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Typical regular graphs are almost Ramanujan

PDF p. 7: Section 1.3, paragraph citing Friedman [20]

Statement

Fix an integer d≥3 and ε>0. A uniformly chosen labelled simple d-regular graph on n vertices is, with high probability, an (n,d,α)-expander with α=(2d−1+ε)/d. The limit is as n→∞ through orders for which such graphs exist, with d and ε fixed.

Here simple means without loops or repeated edges, and d-regular means that every vertex has d neighbors. Uniformly chosen means that every such graph on {1,…,n} has equal probability; with high probability means that the probability tends to one. For adjacency matrix A and all-ones matrix J, the expander condition is ∥A−(d/n)J∥op≤αd, where the operator norm is induced by the Euclidean norm. This is the almost-Ramanujan assertion attributed to Friedman in the article.

Facts & Assumptions

[F1]

For fixed integer d≥3 and fixed ε>0, in the uniform labelled simple d-regular model, the probability that max⁡{λ2(A),∣λn(A)∣}≥2d−1+ε tends to zero along feasible orders. The real adjacency eigenvalues are ordered with multiplicity as d=λ1≥⋯≥λn (Friedman spectral expansion in the uniform simple model).

[F2]

For a labelled simple d-regular graph with d>0 and n≥2, ∥A−(d/n)J∥op=max⁡{λ2(A),∣λn(A)∣} (Ordered eigenvalues and the centered regular norm).

[F3]

In the stated model the expander event is the centered norm inequality above, and high probability is convergence of its probability to one along feasible orders (Adjacency spectral expander — Definition 1.3, Adjacency, degrees, graph Laplacian, and spectral notation, Uniform random regular graph convention, Asymptotic quantifiers and uniform parameter ranges).

Proof

Given: Fixed integer d≥3 and fixed ε>0, with the uniform simple regular graph law at each feasible order.

Goal: Show that the (n,d,α)-expander event has probability tending to one for α=(2d−1+ε)/d.

1.1givenF1

By [F1], the event that max⁡{λ2(A),∣λn(A)∣}<2d−1+ε has probability tending to one. The graph law and feasible-order limit here are exactly those in the statement.

2.1F2F3step 1.1algebra∎

By [F2], on this event ∥A−(d/n)J∥op<2d−1+ε=αd. It therefore satisfies the non-strict expander inequality in [F3]. Its probability is at least that of the event in step 1.1 and at most one, so it tends to one.

Depends on

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