Mm Obius Functions of Lattices

We introduce the concept of a bounded below set in a lattice. This can be used to give a generalization of Rota's broken circuit theorem to any nite lattice. We then show how this result can be used to compute and combinatorially explain the Mobius function in various examples including non-crossing set partitions, shu e posets, and integer partitions in dominance order. Next we present a generalization of Stanley's theorem that the characteristic polynomial of a semimodular supersolvable lattice factors over the integers. We also give some applications of this second main theorem, including the Tamari lattices. 1 Bounded below sets In a fundamental paper [25], Whitney showed how broken circuits could be used to compute the coe cients of the chromatic polynomial of a graph. In another seminal paper [20], Rota re ned and extendedWhitney's theorem to give a characterization of the Mobius function of a geometric lattice. Then one of us [21] generalized Rota's result to a larger class of lattices. In this paper we will present a theorem for an arbitrary nite lattice that includes all the others as special cases. To do so, we shall need to replace the notion of a broken circuit by a new one which we call a bounded below set. Next we present some applications to lattices whose Mobius functions had previously been computed but in a less simple or less combinatorial way: shu e posets [13], non-crossing set partition lattices [15, 19], and integer partitions under dominance order [5, 6, 12]. The second half of the paper is dedicated to generalizing the result of Stanley [23] that the characteristic polynomial of a semimodular supersolvable lattice factors over the integers. We replace both supersolvability and semimodularity by weaker conditions which we call left-modularity and the level condition, respectively, in such a way that the conclusion still holds. Examples of this factorization (not covered by Stanley's theorem) are provided by certain shu e posets and the Tamari lattices [9, 14]. We end with a further generalization of Rota's theorem and a section of comments and questions. Throughout this paper L will denote a nite lattice. Any relevant de nitions not given can be found in Stanley's text [24]. We will use ^ for the meet (greatest lower bound) and _ for the join (least upper bound) in L. Since L is nite it also has a unique minimal element 0̂ and a unique maximal element 1̂. The M obius function of L, : L! Z, is de ned recursively by (x) = ( 1 if x = 0̂, Py 0̂. We let (L) = (1̂). Note that is the unique Z-valued function on L such that Py x (y) = 0̂x (Kronecker delta). Our goal is to give a new combinatorial description of (x). Let A(L) be the set of atoms of L, i.e., those elements covering 0̂. Give A(L) an arbitrary partial order, which we denote to distinguish it from the partial order in L. So can be anything from a total order to the total incomparability order induced by . A nonempty set D A(L) is bounded below or BB if, for every d 2 D there is an a 2 A(L) such that a d and (1) a < _D: (2) 1 So a is simultaneously a strict lower bound for d in the order and for WD in . We will say that B A(L) is NBB if B does not contain any D which is bounded below. In this case we will call B an NBB base for x = WB. We can now state our main result. Theorem 1.1 Let L be any nite lattice and let be any partial order on A(L). Then for all x 2 L we have (x) =XB ( 1)jBj (3) where the sum is over all NBB bases B of x and j j denotes cardinality. Proof. For x 2 L de ne ~ (x) = PB( 1)jBj summed over all NBB bases B of x. To prove that ~ (x) = (x) it su ces to show Py x ~ (y) = 0̂x. If x = 0̂ then x = WB only for B = ; which is NBB. So X y 0̂ ~ (y) = ~ (0̂) = ( 1)0 = 1 as desired. If x > 0̂ then to get Py x ~ (y) = 0 we rst set up a corresponding signed set S. Let S = fB : B an NBB base for some y xg: The sign of B 2 S will be (B) = ( 1)jBj. Then from the de nitions X y x ~ (y) = X B2S (B): If we can nd a sign-reversing involution on S then this last sum will be zero and we will be done. From the set of atoms a x pick one, a0, which is minimal with respect to . Consider the map on S de ned by (B) = B 4 a0 where 4 is symmetric di erence. (Here and below we omit set braces around singletons writing, for example, a0 instead of fa0g.) Clearly is a sign-reversing involution provided it is well de ned, i.e., we must check that B being NBB implies that (B) is NBB. If (B) = B n a0 then clearly (B) is still NBB. We will do the case B 0 := (B) = B [ a0 by contradiction. Suppose B 0 D where D is bounded below. Then a0 2 D since B itself is NBB. Let a be the corresponding element guaranteed by the de nition of a bounded below set. Then a a0 and a < WB0 x which contradicts the de nition of a0. Here is an example to illustrate this result. Suppose the lattice L and partial order are as given in Figure 1. To nd the bounded below sets, note that 2 (L; ) = ŝ 0 s a s b sc s x sy ŝ 1 Q Q Q Q Q Q Q Q Q Q Q Q (A(L); ) = s b s a sc S S S S S Figure 1: A lattice L and partial order on A(L) by (1) no set containing an element minimal in is BB. Furthermore (1) and (2) together imply that no single element set is BB either. (These observations will be important in our other examples.) Thus the only possible BB set is fa; cg, and it does satisfy the de nition since b a; c and b < Wfa; cg = 1̂. So x has one NBB base, namely fa; bg and so (x) = ( 1)2 which is easily checked from the de nition of . Similarly (y) = ( 1)2. Finally 1̂ has no NBB bases and so (1̂) = 0 (the empty sum). We should see why our theorem implies Rota's broken circuit result. To do this we need to recall some de nitions. Let L be a geometric lattice with rank function . It is well known, and easy to prove, that if B A(L) then (WB) jBj. Call B independent if (WB) = jBj and dependent otherwise. A circuit C is a minimal (with respect to inclusion) dependent set. Now let be any total order on A(L). Then each circuit C gives rise to a broken circuit C 0 = C n c where c is the rst element of C under . A set B A(L) is NBC (no broken circuit) if B does not contain any broken circuit and in this case B is an NBC base for x = WB. Rota's NBC theorem [20] is as follows. Theorem 1.2 (Rota) Let L be a nite geometric lattice and let be any total order on A(L). Then for all x 2 L we have = ( 1) (x) (number of NBC bases of x). To derive this result from Theorem 1.1, we rst prove that when L is geometric and is total then the NBB and NBC sets coincide. For this it su ces to show that every broken circuit is bounded below and that every bounded below set contains a broken circuit. If C 0 = C n c is a broken circuit, then for every c0 2 C 0 we have c c0 and c < WC = WC 0 so C 0 is BB. For the other direction, if D is bounded below then consider the rst element d of D and let a 2 A(L) be the element guaranteed by the BB de nition. Then by (2) we have (WD _ a) < jD [ aj. So 3 D [ a is dependent and contains a circuit C. Now (1) and the choice of a and d show that for the corresponding broken circuit we have C 0 D. Thus NBB and NBC sets are the same in this setting. Finally Rota's expression for (x) is obtained from ours by noting that when L is semimodular then all NBB bases for x have the same size, namely (x). Similar arguments show that the main result of [21] is a special case of Theorem 1.1. Another corollary of this theorem is a special case of Rota's Crosscut Theorem [20]. Theorem 1.3 (Rota) Let L be any nite lattice. Then for all x 2 L (x) =XB ( 1)jBj (4) where the sum is over all B A(L) such that WB = x. To obtain (4) it su ces to take to be the total incomparability order in Theorem 1.1. In fact our theorem (where the partial order on A(L) is arbitrary) can be viewed as interpolating between the Crosscut Theorem (where A(L) forms an antichain) and the NBC Theorem (where A(L) forms a chain). This raises the question of why one would want to consider an arbitrary partial order on A(L) when one can always take the one induced by the order in L. The reason is that the number of terms in the sum (4) is generally much larger than the number in (3). From the viewpoint of e cient computation of , the best scenario is the same as the one in the geometric case where (3) has exactly j (x)j terms, all of the same sign. A partial order on A(L) for which this happens for all x 2 L will be called perfect. There are posets where no such exists, such as the k-equal intersection lattices which were introduced by Bjorner, Lov asz and Yao [3, 4]. However, all the examples we will consider in the next sections are perfect. Another thing to note is that if is perfect then so is any linear extension of it. However, to make the combinatorics of as clear as possible it is often best to take a perfect with the least possible number of order relations. 2 Non-crossing partitions The non-crossing partition lattice was rst studied by Kreweras [15] who showed its Mobius function to be a Catalan number. By using NBB sets we can combinatorially explain this fact and relate these bases to the standard NBC bases for the ordinary partition lattice. If is a partition of [n] := f1; 2; : : : ; ng into k subsets, or blocks, then we write = B1= : : : =Bk ` [n]. When it will cause no confusion, we will not explicitly write 4 : 1567=234=8 (noncrossing) 134=256=78 (crossing) G : s 6 s 5 s 7 s 4 s 8 s 3 s 1 s 2 @ @ @ B B B B B B B B A A A A A A @ @ @ s 6 s 5 s 7 s 4 s 8 s 3 s 1 s 2 @ @ @ @ @ H H H H H Figure 2: Partitions and their graphs out any blocks that are singletons. The set of ` [n] form a lattice n under the re nement ordering. We say that is non-crossing if there do not exist i; k 2 B and j; l 2 C for two distinct blocks B;C of with i < j < k < l. Otherwise is crossing. The set of non-crossing partitions of [n] forms a meet-sublattice NCn of n with the same rank function. However unlike n, NCn is not semimodular in general since if = 13 and = 24 then ^ = 0̂ and _ = 1234 so ( ) + ( ) = 2 < 3 = ( ^ ) + ( _ ): Here and throughout this paper, semimodularity refers to upper-semimodularity. Another way to view non-crossing partitions will be useful. Let G = (V;E) be a graph with vertex set V = [n]. Then G is non-crossing if there do not exist edges ik; jl 2 E with i < j < k < l. Equivalently, G is non-crossing if, when the vertices are arranged in their natural order clockwise around a circle and the edges are drawn as straight line segments, no two edges of G cross geometrically. Given a partition we can form a graph G by representing each block B = fi1 < i2 < : : : < ilg by a cycle with edges i1i2; i2i3; : : : ; ili1. (If jBj = 1 or 2 then B is represented by an isolated vertex or edge, repectitively.) Then it is easy to see that is non-crossing as a partition if and only if G is non-crossing as a graph. In Figure 2 we have displayed some partitions and their graphs. The atoms of NCn are the partitions of the form = ij where we will always assume i < j. Then G is a single edge, so we can consider any B A(NCn) as a graph GB with an edge for each 2 B. De ne ij i0j 0 if and only if j < j 0. So the poset (A(NCn); ) will be ranked with the elements at rank j 2 being all atoms of the form ij. An example for n = 4 is given in Figure 3 In order to characterize the NBB sets, we rst need a lemma. Lemma 2.1 Let ; 0 be a pair of atoms such that either and 0 have the same rank in (A(NCn); ) or the graph of D = f ; 0g is crossing. Then D is BB. 5 t 12 t 13 t 23 t 14 t 24 t 34 @ @ @ @ @ @ Figure 3: The partial order for A(NC4) Proof. Suppose rst that the two atoms have the same rank. Then = ij and = i0j where without loss of generality i < i0 < j. Let = ii0. Then ; 0 and _ 0 = ii0j > so D is BB. Suppose instead that the given atoms are crossed. So we have = ij and = i0j0 with i < i0 < j < j 0. Letting = ii0 we get the same inequalities as before, noting that _ 0 = ii0jj 0. Theorem 2.2 The NBB bases of 1̂ in NCn are all B obtained by picking exactly one element from each rank of (A(NCn); ) so that the corresponding graph GB is non-crossing. Proof. First suppose that B is an NBB base of 1̂. Then by the previous lemma we know that B contains at most one element from each rank and that GB is noncrossing. If we do not pick an element from some rank then GB is not connected. But such a non-crossing graph has a block of WB for each component of GB, contradicting WB = 1̂. Conversely, suppose B is picked according to the two given rules. Then GB is connected and so WB = 1̂. IfB D withD a BB set we will derive a contradiction. Let 2 D be -minimal and let = ij be the corresponding element guaranteed by the de nition of bounded below. Then (1) and our choice of shows that for any 0 = i0j0 2 D we have i; j < j 0. Since B is non-crossing, so is D and thus WD is the same in NCn and n. It follows from (2) that there is a path in GD 6 of the form i = i0; i1; : : : ; il = j where each edge fikik+1g is an atom of D. But our remarks about 0 imply that i0 < i1 and il 1 > il so there must be an index m such that im 1 < im > im+1. Thus D has two elements from the same -rank, the promised contradiction. Note that the graphs GB in the previous theorem are certain spanning trees on the vertex set [n]. Furthermore, to get the NBB bases for all elements of NCn it su ces to use the non-crossing restriction but picking at most one element from each rank. Finally, if one removes the non-crossing restriction one gets exactly the standard NBC bases for the geometric lattice n. It is now easy to compute the Mobius function of NCn. It su ces to do this for 1̂ since for any = B1= : : : =Bk 2 NCn, the interval [0̂; ] = QiNCjBij. Recall that the Catalan numbers are de ned by Cn = 1 n+ 1 2n n !: Corollary 2.3 (Kreweras) We have (NCn) = ( 1)n 1Cn 1: Proof. All trees on n vertices have n 1 edges. Furthermore it is easy to see that the number Tn of non-crossing trees on [n] of the given form and Cn 1 satisfy the same initial conditions and recurrence relation, Tn = ( 1 if n = 1 P0 1. The result now follows from Theorems 1.1 and 2.2. There is another natural ordering of A(NCn) for which a result similar to Theorem 2.2 holds, namely ij i0j0 if and only if [i; j] [i0; j 0] as intervals of integers. With this ordering, NBB bases of 1̂ again correspond to trees with vertex set [n], but this time they are the non-crossing trees in which each vertex is either greater than all its neighbors or less than all its neighbors. It follows easily that, in any such tree, f1; ng is an edge, and deletion of this edge leaves two smaller such trees with vertex sets [k] and [k+ 1; n] for some k. This observation easily implies that the numbers of such trees satisfy the recurrence for the Catalan numbers. In fact, there is a simple bijection between these trees and proper parenthesizations P of the product of n factors (one of the most familiar interpretations of the Catalan numbers). To describe the bijection, identify the n factors with the elements of 7 [n], in order. For any particular parenthesization P , consider the sub-products de ned by P . For example, if n = 5 and P is (((12)3)(45)), then the sub-products are (12); ((12)3); (45), and (((12)3)(45)) itself. Now to build the corresponding T , take the vertex set [n] and draw, for each sub-product of P , an edge from the rst to the last element of the subproduct. In the previous example, the edges would be 12; 13; 45, and 15. It is a variant of this ordering which allows us to compute for the non-crossing Bn and Dn lattices. 3 Non-crossing Bn and Dn In this section, we apply Theorem 1.1 to calculate the Mobius invariants (1̂) of the non-crossing Bn and Dn (and intermediate) lattices. These lattices were introduced by Reiner [19] who computed using generating functions. Non-crossing Bn consists of those partitions of f1; 2; : : : ; n; 1; 2; : : : ; ng that satisfy three conditions: First, is invariant under the involution k 7! k. Second, at most one block of is xed by this involution; if there is such a block, it is called the zero-block of . Third, the partition is non-crossing (as in Section 2), with respect to the ordering 1 < 2 < : : : < n < 1 < 2 < : : : < n. (The rst two of these conditions determine a lattice isomorphic to that associated with the hyperplane arrangement Bn.) Non-crossing Dn is the subposet consisting of those for which the zero-block, if present, does not consist of only a single pair fk; kg. An intermediate lattice can be associated to every subset S [n], by allowing the zero-block to be fk; kg only if k = 2 S. We use the notations NCBn, NCDn, and NCBDn(S) for these lattices; thus NCBn = NCBDn(;) and NCDn = NCBDn([n]). We begin by calculating (1̂) for NCBn and afterward indicate the minor changes needed to handle the rest of these lattices. The atoms of NCBn are of three sorts. First, there are the partitions where one block consists of two positive numbers, say fi; jg, another block is f i; jg, and all the remaining blocks are singletons. Second, there are the partitions where one block consists of a positive and a negative number, say fi; jg, another is f i; jg, and the rest are singletons. Third, there are the partitions whose only non-singleton block is of the form fi; ig. Following Zaslavsky [26], we depict atoms as signed edges in a graph whose vertex set is [n]. An atom of the rst sort is depicted as a positive edge ij, one of the second sort is depicted as a negative edge ij, and one of the third sort is depicted as a (negative) half-edge at i. (There is no such thing as a positive half-edge.) To avoid confusion, we emphasize that these signed graphs are quite di erent from the graphs with vertex set f1; 2; : : : ; n; 1; 2; : : : ; ng used in deciding whether a partition is non-crossing. We sometimes identify an atom with the corresponding edge; in particular, we may refer to an atom as positive or negative. 8 We partially order the atoms as follows. Associate to each atom, depicted as a signed edge ij, the interval [i; j] [n]; in the case of a half-edge at i, the interval consists of just i. Then de ne a b to mean that either the interval associated to atom a properly includes that associated to atom b or the two intervals are equal and a is negative while b is positive. (Notice that, apart from signs and half-edges, this matches the ordering described at the end of Section 2.) Regarding atoms as signed edges, we regard sets of atoms as signed graphs with vertex set [n]. It is not di cult (though a bit tedious) to verify that every NBB set B has, as a graph, the following properties. (i) No two signed edges cross. (ii) No vertex i has a neighbor j < i and also a neighbor j 0 > i. (iii) Any path joining two vertices i and j and consisting entirely of vertices k with i k j must be just a single signed edge ij. (iv) There is no cycle of length 3. (A 2-cycle, consisting of a positive edge and a negative edge in the same place, is permitted; we refer to such a pair of signed edges as a double edge.) (v) No two negative atoms in B have disjoint associated intervals. (vi) If the interval associated to one atom in B is properly included in the interval associated to another, and if the former atom is negative, then so is the latter. (vii) If there is a negative atom in B, then there is exactly one, say a, that is -maximal (i.e., its interval is inclusion-minimal); all other atoms in B are negative if their intervals properly include that of a and positive otherwise. (viii) B has at most one half-edge, has at most one double-edge, and cannot have both. Actually, only items (i), (ii), (v), and (vi) in this list directly use the NBB assumption. The other four items follow from these purely graph-theoretically. Conversely, any signed graph satisfying (i) through (viii) is NBB when viewed as a set of atoms of NCBn. We leave the veri cation to the reader, with the hint that items (i), (v), and (vi) ensure that the join of any atoms from this set is the same whether computed in NCBn or in Bn, because none of these atoms cross when regarded as partitions of f1; 2; : : : ; n; 1; 2; : : : ; ng. From this characterization of the NBB sets in NCBn, we easily obtain a characterization of the NBB bases of 1̂. These bases (regarded as signed graphs) are obtainable as follows. First, take a non-crossing tree T with vertex set [n] in which 9 each vertex is either greater than all its neighbors or less than all its neighbors. (This part is just as at the end of Section 2.) Then pick either an edge or a vertex of T . If you picked an edge e, then make it a double edge, give negative signs to all the edges of T whose intervals properly include that of e, and give all remaining edges of T positive signs. If you picked a vertex v, then attach a (negative) half-edge at v, give negative signs to all edges of T whose interval contains v, and give all remaining edges of T positive signs. Finally, to apply Theorem 1.1, we count the NBB bases for 1̂, i.e., we count the signed trees of the sort just described. We already saw in Section 2 that there are Cn 1 = 1 n 2n 2 n 1 ! = 1 2n 1 2n 1 n ! ways to choose T . Then there are 2n 1 ways to choose an edge or vertex, since there are n vertices and n 1 edges. After this choice, the rest of the construction of the NBB signed graph is completely determined. So the number of NBB bases for 1̂ is Cn 1 (2n 1) = 2n 1 n !: Every NBB base of 1̂ has exactly n elements, namely the n 1 edges of T (with signs) plus either an extra edge if you chose an edge and doubled it or an extra half-edge if you chose a vertex. Therefore, (NCBn) = ( 1)n 2n 1 n . The calculation for NCBDn(S) is almost exactly the same. The only di erence is that half-edges can occur only at vertices not in S. Thus, in the description of NBB bases for 1̂, we need only replace \pick an edge or a vertex of T" with \pick an edge of T or a vertex not in S." Thus, the number of options at this step is no longer 2n 1 but only 2n 1 jSj. Therefore, we obtain, in agreement with Reiner's calculation ([19]), the Mobius invariant for NCBDn(S): (NCBDn(S)) = ( 1)nCn 1 (2n 1 jSj): 4 Shu e Posets The poset of shu es was introduced by Greene [13]. We need to recall some of his de nitions and results before applying Theorem 1.1. Let A be a set, called the alphabet of letters. A word over A is a sequence u = u1u2 : : : un of elements of A. All of our words will consist of distinct letters and we will sometimes also use u to stand for the set of letters in the word, depending upon the context. A subword of u is v = ui1 : : : uil where i1 < : : : < il. If u;v are any two words then 10 t de t d t e tDde tdDe tdeD t ; t Dd tDe t dD teD t D Q Q Q Q Q Q Q Q Q @ @ @ @ @ @ A A A A A A !!!!!!!!!!!!!!! A A A A A A !!!!!!!!!!!!!!! A A A A A A A A A A A A @ @ @ @ @ @ Q Q Q Q Q Q Q Q Q Figure 4: The lattice W2;1 the restriction of u to v is the subword uv of u whose letters are exactly those of u \ v. A shu e of u and v is any word s such that s = u ] v as sets (disjoint union) and su = u, sv = v. Given nonnegative integers m;n Greene de ned the poset of shu es Wm;n as follows. Fix disjoint words x = x1 : : : xm and y = y1 : : : yn. The elements of Wm;n are all shu es w of a subword of x with a subword of y. The partial order is v w if vx wx and vy wy. It is easy to see that W has minimal element 0̂ = x, maximal element 1̂ = y and is ranked with rank function (w) = (m jwxj) + jwyj: (5) For example,W2;1 is shown in Figure 4 where x = de and y = D. In order to apply the NBB Theorem we will need to describe the join operation in Wm;n. Greene does this using crossed elements, not to be confused with the crossing partitions discussed in Section 2. Given u;v 2 Wm;n then x 2 x is crossed in u;v if there exist yi; yj 2 y such that i j and x appears before yi in one of the two words but after yj in the other. For example if x = def and y = DEF then for the pair u = dDEe;v = Fdef the only crossed letter is d. The join of u;v is then the unique word w greater than both u;v such that wx = fx 2 ux \ vx : x is not crossedg wy = uy [ vy: (6) 11 In the previous example, u _ v = DEFe. This example also shows that Wm;n is not geometric since the semimodularity law is violated: (u) + (v) = 3 + 1 < 5 = (u _ v) (u _ v) + (u ^ v): The set atoms Am;n = A(Wm;n) consists of two types. An a-atom, respectively b-atom, is one obtained from x by deleting a letter of x, respectively inserting a letter of y. Let Aa denote the set of a-atoms and similarly for Ab. De ne on Am;n to be the poset whose relations are all those of the form a b with a 2 Aa and b 2 Ab Lemma 4.1 Suppose b;b0 2 Ab. If b;b0 have crossed elements then D = fb;b0g is a BB set of (Am;n; ). Proof. Our hypothesis and (6) show that (b _ b0)x x (proper containment of sets). So there is an a-atom a with a (b_ b0)x which forces a < b_ b0 in Wm;n. Since by de nition a b;b0 the element a satis es the de nition of a BB set. The next result will characterize the NBB bases and show that the converse of the previous lemma also holds. Theorem 4.2 Let s be a shu e of x;y and consider Bs = Aa [ fb 2 Ab : b sg: Then the NBB bases of y 2 Wm;n under the given partial order are exactly the Bs: Proof. Suppose rst that B is an NBB base of y. Then for each element y 2 y we must have a corresponding b-atom by in order to get WB = y. In fact there must be exactly one such atom for each y 2 y and these atoms must all lie below a shu e s, for otherwise B would contain a BB pair as in Lemma 4.1. It follows that Wy by = s. So in order to get WB = y we must have Aa B. Thus B is of the form Bs as desired. Conversely, consider any Bs. It is easy to see that WBs = y. We will show that Bs is NBB by contradiction. Suppose that D Bs is a BB set. Then (1) forces D to contain only b-atoms which in turn implies WD s. Now pick any d 2 D and let a 2 Am;n be the atom guaranteed by the de nition of BB. Then a is an a-atom and a < WD s. But this contradicts the fact that there are no a-atoms below any shu e of x and y. Thus Bs is an NBB base of y and our characterization is complete. To determine the Mobius function of Wm;n it su ces to compute (1̂) since for any w 2 Wm;n the interval [0̂;w] is isomorphic to a product of Wp;q's for certain p m and q n. 12 Corollary 4.3 (Greene) We have (Wm;n) = ( 1)m+n m+ n m !: Proof. The bases Bs all have size m+ n and the number of possible shu es s is m+n m . The result now follows from Theorems 1.1 and 4.2. 5 The Dominance Order Bogart [5] and Brylawski [6] rst computed the two-variable Mobius function of the lattice of integer partitions under dominance. Subsequently Greene [12] gave two alternative ways to compute this function. Our NBB set characterization leads to a formula in Corollary 5.2 which is essentially equivalent to, but simpler than, Greene's second description of [12, Theorem 4.1]. We begin by reviewing the relevant de nitions. A partition of n is a weakly decreasing sequence of positive integers 1 2 : : : r whose sum is n. For any such partition and any non-negative integer k, we write j jk for Pki=1 i, where i is interpreted as 0 for i > r, so j jk = n for all k r. A partition dominates another partition (of the same n), written , if j jk j jk for all k. In terms of Ferrers diagrams (in the English orientation) means that the diagram for can be obtained from that of by moving some squares up to earlier rows. It is well known that this ordering makes the set of partitions of n into a lattice Pn; we review the construction of joins in Pn because it will be needed for our NBB calculations. A composition of n is like a partition except that the parts i need not be in weakly decreasing order. The dominance order of compositions is de ned exactly as for partitions, and the result is a lattice in which joins are easily computed since j _ jk = maxfj jk; j jkg. The join of two partitions, computed in this lattice of compositions, need not be a partition. However, for every composition there is a unique smallest partition above , which we call the partition re ection of . Joins of partitions in Pn can be computed by rst forming the join in the dominance lattice of compositions and then forming the partition re ection of the result. We should point out that the partition re ection of a composition need not be simply the result of rearranging the parts into weakly decreasing order. For example, the partition re ection of (1; 3) is not (3; 1) but (2; 2). The bottom element 0̂ of Pn is the partition (1; 1; : : : ; 1) and the only atom is (2; 1; 1; : : : ; 1). By Theorem 1.3, these are the only elements of Pn where the 13 Mobius function has a non-zero value. But we can also consider the M obius function of elements in any upper interval [ ; 1̂]; indeed, the value of this M obius function at some is what is usually called the (two-variable) Mobius function ( ; ). To calculate this Mobius function, we x , we describe the atoms of [ ; 1̂], and we calculate the joins of sets of atoms with particular attention to determining when one atom is below a join of others. Then, we describe an appropriate partial ordering of the atoms, characterize its NBB sets, and use Theorem 1.1 to evaluate the Mobius function. For the rest of this section, let = ( 1; 2; : : : ; r) be a xed but arbitrary partition of n. By a wall we mean a maximal sequence of at least two numbers k for which the corresponding k are equal. (This corresponds to a at in Greene's terminology.) Thus, an interval [i; j] [1; r] is a wall if and only if i < j and i 1 > i = j > j+1. Here and below, we use the convention that 0 = 1 and r+1 = 0, so that 1 (respectively, r) can be part of a wall if the rst (respectively, last) two components of are equal. If [i; j] is a wall then we call i its top and j its bottom, the terminology being suggested by the Ferrers diagram. The atoms of [ ; 1̂] are of two sorts: (i) If 1 i < r and neither i nor i+1 is in a wall (i.e., i 1 > i > i+1 > i+2), then is covered by the partition that agrees with except that i = i+1 and i+1 = i+1 1. We denote this atom of [ ; 1̂] by i + 1 ! i, since its Ferrers diagram is obtained from that of by moving a square from row i+ 1 up to row i. (ii) If [i; j] is a wall, then is covered by the partition that agrees with except that i = i + 1 and j = j 1. We denote this by j ! i. For both types of , we refer to the set of k where j jk 6= j jk as the critical interval of , I . (This is not the same as Brylawski's use of \critical.") It consists of only i for of type (i) and it is [i; j 1] for of type (ii). Note that j jk = j jk+1 if and only if k 2 I Note also that di erent atoms of [ ; 1̂] have disjoint critical intervals and ] I = [1; r 1] n fk : [i; k] or [k + 1; i] is a wall for some ig: (7) So an element of [1; r 1] lies in a critical interval unless it is the bottom of a wall or the immediate predecessor of the top of a wall. Let us consider the join of some subset B A([ ; 1̂]), rst in the sense of compositions and then in the sense of partitions. If is the composition join, then 14 from the previous paragraph j jk j jk = ( 0 if k 62 X 1 if k 2 X (8) where X = ] 2BI . If this is a partition, then it is also the join in the partition sense; otherwise, we must take its partition re ection. So we consider next how could fail to be a partition and how it is changed by re ection. By (8) the components of di er from those of by 1 or 0. It follows that any failure of to be a partition, i.e., any occurrence of k < k+1, must arise in one of two ways: k = k+1 or k+1 + 1. The rst possibility can be excluded, because it requires k and k+1 to be part of a wall, say [i; j]. If the critical interval [i; j 1] X then by (7) and (8) we have i = i + 1, j = j 1, and agrees with on the rest of [i; j]; if, on the other hand, [i; j 1] \X = ;, then agrees with on all of [i; j]. In either case, we cannot have k < k+1. So any occurrence of k < k+1 has k = k+1 + 1, k = k 1 = k+1, and k+1 = k+1 + 1 = k. Thus by (8) X must contain k 1 and k + 1 but not k. Consider the set Y obtained by adjoining to X all those numbers k = 2 X for which k 1; k+1 2 X and k = k+1+1. Let be the composition of n such that j jk j jk = ( 0 if k 62 Y 1 if k 2 Y . (9) The preceding observations show that is a partition, because, by enlarging X to Y , we have corrected all the failures of to be a partition. Furthermore, clearly dominates . We claim that is the partition re ection of , i.e., that every partition dominating also dominates . To see this, it su ces to show that j jk > j jk for k 2 Y X; but if this failed for some k, then would fail to be a partition because k < k+1 just as for . Thus, is the join of B in Pn. We must also determine which atoms of [ ; 1̂] other than members of B are below . Distinct atoms have disjoint critical intervals, so the atoms we are looking for are those whose critical intervals are singletons k where k 2 Y X. Note that there is an atom with critical interval k if and only if either k and k+1 constitute a wall or neither of them belongs to a wall. The former alternative is irrelevant in the present context, since k 2 Y X implies k = k+1 + 1. So we need only consider the second alternative, where k 1 > k > k+1 > k+2. We are ready to begin the description of the NBB bases for an arbitrary . Let A be the set of atoms of [ ; ]. If WA 6= then ( ; ) = 0 by Theorem 1.3, so from now on we assume WA = . Call an atom b 2 A of the form k+1! k special if k = k+1+ 1 and A also contains atoms a; c equal to k+ 2! k+1; k ! k 1, respectively. The preceding discussion shows that b < a_ c; conversely, if an atom 15 t t t t t t t @ @ @ @ @ @ 1 2 3 4 5 0 Figure 5: The partial order on an extended special run in A is under the join of some other atoms in A, then it must be special. We let S A be the set of special atoms. Let us list the atoms in A according to the ordering of their critical intervals. By a special run in this list, we mean a maximal sequence of consecutive 2 S. De ne a partial ordering on A by imposing on each sequence = 0; 1; : : : ; q+1 = 0 (10) where 1; : : : ; q is a special run (so ; 0 62 S) the relations 3i+1 3i; 3i+2 for 1 3i+ 1 q: An example is given in Figure 5. Note that this ordering makes f 3i; 3i+2g a BB set for each i. We let Sj , 0 j 2, be the set of elements in all special runs of the form 3i+j for some i, so S = S0 ] S1 ] S2. Theorem 5.1 Let [ ; ] be an interval in Pn with atom set A such that WA = . Then has an NBB base in A if and only if there is no special run of length 1 modulo 3. If a base exists, then it is unique and equals B = A n S2: (11) Proof. If B is an NBB base of then it must include A n S since these elements are not the join of any others. Now consider any sequence of the form (10). Since 62 S we have 2 B and this forces 2 62 B since f ; 2g is BB. Now we must put 1; 3 2 B since neither is below W(A n 2). Now repeat the argument with 3 in place of to inductively see that the only possible candidate for an NBB base is (11). But if there is a special run (10) of length 3k + 1 for some k then 3k and 0 = 3k+2 form a BB set in B, a contradiction. If no such run exists then we never have such a pair in B and so it is NBB as desired. 16 Corollary 5.2 Let [ ; ] be an interval in Pn with atom set A. If WA 6= then ( ; ) = 0. If WA = then let ri, 0 i 2 be the number of special runs of length i modulo 3. Then ( ; ) = ( 0 if r1 1 ( 1)jAnSj+r2 if r1 = 0. Proof. We have already noted that ( ; ) = 0 if WA 6= . For the other two cases, suppose rst that r1 1. Then has no NBB base in [ ; ] by Theorem 5.1 so ( ; ) = 0 by Theorem 1.1. If r1 = 0 then by the same two results ( ; ) = ( 1)jBj = ( 1)jAnSj+jS0 j+jS1j: If a special run has length congruent to 0 (respectively, 2) modulo 3 then its contribution to jS0j + jS1j is 0 (respectively, 1) modulo 2. The last case now follows from the previous displayed equation. 6 Supersolvable Lattices Stanley [23] de ned a supersolvable lattice to be a pair (L; ) where L is a lattice, : 0̂ = x0 < x1 < : : : < xn 1 < xn = 1̂ is a maximal chain of L, and together with any other chain of L generates a distributive lattice. One often refers to L as supersolvable, being tacitly understood. It is easy to see that a supersolvable lattice has a rank function . He showed that, if such an L is also semimodular, then its characteristic polynomial (L; t) = X x2L (0̂; x)tn (x) (12) factors as (t a1)(t a2) (t an), where ai is the number of atoms of L that are below xi but not below xi 1. Our purpose in this section is to use Theorem 1.1 to prove Stanley's factorization of the characteristic polynomial for a wider class of lattices. For this purpose we will replace both supersolvability and semimodularity by weaker hypotheses. To state the rst of our two hypotheses, we de ne an element x of a lattice L to be left-modular if, for all y z, y _ (x ^ z) = (y _ x) ^ z: It is standard to call (x; z) a modular pair if the preceding equation is satis ed by every y that is z. So x is left-modular if and only if every pair with x on the left is modular. Note that left-modularity, unlike the analogously de ned right-modularity, is a self-dual concept of lattice theory. Our rst hypothesis is that 17 L has a maximal chain , all of whose elements are left-modular. In this case we will call L itself left-modular. Fix a maximal chain : 0̂ = x0 < x1 < : : : < xn 1 < xn = 1̂ whose elements may or may not be left-modular. We partition the set A of atoms of L into pieces Ai = fa 2 A j a xi but a 6 xi 1g, which we call the levels of A, and we partially order A by setting a b if and only if a is in an earlier level than b, where \earlier" means having a smaller subscript. We say that this order is induced by the chain . Note that an atom a cannot be WS for any set S of atoms from strictly lower levels, for there is an xi that is all elements of S but not a. We can now state our second hypothesis. If is induced by and a b1 b2 : : : bk then a 6 Wki=1 bi. A lattice L having a chain with this property will be said to satisfy the level condition. If L is a nite lattice with a maximal chain satisfying both the leftmodular and level conditions then it is called an LL lattice. Proposition 6.1 1. If L is supersolvable then it is left-modular but not conversely. 2. If L is semimodular then it satis es the level condition (for any maximal chain) but not conversely. Proof. Stanley has already pointed out that rst statement holds [23, Proposition 2.2 and .]. For the second, recall that semimodularity implies that if x 2 L and a is an atom not below x then x _ a covers x. Now prove the implication by induction on k, the number of b atoms in the level condition. The cases k = 0 and k = 1 are obvious, so suppose the result holds for k 1 but fails for k. So we have a b1 b2 : : : bk and a Wki=1 bi. Let c = Wk 1 i=1 bi and notice that a 6 c by induction hypothesis. Thus c < c _ a c _ bk. But bk is an atom, so c _ bk covers c. Therefore, c _ a = c _ bk bk. But this is absurd, as c _ a is a join of atoms from levels strictly earlier than that of bk. The implication is not reversible as the shu e posets Wm;1 serve as counterexamples. It is not hard to check that the level condition holds when y = y1 with being the chain where the xi's are rst removed one at a time and then y1 is added. On the other hand, as long as m 1, there are pairs of atoms not covered by their join (atoms that add y1 in di erent places), so Wm;1 is not a semimodular lattice. We intend to generalize to LL lattices Stanley's factorization of the characteristic polynomial. Since LL lattices need not be ranked, we need a suitable substitute 18 for the rank function used in de ning the characteristic polynomial. Let L be a left-modular lattice, with left-modular maximal chain inducing a partition into levels Ai as above. De ne the generalized rank of x 2 L to be (x) = number of Ai's containing atoms x. Later in this section, we shall relate this to lengths of chains, but for now we use it in (12) (with n still denoting the length of the chain ) to de ne the characteristic polynomial of any left-modular lattice. We shall obtain a factorization of this polynomial for any LL lattice. We rst characterize the NBB sets using the following lemma. Lemma 6.2 If a and b are distinct atoms from the same level Ai in a left-modular lattice, then a _ b is above some atom c from an earlier level. Proof. Since b is below xi but not below xi 1, we have xi 1 < xi 1 _ b xi. By maximality of the chain , it follows that xi 1 _ b = xi a _ b. Applying left-modularity of xi 1 with y = b and z = a _ b, we nd that b _ (xi 1 ^ (a _ b)) = (b _ xi 1) ^ (a _ b) = a _ b: But b 6 a so the right side of this equation is not b. Since the left side is also not b, we have that xi 1 ^ (a _ b) is not 0̂ and so is above some atom c satisfying the lemma. Theorem 6.3 In an LL lattice, the NBB sets are exactly those subsets of A that have at most one member in each level Ai. Proof. The level condition immediately implies that, ifB A has no two members from the same level then no subset of it is BB. Conversely, if B has two members at the same level then, by the lemma just proved, those two constitute a BB set. The following lemma is useful not only for our primary goal, factoring the characteristic polynomial of an LL lattice, but also for relating our generalized rank functions to lengths of chains. Lemma 6.4 Let B be an NBB set in an LL lattice. Then every atom a WB is at the same level as some element of B. In particular, any NBB base for x has exactly (x) elements. Proof. It su ces to prove the rst statement, since the second follows from it and Theorem 6.3. So suppose B and a were a counterexample to the rst statement. Let Aj be the level containing a, and let y be the join of all the elements of B 19 of higher level than Aj. Since B has no element at level Aj, we have a WB xj 1 _ y. Setting z = a _ y, we obtain from the left-modularity of xj 1 that (y _ xj 1) ^ z = y _ (xj 1 ^ z). On the left side of this last equation, both sides of the meet are a, and therefore so is the whole left side. On the right side, since z is a join of atoms from levels Aj and higher, the level condition tells us that z is above no atom of lower level than Aj, so no atom is below xj 1 ^ z. Therefore xj 1 ^ z = 0̂ and the right side reduces to y. Combining these results, we have a y. But, in view of the de nition of y, this contradicts the level condition. We can now prove our generalization of Stanley's theorem on semimodular supersolvable lattices. Theorem 6.5 If L is an LL lattice then characteristic polynomial of L factors as (L; t) = n Y i=1(t jAij) Proof. Combining Theorems 1.1 and 6.3 along with the new de nition of we have (L; t) = X x2L X B NBB WB=x ( 1)jBjtn (x) = X 8i: jB\Aij 1( 1)jBjtn jBj = n Y i=1(t jAij): To close this section, we point out two situations where our generalized rank can be described in terms of lengths of chains. The rst of these situations is in a semimodular (hence ranked), left-modular (hence LL) lattice L. As above, let be the partial order of A(L) induced by a maximal chain of left-modular elements and let be the generalized rank function given by the levels induced by the same chain. Also, let be an arbitrary linear extension of . We write NBB and NBB to mean NBB with respect to and , respectively. Notice that every NBB set is also an NBB set. As noted in our discussion of Rota's NBC theorem, the size of any NBB base for x is the ordinary rank of x. By Lemma 6.4, the size of any NBB base for x is (x). So the generalized rank (x) agrees with the ordinary rank of x provided x has at least one NBB base. This proviso can be reformulated as (x) 6= 0, so the characteristic 20 polynomial is the same for both notions of rank. The proviso cannot be omitted. A chain is a distributive lattice (hence also semimodular and left-modular) in which our generalized rank is 1 for all elements except 0̂ and therefore di ers from the ordinary rank if the chain has more than two members. The second situation is described in the following proposition, which connects to lengths of chains even in some unranked lattices. Proposition 6.6 Let L be a left-modular lattice which is atomic. Then for all x 2 L we have (x) = the length of the longest 0̂ to x chain. Proof. Let 0̂ = x0 < x1 < : : : < xn 1 < xn = 1̂ be the left-modular chain used for the de nitions of the sets Ai and thus of . There are, for any x 2 L, exactly (x) values of i such that Ai contains an atom x. The elements x ^ xi for these values of i (along with 0̂) constitute a chain of length (x) from 0̂ to x. It remains to show that no chain from 0̂ to x is longer, and for this it su ces to show that, for all a < b in L, (a) < (b). Let a < b and choose, by atomicity of L, an atom p b such that p 6 a and such that p 2 Aj for the smallest possible j. Then, by atomicity again, xj 1^b a. This and the left-modularity of xj 1 imply a = a _ (xj 1 ^ b) = (a _ xj 1) ^ b: If there were an atom q 2 Aj that is a, then we would have xj 1 < q_xj 1 xj and, by maximality of the chain of xi's, q _ xj 1 = xj. But then (a _ xj 1) ^ b xj ^ b p. This and a 6 p contradict the equation displayed above. So there is no such q. But that means that the Ai's counted by (b) include all those counted by (a) and at least one more, namely Aj. Therefore, (a) < (b). Notice that the hypotheses of the preceding proposition do not imply that L is ranked. A counterexample is given by the six-element lattice obtained from the eight-element Boolean algebra by removing two co-atoms. 7 More examples We will now give two examples where our factorization theorem can be applied but Stanley's cannot because the lattices involved are not semimodular. The rst example is the shu e poset Wm;1 with the maximal chain x = x0 < x1 < : : : < xm+1 = y (13) 21 where x1; : : : ; xm are obtained by deleting the letters of x in some order. (Note that we are using xi to denote an element of the chain rather than a letter of x.) Greene [13] showed that the given chain satis es the supersolvability condition even when extended in Wm;n by adding the letters of y in some order. So by Proposition 6.1 Wm;1 is left-modular. We also mentioned in the proof of the same proposition that this poset satis es the level condition. (However the level condition does not hold inWm;n for general n 2, and this is re ected by the fact that the corresponding characteristic polynomials usually do not factor over the integers.) It is now easy to see that the number of new atoms below xi in (13) is jAij = ( 1 if i m m+ 1 if i = m+ 1. From Theorem 6.5 we immediately get the following. Corollary 7.1 We have (Wm;1; t) = (t 1)m(t m 1): Note thatWm;n is ranked in the ordinary sense and Greene computed (Wm;n; t) using the usual rank function. But, by Proposition 6.6, this rank function coincides with ours. For our second example we will use the Tamari lattices [8, 9, 10, 11, 14]. Consider all proper parenthesizations of the word x1x2 : : : xn+1. It is well known that the number of these is the Catalan number Cn. Partially order this set by saying that covers whenever = : : : ((AB)C) : : : and = : : : (A(BC)) : : : for some subwords A;B;C. The corresponding poset turns out to be a lattice called the Tamari lattice Tn. Figure 6(a) gives a picture of T3. A left bracket vector, (v1; : : : ; vn), is a vector of positive integers satisfying 1. 1 vi i for all i and 2. if Si = fvi; vi + 1; : : : ; ig then for any pair Si; Sj either one set contains the other or Si \ Sj = ;. The number of left bracket vectors having n components is also Cn. In fact given a parenthesized word we have an associated left bracket vector v = (v1; : : : ; vn) de ned as follows. To calculate vi, start at xi in and move left, counting the number of x's and the number of left parentheses you pass (including xi itself) until 22 s (x1(x2(x3x4))) s (x1((x2x3)x4)) s ((x1(x2x3))x4) s (((x1x2)x3)x4) s((x1x2)(x3x4)) HHHH JJJJJJ s (1; 2; 3) s (1; 2; 2) s (1; 2; 1) s (1; 1; 1) s(1; 1; 3) HHHH JJJJJJ (a) Parenthesized version (b) Left bracket version Figure 6: The Tamari lattice T3 these two numbers are equal. Then vi = j where xj is the last x passed before the numbers balance. It is not hard to show that this gives a bijection between parenthesizations and left bracket vectors, thus inducing a partial order on the latter. In fact this induced order is just the component-wise one. Figure 6(b) gives the bracket vector version of T3. Left bracket vectors are also directly related to the trees GB described in Theorem 2.2, but with n + 1 in place of n. Indeed, given a left bracket vector (v1; : : : ; vn), we obtain such a tree GB with vertex set [n+1] by joining i+1 to vi for i = 1; 2; : : : ; n, and all trees as in Theorem 2.2 can be obtained in this way. We should note that we have used the left bracket vector rather than the more traditional right bracket one because when using the former it is easier to describe the join operation than the meet. (The situation is reversed for the right bracket vector.) This makes it simpler to work with some of our conditions which only involve joins. Expressions for the join and meet can be obtained by dualizing results in [14] and [17] respectively. Proposition 7.2 Given left bracket vectors v = (v1; : : : ; vn) and w = (w1; : : : ; wn) then v _ w = (maxfv1; w1g; : : : ;maxfvn; wng): If we let m = (m1; : : : ;mn) where mi = minfvi; wig then v ^w = (l1; : : : ; ln) where the li are computed recursively by li = minfmi; lmi; lmi+1; : : : ; li 1g: To show that Tm is LL, consider the chain : (1; : : : ; 1) < (1; 2; 1; : : : ; 1) < (1; 2; 2; 1; : : : ; 1) < (1; 2; 3; 1; : : : ; 1) < (1; 2; 3; 2; 1; : : : ; 1) < : : : < (1; 2; 3; : : : ; n): 23 It is easy to see that this is of maximum length in Tn. The description of join in Proposition 7.2 also gives a quick proof that the level condition holds. To verify left-modularity will take more work. A typical element of looks like x = (1; 2; : : : ; j 1; i; 1; : : : ; 1) (14) where i j. Take y = (y1; : : : ; yn) z = (z1; : : : ; zn) in Tn. We wish to compute y _ (x ^ z) = (c1; : : : ; cn) so we rst consider x ^ z. Following the notation of Proposition 7.2 we have m = (z1; : : : ; zj 1;minfzj; ig; 1; : : : ; 1): Using the recursive construction on the rst j 1 components of m leaves them unchanged since these are the initial components of a vector in Tj 1. Also the last n j components will still be 1 because mi is in the minimum taken for li. So x ^ z is the same as m except possibly in the jth component which is zj if zj i or minfi; zi; : : : ; zj 1g if zj i. Since y z we have yj zj for all j and so y _ (x ^ z) = (z1; : : : ; zj 1; cj; yj+1; : : : ; yj) (15) where cj = ( zj if zj i maxfyj;minfi; zi; : : : ; zj 1gg if zj i. (16) Now we concentrate on (y _ x) ^ z = (d1; : : : ; dn). Clearly y _ x = (1; 2; : : : ; j 1;maxfi; yjg; yj+1; : : : ; yn) Taking the minimum vector to compute (y _ x) ^ z we get m0 = (z1; : : : ; zj 1; d0j ; yj+1; : : : ; yn) where d0j is zj if zj i and maxfi; yjg if zj i. So in (y_x)^z we have di = zi = ci for i < j by the same reasoning as before. We claim that di = yi for i > j. From the minimization procedure in Proposition 7.2 we have di mi = yi. But in any lattice (y _ x) ^ z y _ (x ^ z) and so di ci for all 1 i n: (17) Since ci = yi for i > j this forces the desired equality. To complete the proof of left-modularity we must show dj = cj and this breaks up into three cases. If zj i then we have that dj = zj = cj. If yj i zj then 24 d0j = i and dj = minfi; zi; : : : ; zj 1g. Comparison of this last equation with (16) gives dj cj and then (17) results in dj = cj. Finally if i yj zj then d0j = yj and dj = minfyj; zyj ; : : : ; zj 1g yj = cj: Using (17) again nishes this last case. It is now an easy matter to compute the characteristic polynomial. Note that the atoms of Tn are all elements of the form (1; : : : ; 1; j; 1; : : : ; 1) with the j 2 in the jth position. If x is as in (14) then it covers no new atoms if i < j and exactly one new atom if i = j. Using Theorem 6.5 this translates as follows. Corollary 7.3 We have (Tn; t) = t(n 1 2 ) (t 1)n 1: 8 A further generalization In this section, we brie y describe a generalization of Theorem 1.1. As before, we work with a nite lattice L, but instead of an additional partial ordering of the atoms we use an arbitrary function M assigning to each x 2 L n 0̂ a non-empty set M(x) of atoms x. For comparison with Theorem 1.1, the reader should regard the ordering used there as inducing the function M(x) = fa 2 A(L) j a is -minimal among atoms xg: But what follows applies also to M 's that do not arise from partial orderings in this way. For any set B of atoms, let S(B) be the subset obtained by deleting from B all members of M(x) for all x WB. By the core of B, we mean the set obtained by starting with B and repeatedly applying S until the decreasing sequence B;S(B); S(S(B)); : : : stabilizes; the nal Sn(B) is the core of B. (This can also be described as the largest subset of B unchanged by S.) We call B coreless if its core is empty. Theorem 8.1 Let L be any nite lattice and let M be any function assigning to every x 2 L n 0̂ a nonempty set M(x) of atoms that are x. Then for all x 2 L we have (x) =XB ( 1)jBj where the sum is over all coreless B A(L) whose join is x. 25 We omit the proof since it is essentially the same as that of Theorem 1.1. The only di erence is that, instead of choosing a -minimal atom a0 x, we choose an a0 2M(x). Notice that, if M happens to be obtained from a partial order as described above, then non-empty cores with respect to M are the same as BB sets with respect to , and therefore coreless sets with respect to M are the same as NBB sets with respect to . Notice also that the theorem of this section is \stronger," in the sense of having fewer summands in the formula for the Mobius function, when M(x) is smaller, for this will make S(B) larger and therefore make B less likely to be coreless. From this point of view, one should always take M(x) to be a singleton. Of course considerations of naturality or clarity may make other choices of M preferable (just as they may make nonlinear orderings preferable to the more e cient linear orderings in applications of Theorem 1.1). 9 Comments and Questions (1) It would be interesting to nd other applications of our two main theorems. The higher Stashe -Tamari posets as recently de ned in [8] are obvious candidates. However, one would rst have to prove that they are lattices. (2) In a previous paper [21] one of us proved a somewhat weaker generalization of Rota's NBC Theorem which, nonetheless, has some interesting connections to our work and others. To state the result we must rst recall some de nitions from [21]. Call B A(L) independent if WB0 < WB for every proper subset B0 B. So C is dependent if WC 0 = WC for some C 0 C. Note that these generalize the corresponding notions for a geometric lattice. The de nitions of base, circuit, and broken circuit remain as before. The main result of [21] can now be stated. Theorem 9.1 Let L be a nite lattice. Let be any total ordering of A(L) such that for all ciruits C we have _C = _(C nminC): (18) Then for all x 2 L we have (x) =XB ( 1)jBj where the sum is over all NBC bases of x. 26 We should note that not every lattice has a total ordering of the atoms satis-fying (18) and so this result is not as strong as Theorem 1.1. On the other hand,there is an interesting relationship between a dual of this restriction and the levelcondition from Section 6.Proposition 9.2 Let L be a nite lattice. Let be an induced ordering of A(L)such that for all ciruits C having a unique maximal element we have_C = _(C nmaxC):Then L satis es the level hypothesis.Proof. Suppose, toward a contradiction, that we have b0 b1 : : : bk andb0Wi 1 bi. Then D = fb0; b1; : : : ; bkg is dependent and so contains a circuit C.Also C must have a unique maximal element bj because D intersects each level atmost once. So we have WC = W(C n bj), or equivalently bj W(C n bj). But thiscannot happen since for some x of the inducing chain we have bi x for i < j butbj 6 x.In his work [1] on Tutte polynomials for hypermatroids, Christos Athanasiadisde nes a generalized lattice to be a triple L = (L;E; f) where L is a nite lattice,E is a set, and f : E ! L is some map. He then proves a theorem about theMobius function of L in the case that the hypermatroid associated with L sati escondition (18). In the case that E = A(L) and f is inclusion his result becomes aspecial case of Theorem 9.1.(3) There are certain to be topological rami cations of our work. In fact YoavSegev [22] has already proved that the order complex of any lattice is homotopyequivalent to the simplicial complex of all NBB sets B with WB < 1̂. This canbe used to demonstrate a recent result of Linusson [16] that the order complex ofan interval in the partition lattice under dominance is homotopy equivalent to asphere or contractible.An important use of NBC sets is to give a basis for the Orlik-Solomon algebraA(L) of a geometric lattice L [18]. If L is the intersection lattice of a complexhyperplane arrangement, then A(L) is isomorphic to the cohomology algebra ofthe complement of the arrangement. Recently there has been a lot of interest insubspace arrangements [2]. The corresponding intersection lattices are no longergeometric, but perhaps NBB sets can be used to give information about the asso-ciated cohomology algebra in this case. Recently, De Concini and Procesi [7] usedalgebraic geometric techniques to show that this algebra is indeed determinedsolely by the lattice and dimension information. This provides some hope that acombinatorial description is also possible.27 References[1] C. A. Athanasiadis, The Tutte polynomial of a hypermatroid, preprint, 1995.[2] A. Bjorner, Subspace arrangements, in \Proc. 1st European Congress Math.(Paris 1992)," A. Joseph and R. Rentschler eds., Birkhauser, Boston, MA, toappear.[3] A. Bjorner and L. Lovasz, Linear decision trees, subspace arrangements andMobius functions, J. Amer. Math. Soc. 7 (1994), 677{706.[4] A. Bjorner, L. Lovasz and A. Yao, Linear decision trees: volume estimates andtopological bounds, in \Proc. 24th ACM Symp. on Theory of Computing,"ACM Press, New York, NY, 1992, 170{177.[5] K. Bogart, The Mobius function of the domination lattice, unpublishedmanuscript, 1972.[6] T. Brylawski, The lattice of integer partitions, Discrete Math. 6 (1973), 201{219.[7] C. De Concini and C. Procesi, Wonderful models of subspace arrangements,preprint, 1995.[8] P. H. Edelman and V. Reiner, The higher Stashe -Tamari posets, preprint.[9] H. Friedman and D. Tamari, Problemes d'associativite: Une treillis nie in-duite par une loi demi-associative, J. Combin. Theory 2 (1967), 215{242.[10] W. Geyer, On Tamari lattices, Discrete Math. 133 (1994), 99{122.[11] G. Gratzer, \Lattice Theory," Freeman and Co., San Francisco, CA, 1971, pp.17{18, problems 26-36.[12] C. Greene, A class of lattices with Mobius function 1, European J. Combin.9 (1988), 225{240.[13] C. Greene, Posets of Shu es, J. Combin. Theory Ser. A 47 (1988), 191{206.[14] S. Huang and D. Tamari, Problems of associativity: A simple proof for thelattice property of systems ordered by a semi-associative law, J. Combin.Theory Ser. A 13 (1972), 7{13.[15] G. Kreweras, Sur les partitions non-croisees d'un cycle, Discrete Math. 1(1972), 333{350.28 [16] S. Linusson, A class of lattices whose intervals are spherical or contractible,preprint.[17] G. Markowsky, Primes, irreducibles and extremal lattices, Order 9 (1992),7{13.[18] P. Orlik and L. Solomon, Combinatorics and topology of complements ofhyperplanes. Invent. Math. 56 (1980), 167{189.[19] V. Reiner, Non-crossing partitions for classical re ection groups, preprint.[20] G.-C. Rota, On the foundations of combinatorial theory I. Theory of Mobiusfunctions, Z. Wahrscheinlichkeitstheorie 2 (1964), 340{368.[21] B. E. Sagan, A generalization of Rota's NBC Theorem, Adv. in Math., 111(1995), 195{207.[22] Y. Segev, The simplicial complex of all NBB nonspanning subsets of the setof atoms of a prelattice L is homotopic to the order complex of L, preprint,1995.[23] R. P. Stanley, Supersolvable lattices, Alg. Univ., 2 (1972), 197{217.[24] R. P. Stanley, \Enumerative Combinatorics, Volume 1," Wadsworth &Brooks/Cole, Paci c Grove, CA, 1986.[25] H. Whitney, A logical expansion in mathematics, Bull. Amer. Math. Soc. 38(1932), 572{579.[26] T. Zaslavsky, The geometry of root systems and signed graphs, Amer. Math.Monthly 88 (1981), 88{105.29