatrix}\|A\|&\|B\|\\\|C\|&\|D\|\end{pmatrix}$.

5.4. Verify the formulas for $u_S$ and $u_T$ in terms of $F_p$ and $\kappa_p$.

6. Solutions

5.1. Put $t=e^y$. The claim is $y^2\le e^y+e^{-y}-2=2(\cosh y-1)$, which holds because $\cosh y-1=\sum_{n\ge1}y^{2n}/(2n)!\ge y^2/2$.

5.2. Since $Q_1Q_2=0$, $Q_1\Pi Q_2=Q_1(\Pi-\frac12I)Q_2$, and $\Pi-\frac12I$ is self-adjoint with eigenvalues $\pm\frac12$, so its norm is $\frac12$. Equality: in $\mathbb C^2$, $Q_1,Q_2$ the projections onto the two coordinate axes and $\Pi$ the projection onto the diagonal line; then $Q_1\Pi Q_2\mathbf e_2=\frac12\mathbf e_1$.

5.3. For a vector $(x,y)$, $\|(Ax+By,Cx+Dy)\|^2$$\le(\|A\|\|x\|+\|B\|\|y\|)^2$$+(\|C\|\|x\|+\|D\|\|y\|)^2$, which is the squared norm of the real matrix applied to $(\|x\|,\|y\|)$, and $\|(\|x\|,\|y\|)\|=\|(x,y)\|$.

5.4. $\sqrt{\kappa_p}F_p=\sqrt{\frac{1-\sigma}\sigma}\cdot\frac{\mathbf 1_S-\sigma}{\sqrt{\sigma(1-\sigma)}}=\frac{\mathbf 1_S-\sigma}\sigma$, so $\mathbf 1+\sqrt{\kappa_p}F_p=\mathbf 1_S/\sigma$ and $(\mathbf 1+\sqrt{\kappa_p}F_p)/p=\mathbf 1_S/|S|$. Similarly $F_p/\sqrt{\kappa_p}=(\mathbf 1_S-\sigma)/(1-\sigma)$, so $\mathbf 1-F_p/\sqrt{\kappa_p}=\mathbf 1_T/(1-\sigma)$.

Sources

[OpenAI-013] OpenAI, The additive indecomposability of the primes, OpenAI Math Release preprint, 24 September 2026, Section 5: Proposition 5.1, a budget for centred tensor coordinates (5.2)–(5.4), a contracting local kernel from a sparse transform, Lemma 5.2, and tensoring and truncating the kernel (5.13). https://github.com/openai/math/tree/main/preprints/the-additive-indecomposability-of-the-primes-September-24-2026