Generalized Forbidden Subposet Problems

Abstract

A subfamily {F_{1},F_{2},dots ,F_{|P|}}subseteq mathcal {F} of sets is a copy of a poset P in mathcal {F} if there exists a bijection phi :Prightarrow {F_{1},F_{2},dots ,F_{|P|}} such that whenever x le _{P} x^{prime } holds, then so does phi (x)subseteq phi (x^{prime }). For a family mathcal {F} of sets, let c(P,mathcal {F}) denote the number of copies of P in mathcal {F}, and we say that mathcal {F} is P-free if c(P,mathcal {F})=0 holds. For any two posets P, Q let us denote by La(n, P, Q) the maximum number of copies of Q over all P-free families mathcal {F} subseteq 2^{[n]}, i.e. max limits {c(Q,mathcal {F}): mathcal {F} subseteq 2^{[n]}, c(P,mathcal {F})=0 }. This generalizes the well-studied parameter La(n, P) = La(n, P, P1) where P1 is the one element poset, i.e. La(n, P) is the largest possible size of a P-free family. The quantity La(n, P) has been determined (precisely or asymptotically) for many posets P, and in all known cases an asymptotically best construction can be obtained by taking as many middle levels as possible without creating a copy of P. In this paper we consider the first instances of the problem of determining La(n, P, Q). We find its value when P and Q are small posets, like chains, forks, the N poset and diamonds. Already these special cases show that the extremal families are completely different from those in the original P-free cases: sometimes not middle or consecutive levels maximize La(n, P, Q) and sometimes the extremal family is not the union of levels. Finally, we determine (up to a polynomial factor) the maximum number of copies of complete multi-level posets in k-Sperner families. The main tools for this are the profile polytope method and two extremal set system problems that are of independent interest: we maximize the number of r-tuples A_{1},A_{2},dots , A_{r} in mathcal {A} over all antichains mathcal {A}subseteq 2^{[n]} such that (i) cap _{i=1}^{r}A_{i}=emptyset , (ii) cap _{i=1}^{r}A_{i}=emptyset and cup _{i=1}^{r}A_{i}=[n].