Random-graph inertia conjecture

Let GG be sampled from the binomial random graph G(n,12)\mathbb G(n,\frac 12), and let AA be a weighted adjacency matrix of GG. Write n0(A)n_{\geq 0}(A) for the number of non-negative eigenvalues of AA. Random-graph inertia conjecture. With probability 1o(1)1-o(1) as nn\to\infty, every weighted adjacency matrix AA of GG satisfies

n0(A)=Ω(nlogn).n_{\geq 0}(A)=\Omega\left(\frac{n}{\log n}\right).

Since the independence number of G(n,12)\mathbb G(n,\frac 12) is O(logn)O(\log n) with probability 1o(1)1-o(1), this would show that the inertia bound is far from tight for almost all graphs. The bound would also be best possible up to constants, by the corresponding clique-cover estimate.

Sources & referencesView supporting material

Primary source

Matthew Kwan and Yuval Wigderson, “The inertia bound is far from tight”, arXiv:2312.04925 (2024).

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