combinatorics / Quasirandom graphs, graph limits

The forcing conjecture for bipartite graphs

Corollary 1.3: for the same connected bipartite HH (which contains cycles) there are p∈(0,1)p\in(0,1) and a nonconstant graphon WW with t(K2,W)=pt(K_2,W)=p and t(H,W)=p66t(H,W)=p^{66}; sampling gives finite graphs with edge and HH-densities tending to pp and p66p^{66} but four-cycle density tending to a limit above p4p^4. So HH is not pp-forcing at that pp, and the conjecture fails. Since the forcing conjecture implies Sidorenko's, this follows from the Sidorenko counterexample; no separate idea is claimed.

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combinatoricsSep 23, 2026Significance 28/100Registry: unreviewed

The forcing conjecture for bipartite graphs

Prior state unknown→disproved

Corollary 1.3: for the same connected bipartite HH (which contains cycles) there are p∈(0,1)p\in(0,1) and a nonconstant graphon WW with t(K2,W)=pt(K_2,W)=p and t(H,W)=p66t(H,W)=p^{66}; sampling gives finite graphs with edge and HH-densities tending to pp and p66p^{66} but four-cycle density tending to a limit above p4p^4. So HH is not pp-forcing at that pp, and the conjecture fails. Since the forcing conjecture implies Sidorenko'…

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Corollary 1.3: for the same connected bipartite HH (which contains cycles) there are p∈(0,1)p\in(0,1) and a nonconstant graphon WW with t(K2,W)=pt(K_2,W)=p and t(H,W)=p66t(H,W)=p^{66}; sampling gives finite graphs with edge and HH-densities tending to pp and p66p^{66} but four-cycle density tending to a limit above p4p^4. So HH is not pp-forcing at that pp, and the conjecture fails. Since the forcing conjecture implies Sidorenko's, this follows from the Sidorenko counterexample; no separate idea is claimed.

Corollary 1.3: for the same connected bipartite HH (which contains cycles) there are p∈(0,1)p\in(0,1) and a nonconstant graphon WW with t(K2,W)=pt(K_2,W)=p and t(H,W)=p66t(H,W)=p^{66}; sampling gives finite graphs with edge and HH-densities tending to pp and p66p^{66} but four-cycle density tending to a limit above p4p^4. So HH is not pp-forcing at that pp, and the conjecture fails. Since the forcing conjecture implies Sidorenko's, this follows from the Sidorenko counterexample; no separate idea is claimed.

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