Turkish Journal of Computer and Mathematics Education Vol.12 No.2 (2021), 1011-1013
1011
Research Article
Simplicial Eulerian Lattices
G.Sheeba Merlina, K. Rebecca Jebaseeli Ednaa and V.Jemmy Joyca
aAssistant Professor in Department of Mathematics, Karunya Institute of Technology and Sciences, Coimbatore, India. Article History: Received: 11 January 2021; Accepted: 27 February 2021; Published online: 5 April 2021
______________________________________________________________________________________________________ AbstractIn this paper we prove that the lattice Ms (Bn) is simplicial and also we prove that S (Cn) and Sm(Cn)
Is simplicial. Keywords:
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1. Introduction
The lattice of convex sublattices of a Boolean algebra Bn of rank n,with respect to the set inclusion relation is a dually simplicial Eulerian Lattice(Alizadeh, 2019; Balamurugan, 2018).This has motivated us to look into similar property. As found in the thesis of Santhi.V.K[1],Sm(Bn) is defined as Sm(Bn) = (Bm̅̅̅̅̅ × B̅̅̅ ) ∪n {1,1}. where B̅̅̅̅ =Bm\{1} m
And B̅̅̅=Bn\{1} where Bnn is a Boolean of rank n and Bm is a Boolean of rank m.We prove that the lattice Sm(Bn) is simplicial for all m and n.The notation of Sm(Bn) is just Sg(Bm, Bn) as used in (Prasad, 2014; Arunkarthikeyan, 2020; Pavan 2020).
In a similar manner we also prove that S(Cn) and Sm(Cn) are simplicial but not dually simplicial.
1.1 Definition: (Simplicial Poset)
A Poset P with 0 is said to be simplicial if for every element t∈ 𝑃,[0,t] is Boolean. Dual simplicial poset is defined dually.
2. Theorem 1:
The lattice Sm(Bn) is simplicial. 2.1 Proof:
Let us prove the theorem by induction on i.Let𝑡 ∈ 𝑆𝑚(𝐵𝑛), the rank of t in Sm(Bn).
Let rank(t)=I,where 1 <I < m+n-1.When i=1,rank of t is 1.[0,t] is the one element chain. So it is Boolean.
We are going to prove that the interval [0,t] contains exactly icoatoms.
We know that t is of the form t=(a,b) where 𝑎 ∈ 𝐵̅̅̅̅ and 𝑏 ∈ 𝐵𝑛𝑚 ̅̅̅ (Garikipati, 2021) The following cases arise.
Case(1)
When rank(a)=i and rank(b)=0,then there are exactly i edges going immediately down a. Case(2)
When rank(a)=i-1 then rank(b)=1 Case(3)
When rank(a)=i-2 then rank(b)=2.
G. Sheeba Merlin, K. Rebecca Jebaseeli Edna, V.Jemmy Joyce
1012
In case(1) there are exactly i edges going immediately down a and since rank of b=0,there is no edge going down from b.
Therefore the number of edges going down from t is exactly i. Therefore [0,t] contains exactly I coatoms.
Therefore [0,t] is Boolean.((Santhi, 1992; Arunkarthikeyan, 2021) Lemma)
Similarly in all the cases from case (2) onwards we can prove that the number of edges going down from t is exactly i and so [o,t] contains exactly I coatoms.
Therefore [0,t] is Boolean in all the cases. Therefore the lattice Sm(Bn) is simplicial. 3.Theorem:2
The Lattice S(Cn) is Simplicial. 3.1 Proof:
We observed that the rank of S(Cn) is 4.
Let 𝑡 ∈ S(Cn)
If rank of t ≤ 2, then [0,t]≈ either B1 or B2.so [0,t] is Boolean.
If rank of t=3,then It is form t=(a,b) where 𝑎 ∈ 𝐵2̅̅̅ and b∈ 𝐶𝑛̅̅̅ and rank of a=1 and rank of b=2. There is only one edge going down in 𝐵̅̅̅ and are exactly 3 edges going down t in S(Cn). 2 Therefore [0.t]∈ 𝐵3.So S(Cn) is simplicial.
4.Theorem 3:
The lattice Sm(Cn) is simplicial. 4.1 Proof:
We observe that the rank of Sm(Cn) is m+2.
It is obvious that [0,t] is Boolean if rank of t is ≤ 2 in Sm(Cn) .
So,let t∈ Sm(Cn) with rank of t≥ 3.
t is of the form,t=(a,b) where 𝑎 ∈ 𝐵̅̅̅̅ and 𝑏 ∈ 𝐶𝑛𝑚 ̅̅̅ Let rank of t=I,wherei≥ 3.
Therefore,we have the following cases. Case(i) rank of a=i and rank of b=0. Case(ii) rank of a=i-1 an d rank of b=1 Case(iii) rank of a=i-2 and rank of b=2 Consider the first case:
Since rank of a is I,there are I edges going down from a in 𝐵𝑚̅̅̅̅ , which when combined with 0,the rank of b,we get the number of edges going down t in Sm(Cn) is i+0=i.So, [0,t]
Contains exactly icoatoms implying that [0,t] is Boolean of rank i. Similar arguments hold for the cases (ii) and (iii)
Simplicial Eulerian Lattices
1013 So Sm(Cn) is simplicial.
5.Conclusion:
To decide on the simplicial property of Sm(L),for a general Eulerian Lattice L,seems to be difficult. So it remains an open problem.
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