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}, we define the nilpotent Lie algebra L_{P} to be the span of all elementary matrices z_{x,y}, such that x is less than y in P. In this 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 corresponds 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, then x = y. (antisymmetry)

3. If x ≤ y and y ≤ z, then x ≤ z. (transitivity)

A chain (or totally ordered set or linearly ordered set) is a poset in which any 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. Björner [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. The map ∂_{r} : C_{r} → C_{r−1}, called the boundary map, is defined by:

∂_{r}(ˆ0 < x_{1} < … < x_{r} < ˆ1) = ∑_{i=1}^{r}(−1)^{i−1}(ˆ0 < x_{1} < … < x̂_{i} < … < x_{r} < ˆ1)

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}(ˆ0 < x_{1} < … < x_{r} < ˆ1)
= ∑_{i=0}^{r} ∑_{x_{i}<y<x_{i+1}}(−1)^{i}(ˆ0 < x_{1} < … < x_{i} < y < x_{i+1} < … < x_{r} < ˆ1),

where x_{0} = ˆ0 and x_{r+1} = ˆ1.

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 ℂ, 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 ℂ, with an operation L × L → L, denoted (x, y) → [x, y] and, called the bracket or commutator of x and y, is a Lie algebra over ℂ if the following axioms are satisfied:

(L1)  The bracket operation is bilinear.
(L2)  [x, x] = 0 for 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. The space Γ_{q}(L; A) of q-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} ∧… x̂_{s}… x̂_{t}… ∧ x_{q})  (1)
+  ∑_{1≤s≤q}(−1)^{s−1}x_{s}a ⊗ (x_{1} ∧… x̂_{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. If y ∈ L, we have:

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:

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 numbers viewed as a trivial L-module (as in our case), the second sum in the formula 1 vanishes. In this case the notations Γ_{q}(L; A) and H_{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. We will restrict our attention to the subspaces of the nilpotent Lie algebra T_{n}(ℂ) 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}(ℂ). Define an inner product for standard basis elements v, w ∈ L by:

⟨v, w⟩ = \{ \begin{matrix}1 & if v = w \\ 0 & otherwise \\ 0 & if v and w have different exterior degrees \end{matrix}

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)}⟩ = \{ \begin{matrix}sgn(σ) & iff v_{i} = w_{σ(i)} for all i \\ 0 & otherwise \end{matrix}

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}, and all w ∈ 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.

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}})
= ∑_{s=1}^{r}(−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, thus δ_{∗} 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}})
= ∑_{s=1}^{r}(−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].

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 Δ. Thus G = H ⊕ (⊕_{α∈R}⟨z_{α}⟩), where H is the torus. Suppose that S ⊂ Δ, and let R_{S} be the set of roots in the ℤ (integer) module spanned by elements of S. Define G_{S} to be G_{S} = H ⊕ ⟨z_{α} : α ∈ R_{S}⟩. Define a G_{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}) as G_{S}-module, where W^{μ} is an irreducible Gmodule. 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}. Let V 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

\frac{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.

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:

∂_{r}^{t}(z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧… ∧ z_{x_{r},y_{r}})
= ∑_{s=1}^{r}(−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}) we begin with the action of ∂∂^{t}. We have,

∂∂^{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}})

+ ∑_{m=1}^{k} |(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}})

+ ∑_{m=1}^{k} |(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}}

and we have the following:

∂∂^{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}})

+ ∑_{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}})
+ ∑_{m=1}^{k} |(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(∗) = \{ \begin{matrix}1, & if ∗ is true \\ 0, & if ∗ is false \end{matrix}

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}})

+ ∑_{m=1}^{k} |(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 :

(9) + (3) = 0
(10) + (7) = 0
(11) + (4) = 0
(12) + (6) = 0
(13) + (5) = 0

After these cancellations we obtain the following expression for the action of the Laplacian L:

L(z_{x_{1},y_{1}} ∧ ⋯ ∧ z_{x_{k},y_{k}}) = (∂∂^{t} + ∂^{t}∂)(z_{x_{1},y_{1}} ∧ ⋯ ∧ z_{x_{k},y_{k}})
= ∑_{m=1}^{k} |(x_{m}, y_{m})|(z_{x_{1},y_{1}} ∧ ⋯ ∧ z_{x_{k},y_{k}})
+ ∑_{i<j}(δ_{x_{i},y_{j}} + δ_{x_{j},y_{i}})(z_{x_{1},y_{1}} ∧ ⋯ ∧ z_{x_{k},y_{k}})
+ ∑_{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}})
+ ∑_{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} 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} 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}})

2.2 The Formula

To further simplify our expressions we will introduce some notation. Define

ζ = z_{x_{1},y_{1}} ∧ ⋯ ∧ z_{x_{k},y_{k}}
ζ_{i,j} = z_{x_{1},y_{1}} ∧ ⋯ ∧ z_{x_{i},y_{j}} ∧ ⋯ ∧ z_{x_{j},y_{i}} ∧ ⋯ ∧ z_{x_{k},y_{k}}
w(ζ) = ∑_{m=1}^{k} |(x_{m}, y_{m})|
Δ(ζ) = ∑_{i<j}(δ_{x_{i},y_{j}} + δ_{x_{j},y_{i}}) = ∑_{i,j} δ_{x_{i},y_{j}}

Thus, we can reformulate the calculations from the previous section into:

Theorem 8 (The Formula) Let P be a linear poset and let L_{P} be the corresponding Lie algebra. The action of the Laplacian L on an element

ζ = z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ ⋯ ∧ z_{x_{k},y_{k}}

is given by the following formula:

L(ζ) = (w(ζ) + Δ(ζ))ζ

+ ∑_{i<j}(χ(x_{i} < y_{j} < y_{i}) + χ(x_{j} < y_{i} < y_{j}) − χ(x_{j} < x_{i} < y_{j}) − χ(x_{i} < x_{j} < y_{i}))ζ_{i,j}

Note that ζ_{i,j} is obtained from ζ by transposing a comparable pair of y’s or a comparable pair of x’s.

3 Linear poset with a ˆ0

Suppose now that the poset P has a ˆ0, the minimum element. That is the assumption under which we will work in the future. In that case, we can further simplify our notation:

Lemma 9

L(ζ) = (w(ζ) + Δ(ζ))ζ

+ ∑_{i<j}(χ(x_{i} < y_{j} < y_{i}) + χ(x_{j} < y_{i} < y_{j}) − χ(x_{j} < x_{i} < y_{j}) − χ(x_{i} < x_{j} < y_{i}))ζ_{i,j}
= (w(ζ) + Δ(ζ))ζ + ∑_{y_{i}<_{P} y_{j}} ζ_{i,j} − ∑_{x_{i}<_{P} x_{j}} ζ_{i,j}

Proof: We need to prove that we can write

χ(y_{i} < y_{j}) + χ(y_{j} < y_{i}) − χ(x_{i} < x_{j}) − χ(x_{j} < x_{i})

instead of

χ(x_{i} < y_{j} < y_{i}) + χ(x_{j} < y_{i} < y_{j}) − χ(x_{j} < x_{i} < y_{j}) − χ(x_{i} < x_{j} < y_{i})

in the expression for the Laplacian above.

Let y_{i} and y_{j} be two comparable distinct y’s. Without loss of generality, assume that y_{i} < y_{j}. Thus x_{i} < y_{i} < y_{j}. The existence of ˆ0 and linearity of the poset implies that the interval [ˆ0, y_{j}] must be a chain, and since x_{i}, x_{j} ∈ [ˆ0, y_{j}], x_{i} and x_{j} must be comparable. There are several possibilities:

1. x_{j} < x_{i} < y_{i} < y_{j}

2. x_{i} < x_{j} < y_{i} < y_{j}

3. x_{i} < y_{i} < x_{j} < y_{j}

In all three possibilities,

χ(x_{i} < y_{j} < y_{i}) + χ(x_{j} < y_{i} < y_{j}) − χ(x_{j} < x_{i} < y_{j}) − χ(x_{i} < x_{j} < y_{i}) = 0,

and at the same time

χ(y_{i} < y_{j}) + χ(y_{j} < y_{i}) − χ(x_{i} < x_{j}) − χ(x_{j} < x_{i}) = 0.

On the other hand, if y_{i} and y_{j} are incomparable, then we have one of:

1. x_{j} < x_{i} < y_{i}, y_{j}

2. x_{i} < x_{j} < y_{i}, y_{j}

3. x_{i} < y_{i}, x_{j} < y_{j}, x_{i} and x_{j} are incomparable. Now in the first two cases

χ(x_{i} < y_{j} < y_{i}) + χ(x_{j} < y_{i} < y_{j}) − χ(x_{j} < x_{i} < y_{j}) − χ(x_{i} < x_{j} < y_{i}) = −1,

with

χ(y_{i} < y_{j}) + χ(y_{j} < y_{i}) − χ(x_{i} < x_{j}) − χ(x_{j} < x_{i}) = −1

too. In the last remaining case both expressions are zero.

Hence, the expression for the Laplacian above can be rewritten in the following form:

L(ζ) = (w(ζ) + Δ(ζ))ζ + ∑_{y_{i}<_{P} y_{j}} ζ_{i,j} − ∑_{x_{i}<_{P} x_{j}} ζ_{i,j}.

In other words, the meaning of the theorem above is that the Laplacian only transposes comparable labels of the element z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ ⋯ ∧ z_{x_{k},y_{k}}, without introducing any new indices. This is the key observation for next section.

Lemma 10 Let ζ = z_{x_{1},y_{1}} ∧ ⋯ ∧ z_{x_{n},y_{n}}, and let ζ_{σ} = z_{x_{1},y_{σ(1)}} ∧ z_{x_{2},y_{σ(2)}} ∧ ⋯ ∧ z_{x_{n},y_{σ(n)}}. If ζ_{σ} ≠ 0, i.e., if x_{i} <_{P} y_{σ(i)} for all i, then

1. w does not depend on σ, i.e.,

w(ζ) = w(ζ_{σ})

2. Δ does not depend on σ, i.e.,

Δ(ζ) = Δ(ζ_{σ})

This lemma actually proves that w and Δ are dependent only on the choice of the (multi–)sets X = {x_{1}, x_{2},…, x_{k}}, Y = {y_{1}, y_{2},…, y_{k}} (and a poset P), and not on the specific pure wedge constructed from those sets.

Proof: First we will check the claim for w.

w(ζ) = ∑_{i} |(x_{i}, y_{i})| = ∑_{i} ht(y_{i}) − ht(x_{i}) − 1,

where the ht(v) is the size of the interval [ˆ0, v]. The sum on the right does not depend on σ, so we can write w(X, Y) instead of w(ζ).

Now we will check the claim for Δ.

Δ(ζ) = ∑_{i<j}(δ_{x_{i},y_{j}} + δ_{x_{j},y_{i}})
= ∑_{i} ( multiplicity of x_{i} in the set Y)
= ∑_{j} ( multiplicity of y_{j} in the set X)
= ∑_{i,j} δ_{x_{i},y_{j}}

which also does not depend on σ. Thus we can write Δ(X, Y) instead of Δ(ζ) too.

We will use both notations, depending whether we want to stress ζ or the sets (X, Y). Note that while Δ is completely determined by the sets (X, Y), w also depends on the poset P globally, i.e., it counts the sizes of intervals (x_{i}, y_{i}) not relative to the sets X and Y, but with respect to the whole poset P.

The simplicity of this formula is in the way the elements to which we are restricting the Laplacian, are obtained one from another, by simply transposing the labels. 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 z_{x_{i},∗} at the ith position of the pure wedge.

4 The eigenvalues of the Laplacian

Let ζ = z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ ⋯ ∧ z_{x_{n},y_{n}} 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 z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ ⋯ ∧ z_{x_{n},y_{n}} 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 block corresponds to all possible permutations of the x’s and y’s for a fixed choice of the element z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ ⋯ ∧ z_{x_{n},y_{n}}, i.e., for the fixed choice of the multisets X = {x_{1}, x_{2},…, x_{n}}, and Y = {y_{1}, y_{2},…, y_{n}}. In other words each block represents the “action” of the Laplacian on the subspace of the n th exterior power of our Lie algebra spanned by the elements {z_{x_{1},y_{σ(1)}} ∧z_{x_{2},y_{σ(2)}} ∧⋯∧z_{x_{n},y_{σ(n)}} : σ ∈ S_{n}}. Here the element z_{x_{1},y_{σ(1)}} ∧z_{x_{2},y_{σ(2)}} ∧ ⋯ ∧ z_{x_{n},y_{σ(n)}} is defined if and only if x_{i} <_{P} y_{σ(i)} for every i = 1, 2,…, n. Thus each block is of size n!, if all the elements are defined, or less, if some of the elements are not defined which is the case in general. The size of the block depends on the structure of the poset, and in particular, it depends on the relations in the subposet of P spanned by the sets X and Y. More formally :

Definition 4.1 The L-block V spanned by the (multi)-sets (X, Y)_{P}, subsets of a poset P, is the vector space with basis

{z_{x_{1},y_{σ(1)}} ∧ z_{x_{2},y_{σ(2)}} ∧ ⋯ ∧ z_{x_{n},y_{σ(n)}} : σ ∈ S_{n}}

where n = |X| = |Y |, σ is a permutation in S_{n}, and the element

z_{x_{1},y_{σ(1)}} ∧ z_{x_{2},y_{σ(2)}} ∧ ⋯ ∧ z_{x_{n},y_{σ(n)}}

is zero unless x_{i} <_{P} y_{σ(i)} for all i = 1,…, n.

If we want to stress the dependence of the L-block V of the sets X and Y and the poset P, we write V (X, Y)_{P}.

The sets X and Y may be multisets since some of the x’s or y’s might appear more than once as a label. In that case the sizes |X| and |Y | are counting multiplicities as well.

Using this division of the chain space into L-blocks, we can use the results of the previous section, and state the theorem:

Theorem 11 Let L_{P} be the Lie algebra corresponding to a linear poset P, and let C_{n}(L_{P}) be the n th chain space. Then

C_{n}(L_{P}) = ⨁_{(X,Y)} V (X, Y)_{P}

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 these blocks.

4.1 Embedding of the L-block in ℂS_{n}

Write the multisets X and Y as X = ∪_{i∈A_{1}}{x_{i}} ∪ … ∪_{i∈A_{l}} {x_{i}}, and Y = ∪_{j∈B_{1}}{y_{j}} ∪ … ∪_{j∈B_{m}} {y_{j}}, where the A_{i}’s contain the sets of indices of equal x’s, and B_{i}’s contain the sets of indices of equal y’s.

For example, if X = {x_{1}, x_{2}, x_{3}, x_{4}, x_{5}}, where x_{1} = x_{2}, x_{3} = x_{4}, then A_{1} = {1, 2}, A_{2} = {3, 4} and A_{3} = {5}.

Switching two of the x’s will displace x_{i} from its original position. To take into account the fact that we have to bring it back (by the choice of our basis) into the i th place, we need a minus sign.

Let

Π_{x} = ∑_{σ_{1}∈Sym(A_{1}),σ_{2}∈Sym(A_{2})...}(∏_{i} sgn(σ_{i}))σ_{1}σ_{2} ⋯ σ_{l}

and

Π_{y} = ∑_{σ_{1}∈Sym(B_{1}),σ_{2}∈Sym(B_{2})...} σ_{1}σ_{2} ⋯ σ_{m}.

Then the L-block V can be identified with a subspace of Π_{x}ℂS_{n}Π_{y}. So Π_{y} symmetrizes over equal y’s and Π_{x} anti-symmetrizes over equal x’s. In other words, Π_{x} permutes the positions, while Π_{y} permutes indices.

4.2 The Laplacian L_{Y}

Let X = {x_{1},…, x_{n}} and Y = {y_{1},…, y_{n}} be two fixed (multi-)sets of vertices of the poset P, and consider the restriction of the Laplacian L to L-block V (X, Y).

To simplify the notation, we will write the Laplacian L as:

L(ζ) = ((w(X, Y) + Δ(X, Y))Id + ∑_{x_{i}<_{P} x_{j}}(x_{i}, x_{j}) + ∑_{y_{i}<_{P} y_{j}}(y_{i}, y_{j})) · ζ,

where the “action” of (y_{i}, y_{j}) or (x_{i}, x_{j}) on z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ ⋯ ∧ z_{x_{n},y_{n}} means switching the corresponding pairs of y’s, or x’s.

To simplify our examination, we will split it into these three parts:

L = L_{D} + L_{X} + L_{Y} ,

where

• L_{D} is the scalar matrix, L_{D} = w(X, Y) + Δ(X, Y)

• L_{X} is the “action” of the Laplacian on the set of the x’s, i.e.

L_{X} = ∑_{x_{i}<_{P} x_{j}}(x_{i}, x_{j})

• L_{Y} is the “action” of the Laplacian on the set of the y’s

L_{Y} = ∑_{y_{i}<_{P} y_{j}}(y_{i}, y_{j})

on our L-block V (X, Y).

In the embedding of this L-block into ℂS_{n}, the notation for the Laplacian L_{Y} would be L_{Y} = ∑_{i<j:y_{i}<_{P} y_{j}}(i, j), where the actual multiplication is from the right. The proper notation for L_{X} in the Π_{x}ℂS_{n}Π_{y} is L_{X} = ∑_{i<j:x_{i}<_{P} x_{j}}(i, j), but the multiplication in this case is from the left.

Lemma 12 L_{Y} and Π_{y} commute, i.e.,

L_{Y} · Π_{y} = Π_{y} · L_{Y} .

Proof: It is sufficient to prove that the Laplacian L_{Y} commutes with every transposition of the form (i, k), where y_{i} = y_{k}, because every permutation in Π_{y} can be written as a product of those permutations. So, let y_{i} = y_{k}. That means that Π_{y} has transposition (i, k) as one of its summands. Let y_{j} ∈ Y be comparable to y_{i} (thus it is comparable to y_{k}). In that case, the Laplacian L_{Y} contains both transpositions, (i, j), and (k, j), i.e., L_{Y} = ⋯ + (i, j) + (k, j) + · · ·.

But, (i, k) · (i, j) = (k, j) · (i, k), which shows that Π_{y} · L_{Y} = L_{Y} · Π_{y}.

Using exactly same argument we see that L_{X} and Π_{x} also commute.

We know from section 2 that L_{D} is a scalar matrix on each block, and thus it commutes with L_{X} and L_{Y}.

As for the L_{X} and L_{Y}, we have the following Lemma:

Lemma 13

L_{X} · L_{Y} · (z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ ⋯ ∧ z_{x_{n},y_{n}}) = L_{Y} · L_{X} · (z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ ⋯ ∧ z_{x_{n},y_{n}})

Proof:

The absence of certain relations in the poset may cause terms in the Laplacian to be missing. That is why this lemma is not obvious, and needs to be proved.

Let ζ = z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ ⋯ ∧ z_{x_{n},y_{n}}. Without loss of 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 (x_{i}, x_{j}) be a transposition of the operator L_{X}, and let (y_{k}, y_{l}) be a transposition of the operator L_{Y}. If all of the numbers i, j, k, l are distinct, we have nothing to prove since it would not make any difference which transposition was applied first. On the other hand, if i = k and j = l, again there is nothing to prove, since their combined action would amount to multiplying with -1 no matter in which order they are applied.

Therefore assume that i ≠ k but j = l, i.e., we have two transpositions, (x_{i}, x_{j}) and (y_{j}, y_{k}) in L_{X} and L_{Y} respectively, which overlap at one position. Without loss of generality assume that n = 3. There are only three elements of the pure wedge, call them z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ z_{x_{3},y_{3}}, i.e., i = 1, j = 2, n = k = 3.

Let

A = (x_{1}, x_{2}) · (y_{2}, y_{3}) · (z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ z_{x_{3},y_{3}})
= (x_{1}, x_{2}) · (z_{x_{1},y_{1}} ∧ z_{x_{2},y_{3}} ∧ z_{x_{3},y_{2}})
= −(z_{x_{1},y_{3}} ∧ z_{x_{2},y_{1}} ∧ z_{x_{3},y_{2}})

and

B = (y_{2}, y_{3}) · (x_{1}, x_{2}) · (z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ z_{x_{3},y_{3}})
= −(y_{2}, y_{3}) · (z_{x_{1},y_{2}} ∧ z_{x_{2},y_{1}} ∧ z_{x_{3},y_{3}})
= −(z_{x_{1},y_{3}} ∧ z_{x_{2},y_{1}} ∧ z_{x_{3},y_{2}}).

Thus

L_{X} · L_{Y} · (z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ z_{x_{3},y_{3}}) = L_{Y} · L_{X} · (z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ z_{x_{3},y_{3}}),

whenever all of the relations used above are present, i.e., whenever every x_{i} is beneath each y_{j}. That can be explained by the fact that L_{X} is acting on the x-indices and L_{Y} is acting on the y-indices.

The question remains whether the answer would be the same if some of the relations needed above were missing, and only one of the expressions above gets annulled. The final expressions in both A and B are 0 unless:

x_{1} < y_{3},  x_{2} < y_{1},  x_{3} < y_{2}.

Suppose (without loss of generality) that B above survives the procedure, i.e., we have the relation x_{1} < y_{2}. On the other hand if A is annulled in the middle step, the only possible conflict left is x_{2} ≮ y_{3}. We have that y_{2} and y_{3} are comparable, otherwise the transposition (y_{2}, y_{3}) wouldn’t be a summand of L_{Y}. If y_{2} < y_{3}, then x_{2} < y_{2} < y_{3}, which is contrary to the just stated assumption. Thus, we must have y_{3} < y_{2}. Also x_{1} < y_{3} by our assumption above, and (x_{1}, x_{2}) is a transposition in L_{X}, so they must also be comparable. By the same argument as above, x_{2} must be larger than x_{1}. Hence in the interval (x_{1}, y_{2}) there are two elements x_{2} and y_{3}. Since the poset is linear – those two elements must be comparable, and since we assumed that x_{2} ≮ y_{3}, it must be that x_{2} > y_{3}.

All together, the relations are:

\begin{matrix}y_{2} \\ y_{1} \end{matrix} > x_{2} > y_{3} > \begin{matrix}x_{3} \\ x_{1} \end{matrix}

So in this case we have y_{1} > y_{3}, and x_{2} > x_{3}. Let

C = (x_{2}, x_{3}) · (y_{1}, y_{3}) · (z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ z_{x_{3},y_{3}})
= (x_{2}, x_{3}) · (z_{x_{1},y_{3}} ∧ z_{x_{2},y_{2}} ∧ z_{x_{3},y_{1}})
= −(z_{x_{1},y_{3}} ∧ z_{x_{2},y_{1}} ∧ z_{x_{3},y_{2}})

and

D = (y_{1}, y_{3}) · (x_{2}, x_{3}) · (z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}} ∧ z_{x_{3},y_{3}})
= −(y_{1}, y_{3}) · (z_{x_{1},y_{1}} ∧ z_{x_{2},y_{3}} ∧ z_{x_{3},y_{2}})
= 0

since x_{2} > y_{3}.

The expressions A and C are two summands of the product L_{X}L_{Y}, while B and D are two summands of the product L_{Y} L_{X}. As we can see, A+C = B+D. Thus L_{X}·L_{Y} = L_{Y} ·L_{X}.

In view of Lemma 13, L_{X}, L_{Y} and L_{D} are commuting linear transformations. So, to analyze the spectrum of their sum, we can compute the eigenvalues and eigenspaces of each separately. We will begin with L_{Y}.

4.3 A poset tableau of type (X, Y)_{P}

Definition 4.2 The diagram of the L-block, P [X, Y], spanned by the sets (X, Y)_{P}, is the Hasse diagram of the subposet X∪Y with order inherited from the poset P. Furthermore every vertex of P, which is in the intersection X ∩Y is split into two nodes, with the x-node above the y-node.

Definition 4.3 Given a node v in P [X, Y] define the repetition number of v , k(v), to be the number of times that v appears in the multiset X if v is an x-node of P [X, Y], or the multiset Y if v is a y-node of P [X, Y].

Let C(v) be the set of covers of node v in P [X, Y]. If v is a maximal node, than C(v) = ∅.

Definition 4.4 A poset tableau of type (X, Y)_{P} (or just of type (X, Y)) is any labeling Λ of the diagram, P [X, Y], of the L-block V spanned by (X, Y), where the labels are partitions Λ(v), such that Λ(v) is a partition of the number ∑_{w≥v} ϵ(w)k(w), where

ϵ(w) = \{ \begin{matrix}+1 & if w is a y-node \\ −1 & if w is an x-node. \end{matrix}

Given a poset tableau Λ we will define the multiplicity of Λ, m(Λ), and the eigenvalues of Λ, e(Λ).

Definition 4.5 • Let v be a y-node of the diagram P [X, Y], labeled with the partition Λ(v) and with repetition number k(v). Let C(v) = {v_{1}, v_{2},…, v_{l}} be the set of covers of v. Let λ_{i} denote Λ(v_{i}), and let k_{i} denote the repetition numbers, k(v_{i}). The multiplicity of Λ at v is defined to be

m_{v}(Λ) = c_{λ_{1},...,λ_{l},k(v)}^{Λ(v)}

• Let v be an x-node of the diagram P [X, Y], labeled with the partition Λ(v) and with repetition number k(v). Let C(v) = {v_{1}, v_{2},…, v_{l}} be the set of covers of v. Let λ_{i} denote Λ(v_{i}), and let k_{i} denote the repetition numbers, k(v_{i}). The multiplicity of Λ at v is defined to be

m_{v}(Λ) = ∑_{μ} c_{λ_{1},...,λ_{l}}^{μ}c_{Λ(v),1^{k(v)}}^{μ}.

If the multiplicity m_{v}(Λ) = 0 then we know that that particular labeling is not valid. Now, we will define the y-eigenvalues for each y-node v of the diagram P [X, Y]. We

want to have as many y-eigenvalues as the value of multiplicity. From the representation theory of the symmetric group, we know that

c_{λ_{1},...,λ_{l},k(v)}^{Λ(v)} = ∑_{Λ(v)/μ = k(v)–horizontal strip} c_{λ_{1},...,λ_{l}}^{μ}.  (14)

The node–eigenvalue, e_{v}(Λ), for each node v, is the set of the sums of the content over all squares in Λ(v)/μ for all possible μ for which Λ(v)/μ is a k(v)–horizontal strip minus

the binomial coefficient (\begin{matrix}k(v) \\ 2 \end{matrix}).

Recall that the content of a square is given by c(i, k) = k − i if the square is at position (i, k) in a partition (i th row and k th column).

This gives m_{v}(Λ) eigenvalues at each y-node v. We now define y-eigenvalue of Λ, e_{y}(Λ), to be the set of numbers obtained by taking a sum of one element of e_{v}(Λ) for each y-node v. So |e_{y}(Λ)| = ∏y-nodes v m_{v}(Λ).

4.4 Example

Let the poset P = {1, 2, 3, 4, 5} with the relations 1 <_{P} 2, 2 <_{P} 3, 3 <_{P} 4 and 4 <_{P} 5. The Hasse diagram of this poset is given in figure 1.

[figure: 1; 2; 3; 4; 5]

Figure 1: Example: poset P

Let X and Y be the sets X = {1, 2, 3} and Y = {4, 4, 5}. So the node 4, is a node with non-trivial repetition number k(4) = 2. The L-block V is spanned by the following pure wedges:

ζ = z_{1,4} ∧ z_{2,4} ∧ z_{3,5}
τ = z_{1,4} ∧ z_{2,5} ∧ z_{3,4}
η = z_{1,5} ∧ z_{2,4} ∧ z_{3,4}.

Thus the L-block V is 3-dimensional. We calculate the Laplacian L_{Y} on these three elements. Note that the Laplacian L_{Y} is in fact L_{Y} = (4, 5), since those are the only two comparable y’s.

L(ζ) = τ + η
L(τ) = ζ + η
L(η) = ζ + τ

The matrix representation of L_{Y} with respect to the basis < ζ, τ, η > is thus

L_{Y} = ( \begin{matrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} )

So the eigenvalues of the Laplacian are −1, −1, +2.

Now we will evaluate the y–eigenvalue for each of the poset tableaux for this L-block. The only nontrivial node is node 4. Thus, the y–eigenvalue, e_{y}(Λ), is the node-eigenvalue,

e_{4}(Λ) = c(1, 2) + c(1, 3) − (\begin{matrix}2 \\ 2 \end{matrix}). The result is given in figure 2.

Note that the y–eigenvalues of this labeling give exactly the same numbers as the eigenvalues of the Laplacian L_{Y}. In the next section we will show that this is not coincidental.

[figure: ∅; e_{y}(Λ) = 2; ∅; e_{y}(Λ) = −1; ∅; e_{y}(Λ) = −1]

Figure 2: Example: the y-eigenvalues

5 Centerpiece Theorem for L_{Y}

Theorem 14 (L_{Y} -Centerpiece) Let P be a linear poset with a minimum element, ˆ0. Let X and Y be two (multi-)sets, subsets of P. For every labeling Λ of positive multiplicity, each element in e_{y}(Λ) is an eigenvalue of L_{Y} with multiplicity ∏x-nodes v m_{v}(Λ).

Proof:

The proof of this theorem will be by induction on the sizes of the (multi)-sets X and Y. So let n = |X| = |Y | (counting multiplicities).

If n = 1 — there is nothing to prove as the Laplacian L_{Y} has no pairs to switch, and the only y-node is the maximal element for the diagram of the L-block. The Laplacian L_{Y} is the one-by-one zero matrix and the eigenvalue of this unique pair is zero.

Suppose n = 2. There are several different possible combinations of relations between sets X = {x_{1}, x_{2}} and Y = {y_{1}, y_{2}}.

• The most obvious one is x_{1} < x_{2} < y_{1} < y_{2}. In that case the Laplacian L_{Y} = (y_{1}, y_{2}), and the two possible elements are ζ_{1} = z_{x_{1},y_{1}} ∧z_{x_{2},y_{2}}, ζ_{2} = z_{x_{1},y_{2}} ∧z_{x_{2},y_{1}}. The Laplacian L_{Y} has the following matrix representation with respect to the basis ⟨ζ_{1}, ζ_{2}⟩:

L_{Y} = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} )

The eigenvalues of L_{Y} are +1 and −1. The eigenvalue of the poset tableau of type (X, Y)_{P} is given in figure 3. It also gives values +1 and −1, so the claim of the theorem holds.

[figure: ∅; e_{y}(Λ) = +1; ∅; e_{y}(Λ) = −1]

Figure 3: poset tableaux

[figure: ∅; ·; e_{y}(Λ) = 0; ∅; ·; e_{y}(Λ) = 0]

Figure 4: poset tableaux

• The second case is when

x_{1} < x_{2} < \begin{matrix}y_{1} \\ y_{2} \end{matrix}

In that case the Laplacian L_{Y} does nothing (since y_{1} and y_{2} are not comparable), thus the eigenvalues of L_{Y} are 0. The dimension of the L-block spanned by (X, Y) is two. The eigenvalues of the poset tableaux give the same values (figure 4), where the ”·” in a box denotes which square was deleted in that step.

Thus in this case the theorem checks too.

• x_{1} < y_{1} < x_{2} < y_{2} or equivalently (for our purpose) x_{1} < y_{1} = x_{2} < y_{2}. There is only one poset tableau spanned by these sets X and Y, namely the one shown on the figure 5.

The y-eigenvalue for the poset tableau is zero in both cases.

The Laplacian L_{Y} can not switch the y’s, since that would produce the element z_{x_{2},y_{1}} which doesn’t exist. So the Laplacian L_{Y} also acts as zero.

[figure: ∅; ∅; e_{y}(Λ) = 0]

Figure 5: poset tableau

[figure: ∅; e_{y}(Λ) = 0]

Figure 6: poset tableau

• x_{1} < x_{2} < y_{1} = y_{2}.

There is only one poset tableau spanned by these sets X and Y, namely the one shown on the figure 6.

The y-eigenvalue is again zero (contents of the partition (2) minus the binomial

coefficient (\begin{matrix}2 \\ 2 \end{matrix})). The Laplacian L_{Y} doesn’t have two distinct y’s to switch, thus, it

is zero.

• x’s are the same.

x_{1} = x_{2} < \begin{matrix}y_{1} \\ y_{2} \end{matrix}

There is only one poset tableau spanned by these sets X and Y, namely the one shown on the figure 7.

The y-eigenvalue is zero. The Laplacian L_{Y} has no comparable y’s to switch - thus L_{Y} = 0.

• x_{1} = x_{2} < y_{1} < y_{2}. There is only one poset tableau spanned by these sets X and Y, namely the one shown on the figure 8.

The y-eigenvalue is equal to 1. The Laplacian L_{Y} can switch y_{1} and y_{2} but the result would be the same element, since the x’s are indistinguishable. Thus the Laplacian

[figure: ∅; e_{y}(Λ) = 0]

Figure 7: poset tableau

[figure: ∅; e_{y}(Λ) = 1]

Figure 8: poset tableau

L_{Y} = 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

\begin{aligned}x_{1} & < & y_{1} \\ x_{2} & < & y_{2} \end{aligned} .

The poset tableau again gives zero as the y-eigenvalue, and since the y’s are not comparable, the Laplacian L_{Y} is also zero. So the theorem holds for the case n = 2.

Now, we will treat the general case n > 2.

• Label the y-nodes of the diagram of the L-block using the depth-first algorithm: 1. Start with a leftmost minimal y-element v.

2. If v is not the maximal unlabeled y-node go to the leftmost unlabeled cover of v, and repeat this step. Otherwise label v with next available number from the set {1, 2,…, |Y |}.

From this labeling we see that, y_{i} > y_{j} ⇒ i < j.

• Let P [X_{1}, Y_{1}], P [X_{2}, Y_{2}],…, P [X_{c}, Y_{c}] be the connected components of P [X, Y]. In that case the L-block V is the tensor product of the L-blocks of the P [X_{i}, Y_{i}]. The Laplacian L_{Y} switches only comparable y’s, and two y’s from different connected components are not comparable. Thus

L_{Y} (v_{1} ⊗ ⋯ ⊗ v_{c}) = ∑_{i=1}^{c} v_{1} ⊗ ⋯ ⊗ (L_{Y_{i}}v_{i}) ⊗ ⋯ ⊗ v_{c}.

So if v_{1},…, v_{c} are eigenvectors of L_{Y} with the eigenvalues e_{1},…, e_{c}, then v_{1} ⊗⋯⊗v_{c} is an eigenvector with eigenvalue e_{1} + ⋯ + e_{c}. Thus, by induction, we can label each of the components of P [X, Y] to get the eigenvalues of L_{Y} on the total L-block V.

• Suppose that P [X, Y] is connected. In that case, there must be a minimal element in P [X, Y], which must be an x-node.

Call the minimal element x_{n}. Then define x_{n−1}, x_{n−2},…, x_{a} by:

1. x_{a} > x_{a+1} > ⋯ > x_{n}, where all > are covering relations.

2. Either

Case 1: There is more than one element covering x_{a}. Case 2: x_{a} has unique cover in P [X, Y] but it is a y-element.

Let B = {x_{a},…, x_{n}}, and let G =Sym(B).

Lemma 15 Let σ ∈ G, and let ζ = z_{x_{i_{1}},y_{1}} ∧ z_{x_{i_{2}},y_{2}} ∧ ⋯ ∧ z_{x_{in},y_{n}} be non-zero. Then

ζ^{σ} = z_{σ(x_{i_{1}}),y_{1}} ∧ z_{σ(x_{i_{2}}),y_{2}} ∧ ⋯ ∧ z_{σ(x_{in}),y_{n}}

is also non-zero.

Proof: It is sufficient to prove the lemma for the transposition (x_{i_{k}}, x_{i_{l}}) ∈ G.

ζ^{(x_{i_{k}} ,x_{i_{l}})} = z_{x_{i_{1}},y_{1}} ∧ ⋯ ∧ z_{x_{i_{l}},y_{l}} ∧ ⋯ ∧ z_{x_{i_{k}} ,y_{k}} ∧ ⋯ ∧ z_{x_{in},y_{n}}.

Now, since (x_{i_{k}}, x_{i_{l}}) ∈ B, i.e., x_{i_{k}}, x_{i_{l}} are both less or equal to x_{a}, which is below all of the y’s – the lemma is clear.

Lemma 16 This action of G commutes with L_{Y}.

Proof: Let ζ = z_{x_{i_{1}},y_{1}} ∧ z_{x_{i_{2}},y_{2}} ∧ ⋯ ∧ z_{x_{in},y_{n}} be in our L-block, V, let y_{k} <_{P} y_{l} and let σ ∈ G. Then

σ · (y_{k}, y_{l}) · ζ = z_{σ(x_{i_{1}}),y_{1}} ∧ ⋯ ∧ z_{σ(x_{i_{k}}),y_{l}} ∧ ⋯ ∧ z_{σ(x_{i_{l}}),y_{k}} ∧ ⋯ ∧ z_{σ(x_{in}),y_{n}}

and

(y_{k}, y_{l}) · σ · ζ = z_{σ(x_{i_{1}}),y_{1}} ∧ ⋯ ∧ z_{σ(x_{i_{k}}),y_{l}} ∧ ⋯ ∧ z_{σ(x_{i_{l}}),y_{k}} ∧ ⋯ ∧ z_{σ(x_{in}),y_{n}}.

[figure: x_{n}; x_{a+1}; x_{a}; μ_{1}; μ_{2}; μ_{r}; T_{1}; T_{2}; T_{r}]

Figure 9: Case 1

So they are equal unless one of the expressions above is zero, and the other is not. The only way for that to happen is in the middle step, i.e., either σ·ζ = 0 or (y_{k}, y_{l})·ζ = 0. But we assumed the ζ ≠ 0, and by our lemma above σ · ζ ≠ 0. So the only possible conflict is (y_{k}, y_{l}) · ζ = 0, and (y_{k}, y_{l}) · σ · ζ ≠ 0. But since σ ∈ G, it only moves elements of B which are allowed to be paired with any y_{k}.

Let C(x_{a}) = {v_{1}, v_{2},…, v_{l}} be the set of covers of x_{a}. We will prove the following generalization of the Centerpiece Theorem:

Theorem 17 Let Λ be a poset tableau of positive multiplicity, and let λ_{i} be the label of v_{i} in Λ. Let λ be a partition such that

c_{λ_{1},...,λ_{l}}^{λ}c_{Λ(x_{a}),1^{k(xa)}}^{λ} ≠ 0.

Then the occurrences of G-irreducibles S^{λ} in V can be indexed by such poset tableaux Λ, and the Laplacian L_{Y} acts on S^{λ} as one of the scalars in e_{Y} (Λ).

5.0.1 Case 1

Suppose there are two or more subtrees above the node x_{a}, in our poset P (as in figure 9). Label the subtrees above the x_{a} by T_{1}, T_{2},…, T_{r}. Let Y = Y_{1} ∪Y_{2} ∪⋯∪Y_{r} (Y_{i} ⊂ T_{i}), where Y_{i} ∩ Y_{j} = ∅. Let k_{i} = |Y_{i}|, and let

Y_{i} = {y_{k_{1}+k_{2}+···+k_{i−1}+1},…, y_{k_{1}+···+k_{i−1}+k_{i}}}.

Because of our labeling, we know that all relations between y’s are contained within the sets Y_{i}, i.e., y_{i} <_{P} y_{j} implies that both y_{i}, y_{j} are in the same Y_{k}.

Let b_{i} be the number of y_{j}’s in T_{i} minus the number of x_{j}’s in T_{i} (note that in general T_{i} will have more y_{j}’s than x_{j}’s). In other words, b_{i} = |Y_{i}| − |X ∩ T_{i}|.

Consider element ζ = z_{x_{i_{1}},y_{1}} ∧ z_{x_{i_{2}},y_{2}} ∧ ⋯ ∧ z_{x_{in},y_{n}}. If we want ζ to be non-zero, there will be exactly b_{i} x_{j}’s from B in ζ paired up with the y_{j}’s of T_{i}.

Split the L-block

V = ⊕_{(S_{1},S_{2},...,S_{r})}V [S_{1}, S_{2},…, S_{r}],  (15)

where S_{1} ∪ S_{2} ∪ ⋯ ∪ S_{r} = B, |S_{i}| = b_{i}, and V [S_{1},…, S_{r}] is the span of all ζ with exactly the elements of S_{i} paired with the y’s from T_{i}.

Lemma 18 1. As a vector space

V [S_{1},…, S_{r}] ≅ V (X_{1} ∪ S_{1}, Y_{1}) ⊗ V (X_{2} ∪ S_{2}, Y_{2}) ⊗ ⋯ ⊗ V (X_{r} ∪ S_{r}, Y_{r}),

where X_{i} and Y_{i} are the multisets of the x and y elements of the subtree T_{i}.

2. With respect to the decomposition in 1., the Laplacian L_{Y} acts as:

L_{Y} (v_{1} ⊗ ⋯ ⊗ v_{r}) = ∑_{i} v_{1} ⊗ ⋯ ⊗ (L_{Y_{i}}v_{i}) ⊗ ⋯ ⊗ v_{r}.

Proof: Statement 1. is clear by the definition of V [S_{1},…, S_{r}]. To prove statement 2., we only need to recall that the Laplacian L_{Y} switches comparable y’s, and that y’s in different subtrees can not be comparable because of linearity of the poset. L_{Y} can switch only y’s in the same subtree T_{i}.

Let G_{i} = Sym(S_{i}). Note that G_{1} × G_{2} ×… × G_{r} acts on V [S_{1}, S_{2},…, S_{r}]. Let s_{i} be the minimal node of subtree T_{i}.

Now apply the induction hypothesis to L-blocks, V_{i} = V (X_{i} ∪ S_{i}, Y_{i}). According to our theorem this gives the decomposition of the L-block V_{i} as G_{i}-module, and the eigenvalues of L_{Y_{i}} are indexed by poset tableau of shape μ_{i} ⊢ |S_{i}|. Moreover each poset tableau of shape μ_{i} with eigenvalue e_{i} represents a copy of the irreducible S^{μ_{i}} in the e_{i}-eigenspace.

Now, as a G_{1} × … × G_{r} module we know the eigenspaces of L_{Y} are given by our labeling up to the points s_{i} where the partitions μ_{i} come together at x_{a}.

Lemma 19 As a G =Sym(B)-module, the space V is

V ≅ ind(Sym(S^{0}_{1}) × ⋯ × Sym(S^{0}_{r}) )^{Sym(B)}(V [S_{1}^{0},…, S_{r}^{0}]),

where (S_{1}^{0},…, S_{r}^{0}) is any fixed ordered partition of B.

Proof: Choose S_{i}^{0} = {x_{a+b_{1}+···+b_{i−1}}, x_{a+b_{1}+···+b_{i−1}+1},…, x_{a+b_{1}+···+b_{i}−1}}. Let Sh denote the set of permutations σ ∈ Sym(B) such that σ(u) < σ(v) whenever u, v are in the same set S_{i}^{0} for some i. There is 1–1 correspondence between the σ ∈ Sh and the sequences indexing the summands in the 15, namely

σ ↔ (S_{1}^{σ},…, S_{r}^{σ})

where S_{i}^{σ} = {σ(u) : u ∈ S_{i}^{0}}. Also Sh is a collection of coset representatives for Sym(S_{1}^{0})×⋯×Sym(S_{r}^{0}) in Sym(B). Thus we have a natural vector space isomorphism between V and

V [S_{1}^{0},…, S_{r}^{0}] ⊗_{(Sym(S^{0}_{1}) × ⋯ × Sym(S^{0}_{r}))} Sym(B).

It is straightforward to check that this isomorphism commutes with the action of Sym(B).

Let Λ_{i} be a poset tableau of type (X_{i} ∪ S_{i}, Y_{i}) , where μ_{i} ⊢ b_{i} is the label of the vertex s_{i}. By our inductive hypothesis the Laplacian L_{Y} acts as a scalar on the irreducible S^{μ_{i}}, i.e., L_{Y} |_{S^{μ_{i}}} = e_{Y} (Λ_{i}). Applying Lemma 18 part 2., we have

L_{Y} (v_{1} ⊗ ⋯ ⊗ v_{r}) = ∑_{i} v_{1} ⊗ ⋯ ⊗ (L_{Y_{i}}v_{i}) ⊗ ⋯ ⊗ v_{r}
= ∑_{i} v_{1} ⊗ ⋯ ⊗ e_{Y} (Λ_{i})v_{i} ⊗ ⋯ ⊗ v_{r}
= (∑_{i} e_{Y} (Λ_{i}))(v_{1} ⊗ ⋯ ⊗ v_{r}).

Now, we will use the fact ([16, 17, 24]) that

(S^{μ_{1}} ⊗ ⋯ ⊗ S^{μ_{r}}) ↑_{G_{1}×···×G_{r}}^{G}= ⊕_{λ⊢|B|}c_{μ_{1},μ_{2},...,μ_{r}}^{λ}S^{λ}.

Thus we have c_{μ_{1},μ_{2},...,μ_{r}}^{λ} copies of the G-module S^{λ} which explains why this is the multiplicity of the label λ on node x_{a} in our labeling.

• Now we have to decide what is the dimension of each eigenspace. But that is something we will have to do in the second case too - so we will do it for both cases at the end.

5.0.2 Case 2

Let A = {y_{n−k+1},…, y_{n}} be the largest possible set so that

x_{a} ≤_{P} y_{n} ≤_{P} y_{n−1} ≤_{P} ⋯ ≤_{P} y_{n−k+1}

and there are no x_{i}’s with y_{n} ≤_{P} x_{i} < y_{n−k+1}. Call A the ”terminal Y -set of V ” (figure 10). Note that |B| ≥ |A|. Split the L-block

V = ⊕_{(a_{1},...,a_{k})}V (a_{1},…, a_{k})

for (a_{1},… a_{k}), a sequence of distinct elements of length k from B, and the vector space V (a_{1}, a_{2},…, a_{k}) represents the span of all ζ which are of the form

ζ = z_{x_{1},y_{1}} ∧ ⋯ ∧ z_{x_{n−k},y_{n−k}} ∧ z_{a_{1},y_{n−k+1}} ∧ ⋯ ∧ z_{a_{k},y_{n}}

[figure: x_{n}; x_{a}; y_{n}; y_{n−k+1}; t; ⟩NB; s; ⟩Â; v_{0}; B; A; T]

Figure 10: Case 2

For the moment assume that all of the elements x_{a},…, x_{n} are distinct and all of y_{n−k+1},…, y_{n} are distinct. Then we will look back at how we must modify the argument when some of the x_{i}’s and y_{j}’s are equal.

Fix the sequence (a_{1},…, a_{k}), let B′ = B\{a_{1},…, a_{k}} and let G′ be the subgroup G′ = Sym(B′) ≤ G. Note that G′ acts on V (a_{1},…, a_{k}).

Lemma 20 V (a_{1},…, a_{k}) is isomorphic to the L-block V_{0} given by the following sets:

X′ = X\{a_{1}, a_{2},…, a_{k}} and Y′ = Y \{y_{n−k+1},…, y_{n}}.

Proof: The isomorphism ψ : V (a_{1},…, a_{k}) → V_{0} is an obvious one

ψ(z_{x_{1},y_{1}} ∧ ⋯ ∧ z_{x_{n−k},y_{n−k}} ∧ z_{a_{1},y_{n−k+1}} ∧ ⋯ ∧ z_{a_{k},y_{n}}) = z_{x_{1},y_{1}} ∧ ⋯ ∧ z_{x_{n−k},y_{n−k}}.

It is clearly a bijective linear map.

• Now let’s examine the L-block given by X′ and Y′. Let Â be the terminal Y -set for X′, Y′. Let B̂ = B′ ∪ {new x_{i}’s} be the x_{i}’s below Â. Let NB denote the set of those new x_{i}’s. Let Ĝ = Sym(B̂). Note that G′ ≤ Ĝ.

If Â = ∅ then there are no y’s in the interval (y_{n−k+1}, v_{0}]. In that case the group Ĝ is the group described in the case 1., i.e., Ĝ = G_{1} × G_{2} × ⋯ × G_{r}, where G_{i} is acting on the subtree T_{i} above x_{a}.

Now apply the induction hypothesis to X′, Y′. This gives the decomposition of the L-block as a Ĝ-module. The theorem says that the irreducible summands are indexed by the (X′, Y′) poset tableaux Λ̂ of shape λ′ (where λ′ ⊢ |B̂|), and ”shape” means that the minimal element s of Â is labeled with λ′. Also the theorem tells us that the Y -Laplacian L_{Y} for (X′, Y′) acts like the scalar e_{Y} (Λ̂) on this copy of the irreducible S^{λ′}. From this we can deduce by restriction from Ĝ to G′ the following Lemma 21 As a G′-module, the space V (a_{1},…, a_{k}) decomposes as a sum over all (X′, Y′) tableaux Λ̂ of shape λ′ of the module

S^{λ′} ↓Ĝ ; G′

Moreover the Laplacian L_{Y} for (X′, Y′) acts like the scalar e_{Y} (Λ̂) on this entire restriction.

Let Ch(λ′, μ′) be the set of chains λ′ ≥ λ_{1} ≥ ⋯ ≥ μ′ where the steps in the chain of the partitions are all of size 1.

Now using the fact [16] that

(S^{λ′}) ↓Ĝ ; G′= ⊕_{μ′⊢|B′|}S^{μ′}|Ch(λ′, μ′)|,

we can rewrite this lemma to say that the sum is over all extensions of Λ̂ to a labeling of the new x_{i}’s of S^{μ′} where the extension gives the label μ′ to t, the minimal element of the set NB.

• Now we need one last lemma.

Lemma 22 As a G-module V ≅ ind_{G′}^{G}(V (x_{n−k+1},…, x_{n})).

Proof: For each sequence \underline{α} = (a_{1},…, a_{k}) let π\underline{α} be the permutation in G which maps a_{i} to x_{n−k+i} and which leaves the elements of B\{a_{1},…, a_{k}} in increasing order. Then π\underline{α} is a set of coset representatives for G\G′. Since there is one for every \underline{α} this shows that as vector spaces

V ≅ V (x_{n−k+1},…, x_{n}) ⊗_{G′} ℂ(G).

Let g ∈ G, let (a_{1},…, a_{k}) = \underline{α} be a sequence, and let b_{i} = g(a_{i}), \underline{β} = (b_{1},…, b_{k}). Then

g(z_{x_{1},y_{1}} ∧ ⋯ ∧ z_{x_{n−k},y_{n−k}} ∧ z_{a_{1},y_{n−k+1}} ∧ ⋯ ∧ z_{a_{k},y_{n}}) =
= z_{g(x_{1}),y_{1}} ∧ ⋯ ∧ z_{g_{(}x_{n−k}),y_{n−k}} ∧ z_{b_{1},y_{n−k+1}} ∧ ⋯ ∧ z_{b_{k},y_{n}}
= (π−1 ; \underline{β}gπ\underline{α})z_{x′_{1},y_{1}} ∧ ⋯ ∧ z_{x′_{n−k},y_{n−k}} ∧ z_{x_{n−k+1},y_{n−k+1}} ∧ ⋯ ∧ z_{x_{n},y_{n}},

where x′_{i} is obtained by replacing the elements of {x_{a},…, x_{n}}\{a_{1},…, a_{k}} by the elements of the set {x_{a},…, x_{n−k}} in order. This computation shows that the vector space isomorphism above is a G-module isomorphism.

Now putting all the claims together with the fact([16]) that

ind_{Sym(B′)}^{Sym(B)}(S^{μ′}) = ⊕_{Ch(λ,μ′)}S^{λ},  (16)

shows that as a module for G, V decomposes as a sum, over all (X, Y) poset tableaux Λ of shape λ, of a copy of S^{λ}. In the expression 16 we know by induction that the Laplacian L_{Y′} involving all switches which do not involve the terminal set, i.e., L_{Y′} = L_{Y} − L_{0} (where L_{0} is the sum of the switches involving the terminal Y -set), acts like the scalar e_{Y} (Λ′), where Λ′ is the labeling as far as the point t (the minimal x above A). The switches involving the terminal Y -set A must be studied. But those y_{i} in A can either be switched with each other or with y_{j} that are above an x_{i} ∈ B. By the choice of the terminal set A, y_{i} is comparable to y_{j} for all y_{i} ∈ A. It follows that L_{0} acts on V = ⊕V (a_{1},…, a_{k}) by the sum of all transpositions (x_{i}, x_{j}) for x_{i} ∈ B, x_{j} ∈ {a_{1},…, a_{k}}. In terms of our induced module, L_{0} acts on

ind_{G′}^{G}(S^{μ′}) = S^{μ′} ⊗_{ℂG′} ℂG

like left multiplication on ℂG by ∑_{x_{i},x_{j}∈B}(x_{i}, x_{j}) − ∑_{x_{i},x_{j}∈B′}(x_{i}, x_{j}).

But ∑_{x_{i},x_{j}∈B′}(x_{i}, x_{j}) passes through ⊗_{ℂG′} to act on S^{μ′} like the scalar ∑_{x∈μ′} c_{x} [24]. And ∑_{x_{i},x_{j}∈B}(x_{i}, x_{j}) acts on each G-irreducible S^{λ} in S^{μ′} ⊗_{ℂG′} ℂG like ∑_{y∈λ} c_{y}. The result is that L_{0} acts on each copy of S^{λ} in ind_{G′}^{G}(S^{μ′}) like ∑_{y∈λ/μ′} c_{y}. This explains the scalars e_{Y} (Λ) and their multiplicity. In order to be able to add the eigenvalues of L_{Y′} and L_{0}, we need the following lemma.

Lemma 23

L_{0} · L_{Y′} = L_{Y′} · L_{0}

Proof: Let (x_{i}, x_{j}) be a transposition in Sym(B), where x_{i} ∈ B′, and let (x_{i}, x_{k}) be a transposition of L_{Y′}. By the choice of B we know that (x_{j}, x_{k}) is also transposition in L_{Y′}, and since

(x_{i}, x_{j})(x_{i}, x_{k}) = (x_{j}, x_{k})(x_{i}, x_{j})

the lemma is clear.

• Now we want to consider the case where some of the elements of the sets A, B, and NB = {new x’s} are equal. So let’s write

A = α_{1} ∪ α_{2} ∪… ∪ α_{l},
B = β_{1} ∪ β_{2} ∪… ∪ β_{m},
NB = γ_{1} ∪ γ_{2} ∪… ∪ γ_{n},

where |α_{i}| = a_{i}, |β_{i}| = b_{i} and |γ_{i}| = c_{i}, and the y’s in each of the α_{i} are equal, the x’s in each of the β_{i} are equal, and the x’s in each γ_{i} are equal. Let

Π_{A} = ∑_{σ_{i}∈S_{α_{i}}} σ_{1} × σ_{2} ×… × σ_{l}
Π_{B} = ∑_{σ_{i}∈S_{β_{i}}}(∏ sgn(σ_{i}))σ_{1} ×… × σ_{m}
Π_{NB} = ∑_{σ_{i}∈S_{γ_{i}}}(∏ sgn(σ_{i}))σ_{1} ×… × σ_{n}.

These projections commute with the Laplacian L_{Y} according to lemma 12.

Let Π be the projection Π = Π_{A} × Π_{B} × Π_{NB}, where the Π_{A} acts on the y’s while the other two projections act on the set of the x’s. Then the projection Π projects our original space (with all of the x’s and y’s distinct) examined in the previous paragraph to the space where y_{i}’s in each α_{i}, x’s in each β_{i} and γ_{i} are equal. Each of the components of Π commutes with L_{Y} so we know that Π and L_{Y} commute.

Lemma 24 The projection Π maps the L-block V with all x’s and y’s distinct to the L-block ΠV, where α_{i}, β_{i} and γ_{i} are sets of equal elements.

Proof: Let ζ = z_{x_{1},y_{1}} ∧ ⋯ ∧ z_{a_{1},y_{i_{1}}} ∧ ⋯ ∧ z_{a_{k},y_{i_{k}}} ∧ ⋯ ∧ z_{x_{n},y_{n}}. Note that if y_{i} = y_{j}, the element ζ^{(i,j)} and ζ are the same. In general, if σ_{i} ∈ S_{α_{i}}, then ζ^{σ_{i}} = ζ. So we need to identify all equal elements. This is accomplished by projecting with a symmetrizer, i.e., we identify the class of elements ∪_{σ∈S_{α_{i}}}ζ^{σ} with the sum ∑_{σ∈S_{α_{i}}} ζ^{σ}. But Π_{A} does exactly this identification, Π_{A} = ∑_{σ∈S_{α_{i}}} ζ^{σ}. The same is true for Π_{B} and Π_{NB}.

NOTE: When we deal with the projection Π_{A} there is one important point we have to make. The Laplacian L_{Y} 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 L_{0}, we have to subtract all switches involving two y’s from the same α_{k}. Each of these transpositions doesn’t move any

of the y’s (or x’s), and there are exactly (\begin{matrix}a_{k} \\ 2 \end{matrix} ) of them. Thus, we have to subtract

(\begin{matrix}a_{k} \\ 2 \end{matrix} ) from the eigenvalue ∑_{y∈λ/μ′} c_{y} of L_{0}.

Thus we can write our space V as a direct sum of the eigenspaces

V = ⊕_{w}V_{w},

where the sum is over all eigenvalues w of L_{Y}.

We want to know the eigenvalues of L_{Y} on the image ΠV. We will use that fact that the multiplicity of w as an eigenvalue on ΠV is the dimension of ΠV_{w}.

So we need to compute the dimension of ΠV_{w}. At present we have V_{w} written in terms of labellings of poset tableaux. So pick such a labeling which at the end has a λ′ at vertex s coming down to a μ′ at vertex t then back up to a λ at vertex y_{n} (using the notation of this proof). This represents a piece of the eigenspace of the corresponding eigenvalue w where Ĝ = Sym(B̂) acts like S^{λ′} and G′ = Sym(B′) acts like S^{μ′}. We need the following lemma.

Lemma 25 Let B′ ⊂ B̂, NB = B̂\B′. Then the multiplicity of S^{μ′} in Π_{NB}S^{λ′} is equal to the number of ways to get μ′ from λ′ by successfully removing vertical strips of lengths c_{1}, c_{2},. . ..

Note: As a consequence of this lemma, when we came to an x-node v of the repetition number k(v) we have to remove a vertical strip of length k(v).

Proof of lemma: Recall that NB = γ_{1}∪γ_{2}∪…∪γ_{n}. The projection Π_{NB} projects onto the sgn_{γ_{1}} ⊗ sgn_{γ_{2}} ⊗ ⋯ ⊗ sgn_{γ_{n}} isotypic component of S^{λ′} considered as a S_{γ_{1}} ×… S_{γ_{n}} module. Thus, for μ′ a partition of |B′|,

⟨Π_{NB}S^{λ′}, S^{μ′}⟩_{S_{B′}} = ⟨S^{λ′}, S^{μ′} ⊗ sgn_{γ_{1}} ⊗ ⋯ ⊗ sgn_{γ_{n}}⟩_{S_{B′}×S_{γ_{1}}×...×S_{γn}}

= (# of ways to get μ′ from λ′ by successively removing vertical strips of lengths c_{1}, c_{2},…)

Now consider the next step of going up from μ′ to λ. At this point we have a piece of the w-eigenspace on which B′ acts like S^{μ′}.

The multiplicity coming from this S^{μ′} is dim(Π_{A}ind_{G′}^{G}(S^{μ′})Π_{B}). We need to check how Π_{A} acts on the induction ind_{G′}^{G}(S^{μ′}) = S^{μ′} ⊗_{ℂG′} ℂG. The σ ∈ Π_{A} permutes the y’s. We identified the induction by identifying the sequence of the x_{i}’s that are paired with the set A. So switching y_{i}’s has the effect in Sym(B) of switching the positions corresponding to B\B′. In other words, Π_{A} −Π_{B} has the effect of projecting onto the (trivial ⊗ sgn) characters of (S_{α_{1}} ×… × S_{α_{l}}) × (S_{β_{1}} ×… × S_{β_{m}}) ⊂ S_{B\B′} × S_{B} where S_{B\B′} × S_{B} is acting on S^{μ′} ⊗_{ℂG′} ℂG via left multiplication on ℂG by S_{B\B′} and right multiplication on ℂG by S_{B}. So to determine dim(Π_{A}ind_{G′}^{G}(S^{μ′})Π_{B}) it will be helpful to know the decomposition of the induction ind_{G′}^{G}(S^{μ′}) as a S_{B\B′} × S_{B}-module.

Lemma 26 ([12]) Let μ′ ⊢ m, G = S_{r}, G′ = S_{m} and H = S_{r−m} (acting on {m + 1,…, r}). Then as a H × G module, the induced representation S^{μ′} ⊗_{ℂG′} ℂG decomposes as

ind_{G′}^{G}(S^{μ′}) = ⊕_{λ⊢r,μ⊂λ}S^{λ/μ} ⊗ S^{λ}.

Now armed with that lemma, let us return to the dimension count.

dim(Π_{A}ind_{G′}^{G}(S^{μ′})Π_{B}) =

= ⟨(ind_{G′}^{G}(S^{μ′})) ↓, (ϵ_{α_{1}} ⊗ ⋯ ⊗ ϵ_{α_{l}}) ⊗ (sgn_{β_{1}} ⊗ ⋯ ⊗ sgn_{β_{m}})⟩
= ∑_{λ⊢n,μ′⊂λ}⟨(S^{λ/μ′}) ↓, (ϵ_{α_{1}} ⊗ ⋯ ⊗ ϵ_{α_{l}})⟩⟨S^{λ} ↓, sgn_{β_{1}} ⊗ ⋯ ⊗ sgn_{β_{m}}⟩
= ∑_{μ′⊂λ⊢n}⟨(S^{λ/μ′} ⊗ S^{λ}) ↓, (ϵ_{α_{1}} ⊗ ⋯ ⊗ ϵ_{α_{l}}) ⊗ (sgn_{β_{1}} ⊗ ⋯ ⊗ sgn_{β_{m}})⟩
= ∑_{λ⊢n} (# of ways to get λ from μ′ by adding horizontal strips of lengths α_{1}, α_{2},…) · (# of ways to get λ from ∅ by removing a vertical strips of lengths β_{1}, β_{2},…)

= (# of poset tableaux labellings from μ′ up to λ then down to ∅, which add a horizontal strip of length k for every y-vertex of repetition number k and subtract a vertical strip of length k for every x-vertex of repetition number k.)

This completes the proof of the theorem.

5.1 Adding the L_{X}

Consider the Laplacian L_{X}. Since we have identified the L-block V with a subspace of the symmetric group algebra ℂS_{n}, by fixing the order on the x’s, every time the Laplacian L_{X} switches a pair of x’s, it is actually putting a minus sign in front of the corresponding basis element, with the x’s ordered. Since L_{X} acts as a sum of transpositions, every eigenvalue we obtain from the L_{X}, will have a minus sign.

Recall the multiplicity of the x-node. Let v be an x-node of the diagram P [X, Y], labeled with the partition Λ(v) and with repetition number k(v). Let C(v) = {v_{1}, v_{2},…, v_{l}} be the set of covers of v. Let λ_{i} denote Λ(v_{i}), and let k_{i} denote the repetition numbers, k(v_{i}). The multiplicity of Λ at v is defined to be

m_{v}(Λ) = ∑_{μ} c_{λ_{1},...,λ_{l}}^{μ}c_{Λ(v),1^{k(v)}}^{μ}.

The node–eigenvalue, e_{v}(Λ), for each node v, is the set of sums of the content over

all squares in μ/Λ(v) above for a given μ minus the binomial coefficient (\begin{matrix}k(v) \\ 2 \end{matrix}).

This gives m_{v}(Λ) eigenvalues at each x-node v. We now define the x-eigenvalue of Λ, e_{x}(Λ), to be the set of numbers obtained by taking a sum of one element of e_{v}(Λ) for each x-node v. So |e_{x}(Λ)| = ∏x-nodes v m_{v}(Λ).

Theorem 27 (L_{X}-Centerpiece) Let P be a linear poset with a minimum element, ˆ0. Let X and Y be two (multi-)sets, subsets of P. For every labeling Λ of positive multiplicity, each element in e_{x}(Λ) is an eigenvalue of L_{X}.

Proof:

The proof of this theorem is similar to the proof of the L_{Y} -Centerpiece Theorem. We will omit the details here.

Since L_{X} and L_{Y} commute (as established in Lemma 13 ), the eigenvalues of L_{X} + L_{Y} will be the sum of the eigenvalues on the corresponding irreducibles of the eigenspaces.

Recall that the complete Laplacian L is the sum of three things (from the beginning of this section):

L = L_{D} + L_{X} + L_{Y}

The L_{D} component is the diagonal matrix, which on the L-block V spanned by the sets (X, Y), has value:

e_{D}(X, Y) = w(X, Y) + Δ(X, Y)

[figure: ∅]

Figure 11: A diagram of the L-block

w(X, Y) = ∑_{m=1}^{k} |(x_{m}, y_{m})|
Δ(X, Y) = ∑_{i,j} δ_{x_{i},y_{j}}

The computer evidence strongly supports the following conjecture:

6 Complete Centerpiece Conjecture

Conjecture 28 Let P be a linear poset with a minimum element, ˆ0. Let X and Y be two multisets of vertices of P. Let e_{D}(X, Y) be defined as above. For every poset tableau Λ of positive multiplicity, let τ_{Y} (Λ) ∈ e_{Y} (Λ) and let τ_{X}(Λ) ∈ e_{X}(Λ). Then the scalars e(Λ) = τ_{Y} (Λ) − τ_{X}(Λ) + e_{D}(X, Y) are the complete set of eigenvalues of the Laplacian L.

This conjecture claims that the same poset tableau will work simultaneously for both Laplacians (L_{X} and L_{Y}),i.e., that the eigenvalues of the Laplacian L are the sum of the eigenvalues of L_{Y} and the eigenvalues of L_{X} 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 a linear poset, using only combinatorial properties of the poset. In these two small cases (n=1 and n=2) we had no difficulty. For larger n, we need some extra results.

7.1 H_{1}

For example, if we want to evaluate the homology H_{1}(L_{P}) of a Lie algebra L_{P} corresponding to a linear poset P, with ˆ0, our construction gives an immediate answer.

An L-block V of size 1, is determined by the sets (X, Y), X = {x}, and Y = {y}. Obviously, if we want V to be non-zero, x <_{P} y. So the corresponding diagram of this L-block is given in figure 11.

Both indicators e_{Y} and e_{X} are zero, so the eigenvalues are given by e_{D}. But Δ(X, Y) = 0 too, since X ∩ Y = ∅. Thus L(z_{x,y}) = w(X, Y)z_{x,y} = |(x, y)|z_{x,y}. In other words, the

[figure: ∅; e(Λ) = 0; ∅; e(Λ) = 0]

Figure 12: poset tableaux

eigenvectors of the Laplacian L_{1} are the basis vectors z_{x,y}, and the corresponding eigenvalues are |(x, y)|, i.e., the number of the vertices in the poset P, between x and y.

The dimension of the homology is the number of zero eigenvalues, i.e., the number of the intervals z_{x,y}, such that y covers x.

Thus dim(H_{1}(L_{P})) = (# of covering relations in P).

7.2 H_{2}

In this case the L-block V in question is spanned by the (multi–)sets (X, Y), each of size 2, i.e., X = {x_{1}, x_{2}} and Y = {y_{1}, y_{2}}. There are several possibilities for the L-block.

1. All four elements are comparable, and x’s are below the y’s.

x_{1} < x_{2} < y_{1} < y_{2}.

All possible poset tableaux are shown in figure 12.

As we can see, both e_{Y} and e_{X} eigenvalues are zero. So we don’t have to worry how to add them up - we will always get zero. Δ is also zero, since the sets X and Y are disjoint. Thus again, the Laplacian is L(z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}}) = w(X, Y)(z_{x_{1},y_{1}} ∧ z_{x_{2},y_{2}}). But in this case, both intervals contain at least one element, so w(X, Y) > 0. Thus in this case we never get a zero eigenvalue, which might contribute to the homology H_{2}.

2. y’s are not comparable.

x_{1} < x_{2} < \begin{matrix}y_{1} \\ y_{2} \end{matrix} .

All possible poset tableaux are shown in figure 13.

The value of e_{Y} is zero, while the values of e_{X} are +1 and -1. Since e_{D} is always non-negative, the value +1 cannot contribute to homology H_{2}. The other value can, but only if e_{D} is +1. That means that all of the relations indicated: x_{1} < x_{2}, x_{2} < y_{1} and x_{2} < y_{2} are covering relations in the poset P. Whenever we have a four-element

[figure: ∅; e(Λ) = −1; ∅; e(Λ) = +1]

Figure 13: poset tableaux

[figure: ∅; ∅; e(Λ) = 0]

Figure 14: poset tableau

subset of the poset P, with covering relations x_{1} < x_{2}, x_{2} < y_{1} and x_{2} < y_{2}, we will call that a ”Y“-configuration. Thus in this case every occurrence of ”Y“-configuration (described as above) in the Hasse diagram of the poset contributes to the homology H_{2}.

3. x_{1} < y_{1} < x_{2} < y_{2} or equivalently (for our purpose) x_{1} < y_{1} = x_{2} < y_{2}. There is only one poset tableau spanned by these sets X and Y, namely the one shown in figure 14.

Since the space is one dimensional, we will add the eigenvalues in the only possible way. In the first case, the eigenvalue will be zero, if and only if both relations, x_{1} < y_{1} and x_{2} < y_{2}, are covering relations, i.e., every occurrence of a distinct (all vertices are distinct) pair of covering relations contributes to the dimension of the homology H_{2}. If the second case occurs, i.e., if y_{1} = x_{2} then Δ will contribute to the eigenvalue, and it won’t be zero anymore.

4. x_{1} < x_{2} < y_{1} = y_{2}.

There is only one poset tableau spanned by these sets X and Y, namely the one shown on the figure 15.

[figure: ∅; e(Λ) = −1]

Figure 15: poset tableau

[figure: ∅; e(Λ) = 0]

Figure 16: poset tableau

Since the space is one dimensional, we will add the eigenvalues in the only possible way. The eigenvalue will be zero, if and only if x_{2} covers x_{1}, and both y_{1} and y_{2} cover x_{2}. Note that this is a degenerate letter ”Y“, with a joint top vertex. Every such occurrence contributes to the homology H_{2}.

5. x’s are the same.

x_{1} = x_{2} < \begin{matrix}y_{1} \\ y_{2} \end{matrix} .

There is only one poset tableau spanned by these sets X and Y, namely the one shown in figure 16.

Since the space is one dimensional, we will add the eigenvalues in the only possible way. The eigenvalue will be zero, if and only if both y_{1} and y_{2} cover x_{1} = x_{2}. Again, it is a pair of distinct covering relations with joint vertex, this time the x-vertex. Every such occurrence contributes to the homology H_{2}.

6. x_{1} = x_{2} < y_{1} < y_{2}. There is only one poset tableau spanned by these sets X and Y, namely the one shown on the figure 17.

Since the space is one dimensional, we will add the eigenvalues in the only possible way. Since e_{D} is always non-negative, the value +1 cannot contribute to homology H_{2}.

7. The trivial case when x’s are not comparable (so because of linearity and the minimum element neither are y’s).

\begin{aligned}x_{1} & < & y_{1} \\ x_{2} & < & y_{2} \end{aligned} .

[figure: ∅; e(Λ) = 1]

Figure 17: poset tableau

This is equivalent to case 3. We have two distinct covering relations, so it contributes to the homology H_{2}.

All together, the dimension of the homology H_{2} is in fact the number of distinct pairs of covering relations + the number of occurrences of letter “Y” in the poset P. In other words, if H is a Hasse diagram of a poset P, with e edges, and γ letter “Y”’s (degenerate

or not), we have H_{1}(L_{P}) = ℂ^{e} and H_{2} = ℂ^{(\begin{matrix}e \\ 2 \end{matrix})+γ}.

7.3 Examples

• For example, suppose that we are dealing with the chain poset on n vertices (1 < 2 < ⋯ < n), T_{n}. The number of non-degenerate Y’s is zero. The number of degenerate letters Y is (n-2). The number of edges in the Hasse diagram is (n-1). Hence

H_{1}(T_{n}) = ℂ^{n−1},

and

H_{2}(T_{n}) = ℂ^{n−2+(\begin{matrix}n − 1 \\ 2 \end{matrix})} = ℂ^{\frac{(n+1)(n−2)}{2}}

• Let the poset P be given in figure 18, where the length of the chains are m and n. This is in fact equivalent to having two disjoint chains, of length m and n. Thus the corresponding homologies will be H_{1} = ℂ^{n−1+m−1}, and H_{2}(T_{n}) = ℂ^{\frac{(n+1)(n−2)}{2} + \frac{(m+1)(m−2)}{2}}

8 Conclusion

These results have several interesting corollaries that are of a combinatorial nature. We will state one. Let P be a rooted tree on n nodes and let Σ be the sum in the group algebra of S_{n} of all transpositions (i, j) such that i is on the unique path from j to the root in P. Then Σ acting on ℂS_{n} by left multiplication has non-negative integer eigenvalues and the corresponding eigenspaces can be identified in representation-theoretic terms.

There is still a lot to do in this area. Although my work provides partial answers and a conjecture for all rooted trees, the question is still open for other posets. What

[figure: 1; n; m]

Figure 18: poset P

happens in those cases is very difficult to control, since the expression for the Laplacian becomes more complicated. In the tree case I wish to examine a twisting of the Laplacian by a parameter α which, my advisor has shown, is related to Jack polynomials and the Krawtchouk polynomials in certain special cases. Lastly I would like to see if the more algebraic consequences of Kostant’s theorem have sensible analogues in my case.

Acknowledgement. I would like to thank my advisor, Professor Phil Hanlon, for the guidance, help, and for his invaluable suggestions.

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