Proof. Conditional on all types, sample an activation law independently at each incidence when expanding the product in Equation (8). Conditional on these samples, the matrix integral factors across coordinates. At coordinate \(\kappa\) apply Lemma 5 with the active set of incidences for which \(\kappa\in S\), and with their sampled signs. Use Equation (7) at each coordinate. There are only finitely many activation and sign configurations. The model expectations are bounded, and their approximating matrix expectations are consequently bounded for large \(q\). A finite product and then the finite average therefore preserve the \(o(1)\) error.

Expand these finite parity products, and let \(\alpha_e\) be the set of coordinates in which the four-sign term for \(e\) is selected. A coordinate selected for \(e\) requires activation at all four corners in \(U_e\) and contributes the product of their signs. At \((i,j)\) the two subsets on its incident pairs give exactly \(\phi_{\alpha_{e(i,j)}}\phi_{\alpha_{f(i,j)}}\); averaging that incidence’s activation and signs gives its factor \(A\). Thus the limiting contribution at the fixed type assignment is \[ \sum_{\alpha_e\subseteq\mathcal K\ (e\in E)} \prod_{(i,j)\in\mathscr C} A_{a(i)b(j)} \bigl(\alpha_{e(i,j)},\alpha_{f(i,j)}\bigr). \tag{14}\] Repeated vertex types leave the matrix draws at distinct source vertices independent. Repeated type pairs use independent samples from the same activation law at their distinct source incidences.

The all-empty labeling contributes one. In any other nonzero term, no face can have both an empty and a nonempty pair label: the two meet at a corner, where Equation (11) makes the factor zero. Hence any face with one nonempty label has all three labels nonempty. The shared-pair labels and connectedness of the face-neighbor graph propagate this conclusion to every face. Thus the remaining labels are exactly those summed in Equation (12). This proves Equation (13). For the transpose, the same argument uses the last assertion of Lemma 5. ◻

Proof. We first prove the support identity \[ \bigcap_{\kappa\in U_g}M_\kappa=U_g \qquad(g\in E). \tag{19}\] The same set \(\mathscr C\) indexes coordinates \(\gamma\) and activation corners \(\kappa\). Availability is symmetric: \(\gamma\in M_\kappa\) if and only if \(\kappa\in M_\gamma\). Both conditions mean that the two corners lie in a common set \(U_e\). Fix a coordinate \(\gamma=(v,j)\), write \(j=\{v,w,z\}\), and let \(e=\{v,w\}\), \(f=\{v,z\}\) have opposite faces \(l,m\), respectively. This coordinate belongs to precisely \(U_e\) and \(U_f\). Consequently it is available at exactly the following activation corners: all three corners of \(j\), the two \(e\)-endpoint corners of \(l\), and the two \(f\)-endpoint corners of \(m\). To be available at all four corners of \(g\), its two faces must therefore be among \(j,l,m\). The choice \(j,l\) forces \(g=e\), and \(j,m\) forces \(g=f\), since distinct triples share at most one pair. The choice \(l,m\) cannot work: its two endpoints would both have to lie in \(\{v,w\}\cap\{v,z\}=\{v\}\). Thus the coordinate \(\gamma\) is available at every corner of \(g\) if and only if \(g\in\{e,f\}\), equivalently if and only if \(\gamma\in U_g\). This proves Equation (19).

In a nonzero identity-type term, \(\alpha_g\) must lie in the maximal activation support at each of the four corners of \(g\). Hence \(\alpha_g\subseteq U_g\). At a corner \(\kappa\) with pairs \(e,f\), the two nonempty labels cannot be disjoint by Equation (11). Since \(U_e\cap U_f=\{\kappa\}\), both labels contain \(\kappa\). Applying this at all four corners of \(g\) gives \(\alpha_g=U_g\).

For these labels, Equation (16) gives the local factor \(\sigma_\kappa\theta/2\). Multiplying the \(66\) factors, with exactly one negative \(\sigma_\kappa\), proves the assertion. ◻

Proof. Let \(\ell_0\) be the two part colors and, at round \(t\), let \(h_t(u)\) be the histogram of \(\ell_{t-1}\)-colors among the neighbors of \(u\). The new color is \(\ell_t(u)=(\ell_{t-1}(u),h_t(u))\). Choose \(R\ge1\) rounds giving a discrete coloring. At round \(t\) assign the vertex score \(-\|h_t(a)\|^2\) to type \(a\) and the pair score \[2h_t(a)(\ell_{t-1}(b))+2h_t(b)(\ell_{t-1}(a))\] to every \((a,b)\in P\times Q\), including nonedges. The total round score of \(f\) is \[F_t(f)=-\sum_u\|h_t(f(u))\|^2 +2\sum_u\sum_{v\in N_F(u)} h_t(f(u))\bigl(\ell_{t-1}(f(v))\bigr).\] Here \(N_F(u)\) is the neighbor set of \(u\). Both sums are indexed by the vertices and directed incidences of the source graph \(F\), even when \(f\) repeats a type.

If \(\ell_{t-1}(f(u))=\ell_{t-1}(u)\) for every \(u\), the inner neighbor sum equals \(\langle h_t(f(u)),h_t(u)\rangle\). Thus \[\begin{align*} F_t(f) &=\sum_u\bigl(2\langle h_t(f(u)),h_t(u)\rangle -\|h_t(f(u))\|^2\bigr)\\ &=\sum_u\|h_t(u)\|^2 -\sum_u\|h_t(f(u))-h_t(u)\|^2. \tag{21}\end{align*}\] The first equality uses only preservation of previous colors. It therefore applies to noninjective maps and to maps taking edges to nonedges. In this situation \(F_t(\mathrm{id})-F_t(f)\) is a nonnegative integer, zero precisely when \(f\) also preserves all new colors.

To combine the rounds into a single score, set \[M=\max_{f,\,1\le t\le R}|F_t(f)-F_t(\mathrm{id})|, \qquad L_0=M+2, \qquad \mathcal Q(f)=\sum_{t=1}^R L_0^{R-t}F_t(f).\] The maximum defining \(M\) is finite. For a nonidentity map, let \(t\) be the first round at which a color fails to be preserved. All earlier scores tie the identity; the loss at round \(t\) is at least \(L_0^{R-t}\). Its possible later advantage is at most \[M\sum_{j=0}^{R-t-1}L_0^j<L_0^{R-t}.\] Hence its total score is strictly smaller. A nonidentity map must have such a first round because the final colors are discrete. The weighted sums of the round vertex scores and pair scores are the required \(s_t,s_{ab}\). Finiteness of the set of maps gives the positive gap. ◻

Proof. Use the kernel \(W:\mathcal X\times\mathcal Y\to[0,\infty)\) just constructed and set \(\Omega=\mathcal X\times\mathcal Y\) with product probability. Define \[ V\bigl((x,y),(x',y')\bigr)=W(x,y')W(x',y). \tag{31}\] This kernel is symmetric, nonnegative, and has mean one: the two factors on the right use independent variables and each has mean one. In its product over \(H\), the factors using \(x_i\) for \(i\in I\) and \(y_j\) for \(j\in J\) are independent of those using \(y_i\) for \(i\in I\) and \(x_j\) for \(j\in J\). Therefore \[t(H,V)=t_{I,J}(W)t_{I,J}(W^\top)<1.\] Let \(L=\max V\), which is finite and at least one, and set \(U=V/L\). Then \(0\le U\le1\), \(\mu=\mathbb EU=1/L>0\), and homogeneity gives \(t(H,U)=L^{-66}t(H,V)<L^{-66}=\mu^{66}\). ◻

Proof. Sample labels \(z_1,\ldots,z_n\) independently from the probability space of \(U\). Conditional on these labels, put an edge on each unordered pair of distinct vertices \(\{a,b\}\) independently with probability \(U(z_a,z_b)\). No loops are drawn. This defines a random finite simple undirected graph \(G_n\), and \[ \mathbb Ep(G_n)=(1-1/n)\mu. \tag{33}\]

For every injective map \(f:V(H)\to\{1,\ldots,n\}\), its image vertices have independent labels, and its \(66\) source edges use distinct unordered host pairs. Its probability of being a homomorphism is exactly \(\tau\). Among all \(n^{35}\) maps, the proportion that identify any two source vertices is at most \(\binom{35}{2}/n=595/n\), by the union bound. Each such map has a homomorphism indicator between zero and one; a collision on a source edge gives zero, while any repeated host edges are counted by their actual single Bernoulli trials. No independence claim for these maps is needed. Comparing their contribution with \(\tau\) times their proportion gives \[ |\mathbb Et(H,G_n)-\tau|\le\frac{595}{n}. \tag{34}\] Thus the density still counts all homomorphisms, including every noninjective one.

Using \((1-1/n)^{66}\ge1-66/n\), Equations (32) and (34) give \[\begin{align*} \mathbb Et(H,G_n) &\le\mu^{66}-\Delta+595/n\\ &<\mu^{66}-66\mu^{66}/n\\ &\le\bigl((1-1/n)\mu\bigr)^{66} =\bigl(\mathbb Ep(G_n)\bigr)^{66} \le\mathbb E\bigl[p(G_n)^{66}\bigr]. \end{align*}\] The last inequality is convexity. Hence some realization has \(t(H,G_n)<p(G_n)^{66}\). Such a realization cannot be edgeless. ◻

Proof. Count ordered linearly independent \(k\)-tuples in \(\mathbb F_q^n\) and divide by the number of ordered bases of \(\mathbb F_q^k\) to obtain Equation (35). Its asymptotic follows by factoring out the displayed powers of \(q\); all products have fixed length.

For the nonsingularity and sign assertions, use induction on \(n\). On the event that the leading \((n-1)\)-dimensional principal minor \(C\) is nonsingular, write \[M=\begin{pmatrix}C&z\\z^{\mathsf T}&a\end{pmatrix},\qquad \det M=\det C\,(a-z^{\mathsf T}C^{-1}z).\] Given \(C,z\), the final scalar factor is uniform on \(\mathbb F_q\). It is zero with probability \(q^{-1}\) and has either nonzero character with probability \((q-1)/(2q)\). The exceptional principal minor has probability \(O(q^{-1})\) by induction. The same argument starts at \(n=1\) with the empty principal minor.

For the rank assertion, fix a \(k\)-subspace \(U\) with basis-column matrix \(P\). Every symmetric matrix with image \(U\) is uniquely \(PRP^{\mathsf T}\) for a nonsingular symmetric \(k\)-by-\(k\) matrix \(R\). Indeed, if \(Q P\) is the identity, symmetry and the image condition give \(M=P(QMQ^{\mathsf T})P^{\mathsf T}\), and multiplication by \(Q,Q^{\mathsf T}\) also proves uniqueness. The surjection \(P^{\mathsf T}:\mathbb F_q^n\to\mathbb F_q^k\) identifies the nondegenerate quotient form with \(R\), so the discriminant sign is \(\chi(\det R)\). Consequently the desired count is \[\genfrac{[}{]}{0pt}{}{n}{k}_q\,q^{b(k)} \left(\frac12+O(q^{-1})\right).\] Divide by \(q^{b(n)}\) and use \(k(n-k)+b(k)-b(n)=-b(n-k)\).

For Equation (37), sum over \(u\)-subspaces \(U\) contained in the radical. In a basis beginning with \(U\), this containment sets to zero the first \(u\) rows and columns, imposing \(nu-\binom u2\) independent linear conditions. A uniform symmetric form therefore has \(U\) in its radical with probability \(q^{-nu+\binom u2}\). Multiplication by \(\genfrac{[}{]}{0pt}{}{n}{u}_q=O(q^{u(n-u)})\) gives the assertion. ◻

Proof. First, a nondegenerate symmetric form has an orthogonal basis. A nonzero form has a vector of nonzero norm, since otherwise polarization would make the form zero. Split off this nondegenerate line and continue on its orthogonal complement.

For any \(a,b\in\mathbb F_q^\times\) and \(t\in\mathbb F_q\), the two sets \(\{ax^2:x\in\mathbb F_q\}\) and \(\{t-by^2:y\in\mathbb F_q\}\) each have size \((q+1)/2\) and therefore meet. Thus \(ax^2+by^2=t\) is solvable. A diagonal nondegenerate form in at least three variables consequently has a nonzero isotropic vector: solve \(ax^2+by^2=-c\) and set the third coordinate to \(1\). Given an isotropic vector \(e\ne0\), choose \(f\) with \(\langle e,f\rangle=1\) and replace \(f\) by \(f-\langle f,f\rangle e/2\). The resulting plane has Gram matrix \(\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)\) and splits off orthogonally. Repetition leaves an anisotropic space of dimension at most two. In dimension two, \(ax^2+by^2\) is isotropic exactly when \(\chi(-ab)=1\), or equivalently when its determinant character is \(\chi(-1)\). Each hyperbolic plane also has determinant character \(\chi(-1)\). This proves that the determinant condition is precisely the condition that the entire \(2r\)-space split into hyperbolic planes.

A split form has the required isotropic subspace, spanned by one vector from each hyperbolic pair. Conversely, if such a subspace \(U\) is given, split off a hyperbolic plane using \(e\in U\). Its orthogonal complement has an isotropic \((r-1)\)-subspace: for \(u\in U\), replace \(u\) by \(u-\langle u,f\rangle e\). This linear map on \(U\) has kernel \(\langle e\rangle\), so its image has dimension \(r-1\). Induction shows that the entire form is split.

In split coordinates \((x,y)\in\mathbb F_q^r\oplus\mathbb F_q^r\), isotropy is the equation \(x^{\mathsf T}y=0\). There are \[(q^r-1)q^{r-1}+(q^r-1) =(q^r-1)(q^{r-1}+1)\] nonzero isotropic vectors. The quotient \(\ell^\perp/\ell\) by an isotropic line has dimension \(2r-2\). Splitting off a hyperbolic plane through \(\ell\) identifies this quotient with its nondegenerate orthogonal complement, whose determinant character is \(\chi((-1)^{r-1})\). The criterion just proved makes the quotient split. Count pairs \((\ell,U)\) with \(\ell\subset U\) and \(\dim U=r\) to get \[N_r\frac{q^r-1}{q-1} =\frac{(q^r-1)(q^{r-1}+1)}{q-1}N_{r-1}.\] With \(N_0=1\), this gives the product in Equation (39). In particular \(N_1=2\); no choice of one isotropic family has been imposed in this count. ◻

Proof. Restriction onto a fixed subspace is a surjective linear map on spaces of symmetric forms, so each summand has expectation \(1/2+O(q^{-1})\) by Lemma 11. For distinct \(U,V\in\mathcal U\) write \(T=U\cap V\). Extend a basis of \(T\) separately to bases of \(U\) and \(V\), and then to an ambient basis. In this basis, the coefficients defining the two restricted forms overlap exactly in the symmetric block on \(T\). Thus any two forms on \(U,V\) agreeing on \(T\) extend simultaneously to the ambient space; given \(Q|_T\), their remaining coefficients are independent and uniform.

If \(T\) has positive dimension, its restriction is nonsingular outside an event of probability \(O(q^{-1})\); for \(T=0\) this holds with probability one. Given a nonsingular restriction on \(T\), the determinant on \(U\) is the determinant on \(T\) times that of a uniform symmetric Schur complement of size \(r-\dim T>0\). The corresponding Schur complement on \(V\) is independent. Both signs are balanced up to \(O(q^{-1})\), uniformly over the fixed common block and mixed coefficients. The covariance of the two indicator summands is therefore \(O(q^{-1})\). The diagonal covariance is at most \(1\), giving \[\mathop{\mathrm{Var}}(f_\xi)=O(q^{-1}+|\mathcal U|^{-1}).\] Cauchy–Schwarz and the expectation estimate prove Equation (40).

For the last assertion, each of \(Q\) and \(Q-L\) is marginally uniform. The event \(\mathcal A\) that the specified forms are nonsingular has probability \(1-O(q^{-1})\), uniformly in \(L\). Since the error being averaged is nonnegative, \[\mathbb E\left[\left|f_\xi-\frac12\right|\,\middle|\,\mathcal A\right] \leq \frac{\mathbb E|f_\xi-1/2|}{\mathbb P(\mathcal A)}.\] This proves the same bound under either conditioning. ◻

Proof. If \(A\) and \(B-A\) both have rank \(r\), their images span the image of \(B\), so these images are complementary. Since \(AB^{-1}+(B-A)B^{-1}\) is the identity, uniqueness of the corresponding direct-sum decomposition makes \(AB^{-1}\) the projection onto the first image. In particular \(AB^{-1}A=A\). Conversely, this identity makes \(AB^{-1}\) an idempotent of rank \(r\). Its complementary idempotent has rank \(r\) and, after multiplication by \(B\), equals \(B-A\).

Choose basis-column matrices \(P_1,P_2\) of the two complementary images and write \(A=P_1R_1P_1^{\mathsf T}\) and \(A-B=P_2R_2P_2^{\mathsf T}\). Then \[B=(P_1\ P_2)\mathop{\mathrm{diag}}(R_1,-R_2)(P_1\ P_2)^{\mathsf T}.\] Taking determinant characters gives Equation (42). ◻

Proof. First suppose all three incidences are active, and locally write their point matrices as \(X_1,X_2,X_3\) and their signs as \(\xi_1,\xi_2,\xi_3\). Set \[B_0=X_1-X_2,\quad B_1=X_1-X_3,\quad A=X_1-Y, \qquad Q=B_0^{-1},\quad L=B_0^{-1}-B_1^{-1}.\] On \(\mathcal T_j\), the form \(L=B_0^{-1}(B_1-B_0)B_1^{-1}\) is symmetric and nonsingular. Write a rank-\(r\) matrix \(A\) as \(PRP^{\mathsf T}\). By Lemma 14, its other two rank conditions are equivalent to \[R(P^{\mathsf T}B_\nu^{-1}P)R=R\quad(\nu=0,1),\] or to \[ P^{\mathsf T}LP=0,\qquad P^{\mathsf T}QP\text{ nonsingular},\qquad R=(P^{\mathsf T}QP)^{-1}. \tag{47}\] Thus the centers satisfying the three rank conditions are in bijection with the maximal \(L\)-isotropic subspaces \(U\) on which \(Q\) restricts nonsingularly. Changing the basis of \(U\) leaves the resulting \(A\) unchanged, and every \(A\) recovers \(U\) as its image, proving the stated bijection.

Let \(\mathcal M\) be the event that all three required pair matches \(\eta^X_{\{a,b\}}=c_0\xi_a\xi_b\) hold. If any match fails, the signed count is zero by Equation (42). If they all hold, then \[\chi(\det L) =\eta^X_{\{1,2\}}\eta^X_{\{1,3\}}\eta^X_{\{2,3\}} =c_0,\] so \(L\) is split by Lemma 12. In Equation (47), the sign of \(A\) is the sign of \(Q|_U\), since inversion preserves a nonzero determinant character. Once it equals \(\xi_1\), the pair matches with point \(1\) force the other two signs to equal \(\xi_2,\xi_3\).

For split \(L\), let \(f_{\xi_1}(Q;L)\) denote the fraction of its \(N_r\) maximal isotropic subspaces having a nonsingular restriction of sign \(\xi_1\). On a matching transverse triple the exact normalized count is \[ F_j=\frac{q^{-b(D)}N_r}{\pi_{\xi_1}\pi_{\xi_2}\pi_{\xi_3}} f_{\xi_1}(Q;L) =(16+O(q^{-1}))f_{\xi_1}(Q;L). \tag{48}\] Here the exponent cancels because \(r(r-1)/2-b(2r)+3b(r)=0\). The factor \(16\) combines the leading factor \(2\) in \(N_r\) with the three signed-layer normalization factors \(2\); the restriction-sign fraction will supply a further factor \(1/2\).

We specify the probability law used for that fraction. Before any invertibility conditioning, \(B_0,B_1\) are independent uniform symmetric matrices: this follows by conditioning on \(X_1\) and translating the independent matrices \(X_2,X_3\). After requiring \(B_0,B_1\) to be nonsingular, inversion preserves their independent uniform laws on nonsingular symmetric matrices. Conditional also on a fixed value of \(L=B_0^{-1}-B_1^{-1}\), the matrix \(Q\) is therefore uniform on \[\{Q\in\mathcal S_D:Q\text{ and }Q-L\text{ are nonsingular}\}.\] Each admissible \(Q\) corresponds to exactly one ordered inverse pair \((Q,Q-L)\), which verifies the conditional uniformity directly. Requiring \(L\) to be nonsingular is exactly the remaining transverse condition. Lemma 13, uniformly for every fixed nonsingular split \(L\), now gives \[ \mathbb E\left[\mathbf 1_{\mathcal M} \left|f_{\xi_1}(Q;L)-\frac12\right| \,\middle|\,\mathcal T_j,L\right] \leq \mathbb E\left[\left|f_{\xi_1}(Q;L)-\frac12\right| \,\middle|\,\mathcal T_j,L\right] =O(q^{-1/2}). \tag{49}\] We used \(r>1\), for which \(N_r^{-1/2}=O(q^{-1/2})\). The matching indicator is kept inside the expectation; no additional conditioning on the matches is used. For nonsplit \(L\), the matching event is empty and both the signed count and its model vanish. Equations (48) and (49) prove the local error estimate with limiting value \(8\) on matches. The same exact formula, with \(0\leq f_{\xi_1}\leq1\), gives a uniform bound on every transverse triple, including the exceptional restrictions.

For two active incidences, relabel them \(1,2\) and put \(B=X_1-X_2\) and \(Q=B^{-1}\). The same projection argument now parametrizes centers by all \(r\)-subspaces \(U\) for which \(Q|_U\) is nonsingular. On the pair match \(\chi(\det B)=c_0\xi_1\xi_2\), only the sign \(\xi_1\) of this restriction needs to be imposed; off the match the count is zero. If \(f_{\xi_1}(Q)\) is the fraction among all \(\genfrac{[}{]}{0pt}{}{2r}{r}_q\) subspaces, the exact count on the match is \[ F_j=\frac{q^{-b(D)}\genfrac{[}{]}{0pt}{}{2r}{r}_q} {\pi_{\xi_1}\pi_{\xi_2}} f_{\xi_1}(Q) =(4+O(q^{-1}))f_{\xi_1}(Q). \tag{50}\] The exponent identity here is \(r^2-b(2r)+2b(r)=0\). Conditional on \(B\) being nonsingular, \(Q\) is uniform nonsingular symmetric. Apply Lemma 13 with this single conditioning and keep the matching indicator inside the expectation as before. This gives limiting value \(2\), error \(O(q^{-1/2})\), and a uniform bound on the nonsingular-pair event. Restricting further to \(\mathcal T_j\) can only decrease the nonnegative error expectation. The remaining cases follow from the mean-one normalization of each \(K_\xi\) and translation invariance. ◻

Proof. Fix a nonempty subset \(F\subseteq E\), and choose a vertex \(i\) incident with \(s\geq1\) of its pairs. Write \(N=D-1\) and, for every point \(v\), write its matrix in block form \[X_v=\begin{pmatrix}a_v&z_v^{\mathsf T}\\z_v&W_v\end{pmatrix}, \qquad W_v\in\mathcal S_N,\quad z_v\in\mathbb F_q^N.\] Let \(k_1,\ldots,k_s\) be the selected neighbors of \(i\), and put \(M_\ell=W_i-W_{k_\ell}\). We first establish a high-probability event using these lower-right blocks, before conditioning on the other entries. Given \(W_i\), the \(M_\ell\) are independent uniform symmetric matrices. All are nonsingular except with probability \(O(q^{-1})\). Conditional on their nonsingularity, their inverses are independent uniform nonsingular symmetric matrices.

Let \(Q_1,\ldots,Q_s\) instead be independent uniform matrices in \(\mathcal S_N\). For any \(t\in\mathbb F_q^s\setminus\{0\}\), the matrix \(\sum_\ell t_\ell Q_\ell\) is uniform in \(\mathcal S_N\): condition on all but a coordinate with nonzero coefficient. By Equation (37), its probability of rank at most \(s\) is \(O(q^{-b(N-s)})\). The union bound over all nonzero \(t\) gives \[\mathbb P\left(\exists t\ne0: \mathop{\mathrm{rank}}\left(\sum_\ell t_\ell Q_\ell\right)\leq s\right) =O(q^{s-b(N-s)}).\] Conditioning all the \(Q_\ell\) to be nonsingular divides this upper bound by a probability \(1-O(q^{-1})\). It follows that the event \[ \mathcal G=\left\{\begin{array}{l} M_\ell\text{ is nonsingular for every }\ell,\\ \mathop{\mathrm{rank}}(\sum_\ell t_\ell M_\ell^{-1})\geq s+1 \text{ for every }t\in\mathbb F_q^s\setminus\{0\} \end{array}\right\} \tag{51}\] satisfies \[ \mathbb P(\mathcal G^c)=O(q^{-1}+q^{s-b(N-s)})=O(q^{-1}). \tag{52}\] Indeed \(s\leq6\) and \(D\geq12\) imply \(N-s\geq5\) and \(b(N-s)\geq15>s\). These probability bounds hold given every \(W_i\) and hence hold without conditioning as well.

Only now condition on all entries of all point matrices except \(z_i\). The event \(\mathcal G\) is determined by this conditioning, whereas \(z_i\) remains uniform on \(\mathbb F_q^N\). On every such fixed configuration in \(\mathcal G\), the determinant formula gives \[ \det(X_i-X_{k_\ell})=\det(M_\ell)h_\ell(z_i),\qquad h_\ell(z)=a_i-a_{k_\ell} -(z-z_{k_\ell})^{\mathsf T}M_\ell^{-1}(z-z_{k_\ell}). \tag{53}\] Let \(\mu\) be the law of \((h_1(z),\ldots,h_s(z))\) for uniform \(z\). Fix a nontrivial additive character \(\psi\) of \(\mathbb F_q\) and define \(\widehat\mu(t)=\mathbb E_z\psi(\sum_\ell t_\ell h_\ell(z))\). The quadratic part at every frequency \(t\ne0\) has rank at least \(s+1\) by Equation (51).

We use the standard rank bound for a quadratic Gauss sum; see Gowers and Wolf (Gowers and Wolf 2011, Lemma 3.3). The short differencing argument also fixes our normalization. A quadratic polynomial \(p(z)=z^{\mathsf T}Hz+\lambda^{\mathsf T}z+c\) with symmetric quadratic matrix \(H\) obeys \[\begin{align*} \left|\mathbb E_z\psi(p(z))\right|^2 &=\mathbb E_{z,u}\psi(p(z+u)-p(z))\\ &=\mathbb E_u\psi(u^{\mathsf T}Hu+\lambda^{\mathsf T}u) \mathbb E_z\psi(2z^{\mathsf T}Hu), \end{align*}\] whose absolute value is at most \(\mathbb P_u(Hu=0)=q^{-\mathop{\mathrm{rank}}H}\). Here the inner average vanishes unless \(Hu=0\), by additive-character orthogonality and the fact that \(q\) is odd. Therefore \(|\widehat\mu(t)|\leq q^{-(s+1)/2}\) for \(t\ne0\), uniformly over all the fixed linear and constant terms in Equation (53).

Writing \(\nu\) for uniform measure on \(\mathbb F_q^s\), finite Fourier orthogonality and Cauchy–Schwarz yield the explicit bound \[\begin{align*} \mathop{\mathrm{TV}}(\mu,\nu) &=\frac12\sum_x|\mu(x)-q^{-s}|\tag{54}\\ &\leq\frac12\left(q^s\sum_x|\mu(x)-q^{-s}|^2\right)^{1/2} =\frac12\left(\sum_{t\ne0}|\widehat\mu(t)|^2\right)^{1/2} \leq\frac12q^{-1/2}. \end{align*}\] The identity in the middle follows by expanding the square and using \(\sum_t\psi(t\cdot(x-y))=q^s\mathbf 1_{\{x=y\}}\).

Under the same fixed conditioning, all factors \(\eta_e^X\) for \(e\in F\) not incident to \(i\) are constant. They and the characters of the minor determinants have absolute value at most one. By Equation (53), the remaining part of the product is \(\prod_{\ell=1}^s\chi(h_\ell(z_i))\). Its expectation under \(\nu\) is zero, because \(\mathbb E_{x\in\mathbb F_q}\chi(x)=0\) and \(s\geq1\). The difference of expectations of a function bounded by one is at most twice total variation, so on \(\mathcal G\) the absolute conditional expectation of the entire selected product is at most \(q^{-1/2}\). On \(\mathcal G^c\) it is at most one. Averaging and using Equation (52) proves \[ \mathbb E\prod_{e\in F}\eta_e^X=O(q^{-1/2})\qquad(F\ne\varnothing). \tag{55}\]

Finally, replace any zero coordinate of \(\eta^X\) by \(+1\), obtaining a sign vector \(\widetilde\eta^X\). A union bound and Lemma 11 show that this replacement changes the vector with probability \(O(q^{-1})\) and changes each subset moment by \(O(q^{-1})\). For every \(\varepsilon\in\{+1,-1\}^E\), the identity \[\mathbf 1_{\{\widetilde\eta^X=\varepsilon\}} =2^{-|E|}\prod_{e\in E}(1+\varepsilon_e\widetilde\eta_e^X)\] expresses its probability through the subset moments. The empty subset contributes \(2^{-|E|}\) and every other subset is \(O(q^{-1/2})\) by Equation (55). Since \(E\) is fixed, summing over all sign vectors proves convergence in total variation. ◻

Proof. Given the point matrices, the center variables are independent, so the left side is \(\mathbb E[\mathbf 1_{\mathcal T}\prod_jF_j]\). Increase \(C_D\) from Lemma 15 to at least \(8\). The deterministic telescoping identity for two products gives, on \(\mathcal T\), \[\left|\prod_jF_j-\prod_jG_j\right| \leq C_D^{|J|-1}\sum_j|F_j-G_j|.\] Since \(\mathcal T\subseteq\mathcal T_j\), taking expectations and applying Equation (46) bounds this difference by \(O(q^{-1/2})\). This argument applies to faces sharing point matrices; it uses no independence of their local errors.

The bound \(\prod_jG_j\leq8^{|J|}\) and \(\mathbb P(\mathcal T^c)=O(q^{-1})\) allow removal of \(\mathbf 1_{\mathcal T}\) from the model expectation. Lemma 16 then replaces \(\eta^X\) by independent uniform signs. Total variation controls every bounded function, including the displayed model even when \(c_0\) varies with the prime \(q\). Finally, there are only finitely many activity and sign choices, so the largest of their errors also tends to zero. ◻

Proof. Every line is isotropic. Count pairs consisting of a line and a Lagrangian containing it. Each Lagrangian has \((q^N-1)/(q-1)\) lines, and there are \((q^{2N}-1)/(q-1)\) lines in \(\mathcal V\). Reduction by a line gives \(L_N(q^N-1)=(q^{2N}-1)L_{N-1}\). Starting at \(L_0=1\) yields the product in Equation (61). Equation (58) gives the containment count. Counting pairs \((S,L)\) with \(\dim S=t\) and \(S\subseteq L\) gives \(I_N(t)L_{N-t}=L_N\genfrac{[}{]}{0pt}{}{N}{t}_q\).

For the pair count, first choose \(H=A\cap B\) inside \(A\). In \(H^\perp/H\), the image of \(B\) must be transverse to the image of \(A\), where transverse means having zero intersection. There are exactly \(q^{b(N-h)}\) such Lagrangians: after choosing complementary coordinate Lagrangians, they are graphs of symmetric linear maps from one coordinate space to the other. Symmetry follows by evaluating the alternating form on two graph vectors. This proves Equation (64). To prove pair transitivity, first identify the common subspaces by symplectic bases. In their common-space quotient, identify the transverse pairs by choosing a basis of one member and its dual basis in the other. The resulting quotient map extends by Equation (60); containment reduction is a bijection, so the extension identifies the original pairs. Finally, factoring the leading powers of \(q\) out of the finite products proves all the displayed estimates, including their endpoint cases. ◻

Proof. On the chart, dropping the sign condition gives the pointwise inequality \[ K_\xi(X-Y)\le \frac{\pi_{D,r}}{p_\xi}J(\Gamma_X,\Gamma_Y). \tag{73}\] The graph variables have exactly the law of the uniform Lagrangian variables conditioned on all \(35\) variables belonging to \(\mathcal A\). That event has probability \(\beta_D^{35}\). Apply Equation (73), express the conditional expectation by inserting the event’s indicator and dividing by its probability, and then remove the indicator by nonnegativity. This gives Equation (72) without any boundedness assumption on \(F\) or on the normalized kernels. Finally \(\beta_D\to1\) and \(\pi_{D,r}/p_\xi\to2\). ◻

Proof. We divide the argument into the common-space count, the classification above the pair intersections, and the orbit lower bound.

Proof. Write \(n=d-v\). First choose a prospective span \(Z\) of dimension \(n\). There are \(\asymp q^{nv}\) choices, and the probability that all three independent \(r'\)-subspaces lie in it is \(\asymp q^{-3r'v}\). Conditional on these separate containments, the three subspaces are independent uniform \(r'\)-subspaces of \(Z\).

The number of choices of three pair spaces in direct sum inside \(Z\) is \(O(q^{nP-\sum_i p_i^2})\). Indeed, choose their \(P\)-dimensional sum and then an ordered direct-sum decomposition; the two exponents are \(P(n-P)\) and \(P^2-\sum_i p_i^2\). Each \(R_i\) must contain the sum of the other two pair spaces, of dimension \(P-p_i\). In \(Z\) this costs exponent \((n-r')(P-p_i)\), for a total of \(2(n-r')P\). We may drop the requirements that \(Z\) be exactly the span and that these containments give exactly the pair intersections. Thus the total cost exponent is \[\begin{align*} (3r'-d+v)v-nP+\sum_i p_i^2+2(n-r')P &= (3r'-d+v)v+(d-v-2r')P+\sum_i p_i^2\\ &=\delta P+\sum_i p_i^2+v(3r'-d-P+v), \end{align*}\] which proves the claim. Infeasible containments have probability zero. ◻

Proof. The displayed constraint is necessary by cancellation. Conversely, take any three target forms satisfying it, and prescribe the values of the three original forms on \(V\) to be \[H_1=0,\qquad H_2=-D_3|_V,\qquad H_3=D_2|_V.\] They have the required three cyclic differences on \(V\). On each \(Q_i\) choose any symmetric extension of the prescribed \(H_j\) for the positive term of its cyclic difference, and prescribe the negative term there by subtracting \(D_i\).

Each \(F_i\) has now received prescriptions on exactly two subspaces of \(A_i\). Their intersection is \(V\), and the prescriptions agree there. Two symmetric forms on subspaces \(B,C\) agreeing on \(B\cap C\) always extend simultaneously: write \(B+C=(B\cap C)\oplus B_0\oplus C_0\), retain the prescribed blocks, choose the mixed \(B_0\)–\(C_0\) block arbitrarily, and then extend to the ambient space. Apply this construction separately for each \(F_i\). It proves sufficiency even if the three \(A_i\) or \(Q_i\) have further linear relations.

Finally, the map taking a target triple to the sum of its restrictions on \(V\) is onto the form space on \(V\): extend any prescribed form from \(V\) to one \(Q_i\) and take the other two forms to be zero. Its kernel therefore has codimension \(b(\dim V)\). A linear map between finite vector spaces sends uniform measure to uniform measure on its image, proving the last assertion. ◻

Proof. Fix an actual ordered orbit \(\mathcal O\). Start with four independent ambient Lagrangians and the normalizer \(\pi_{D,r}^{-3}\) in Equation (91). Partition its contribution according to the unique common space \(C_0\), and then according to \(u,p_i,t_i,v\). Equations (94)–(98), Lemma 21, and Equation (105) give the numerator bound \[\nu(\mathcal O)\le O\left(\sum_{\text{actual parameter tuples}} q^{-(D+1)c+g_C-\delta P-\sum_i p_i^2-vw -\sum_i b(t_i)+b(v)}\right).\] At intermediate counting steps we have discarded exactness or orbit conditions only to enlarge the count. The parameter tuples indexing the original partition remain actual ones. Dividing by the lower bound \(\mu(\mathcal O)\gg q^{-(D+1)c-\sum_i b(s_i)}\) from Proposition 20 proves Equation (106). There are only boundedly many dimension tuples for fixed \(D\). The tuples from the current field belong to \(\mathcal U\), so its maximum gives a bound uniform in \(q\).

The bounds on \(u,r',\delta\) follow from their definitions and Equation (96). The reduced pair intersections are in isotropic direct sum, giving \(\sum_i s_i\le d\) as in Proposition 20. Equation (103) gives the bounds on \(t_i\), and \(V\subseteq Q_i\) gives those on \(v\). We already proved \(w\ge0\). Summing \(t_i\le p_i+\delta\) gives \(T\le P+3\delta\), and summing \(v\le p_i+\delta\) gives \(v\le P/3+\delta\). Thus \(P+3\delta\ge0\) and \[d+3\delta\ge 6\delta+2P-2v \ge\frac43(P+3\delta)\ge0,\] where the first inequality uses \(w\ge0\). There are further necessary conditions, such as \(p_i\le r'\) and \(P-p_i\le r'\). The later estimates need only Equation (107); maximizing over actual tuples preserves all the others automatically. ◻

Proof. Use the two-point version of the planted density identity. Under this law the two \(R_i=L_i\cap Y\) are independent uniform \(r\)-subspaces of the \(2r\)-space \(Y\). Their intersection dimension \(p\) has probability \(O(q^{-p^2})\): for fixed \(R_1\), choose a \(p\)-subspace of it, in \(O(q^{p(r-p)})\) ways, and require its containment in \(R_2\), at cost \(q^{-pr}\). Conditional on the two individual \(R_i\), the two lift forms are independent uniform. Their difference on \((R_1+R_2)^\circ\), of dimension \(p\), is a uniform symmetric form, since restriction from either original form space is surjective. Equation (103) now gives \(h=p+t\) with \(0\le t\le p\), and the nullity event costs at most \(O(q^{-b(t)})\). The ambient pair orbit has mass \(\asymp q^{-b(h)}\) by Lemma 18. Dividing gives the maximand in Equation (108).

For fixed \(h\), let \(f(p)=b(h)-p^2-b(h-p)\). On its integer domain \(\lceil h/2\rceil\le p\le\min(h,r)\), \[f(p+1)-f(p)=h-3p-1<0.\] Its maximum is therefore at \(p=\lceil h/2\rceil\), which is allowed since \(h\le2r\). If \(h=2a\) the value is \(b(2a)-a^2-b(a)=b(a)\); if \(h=2a+1\) it is \(b(2a+1)-(a+1)^2-b(a)=b(a)\). The maximizing parameters are realizable: take two \(r\)-subspaces with intersection \(p\), then a symmetric difference form of nullity \(t=h-p\). Pair transitivity makes this applicable to every pair of intersection dimension \(h\). Finally, an empty product is one, and for one incidence the mean-one normalization of \(J(L,Y)\) gives exactly one for every \(L\). ◻

Proof. Fix one of the six classes obtained by prescribing one bit position and its value. It consists of eleven faces, is connected in the face adjacency graph, and uses all thirteen points. Choose a spanning tree and a parent-first order. The first face uses three points. Every subsequent face shares a pair with its parent and can add at most one point. Since all thirteen points occur, the ten subsequent faces must each add exactly one fresh point.

Expose all the point Lagrangians occurring in the first face. Its prescribed profile has probability \(O_D(q^{-C_j})\) by Proposition 20. At a subsequent face \(j\), let \(e\) be its pair shared with its parent, and condition on the sigma-field generated by all previously exposed point Lagrangians. Earlier face-profile events are measurable in this sigma-field. The fresh point remains independent and uniform. If the already exposed pair has the prescribed intersection dimension \(h_e\), pair transitivity and Proposition 20 give the conditional probability \[\frac{\mathbb P(\text{the prescribed profile on }j)} {\mathbb P(\dim(L_i\cap L_k)=h_e)} =O_D(q^{-C_j+b(h_e)}).\] Here pair transitivity makes the ratio the same for every specific pair of that dimension. Thus the bound holds after conditioning on the full earlier configuration, however its older points are related. If the pair has the wrong dimension, the conditional probability is zero. Iterated conditional expectation proves a class probability bound with cost \(\sum_{j\text{ in class}}C_j\) minus \(b(h_e)\) for its tree pairs. Subtracting \(b(h_e)\) for every internal pair of the class only weakens this bound.

Let \(A\) be the complete profile event and let \(A_1,\ldots,A_6\) be the six class profile events. Because \(A\subseteq A_t\) for each \(t\), \[\mathbb P(A)^6\le\prod_{t=1}^6\mathbb P(A_t).\] This inequality does not require independence among the classes, all of which use all thirteen points. Each face appears in three class bounds. A pair \(e\) is internal in exactly \(3(1-a_e)\) classes, one for each position where its two faces have the same bit. Taking sixth roots gives Equation (109). ◻

Proof. Put \(m=\max_\ell s_\ell\) and choose \(\ell_0\) with \(s_{\ell_0}=m\). Using \(h=c+m\) and \(\sum s_\ell\le D-c\), expand \[\begin{align*} 3C-\sum_\ell b(c+s_\ell)-2b(c+m) &=3(D+1)c-5b(c) -c\left(\sum_\ell s_\ell+2m\right) +2\sum_{\ell\ne\ell_0}b(s_\ell)\\ &\ge b(c)+2\sum_{\ell\ne\ell_0}b(s_\ell). \end{align*}\] The last step uses \(\sum s_\ell+2m\le3(D-c)\); substituting this upper bound leaves exactly \(b(c)\) in the terms involving \(c\). Consequently \[ \beta_j\ge\frac{b(h)}3+\frac{b(c)}6 +\frac13\sum_{\ell\ne\ell_0}b(s_\ell). \tag{114}\] The gain is zero with zero or one active incidence. With two, it is \(b(\lfloor h'/2\rfloor)\) for their pair dimension \(h'\le h\). For an integer \(h\ge0\), \[b(\lfloor h/2\rfloor)\le\frac{b(h)}3, \qquad b(\lfloor h/2\rfloor)\le\frac{h(h+2)}8.\] The first follows separately for \(h=2k\) and \(h=2k+1\); its differences are \(k(k-1)/6\) and \((k+1)(k+2)/6\), respectively, so equality holds exactly for \(h=0,2\). Combining these inequalities with Equation (114) proves nonnegativity and \(S_j\ge b(h)/3-h(h+2)/8=h^2/24-h/12\). Equality forces \(c=0\), \(s_\ell=0\) for \(\ell\ne\ell_0\), and \(h=0\) or \(2\). When \(h=2\), the gain must equal \(1\), which requires the stated two active endpoints. With at most one active incidence the positive baseline for \(h>0\) excludes equality. ◻

Proof. Set \[a=\left(P-\frac{d-3\delta}{2}\right)_+, \qquad v_0=\min\left(\frac P3+\delta, \frac{P-T+3\delta}{2}\right).\] Both arguments of the minimum are nonnegative, by Equation (115). Completing the square gives \[vw-\frac{v^2}{2} =\left(\frac{d-3\delta}{2}-P\right)v+\frac{v^2}{2} \ge-\frac{a^2}{2}.\] The condition \(w\ge0\) gives \(a\le v\); furthermore \(v\le\min_\ell(p_\ell+\delta)\le P/3+\delta\). From \(d\ge P+T\) we also obtain \(a\le(P-T+3\delta)/2\). Therefore \(a\le v_0\), and \[ \begin{split} Q\ge{}&\frac{c(d+\delta)-\delta d+2\delta^2}{4} +\frac{(2P-T)^2}{18}+\delta P -\frac{c(P+T)}6-\frac{v_0^2}{2}. \end{split} \tag{120}\] Here we also used \(\sum_\ell(2p_\ell-t_\ell)^2\ge(2P-T)^2/3\). The coefficient of \(c\) on the right is \[\frac{d+\delta}{4}-\frac{P+T}{6} \ge\frac{d+3\delta}{12}\ge0.\] Because \(c\ge|\delta|\), replacing \(c\) by \(|\delta|\) can only decrease the right side. We do so and treat the two signs of \(\delta\).

First suppose \(\delta\ge0\). The remainder in Equation (120) becomes \[F(T)=\frac{3\delta^2}{4}+\frac{(2P-T)^2}{18} +\frac{(5P-T)\delta}{6}-\frac12 \min\left(\frac P3+\delta, \frac{P-T+3\delta}{2}\right)^2, \qquad 0\le T\le P+3\delta.\] On \(0\le T\le P/3+\delta\), the first argument attains the minimum, and \(F'(T)=(T-2P)/9-\delta/6\le0\); at the right endpoint this derivative is \(-5P/27-\delta/18\). On the remaining interval the coefficient of \(T^2\) is \(1/18-1/8=-5/72\), so \(F\) is concave. Its minimum is therefore at one of the following two endpoints: \[\begin{align*} F(P/3+\delta)&=\frac{32P^2+84P\delta+45\delta^2}{324},\\ F(P+3\delta)&=\frac{2P^2+12P\delta+27\delta^2}{36}. \end{align*}\] Both are at least \((P^2+\delta^2)/18\). Moreover \[P^2+T^2+\delta^2 \le P^2+(P+3\delta)^2+\delta^2 \le3P^2+19\delta^2\le19(P^2+\delta^2).\] This proves the desired coefficient \(1/(18\cdot19)\) in this case, including \(P=\delta=0\).

Next suppose \(\delta=-y<0\). Put \(x=P-3y\ge0\); then \(0\le T\le x\). After setting \(c=y\), the coefficient of \(d\) is \(y/2\). Using \(d\ge P+T=x+3y+T\) in Equation (120) leaves \[G(T)=\frac{y^2}{4}+\frac{(2x-T)y}{3} +\frac{(2x-T)^2}{18} -\frac12\min\left(\frac x3,\frac{x-T}{2}\right)^2.\] Its derivative up to \(T=x/3\) is \((T-2x)/9-y/3\le0\); after that breakpoint its quadratic coefficient is again \(-5/72\). Thus its minimum is at least the smaller of its breakpoint and last-endpoint values, \[\begin{align*} G(x/3)&=\frac{32x^2+180xy+81y^2}{324},\\ G(x)&=\frac{2x^2+12xy+9y^2}{36}. \end{align*}\] Each is at least \((x^2+y^2)/18\), including the degenerate interval \(x=0\). Finally, \[P^2+T^2+y^2\le(x+3y)^2+x^2+y^2 \le3x^2+19y^2\le19(x^2+y^2).\] The same constant \(1/342\) follows, completing the proof. ◻

Proof. Assume \(\sum_j S_j+\mathcal E\le0\). We first bound all pair dimensions independently of \(D\), the activity set, and the choices of maximizing tuples.

Proof. For any ordering of the spaces, let \(\Delta_e\) be the increment in the dimension of the global span when \(H_e\) is added. At each endpoint \(i\) of \(e\), let \(\Delta_{e,i}\) be its increment in the local span. The earlier local span is contained in the earlier global span, so \(\Delta_{e,i}\ge\Delta_e\). Summing gives \(\sum_i k_i\ge2\sum_e\Delta_e=2k\).

If the family is not direct, choose an inclusion-minimal subfamily \(\mathcal T\) which is not direct. Order its members first and let \(e\) be its last member. Put \[\delta_*=\dim\left(H_e\cap \sum_{f\in\mathcal T\setminus\{e\}}H_f\right)>0.\] At least one endpoint \(i\) of \(e\) fails to belong to some other member of \(\mathcal T\): otherwise every other pair in \(\mathcal T\) would contain both endpoints of \(e\) and would equal \(e\). The members of \(\mathcal T\) visible at this \(i\), including \(e\), are therefore a proper subfamily and form a direct sum by minimality. Consequently \[\Delta_{e,i}=\dim H_e=\Delta_e+\delta_*.\] All other increment comparisons remain nonnegative, and hence \(\sum_i k_i\ge2k+\delta_*\ge2k+1\). The argument uses the actual dimension \(\delta_*\) and in particular includes a partial intersection of \(H_e\) with the earlier span. ◻

Proof. Write \(H_e=L_i\cap L_k\) for \(e=\{i,k\}\), and use \(k,k_i\) as in Lemma 29. We count possible \(H_e\) and impose their containment in the independent \(L_i\). Discarding exactness of the intersections can only enlarge the count. A nonisotropic local span cannot be contained in a Lagrangian and contributes zero. For every isotropic local span of dimension \(k_i\), its containment probability is \(O_D(q^{-Dk_i+\binom{k_i}{2}})\) by Lemma 18.

For the nondirect case, first choose an arbitrary \(k\)-space in the \(2D\)-space, in \(O_D(q^{k(2D-k)})\) ways, and then choose the \(m\) two-spaces inside it, in \(O_D(q^{2m(k-2)})\) ways. Requiring their span to be the chosen space would only reduce this upper bound. On the stratum with fixed \(k\) and \(k_i\), multiplying the containment probabilities gives exponent \[-D\left(\sum_i k_i-2k\right) -k^2+2mk-4m+\sum_i\binom{k_i}{2}.\] The point degrees in the pair graph, given by Proposition 4, are \((4,6,4,6,5,4,5,5,5,6,5,6,5)\). Therefore \[\sum_i\binom{k_i}{2} \le\sum_i\binom{2\deg(i)}2=618.\] Since \(m\le33\) and \(-k^2+2mk\le m^2\), the part of the exponent independent of \(D\) is at most \[m^2-4m+618\le1575.\] Lemma 29 makes the coefficient of \(-D\) at least one. There are only boundedly many \(k,k_i\), independently of \(D\), because \(k\le2m\le66\) and \(k_i\le2\deg(i)\). Thus the probability of this part of the profile is \(O_D(q^{-D+1575})\). Its total gain is at most \(\tfrac23\sum_e b(h_e)=2m\le66\), proving the first bound. The ambient span in this argument was arbitrary; only the local spans were required to be isotropic.

For the direct case, the following count applies to arbitrary fixed dimensions \(h_e\). Order the pairs and choose an ordered basis of each \(H_e\). Each new basis vector must be orthogonal to the previous vectors of its own space and to all vectors of earlier spaces sharing an endpoint. In a direct family these previous vectors are linearly independent. Nondegeneracy of the symplectic form identifies ambient vectors with linear functionals, so their orthogonality equations are independent, even if the previous vectors are mutually orthogonal. If there are \(a\) such equations, there are exactly \(q^{2D-a}\) vectors satisfying them. Excluding choices which would destroy directness only decreases this count.

Let \(e\sim f\) mean distinct pairs sharing an endpoint, and count each unordered pair \(\{e,f\}\) once. Dividing the sequential basis count by the number of ordered bases in each space gives at most \[O_D\left(q^{\,2D\sum_eh_e-\sum_eh_e^2 -\sum_e\binom{h_e}{2}-\sum_{e\sim f}h_eh_f}\right)\] possible direct families satisfying all local orthogonality conditions. Here \(k_i=\sum_{e\ni i}h_e\), so \[\sum_i k_i=2\sum_eh_e, \qquad \sum_i\binom{k_i}{2} =2\sum_e\binom{h_e}{2}+\sum_{e\sim f}h_eh_f.\] Distinct pairs in a simple graph share at most one endpoint, which explains the coefficient of the last sum. Multiplying by the containment probabilities cancels both the \(D\) terms and the cross terms. The remaining exponent is exactly \[-\sum_eh_e^2+\sum_e\binom{h_e}{2}=-\sum_e b(h_e).\] The gain bound in Equation (123) now leaves \(-\tfrac13\sum_e b(h_e)\), as asserted. ◻

Proof. Choose an even \(D\) above the threshold in Lemma 28 and also with \(D>1641\). These thresholds refer only to dimensions and the fixed complex. For any profile with some \(h_e>0\), either \(\mathcal C>\sum_jg_j\), or it satisfies Equation (123). In the first case, Equation (110) gives a strictly negative \(q\)-exponent. In the second, the two bounds in Lemma 30 are strictly negative because \(D>1641\) and \(\sum_e b(h_e)>0\).

At this fixed \(D\), every selected gain is an integer and \(6\mathcal C\) is an integer, because \(3a_e\) is an integer. A strict first-cost saving is therefore at least \(1/6\). A singular direct residual profile saves at least \(1\), and the nondirect case also saves at least \(1\) because even \(D>1641\) implies \(D\ge1642\). There are finitely many profiles and activity sets, and the gain exponents were selected from the fixed finite unions in Proposition 23. Summing their vanishing contributions proves Equation (126). The local maximizing tuples need not be realizable simultaneously: the residue argument uses only their necessary dimension inequalities and the shared profile dimensions. In particular, no simultaneous limit in \(D\) and \(q\) has been used. The bound \(D>1641\) is needed here only for the nondirect residual case and is not asserted to be a threshold for the whole proof. ◻

Proof of Lemma 5. Choose an even \(D\) large enough for both Proposition 17 and Proposition 31. For the matrix variables let \[\mathcal T=\{\det(X_i-X_k)\ne0\text{ for every }\{i,k\}\in E\}.\] On \(\mathcal T\), Proposition 17 gives the right-hand side of Equation (5), up to \(o(1)\), for every fixed activity set and every choice of incidence signs. The integrand is nonnegative. Lemma 19 bounds its contribution on \(\mathcal T^c\) by a constant depending on \(D\) times the unsigned Lagrangian contribution in Equation (126): graph intersections project isomorphically onto the kernels of matrix differences, and dropping the prescribed signs only enlarges the integrand after the comparable normalizations. This contribution is \(o(1)\) by Proposition 31. Adding the transverse and singular parts proves Equation (5). There are only finitely many activity sets and sign assignments, so the convergence is uniform over the choices required in Section 3. This proves Lemma 5. The construction in Section 3 therefore yields the strict symmetric-kernel inequality, and Lemma 10 gives a finite simple host \(G\) with \(t(H,G)<p(G)^{66}\), proving Theorem 1. ◻

Proof of Corollary 2. Take \(G\) from Theorem 1 and put \(N=|V(G)|\), \(p_0=p(G)\), \(t_0=t(H,G)\), and \(\rho=t_0/p_0^{66}\). Simplicity and an edge give \(0<p_0<1\), and mapping the two parts of \(H\) to the endpoints of an edge gives \(t_0>0\). Thus \(0<\rho<1\).

Let \(G^{\times r}\) be the categorical power, whose adjacency is coordinatewise. It is finite, simple and undirected, and homomorphisms into it factor coordinatewise. Hence \[p(G^{\times r})=p_0^r,\qquad t(H,G^{\times r})=t_0^r,\qquad \frac{t(H,G^{\times r})}{p(G^{\times r})^{66}}=\rho^r\longrightarrow0,\] which proves (i). This is the standard tensor-power principle (Conlon et al. 2010, sec. 2).

For (ii), use instead the balanced independent blow-ups \(F_s=G[s]\): replace each vertex by an independent set of size \(s\), and each edge by the complete bipartite graph between its two sets. Then \(n_s=Ns\), \(p(F_s)=p_0\), and \(t(H,F_s)=t_0\), since each homomorphism into \(G\) has exactly \(s^{35}\) lifts. For \(n_s\ge35\), set \(q_s=\operatorname{Inj}(H,F_s)/(n_s)_{35}\). This is the homomorphism probability for a uniform vertex map conditioned on injectivity, so the collision union bound gives \[|q_s-t_0|\le1-\frac{(n_s)_{35}}{n_s^{35}} \le\frac{\binom{35}{2}}{n_s}.\] Meanwhile \(d_s=p_0n_s/(n_s-1)\to p_0\). Therefore \(q_s/d_s^{66}\to\rho<1\), proving (ii) with \(p_*=p_0\) and \(\delta_*=(1-\rho)/2\). ◻

Proof of Corollary 3. Let \(G\) be the host from Theorem 1, put \(N=|V(G)|\), and set \(p=p(G)\). Since \(G\) has an edge and is simple, \[0<p\le\frac{N-1}{N}<1.\] Let \(W_0\) be its adjacency kernel on the uniform probability space \(V(G)\). Independent vertex samples count all homomorphisms, so \[t(K_2,W_0)=p,\qquad t(H,W_0)=t(H,G)<p^{66}.\] Moreover, \(W_0\) is \(\{0,1\}\)-valued, and hence \(\|W_0-p\|_2^2=p(1-p)>0\).

On the balanced sign space \(\{-1,1\}\), choose \(0<\delta\le\min(p,1-p)\) and define the rank-one perturbation \[W_1(a,b)=p+\delta ab.\] It takes values in \([0,1]\) and has mean \(p\). For \(A\subseteq E(H)\), let \(d_A(v)\) be the degree of \(v\) in the edge set \(A\). Expanding the product defining \(t(H,W_1)\) and averaging the independent signs gives \[t(H,W_1)= \sum_{\substack{A\subseteq E(H)\\ d_A(v)\text{ even for every }v}} p^{66-|A|}\delta^{|A|}>p^{66}.\] Indeed, the empty edge set contributes \(p^{66}\), every surviving term is nonnegative, and the edge set of a cycle contributes a strictly positive term. Such a cycle exists because a forest on \(35\) vertices has at most \(34\) edges, whereas \(H\) has \(66\).

Lift the two kernels to the independent coordinates of \(\Omega=V(G)\times\{-1,1\}\), with product probability, and set \[W_s\bigl((x,a),(y,b)\bigr) =(1-s)W_0(x,y)+sW_1(a,b),\qquad 0\le s\le1.\] Each \(W_s\) is symmetric, takes values in \([0,1]\), and has mean \(p\). The endpoint lifts preserve their \(H\)-densities. The two centered summands depend on independent coordinate pairs and have mean zero, so they are orthogonal in \(L^2(\Omega^2)\). Thus \[\|W_s-p\|_2^2=(1-s)^2p(1-p)+s^2\delta^2>0 \qquad(0\le s\le1).\] The function \(s\mapsto t(H,W_s)\) is a polynomial. Its endpoint values lie strictly below and above \(p^{66}\), respectively, so some \(s_*\in(0,1)\) satisfies \(t(H,W_{s_*})=p^{66}\). The norm identity shows that \(W=W_{s_*}\) is nonconstant.

For completeness, this gives failure of asymptotic quasirandomness, not just a graphon equality. On the finite atom space of \(W\), put \[r(x)=\mathbb E_z W(x,z),\qquad c(x,y)=\mathbb E_z W(x,z)W(y,z).\] Two applications of Cauchy–Schwarz give \[\begin{align*} t(C_4,W)&=\mathbb E_{x,y}c(x,y)^2\\ &\ge\bigl(\mathbb E_{x,y}c(x,y)\bigr)^2 =\bigl(\mathbb E_z r(z)^2\bigr)^2\ge p^4. \end{align*}\] Equality would force \(r(x)=p\) for every \(x\) and \(c(x,y)=p^2\) for every pair \((x,y)\): every atom pair has positive mass, including the diagonal pairs. Then, for every \(x\), \[\mathbb E_z\bigl(W(x,z)-p\bigr)^2=c(x,x)-2pr(x)+p^2=0,\] contrary to nonconstancy. Hence \(t(C_4,W)>p^4\).

Use the independent sampling construction for this finite kernel, as in Lemma 10 and (Lovász and Szegedy 2006, sec. 2.6, Lemma 2.4). For each fixed finite graph \(F\), the collision count gives \(\mathbb Et(F,G_n)=t(F,W)+O_F(n^{-1})\). The covariance of the indicators for two vertex maps vanishes when their images are disjoint, and the proportion of pairs with overlapping images is \(O_F(n^{-1})\). Therefore \(\operatorname{Var}(t(F,G_n))=O_F(n^{-1})\). With tolerance \(n^{-1/4}\), Chebyshev’s inequality and a union bound for \(F=K_2,H,C_4\) give total failure probability \(O(n^{-1/2})<1\) for all sufficiently large \(n\). Choose one successful realization for each such \(n\). The expectation estimate then gives a deterministic sequence of finite simple graphs with \[p(G_n)\to p,\qquad t(H,G_n)\to p^{66},\qquad t(C_4,G_n)\to t(C_4,W)>p^4.\] This sequence is not quasirandom of density \(p\), proving the claim. ◻