2018
DOI: 10.1112/jlms.12142
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Some advances on Sidorenko's conjecture

Abstract: A bipartite graph H is said to have Sidorenko's property if the probability that the uniform random mapping from V (H) to the vertex set of any graph G is a homomorphism is at least the product over all edges in H of the probability that the edge is mapped to an edge of G. In this paper, we provide three distinct families of bipartite graphs that have Sidorenko's property. First, using branching random walks, we develop an embedding algorithm which allows us to prove that bipartite graphs admitting a certain t… Show more

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Cited by 54 publications
(86 citation statements)
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“…where χ L is the colouring obtained by doubling the 'left half' of the rainbow cycle, represented by the vector (1, 2, 3, 3, 2, 1). A similar equation also holds for h, namely, h 2 = F; χ R C 6 , where χ R = (6,5,4,4,5,6) is the colouring obtained by doubling the 'right half' of the rainbow cycle. We have therefore bounded F; χ C 6 from above by the geometric mean of two functions of the same form, but simpler in the sense that they both contain fewer colours.…”
Section: A Motivating Examplementioning
confidence: 84%
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“…where χ L is the colouring obtained by doubling the 'left half' of the rainbow cycle, represented by the vector (1, 2, 3, 3, 2, 1). A similar equation also holds for h, namely, h 2 = F; χ R C 6 , where χ R = (6,5,4,4,5,6) is the colouring obtained by doubling the 'right half' of the rainbow cycle. We have therefore bounded F; χ C 6 from above by the geometric mean of two functions of the same form, but simpler in the sense that they both contain fewer colours.…”
Section: A Motivating Examplementioning
confidence: 84%
“…Let us mention another example of a norming graph. Following [6], we say that a graph is a K 2,t -replacement of H if each edge of H is replaced with a copy of K 2,t by identifying the two vertices of the edge with the two vertices on the smaller side of K 2,t . Finally, we remark that one may presumably use the exceptional reflection groups E 6 , E 7 , and E 8 to build some more exotic (weakly) norming graphs, though we have not pursued this further.…”
Section: Proof Of Theorem 12mentioning
confidence: 99%
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“…Combining this with Claim 8.11 (2), we see that again A k = R k ∪X ⊆ A * k . Then Claim 8.11 (3) implies that for all i ∈ [k−1] we have A * σ(i) \A i ⊆ A k ⊆ A * k and so A * σ(i) ⊆ A i . By Claim 8.11 (4), there is j…”
Section: 2mentioning
confidence: 99%