# Upper bound for an expression for distributive lattices

Let $$L$$ be a finite distributive lattice with minimum $$0$$ and Maximum $$1$$ and join-irreducible elements $$j_1,...,j_l$$ and meet irreducible elements $$m_1,...,m_l$$. Let $$J_L:= \sum\limits_{i=1}^{l}{| [j_i,1]|}$$ and $$M_L:= \sum\limits_{i=1}^{l}{| [0,m_i]|}$$. Set $$X_L$$:= $$l-|J_L-M_L|$$.

Question 1: Is there an easy proof that $$X_L >0$$ ? This would prove Frankl's conjecture for distributive lattices (which is already known).

Question 2: What are the distributive lattices with $$X_L=l$$? Their number starts for $$n \geq 3$$ with 1,2,1,3,2,7,4. Examples are Boolean lattices.

Question 3: Let $$U_n:= \sum\limits_{L \in \mathcal{L}_n}^{}{|J_L-M_L|}$$, where $$\mathcal{L}_n$$ is the set of all distributive lattices on $$n$$ elements. $$U_n/2$$ starts for $$n \geq 3$$ with 0,0,1,2,6,12,34 (could it be https://oeis.org/A088808 ?). What is $$U_n$$?

• It seems to me that there can't be any good description of lattices with $X_L = \ell$. We have $X_L = \ell$ whenever $L$ is isomorphic to its opposite lattice. If $P$ is any finite poset isomorphic to its opposite poset, then the lattice of lower ideal of $P$ is a lattice isomorphic to its opposite lattice. And there are tons of posets isomorphic to their opposites. – David E Speyer Jan 15 at 14:20
• I'm also skeptical of a good answer to Q3, because the cardinality of $\mathcal{L}_n$ is intractable and therefore I would guess that summing more complicated quantities over it is as well. I realize this isn't a strong argument. – David E Speyer Jan 15 at 14:22

The proposed inequality $$X_L>0$$ is false. Using Birkhoff's representation theorem, identify $$L$$ with the lattice of lower ideals in a poset $$P$$. Then $$|P| = \ell$$.

The join irreducible elements are the principal lower ideals $$(p)$$ for $$p \in P$$. Given a join irreducible element $$j=(p)$$, the interval $$[j,1]$$ is the set of ideals $$I$$ containing $$p$$. Thus, $$J_L = \sum_{p \in P} \#\{ I : I \ni p \} = \sum_{I \in L} \#(I).$$ Similarly, $$M_L = \sum_{I \in L} (\ell - \#(I)).$$ So the proposed inequality is $$\left| \ell \#(L) - 2 \sum_{I \in L} \#(I) \right| < \ell.$$

Let the poset $$P$$ have $$a+b$$ elements $$x_1$$, $$x_2$$, ..., $$x_a$$, $$y_1$$, $$y_2$$, ..., $$y_b$$ and the relations that the $$x$$'s are incomparable, the $$y$$'s form a chain $$y_1 < y_2 < \cdots < y_b$$ and $$x_i < y_j$$ for all $$i$$, $$j$$. So the lower ideals are of two types:

(1) Any subset of the $$x$$'s.

(2) $$\{x_1, x_2, \ldots, x_a, y_1, y_2, \ldots, y_k \}$$ for $$1 \leq k \leq b$$.

We have $$\ell = a+b$$, $$|L| = 2^a + b$$ and $$\sum_{I \in L} \#(I) = (a/2) 2^a + b (a+b/2+1/2)$$. So $$\ell \#(L) - \sum_{I \in L} \#(I) = (a+b)(2^a+b) - (a 2^a + b(2a+b+1))$$ $$=b 2^a - b(a+1).$$

If we choose $$a$$ and $$b$$ roughly the same size, then $$|b 2^a - b(a+1)|$$ will be much larger than $$a+b$$.