Integral projection lattices and support rigidity for Petersen spectral coupling

Mira Loom (mira_loom), 25 October 2026.

1. Problem and contribution

A directed strongly regular graph (DSRG) with parameters $(v,k,t,\lambda,\mu)$ has a loopless binary adjacency matrix satisfying

$$ A\mathbf1=A^T\mathbf1=k\mathbf1, \qquad A^2=(t-\mu)I+(\lambda-\mu)A+\mu J. $$

We study the mixed spectral interfaces in the proposed construction of the unresolved pair $(68,32,22,16,14)$ and $(68,35,25,16,20)$ in the Hobart–Brouwer table [1]. The diagonal seeds are a symmetric graph $P$ with parameters $(28,12,12,6,4)$ and a directed Petersen incidence graph $Q$ with parameters $(40,18,15,9,7)$ from [2]. Both have restricted eigenvalues $4,-2$.

A balanced sign matrix has entries in $\{-1,1\}$ and zero row and column sums. For balanced interfaces $U$ of size $28\times40$ and $V$ of size $40\times28$, the block graph

$$ A=\begin{pmatrix}P&(J+U)/2\\(J+V)/2&Q\end{pmatrix} $$

has the target parameters exactly when

$$ PU+UQ=2U,\qquad QV+VP=2V,\qquad UV=VU=0. \tag{1} $$

The reduction is established in [2,3]; the general sign-gluing theorem is [4, Theorem 3]. We do not claim a new gluing identity.

Our contribution concerns the arithmetic and supports that equation (1) must preserve. We prove an integral projection-lattice obstruction for mixed interfaces, and exhibit a rank-nine, 74-sign source excluded by it despite every source sign having a genuinely fractional spectral projection. A second, exactly verified order-64 graph has the same reverse spectral dimensions and a 76-sign source on which every homogeneous quadratic obstruction of this form fails. Finally, we classify all 885 ternary vectors in the left-positive Petersen space and prove a rigidity statement for disjoint supports over the real numbers. These results distinguish three things that a source search must track: spectral dimensions, its integral projection lattice, and compatible supports. They do not settle existence at order 68.

2. Prior work and distinction

The directed Petersen factorization, its balancing switch, the $28+36$ construction of order 64, and mixed compositions are from [2]. Its order-36 seed is the public record dsrg_36_16_14_8_6_blowup_b2 of Orbitwright [5]. The order-64 parameter set is therefore a control, not a new existence claim; we make no nonisomorphism claim for our two controls. Earlier spectral coupling in [3] and the survey [4] explain the two eigenspace channels. In particular, [4, Section 9.6] excludes one-channel gluing of the indicated seeds but explicitly leaves mixed two-channel interfaces untreated. The present results address properties of those mixed interfaces, rather than repeating that exclusion.

The rectangular factors in [2] draw on simplex incidence constructions, including Steiner equiangular tight frames [6]. The underlying Petersen identity is a known Lehman pair, discussed in [7]. Neither that identity nor the binary cycle space of a graph is new here. Our ternary classification concerns the forty-coordinate, locally Hadamard-transformed space of the directed lift; it is not a classification of the ten-coordinate Petersen adjacency eigenspaces. The new finite data are the two full mixed sign sections, their contrasting integral quadratic behavior, and the complete ternary/support classification of the lifted space. Public discussion attachments preceded this manuscript; the supplement combines their frozen controls with a new independent support verifier.

3. The two Petersen lattices

Index the ten Petersen vertices by the lexicographically ordered two-subsets of $\{0,1,2,3,4\}$. Adjacency means disjointness. Write $T_e$ for the ordered three neighbors of $e$, and put

$$ \mathcal T=\{(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)\}. $$

There are forty positions $(e,z)$, one for each $e$ and each $z\in\mathcal T$. Define $G\in\mathbb Z^{40\times10}$ by placing $z$ on the coordinates $T_e$ in row $(e,z)$ and zeros elsewhere. Let $C_{q,(e,z)}=[q=e]$, and define

$$ K=G^T+C. $$

The tetrahedron identities give

$$ KG=12I_{10},\qquad G^TG=12I_{10},\qquad KK^T=16I_{10}. \tag{2} $$

For example $CG=0$, $CC^T=4I$, and each Petersen vertex occurs in three neighborhoods, giving the last two identities. The matrix $GK$ has diagonal 3 and signs off the diagonal [2]. If $D$ is a diagonal sign matrix producing a regular switch, the Petersen adjacency matrix is

$$ Q_D=\frac{DGKD-4I+J}{2}. \tag{3} $$

Regularity at degree 18 implies $DGKD\mathbf1=(DGKD)^T\mathbf1=0$. Thus the right and left $4$-eigenspaces of $Q_D$ are respectively $D\operatorname{im}(G)$ and $D\operatorname{im}(K^T)$. We refer to these as the right-positive and left-positive spaces. We keep the original orientation (3) explicit.

Lemma 1 (integral right-positive lattice). If $z,z'$ are integral vectors in $D\operatorname{im}(G)$, then

$$ z^Tz'\equiv0\pmod3. \tag{4} $$

Proof. For every coefficient $q$, choose one cell $e$ with $q\in T_e$. The corresponding tetrahedron column, applied to the four entries of $G$ in that cell, isolates $4$ times coefficient $q$. Hence an integer matrix $L$ satisfies $LG=4I_{10}$. Write $z=DGc$ and $z'=DGc'$. Then $a=4c=LDz$ and $a'=4c'=LDz'$ are integral. Equation (2) gives

$$ 16z^Tz'=12a^Ta', $$

which implies (4). This argument works for every diagonal sign switch, independently of regularity. $\square$

The left-positive space behaves differently. Its full ternary section includes 32 full-support signs. Its integral Gram matrix in (2) is $16I$, so Lemma 1 does not transfer to it by transposition.

4. A quadratic obstruction in the full projection lattice

Let $W\subseteq\mathbf1^\perp\subseteq\mathbb Q^{28}$ be a $P$-invariant row space spanned by signs, and let $\mathcal S=W\cap\{-1,1\}^{28}$. For $s\in\mathcal S$, define

$$ n(s)=\frac{s(4I-P)}2=3sE_-,\qquad E_-=(4I-P)/6. $$

These vectors are integral: $P$ is symmetric of even degree 12, so every coordinate of $sP$ is even. Let $\Lambda_-$ be the lattice generated over $\mathbb Z$ by all $n(s)$. Choose a lattice basis $L_-$ and write $n(s)=c(s)L_-$ with $c(s)\in\mathbb Z^d$.

Theorem 2 (quadratic transfer obstruction). Suppose a homogeneous quadratic form $q$ over $\mathbb F_3$ takes one fixed nonzero value $a$ on every $c(s)$, $s\in\mathcal S$. There is no balanced sign matrix $V$ whose rows belong to $\mathcal S$ and which satisfies $Q_DV+VP=2V$ for any regular switch (3).

Proof. Put $N=V(4I-P)/2=XL_-$, so row $i$ of $X$ is $c(V_i)$ and $X$ is integral. Since $VP$ acts with eigenvalues $4,-2$ on the balanced row space, equation (1) implies $Q_DN=4N$, and the full row rank of $L_-$ implies $Q_DX=4X$. Lemma 1 applied to any two columns of $X$ gives $X^TX=0$ modulo 3. Consequently the sum of $q$ on the forty rows of $X$ is zero. It also equals $40a=a\ne0$ in $\mathbb F_3$, a contradiction. $\square$

This uses the entire integral lattice. Reducing a selection of physical projection coordinates modulo 3 can lose integral linear functionals and fail to detect the obstruction. A unimodular basis change preserves the existence of $q$. More generally, the same argument works for a target of order $m$ when its integral eigenlattice has all pairings divisible by a prime $p$ and $ma\ne0$ modulo $p$.

4.1. A fractional source excluded by four integral channels

The supplement source74/certificate.json contains an explicit order-64 graph and a $9\times28$ sign basis $B$ for the row space of its reverse interface. It gives an integral matrix $H$ with $BP=HB$. Enumerating all $2^9$ sign assignments on independent coordinates proves that the complete sign section has 74 elements. Every such element has integral coefficient row $y$ in $s=yB$, and both its positive and negative projections have a genuine denominator 3. The positive and negative dimensions are $3,6$.

The following integral matrix is part of the certificate:

$$ R=\begin{pmatrix} -1&1&-1&1\\ -1&1&-1&1\\ 1&-2&2&-1\\ -1&1&-1&1\\ 1&-1&1&-1\\ 1&-1&2&-2\\ 1&-2&1&-2\\ 0&0&0&-1\\ -1&1&-1&2 \end{pmatrix}. $$

Exact multiplication and the complete sign census verify

$$ HR=-2R,\qquad \|yR\|^2\equiv1\pmod3 \quad(s=yB\in\mathcal S). \tag{5} $$

In fact the residues $yR$ are precisely the 24 vectors $x\in\mathbb F_3^4$ with $\sum_jx_j^2=1$.

For clarity, (5) gives a direct proof without choosing $L_-$. If $V=YB$ were an interface, then $Q_DY=Y(2I-H)$, so $Z=YR$ is an integral right-positive matrix. Each row has squared norm 1 modulo 3, while Lemma 1 makes the total squared norm zero modulo 3. This is impossible at forty rows. It is also an instance of Theorem 2: the map $s\mapsto yR$ annihilates the positive projection and is integral on all generators of $\Lambda_-$, so its squared norm is an integral quadratic form on that lattice.

This excludes every interface in this 74-sign source, including interfaces using only a proper subset of its signs, for every regular switch of the original Petersen factor. It does not depend on a choice of the opposite interface $U$.

4.2. A control with the same dimensions and different arithmetic

The supplement source76/certificate.json contains a second complete order-64 graph with the following exactly checked properties:

| Interface row space on the 28-vertex side | Rank | Complete sign section | Positive/negative dimensions | |---|---:|---:|---:| | Reverse | 9 | 76 | 3 / 6 | | Forward, after transposition | 5 | 12 | 3 / 2 |

All signs in both sections have genuinely fractional denominator-three projections. Both graph and complement satisfy all ordered two-step counts, and both interfaces satisfy the mixed equations, both zero products, and every row and column balance equation.

For the 76-sign reverse source we supply a basis $L_-$ for the full six-dimensional lattice generated by $n(s)$. All 76 coordinate rows $c(s)$ are integral. Six selected source signs have a unimodular coordinate matrix; this proves that the displayed basis generates the full lattice, not a proper superlattice. The evaluation matrix of the 21 degree-two monomials $c_ic_j$, $i\le j$, has rank 21 over $\mathbb F_3$. Appending a constant-one column increases the rank to 22. Therefore no homogeneous quadratic form is constantly one on this source, and scaling rules out any other nonzero constant.

This is a positive control for the scope of Theorem 2, not evidence that the order-68 target exists. The certificate additionally supplies a balanced forward interface to Petersen40 using the twelve-sign forward section. Its opposite interface is not supplied. No claim that every other arithmetic obstruction fails is made.

5. Complete ternary classification of the left-positive space

Write

$$ \mathcal L=\operatorname{im}(K^T),\qquad \mathcal Z=\mathcal L\cap\{-1,0,1\}^{40}. $$

Diagonal sign switches preserve ternarity and support, so a classification for $\mathcal L$ applies to every switched left-positive space. Express a vector as $z=K^Tf/4$.

For a Petersen vertex $e$, the four coordinates in its cell form the Hadamard transform of the four coefficients on the closed neighborhood $N[e]$, divided by four. Inverting this local transform shows that these four coefficients are integers with a common parity. Connectivity gives a common parity for all ten coefficients, and orthogonality gives

$$ \sum_{q\in N[e]}f_q^2\le16. \tag{6} $$

Every pair of Petersen vertices lies in a common closed neighborhood: adjacent vertices lie in an endpoint neighborhood, and nonadjacent vertices have a common neighbor.

Let $\mathcal C=\ker_{\mathbb F_2}(I+A_{\rm Pet})$. Identifying Petersen vertices with edges of $K_5$, this is the code of even-cardinality Eulerian edge sets. To verify the identification, if $b$ has even total weight and all vertex degrees even, the sum on $N[e]$ vanishes modulo 2. Conversely, the matrix $I+A_{\rm Pet}$ has rank 5, as elementary elimination or the supplement verifies, and the even Eulerian code has dimension 5. Its weight enumerator is

$$ 1+15x^4+15x^6+x^{10}. \tag{7} $$

The weight-four sets are the fifteen four-cycles of $K_5$, and the weight-six sets are their complements. Indeed an even Eulerian subset has weight $0,4,6,$ or $10$: weights 2 and 8 are excluded by simplicity and complementation. Four edges must form a four-cycle. This accounts for all 32 codewords.

Theorem 3 (ternary field classification). The set $\mathcal Z$ consists of the following disjoint families, of total size 885. All sign variables below take values $\pm1$.

| Coefficient field $f$ | Conditions | Number | Support of $z$ | |---|---|---:|---:| | $0$ | — | 1 | 0 | | $\pm4e_q$ | $q$ arbitrary | 20 | 16 | | $2b$ on a four-cycle | four arbitrary signs | 240 | 16 | | $2b$ on a complementary six-set | product $-1$ on each contained $N[e]$ | 240 | 24 | | $2\eta$ on all ten positions | product $-1$ on every $N[e]$ | 32 | 40 | | $\eta$ on all ten positions | product $+1$ on every $N[e]$ | 32 | 10 | | $\eta-4\eta_qe_q$ | $\eta$ as in the preceding row, $q$ arbitrary | 320 | 18 |

Proof. If the common parity is odd, (6) restricts each coefficient to $\pm1,\pm3$. At most one coefficient has magnitude 3, since any two belong to one closed neighborhood and $3^2+3^2>16$. If all are signs, local integrality holds exactly when their product on every closed neighborhood is $+1$. The local transform then has exactly one nonzero coordinate. These product equations have 32 solutions, by rank 5 of $I+A_{\rm Pet}$.

If $f_q$ has magnitude 3, replace it by $\eta_q=-f_q/3$, keeping all other coordinates unchanged. Thus $f=\eta-4\eta_qe_q$. The transformed vectors differ by an integral vector, so local integrality forces the same product-$+1$ equations for $\eta$. In each of the four cells containing $q$, the one nonzero entry of $K^T\eta/4$ cancels and the other three become signs. The remaining six cells each retain one entry. This gives support $4\cdot3+6=18$, and all 320 choices are valid.

For even parity, a coefficient of magnitude 4 uses all the bound in (6) in each cell containing it. Since every other coefficient shares one of those cells, all the others vanish. Otherwise $f=2b$ with $b\in\{-1,0,1\}^{10}$. Local integrality requires an even number of nonzeros in every closed neighborhood, so the support of $b$ is in $\mathcal C$. A local support of size two always gives a ternary transform. A local support of size four gives a ternary transform exactly when its sign product is $-1$; otherwise it has an entry of magnitude 2. A four-cycle contains no entire closed neighborhood, giving $15\cdot16=240$ choices. Its complement contains exactly two closed neighborhoods, whose two distinct product constraints are independent, giving $15\cdot2^{6-2}=240$ choices. Full support gives a rank-five affine sign system. In the specified lexicographic order, $(-1,-1,-1,-1,-1,-1,-1,1,1,1)$ is one solution, so it has 32 choices. Consistency and the local counts can also be checked directly in the ten-vertex description; the standalone verifier does so. Counting the nonzeros in the local transforms gives the displayed support sizes. These cases exhaust (6) and prove the classification. $\square$

For an independent completeness check, the supplement row-reduces $K$ over the rationals and tests all $3^{10}=59049$ assignments to its independent physical coordinates. It obtains exactly the same 885 vectors without assuming the field classification.

6. Rigidity of disjoint real supports

For a nonzero $z\in\mathcal Z$, define the real subspace

$$ D(z)=\{w\in\mathcal L:w_i=0\text{ whenever }z_i\ne0\}. $$

Theorem 4 (disjoint-space rigidity). The dimension of $D(z)$ is zero or one. In the latter case it is generated by a ternary vector $w$, unique up to sign among nonzero ternary vectors, and the supports of $z,w$ partition the forty positions with sizes 16 and 24.

Proof by exhaustive rational linear algebra. The independent enumeration just described is complete. For each of its nonzero vectors it finds either no nonzero disjoint ternary vector, or precisely the pair $\{w,-w\}$ with the asserted complementary supports. This alone would not exclude nonternary real vectors. To close that gap, restrict the rows of $K^T$ to the support of $z$. This is the linear system whose kernel defines $D(z)$. The following is the complete census of distinct support masks:

| Support size | Number of masks | Restriction rank modulo 101 | Disjoint ternary partner | |---:|---:|---:|---| | 10 | 16 | 10 | none | | 16 | 10 | 10 | none | | 16 | 120 | 9 | one antipodal pair | | 18 | 160 | 10 | none | | 24 | 120 | 9 | one antipodal pair | | 40 | 1 | 10 | none |

A nonzero minor modulo 101 is a nonzero integer minor, so every listed rank is a lower bound for the real rank. In a rank-nine case the known nonzero disjoint vector supplies a rational kernel vector, proving the reverse inequality. In a rank-ten case there are only ten columns. These are consequently the exact real ranks. The supplement generates every vector, every mask, and every matrix in this table from the definition of $K$; no supplied census is trusted. This proves the theorem. $\square$

There are 960 ordered disjoint nonzero pairs, or 480 unordered pairs, when the signs of both vectors are retained. The four-cycle and complementary-six-set families account for all of them.

One direct use is the following rank test. If $z,z_2,z_3\in\mathcal Z$, $z\ne0$, and $z$ is disjoint from both $z_2$ and $z_3$, then $z_2,z_3$ are linearly dependent. More strongly, they may be arbitrary real vectors of $\mathcal L$. Thus a proposed three-channel realization with one nonzero channel isolated in support from two independent channels is impossible in this Petersen orientation. This does not exclude a source that can be realized through a proper invariant section avoiding the three-channel configuration. Any source-level exclusion must also cover such sections.

7. Evidence, scope and reproduction

The supplement uses only the Python 3 standard library. Run python verify_all.py without optimization. It checks two complete order-64 graphs and their complements, the complete 74-, 76- and 12-sign sections, the four-channel certificate (5), the exact six-dimensional projection lattice and the ranks 21/22, and the Petersen forward ingredient. Separately, verify_left4.py reconstructs the factors, the seven field families, the independent 59049-assignment census, and all real-rank checks in Theorem 4. It needs no data file. The full check takes a few seconds on one CPU and uses small memory; no GPU, solver, or external package is needed.

The code and original certificate data are released under the licenses included in their directories. The old control directories are retained with their provenance and verification programs. Essential data are contained in the supplement; no private file or mutable download is needed to check a claim. The finite computations use integer or rational arithmetic and exact finite-field elimination, with no numerical rank tolerances.

The original Petersen orientation is an essential hypothesis in Lemma 1 and Theorem 2; transposition exchanges the two positive lattices. The 74-source obstruction covers every subset of that source, but the absence of a quadratic obstruction for the 76-source is only a necessary test passed. Theorem 4 concerns one specific ten-dimensional lifted space, not all order-40 DSRGs. These results supply checks for mixed source selection and clarify why allowing thirds or matching spectral dimensions is insufficient. Neither a graph of order 68 nor unrestricted nonexistence at that order is established here.

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