The eigenvalues of the Laplacian for the homology of the Lie algebra corresponding to a poset Iztok Hozo Department of Mathematics Indiana University Northwest Gary, In 46408 email: ihozo@iunhaw1.iun.indiana.edu Submitted: April 6, 1995; Accepted: July 21, 1995. Abstract In this paper we study the spectral resolution of the Laplacian L of the Koszul complex of the Lie algebras corresponding to a certain class of posets. Given a poset P on the set {1, 2, ,n},wedefinethenilpotentLiealgebraL P to be the span of all elementary matrices z x,y , such that x is less than y in P .Inthis paper, we make a decisive step toward calculating the Lie algebra homology of L P in the case that the Hasse diagram of P is a rooted tree. We show that the Laplacian L simplifies significantly when the Lie algebra cor- responds to a poset whose Hasse diagram is a tree. The main result of this paper determines the spectral resolutions of three commuting linear operators whose sum is the Laplacian L of the Koszul complex of L P in the case that the Hasse diagram is a rooted tree. We show that these eigenvalues are integers, give a combinatorial indexing of these eigenvalues and describe the corresponding eigenspaces in representation-theoretic terms. The homology of L P is represented by the nullspace of L, so in future work, these results should allow for the homology to be effectively computed. AMS Classification Number: 17B56 (primary) 05E25 (secondary) 1 Preliminaries 1.1 Definitions A partially ordered set P (or poset, for short) is a set (which by abuse of notation we also call P ), together with a binary relation denoted ≤ (or ≤ P when there is a possibility of confusion), satisfying the following three axioms: 1. For all x ∈ P , x ≤ x. (reflexivity) 2. If x ≤ y and y ≤ x,thenx = y. (antisymmetry) 3. If x ≤ y and y ≤ z,thenx ≤ z. (transitivity) the electronic journal of combinatorics 2 (1995), #R14 2 A chain (or totally ordered set or linearly ordered set)isaposetinwhichany two elements are comparable. A subset C of a poset P is called a chain if C is a chain when regarded as a subposet of P . Definition 1.1 A poset P is linear if for any two comparable elements x, y ∈ P ,the interval [x, y] is a chain, i.e., if every interval has the structure of a chain. The length l(C) of a finite chain C is defined by l(C)=|C|−1. 1.2 The homology of a poset The combinatorial approach to a homology theory for posets was developed by Rota [29], Farmer [8], Lakser [22], Mather [25], Crapo [5] and others (more references can be found in [33]). A systematic development of the relationship between the combinatorial and topological properties of posets was begun by K. Baclawski [1] and A. Bj¨orner [2] and continued by J. Walker [33]. Define the set C r (P ) to be the set of 0-1 chains of length r in the poset P . By abuse of notation we will use the same name for the complex vector space C r or C r (P ), with basis the set of r-chains. The C r ’s are called chain spaces.Themap∂ r : C r → C r−1 ,calledthe boundary map, is defined by: ∂ r ( ˆ 0 <x 1 < <x r < ˆ 1) = r  i=1 (−1) i−1 ( ˆ 0 <x 1 < < x i < <x r < ˆ 1) It is easy to check that: Lemma 1 ∂ r−1 ◦ ∂ r =0. This allows us now to define the homology of a poset to be: H r (P )=Ker(∂ r )/Im(∂ r+1 ) Later in this work we will talk about an operator, called the Laplacian of a complex, for which we need to identify the transpose of the boundary map. We are in fact transposing the matrix of the boundary map with respect to the basis of r-chains. In this case - the case of the poset homology, the transpose of the boundary map is not so difficult to evaluate. Lemma 2 The transpose of the boundary operator (viewed as a linear map), is given by the following expression: ∂ t ( ˆ 0 <x 1 < <x r < ˆ 1) = r  i=0  x i <y<x i+1 (−1) i ( ˆ 0 <x 1 < <x i <y<x i+1 < <x r < ˆ 1), where x 0 = ˆ 0 and x r+1 = ˆ 1. the electronic journal of combinatorics 2 (1995), #R14 3 1.3 Lie Algebras In this section we will introduce some basic notions from the theory of Lie algebras, and the homology of Lie algebras. We will always work over , the field of complex numbers. Lie algebras arise “in nature” as vector spaces of linear transformations endowed with an operation which is in general neither commutative nor associative: [x, y]=xy − yx. It is possible to describe this kind of system abstractly in a few axioms. Definition 1.2 A vector space L over a field , with an operation L × L → L,denoted (x, y) → [x, y] and, called the bracket or commutator of x and y,isaLie algebra over if the following axioms are satisfied: (L1) The bracket operation is bilinear. (L2) [x, x]=0for all x ∈ L. (L3) [x, [y, z]] + [y, [z,x]] + [z, [x, y]] = 0 (x, y, z ∈ L). Axiom (L3) is called Jacobi identity. The axioms (L1) and (L2) imply (L2’): [x, y]= −[y,x]. In the field of complex numbers (L2’) implies (L2). 1.4 Homology of a Lie algebra Suppose L is a Lie algebra and A is a module over L.ThespaceΓ q (L; A)ofq-dimensional chains of the Lie algebra L with coefficients in A is defined as A ⊗ Λ q L. The boundary operator ∂ = ∂ q :Γ q (L; A) → Γ q−1 (L; A) acts in accordance with the formula ∂(a ⊗ (x 1 ∧ ∧ x q )) = =  1≤s<t≤q (−1) s+t−1 a ⊗ ([x s ,x t ] ∧ x 1 ∧ ˆx s ˆx t ∧ x q )(1) +  1≤s≤q (−1) s−1 x s a ⊗ (x 1 ∧ ˆx s ∧ x q ) Lemma 3 ∂ r−1 ∂ r =0 The proof of this lemma is straightforward. Let θ be the representation of L on A ⊗ Λ q L.Ify ∈ L,wehave: θ(y)(a ⊗ x 1 ∧ ∧ x q ) =(y · a ⊗ x 1 ∧ ∧ x q )+  i (a ⊗ x 1 ∧ ∧ [y, x i ] ∧ ∧ x q ) It is easy to check: the electronic journal of combinatorics 2 (1995), #R14 4 Lemma 4 For y ∈ L: ∂ q ◦ θ(y)=θ(y) ◦ ∂ q The homology of the complex {Γ q (L; A),∂ q } is referred to as the homology of the Lie algebra L with coefficients in A and denoted by H q (L; A); if A is the field of complex numbersviewedasatrivialL-module (as in our case), the second sum in the formula 1 vanishes. In this case the notations Γ q (L; A)andH q (L; A) are abbreviated to Γ q (L)and H q (L). 1.5 The Laplacian operator Suppose that {Γ r (L),∂ r } is a finite dimensional complex. We will first define an orthogonal inner product ·, · on the product ⊕Γ r , such that Γ r , Γ s  = 0 whenever r = s.Wewill restrict our attention to the subspaces of the nilpotent Lie algebra T n ( ) of all strictly upper triangular matrices over the complex numbers, with standard basis {z i,j :1≤ i<j≤ n}, so we can define this product naturally: Definition 1.3 Let L be a Lie algebra, L ⊂ T n ( ). Define an inner product for standard basis elements v, w ∈ L by: v, w =    1 if v = w 0 otherwise 0 if v and w have different exterior degrees Extend this to the exterior algebra, i.e., to the complexes mentioned above. Definition 1.4 Suppose that v = v 1 ∧···∧v k and w = w 1 ∧···∧w k . Then define the inner product: v, w = det(v i ,w j ) 1≤i,j≤k Note that this can be written also as v, w =  σ∈S n sgn(σ)  i v i ,w σ(i)  =  sgn(σ)iffv i = w σ(i) for all i 0otherwise In other words, the product of two pure wedges of basis elements is nonzero if and only if two pure wedges differ only in the order of the elements, and in that case, the product is just the sign of the permutation that changes one into another. Define δ r mapping Γ r into Γ r+1 by δ r v, w = v, ∂ r+1 w over all v ∈ Γ r ,andallw ∈ Γ r+1 . It is enough to calculate δ on pure wedges (as in our definitions), since the inner product and δ are both linear functions. the electronic journal of combinatorics 2 (1995), #R14 5 Lemma 5 The map δ is given by δ r (z x 1 ,y 1 ∧ z x 2 ,y 2 ∧ ∧ z x r ,y r ) = r  s=1 (−1) s−1  x s <l<y s z x 1 ,y 1 ∧ ∧ z x s ,l ∧ z l,y s ∧ ∧ z x r ,y r Note: It is easy to check that δ r+1 δ r =0,thusδ ∗ defines a coboundary operator, and so we can define the cohomology to be H r (L)=Ker(δ r )/Im(δ r−1 ) Proof: But to prove that, it is enough to show that the coefficient of the pure wedge z x 1 ,y 1 ∧ z x 2 ,y 2 ∧ ∧ z x r ,y r in ∂(z x 1 ,y 1 ∧ ···∧z x s ,l ∧ z l,y s ∧···∧ z x r ,y r )is(−1) s−1 for any l ∈ (x s ,y s ), i.e., ∂(z x 1 ,y 1 ∧ ∧ z x s ,l ∧ z l,y s ∧ ∧ z x r ,y r ) = +(−1) s−1 (z x 1 ,y 1 ∧ z x 2 ,y 2 ∧ ∧ z x r ,y r )+ and this is not difficult by the definition of ∂. Note that we can change the order of the elements in the pure wedges, and obtain a slightly different form for δ: δ r (z x 1 ,y 1 ∧ z x 2 ,y 2 ∧ ∧ z x r ,y r ) = r  s=1 (−1) s−1  x s <l<y s z x 1 ,y 1 ∧ ∧ z x s ,l ∧ z l,y s ∧ ∧ z x r ,y r =  m  x m <l<y m (z x m ,l ∧ z x 1 ,y 1 ∧···∧z l,y m ∧···∧z x k ,y k ) This is the form for the δ = ∂ t we will use. Definition 1.5 Define the Laplacian operator L r :Γ r → Γ r by L r = δ r−1 ∂ r + ∂ r+1 δ r Theorem 6 (Kostant, [19] ) Let B = {β 1 , ,β d } be a basis for Ker(L r ). Then B is simultaneously a complete set of representatives of H r (L) and H r (L). In particular dim(H r (L)) = dim(H r (L)) = dim(Ker(L r )). Sometimes, the Laplacian L r will turn out to be very simple. In these cases, Theorem 6 is a very efficient method for evaluating the homology and cohomology of a Lie algebra. One famous result obtained in this way is given by Kostant [19]. the electronic journal of combinatorics 2 (1995), #R14 6 1.6 Kostant’s Theorem We need some preliminary definitions. Suppose G is a semisimple Lie algebra, with the root system R, whose basis is ∆. Thus G = H ⊕ (⊕ α∈R z α ), where H is the torus. Suppose that S ⊂ ∆, and let R S be the set of roots in the (integer) module spanned by elements of S.DefineG S to be G S = H ⊕z α : α ∈ R S .DefineaG S module N S to be N S = z α : α ∈ R + \ R + S . We will state a couple of facts without proof: • N S is a nilpotent subalgebra of G. • Let W be a G-module. Then W is also a N S -module and a G S -module. • Thus we can compute H(N S ; W µ )asG S -module, where W µ is an irreducible G- module. Kostant used the Laplacian operator to prove the following theorem: Theorem 7 (Kostant, Theorem 5.7,[19]) Let λ be a dominant weight for G, and let µ be a minimal weight for G S .LetV be a G S -invariant subspace of W λ ⊗  r N S isomorphic to the G S -irreducible (indexed by µ) with minimal weight µ. • The Laplacian L = δ∂ + ∂δ preserves V . • Then, L| V is a scalar, and the scalar is given by 1 2 (|ρ + λ| 2 −|ρ − µ| 2 ) where ρ is half of the sum of the positive roots of G. 1.7 The Lie Algebra corresponding to a Poset Definition 1.6 A standard labeling of the poset P is a total ordering of the elements of P such that whenever x< P y, x precedes y in that total ordering. Since P is a partial order, i.e. transitive , there always is such labeling. Fix a standard labeling of the poset P . We can define a Lie algebra L P corresponding to the poset P in the following way. First, for every relation x< P y in the poset P , i.e., for every two elements x, y ∈ P such that x< P y we can define the matrix z x,y having all entries equal to zero, except for exactly one entry equal to 1, namely the entry at the position x, y in the standard labeling of the poset P . All matrices z x,y are strictly upper triangular because of our labeling. So L P is a subalgebra of T n . The Lie algebras L P obtained from distinct labellings are isomorphic – the labeling only specifies embedding of L P in the n × n matrices. the electronic journal of combinatorics 2 (1995), #R14 7 2 The Formula for Laplacian of a Linear Poset In this section we will present a significant simplification of the Lie algebra Laplacian in the case of linear posets. That will allow us to prove our main result on the eigenvalues of those Laplacians. 2.1 Simplification Recall the Lie algebra boundary map: ∂(z x 1 ,y 1 ∧ ∧ z x k ,y k ) =  i<j (−1) i+j−1 [z x i ,y i ,z x j ,y j ] ∧ z x 1 ,y 1 ∧ ∧ z x i ,y i ∧ ∧ z x j ,y j ∧ ∧ z x k ,y k The transpose, ∂ t , is given by the following formula: ∂ t r (z x 1 ,y 1 ∧ z x 2 ,y 2 ∧ ∧ z x r ,y r ) = r  s=1 (−1) s−1  x s <l<y s z x 1 ,y 1 ∧ ∧ z x s ,l ∧ z l,y s ∧ ∧ z x r ,y r =  m  x m <l<y m (z x m ,l ∧ z x 1 ,y 1 ∧···∧z l,y m ∧···∧z x k ,y k ) To compute the action of L on a basis vector z x 1 ,y 1 ∧···∧z x k ,y k of Γ k (L P )webegin with the action of ∂∂ t .Wehave, ∂∂ t (z x 1 ,y 1 ∧···∧z x k ,y k ) =  m  x m <l<y m ∂(z x m ,l ∧ z x 1 ,y 1 ∧···∧z l,y m ∧···∧z x k ,y k ) =  i<j  m=i,j  x m <l<y m (−1) i+1+j ([z x i ,y i ,z x j ,y j ] ∧ z x m ,l ∧ z x 1 ,y 1 ∧ ∧ z x i ,y i ∧···∧z l,y m ∧···∧z x j ,y j ∧···∧z x k ,y k ) +  m  j=m  x m <l<y m (−1) 1+j+1−1 ([z x m ,l ,z x j ,y j ] ∧ z x 1 ,y 1 ∧ ∧ z x j ,y j ∧···∧z l,y m ∧···∧z x k ,y k ) +  i<m  x m <l<y m (−1) i+1+m+1−1 ([z x i ,y i ,z l,y m ] ∧ z x m ,l ∧ z x 1 ,y 1 ∧ ∧ z x i ,y i ∧···∧z x m ,y m ∧···∧z x k ,y k ) +  m<j  x m <l<y m (−1) m+1+j+1−1 ([z l,y m ,z x j ,y j ] ∧ z x m ,l ∧ z x 1 ,y 1 ∧ ∧ z x m ,y m ∧···∧z x j ,y j ∧···∧z x k ,y k ) + k  m=1 |(x m ,y m )|(z x 1 ,y 1 ∧···∧z x k ,y k ) the electronic journal of combinatorics 2 (1995), #R14 8 which is equal to: =  i<j  m=i,j  x m <l<y m (−1) i+j−1 ([z x i ,y i ,z x j ,y j ] ∧ z x m ,l ∧ z x 1 ,y 1 ∧ ∧ z x i ,y i ∧···∧z l,y m ∧···∧z x j ,y j ∧···∧z x k ,y k ) +  i<m  x m <l<y m (−1) i+1 ([z x m ,l ,z x i ,y i ] ∧ z x 1 ,y 1 ∧ ∧ z x i ,y i ∧···∧z l,y m ∧···∧z x k ,y k ) +  m<j  x m <l<y m (−1) j+1 ([z x m ,l ,z x j ,y j ] ∧ z x 1 ,y 1 ∧ ∧ z l,y m ∧···∧z x j ,y j ∧···∧z x k ,y k ) +  i<m  x m <l<y m (−1) i+m+1 ([z x i ,y i ,z l,y m ] ∧ z x m ,l ∧ z x 1 ,y 1 ∧ ∧ z x i ,y i ∧···∧z x m ,y m ∧···∧z x k ,y k ) +  m<j  x m <l<y m (−1) m+j+1 ([z l,y m ,z x j ,y j ] ∧ z x m ,l ∧ z x 1 ,y 1 ∧ ∧ z x m ,y m ∧···∧z x j ,y j ∧···∧z x k ,y k ) + k  m=1 |(x m ,y m )|(z x 1 ,y 1 ∧···∧z x k ,y k ) Now use the definition of bracket in this Lie algebra: [z x i ,y i ,z x j ,y j ]=δ y i ,x j z x i ,y j − δ x i ,y j z x j ,y i andwehavethefollowing: ∂∂ t (z x 1 ,y 1 ∧···∧z x k ,y k ) =  i<j  m=i,j  x m <l<y m (−1) i+j−1 ([z x i ,y i ,z x j ,y j ] ∧ z x m ,l ∧ z x 1 ,y 1 ∧···∧z x i ,y i ∧ ∧ z l,y m ∧···∧z x j ,y j ∧···∧z x k ,y k ) +  i<m  x m <l<y m δ l,x i (z x 1 ,y 1 ∧···∧z x m ,y i ∧···∧z l,y m ∧···∧z x k ,y k ) − δ x m ,y i (z x 1 ,y 1 ∧···∧z x i ,l ∧···∧z l,y m ∧···∧z x k ,y k ) +  m<j  x m <l<y m δ l,x j (z x 1 ,y 1 ∧···∧z l,y m ∧···∧z x m ,y j ∧···∧z x k ,y k ) − δ x m ,y j (z x 1 ,y 1 ∧···∧z l,y m ∧···∧z x j ,l ∧···∧z x k ,y k ) +  i<m  x m <l<y m δ l,y i (z x 1 ,y 1 ∧···∧z x i ,y j ∧···∧z x m ,l ∧···∧z x k ,y k ) − δ x i ,y m (z x 1 ,y 1 ∧···∧z l,y i ∧···∧z x m ,l ∧···∧z x k ,y k ) the electronic journal of combinatorics 2 (1995), #R14 9 +  m<j  x m <l<y m δ l,y j (z x 1 ,y 1 ∧···∧z x m ,l ∧···∧z x j ,y m ∧···∧z x k ,y k ) − δ x j ,y m (z x 1 ,y 1 ∧···∧z x m ,l ∧···∧z l,y j ∧···∧z x k ,y k ) + k  m=1 |(x m ,y m )|(z x 1 ,y 1 ∧···∧z x k ,y k ) Note that every sum over x m <l<y m which has an occurrence of δ l,∗ has only one summand if ∗ really is between x m and y m , and is zero otherwise. We will use the symbol χ for denoting the truth of some statement, i.e., χ(∗)=  1, if ∗ is true 0, if ∗ is false We label some of the resulting sums: ∂∂ t (z x 1 ,y 1 ∧···∧z x k ,y k )(2) =  i<j  m=i,j  x m <l<y m (−1) i+j−1 ([z x i ,y i ,z x j ,y j ] ∧ z x m ,l ∧ z x 1 ,y 1 ∧ ∧ z x i ,y i ∧···∧z l,y m ∧···∧z x j ,y j ∧···∧z x k ,y k )(3) −  i<j χ(x j <x i <y j )(z x 1 ,y 1 ∧···∧z x i ,y j ∧···∧z x j ,y i ∧···∧z x k ,y k ) −  i<j  x j <l<y j δ x j ,y i (z x 1 ,y 1 ∧···∧z x i ,l ∧···∧z l,y j ∧···∧z x k ,y k )(4) −  i<j χ(x i <x j <y i )(z x 1 ,y 1 ∧···∧z x i ,y j ∧···∧z x j ,y i ∧···∧z x k ,y k ) −  i<j  x i <l<y i δ x i ,y j (z x 1 ,y 1 ∧···∧z l,y i ∧···∧z x j ,l ∧···∧z x k ,y k )(5) +  i<j χ(x j <y i <y j )(z x 1 ,y 1 ∧···∧z x i ,y j ∧···∧z x j ,y i ∧···∧z x k ,y k ) −  i<j  x j <l<y j δ x i ,y j (z x 1 ,y 1 ∧···∧z l,y i ∧···∧z x j ,l ∧···∧z x k ,y k )(6) −  i<j  x i <l<y i δ x j ,y i (z x 1 ,y 1 ∧···∧z x i ,l ∧···∧z l,y j ∧···∧z x k ,y k )(7) +  i<j χ(x i <y j <y i )(z x 1 ,y 1 ∧···∧z x i ,y j ∧···∧z x j ,y i ∧···∧z x k ,y k ) + k  m=1 |(x m ,y m )|(z x 1 ,y 1 ∧···∧z x k ,y k ) On the other hand: the electronic journal of combinatorics 2 (1995), #R14 10 ∂ t ∂(z x 1 ,y 1 ∧···∧z x k ,y k ) =  i<j (−1) i+j−1 ∂ t ([z x i ,y i ,z x j ,y j ] ∧ z x 1 ,y 1 ∧ z x i ,y i ∧···∧z x j ,y j ∧···∧z x k ,y k ) =  i<j  m=i,j  x m <l<y m (−1) i+j−1 (z x m ,l ∧ [z x i ,y i ,z x j ,y j ] ∧ z x 1 ,y 1 ∧ ∧ z x i ,y i ∧···∧z l,y m ∧···∧z x j ,y j ∧···∧z x k ,y k ) +  i<j  x m <l<y m (−1) i+j−1 δ x j ,y i (z x i ,l ∧ z l,y j ∧ z x 1 ,y 1 ∧ ∧ z x i ,y i ∧···∧z x j ,y j ∧···∧z x k ,y k ) −  i<j  x m <l<y m (−1) i+j−1 δ x i ,y j (z x j ,l ∧ z l,y i ∧ z x 1 ,y 1 ∧ ∧ z x i ,y i ∧···∧z x j ,y j ∧···∧z x k ,y k ) Now use the fact that we are dealing with a linear poset. This implies that for every interval (x m ,y m ) and every l , x m <l<y m we have (x m ,y m )=(x m ,l) ∪{l}∪(l, y m ) Hence ∂ t ∂(z x 1 ,y 1 ∧···∧z x k ,y k )(8) =  i<j  m=i,j  x m <l<y m (−1) i+j−1 (z x m ,l ∧ [z x i ,y i ,z x j ,y j ] ∧ z x 1 ,y 1 ∧ ∧ z x i ,y i ∧···∧z l,y m ∧···∧z x j ,y j ∧···∧z x k ,y k )(9) +  i<j  x i <l<y i δ x j ,y i (z x 1 ,y 1 ∧···∧z x i ,l ∧···∧z l,y j ∧···∧z x k ,y k )(10) +  i<j  l=x j =y i δ x j ,y i (z x 1 ,y 1 ∧···∧z x i ,l ∧···∧z l,y j ∧···∧z x k ,y k ) +  i<j  x j <l<y j δ x j ,y i (z x 1 ,y 1 ∧···∧z x i ,l ∧···∧z l,y j ∧···∧z x k ,y k )(11) +  i<j  x j <l<y j δ x i ,y j (z x 1 ,y 1 ∧···∧z l,y i ∧···∧z x j ,l ∧···∧z x k ,y k )(12) +  i<j  l=x i =y j δ x i ,y j (z x 1 ,y 1 ∧···∧z l,y i ∧···∧z x j ,l ∧···∧z x k ,y k ) +  i<j  x i <l<y i δ x i ,y j (z x 1 ,y 1 ∧···∧z l,y i ∧···∧z x j ,l ∧···∧z x k ,y k )(13) Then we have : [...]... Laplacian L This conjecture claims that the same poset tableau will work simultaneously for both Laplacians (LX and LY ),i.e., that the eigenvalues of the Laplacian L are the sum of the eigenvalues of LY and the eigenvalues of LX evaluated simultaneously with the same poset tableau 7 Homology The object of the paper is to get a step closer to evaluating the homology of any Lie algebra corresponding to. .. be an element of the exterior algebra of the Lie algebra of P In the last section we saw that the Laplacian acts on pure wedges of Lie algebra elements zx1 ,y1 ∧ zx2 ,y2 ∧ · · · ∧ zxn ,yn by summing the action of switching pairs of comparable x’s, and pairs of comparable y’s among themselves (plus a scalar) That fact gives us the opportunity to divide our Laplacian into diagonal blocks where each... state the theorem: Theorem 11 Let LP be the Lie algebra corresponding to a linear poset P , and let Cn (LP ) be the n th chain space Then Cn (LP ) = V (X, Y )P (X,Y ) where the direct sum is over all possible choices of (multi)-sets X and Y of equal cardinality, and each summand V (X, Y )P is invariant under the action of the Laplacian Thus we can now concentrate on the action of the Laplacian on each of. .. tableau r r r ∅ ey (Λ) = 1 Figure 8: poset tableau LY = Id, with the eigenvalue 1 • In the trivial case when the x’s are not comparable the y’s are not comparable because of linearity and the existence of a minimal element So we have x1 < y1 x2 < y2 The poset tableau again gives zero as the y-eigenvalue, and since the y’s are not comparable, the Laplacian LY is also zero So the theorem holds for the. .. have to subtract all switches involving two y’s from the same αk Each of these transpositions doesn’t move any of the y’s (or x’s), and there are exactly ak 2 from the eigenvalue y∈λ/µ ak 2 of them Thus, we have to subtract cy of L0 Thus we can write our space V as a direct sum of the eigenspaces V = ⊕w Vw , where the sum is over all eigenvalues w of LY We want to know the eigenvalues of LY on the. .. In general, this example shows that the Laplacian L can be broken down into diagonal blocks, which are generated by a pure wedge ζ, and all pure wedges obtained by permutations of the labels of ζ Furthermore, since a ∧ b = −b ∧ a, we can always keep the x-labels in order, i.e, we will always put the element zxi ,∗ at the ith position of the pure wedge 4 The eigenvalues of the Laplacian Let ζ = zx1 ,y1... generality we can assume that all of the x’s and all of the y’s are distinct, because if they were not, we would just apply the same reasoning to each appearance of an observed element Let (xi , xj ) be a transposition of the operator LX , and let (yk , yl ) be a transposition of the operator LY If all of the numbers i, j, k, l are distinct, we have nothing to prove since it would not make any difference... is acting on the subtree Ti above xa Now apply the induction hypothesis to X , Y This gives the decomposition of the ˆ L-block as a G-module The theorem says that the irreducible summands are indexed ˆ ˆ by the (X , Y ) poset tableaux Λ of shape λ (where λ |B|), and ”shape” means ˆ that the minimal element s of A is labeled with λ Also the theorem tells us that the ˆ Y -Laplacian LY for (X , Y ) acts... labeling Λ of positive multiplicity, each element in ex(Λ) is an eigenvalue of LX Proof: The proof of this theorem is similar to the proof of the LY -Centerpiece Theorem We will omit the details here Since LX and LY commute (as established in Lemma 13 ), the eigenvalues of LX + LY will be the sum of the eigenvalues on the corresponding irreducibles of the eigenspaces Recall that the complete Laplacian. .. identify the class of elements ∪σ∈Sαi ζ σ with the sum σ∈Sα ζ σ But A does i exactly this identification, A = σ∈Sα ζ σ The same is true for ΠB and ΠN B i NOTE: When we deal with the projection A there is one important point we have to make The Laplacian LY switches all comparable pairs of y’s If two of the y’s are the same – they would not get switched Therefore, when we observed the Laplacian L0, . The eigenvalues of the Laplacian for the homology of the Lie algebra corresponding to a poset Iztok Hozo Department of Mathematics Indiana University Northwest Gary, In 46408 email: ihozo@iunhaw1.iun.indiana.edu Submitted:. of (multi)-sets X and Y of equal cardinality, and each summand V (X, Y ) P is invariant under the action of the Laplacian. Thus we can now concentrate on the action of the Laplacian on each of. present a significant simplification of the Lie algebra Laplacian in the case of linear posets. That will allow us to prove our main result on the eigenvalues of those Laplacians. 2.1 Simplification Recall