# C35 referee review: `milgram_obstruction.tex` v2

## Verdict

**PASS WITH MINOR CAVEATS.** I found no fatal error in Lemma 3.5 (odd representations), the
parity corollary, or the stated finite-form logic. My independent decider processed all 48 table
rows in the paper (both orientations: 96 representations): 81 returned `NO_FORM`, 15 hit the
five-second safety timeout, and none returned an admissible module. The timeouts are reported as
timeouts, never interpreted as exclusions.

For the two requested C1 rows:

| Row | V4 result |
| --- | --- |
| (508,247,126,114), representation `(-7,1,1)` | `NO_FORM` (40 candidate images considered) |
| (508,247,126,114), other orientation `(19,-1,1)` | timeout |
| (1072,391,150,138), representation `(-11,1,1)` | `NO_FORM` (72 candidate images considered) |
| (1072,391,150,138), other orientation `(23,-1,1)` | timeout |

Thus the paper's C1 claims agree where the displayed example representation is decidable; the
other orientation is not independently certified by this short run. Full machine output is
`out/codex/c35/table_results.json`, generated by `scripts/codex_milgram_v2_table.py`.

## Lemma 3.5 (odd representations)

The proof is correct. Since every `v_i` has odd norm, the norm modulo two is a surjective
homomorphism on L; its kernel L0 is even of index two, and `2L` is contained in L0. For x in
L0#, differences `v_i-v_j` and vectors `2v_i` lie in L0, hence all pairings are congruent
modulo integers and have denominator at most two. If they are integral then x lies in L#.

In the half-integral case `w_i=2z_i` are odd. The congruences
`2sigma = sum w_i = v (mod 2)`, `2B sigma = k-theta (mod 2)`, and
`4 sum z_i^2 = sum w_i^2 = v (mod 8)` follow exactly as printed. The spectral identity and
the norm identity are rational identities independent of integrality. Maximality of M supplies
anisotropy and Milgram phase; allowing the half-integral value set `S_{1/2}` is necessary and
correct. V3's weaker omission of the second congruence is a safe relaxation for exclusions.

One small exposition improvement: state explicitly that `L0#` pairings with all `v_i` being
half-integral is the only alternative because `z_i-z_j` and `2z_i` are integral. Also explain
that the rank of L0 remains d because finite-index sublattices preserve rank.

The centred odd conclusion is sound: `1^T G 1=vT=0` and the off-diagonal contribution is even,
so `v nu` is even; odd nu forces even v.

## Parity corollary and NL family

The centred representation has `T=0`, `ell L# subset L`, so `det L` divides `ell^n`. Applying
the standard odd-determinant congruence to the even lattice gives the displayed phase constraint.
The NL_m(q) algebra is consistent: after complementing if needed, `a=q`, `ell=q^2`, `nu` is
even, and the relevant multiplicity is 2 modulo 4, contradicting the all-primes-1-mod-4 case.
The caveat that this is a necessary condition, not a complete SRG test, is appropriate.

## Table and scope

The table's distinction between H, C2 and C1 is important and honest. A `NO_FORM` in one
representation suffices for nonexistence, while a timeout or a passing representation does not
prove existence. The paper explicitly records partial scans, known-graph controls, and the lack
of human expert review. Keep these caveats in the abstract and Zenodo metadata.

No disagreement appeared on 508 or 1072 for the paper's displayed representations. The remaining
timeouts should not be silently converted to “no module” in any future table regeneration.
