An upper bound on the largest eigenvalue of the Helmholtzian of a graph

Abstract

The Helmholtzian of a graph $G$ is the Hodge $1$-Laplacian $L_1=L_1^{\mathrm{up}}+L_1^{\mathrm{down}}$ of its clique complex, built from the triangle--edge and edge--vertex boundary operators $\partial_2$ and $\partial_1$. Problem~5.5 of Lu, Shi, Stanić, Wang and Wang asks whether $λ_{\max}(L_1)=μ_1(G)$ for every graph $G$, where $μ_1(G)$ is the largest Laplacian eigenvalue; by the Hodge decomposition this is equivalent to $λ_{\max}(L_1^{\mathrm{up}})\leμ_1(G)$. We recast it as a question about the complement of $G$: localizing $L_1^{\mathrm{up}}$ on the cycle space of $K_n$ turns it into the inequality $λ_{\min}(\bar L|_{Z_1})\ge a(\overline{G})$, where $\bar L$ is the up Laplacian of the missing triangles of $G$ and $a(\overline{G})=n-μ_1(G)$ is the algebraic connectivity of the complement. From this viewpoint, we prove the unconditional bound \[ λ_{\max}\!\big(L_1^{\mathrm{up}}(G)\big)\ \le\ μ_1(G)+\frac13\big(n-μ_1(G)\big), \] which refines the integrality ceiling $λ_{\max}(L_1^{\mathrm{up}})\le n$ of Duval and Reiner and is sharp exactly when that ceiling is attained. We then isolate the single sharp inequality, on the dense part of $\overline{G}$, that stops the method short of Problem~5.5, and we show that the localization, the bound, and this obstruction all persist for the up Laplacian of an arbitrary finite simplicial complex, in every dimension.