An obstruction to Ore-heavy mass localization

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This page preserves an earlier research route and its local standing. The completed threshold proof is in the solution note.

The construction below tests an auxiliary Ore-heavy endpoint lemma; it leaves the $C_7$ color bound open.

The Ore-heavy endpoint set

$$S_0=\{v:\text{some edge }vu\text{ has }d(v)+d(u)>n\}$$

need not contain half the vertices of a super-Turan graph.

A finite simple-graph construction

For an integer $k\ge3$, use $n=54k$ vertices:

• a clique $Z$ of order $24k$, partitioned into three sets $Z_i$ of size $8k$;
• an independent set $S$, partitioned into three sets $S_i$ of size $4k$;
• a complete six-partite graph $R$, with parts of size $3k$.

Add all $R$-$S$ edges and all $S_i$-$Z_i$ edges, and no others. The vertex degrees are

$$d_R=27k,\qquad d_S=26k,\qquad d_Z=28k-1.$$

Thus exactly the edges inside $Z$ have endpoint-degree sum greater than $54k$: their sum is $56k-2>54k$, whereas the sums on $RR,RS,S_iZ_i$ are respectively $54k,53k,54k-1$. Consequently

$$|S_0|=24k=4n/9<n/2.$$

Nevertheless

$$\begin{aligned} e(G)&=\binom{24k}{2}+15(3k)^2+(18k)(12k)+3(8k)(4k)\\ &=735k^2-12k =n^2/4+6k(k-2)>n^2/4. \end{aligned}$$

This is an unbounded-order obstruction, not a floating-point example.

Weighted formulation and its limitation

The corresponding twelve-type template has three looped, mutually joined $Z_i$ of weight $4/27$, three independent $S_i$ of weight $2/27$, and six independent $R$-parts of weight $1/18$, joined as above. Its parameters are

$$d_Z=14/27,\quad d_S=13/27,\quad d_R=1/2,\qquad q=\frac{245}{972}=\frac14+\frac1{486},\quad w(S_0)=4/9.$$

This refutes both the weak conjecture $w(S_0)>1/2$ and the stronger attempted analogue $w(S_0)\ge1/2+\sqrt{q-1/4}$ of the triangle-vertex mass theorem.

It does not obstruct the unrestricted three-walk clique: every type pair has a two-walk. For $Z_i,R_j$, use $S_i$; for $Z_i,S_j$, use $Z_j$; for $S_i,S_j$ and $R_i,R_j$, use respectively $R$ and $S$; for $S_i,R_j$, use a different $R$-part. The $Z$-$Z$ case is immediate. There are no isolated types, so all pairs also have three-walks. Every type is triangular. Therefore $J_{23}$ is complete and its palette cost is $\rho=q$, well above the desired $q/2$.

The proved fact that $S_0$ is a three-walk clique remains valid. Its mass, however, cannot serve as a universal half-vertex certificate.