Counterexample dossier · spectral graph theory

Graffiti conjecture 284

A familiar 50-vertex graph makes the claimed dual-degree bound fail by three.

§1

The conjecture

Write d(v) for the degree of v, and define its dual degree d*(v) as the average of the degrees of its neighbors. For a connected graph with girth at least 5, Graffiti 284 asserted the following inequality, where λmin(D) is the least eigenvalue of the graph's distance matrix.

minvV(G)d*(v)λmin(D(G))

The report that prompted this entry identifies the Hoffman–Singleton graph as a counterexample. The calculation below independently verifies the implication; the original Graffiti-record attribution remains flagged for bibliographic audit.

§2

The witness

The Hoffman–Singleton graph H is the standard strongly regular graph with parameters (50, 7, 0, 1). In particular, it has 50 vertices, is 7-regular, has diameter 2, and has girth 5. Regularity immediately gives the left side: every neighbor of every vertex has degree 7.

d*(v)=17uvd(u)=7
§3

Distance spectrum — exact reduction

Let A be H's adjacency matrix, I the identity, and J the all-ones matrix. Because H has diameter 2, distinct vertices are at distance 1 precisely on edges and at distance 2 otherwise. Therefore its distance matrix is:

D=2(JI)A

The adjacency spectrum of H is 71, 228, (−3)21. On the all-ones line, D has eigenvalue 2(50−1)−7=91. On the orthogonal complement, J vanishes, so an A-eigenvalue θ becomes −2−θ for D. Hence:

Spec(D)=911,121,(4)28
§4

Contradiction

Thus λmin(D)=−4 and the conjectured right side is 4, while the left side is 7. H has the required girth, so it is a valid witness — no numerical eigensolver or graph search is involved.

7=minvd*(v)λmin(D)=4
§5

References