A Survey of the Valuation Algebra motivated by a Fundamental Application to Dissection Theory
AA Survey of the Valuation Algebramotivated by a Fundamental Applicationto Dissection Theory
Hery Randriamaro This research was funded by my mother
Lot II B 32 bis Faravohitra, 101 Antananarivo, Madagascar [email protected]
Abstract
A lattice L is said lowly finite if the set [ , a ] is finite for every element a of L . We mainlyaim to provide a complete proof that, if M is a subset of a complete lowly finite distributivelattice L containing its join-irreducible elements, and a an element of M which is not join-irreducible, then (cid:88) b ∈ M ∩ [ ,a ] µ M ( b, a ) b belongs to the submodule (cid:104) a ∧ b + a ∨ b − a − b | a, b ∈ L (cid:105) of Z L . That property was originally established by Zaslavsky for finite distributive lattice. Itis essential to prove the fundamental theorem of dissection theory as will be seen. We finishwith a concrete application of that theorem to face counting for submanifold arrangements. Keywords : Lattice, Valuation, M¨obius Algebra, Subspace Arrangement
MSC Number : 06A07, 06A11, 06B10, 57N80
A distributive lattice is a partially order set with join and meet operations which distributeover each other. The prototypical example of a such structure are the sets where join andmeet are the usual union and intersection. Further examples include the Lindenbaum alge-bra of logics that support conjunction and disjunction, every Heyting algebra, and Young’slattice whose join and meet of two partitions are given by the union and intersection of thecorresponding Young diagrams. This survey mainly aims to provide a complete proof that, if L is a complete lowly finite distributive lattice, M a subset of L containing its join-irreducibleelements, f : L → G a valuation on L to a module G , and a an element of M which is notjoin-irreducible, then (cid:88) b ∈ [ ,a ] ∩ M µ M ( b, a ) f ( b ) = 0 . (1)The proof is carried out in several stages. We first have to consider the general case of posetsin Section 2. A proof of Zorn’s lemma and an introduction to the M¨obius algebra M¨ob( L ) a r X i v : . [ m a t h . C O ] J a n ¨obius Algebra L are namely provided. The lemma bearing his name was proved byZorn in 1935 [19]. Although we can easily find diverse proofs of that lemma in the literature,new ones are still proposed other time like that of Lewin in 1991 [11]. Ours is inspired by thelecture notes of Debussche [5, § § (cid:26) (cid:88) b ∈ [ ,a ] µ L ( b, a ) b (cid:12)(cid:12)(cid:12)(cid:12) a ∈ L (cid:27) is a complete set of orthogonal idempotents in M¨ob( L ).We study the special case of lattices in Section 3. After viewing some essential generalities,we focus on the distributive lattices, and establish diverse properties like the distributivity ofa lattice L if and only if, for all a, b, c ∈ L , c ∨ a = c ∨ b and c ∧ a = c ∧ b imply a = b .Those properties are necessary to investigate the valuation algebra in Section 4. It is thecentral part of this survey, and principally inspired from the articles of Geissinger [7], [8, § § M is a subset ofa complete lowly finite distributive lattice L containing its join-irreducible elements, and a anelement of M which is not join-irreducible, then (cid:88) b ∈ M ∩ [ ,a ] µ M ( b, a ) b belongs to the submodule (cid:104) a ∧ b + a ∨ b − a − b | a, b ∈ L (cid:105) of Z L . It is the property that allows to deduce Equation 1.Thereafter, we use Equation 1 to deduce the fundamental theorem of dissection theory inSection 5, that is, if A is a subspace arrangement in a topological space T with | χ ( T ) | < ∞ ,and L a meet-refinement of L A , then (cid:88) C ∈ C A χ ( C ) = (cid:88) X ∈ L µ L ( X, T ) χ ( X ). In its original form of1977 [18, Theorem 1.2], Zaslavsky expressed it for CW complexes. The number of chambersis consequently C A = 1 c (cid:88) X ∈ L µ L ( X, T ) χ ( X ) if c is the Euler characteristic of every chamber.Deshpande showed a similar result in 2014 for the special case of a submanifold arrangementwith chambers having the same Euler characteristic ( − l [6, Theorem 4.6].We finally compute the f-polynomial of submanifold arrangements from the dissection theoremof Zaslavsky in Section 6. Face counting of topological space has doubtless its origins in theformulas for the numbers of the i -dimensional faces if planes in R fall into k parallel families ingeneral position established by Steiner in 1826 [15]. About 150 years later, Alexanderson andWetzel computed the same numbers but for an arbitrary set of planes [1], and Zaslavsky forhyperplane arrangements in a Euclidean space of any dimension [17, Theorem A]. One of ourformulas is a generalization of those results as it considers a submanifold arrangement A suchthat χ ( X ) = ( − dim X for every X ∈ L A ∪ F A , and states that f A ( x ) = ( − rk A M A ( − x, − A is the M¨obius polynomial of A . Moreover, Pakula computed the number ofchambers of a pseudosphere arrangement with simple complements in 2003 [12, Corollary 1].Another formula is a generalization of his result considering a submanifold arrangement A such that ∀ C ∈ F A : χ ( C ) = ( − dim C and ∀ X ∈ L A : χ ( X ) = (cid:40) X ≡ , and statesf A ( x ) = ( − n − rk A (cid:0) M A ( x, −
1) + γ n M A ( − x, − (cid:1) with γ n := (cid:40) X ≡ − . H. Randriamaro
Page 2 of 25 ¨obius Algebra
We begin with the general case of posets. The Zorn’s lemma is especially proved, and theM¨obius algebra described as it plays a key role in this survey.
Definition 2.1. A partial order is a binary relation (cid:22) over a set L such that, for a, b, c ∈ L , • a (cid:22) a , • if a (cid:22) b and a (cid:23) b , then a = b , • if a (cid:22) b and b (cid:22) c , then a (cid:22) c .The set L with a partial order is called a partially ordered set or poset , and two elements a, b ∈ L are said comparable if a (cid:22) b or a (cid:23) b . Definition 2.2.
A poset L has an uppest resp. lowest element resp. ∈ L if, for every a ∈ L , one has a (cid:22) resp. a (cid:23) . The poset is said complete if it has an uppest and alowest element. Definition 2.3.
A subset C of a poset P is a chain if any two elements in C are comparable.Denote by C L the set formed by the chains of a poset L . A subset S of L has an upper resp.lower bound if there exists u resp. l ∈ L such that s (cid:22) u resp. l (cid:22) s for each s ∈ S . Theupper resp. lower bound u resp. l is said strict if u resp. l / ∈ S . Definition 2.4.
A poset L is said inductive if every chain included in L has an upper bound.For an inductive poset L , and C ∈ C L , let C ≺ be the set formed by the strict upper bound of C , and denote by E L the set { C ∈ C L | C ≺ (cid:54) = ∅} . The axiom of choice allows to deduce theexistence of a function c : 2 L \ {∅} → L such that, for every A ∈ L \ {∅} , we have c( A ) ∈ A .Define the function m : E L → L , for C ∈ E L , by m( C ) := c( C ≺ ). Definition 2.5.
Let
S, A be subsets of a poset L . The set S is called a segment of A if S ⊆ A and ∀ s ∈ S, ∀ a ∈ A : s (cid:23) a ⇒ a ∈ S. Definition 2.6.
An upper resp. lower bound u resp. l of the subset S of a poset L is calleda join resp. meet if u (cid:22) a resp. b (cid:22) l for each upper resp. lower bound a resp. b of S . Definition 2.7.
A chain C of an inductive poset L is called a good set if, for every segment S of C with S (cid:54) = C , we have S ≺ ∩ C (cid:54) = ∅ and m( S ) is the meet of S ≺ ∩ C .For elements a, b of a poset, by a ≺ b we mean that a (cid:22) b and a (cid:54) = b . Lemma 2.8.
Let
A, B be nonempty good sets of an inductive poset L . Then, either A is asegment of B or vice versa. H. Randriamaro
Page 3 of 25 ¨obius Algebra
Proof.
Note first that ∅ is a chain of L . As L is inductive, ∅ has then an upper bound in L which is necessary a strict upper bound, hence ∅ ∈ E L . Moreover, since ∅ is obviously asegment of both A and B which are good sets, then m( ∅ ) ∈ ∅ ≺ ∩ A ∩ B and A ∩ B (cid:54) = ∅ .For a ∈ A ∩ B , the sets S a,A := { s ∈ A | s ≺ a } and S a,B := { s ∈ B | s ≺ a } are clearlysegments of A and B respectively. Set C := { a ∈ A ∩ B | S a,A = S a,B } , and let b ∈ C , c ∈ A ,with b (cid:31) c . We have c ∈ S b,A = S b,B , then c ∈ B which implies c ∈ A ∩ B . If d ∈ S c,A ,then d ≺ c ≺ b implies d ∈ S b,A = S b,B , hence b ∈ S c,B and S c,A ⊆ S c,B . Similarly, we have S c,B ⊆ S c,A , then c ∈ C . Therefore, C is a segment of A and B .Suppose now that C (cid:54) = A and C (cid:54) = B . As A, B are good sets, then m( C ) ∈ A ∩ B . Remarkthat C (cid:116) (cid:8) m( C ) (cid:9) = S m( C ) ,A = S m( C ) ,B , then m( C ) ∈ C which is absurd. Hence C = A or C = B , in other words, A is a segment of B or vice versa.Denoting by G L the set formed by the good sets of an inductive poset L , set U L := (cid:91) A ∈G L A . Theorem 2.9. If L is an inductive poset, then U L is a good set.Proof. For a, b ∈ U L , there exist good sets S a , S b such that a ∈ S a and b ∈ S b . UsingLemma 2.8, we get either S a ⊆ S b or S b ⊆ S a . That means either a (cid:22) b or a (cid:23) b , and U L isconsequently a chain.Let A ∈ G L , a ∈ A , and b ∈ U L with a (cid:23) b . There is B ∈ G L with b ∈ B . From Lemma 2.8, • if A is a segment of B , then A is a segment and b ∈ A , • if B is a segment of A , then B ⊆ A and b ∈ A .In any case, we have b ∈ A , then A is a segment of U L .Consider a segment S of U L such that S (cid:54) = U L . Since U L is a chain, necessarily U L \ S ⊆ S ≺ .Let a ∈ U L \ S , and A ∈ G L such that a ∈ A . As A is a segment of U L , then S (cid:32) A and S isa segment of A . Moreover, m( S ) is the meet of S ≺ ∩ A . If there exists b ∈ S ≺ ∩ U L such that b ≺ m( S ), we would get b ∈ A , which is absurd. Therefore, m( S ) is the meet of S ≺ ∩ U L , and U L is a good set. Definition 2.10.
An element a of a poset L is said maximal if there does not exist anelement b ∈ L \ { a } such that b (cid:31) a . Corollary 2.11 (Zorn’s Lemma) . Every inductive poset L has a maximal element.Proof. Recall that, since U L is a chain, it consequently possesses an upper bound. Suppose U L ≺ (cid:54) = ∅ , and let u ∈ U L ≺ . Then U L (cid:116) { u } is a good set which is absurd. Hence, U L has aunique upper bound, contained in U L , which is a maximal element of L . For two elements a, b of a poset L such that a (cid:22) b , denote by [ a, b ] the set { c ∈ L | a (cid:22) c (cid:22) b } . Definition 2.12.
A poset L is locally finite if, for all a, b ∈ L such that a (cid:22) b , [ a, b ] is finite.For a locally finite poset L , denote by Inc ( L ) the module of the functions f : L → Z withthe property that, if x, y ∈ L , then f ( x, y ) = 0 if x (cid:14) y .H. Randriamaro Page 4 of 25 ¨obius Algebra
Definition 2.13.
The incidence algebra
Inc( L ) of a locally finite poset L is the module offunctions f : L → Z , having the property f ( a, b ) = 0 if a (cid:14) b , with distributive multiplication h = f · g defined, for f, g ∈ Inc( L ), by h ( a, b ) := 0 if a (cid:14) b and h ( a, b ) := (cid:88) c ∈ [ a,b ] f ( a, c ) g ( c, b ) otherwise . Its multiplicative identity is the Kronecker delta δ : L → Z with δ ( a, b ) := (cid:40) a = b, Definition 2.14.
For a locally finite poset L , the zeta function ζ L and the M¨obius func-tion µ L in the incidence algebra Inc( L ) are defined, for a, b ∈ L with a (cid:22) b , by ζ L ( a, b ) := 1 and µ L ( a, b ) := a = b − (cid:88) c ∈ [ a,b ] c (cid:54) = b µ L ( a, c ) = − (cid:88) c ∈ [ a,b ] c (cid:54) = a µ L ( c, b ) otherwise . Lemma 2.15.
For a locally finite poset L , the zeta function is the multiplicative inverse ofthe M¨obius function in the incidence algebra Inc( L ) .Proof. For a, b ∈ L with a (cid:22) b , we have ζ L · µ L ( a, a ) = µ L · ζ L ( a, a ) = 1 = δ ( a, a ), but also ζ L · µ L ( a, b ) = (cid:88) c ∈ [ a,b ] µ L ( c, b ) = 0 = δ ( a, b ) and µ L · ζ L ( a, b ) = (cid:88) c ∈ [ a,b ] µ L ( a, c ) = 0 = δ ( a, b ) . The proof of this proposition is inspired from the original proof of Rota [13, Proposition 2].
Proposition 2.16 (M¨obius Inversion Formula) . Let L be a locally finite poset, a, b ∈ L with a (cid:22) b , and f, g two functions from L onto a module M over Z . Then, ∀ x ∈ [ a, b ] : g ( x ) = (cid:88) c ∈ [ a,x ] f ( c ) ⇐⇒ ∀ x ∈ [ a, b ] : f ( x ) = (cid:88) c ∈ [ a,x ] g ( c ) µ L ( c, x ) . Proof.
Assume first that, for every x ∈ [ a, b ], g ( x ) = (cid:88) c ∈ [ a,x ] f ( c ). Using Lemma 2.15, we get (cid:88) c ∈ [ a,x ] g ( c ) µ L ( c, x ) = (cid:88) c ∈ [ a,x ] (cid:88) d ∈ [ a,c ] f ( d ) µ L ( c, x ) = (cid:88) c ∈ [ a,x ] (cid:88) d ∈ [ a,c ] f ( d ) ζ L ( d, c ) µ L ( c, x )= (cid:88) d ∈ [ a,c ] (cid:88) c ∈ [ a,x ] f ( d ) ζ L ( d, c ) µ L ( c, x ) = (cid:88) d ∈ [ a,c ] f ( d ) (cid:88) c ∈ [ a,x ] ζ L ( d, c ) µ L ( c, x )= (cid:88) d ∈ [ a,c ] f ( d ) ζ L · µ L ( d, x ) = (cid:88) d ∈ [ a,c ] f ( d ) δ ( d, x )= f ( x ) . H. Randriamaro
Page 5 of 25 ¨obius Algebra f ( x ) = (cid:88) c ∈ [ a,x ] g ( c ) µ L ( c, x ) for every x ∈ [ a, b ], we obtain (cid:88) c ∈ [ a,x ] f ( c ) = (cid:88) c ∈ [ a,x ] (cid:88) d ∈ [ a,c ] g ( d ) µ L ( d, c ) = (cid:88) c ∈ [ a,x ] (cid:88) d ∈ [ a,c ] g ( d ) µ L ( d, c ) ζ L ( c, x ) = (cid:88) d ∈ [ a,c ] g ( d ) δ ( d, x )= g ( x ) . Definition 2.17.
We say that a poset L is lowly finite if the set { b ∈ L | b (cid:22) a } is finitefor any a ∈ L .For a lowly finite poset L and a ∈ L , let u L ( a ) be the element (cid:88) c ∈ Lc (cid:22) a µ L ( c, a ) c of the module Z L . Definition 2.18.
The
M¨obius Algebra
M¨ob( L ) of a lowly finite poset L is the module Z L with distributive multiplication defined, for a, b ∈ L , by a · b := (cid:88) c ∈ Lc (cid:22) a, c (cid:22) b u L ( c ) . Remark that the M¨obius algebra was initial defined for finite posets [14, § L with a lowest element, define the algebra A L := (cid:104) α a | a ∈ L (cid:105) over Z withmultiplication α a α b := (cid:40) α a if a = b, . To each a ∈ L , associate an element a (cid:48) ∈ A L by setting a (cid:48) := (cid:88) b ∈ [ ,a ] α b . Lemma 2.19.
For a lowly finite poset L with a lowest element, the set { a (cid:48) | a ∈ L } forms abasis of the algebra A L .Proof. From the M¨obius inversion formula, we get α a = (cid:88) b ∈ [ ,a ] µ ( b, a ) b (cid:48) . The set { a (cid:48) | a ∈ L } consequently generates A L . Suppose that there exists a finite set I ⊆ L and an integer set { i a } a ∈ I such that (cid:88) a ∈ I i a a (cid:48) = 0. If b is a maximal element of I , then α b (cid:88) a ∈ I i a a (cid:48) = i b α b = 0,hence i b = 0. Inductively, we deduce that i a = 0 for every a ∈ I . The set { a (cid:48) | a ∈ L } istherefore independent.The following results were initially established by Greene for finite lattice [10, § Theorem 2.20.
For a lowly finite poset L with a lowest element, the map φ : L → A L , a (cid:55)→ a (cid:48) extends to an algebra isomorphism from M¨ob( L ) to A L . H. Randriamaro
Page 6 of 25 ¨obius Algebra
Proof.
The map φ clearly becomes a module homomorphism by linear extension, and anisomorphism by Lemma 2.19. Moreover, for a, b ∈ L , φ ( a · b ) = φ (cid:16) (cid:88) c ∈ [ ,a ] ∩ [ ,b ] u L ( c ) (cid:17) = (cid:88) c ∈ [ ,a ] ∩ [ ,b ] φ (cid:16) (cid:88) d ∈ [ ,c ] µ ( d, c ) d (cid:17) = (cid:88) c ∈ [ ,a ] ∩ [ ,b ] (cid:88) d ∈ [ ,c ] µ ( d, c ) d (cid:48) = (cid:88) c ∈ [ ,a ] ∩ [ ,b ] α c , and φ ( a ) φ ( b ) = a (cid:48) b (cid:48) = (cid:88) c ∈ [ ,a ] α c × (cid:88) d ∈ [ ,b ] α d = (cid:88) c ∈ [ ,a ] ∩ [ ,b ] α c . Then φ ( a · b ) = φ ( a ) φ ( b ), and φ isconsequently an algebra isomorphism. Corollary 2.21.
For a lowly finite poset L with a lowest element, the set { u L ( a ) | a ∈ L } isa complete set of orthogonal idempotents in M¨ob( L ) .Proof. Since φ (cid:0) u L ( a ) (cid:1) = (cid:88) b ∈ [ ,a ] µ L ( b, a ) b (cid:48) = α a , then { u L ( a ) | a ∈ L } is a basis of M¨ob( L ).Moreover φ (cid:0) u L ( a ) · u L ( a ) (cid:1) = α a = φ (cid:0) u L ( a ) (cid:1) , so the u L ( a )’s are idempotents.Finally φ (cid:0) u L ( a ) · u L ( b ) (cid:1) = α a α b = 0 if a (cid:54) = b , hence the u L ( a )’s are orthogonal. Corollary 2.22.
Let L be a lowly finite poset with a lowest element, and M a subset of L containing . Then, the linear map j : M¨ob( L ) → M¨ob( M ) , which on the basis { u L ( a ) | a ∈ L } has the values j (cid:0) u L ( a ) (cid:1) := (cid:40) u M ( a ) if a ∈ M, otherwise , is an algebra homomorphism.Proof. Using Corollary 2.21, j (cid:0) u L ( a ) · u L ( a ) (cid:1) = j (cid:0) u L ( a ) (cid:1) = u M ( a ) = j (cid:0) u L ( a ) (cid:1) · j (cid:0) u L ( a ) (cid:1) if a ∈ M . Otherwise, j (cid:0) u L ( a ) · u L ( a ) (cid:1) = 0 = j (cid:0) u L ( a ) (cid:1) · j (cid:0) u L ( a ) (cid:1) . For a, b ∈ L with a (cid:54) = b ,j (cid:0) u L ( a ) · u L ( b ) (cid:1) = 0 = j (cid:0) u L ( a ) (cid:1) · j (cid:0) u L ( b ) (cid:1) . We study the special but important case of lattices. After viewing some generalities, we focuson distributive ones, and establish diverse properties which are necessary to investigate thevaluation algebra in the next section.
Definition 3.1.
A poset L is a join-semilattice resp. meet-semilattice if each 2-element subset { a, b } ⊆ L has a join resp. meet denoted by a ∨ b resp. a ∧ b . It is called a lattice if L isboth a join- and meet-semilattice, moreover ∨ and ∧ become binary operations on L . Proposition 3.2.
If a lattice L is lowly finite, then it has a lowest element.Proof. For any a ∈ L , the principal ideal id( a ) has a lowest element which is a := (cid:94) x ∈ id( a ) x .Consider b ∈ L \ { a } and the lowest element b of id( b ). The fact a ∧ b (cid:54) = a would contradictthe fact that a is the lowest element of id( a ). Hence, L has a lowest element .H. Randriamaro Page 7 of 25 ¨obius Algebra
Definition 3.3. A sublattice of a lattice L is a nonempty subset M ⊆ L such that, for all a, b ∈ M , we have a ∨ b ∈ M and a ∧ b ∈ M . Definition 3.4. A lattice homomorphism is a function ϕ : L → L between two lattices L and L such that, for all a, b ∈ L , ϕ ( a ∨ b ) = ϕ ( a ) ∨ ϕ ( b ) and ϕ ( a ∧ b ) = ϕ ( a ) ∧ ϕ ( b ) . Definition 3.5. An ideal of a lattice L is a sublattice I ⊆ L such that, for any a ∈ I and b ∈ L , we have a ∧ b ∈ I . If in addition I (cid:54) = L and, for any a ∧ b ∈ I , either a ∈ I or b ∈ I ,then I is a prime ideal . Definition 3.6.
Dually, a filter of a lattice L is a sublattice F ⊆ L such that, for any a ∈ F and b ∈ L , we have a ∨ b ∈ F . If in addition F (cid:54) = L and, for any a ∨ b ∈ F , either a ∈ F or b ∈ F , then F is a prime filter . Proposition 3.7.
A subset M of a lattice L is a prime ideal if and only if the subset L \ M is a prime filter.Proof. Assume that M is a prime ideal: • If a, b ∈ L \ M , clearly a ∧ b ∈ L \ M and a ∨ b ∈ L \ M since ( a ∨ b ) ∧ b ∈ L \ M , then L \ M is a sublattice. • If a ∈ M and b ∈ L \ M , once again a ∨ b ∈ L \ M since ( a ∨ b ) ∧ b ∈ L \ M , then L \ M is a filter. • If a ∨ b ∈ L \ M , it is clear that both a, b cannot be all in M , then L \ M is prime.One similarly proves that if M is a prime filter, then L \ M is a prime ideal. Definition 3.8.
Let L be a lattice, and a ∈ L . The principal ideal generated by a isthe ideal id( a ) := { b ∈ L | b (cid:22) a } , dually the principal filter generated by a is the filterfil( a ) := { b ∈ L | b (cid:23) a } . Definition 3.9.
An element a of a lattice L is join-irreducible if, for any subset S ⊆ L , a = (cid:95) b ∈ S b implies a ∈ S . Denote by ji( L ) the set formed by the join-irreducible elements of L . Lemma 3.10.
Let L be a lattice, and a ∈ L . Then, a ∈ ji( L ) if and only if a (cid:54) = (cid:95) b ∈ Lb ≺ a b .Proof. If a ∈ ji( L ), as a / ∈ { b ∈ L | b ≺ a } , then a (cid:54) = (cid:95) b ∈ Lb ≺ a b .Assume now that a (cid:54) = (cid:95) b ∈ Lb ≺ a b , and let S ⊆ L such that a = (cid:95) b ∈ S b . Since b (cid:22) a for every b ∈ S ,the only possibility is a ∈ S , and consequently a ∈ ji( L ).The proof of the following proposition takes inspiration from that of Bhatta and Ramananda[2, Proposition 2.2].H. Randriamaro Page 8 of 25 ¨obius Algebra
Proposition 3.11.
Let L be a lowly finite lattice, and a ∈ L . Then, a = (cid:95) b ∈ id( a ) ∩ ji( L ) b .Proof. It is obvious if a ∈ ji( L ). Now, assume that a ∈ L \ ji( L ) and a (cid:54) = (cid:95) b ∈ id( a ) ∩ ji( L ) b . Theset S = (cid:110) x ∈ L (cid:12)(cid:12)(cid:12) x (cid:54) = (cid:95) b ∈ id( x ) ∩ ji( L ) b (cid:111) is nonempty and has a minimal element c as L is lowlyfinite. Since c (cid:54) = (cid:95) b ∈ id( c ) ∩ ji( L ) b , then c / ∈ ji( L ), and it follows from Lemma 3.10 that c = (cid:95) b ∈ Lb ≺ c b .Clearly, c is an upper bound of the set X = (cid:91) b ∈ Lb ≺ c id( b ) ∩ ji( L ). If u is another upper bound of X , then u is an upper bound of id( x ) ∩ ji( L ) for every x ∈ L with x ≺ c . As c is minimal in S , then x = (cid:95) b ∈ id( x ) ∩ ji( L ) b if x ≺ c , hence u is an upper bound of { b ∈ L | b ≺ c } implying u (cid:23) c . Observe that X = (cid:91) b ∈ Lb ≺ c (cid:0) id( b ) ∩ ji( L ) (cid:1) = ji( L ) ∩ (cid:91) b ∈ Lb ≺ c id( b ) = ji( L ) ∩ id( c ). Therefore, c isa minimal upper bound for id( c ) ∩ ji( L ) which is a contradiction.For two elements a, b of a lattice L such that a (cid:22) b , let j a,b : [ a ∧ b, b ] → [ a, a ∨ b ] andm a,b : [ a, a ∨ b ] → [ a ∧ b, b ] be functions respectively defined byj a,b ( x ) := a ∨ x and m a,b ( x ) := x ∧ b. Definition 3.12.
A lattice L is modular if, for all a, b ∈ L , x ∈ [ a ∧ b, b ], and y ∈ [ a, a ∨ b ],we have x = m a,b j a,b ( x ) and y = j a,b m a,b ( y ) . Proposition 3.13.
A lattice L is modular if and only if, for all a, b, z ∈ L , we have ( a ∨ z ) ∧ ( a ∨ b ) = a ∨ (cid:0) z ∧ ( a ∨ b ) (cid:1) and ( a ∧ z ) ∨ ( a ∧ b ) = a ∧ (cid:0) z ∨ ( a ∧ b ) (cid:1) . Proof.
Assume first that L is modular. We have a (cid:22) ( a ∨ z ) ∧ ( a ∨ b ) (cid:22) a ∨ b . Letting u = ( a ∨ z ) ∧ ( a ∨ b ), we get u = j a,b m a,b ( u ) = a ∨ (cid:0) ( a ∨ z ) ∧ ( a ∨ b ) ∧ b (cid:1) = a ∨ (cid:0) ( a ∨ z ) ∧ b (cid:1) . Since it is true for all a, b, z ∈ L , interchanging z and b , we obtain u = a ∨ (cid:0) z ∧ ( a ∨ b ) (cid:1) .Likewise, we have a ∧ b (cid:22) ( z ∧ b ) ∨ ( a ∧ b ) (cid:22) b . Letting v = ( z ∧ b ) ∨ ( a ∧ b ), we get v = m a,b j a,b ( v ) = b ∧ (cid:0) ( z ∧ b ) ∨ ( a ∧ b ) ∨ a (cid:1) = b ∧ (cid:0) ( b ∧ z ) ∨ a (cid:1) . Since it is true for all a, b, z ∈ L , interchanging z and a , we obtain v = b ∧ (cid:0) z ∨ ( a ∧ b ) (cid:1) .Assume now that ( a ∨ z ) ∧ ( a ∨ b ) = a ∨ (cid:0) z ∧ ( a ∨ b ) (cid:1) and ( a ∧ z ) ∨ ( a ∧ b ) = a ∧ (cid:0) z ∨ ( a ∧ b ) (cid:1) for all a, b, z ∈ L . If a (cid:22) z (cid:22) a ∨ b , then z = ( a ∨ b ) ∧ z = ( a ∨ b ) ∧ ( a ∨ z ) = a ∨ (cid:0) b ∧ ( a ∨ z ) (cid:1) . Since a ∨ z = z , then z = a ∨ ( z ∧ b ) = j a,b m a,b ( z ). Likewise, if a ∧ b (cid:22) z (cid:22) b , then z = ( a ∧ b ) ∨ z = ( b ∧ a ) ∨ ( z ∧ b ) = b ∧ (cid:0) a ∨ ( z ∧ b ) (cid:1) . And since z ∧ b = z , then z = b ∨ ( a ∧ z ) = m a,b j a,b ( z ).H. Randriamaro Page 9 of 25 ¨obius Algebra
Proposition 3.14.
Let L be a poset, and a, b, c ∈ L . The condition a ∧ ( b ∨ c ) = ( a ∧ b ) ∨ ( a ∧ c ) is equivalent to a ∨ ( b ∧ c ) = ( a ∨ b ) ∧ ( a ∨ c ) . Proof.
Assume that a ∧ ( b ∨ c ) = ( a ∧ b ) ∨ ( a ∧ c ). Then,( a ∨ b ) ∧ ( a ∨ c ) = (cid:0) ( a ∨ b ) ∧ a (cid:1) ∨ (cid:0) ( a ∨ b ) ∧ c (cid:1) = a ∨ (cid:0) ( a ∨ b ) ∧ c (cid:1) = a ∨ ( a ∧ c ) ∨ ( b ∧ c )= a ∨ ( b ∧ c ) . Similarly, if we assume a ∨ ( b ∧ c ) = ( a ∨ b ) ∧ ( a ∨ c ), then we obtain( a ∧ b ) ∨ ( a ∧ c ) = (cid:0) ( a ∧ b ) ∨ a (cid:1) ∧ (cid:0) ( a ∧ b ) ∨ c (cid:1) = a ∧ ( b ∨ c ) . Definition 3.15.
A lattice L is distributive if, for all a, b, c ∈ L , a ∧ ( b ∨ c ) = ( a ∧ b ) ∨ ( a ∧ c ).Denote by I L the poset formed by the ideals of a lattice L with inclusion as partial order. Itis a lattice such that, for I, J ∈ I L , I ∨ J := (cid:92) K ∈I L I,J ⊆ K K and I ∧ J := (cid:91) K ∈I L K ⊆ I ∩ J K . Theorem 3.16.
Let L be a distributive lattice, I an ideal of L , and F a filter of L such that I ∩ F = ∅ . Then, there exists a prime ideal P of L such that I ⊆ P and P ∩ F = ∅ .Proof. Set X I,F := { M ∈ I L | I ⊆ M, M ∩ F = ∅} . It is a poset with inclusion as partialorder, and is nonempty since I ∈ X I,F . Consider a chain
E ∈ C X I,F , and let E = (cid:91) C ∈E E . If a, b ∈ E , then a ∈ A and b ∈ B for some A, B ∈ E . Since E is a chain, either A ⊆ B or A ⊇ B hold, so let assume A ⊆ B . Then, a ∈ B , and a ∨ b ∈ B ⊆ E , as B is an ideal. Moreover, if c ∈ L , then a ∧ c ∈ A ⊆ E , as A is also an ideal. We deduce that E ∈ I L . Besides, I ⊆ E and E ∩ F = ∅ ≺ obviously hold. Hence, E is an upper bound of E in X I,F . Therefore, X I,F is an inductive poset, and Zorn’s lemma allows to state that it has a maximal element P .Suppose that P is not prime. Then, there exists a, b ∈ L such that a, b / ∈ P but a ∧ b ∈ P .The maximality of P yields (cid:0) P ∨ id( a ) (cid:1) ∩ F (cid:54) = ∅ and (cid:0) P ∨ id( b ) (cid:1) ∩ F (cid:54) = ∅ . Thus, there are p, q ∈ P such that p ∨ a ∈ F , q ∨ b ∈ F , and ( p ∨ a ) ∧ ( q ∨ b ) ∈ F since F is a filter. Expandingby distributivity, we obtain( p ∨ a ) ∧ ( q ∨ b ) = (cid:0) ( p ∨ a ) ∧ q (cid:1) ∨ (cid:0) ( p ∨ a ) ∧ b (cid:1) = ( p ∧ q ) ∨ ( a ∧ q ) ∨ ( p ∧ b ) ∨ ( a ∧ b )which belongs to P . That means P ∩ F (cid:54) = ∅ or a contradiction. Corollary 3.17.
Let L be a distributive lattice, I ∈ I L , and a ∈ L such that a / ∈ I . Then,there exists a prime ideal P of L such that I ⊆ P and a / ∈ P .Proof. Remark that I (cid:54) = fil( a ) = ∅ , otherwise, if b ∈ I (cid:54) = fil( a ), then b ∧ a = a ∈ I , which isabsurd. Now, for the proof, we apply Theorem 3.16 to I and F = fil( a ).H. Randriamaro Page 10 of 25 ¨obius Algebra
Corollary 3.18.
Let L be a distributive lattice, and a, b ∈ L such that a (cid:54) = b . Then, L has aprime ideal containing exactly one of a and b .Proof. If a and b are not comparable or b ≺ a , then a / ∈ id( b ). It remains to apply Corol-lary 3.17 to I = id( b ). Theorem 3.19.
A lattice L is distributive if and only if, for all a, b, c ∈ L , c ∨ a = c ∨ b and c ∧ a = c ∧ b imply a = b .Proof. Suppose first that L is distributive and that there exist a, b, c ∈ L such that a ∨ c = b ∨ c and a ∧ c = b ∧ c . Then, a = a ∨ ( a ∧ c ) = a ∨ ( b ∧ c ) = ( a ∨ b ) ∧ ( a ∨ c ) = ( a ∨ b ) ∧ ( b ∨ c ) = b ∨ ( a ∧ c ) , which implies a (cid:22) b , and similarly we have b (cid:22) a .Suppose now that a ∨ c = b ∨ c and a ∧ c = b ∧ c imply a = b . If x ∈ [ a ∧ b, b ], • as x (cid:22) b ∧ ( a ∨ x ) then a ∨ x (cid:22) a ∨ (cid:0) b ∧ ( a ∨ x ) (cid:1) , as a ∨ x (cid:23) b ∧ ( a ∨ x ) then a ∨ x (cid:23) a ∨ (cid:0) b ∧ ( a ∨ x ) (cid:1) ,hence a ∨ x = a ∨ (cid:0) b ∧ ( a ∨ x ) (cid:1) on one side, • on the other side, a ∧ x = a ∧ b ∧ x = a ∧ b ∧ ( a ∨ x ).By canceling a , we obtain x = m a,b j a,b ( x ). If y ∈ [ a, a ∨ b ], as y (cid:23) a ∨ ( b ∧ y ) then b ∧ y (cid:23) b ∧ (cid:0) a ∨ ( b ∧ y ) (cid:1) , as b ∧ y (cid:22) a ∨ ( b ∧ y ) then b ∧ y (cid:22) b ∧ (cid:0) a ∨ ( b ∧ y ) (cid:1) , hence b ∧ y = b ∧ (cid:0) a ∨ ( b ∧ y ) (cid:1) on one side, and b ∨ y = a ∨ b ∨ y = a ∨ b ∨ ( b ∧ y ) on the other side. By canceling b , we obtain y = j a,b m a,b ( y ). Therefore, L is modular.Let a ∗ = a ∧ ( b ∨ c ), b ∗ = b ∧ ( c ∨ a ), and c ∗ = c ∧ ( a ∨ b ). Then, a ∗ ∧ b ∗ = a ∧ ( c ∨ a ) ∧ b ∧ ( b ∨ c ) = a ∧ b , a ∗ ∧ c ∗ = a ∧ c , and b ∗ ∧ c ∗ = b ∧ c . Set d = ( a ∨ b ) ∧ ( b ∨ c ) ∧ ( c ∨ a ). Using twice Proposition 3.13,we get a ∗ ∨ b ∗ = a ∗ ∨ (cid:0) b ∧ ( a ∨ c ) (cid:1) = ( a ∗ ∨ b ) ∧ ( a ∨ c )= (cid:16)(cid:0) ( b ∨ c ) ∧ a (cid:1) ∨ b (cid:17) ∧ ( a ∨ c ) = ( b ∨ c ) ∧ ( a ∨ b ) ∧ ( a ∨ c )= d. By symmetry, we also have a ∗ ∨ c ∗ = b ∗ ∨ c ∗ = d . Hence, • c ∗ ∨ a ∗ ∨ ( b ∧ c ) = c ∗ ∨ b ∗ ∨ ( a ∧ c ) = d , • and c ∗ ∧ (cid:0) a ∗ ∨ ( b ∧ c ) (cid:1) = ( c ∗ ∧ a ∗ ) ∨ ( b ∧ c ) = ( c ∗ ∧ b ∗ ) ∨ ( a ∧ c ) = c ∗ ∧ (cid:0) b ∗ ∨ ( a ∧ c ) (cid:1) .By canceling c ∗ , we obtain a ∗ ∨ ( b ∧ c ) = b ∗ ∨ ( a ∧ c ), whence a ∗ ∨ ( b ∧ c ) = a ∗ ∨ ( b ∧ c ) ∨ b ∗ ∨ ( a ∧ c ) = a ∗ ∨ b ∗ = d. It follows that ( a ∨ b ) ∧ c = c ∗ = c ∗ ∧ d = c ∗ ∧ (cid:0) a ∗ ∨ ( b ∧ c ) (cid:1) = ( a ∧ c ) ∨ ( b ∧ c ), hence L isconsequently distributive.H. Randriamaro Page 11 of 25 ¨obius Algebra
This section is the central part of this survey. After defining the valuation algebra and showingsome important properties, we prove that if M is a subset of a complete lowly finite distributivelattice L containing its join-irreducible elements, and a an element of M which is not join-irreducible, then (cid:88) b ∈ M ∩ [ ,a ] µ M ( b, a ) b belongs to the submodule (cid:104) a ∧ b + a ∨ b − a − b | a, b ∈ L (cid:105) of Z L . It would not have been possible to write the first two subsections without the articlesof Geissinger [7], [8, § § Definition 4.1. A valuation on a lattice L is a function f from L to a module G such that,for all a, b ∈ L , f ( a ∧ b ) + f ( a ∨ b ) = f ( a ) + f ( b ) . Definition 4.2.
The valuation module of a lattice L is the module Val( L ) := Z L/ N( L ),where N( L ) is the submodule (cid:104) a ∧ b + a ∨ b − a − b | a, b ∈ L (cid:105) of the module Z L . Proposition 4.3.
Let i : L → Val( L ) be the natural induced map for a lattice L . Then, i is avaluation, and, for every valuation f : L → G , there exists a unique module homomorphism h : Val( L ) → G such that the following diagram is commutative L Val( L ) G f i h Proof.
It is clear that i is a valuation as i( a ∧ b ) + i( a ∨ b ) − i( a ) − i( b ) = a ∧ b + a ∨ b − a − b = 0.Besides, we get the homomorphism h by setting ∀ a ∈ L : h ( a ) := f ( a ) and ∀ x, y ∈ Val( L ) : h ( x + y ) = h ( x ) + h ( y ) . Proposition 4.4.
For lattices L , L with natural induced maps i , i respectively, a latticehomomorphism ϕ : L → L induces a unique module homomorphism ψ : Val( L ) → Val( L ) such that, for every a ∈ L , ψ i ( a ) = i ϕ ( a ) .Proof. We obtain the homomorphism ψ by setting ∀ a ∈ L : ψ ( a ) := ϕ ( a ) and ∀ x, y ∈ Val( L ) : ψ ( x + y ) = ψ ( x ) + ψ ( y ) . Proposition 4.5.
For any prime ideal or prime filter M of a lattice L with natural inducedmap i , each element of i( M ) is linearly independent of those in i( L \ M ) and vise versa.Proof. Assume that M is a prime ideal, and consider the indicator function 1 M : L → Z defined as 1 M ( a ) := (cid:40) a ∈ M a, b ∈ L ,H. Randriamaro Page 12 of 25 ¨obius Algebra • if a, b ∈ M , we clearly have 1 M ( a ∧ b ) + 1 M ( a ∨ b ) = 1 M ( a ) + 1 M ( b ) = 2, • if a ∈ M and b / ∈ M , since ( a ∨ b ) ∧ b = b / ∈ M , then a ∨ b / ∈ M and 1 M ( a ∧ b )+1 M ( a ∨ b ) =1 M ( a ) + 1 M ( b ) = 1, • if a, b / ∈ M , then a ∧ b / ∈ M , the fact ( a ∨ b ) ∧ b = b / ∈ M implies a ∨ b / ∈ M , and1 M ( a ∧ b ) + 1 M ( a ∨ b ) = 1 M ( a ) + 1 M ( b ) = 0.Therefore, 1 M is a valuation on L . One similarly proves that if M is prime filter, then 1 M isalso a valuation on L . We know from Proposition 4.3 that there exists a unique homomorphism h : Val( L ) → Z such that the diagram L Val( L ) Z M i h is commutative. As h i( a ) = 1, forevery a ∈ M , and (cid:10) i( b ) (cid:12)(cid:12) b ∈ L \ M (cid:11) ⊆ ker h , each element of i( M ) is then linearly independentof those in i( L \ M ). Likewise, Proposition 3.7 allows to state that 1 L \ M is a valuation, thenone also proves that each element of i( L \ M ) is linearly independent of those in i( M ). Proposition 4.6.
The natural induced map i : L → Val( L ) of a lattice L is an injection ifand only if L is distributive.Proof. If L is distributive, we know from Corollary 3.18 that any two different elements a, b ∈ L can be separated by a prime ideal, hence Proposition 4.5 allows to deduce that i( a )and i( b ) are independent in Val( L ).If L is not distributive, then, by Theorem 3.19, it contains distinct elements a, b, c with c ∨ a = c ∨ b and c ∧ a = c ∧ b . Hence, i( a )+i( c ) = i( c ∨ a )+i( c ∧ a ) = i( c ∨ b )+i( c ∧ b ) = i( b )+i( c ),and i( a ) = i( b ). Proposition 4.7.
Let L be a distributive lattice, and a , . . . , a n , b ∈ L with b / ∈ (cid:2) (cid:94) i ∈ [ n ] a i , (cid:95) i ∈ [ n ] a i (cid:3) .Then, b is linearly independent of { a , . . . , a n } in Val( L ) .Proof. If b / ∈ id (cid:16) (cid:95) i ∈ [ n ] a i (cid:17) , then there exists a prime ideal P such that { a , . . . , a n } ⊆ P and b / ∈ P by Corollary 3.17, and b is linearly independent of { a , . . . , a n } by Proposition 4.5.If b ∈ id (cid:16) (cid:95) i ∈ [ n ] a i (cid:17) , then b / ∈ fil (cid:16) (cid:94) i ∈ [ n ] a i (cid:17) , otherwise b ∈ (cid:2) (cid:94) i ∈ [ n ] a i , (cid:95) i ∈ [ n ] a i (cid:3) which is a contradic-tion. Hence, id( b ) ∩ fil (cid:16) (cid:94) i ∈ [ n ] a i (cid:17) = ∅ , and there exists a prime ideal P such that id( b ) ⊆ P and P ∩ fil (cid:16) (cid:94) i ∈ [ n ] a i (cid:17) = ∅ by Theorem 3.16. As { a , . . . , a n } ⊆ fil (cid:16) (cid:94) i ∈ [ n ] a i (cid:17) , we once again obtainthe independence of b by Proposition 4.5.As the lattice L with either the operation ∨ or ∧ form a semigroup, the module Z L mayconsequently be considered as an algebra with either ∨ or ∧ as multiplication. Besides, if L is distributive, Proposition 4.6 allows to identify L with i( L ). Proposition 4.8. If L is a distributive lattice, then N( L ) is an ideal of the algebra Z L forboth ∨ and ∧ as multiplication. H. Randriamaro
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Proof.
For a, b, c ∈ L , we have( a ∧ b + a ∨ b − a − b ) ∧ c = ( a ∧ b ) ∧ c + ( a ∨ b ) ∧ c − a ∧ c − b ∧ c = ( a ∧ c ) ∧ ( b ∧ c ) + ( a ∧ c ) ∨ ( b ∧ c ) − a ∧ c − b ∧ c which belongs to N( L ). Then, by linearly extension, we get ( a ∧ b + a ∨ b − a − b ) ∧ t ∈ N( L )for any t ∈ Z L . Similarly, we have( a ∧ b + a ∨ b − a − b ) ∨ c = ( a ∨ c ) ∧ ( b ∨ c ) + ( a ∨ c ) ∨ ( b ∨ c ) − a ∨ c − b ∨ c ∈ N( L ) . If the lattice L is distributive, Proposition 4.8 allows to state that the valuation module Val( L )becomes a commutative algebra for either ∨ or ∧ as multiplication. Definition 4.9.
The valuation algebra is the algebra (cid:0)
Val( L ) , ∨ (cid:1) or (cid:0) Val( L ) , ∧ (cid:1) for adistributive lattice L . Lemma 4.10.
Let L be a complete distributive lattice, and define the map τ : Val( L ) → Val( L ) by τ ( x ) := + − x . Then, for a, b ∈ L , we have τ ( a ∨ b ) = τ ( a ) ∧ τ ( b ) .Proof. We have + − a ∨ b = + + a ∧ b − a − b = ( + − a ) ∧ ( + − b ). Proposition 4.11.
Let L be a complete distributive lattice, n ∈ N ∗ , and a , . . . , a n ∈ L .Then, we have − (cid:95) i ∈ [ n ] a i = (cid:94) i ∈ [ n ] ( − a i ) , that is (cid:95) i ∈ [ n ] a i = n (cid:88) k =1 ( − k − (cid:88) I ⊆ [ n ] I = k (cid:94) i ∈ I a i . Proof.
Using Lemma 4.10 and ∧ ( − a i ) = , we obtain τ (cid:16) (cid:95) i ∈ [ n ] a i (cid:17) = + − (cid:95) i ∈ [ n ] a i = (cid:94) i ∈ [ n ] τ ( a i ) = (cid:94) i ∈ [ n ] ( + − a i ) = + (cid:94) i ∈ [ n ] ( − a i ) . Then − (cid:95) i ∈ [ n ] a i = (cid:94) i ∈ [ n ] τ ( a i ) = (cid:94) i ∈ [ n ] ( − a i ) = + n (cid:88) k =1 ( − k (cid:88) I ⊆ [ n ] I = k (cid:94) i ∈ I a i . Corollary 4.12.
Let L be a complete distributive lattice, n ∈ N ∗ , a , . . . , a n ∈ L , and f avaluation on L . Then, f (cid:16) (cid:95) i ∈ [ n ] a i (cid:17) = n (cid:88) k =1 ( − k − (cid:88) I ⊆ [ n ] I = k f (cid:16) (cid:94) i ∈ I a i (cid:17) . H. Randriamaro
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Proof. If f is a valuation to module G , we know from Proposition 4.3 that a unique modulehomomorphism h : Val( L ) → G such that h i = f exists. Then, using Proposition 4.11, weobtain f (cid:16) (cid:95) i ∈ [ n ] a i (cid:17) = h (cid:16) (cid:95) i ∈ [ n ] a i (cid:17) = n (cid:88) k =1 ( − k − (cid:88) I ⊆ [ n ] I = k h (cid:16) (cid:94) i ∈ I a i (cid:17) = n (cid:88) k =1 ( − k − (cid:88) I ⊆ [ n ] I = k f (cid:16) (cid:94) i ∈ I a i (cid:17) . Theorem 4.13.
Let L be a complete lowly finite distributive lattice. Then, Val( L ) is equalto Z ji( L ) as modules.Proof. We obviously have ∈ ji( L ). Let a ∈ L , and assume that every b ∈ L such that a (cid:31) b is a linear combination in Val( L ) of a finite number of elements in ji( L ). We know fromProposition 3.11 that there exists a subset { b , . . . , b n } of ji( L ) such that a = (cid:95) i ∈ [ n ] b i . UsingProposition 4.11, we get a = n (cid:88) k =1 ( − k − (cid:88) I ⊆ [ n ] I = k (cid:94) i ∈ I b i with a (cid:31) (cid:94) i ∈ I b i for each I ⊆ [ n ]. Thusji( L ) generates Val( L ).Assume now that every subset with cardinality n − L ) is independent, and considera subset of n elements { a , . . . , a n } ⊆ ji( L ). We can suppose that a n is a maximal elementin that set. Since a n (cid:54) = (cid:95) i ∈ [ n − a i , then a n / ∈ (cid:2) (cid:94) i ∈ [ n − a i , (cid:95) i ∈ [ n − a i (cid:3) . We deduce from Proposi-tion 4.7 that { a , . . . , a n } is independent. Hence ji( L ) is an independent set in Val( L ). Corollary 4.14. If L is a complete lowly finite distributive lattice, then every valuation of L is determined by its values on ji( L ) which can be assigned arbitrarily.Proof. If f is a valuation to a module G , we know from Proposition 4.3 that a unique modulehomomorphism h : Val( L ) → G such that h i = f exists. We know from Theorem 4.13 that, if a ∈ L , there exist subsets { λ , . . . , λ n } ⊆ Z and { a , . . . , a n } ⊆ ji( L ) such that a = (cid:88) i ∈ [ n ] λ i a i .Then, f ( a ) = h ( a ) = h (cid:16) (cid:88) i ∈ [ n ] λ i a i (cid:17) = (cid:88) i ∈ [ n ] λ i h ( a i ) = (cid:88) i ∈ [ n ] λ i f ( a i ).For a poset L , and a, b ∈ L , we write a (cid:108) b if a ≺ b and { c ∈ L | a ≺ c ≺ b } = ∅ . Proposition 4.15.
Let L be a distributive lattice, and a ∈ ji( L ) such that a is not minimal.Then, there exists a unique element a ∗ ∈ L such that a ∗ (cid:108) a .Proof. Suppose that there exist two different elements b, c ∈ L such that b (cid:108) a and c (cid:108) a .Then, b ∨ c (cid:23) b , b ∨ c (cid:23) c , and b ∨ c / ∈ { b, c } . The only possibility is b ∨ c = a which contradictsthe join-irreducibility of a .Let L be a distributive lattice having a lowest element . Define e := ∈ Val( L ) and e a := a − a ∗ ∈ Val( L ) for each a ∈ ji( L ) \ { } . H. Randriamaro
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Theorem 4.16.
Let L be a complete lowly finite distributive lattice. Then, (cid:8) e a | a ∈ ji( L ) (cid:9) is an orthogonal idempotent basis of Val( L ) .Proof. For a, b ∈ ji( L ) \ { } with a (cid:54) = b , we have e ∧ e = e and e a ∧ e = a ∧ − a ∗ ∧ = 0, e a ∧ e a = a ∧ a − a ∧ a ∗ − a ∗ ∧ a + a ∗ ∧ a ∗ = a − a ∗ − a ∗ + a ∗ = e a , and e a ∧ e b = a ∧ b − a ∧ b ∗ − a ∗ ∧ b + a ∗ ∧ b ∗ = (cid:40) b − b ∗ − b + b ∗ if a ∗ = ba ∗ ∧ b ∗ − a ∗ ∧ b ∗ − a ∗ ∧ b ∗ + a ∗ ∧ b ∗ otherwise= 0 . Then, (cid:8) e a | a ∈ ji( L ) (cid:9) is orthogonal idempotent. Assume now that every subset withcardinality n − (cid:8) e a | a ∈ ji( L ) (cid:9) is independent, and consider a subset of n elements { e a , . . . , e a n } . We can suppose that a n is a maximal element in the set { a , . . . , a n } . Since a n (cid:54) = (cid:95) i ∈ [ n − a i ∨ (cid:95) i ∈ [ n ] a ∗ i , then a n / ∈ (cid:2) (cid:94) i ∈ [ n − a i ∧ (cid:94) i ∈ [ n ] a ∗ i , (cid:95) i ∈ [ n − a i ∨ (cid:95) i ∈ [ n ] a ∗ i (cid:3) . We deduce fromProposition 4.7 that a n is independent of { a , . . . , a n − , a ∗ , . . . , a ∗ n } . Hence e a n is independentof { e a , . . . , e a n − } , and { e a , . . . , e a n } is consequently an independent set in Val( L ). Finally,since there is a natural bijection a (cid:55)→ e a between ji( L ) and (cid:8) e a | a ∈ ji( L ) (cid:9) , by Theorem 4.13the latter is also a basis of Val( L ). Theorem 4.17.
Let L be a complete lowly finite distributive lattice. Then, ∀ x ∈ L : x = (cid:88) a,b ∈ ji( L ) b (cid:22) a (cid:22) x µ ji( L ) ( b, a ) b. Proof. If a ∈ ji( L ), then a = e a + a ∗ , particularly = e . Now, consider any x ∈ L \ ji( L ), andassume that, for every b ∈ L such that b ≺ x , we have b = (cid:88) d ∈ ji( L ) d (cid:22) b e d . There exist b, c ∈ L \ { x } such that x = b ∨ c . Note that b ∧ c = (cid:88) d ∈ ji( L ) d (cid:22) b ∧ c e d as the e d ’s are orthogonal idempotent. Hence, b ∨ c = b + c − b ∧ c = (cid:88) d ∈ ji( L ) ∩ (id( b ) ∪ id( c )) e d . Besides, remark that, for any y ∈ ji( L ) ∩ id( x ),there exist b, c ∈ L \ { x } such that y (cid:22) b and b ∨ c = x . Therefore, x = (cid:88) d ∈ ji( L ) d (cid:22) x e d .Let b be the natural bijection a (cid:55)→ e a between ji( L ) and (cid:8) e a | a ∈ ji( L ) (cid:9) . For a ∈ ji( L ), wehave i( a ) = (cid:88) d ∈ ji( L ) ∩ [ ,a ] b( d ). Then, using the M¨obius inversion formula, we obtainb( a ) = (cid:88) d ∈ ji( L ) ∩ [ ,a ] µ ji( L ) ( d, a )i( d ) or e a = (cid:88) d ∈ ji( L ) d (cid:22) a µ ji( L ) ( d, a ) d. H. Randriamaro
Page 16 of 25 ¨obius Algebra x = (cid:88) a ∈ ji( L ) a (cid:22) x e a with e a = (cid:88) d ∈ ji( L ) d (cid:22) a µ ji( L ) ( d, a ) d . Lemma 4.18. If L is a lowly finite distributive lattice, then (cid:0) Z L, ∧ (cid:1) is naturally isomorphicto the M¨obius algebra (cid:0) M¨ob( L ) , · (cid:1) .Proof. For a ∈ L , we have u L ( a ) = (cid:88) c ∈ [ ,a ] µ L ( c, a ) c . The M¨obius inversion formula conse-quently allows to state that a = (cid:88) c ∈ [ ,a ] u L ( c ). Then, for a, b ∈ L , we have a · b = (cid:88) c ∈ [ ,a ] ∩ [ ,b ] u L ( c ) = (cid:88) c ∈ [ ,a ∧ b ] u L ( c ) = a ∧ b. Lemma 4.19. If L is a complete lowly finite distributive lattice, then (cid:0) M¨ob( L ) / N( L ) , · (cid:1) isisomorphic to the M¨obius algebra (cid:16) M¨ob (cid:0) ji( L ) (cid:1) , · (cid:17) .Proof. By Lemma 4.18, we get M¨ob( L ) / N( L ) (cid:39) Z L/ N( L ) (cid:39) Val( L ). We know from Theo-rem 4.13 that Val( L ) is isomorphic to Z ji( L ) as modules. Now, as algebras, (cid:0) Val( L ) , ∧ (cid:1) isnaturally isomorphic to (cid:16) M¨ob (cid:0) ji( L ) (cid:1) , · (cid:17) since, for a, b ∈ ji( L ), Theorem 4.17 allows to statethat a · b = (cid:88) c ∈ [ ,a ] ∩ [ ,b ] ∩ ji( L ) u ji( L ) ( c ) = (cid:88) c ∈ [ ,a ∧ b ] ∩ ji( L ) u ji( L ) ( c ) = a ∧ b. The following theorem is the main result of this survey. Zaslavsky originally proved it forfinite distributive lattice [18, Theorem 2.1].
Theorem 4.20.
Let L be a complete lowly finite distributive lattice, and M a subset of L such that ji( L ) ⊆ M . If a ∈ M \ ji( L ) , then u M ( a ) ∈ N( L ) . Proof.
Consider the linear maps j : M¨ob( L ) → M¨ob (cid:0) ji( L ) (cid:1) , j : M¨ob( L ) → M¨ob( M ), andj : M¨ob( M ) → M¨ob (cid:0) ji( L ) (cid:1) which on the basis { u L ( a ) | a ∈ L } , and { u M ( a ) | a ∈ M } respectively have the valuesj (cid:0) u L ( a ) (cid:1) := (cid:40) u ji( L ) ( a ) if a ∈ ji( L ) , , j (cid:0) u L ( a ) (cid:1) := (cid:40) u M ( a ) if a ∈ M, , and j (cid:0) u M ( a ) (cid:1) := (cid:40) u ji( L ) ( a ) if a ∈ ji( L ) , . Then, j, j , and j are algebra homomorphisms by Corollary 2.22. Moreover, as the diagramH. Randriamaro Page 17 of 25 ¨obius Algebra L ) M¨ob( M )M¨ob (cid:0) ji( L ) (cid:1) jj j is commutative, then u M ( a ) ∈ ker j ⊆ ker j if a ∈ M \ ji( L ).Finally, since M¨ob (cid:0) ji( L ) (cid:1) (cid:39) M¨ob( L ) / ker j [4, II-Theorem 6.12], we obtain ker j = N( L ) usingLemma 4.19, and consequently u M ( a ) ∈ N( L ) if a ∈ M \ ji( L ). Corollary 4.21.
Let L be a complete lowly finite distributive lattice, M a subset of L suchthat ji( L ) ⊆ M , and f : L → G a valuation on L . If a ∈ M \ ji( L ) , then (cid:88) b ∈ [ ,a ] ∩ M µ M ( b, a ) f ( b ) = 0 . Proof.
Let h : Val( L ) → G be the module homomorphism associated to f as in Proposi-tion 4.3. We already know from Lemma 4.18 that Val( L ) (cid:39) M¨ob( L ) / N( L ). By Theorem 4.20,we then obtain (cid:88) b ∈ [ ,a ] ∩ M µ M ( b, a ) b = 0 h (cid:16) (cid:88) b ∈ [ ,a ] ∩ M µ M ( b, a ) b (cid:17) = h (0) (cid:88) b ∈ [ ,a ] ∩ M µ M ( b, a ) h ( b ) = 0 (cid:88) b ∈ [ ,a ] ∩ M µ M ( b, a ) f ( b ) = 0 . We use Corollary 4.21 to prove the fundamental theorem of dissection theory.
Definition 5.1.
Let us call subspace arrangement in a topological space T a finite set ofsubspaces in T .For a subspace arrangement A in T , let L A := (cid:110) (cid:92) H ∈ B H ∈ T \ {∅} (cid:12)(cid:12)(cid:12) B ⊆ A (cid:111) be the posetwith partial order (cid:22) defined, for A, B ∈ L A , by A (cid:22) B if and only if A ⊆ B . Definition 5.2.
Let A be a subspace arrangement in a topological space T . A meet-refinement of L A is a finite poset L ⊆ T \ {∅} with the same partial order as that definedfor L A such that (cid:91) X ∈ L X = (cid:91) H ∈ A H and • any element in L A is a union of elements in L , • any nonempty intersection of elements in L is also a union of elements in L .H. Randriamaro Page 18 of 25 ¨obius Algebra X ) the set formed by the connected components of a topological space X , and let A be a subspace arrangement of T . The set L c A := L A (cid:116) (cid:26) C (cid:16) (cid:92) H ∈ B H (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) B ⊆ A , (cid:92) H ∈ B H (cid:54) = ∅ (cid:27) is for instance a meet-refinement of L A . Definition 5.3.
Let A be a subspace arrangement in a topological space T , and denote by C A the set formed by the connected components of T \ (cid:91) H ∈ A H . An element of C A is calleda chamber of A .Consider a subspace arrangement A , and a meet-refinement L of L A . Let D( L ) be the finitedistributive lattice of sets generated by L (cid:116) C A through unions and intersections, that isD( L ) := (cid:110) (cid:91) A ∈ M A (cid:116) (cid:91) X ∈ D X (cid:12)(cid:12)(cid:12) M ⊆ L, D ⊆ C A (cid:111) . In that case, for
A, B ∈ D( L ), we have A ∨ B = A ∪ B and A ∧ B = A ∩ B . Lemma 5.4.
Let A be a subspace arrangement in a topological space T , and L a meet-refinement of L A . Then, ji (cid:0) D( L ) (cid:1) ⊆ {∅} (cid:116) L (cid:116) C A .Proof. Every element of D( L ) \ ( {∅} (cid:116) L (cid:116) C A ) is the union of at least two elements of L (cid:116) C A .Then, none of them can be join-irreducible. Theorem 5.5.
Let A be a subspace arrangement in a topological space T , L a meet-refinementof L A , and f a valuation on D( L ) . Then, (cid:88) C ∈ C A f ( C ) = (cid:88) X ∈ L (cid:116){∅} µ L (cid:116){∅} ( X, T ) f ( X ) . Proof.
Note first that T ∈ L but T / ∈ ji (cid:0) D( L ) (cid:1) as T = (cid:91) H ∈ A H (cid:116) (cid:91) C ∈ C A C . From Corollary 4.21and Lemma 5.4, we get (cid:88) A ∈{∅}(cid:116) L (cid:116) C A µ {∅}(cid:116) L (cid:116) C A ( A, T ) f ( A ) = 0 . The result is finally obtained after taking into account the following remarks: • if C ∈ C A , then µ {∅}(cid:116) L (cid:116) C A ( C, T ) = − µ {∅}(cid:116) L (cid:116) C A ( C, C ) = − • if X ∈ {∅} (cid:116) L , then [ X, T ] ∩ C A = ∅ , hence µ {∅}(cid:116) L (cid:116) C A ( X, T ) = µ {∅}(cid:116) L ( X, T ). Definition 5.6.
Let T be a topological space, and denote by H n ( T ) the n th singular homologygroup of T for n ∈ N . The Euler characteristic of T is χ ( T ) := (cid:88) n ∈ N ( − n rank H n ( T ) . We can now state the fundamental theorem of dissection theory.H. Randriamaro
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Corollary 5.7 (Fundamental Theorem of Dissection Theory) . Let A be a subspace arrange-ment in a topological space T with | χ ( T ) | < ∞ , and L a meet-refinement of L A . Then, (cid:88) C ∈ C A χ ( C ) = (cid:88) X ∈ L µ L ( X, T ) χ ( X ) . Proof.
It is known that χ ( A ) + χ ( B ) = χ ( A ∪ B ) + χ ( A ∩ B ), for A, B ⊆ T , like stated at theend of [16, § L ). Moreover, χ ( ∅ ) = 0by definition. We consequently obtain the result by using Theorem 5.5 with χ . Example . Consider the arrangement A of parametric 1-spheres H : x = cos( π ) y = sin( π ) cos( t ) ,z = sin( π ) sin( t ) H : x = − cos( π ) y = sin( π ) cos( t ) ,z = sin( π ) sin( t ) H : x = cos( π ) sin( t ) y = cos( π ) cos( t ) ,z = sin( π ) H : x = cos( π ) sin( t ) y = cos( π ) cos( t ) ,z = − sin( π ) where t ∈ [0 , π ], in S represented on Figure 1. On one side, C A has 6 chambers having Eulercharacteristic 1, and 1 with Euler characteristic 0, then (cid:88) C ∈ C A χ ( C ) = 6. On the other side, (cid:88) X ∈ L A µ L A ( X, S ) χ ( X ) = µ L A ( S , S ) χ ( S ) + (cid:88) i ∈ [4] µ L A ( H i , S ) χ ( H i )+ µ L A ( H ∩ H , S ) χ ( H ∩ H ) + µ L A ( H ∩ H , S ) χ ( H ∩ H )= 1 × × ( − × × ×
2= 6 . Corollary 5.8.
Let A be a subspace arrangement in a topological space T with | χ ( T ) | < ∞ ,and L a meet-refinement of L A . Suppose that every chamber of A has the same Eulercharacteristic c (cid:54) = 0 . Then, C A = 1 c (cid:88) X ∈ L µ L ( X, T ) χ ( X ) . Proof.
It is obviously a consequence of the fundamental theorem of dissection theory where χ ( C ) = c for C ∈ C A . We use the fundamental theorem of dissection theory to compute the f-polynomial of sub-manifold arrangements having specific face properties.
Definition 6.1.
Let A be a subspace arrangement in topological space T , and X ∈ L A .The induced subspace arrangement on X is the subspace arrangement in X defined by A X := (cid:8) H ∩ X (cid:12)(cid:12) H ∈ A , H ∩ X / ∈ {∅ , X } (cid:9) . Let F A := (cid:71) X ∈ L A C A X , and call an element of F A a face of A .H. Randriamaro Page 20 of 25 ¨obius Algebra
Definition 6.2.
Recall that a n -dimensional manifold or n -manifold is a topologicalspace with the property that each point has a neighborhood that is homeomorphic to R n ,and a submanifold of a n -manifold T is a k -manifold included in T where k ∈ [0 , n ]. Definition 6.3.
Let us call submanifold arrangement in the n -manifold T a finite set ofsubmanifolds A in T such that every element of L A ∪ F A is a submanifold of T . Example . Consider the arrangement A of 1-manifolds H : y = 6 sin( x ), H : y = x +cos( x ), H : x
64 + y
25 = 1 in R represented on Figure 2. We see that (cid:88) X ∈ L A µ L A ( X, R ) χ ( X ) = µ L A ( R , R ) χ ( R ) + µ L A ( H , R ) χ ( H ) + µ L A ( H , R ) χ ( H )+ µ L A ( H , R ) χ ( H ) + µ L A ( H ∩ H , R ) χ ( H ∩ H )+ µ L A ( H ∩ H , R ) χ ( H ∩ H ) + µ L A ( H ∩ H , R ) χ ( H ∩ H )= 1 × − × ( −
1) + ( − × ( −
1) + ( − × × ×
10 + 1 ×
2= 18is the number of chamber in C A . Definition 6.4.
Let A be a submanifold arrangement in a n -manifold T , and x a variable.For k ∈ [0 , n ], denote by f k ( A ) the number of k -dimensional faces of A . The f -polynomial H. Randriamaro
Page 21 of 25 ¨obius Algebra A is f A ( x ) := (cid:88) k ∈ [0 ,n ] f k ( A ) x n − k . Proposition 6.5.
Let A be a submanifold arrangement in a n -manifold T with | χ ( T ) | < ∞ .Suppose that ∀ k ∈ [0 , n ] , ∀ X ∈ L A , dim X = k : χ ( X ) = l k , ∀ k ∈ [0 , n ] , ∀ C ∈ F A , dim C = k : χ ( C ) = c k (cid:54) = 0 . Then, f A ( x ) = (cid:88) i ∈ [0 ,n ] (cid:88) Y ∈ L A dim Y = i (cid:88) k ∈ [0 ,i ] (cid:88) X ∈ L A Y dim X = k l k c i µ L A ( X, Y ) x n − k . H. Randriamaro
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Proof.
Using the fundamental theorem of dissection theory, we getf i ( A ) = (cid:88) Y ∈ L A dim Y = i C L A Y = 1 c i (cid:88) Y ∈ L A dim Y = i (cid:88) X ∈ L A Y µ L A Y ( X, Y ) χ ( X )= (cid:88) Y ∈ L A dim Y = i (cid:88) k ∈ [0 ,i ] (cid:88) X ∈ L A Y dim X = k l k c i µ L A Y ( X, Y )= (cid:88) Y ∈ L A dim Y = i (cid:88) k ∈ [0 ,i ] (cid:88) X ∈ L A Y dim X = k l k c i µ L A ( X, Y ) . Definition 6.6.
Let A be a submanifold arrangement in a n -manifold T . The rank of X ∈ L A is rk X := n − dim X , and that of A is rk A := max { rk X | X ∈ L A } . Definition 6.7.
Let A be a submanifold arrangement in a n -manifold T , and x, y twovariables. The M¨obius Polynomial of A isM A ( x, y ) := (cid:88) X,Y ∈ L A µ L A ( X, Y ) x rk X y rk A − rk Y . Corollary 6.8.
Let A be a submanifold arrangement in a n -manifold T with | χ ( T ) | < ∞ .Suppose that χ ( X ) = ( − dim X for every X ∈ L A ∪ F A . Then, f A ( x ) = ( − rk A M A ( − x, − . Proof.
From Proposition 6.5, we obtainf A ( x ) = (cid:88) i ∈ [0 ,n ] (cid:88) Y ∈ L A dim Y = i (cid:88) k ∈ [0 ,i ] (cid:88) X ∈ L A Y dim X = k ( − k − i µ L A ( X, Y ) x n − k = (cid:88) Y ∈ L A (cid:88) X ∈ L A Y ( − dim X − dim Y µ L A ( X, Y ) x n − dim X = (cid:88) Y ∈ L A (cid:88) X ∈ L A Y ( − n − dim Y µ L A ( X, Y )( − dim X − n x n − dim X = (cid:88) Y ∈ L A (cid:88) X ∈ L A Y ( − rk Y µ L A ( X, Y )( − x ) rk X = ( − rk A (cid:88) Y ∈ L A (cid:88) X ∈ L A Y µ L A ( X, Y )( − x ) rk X ( − rk Y − rk A = ( − rk A M A ( − x, − . H. Randriamaro
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REFERENCES
Corollary 6.9.
Let A be a submanifold arrangement in a n -manifold T with | χ ( T ) | < ∞ .Suppose that ∀ C ∈ F A : χ ( C ) = ( − dim C and ∀ X ∈ L A : χ ( X ) = (cid:40) if dim X ≡ otherwise . Moreover, define γ n := (cid:40) if dim X ≡ − otherwise . Then, f A ( x ) = ( − n − rk A (cid:0) M A ( x, −
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