# C31: independent audit of the claimed Milgram obstruction

Date: 2026-10-09. Reviewer: Codex, working alone. Source reviewed: `docs/s4lat_2026_10_08.md`, section 2, and `scripts/s4lat_milgram.py`. No coordinator/other-lane files modified; nothing published.

## Verdict

**PASS for the two requested hand arguments, with an explicit repair of a missing justification in the uniform lemma.** I independently obtain the stated signature contradictions for SRG(96,35,10,14) and SRG(176,25,0,4). I did not find a false exclusion or a sign/2-adic defect that invalidates them.

My independent exact decider also finds no admissible form for the chosen A+4I representation in **all seven named candidate rows**, even using a weaker, independently justified exponent bound. The positive-control run completes both representations for **110 existing integral-spectrum parameter rows** with zero false kills. There are 224 attempted representations: 222 pass and two hit a deliberately short three-second limit. Those timeouts are not counted as exclusions or as completed two-representation rows.

This is an audit finding, not a publication or a novelty claim. The seven hand arguments beyond the two explicitly requested do not all receive a full prose re-proof here; their chosen representations are independently checked by exact finite enumeration. The additional section-2.8 scan is not certified wholesale by this report.

## 1. Precise statement and hypotheses

Let A be the adjacency matrix of an SRG(v,k,lambda,mu) with integral restricted eigenvalues rho and theta. Choose integers b,c with **b in {1,-1}**, and put

`G=b(A-rho I)+cJ`, `e=b(theta-rho)>0`, `T=b(k-rho)+cv>0`.

Assume the vertex norm `m=c-b*rho` is even. Then G is positive semidefinite, its nonzero eigenvalues are T once and e with multiplicity f=mult(theta), and its rank is d=f+1. In the following the letter s denotes the sum of vertex vectors, not an eigenvalue.

This is not literally a statement about every arbitrary expression aA+bI+cJ: the killed-eigenvalue and positivity hypotheses matter. Centered representations T=0 need a separate statement. No completeness of a search over representations is needed for a nonexistence proof: **one valid representation with an impossible necessary condition suffices**.

Take vectors v_i with Gram G, and let L be their integer span in their d-dimensional real span. This is a discrete even integral lattice: an independent subset generates a full-rank integral sublattice L0, and all the other v_i lie in L0's dual. Every integral overlattice lies between L and L#, so only finitely many can occur. A maximal even overlattice M therefore exists. Its finite discriminant quadratic module D=M#/M has q(x+M)=x^2 mod 2Z and bilinear pairing modulo Z. D is nondegenerate and anisotropic, and its Brown invariant is d modulo eight.

Put `a=(T-e)/(vT)` and `B=(k-theta)/v`. For x in M#, set z_i=(x,v_i), sigma=(x,s). The frame identity gives

`sum z_i^2 = e*x^2+a*sigma^2`,

`sum_{j adjacent to i} z_j = theta*z_i+B*sigma`,

`e*x = sum z_i*v_i-a*sigma*s`.

All z_i are integers. Hence `B*sigma` is integral and `sum z_i^2 = sigma mod 2`.

Let H be the nonempty set of positive divisors h of T satisfying

`B*h in Z`, `a*h^2 in Z`, `a*h^2-h in 2Z`,

and let g=gcd(H). Then y=s/g belongs to M#, with norm Y=vT/g^2. Its class has order

`r=min{j>0 : j^2*Y in 2Z}`.

For every class xbar of D there is a rational t congruent to b(xbar,ybar) modulo one such that

`g*t in Z`, `B*g*t in Z`, `e*q(xbar)+C*t^2-g*t in 2Z`, where `C=a*g^2`.

The last congruence means that a rational representative of q is chosen; its validity is unchanged on adding an even integer because e is integral. Thus it is a well-defined necessary condition on the discriminant data.

## 2. Details which must not be skipped

### The gcd step and the order of y

The actual set of pairings (s,M) is h0 Z, with h0 dividing T since (s,v_i)=T. The preceding identities imply h0 is in H, so g divides every pairing with M and y is dual.

The integers satisfying the three h-conditions form a subgroup. For the quadratic condition, if the denominator of a divides h1^2 and h2^2, it divides h1*h2 as well, prime by prime. Consequently the difference of two allowed integers remains allowed, using

`a(h1-h2)^2-(h1-h2) = (a*h1^2-h1)+(a*h2^2-h2)-2a*h1*h2+2h2`.

Thus **C is an integer and C congruent to g modulo two**. This supplies a justification that is implicit in the source.

Since y is already in M#, the lattice M+Z*j*y is even integral exactly when j^2*Y is even. Maximality then puts j*y in M. Conversely j*y in M has even norm. Hence the stated minimal j is exactly the class order; it is not only an upper bound.

### Exponent: the source needs an integrality argument

Writing `e*x congruent to -C*t*y mod M` does **not by itself** show that the right side is an integer multiple of y. One must justify `r divides C`, since r*t is integral. For these representations that justification is available:

1. `C*Y=T-e` is integral and C is integral.
2. At odd primes, the denominator of Y already contains the required prime power of r, so it divides C. The same argument works at 2 whenever the 2-adic valuation of Y is negative; when it is positive there is no 2-part of r.
3. The remaining case is Y a 2-adic unit, when r has a single factor 2. If C were odd then g would be odd. Since Y=vT/g^2 is a 2-adic unit, v and T would be odd. From T-e=C*Y, e would be even. But `T+f*e=v*m` is even because the vertex norm m is even, a contradiction.

Therefore C/r is integral. It follows that e*D is contained in the cyclic subgroup generated by ybar, and `exp(D) divides e*r` as claimed. **I classify this as a repairable missing proof step, not a counterexample to the lemma.**

As an additional robustness check, my independent program does not need this sharper step: it uses the automatically safe bound `e*den(C)*r^2`. Multiplying the frame identity by den(C)*r^2 puts the correction term in Z*r*y, hence in M. This admits at least all true candidates and still excludes the seven chosen representations.

### Anisotropy and completeness

An isotropic class gives an even integral overlattice, contrary to maximality. For odd p an anisotropic finite quadratic p-module has exponent p and dimension at most two. At p=2 the possible underlying groups are 0, Z2, Z2^2, Z2^3, Z4 and Z4+Z2. The quadratic, not merely bilinear, form matters; odd lattices are never substituted here. My program enumerates every quadratic assignment on these small groups and tests both anisotropy and nondegeneracy. It does not deduplicate by a possibly incomplete invariant and has no discriminant-order cutoff.

The signature is **positive rank d**, not minus d, and Milgram uses exp(pi*i*q). The 2-primary phases are verified exactly in Z[zeta8]/(zeta8^4+1); odd-primary phases use the exact quadratic Gauss-sum/Legendre-symbol formula. There is no floating phase rounding.

## 3. Hand check: SRG(96,35,10,14)

Pass to the complement (96,60,38,36), and use G=A+4I. Its nonzero eigenvalues are 64 once and 10 with multiplicity 32, so d=33. Directly, `G^2=10G+36J`.

For x in a maximal even M, sigma=(x,s) satisfies `9*sigma^2/1024 in Z` and the same expression is congruent to sigma modulo two. Integrality first forces 32|sigma; putting sigma=32h gives parity h^2=0 modulo two, so **64|sigma**. Since sigma(v_i)=64, the greatest common necessary divisor is exactly 64. Therefore

`y=s/64`, `y^2=3/2`, `2y in M`, `ord(ybar)=2`.

For x in M#, write (x,y)=h/2. The frame identity gives

`10x = sum z_i*v_i-18h*y in M`.

Thus the discriminant group is killed by 10. It splits into its 2- and 5-parts. The class ybar lies in the 2-part, has norm 3/2, and has self-pairing 1/2, so it splits off orthogonally. The congruence is

`10q(xbar)+36t^2-64t = 0 mod 2`.

For a class in the orthogonal complement of ybar inside the 2-part, t is integral. Hence 10q is even; because this part has exponent two, q takes half-integer values, and the condition forces q to be integral. Its nondegenerate anisotropic even-type form is either zero or the anisotropic binary form, of Brown invariant 0 or 4. Adding <3/2> gives Brown invariant **7 or 3**.

Every 5-primary quadratic Gauss sum has real phase, so its Brown invariant is 0 or 4. The complete module therefore has Brown invariant **3 or 7**, never `33=1 mod 8`. This contradicts Milgram. All steps concern even lattices; no odd-overlattice assumption is hidden.

## 4. Hand check: SRG(176,25,0,4)

Use its complement (176,150,128,126) and G=A+4I. The nonzero eigenvalues are 154 once and 10 with multiplicity 55, giving d=56. Here `G^2=10G+126J`.

For x in M, `9*sigma^2/1694` is integral, where 1694=2*7*11^2. Therefore **154|sigma**. This value satisfies the parity condition and occurs on each v_i. Hence

`y=s/154`, `y^2=8/7`, `ord(ybar)=7`, `C=126`.

The frame identity shows the discriminant is killed by 70. On the 2-primary part, t=(x,y) is integral, and the congruence forces q integral; its Brown invariant is 0 or 4. The 5-primary part likewise has Brown invariant 0 or 4.

On the 7-primary part write `q=2c/7` and `t=u/7 mod Z`. The congruence becomes `c=4u^2 mod 7`. In the kernel of b(-,ybar), u=0, hence q=0; anisotropy makes that kernel trivial. Therefore the 7-part is exactly the cyclic group generated by ybar, with quadratic norm **8/7**. Since 4 is a square modulo 7, its Gauss sum is +i*sqrt(7), of Brown invariant **2**.

Total Brown invariant is **2 or 6**, not `56=0 mod 8`. This is again a genuine signature contradiction.

## 5. Independent computation and controls

Own code: `scripts/codex_audit_milgram.py`. It imports **none** of `s4lat_milgram.py`, the inherited lattice survey, the shell LP, or their finite-form classes. It uses Python's standard library only. Input controls are parsed directly from the cached primary Brouwer pages, including inherited complement status cells.

Outputs: `out/codex/c31/targets.json`, `controls.json`, and corresponding logs. Reproduction:

```bash
python scripts/codex_audit_milgram.py targets
python scripts/codex_audit_milgram.py controls
```

The positive-control sample contains 112 distinct existing parameter rows. For 110 rows both minimal even representations complete successfully; two rows have one three-second timeout, recorded separately. **222 admissible representations, zero false kills.** This checks parameter-level feasibility on existing rows; it is not an independent reconstruction of 110 actual maximal-overlattice Gram matrices. Both orientations count as parameter rows, not as nonisomorphic graph constructions.

| Chosen complement/orientation | Rank | y norm | y order | Exact decider |
|---|---:|---:|---:|---|
| (96,60,38,36) | 33 | 3/2 | 2 | NO FORM |
| (176,150,128,126) | 56 | 8/7 | 7 | NO FORM |
| (169,126,95,90) | 43 | 13/10 | 10 | NO FORM |
| (616,205,90,57) | 56 | 56/19 | 19 | NO FORM |
| (300,26,4,2) | 118 | 10 | 1 | NO FORM |
| (430,39,8,3) | 130 | 10 | 1 | NO FORM |
| (690,371,220,175) | 46 | 46 | 1 | NO FORM |

Two *other-side* target representations time out; this does not affect the table, since every row already has one completed obstruction. The finite lift test exhausts a rigorously sufficient period: `2*lcm(den C, den(C*bxy), den(B*g))`. Adding this period leaves the integrality conditions unchanged and changes the parity polynomial by an even integer.

## 6. Literature and novelty boundary

- The underlying discriminant-form and overlattice machinery is classical, not a new theorem: [Nikulin, Integral symmetric bilinear forms and some of their applications (1980)](https://www.mathnet.ru/eng/im1677).
- For the finite-module normalization and Gauss-sum theory, see [Stromberg, Weil representations associated to finite quadratic modules](https://arxiv.org/abs/1108.0202). That paper uses Q with values modulo one; here q=2Q, so exp(2*pi*i*Q)=exp(pi*i*q). Mixing these conventions would reverse or otherwise corrupt the calculation; the audit does not mix them.
- The [Brouwer table for 51-100 vertices](https://aeb.win.tue.nl/graphs/srg/srgtab51-100.html) still lists (96,35,10,14) as unresolved. This is a warning to audit carefully, not a mathematical refutation of the argument.
- [Nelson and Swartz, Character theoretic techniques for nonabelian partial difference sets](https://arxiv.org/abs/2507.23039) concerns PDS/Cayley restrictions. Such results must not be mistaken for unrestricted SRG nonexistence.

I did not locate a primary source identifying this exact SRG parameter obstruction, or proving it equivalent to a named Bruck-Ryser-Chowla/Hasse-Minkowski criterion. **Novelty remains unestablished.** The method is recognizably an application of classical quadratic-form arithmetic; that alone does not establish whether these particular consequences were previously known. The author should compare the separate blind audit and seek external review before circulating new nonexistence claims.

## 7. Recommended source edits

1. Insert the proof that C/r is integral before claiming eD is contained in <ybar>.
2. Restrict the headline to the explicitly defined even, positive, killed-eigenvalue representations; distinguish T=0 and odd lattices.
3. Keep the inherited order-20000 classifier out of any completeness claim. The independent exhaustive primary-component argument is the relevant one.
4. Separate hand-proved cases, exact finite checks, and the unfinished whole-table scan. Do not promote timeouts or an empty partial output to nonexistence.
5. Preserve the publication/novelty caveat even though the two requested hand proofs and the exact control sample pass this audit.