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Keevash 2006 sparse halves triangle free graphs

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proposition_1_2: Keevash and Sudakov's proposition that every triangle-free graph on n vertices with at most n^2/12 edges has a set of floor(n/2) vertices spanning at most n^2/50 edges.

theorem_1_1: Keevash and Sudakov's theorem that a triangle-free graph on n vertices with at least n^2/5 edges in which every floor(n/2) vertices span at least n^2/50 edges has n = 10m and is the blow-up C_5(2m) of the 5-cycle.


Keevash, Peter and Sudakov, Benny, Sparse halves in triangle-free graphs. J. Combin. Theory Ser. B 96 (2006), 614-620. DOI 10.1016/j.jctb.2005.11.003. The file prints "© 2005 Elsevier Inc. All rights reserved.", every other right reserved.

Erdos conjectured, offering a prize, that every triangle-free graph on n vertices contains a set of n/2 vertices spanning at most n^2/50 edges; Krivelevich had proved this with n^2/36 in place of n^2/50. Theorem 1.1 establishes the conjecture for dense graphs in a sharp form: if G is triangle-free on n vertices with at least n^2/5 edges and every set of floor(n/2) vertices spans at least n^2/50 edges, then n = 10m and G is exactly the balanced blow-up C_5(2m) of the 5-cycle, identifying C_5(n/5) as the unique extremal example in that range. Proposition 1.2 handles the sparse side: any triangle-free graph with at most n^2/12 edges has a set of floor(n/2) vertices spanning at most n^2/50 edges. The arguments are averaging over random subsets, with Cauchy-Schwarz on degrees and the Andrasfai-Erdos-Sos theorem in the dense case. The setting is the Erdos-Faudree-Rousseau-Schelp local-density problem, where the largest beta with every alpha-n set spanning at least beta n^2 edges is conjectured to be (2alpha-1)/4 from alpha = 17/30 upward and (5alpha-2)/25, the C_5(n/5) value, for a certain range of alpha below 17/30 that includes alpha = 1/2. The authors also discuss the companion Erdos conjecture that n^2/25 edge deletions suffice to make a triangle-free graph bipartite, and Krivelevich's observation that for regular graphs a sparse-half bound implies such a bound, noting the Petersen blow-up P(n/10) shows the converse fails. This is the reference for the n^2/50 sparse-halves problem (problem 128).

Source: https://people.math.ethz.ch/~sudakovb/papers.html.

Read status: claims checked for Theorem 1.1, Proposition 1.2 and the remarks of pp. 615--616, read clause by clause on the page images of the print; the proof of Proposition 1.2 (Section 2, pp. 616--617) followed and the proof of Theorem 1.1 (Section 3, pp. 617--619) followed in outline. Nothing here is independently reviewed. Result pages: theorem_1_1 and proposition_1_2.

Bears on. #128: Proposition 1.2 and Theorem 1.1 (p. 615) give a set of ⌊n/2⌋\lfloor n/2\rfloor vertices spanning at most n2/50n^2/50 edges in every triangle-free graph on nn vertices with at most n2/12n^2/12 or at least n2/5n^2/5 edges, so in those two edge ranges a graph whose every ⌊n/2⌋\lfloor n/2\rfloor vertices span more than n2/50n^2/50 edges contains a triangle; the range between n2/12n^2/12 and n2/5n^2/5 is not covered, as the problem's claim page records.

Results.

  • Theorem 1.1 (p. 615): a triangle-free graph on nn vertices with at least n2/5n^2/5 edges in which every set of ⌊n/2⌋\lfloor n/2\rfloor vertices spans at least n2/50n^2/50 edges has n=10mn=10m and is C5(2m)C_5(2m).
  • Proposition 1.2 (p. 615): every triangle-free graph on nn vertices with at most n2/12n^2/12 edges has a set of ⌊n/2⌋\lfloor n/2\rfloor vertices spanning at most n2/50n^2/50 edges.

Context recorded without result pages. Local density (p. 615): in C5(n/5)C_5(n/5), for 2/5≤α≤3/52/5\le\alpha\le3/5, every αn\alpha n vertices span at least 5α−225n2\frac{5\alpha-2}{25}n^2 edges, more than the value 2α−14n2\frac{2\alpha-1}{4}n^2 for T2(n)T_2(n) when α<17/30\alpha<17/30; Krivelevich's bounds c3<3/5c_3<3/5 and n2/36n^2/36 are cited. Petersen blow-up (pp. 615--616): in P(n/10)P(n/10) every n/2n/2 vertices span at least n2/50n^2/50 edges, yet deleting 3n2/1003n^2/100 edges makes it bipartite, which the paper gives to show that Krivelevich's reasoning from sparse halves to bipartite deletion for regular graphs does not run in reverse.

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