Vol. 20 (2019), No. 2, pp. 839–848 DOI: 10.18514/MMN.2019.2821
A NOTE ON LATTICES WITH MANY SUBLATTICES
G ´ABOR CZ ´EDLI AND ESZTER K. HORV ´ATH Received 20 January, 2019
Abstract. For every natural numbern5, we prove that the number of subuniverses of ann- element lattice is2n,132n 4,232n 5, or less than232n 5. Also, we describe then-element lattices with exactly2n,132n 4, or232n 5subuniverses.
2010Mathematics Subject Classification: 06B99
Keywords: finite lattice, sublattice, number of sublattices, subuniverse
1. INTRODUCTION AND OUR RESULT
For a latticeL, Sub.L/will denote itssublattice lattice. In spite of this standard terminology, Sub.L/consists of allsubuniversesofL. That is, a subsetX ofLis in Sub.L/iffX is closed with respect to join and meet. In particular,¿2Sub.L/. All lattices occurring in this paper will be assumed to befiniteeven if this is not always emphasized. For a natural numbern2NCWD f1; 2; 3; : : :g, let
NS.n/WD fjSub.L/j WLis a lattice of sizejLj Dng:
That is,k2NS.n/if and only if somen-element lattice has exactlyksubuniverses.
Although the acronym NS comes from Number of Sublattices,Lhas onlyjSub.L/j 1 sublattices. If K andL are finite lattices, then theirglued sum KCgluL is the ordinal sum of the posetsKn1K, the singleton lattice, andLn f0Lg, in this order. In other words, we putLatopKand identify the elements1Kand0L; see Figure1. For example, if each ofK andLis the two-element chain, then KCgluLis the three- element chain. Note that Cglu is an associative but not a commutative operation.
The following fact is trivial:
The largest number in NS.n/ is 2nD322n 5. Furthermore, ann-element latticeLhas exactly2n subuniverses if an only ifLis a chain.
(1.1)
This research was supported by the Hungarian Research Grants KH 126581.
c 2019 Miskolc University Press
FIGURE1. The glued sumKCgluLofK andL
Thefour-element boolean latticeB4and thepentagon(lattice)N5are given in Fig- ure2, together with other important lattices. Fork2NC, thek-element chain will be denoted byC.k/. Our goal is to prove the following result:
Theorem 1. If5n2NC, then the following three assertions hold.
(1) The second largest number inNS.n/is262n 5. Furthermore, ann-element latticeL has exactly262n 5 subuniverses if and only ifLŠC1CgluB4
CgluC2, whereC1andC2are chains.
(2) The third largest number inNS.n/is232n 5. Furthermore, ann-element latticeLhas exactly 232n 5 subuniverses if and only if LŠC0CgluN5
CgluC1, whereC0andC1are chains.
FIGURE 2. Lattices for Theorem1and Lemma2
Since NS.n/D f2ngforn2 f1; 2; 3gand NS.4/D f13; 16g, we have formulated this theorem only forn5. To make the comparison of the numbers occurring in the paper easier, we often givejSub.L/jas a multiple of2jLj 5. Next, we repeat the first sentence of the Abstract, which is a trivial consequence of Theorem1.
Corollary 1. For5n2NC, the number of subuniverses of ann-element lattice is2n,132n 4,232n 5, or less than232n 5.
Remark 1. Let Con.L/and NC.n/ stand for the lattice of congruences of a lat- ticeLandfjCon.L/j WLis a lattice withjLj Dng, respectively. Forn5, the five largest numbers in NC.n/are162n 5,82n 5,52n 5,42n 5, and3:52n 5by Freese [4], Cz´edli [2] and, mainly, Kulin and Mures¸an [6].
Remark2. Interestingly, the first three of the five numbers mentioned in Remark1 are witnessed exactly by the same lattices that occur in Theorem1and (1.1). How- ever, we will show at the end of Section2that
jSub.N5CgluC.3//j D2327 5> 21:2527 5 D jSub.B4CgluB4/j> 1927 5
D jSub..C.2/C.3//CgluC.2//j; (1.2) which indicates that jSub..C.2/C.3//CgluC.2//jis not the fourth largest num- ber in NS.7/, although we know from Kulin and Mures¸an [6] that jCon..C.2/
C.3//CgluC.2//jis the fourth largest number in NC.7/.
While there are powerful tools to determine the first few large numbers in NC.n/, see the above-mentioned papers and, for additional tools, Cz´edli [3], the analogous task for NS.n/ seems to be more tedious. This together with Remark2are our ex- cuses that we do not determine the fourth and fifth largest numbers in NS.n/.
The rest of the paper is devoted to the proof of Theorem1.
2. TWO PREPARATORY LEMMAS
Our notation and terminology is standard, see, for example, Gr¨atzer [5]. However, we recall some notation and introduce some auxiliary concepts. For elements u; v in a lattice L, the interval Œu; vWD fx2LWux vg is defined only if uv, but thesublatticeŒfu; vggenerated byfu; vgalways makes sense. In order to avoid confusion, the curly brackets are never omitted fromŒfa1; : : : ; akgwhen a generated sublattice is mentioned. Foru2L, the principal ideal and the principal filter gener- ated byuare#uWD fx2LWxugand"uWD fx2LWuxg, respectively. We can also write#Luand"Lvto specify the latticeL. Foru; v2L, we writeukvifuand v areincomparable, that is,u6v andv6u. We say thatuisjoin-irreducibleifu has at most one lower cover; note that0D0Lis join-irreducible by our convention.
Meet-irreducibilityis defined dually, and an element isdoubly irreducibleif it is both join-irreducible and meet-irreducible. Next, let us call an elementu2Lisolatedifu is doubly irreducible andLD #u[ "u. That is, ifuis doubly irreducible andxku holds for nox2L. Finally, an intervalŒu; vwill be called anisolated edgeif it is a prime interval, that is,uv, andLD #u[ "v.
Lemma 1. IfK is a sublattice andH is a subset of a finite latticeL, then the following three assertions hold.
(1) With the notationt WD jfH\S WS 2Sub.L/gj, we have that jSub.L/j t2jLj jHj.
(2) jSub.L/j jSub.K/j 2jLj jKj.
(3) Assume, in addition, thatKhas neither an isolated element, nor an isolated edge. ThenjSub.L/j D jSub.K/j 2jLj jKj if and only ifL is (isomorphic to)C0CgluKCgluC1for some chainsC0andC1.
Proof. With respect to the map'WSub.L/! fH\SWS 2Sub.L/g, defined by X 7!H\X, eachY 2 fH\SWS2Sub.L/ghas at most2jLj jHjpreimages. This yields part (1). Clearly, (1) implies (2). The argument above yields a bit more than stated in (1) and (2); namely, for later reference, note the following.
IfjSub.L/j D jSub.K/j 2jLj jKj, then for every S2Sub.K/and every subsetX ofLnK, we have thatS[X 2Sub.L/.
(2.1) Next, we claim that for an elementu2L,
uis isolated if and only if for everyX 2Sub.L/, we have
thatX[ fug 2Sub.L/andXn fug 2Sub.L/. (2.2) Assume thatuis isolated andX 2Sub.L/. Sinceuis doubly irreducible,Xn fug 2 Sub.L/. Sinceuis comparable with all elements ofX,X[ fug 2Sub.L/, proving the “only if” part of (2.2). To show the converse, assume thatuis not isolated. Ifuis not doubly irreducible, thenuDa_bwitha; b < uor dually, andXWD fa; b; u; a^ bg 2Sub.L/butXn fug …Sub.L/. Ifukvfor somev2L, thenfvg 2Sub.L/but fvg[fug …Sub.L/. This proves the “if” part, and (2.2) has been verified.
Next, to prove part (3), assume that K has neither an isolated element, nor an isolated edge. First, letLDC1CgluKCgluC2. Since everyuinLnKis clearly an isolated element ofL, it follows from a repeated application of (2.2) that whenever XLnKandS2Sub.K/, thenS[X2Sub.L/. SinceLnKhas2jLj jKjsubsets, jSub.L/j jSub.K/j 2jLj jKj, and we obtain the required equality by the converse inequality given in part (2).
Conversely, assume the equality given in (3). Let x be an arbitrary element of LnK. Applying (2.1) tof0Kg 2Sub.K/ andf1Kg 2Sub.K/, we obtain that both f0K; xgandf1K; xgare in Sub.L/, whence neitherxk0K, norxk1K. So exactly one of the cases0K< x < 1K,x < 0K, and1K< x holds; we are going to exclude the first one. Suppose for a contradiction that0K< x < 1K. Thenxis comparable to everyy2K, because otherwiseSWD fygandfxgwould violate (2.1). By finiteness, we can takeuWDW
.K\ #x/andvWDV
.K\ "x/. Now ify2K, then eithery > x and so y 2 "Kv, or y < x and so y 2 #Ku, which means that K D #Ku[ "Kv.
Hence,Œu; vK is a prime interval ofK, and so it is an isolated edge ofK. This is a contradiction, which excludes that0K< x < 1K. Therefore, with the notationC0WD
#L0K and C1 WD "L1K, we obtain that L is (isomorphic to) C0CgluKCgluC1. Consequently, in order to show thatC0andC1are chains and to complete the proof, it suffices to show that every u2LnK is an isolated element of L. Suppose the contrary. Then (2.2) yields a subuniverseY 2Sub.L/such that
Y [ fug …Sub.L/ or Y n fug …Sub.L/: (2.3)
Sinceu…K, we have thatY \KD.Y [ fug/\KD.Y n fug/\K; we denote this set byS. ThenS2Sub.K/sinceY 2Sub.L/. It follows from (2.1) that
Y [ fug DS[..Y[ fug/nK/2Sub.L/and Yn fug DS[..Yn fug/nK/2Sub.L/;
which contradicts (2.3) and completes the proof of Lemma1 The following lemma is easier and even a computer program could prove it. For the reader’s convenience, we give its short proof.
Lemma 2. For the lattices given in Figure2, the following seven assertions hold.
(1) jSub.B4/j D13D2624 5. (2) jSub.N5/j D23D2325 5.
(3) jSub.C.2/C.3//j D38D1926 5. (4) jSub.B4CgluB4/j D85D21:2527 5.
(5) jSub.B4CgluC.2/CgluB4/j D169D21:12528 5. (6) jSub.M3/j D20D2025 5.
(7) jSub.B8/j D74D9:2528 5.
Proof. The notation given by Figure2will extensively be used.
Among all subsets ofB4, onlyfa; bg,fa; b; 0g, andfa; b; 1garenotsubuniverses;
this proves (1). Implicitly, Lemma1(3) will often be used below. Observe that jfS2Sub.N5/W fa; cg \SD¿gj D8; by (1.1),
jfS2Sub.N5/W fa; cg \S¤¿; b…Sgj D34D12; and jfS2Sub.N5/W fa; cg \S¤¿; b2Sgj D3;
whereby jSub.N5/j D 8C12C3 D 23 proves (2). Next, S will belong to Sub.C.2/C.3//even if this is not indicated. Let us compute:
jfSWa…Sgj D26; by Lemmas1(3) and2(1), jfSWa; b2Sgj D3; since0; d2Sandc2S)12S;
jfSWa2S; b…S; c2Sgj D1; since0; 12Sandd…S;
jfSWa2S; b…S; c…Sgj D8;by (1.1):
Hence, 26C3C1C8 D38 proves (3). Next, S will automatically belong to Sub.B4CgluB4/. We have thatjfS W fa; bg Sjg D7, because thenfc; dg S ) 12S and0; e2S. Also, jfSW fa; bg 6Sjg D1332D78, because Lemma 2(1) applies to the upperB4, there are 3 possibilities foraandb, and two for 0. Hence, 78+7=85 proves (4). ForS2Sub.B4CgluC.2/CgluB4/, the intersection ofS with the lowerB4and that with the upperB4can independently be chosen. Therefore, (5) follows from (1).
Next, we count the subuniversesSofM3. There are 4 with the propertyjfa; b; cg\
Sj 2, because they contain 0 and 1. There are34D12withjfa; b; cg \Sj D1, and 4 withjfa; b; cg \Sj D0. Thus,jSub.M3/j D4C12C4D20, proving (6).
The argument forB8 is more tedious. It has 9 at most one-element subuniverses.
There are 12 edges. We have 6 two-element subuniverses in which the heights of the two elements differ by two and 1 in which this difference is three. We have 12 three-element covering chains and 6 non-covering ones. The number of four-element (necessarily covering) chains is32D6,B4is embedded in 6 cover-preserving ways and (thinking of pairs of complementary elements) in 3 additional ways. The five- element sublattices are obtained from cover-preservingB4-sublattices by adding the unique one of 0B8 or1B8 that is missing; their number is 6. To each of the B4- sublattices at the bottom we can glue aB4-sublattice at the top in two ways, whence there are exactly 6 six-element subuniverses. In absence of doubly irreducible ele- ments, there is no seven-element sublattice, and there is 1 eight-element one. The sum of the numbers we have listed is 74, proving (7) and Lemma2
Now, we are in the position to prove (1.2), mentioned in Remark2.
Proof of Remark2. We obtain the statement by combining Lemma1(3) with parts
(2), (3), and (4) of Lemma2.
3. THE REST OF THE PROOF
For brevity, ak-element antichain will be called ak-antichain. First, we recall two well-known facts from the folklore.
Lemma 3. For every join-semilatticeS generated byfa; b; cg, there is a unique surjective homomorphism' from the free join-semilatticeFjsl.a;Q b;Q c/, given in Fig-Q ure3, ontoSsuch that'.a/Q Da,'.b/Q Db, and'.c/Q Dc.
FIGURE 3. Fjsl.a;Q b;Q c/Q andFlat.a;Q b;Q c/Q
Lemma 4(Rival and Wille [7, Figure 2]). For every latticeKgenerated byfa; b; cg such thata < c, there is a unique surjective homomorphism'from the finitely presen- ted latticeFlat.a;Q b;Q c/, given in FigureQ 3, ontoKsuch that'.a/Q Da,'.b/Q Db, and '.c/Q Dc.
The subscripts in the notations are explained by the facts that Fjsl.a;Q b;Q c/Q is the free join-semilattice on 3 generators, whileFlat.a;Q b;Q c/Q is the free lattice generated by its subposetfa; b; cg:
We are going to use the two lemmas above in the proof of the following lemma.
Implicitly, we will often use the well-known Homomorphism Theorem; see, e.g., Burris and Sankappanavar [1, Theorem 6.12].
Lemma 5. If an n-element lattice L has a 3-antichain, then we have that jSub.L/j 202n 5.
Proof. Letfa; b; cgbe a3-antichain in L. Lemma3 yields a unique join-homo- morphism fromFjsl..a;Q b;Q c//Q toSWD fa; b; c; a_b; a_c; b_c; a_b_cgsuch that 'maps toa,Q b, andQ cQtoa,b, andc, respectively. Sincefa; b; cgis an antichain, none of the six lower edges ofFjsl..a;Q b;Q c//Q is collapsed by the kernelWDker.'/of'.
Hence, there are only four cases for the join-subsemilatticeSŠFjsl..a;Q b;Q c//=Q of L, depending on the number the upper edges collapsed by.
Case 1: [none of the three upper edges is collapsed by] ThenSis isomorphic toFjsl..a;Q b;Q c//, wherebyQ fa_b; a_c; b_cgis a3-antichain. We know from, say, Gr¨atzer [5, Lemma 73], that this3-antichain generates a sublattice iso- morphic toB8. Hence,jSub.L/j 9:252n 5202n 5 by Lemmas1(2) and2(7), as required.
FIGURE 4. Cases 2 and 3
Case 2: [collapses exactly one upper edge] Apart from notation, we have that dWDa_b < a_cDWi andeWDb_c < i; see Figure4on the left. Letting b0WDd^e, we have thata_b0Dd andb0_cDe. Sincebb0andb6a, we have thatb06a. If we hadab0, theniDd_eDa_b0_eDb0_eDe would be a contradiction. Hence,akb0, andfa; b0; cgis a3-antichain bya–c symmetry. We can count the subuniverses T of the join-semilatticeH WD fa; b0; c; d; e; igas follows. We have thatjfT Wb0…Tgj 37D21, because fd; eg 6T allows only three possibilities forT\ fd; eganda_cDi at most seven possibilities forT \ fa; c; ig. Similarly,
jfT Wb02T; a…T; c…Tgj 7; becausefd; eg T )i2T;
jfT Wb02T; a2T; c…Tgj 3; sinced 2T;soe2T )i 2T;
jfT Wb02T; a…T; c2Tgj 3; bya–c symmetry, and jfT Wb02T; a2T; c2Tgj D1; becauseT DH:
Note that some of the inequalities above are equalities, but we do not need this fact. Forming the sum of the above numbers, the join-semilattice H has at most35D17:526 5subuniverses. Hence, Lemma1(1) yields that Sub.L/17:52n 5202n 5, as required.
Case 3: [ collapses two of the upper edges] Apart from notation, we have thatd WDa_b < a_cDb_cDWi. LetuWDa^b; see Figure4. We focus on possible intersections of subuniverses of L with H WD fa; b; c; d; u; ig. Denoting such an intersection byS, we can compute as follows.
jfSWc…Sgj 26; by Lemmas1(2) and2(1), jfSWc2S; a2S; b2Sgj 1 sinceH Œfa; b; cg;
jfSWc2S; a2S; b…Sgj 4 sincei 2Œfa; cg;
jfSWc2S; a…S; b2Sgj 4; bya–b-symmetry, jfSWc2S; a…S; b…S; d2Sgj 2; becausei 2S;
jfSWc2S; a…S; b…S; d…Sgj 3 sinceu2S)i2S:
Since the sum of these numbers is 40, we obtain from Lemma 1(1) that jSub.L/j 402n 6D202n 5, as required.
Case 4: [all the three upper edges are collapsed] Clearly,a_bDa_cDb_ cDa_b_cDWi. Ifa^bDa^cDb^cDa^b^cfailed, then the dual of one of the previous three cases would apply. Hence, we can assume that the sublatticeŒfa; b; cggenerated byfa; b; cgis isomorphic toM3; see Figure2.
Therefore,jSub.L/j 202n 5 by Lemmas 1(2) and 2(6), completing the
proof of Case 4 and that of Lemma5.
Proof of Theorem1. From Lemmas1and2(1), we conclude part (1). So, we are left only with part (2).
In what follows, letLbe ann-element lattice. We obtain from Lemmas1(3) and 2(2) that if
LŠC0CgluN5CgluC1for finite chainsC0andC1, (3.1) thenjSub.L/j D232n 5. In order the complete the proof of Theorem1, it suffices to exclude the existence of a latticeLsuch that
jLj Dn,232n 5 jSub.L/j< 262n 5, butLis
not of the form given in (3.1). (3.2)
Suppose, for a contradiction, thatLis a lattice satisfying (3.2). Then, by (1.1) and Theorem1(1) and Lemma5,
Lhas at least two2-antichains but it has no3-antichain. (3.3)
We claim that
Lcannot have twonon-disjoint2-antichains. (3.4) Suppose the contrary. Then we can pick two distinct antichainsfa; bgandfc; bginL.
Since there is no3-antichain inL, we can assume thata < c. WithK WDŒfa; b; cg, let'WFlat.a;Q b;Q c/Q !Kbe the unique lattice homomorphism from Lemma4, and let be the kernel of '. We claim that collapsese1; see Figure 3. Suppose to the contrary that does not collapsee1. Sincee1 generates the monolith congruence, that is, the smallest nontrivial congruence of the N5 sublattice of Flat.a;Q b;Q c/, noQ other edge of thisN5 sublattice is collapsed. Hence,N5 is a sublattice ofL, and it follows from Lemmas1(2) and2(2) thatjSub.L/j 232n 5. Thus, (3.2) yields that jSub.L/j D232n 5. Applying Lemma1(3) forKWDN5andL, we obtain thatL is of the form (3.1). This contradicts (3.2), and we have shown thatdoes collapse e1. On the other hand, since akb and c kb, none of the thick edges e8; : : : ; e11
is collapsed by . Observe that at least one of e4 and e6 is not collapsed by , since otherwiseh Qa;cQiwould belong toDker.'/by transitivity andaDc would be a contradiction. By duality, we can assume thate4 is not collapsed by. Since e2, e3, ande5 are perspective toe10, e9, ande4, respectively, these edges are not collapsed either. So, with the exception of e1, no edge among the “big” elements in Figure3is collapsed. Thus, the'-images of the elements denoted by big circles form a sublattice (isomorphic to)C.2/C.3/inL. Hence,jSub.L/j 192n 5by Lemmas1(2) and2(3), which contradicts our assumption thatLsatisfies (3.2). This proves (3.4).
To provide a convenient tool to exploit (3.3) and (3.4), we claim that ifx; y; ´2Lsuch thatjfx; y; ´gj D3andxky, then either
fx; yg #´, orfx; yg "´. (3.5) To see this, assume the premise. SinceLhas no3-antichain,´is comparable to one ofxandy. By duality and symmetry, we can assume thatx < ´. Since´ < ywould implyx < yand´kytogether withxkywould contradict (3.4), we have thaty < ´.
This proves (3.5).
Next, by (3.3) and (3.4), we have a four-element subsetfa; b; c; dgofLsuch that akbandckd. By duality and (3.5), we can assume thatfa; bg #c. Applying (3.5) also tofa; b; dg, we obtain thatfa; bgis included either in"d, or in#d. Since the first alternative would lead tod < a < cand so it would contradictckd, we have that fa; bg #d. Thus,fa; bg #c\ #d D #.c^d /, and we obtain thatuWDa_b c^d DWv. LetSWD fa^b; a; b; u; v; c; d; c_dg. Depending onuDvoru < v,S is a sublattice isomorphic toB4CgluB4orB4CgluC.2/CgluB4. Using Lemma2.1 together with (4) and (5) of Lemma2, we obtain thatjSub.L/j 21:252n 5. This inequality contradicts (3.2) and completes the proof of Theorem1.
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Authors’ addresses
G´abor Cz´edli
University of Szeged, Bolyai Institute, Aradi v´ertan´uk tere 1, 6720 Szeged, Hungary E-mail address:czedli@math.u-szeged.hu
Eszter K. Horv´ath
University of Szeged, Bolyai Institute, Aradi v´ertan´uk tere 1, 6720 Szeged, Hungary E-mail address:horeszt@math.u-szeged.hu
