EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 10, No. 4, 2017, 916-928 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global On the Lattice of Convex Sublattices of S(Bn) and S(Cn) G. Sheeba Merlin1,∗, A. Vethamanickam2 1 Department of Mathematics, Karunya University, Coimbatore, India 2 Department of Mathematics, Rani Anna College for Women, Tirunelveli, India Abstract. In this paper we prove that CS[S(Bn)] and CS[S(Cn)] are Eulerian lattices under the set inclusion relation but they are neither simplicial nor dual simplicial. 2010 Mathematics Subject Classifications: 06A06, 06A07, 06B10 Key Words and Phrases: Lattices, Convex Sublattices, Dual Simplicial Lattices, Eulerian Lattices 1. Introduction The study of lattice of convex sublattices of a lattice was started by K. M. Koh[3], in the year 1972. He investigated the internal structure of a lattice L, in relation to CS(L), like so many other authors for various algebraic structures such as groups, Boolean algebras, directed graphs and so on. In [3], several basic properties of CS(L) have been studied where one of the results proved is “If L is complemented then CS(L) is complemented”. Also, the connection of the structure of CS(L) with those of the ideal lattice I(L) and the dual ideal lattice D(L) are examined by K. M. Koh. He also derived the best lower bound and upper bound for the cardinality of CS(L), where L is finite. In a subsequent paper[1], Chen C. K., Koh K. M., proved that CS(L×K) ∼= [(CS(L)− {∅})× (CS(K)− {∅})] ∪ {∅}. Finally they proved that when L is a finite lattice and CS(L) ∼= CS(M) and if L is rela- tively complemented(complemented) then M is relatively complemented(complemented). This is true for Eulerian lattices, since an Eulerian lattice is relatively complemented. These results gave motivation for us to look into the connection between L and CS(L) for Eulerian lattices which are a class of lattices not defined by identities. A construction of a new Eulerian lattice S(Bn) from a Boolean algebra Bn of rank n is found in the thesis ∗Corresponding author. Email addresses: sheebamerlin@karunya.edu (G. Sheeba Merlin), dr vethamanickam@yahoo.co.in (A. Vethamanickam) http://www.ejpam.com 916 c© 2017 EJPAM All rights reserved. G. Sheeba Merlin, A. Vethamanickam / Eur. J. Pure Appl. Math, 10 (4) (2017), 916-928 917 of V. K. Santhi in 1992[11]. In 2012, Subbarayan had proved in his paper that the lattice of convex sublattices of a boolean algebra Bn, of rank n, CS(Bn) with respect to the set inclusion relation is a dual simplicial Eulerian lattice. In this paper, we are going to look at the similar structure of CS(S(Bn)). S(B4) is shown in the following diagram. Figure 1: S(B2) Figure 2: S(B4) 2. Preliminaries Throughout this section CS(L) is equipped with the partial order of set inclusion rela- tion. G. Sheeba Merlin, A. Vethamanickam / Eur. J. Pure Appl. Math, 10 (4) (2017), 916-928 918 Definition 2.1. A finite graded poset P is said to be Eulerian if its Möbius function assumes the value µ(x, y) = (−1)l(x,y) for all x ≤ y in P , where l(x, y) = ρ(y)− ρ(x) and ρ is the rank function on P . An equivalent definition for an Eulerian poset is as follows: Lemma 2.2. [5] A finite graded poset P is Eulerian if and only if all intervals [x, y] of length l ≥ 1 in P contain an equal number of elements of odd and even rank. Example 2.3. Every Boolean algebra of rank n is Eulerian and the lattice C4 of Figure 2 is an example for a non-modular Eulerian lattice. Also, every Cn is Eulerian for n ≥ 4. Figure 3: Non-modular Eulerian lattice Lemma 2.4. [12] If L1 and L2 are two Eulerian lattices then L1 × L2 is also Eulerian. We note that any interval of an Eulerian lattice is Eulerian and an Eulerian lattice cannot contain a three element chain as an interval. Definition 2.5. A poset P is called Simplicial if for all t 6= 1 ∈ P , [0, t] is a Boolean algebra and P is called Dual Simplicial if for all t 6= 0 ∈ P , [t, 1] is a Boolean algebra. Lemma 2.6. [1] Let L and K be any two lattices. Then CS(L×K) u [(CS(L)− {∅})× (CS(K)− {∅})] ∪ {∅}. Lemma 2.7. [14] Let Bn be a Boolean lattice of rank n. Then CS(Bn) is a dual simplicial Eulerian lattice. 3. Convex Sublattices of S(Bn) Theorem 3.1. The lattice of convex sublattices of S(Bn), CS(S(Bn)) with respect to the set inclusion relation is an Eulerian lattice. Proof. It is clear that the rank of CS(S(Bn)) is n+ 2. We are going to prove that CS(S(Bn)) is Eulerian. That is, to prove that this interval [∅, Bn] has the same number of elements of odd and even rank. G. Sheeba Merlin, A. Vethamanickam / Eur. J. Pure Appl. Math, 10 (4) (2017), 916-928 919 b b b b b b b b b b b b b bbb b b b b b b b b b b b b b b b b b b b b {0} {w} {x} {y} {z} {p} {q} {r} {s} {1} {0, w} {0, x} {0, y} {0, z} {w, p} {w, q} {x, p} {x, r} {y, q} {y, s} {z, r} {z, s} {p, 1} {q, 1} {r, 1} {s, 1} {0, p} {0, q} {0, r} {0, s} {w, 1} {x, 1} {y, 1} {z, 1} S(B2) φ Figure 4: CS[S(B2)] Let Ai be the number of elements of rank i in CS(S(Bn)). A1 = The number of singleton subsets of CS[S(Bn)] = 2 + n+ 2 + 2n+ ( n 2 ) + 2 ( n 2 ) + ( n 3 ) + 2 ( n 3 ) + ( n 4 ) + . . .+ 2 ( n n− 2 ) + ( n n− 1 ) + 2 ( n n− 1 ) = 2 + ( n 1 ) + 2 ( n 0 ) + 2 ( n 1 ) + ( n 2 ) + 2 ( n 2 ) + ( n 3 ) +2 ( n 3 ) + ( n 4 ) + . . .+ 2 ( n n− 2 ) + ( n n− 1 ) + 2 ( n n− 1 ) (1) A2 = The number of rank 2 elements in CS(S(Bn)) = The number of edges in S(Bn) = number of edges containing 0 + number of edges containing the atoms +number of edges from the rank 2 elements + . . .+ number of edges containing the coatoms of S(Bn). Number of edges containing 0 =n+ 2 (2) G. Sheeba Merlin, A. Vethamanickam / Eur. J. Pure Appl. Math, 10 (4) (2017), 916-928 920 Number of edges containing an extreme atom =n There are 2 such extreme atoms. Therefore total number of such edges = 2 ( n 1 ) . From an atom of a middle copy, the number of edges = n− 1 + 2 = n+ 1. There are n such atoms. Therefore total number of such type of edges = n(n+ 1). Totally from the atoms, the number of edges is equal to 2 ( n 1 ) + ( n 1 ) (n+ 1). (3) Number of edges from a rank 2 element in an extreme copy = n− 1. There are 2n such elements. Therefore the number of edges from these elements = 2 ( n 1 ) (n− 1). The number of edges from the rank 2 elements in the middle copy = ( n 2 ) × (n − 2 + 2) = ( n 2 ) × n. The total number of edges from rank 2 elements is 2 ( n 1 ) (n− 1) + ( n 2 ) × n. (4) The number of edges from the rank 3 elements in the middle copy is n− 3 + 2 = n− 1. There are ( n 3 ) such elements. Therefore the number of edges from the rank 3 elements in the middle copy = ( n 3 ) (n− 1). The number of edges from a rank 3 element in an extreme copy is n− 2. There are 2 ( n 2 ) such elements. Therefore number of edges from rank 3 elements in the extreme copies = 2 ( n 2 ) (n− 2). Therefore total number of edges from rank 3 elements of CS(S(Bn)) is 2 ( n 2 ) (n− 2) + ( n 3 ) (n− 1) (5) Proceeding like this we get the number of edges from the co-atoms = 2n = 2 ( n n− 1 ) (n− n− 1) G. Sheeba Merlin, A. Vethamanickam / Eur. J. Pure Appl. Math, 10 (4) (2017), 916-928 921 From (2), (3), (4) and (5), the total number of edges in S(Bn) is A2 = n+ 2 + 2 ( n 1 ) + ( n 1 ) (n+ 1) + 2 ( n 1 ) (n− 1) + ( n 2 ) n+ ( n 3 ) (n− 1) + 2 ( n 2 ) (n− 2) + . . .+ 2 ( n n− 1 ) (n− n− 1) = 2 + ( n 1 ) + 2 ( n 1 ) + ( n 1 ) (n+ 2− 1) +2 ( n 1 ) (n− 1) + ( n 2 ) (n+ 2− 2) + 2 ( n 2 ) (n− 2) + ( n 3 ) (n+ 2− 3) + . . .+ 2 ( n n− 1 ) (n− n− 1) (6) A3 = The number of 4-element sublattices. The number of 4-element sublattices from 0 = 2 ( n 1 ) + ( n 2 ) . (7) Fix an atom a ∈ S(Bn). If a is the bottom element of the left copy of S(Bn) then [a, 1] ' Bn. Therefore the number of B2’s containing a is ( n 2 ) . Similarly the number of B2’s containing the bottom element of the right copy is ( n 2 ) . If a is in the middle copy of S(Bn) then [a, 1] ' S(Bn−1). In this S(Bn−1), we have two extreme copies and a middle copy. Therefore the number of B2’s containing a is 2(n− 1) + ( n− 1 2 ) . There are ( n 1 ) such atoms. Therefore the total number of B2’s containing all the atoms in the middle copy is ( n 1 )[ 2(n− 1) + ( n− 1 2 )] . Therefore the number of B2’s containing all the atoms of S(Bn) is 2 ( n 2 ) + ( n 1 )[ 2(n− 1) + ( n− 1 2 )] . (8) Fix a rank 2 element x in S(Bn). If x is in the left copy of S(Bn), we have, [x, 1] ' Bn−1. G. Sheeba Merlin, A. Vethamanickam / Eur. J. Pure Appl. Math, 10 (4) (2017), 916-928 922 A B2 containing x emanates from a rank 2 element in that Bn−1. There are ( n− 1 2 ) rank 2 elements in Bn−1. Therefore the number of B2’s containing x in the left copy is ( n− 1 2 ) . There are n such rank 2 elements x in the left copy. The number of B2’s in the left copy containing all the rank 2 elements is ( n− 1 2 ) n. Similarly the same number in the right copy. If x is in the middle copy of S(Bn), then [x, 1] ' S(Bn−2). The number of B2’s containing x in the left copy of that S(Bn−2) is n − 2. Similarly the number in the right copy is n− 2. Come to the middle copy ' Bn−2. Therefore the number of B2’s containing x in the middle copy of that S(Bn−2) is ( n− 2 2 ) . Therefore the total number of B2’s containing x in this S(Bn−2) is 2(n−2)+ ( n− 2 2 ) . There are ( n 2 ) such x’s. Therefore the total number of B2’s containing all the rank 2 elements in the middle copy is( n 2 )[ 2(n− 2) + ( n− 2 2 )] + 2 ( n 1 )( n− 1 2 ) . (9) Fix a rank 3 element x of S(Bn). If x is in the left copy of S(Bn), then [x, 1] ' Bn−2. A B2 containing x emanates from a rank 4 element in that Bn−2. Therefore the number of B2’s containing x in the left copy of S(Bn) is( n 2 )( n− 2 2 ) . Similarly to the right copy. Come to the middle copy. If x is in the middle copy of S(Bn), then ∴ [x, 1] ' S(Bn−3). In this S(Bn−3) we have to calculate the number of B2’s containing x. The number of B2’s containing x in the left copy of S(Bn−3) is n− 3. Similarly the number in the right copy of this S(Bn−3) is n− 3. The number of B2’s containing x in the middle copy of this S(Bn−3) is ( n− 3 2 ) . G. Sheeba Merlin, A. Vethamanickam / Eur. J. Pure Appl. Math, 10 (4) (2017), 916-928 923 Therefore, the number of B2’s in this S(Bn−2) containing x is 2(n− 3) + ( n− 3 2 ) . There are ( n 3 ) such rank 3 elements in x in the middle copy of S(Bn). Therefore the total number of B2’s containing x in S(Bn) is( n 3 )[ 2(n− 3) + ( n− 3 2 )] . Therefore the total number of B2’s containing all the rank 3 elements is 2 ( n 2 )( n− 2 2 ) + ( n 3 )[ 2(n− 3) + ( n− 3 2 )] (10) Continuing like this, we get, the number of B2’s containing all the rank (n−2) elements in S(Bn) is 2 ( n n− 3 ) × 3 + ( n n− 2 ) × 4. (11) The number of B2’s containing rank (n− 1) elements is 2 ( n n− 2 ) + ( n n− 1 ) . (12) From (7), (8), (9), (10), (11) and (12) we get, A3 = 2 ( n 1 ) + ( n 2 ) + 2 ( n 2 ) + ( n 1 )[ 2(n− 1) + ( n− 1 2 )] + ( n 2 )[ 2(n− 2) + ( n− 2 2 )] + 2 ( n 1 )( n− 1 2 ) + 2 ( n 2 )( n− 2 2 ) + ( n 3 )[ 2(n− 3) + ( n− 3 2 )] + . . . + 2 ( n n− 3 ) × 3 + ( n n− 2 ) × 4 + 2 ( n n− 2 ) + ( n n− 1 ) . That is, A3 = 2 ( n 1 ) + ( n 2 ) + 2 ( n 2 ) + 2 ( n 1 )( n− 1 1 ) + ( n 1 )( n− 1 2 ) + 2 ( n 1 )( n− 1 2 ) + 2 ( n 2 )( n− 2 1 ) + ( n 2 )( n− 2 2 ) + 2 ( n 2 )( n− 2 2 ) + 2 ( n 3 )( n− 3 1 ) + ( n 3 )( n− 3 2 ) + . . .+ 2 ( n n− 2 ) + ( n n− 1 ) . (13) G. Sheeba Merlin, A. Vethamanickam / Eur. J. Pure Appl. Math, 10 (4) (2017), 916-928 924 Similar argument will give, A4 = the number of rank 3 sublattices. A4 = [ 2 ( n 2 ) + ( n 3 )] + 2 ( n 3 ) + ( n 1 )[ 2 ( n− 1 2 ) + ( n− 1 3 )] + ( n 2 )[ 2 ( n− 2 2 ) + ( n− 2 3 )] + 2 ( n 2 )( n− 2 3 ) + 2 ( n 1 )( n− 1 3 ) + ( n 3 )[ 2 ( n− 3 2 ) + ( n− 3 3 )] + . . .+ 2 ( n n− 3 ) + ( n n− 2 ) . That is, A4 = 2 ( n 2 ) + ( n 3 ) + 2 ( n 2 ) + 2 ( n 1 )( n− 1 2 ) + ( n 1 )( n− 1 3 ) + 2 ( n 1 )( n− 1 3 ) + 2 ( n 2 )( n− 2 2 ) + ( n 2 )( n− 2 3 ) + 2 ( n 2 )( n− 2 3 ) + 2 ( n 3 )( n− 3 2 ) + ( n 3 )( n− 3 3 ) + . . .+ 2 ( n n− 3 ) + ( n n− 2 ) . (14) A5 = the number of rank 4 sublattices. A5 = 2 ( n 3 ) + ( n 4 ) + 2 ( n 4 ) + 2 ( n 1 )( n− 1 3 ) + ( n 1 )( n− 1 4 ) + 2 ( n 1 )( n− 1 4 ) + 2 ( n 2 )( n− 2 3 ) + ( n 2 )( n− 2 4 ) + 2 ( n 2 )( n− 2 4 ) + 2 ( n 3 )( n− 3 3 ) + ( n 3 )( n− 3 4 ) + . . .+ 2 ( n n− 4 ) + ( n n− 3 ) (15) and so on. Finally, we get An = 2 ( n n− 2 ) + ( n n− 1 ) + 2 ( n n− 1 ) + 2 ( n 1 )( n− 1 n− 2 ) . (16) An+1 = 2 ( n n− 1 ) + (n+ 2). (17) G. Sheeba Merlin, A. Vethamanickam / Eur. J. Pure Appl. Math, 10 (4) (2017), 916-928 925 C ase (i): Suppose n is even. A1 −A2 +A3 −A4 + . . .+An+1 = ( n 0 ) [2 + 2− 2] + ( n 1 )[ 1 + 2− 1− 2− n− 2 + 1− 2 ( n− 1 1 ) + 2 + 2 ( n− 1 1 ) + ( n− 1 2 ) + 2 ( n− 1 2 ) − 2 ( n− 1 2 ) − ( n− 1 3 ) − 2 ( n− 1 3 ) + 2 ( n− 1 3 ) + ( n− 1 4 ) + 2 ( n− 1 4 ) + . . . + ( n− 1 n− 1 )] + ( n 2 )[ 1 + 2− n− 2 + 2 ( n− 2 1 ) + 1 + 2 + 2 ( n− 2 1 ) + ( n− 2 2 ) + 2 ( n− 2 2 ) − 2− 2 ( n− 2 2 ) − ( n− 2 3 ) − 2 ( n− 2 3 ) + 2 ( n− 2 3 ) + ( n− 2 4 ) + 2 ( n− 2 4 ) + . . .+ ( n− 2 n− 2 )] + ( n 3 )[ 1 + 2− n− 2 + 3 + 2 ( n− 3 1 ) + ( n− 3 2 ) − 1− 2− 2 ( n− 3 2 ) + 2 − ( n− 3 3 ) + 1 + 2 ( n− 3 3 ) + ( n− 3 n− 3 )] + . . .− ( n n− 1 )[ 1 + 2− 2 − 1 + 1− 1− 2 + 2− ( n− n− 1 n− n− 1 )] + ( n n ) [2] =2[2n−1]− [( n 1 ) + ( n 2 ) + . . .+ ( n n− 1 )] =2n − [2n − 2] =2. C ase (ii): Suppose n is odd. A1 −A2 +A3 −A4 + . . .−An+1 = ( n 0 ) [2 + 2− 2] + ( n 1 )[ 1 + 2− 1− 2− n− 2 + 1− 2 ( n− 1 1 ) + 2 + 2 ( n− 1 1 ) + ( n− 1 2 ) + 2 ( n− 1 2 ) − 2 ( n− 1 2 ) − ( n− 1 3 ) − 2 ( n− 1 3 ) + 2 ( n− 1 3 ) + ( n− 1 4 ) + 2 ( n− 1 4 ) + . . . + ( n− 1 n− 1 )] + ( n 2 )[ 1 + 2− n− 2 + 2 ( n− 2 1 ) + 1 + 2 + 2 ( n− 2 1 ) + ( n− 2 2 ) + 2 ( n− 2 2 ) − 2− 2 ( n− 2 2 ) − ( n− 2 3 ) − 2 ( n− 2 3 ) G. Sheeba Merlin, A. Vethamanickam / Eur. J. Pure Appl. Math, 10 (4) (2017), 916-928 926 + 2 ( n− 2 3 ) + ( n− 2 4 ) + 2 ( n− 2 4 ) + . . .+ ( n− 2 n− 2 )] + ( n 3 )[ 1 + 2− n− 2 + 3 + 2 ( n− 3 1 ) + ( n− 3 2 ) − 1− 2− 2 ( n− 3 2 ) + 2 − ( n− 3 3 ) + 1 + 2 ( n− 3 3 ) + ( n− 3 n− 3 )] + . . .− ( n n− 1 )[ 1 + 2− 2 − 1 + 1− 1− 2 + 2− ( n− n− 1 n− n− 1 )] − ( n n ) [2] =2 [( n 0 ) + ( n 2 ) + . . .+ ( n n− 1 )] − [( n 1 ) + ( n 2 ) + . . . + ( n n− 1 )] =2[2n−1 − 1]− [2n − 2] =0. Hence the interval [∅, S(Bn)] has the same number of elements of odd and even rank. Though in the above theorem we have proved that CS(S(Bn)) is Eulerian, it is not dual simplicial. For example, CS(S(B2)) itself is not dual simplicial. For a general non- Boolean Eulerian lattice it seems difficult to decide the structure, but for Cn we give the proof in the next section. 4. Convex Sublattices of S(Cn) Theorem 4.1. The lattice of convex sublattices of S(Cn) with respect to the set inclusion relation is an Eulerian lattice. Proof. We are going to prove that CS(S(Cn)) is Eulerian. That is to prove the interval [5, S(Cn)] has the same number of elements of odd and even rank. Let Ai be the number of elements of rank i in CS(S(Cn)). A1 = The number of singleton subsets of CS(S(Cn)) = 1 + n+ 2 + 3n+ 2n+ 1 = 6n+ 4. (18) A2 = The number of rank 2 elements in CS(S(Cn)) = 2 + n+ 2n+ 4n+ 4n+ 2n+ 2n = 15n+ 2. (19) A3 = The number of 4-element sublattices REFERENCES 927 b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b 1 0 Figure 5: S(C5) = 2n+ n+ 2n+ 2n+ 2n+ 2n+ n = 12n. (20) A4 = The number of rank 3 sublattices = 2n+ n+ 2 = 3n+ 2. (21) Therefore, A1 −A2 +A3 −A4 = 6n+ 4− 15n− 2 + 12n− 3n− 2 = 0. Hence the interval [5, S(Cn)] has a same number of elements of odd and even rank. References [1] Chen C. K., Koh K. M., On the lattice of convex sublattices of a finite lattice, Nanta Math., 5 (1972), 92–95. [2] Gratzer G., General Lattice Theory, Birkhauser Verlag, Basel, 1978. [3] Koh K. M., On the lattice of convex sublattices of a finite lattice, Nanta Math., 5 (1972), 18–37. [4] Lavanya S., Parameshwara Bhatta S., A new approach to the lattice of convex sublat- tices of a lattice, Algebra Univ., 35 (1996), 63–71. 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