Abstract

Four distinct elements a, b, c, and d of a poset form a diamond if \(a< b<d\) and \(a<c<d\). A subset of a poset is diamond-free if no four elements of the subset form a diamond. Even in the Boolean lattices, finding the size of the largest diamond-free subset remains an open problem. In this paper, we consider the linear lattices—poset of subspaces of a finite dimensional vector space over a finite field of order q—and extend the results of Griggs et al. (J. Combin. Theory Ser. A 119(2):310–322, 2012) on the Boolean lattices, to prove that the number of elements of a diamond-free subset of a linear lattice can be no larger than \(2+\frac {1}{q+1}\) times the width of the lattice, so that this fraction tends to 2 as \(q \longrightarrow \infty \). In addition, using an algebraic technique, we introduce so-called diamond matchings, and prove that for linear lattices of dimensions up to 5, the size of a largest diamond-free subset is equal to the sum of the largest two rank numbers of the lattice.

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