2018
DOI: 10.1142/s1793525319500468
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Freeness of the random fundamental group

Abstract: Let Y (n, p) denote the probability space of random 2-dimensional simplicial complexes in the Linial-Meshulam model, and let Y ∼ Y (n, p) denote a random complex chosen according to this distribution. In a paper of Cohen, Costa, Farber, and Kappeler, it is shown that for p = o(1/n) with high probability π 1 (Y ) is free. Following that, a paper of Costa and Farber shows that for values of p which satisfy 3/n < p ≪ n −46/47 , with high probability π 1 (Y ) is not free. Here we improve on both of these results t… Show more

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Cited by 5 publications
(6 citation statements)
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“…We further believe it is possible that for p above this threshold, is free with high probability. For the case of with , the sharp threshold is found by Newman in [27], improving on previous work of [7].…”
Section: Introductionsupporting
confidence: 74%
“…We further believe it is possible that for p above this threshold, is free with high probability. For the case of with , the sharp threshold is found by Newman in [27], improving on previous work of [7].…”
Section: Introductionsupporting
confidence: 74%
“…Note that the algebraic structure of the fundamental group is more intricate than H 1 (X), and thus several other properties have been studied in the literature. For example: its geometric and cohomological dimension [90], torsion [91], property T [33], freeness [92], and more.…”
Section: The Fundamental Groupmentioning
confidence: 99%
“…• X ∼ X(n, p) to go from π 1 (X) free to π 1 (X) nonfree. The constants c d and γ d are described fully in [16], where the following table of the approximation of the first few values also appears: Regarding the fundamental group, [18] proves the following for Y ∼ Y 2 (n, p). These are still the best known bounds on a sharp threshold for π 1 (Y 2 (n, p)) to be free.…”
Section: Discussionmentioning
confidence: 98%
“…[18] If c < γ 2 and Y ∼ Y 2 (n, c/n) then with high probability π 1 (Y ) is free, while if c > c 2 and Y ∼ Y 2 (n, c/n) then with high probability π 1 (Y ) is not free.Naturally Conjecture 22 together with Theorem 16 would imply that π 1 (X(n, p)) is free if p < √ γ n ≈ 1.567 √ n and if not free if p > √ c 2…”
mentioning
confidence: 98%