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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 $\lambda_{\max}(L_1)=\mu_1(G)$ for every graph $G$, where $\mu_1(G)$ is the largest Laplacian eigenvalue; by the Hodge decomposition this is equivalent to $\lambda_{\max}(L_1^{\mathrm{up}})\le\mu_1(G)$. 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 $\lambda_{\min}(\bar L|_{Z_1})\ge a(\overline{G})$, where $\bar L$ is the up Laplacian of the missing triangles of $G$ and $a(\overline{G})=n-\mu_1(G)$ is the algebraic connectivity of the complement. From this viewpoint, we prove the unconditional bound \[ \lambda_{\max}\!\big(L_1^{\mathrm{up}}(G)\big)\ \le\ \mu_1(G)+\frac13\big(n-\mu_1(G)\big), \] which refines the integrality ceiling $\lambda_{\max}(L_1^{\mathrm{up}})\le n$ 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.

Submission history

From: Suil O [view email]
[v1] Thu, 18 Jun 2026 03:14:19 UTC (11 KB)