Conway’s 99-Graph: An Exact Modular Completion Criterion

For strongly regular graph lifts with free order-three symmetry.

Unreviewed research note — 5 September 2026

For a legal graph lift with a fixed-point-free cyclic action of order three, an exact orbit quotient and a small Fourier congruence can force the full strongly regular graph equation. For the family $\operatorname{SRG}(v,k,1,2)$, the required modulus is $(1-\omega)^3$, independently of $v$ and $k$. At the Conway parameters, this leaves one linear and one quadratic matrix equation over $\mathbb F_3$.

This AI-assisted note has received automated mathematical checks but no independent human peer review or formal proof-assistant verification. No worldwide novelty claim is made. It neither constructs a Conway 99-graph nor excludes its existence or order-three automorphisms.

1. A nonnegative coefficient lemma

Let $\mathcal E=\mathbb Z[\omega]$, where $\omega^2+\omega+1=0$, and put $\pi=1-\omega$. Since $\pi^2=-3\omega$, the ideals $(\pi^2)$ and $(3)$ coincide.

Lemma. If $c_0,c_1,c_2$ are nonnegative integers with sum six, then

$$ \pi^3\mid c_0+c_1\omega+c_2\omega^2 \quad\Longleftrightarrow\quad c_0=c_1=c_2=2. $$

Proof. Set $a=c_0-c_2$ and $b=c_1-c_2$. Divisibility by $\pi^3$ is equivalent to

$$ 3\mid a,\qquad 3\mid b,\qquad 9\mid a+b. $$

Indeed, first divide by three, then use $\omega\equiv1\pmod\pi$. Since $a+b=6-3c_2$, the last condition gives $c_2\equiv2\pmod3$. All three coefficients are therefore congruent to two modulo three. Nonnegativity makes each at least two; their sum forces equality. The converse follows from $1+\omega+\omega^2=0$. $\square$

The modulus cannot be reduced to $\pi^2$ under these scalar hypotheses: $(c_0,c_1,c_2)=(1,1,4)$ gives $3\omega^2$, which has valuation exactly two. This does not establish minimal precision for every global graph problem.

2. Exact Fourier completion

Let $A$ be the adjacency matrix of a finite simple undirected $k$-regular graph on $3m$ vertices, with a specified fixed-point-free automorphism of order three. Order each vertex orbit cyclically. Then $A$ consists of $3\times3$ circulant blocks.

Evaluating each circulant polynomial at $1$ gives the ordinary orbit quotient $R$; evaluation at $\omega$ gives a Hermitian matrix $W$ over $\mathcal E$. Evaluation preserves matrix products. Identity and all-one matrices below have the dimensions indicated by their equations.

Theorem. Assume the exact quotient identity

$$ R^2+R-(k-2)I=6J. $$

Then

$$ A^2+A-(k-2)I=2J \quad\Longleftrightarrow\quad W^2+W-(k-2)I\equiv0\pmod{\pi^3}. $$

Proof. Put $K=A^2+A-(k-2)I$. In a block between distinct vertex orbits, its three coefficients count common neighbors plus adjacency. They are nonnegative, and the quotient equation makes their sum six. The Fourier congruence and the lemma force each coefficient to be two.

In a diagonal orbit block, the identity coefficient is $k-(k-2)=2$. Symmetry makes the other two coefficients equal, and their total with the identity coefficient is six. They too are two. Thus every entry of $K$ equals two. Conversely, evaluating a constant block at $\omega$ gives zero. $\square$

The theorem assumes a legal simple graph lift, not an arbitrary Hermitian matrix. The Conway alphabet below ensures this hypothesis directly. For nontrivial strongly regular parameters, the full adjacency equation is exactly the required common-neighbor condition.

3. Two equations over $\mathbb F_3$ for Conway quotients

Let $R$ be symmetric of size $33\times33$, with

$$ R\mathbf1=14\mathbf1,\qquad R^2+R-12I=6J, $$

$$ R_{ii}\in{0,2},\qquad R_{ij}\in{0,1,2}\quad(i\ne j). $$

An internal triangular orbit has Fourier diagonal $-1$; an independent orbit has diagonal zero. Between distinct orbits, quotient entry one specifies one cyclic matching; entry two specifies the complement of one matching. Consequently the legal entries are

Position	$W_{ij}$	$Z_{ij}=z$	$U_{ij}$
$i\ne j$, $R_{ij}=0$	$0$	$0$	$0$
$i\ne j$, $R_{ij}=1$	$\omega^g$	$-g$	$z(z+1)/2$
$i\ne j$, $R_{ij}=2$	$-\omega^g$	$g$	$1-z(z-1)/2$
$i=j$, $R_{ii}=2$	$-1$	$0$	$0$
$i=j$, $R_{ii}=0$	$0$	$0$	$2$

Here $g,z\in\mathbb F_3$, and division by two uses its inverse in $\mathbb F_3$. Require $Z^T=-Z$. Every supported skew-symmetric $Z$ determines a unique legal Hermitian $W$ and hence a symmetric binary circulant lift. Its quotient row sums make it 14-regular.

Set

$$ T=R+I,\qquad B=T\pmod3,\qquad H=W+4I, \qquad L_B(Y)=BY+YB-Y. $$

Then $B^2=B$. With least-residue integer lifts $\widehat Z,\widehat U$, the table gives

$$ H\equiv T+\pi\widehat Z+\pi^2\widehat U\pmod{\pi^3}. $$

This scalar expansion follows by substituting $\omega=1-\pi$ for each of the three phases.

Corollary. A legal phase assignment completes the quotient to a Conway graph exactly when

$$ \boxed{L_B(Z)=0,\qquad L_B(U)=B+I+2J-Z^2.} $$

Here $Z^2$ is a matrix product; the table defines $U$ entrywise.

Proof. Write $Q=-T+2I+J$. The quotient equation gives $T^2-7T=6Q$. Using $6\equiv-2\pi^2\pmod{\pi^3}$, expansion yields

$$ H^2-7H\equiv\pi P+\pi^2 C\pmod{\pi^3}, $$

where

$$ P=T\widehat Z+\widehat ZT-\widehat Z, $$

$$ C=T\widehat U+\widehat UT-\widehat U+\widehat Z^2-2Q. $$

The congruence first forces $P\equiv0\pmod3$. Then $\pi P$ is divisible by $\pi^3$, leaving $C\equiv0\pmod3$. Conversely these two conditions suffice. Reducing the displayed formulas gives the boxed equations, since $2Q\equiv B+I+2J$. Finally, $H^2-7H=W^2+W-12I$, so the theorem upgrades the congruence to equality. $\square$

No third digit or integer carry is required for completion. This does not justify dropping carries from a higher-order expansion; it avoids needing that expansion.

4. Generalization and computational scope

For nonnegative integers $\mu,s$ with $q=3^s>\mu$, the coefficient lemma generalizes to

$$ c_i\ge0,\quad\sum_i c_i=3\mu: \qquad \pi^{2s+1}\mid\sum_i c_i\omega^i \Longleftrightarrow c_i=\mu\text{ for all }i. $$

The same argument gives $q\mid a,b$ and $3q\mid a+b$, hence $c_i\equiv\mu\pmod q$. Nonnegativity and $\mu<q$ force equality. Therefore the graph theorem extends to parameters $0\le\lambda\le\mu$, replacing the residual by $A^2+(\mu-\lambda)A-(k-\mu)I$ and the target by $\mu J$. The quotient target is $3\mu J$. The inequality $\lambda\le\mu$ ensures coefficient nonnegativity.

For Conway quotients, diagonal identities force twelve nonzero off-diagonal positions per row, hence 198 undirected phase variables. The first boxed equation is linear and the second quadratic over $\mathbb F_3$. This simplifies the exact formulation; it establishes no runtime gain or rejected quotient.

5. Reproduction and context

Run the standard-library checker:

python -B verify.py

See verify.py and results.json. Checks cover coefficient triples, all phases of two Paley-9 quotients, two free cyclic actions on the 243-vertex Berlekamp–van Lint–Seidel graph, and six deliberately changed phases. Full adjacency calculations are compared with independently implemented Eisenstein arithmetic. These controls do not supply a completed Conway quotient or test its boxed system on a real instance.

Orbit quotients and automorphism restrictions are established research tools; relevant context includes Behbahani–Lam, Strongly regular graphs with non-trivial automorphisms (2011) and Cesarz–Woldar, On the automorphism group of a putative Conway 99-graph (2025). The exact criterion here requires specialist priority checking under cyclic lifts, group rings, voltage graphs, and modular reconstruction terminology.

Published by Lundle22. Claude/Codex assisted the underlying investigation and this note's derivation, checking, and presentation. The release is offered for mathematical scrutiny, with no claim that automated review substitutes for human review.