J Algebr Comb (2014) 40:45–74DOI 10.1007/s10801-013-0478-1

The hard Lefschetz property for Hamiltonian GKMmanifolds

Shisen Luo

Received: 17 December 2012 / Accepted: 10 September 2013 / Published online: 2 October 2013© Springer Science+Business Media New York 2013

Abstract We introduce characteristic numbers of a graph and demonstrate that theyare a combinatorial analogue of topological Betti numbers. We then use characteris-tic numbers and related tools to study Hamiltonian GKM manifolds whose momentmaps are in general position. We study the connectivity properties of GKM graphsand give an upper bound on the second Betti number of a GKM manifold. When themanifold has dimension at most 10, we use this bound to conclude that the manifoldhas nondecreasing even Betti numbers up to half the dimension, which is a weakversion of the Hard Lefschetz Property.

Keywords Graph cohomology · Hard Lefschetz property · Hamiltonian GKM

1 Introduction

In their landmark paper [2], Goresky, Kottwitz, and MacPherson provided a meansto turn the computation of equivariant cohomology into a combinatorial one. Moreconcretely, we assume that a compact torus T

l = S1 × · · · × S1, l ≥ 2, acts on asmooth compact manifold M . If M is a GKM manifold (the precise definition willbe given in Sect. 2), we can assign to M a simple graph Γ = (V ,E), called the GKMgraph of M , and a map

α : E →C[x1, x2, . . . , xl]1

S. Luo (B)Department of Mathematics, Cornell University, Ithaca, NY 14853-4201, USAe-mail: ssluo@math.cornell.edu

46 J Algebr Comb (2014) 40:45–74

where C[x1, x2, . . . , xl]1 denotes the set of nonzero homogeneous linear polynomialsin x1, x2, . . . , xl . The GKM theorem establishes the following isomorphism:

H ∗Tl (M;C) ∼=

{(f1, f2, . . . , f|V |) ∈

|V |⊕i=1

C[x1, x2, . . . , xl]∣∣∣∣α(eij )|fi − fj ∀eij ∈ E

}.

(1.1)We denote the right-hand side of the isomorphism by H ∗

Tl (Γ,α), or H ∗Tl (Γ ), when

there is no chance of confusion. It is called the (equivariant) graph cohomology of Γ .The graph cohomology H ∗

Tl (Γ ) is naturally graded by assigning degree 1 to eachvariable xi . The isomorphism (1.1) divides the degrees in half. From this isomor-phism we can see that H ∗

Tl (M,C) only has elements of even degree.Inspired by [2], Gullemin and Zara [5] studied the combinatorial properties of

H ∗Tl (Γ ) and showed that many familiar theorems in geometry have natural counter-

parts in graph theory. In this paper, we take the opposite approach. We study thecombinatorics of GKM graphs to deduce properties of Hamiltonian GKM manifolds.

We now introduce the general setup for the paper and some definitions necessaryfor it. The motivation is from symplectic geometry, but no knowledge of geometry isnecessary for understanding the combinatorial object of our study. A more detailedexposition of the geometric background and motivation will be given in Sect. 2.

Definition 1.1 Given an injective map φ : V → R2 where φ(vi) = (zi1, zi2), we de-

fine the induced map α : E →C[x, y]1 from φ by

α(eij ) = (zj1 − zi1)x + (zj2 − zj2)y

for any eij ∈ E. As we are dealing with an undirected graph, the above definition isonly well defined up to sign, but this ambiguity does not cause problems with thedivision condition in (1.1).

Definition 1.2 We say that φ : V → R2 is in general position if no three points in V

are colinear. In particular, this implies that φ is injective.

We assume the following setup for the remainder of the paper.

Setup 1.3 Let Γ = (V ,E) be a simple graph, and φ : V → R2 a map in general

position. For simplicity, we will always use m to stand for |V |. Let α : E →C[x, y]1be induced by φ, and

H ∗T(Γ ) =

{(f1, f2, . . . , fm) ∈

m⊕i=1

C[x, y]∣∣∣∣α(eij )|fi − fj ∀eij ∈ E

}.

Remark 1.4 Equivalently, one may define the graph cohomology as the kernel of themap

δ :⊕v∈V

Sym•(C

2) →⊕e∈E

Sym•(C

2)/α(e),

where the component of δ(f1, f2, . . . , fm) corresponding to eij is fi − fj .

J Algebr Comb (2014) 40:45–74 47

We show in Sect. 3 that H ∗T(Γ ) is always a free module over C[x, y] of dimen-

sion m. For any homogeneous basis of H ∗T(Γ ) as a module over C[x, y],

γ1, γ2, . . . , γm,

we set ci(Γ ) = |{j ∈ N|degree(γj ) = i}| to be the number of degree i elements inthe basis. Although the basis is not unique, the number of basis elements in a fixeddegree is independent of the choice of the basis, hence ci(Γ ) is well defined.

Definition 1.5 We call ci(Γ ) the ith characteristic number of Γ .

These cis will serve as the combinatorial counterpart for the geometric Betti num-bers of the manifolds in this paper. They are different from the combinatorial Bettinumbers in [5], whose definition will be given as follows.

Choose a generic direction ξ ∈ R2, namely one that ensures φ(vi) · ξ �= φ(vj ) · ξ

for i �= j . For any vi ∈ V , define the index of vi to be

σ(vi) = ∣∣{vj ∈ V |eij ∈ E, φ(vj ) · ξ < φ(vi) · ξ}∣∣.The kth (combinatorial) Betti number of Γ is defined as the number of vertices ofindex k and denoted by βk(Γ ). Guillemin and Zara showed in [5] that with a fixedaxial function, although the indices do depend on the choice of ξ , the combinatorialBetti numbers do not. We do not need an axial function in this paper and hence willnot give the definition here. Interested readers are referred to [5] for more details.

When Γ is the GKM graph of a manifold, then these three notions, characteris-tic numbers, combinatorial Betti numbers, and geometric Betti numbers, all agree.Guillemin and Zara introduced the βis as the combinatorial counterpart of geomet-ric Betti numbers [5]. They do exhibit some properties that resemble the geometricones, including the Poincaré duality. But there are certain shortfalls. First of all, theyare well defined only for a very restrictive class of graphs, regular graphs with axialfunctions. This makes inductive arguments difficult. Secondly, for a connected regu-lar graph equipped with an axial function, one would expect β0 to be 1. This is notthe case, as the following simple example shows.

Example 1.6 Consider a cycle on five vertices, mapped into R2 as in Fig. 1. The axial

function on Γ is induced by the given embedding. The arrow points in the ξ direction,and the number beside a vertex indicates the index of that vertex. We can see thatalthough the Poincaré duality still holds, the graph has the undesirable property thatβ0 = 2.

In contrast, the characteristic numbers are defined in a much more general setting,and c0 is equal to the number of connected components of the graph. In Sect. 4, westudy the basic properties of characteristic numbers and give comparisons with thegeometric Betti numbers. In the remaining two sections, we study the properties ofgraphs that are in fact the GKM graphs of Hamiltonian GKM manifolds. In Sect. 5,using our tools of characteristic numbers and the fact that the top Betti number of a

48 J Algebr Comb (2014) 40:45–74

Fig. 1 A regular graph withaxial function

compact oriented manifold is 1, we deduce some connectivity properties of (actual)GKM graphs, under the assumption that the moment map is in general position.

Section 6 is largely motivated by a question posed by Yael Karshon.

Question 1.7 [6, Problem 4.2] Suppose that a symplectic manifold (M,ω) admitsa Hamiltonian S1 action with isolated fixed points. Does (M,ω) satisfy the hardLefschetz property?

Susan Tolman pointed out an easier version of the question, which we have simplifiedto our setting.

Question 1.8 [6, Special case of Problem 4.3] Suppose M is a Hamiltonian GKMmanifold of dimension 2d . Do the Betti numbers of M satisfy

β0 ≤ β2 ≤ · · · ≤ β2 d2 �,

where 2 d2 � is the largest even integer no greater than d?

Using the tools and results we develop in Sects. 4 and 5, in Sect. 6 we give anupper bound for β2 of Hamiltonian GKM manifold whose moment map is in generalposition, which is a subclass of the manifolds in Question 1.7. This is the contentof Theorem 6.1. In dimensions 8 and 10, this implies that the even Betti numbers ofthese manifolds must be nondecreasing up to half the dimension. So our work pro-vides a class of examples for which there is a positive answer to Tolman’s question.We hope this framework can be developed to give further results in higher dimen-sions.

2 Geometric background

The isomorphism (1.1) established in [2] applies in more general settings, but for thepurpose of this paper, we will only give the definition of Hamiltonian GKM mani-folds. We will also explain the geometric motivation behind Setup 1.3.

Assume that ω is a symplectic form on a compact manifold M and Tl = S1 ×

· · ·× S1, l ≥ 2 acts on M Hamiltonianly (we refer the readers to [1] for definitions ofbasic notions in symplectic geometry). Let φ be a moment map for Tl ýM .

J Algebr Comb (2014) 40:45–74 49

Every Hamiltonian manifold (M,ω,Tl , φ) has a Tl-invariant almost complex

structure, and the space of such structures is contractible, so the weights of theisotropy representation of Tl on TpM at each fixed point p is well defined.

The Hamiltonian manifold (M,ω,Tl , φ) is called GKM if

• MTl

consists of isolated fixed points {p1,p2, . . . , pm}.• For every p ∈ MT

l, the weights

αi,p ∈ (tl)∗

, i = 1, . . . , d = dimM

2,

of the isotropy representation of Tl on TpM are pairwise linearly independent,

where we used (tl)∗ to denote the dual of the Lie algebra of Tl .

Given (M,ω,Tl , φ) a compact Hamiltonian GKM manifold, its GKM graphΓ = (V ,E) is defined by

• V = MTl, and

• any two vertices p1 and p2 are incident if and only if there exists a codimension-1subtorus K of Tl such that p1 and p2 are in the same connected component of MK .

The GKM graph of a compact Hamiltonian GKM manifold is a simple regulargraph of degree dimM

2 .Note that the moment map φ : M → (tl)∗ � R

l induces a map from V to Rl by

restriction. This map is also denoted by φ. When we talk about a GKM graph, wealways think of it as a triple (V ,E,φ : V → R

l).We define an induced map

α : E → C[x1, . . . , xl]1,

where C[x1, . . . , xl]1 stands for the set of nonzero homogeneous linear polynomialsin x1, . . . , xl , by α(eij ) = (φ(vi)−φ(vj )) ·(x1, x2, . . . , xl). In other words, if φ(vi) =(p1,p2, . . . , pl) and φ(vj ) = (q1, q2, . . . , ql), then

α(eij ) = (p1 − q1)x1 + (p2 − q2)x2 + · · · + (pl − ql)xl.

The map α is only well defined up to sign, but this is sufficient for our purpose.In the case we are concerned about in this paper, the map α in isomorphism (1.1)

is induced by the moment map φ. For any l ≥ 2, we may restrict the torus actionT

l ýM to a smaller torus action T2 ýM that is still GKM. So we only consider the

case l = 2 in the paper.The second one of the GKM conditions, i.e., the pairwise linear independency of

isotropy weights, requires that α(eij ) and α(eik) to be linearly independent for j �= k.This is a weaker assumption than φ being in general position. However, in this paper,to make some arguments valid, we always assume φ to be in general position.

3 Graph cohomology is a free module in dimension two

The main theorem in this section, Theorem 3.1, holds in a much more general settingthan we need in the rest of the paper. In particular, it does not require the existence of

50 J Algebr Comb (2014) 40:45–74

the map φ, and there is no constraint on the map α. The proof uses some basic tech-niques from algebraic geometry. The corresponding statement in higher dimensionsdoes not hold.

Theorem 3.1 Given Γ = (V ,E) a simple graph and α : E → C[x, y]1, we haveH ∗

T(Γ ) ∼= (C[x, y])m as modules over C[x, y], where m = |V |.

Proof Let M = H ∗T(Γ ). As a graded module over C[x, y], M defines a quasicoherent

sheaf over ProjC[x, y] = CP 1. We denote this sheaf by F . As a module over C[x, y],M also defines a quasicoherent sheaf over SpecC[x, y] = C

2. We denote this sheafby G . The restriction of G to C

2\{0} will be denoted by H , which is a quasicoherentsheaf over C2\{0}.

Lemma 3.2 The sheaf F is locally free.

Proof of Lemma 3.2 CP 1 is covered by D(x) = SpecC[ yx] and D(y) = SpecC[ x

y].

Now that F (D(x)) = (Mx)0, the degree 0 part of the localized module Mx , it followsfrom the definition of M that F (D(x)) is a torsion-free C[ y

x]-module and hence a

free C[ yx]-module. Similarly, F (D(y)) is a free C[ x

y]-module. �

Lemma 3.3 Let π :C2\{0} → CP 1 be the natural projection. Then π∗F = H .

Proof of Lemma 3.3 This can be proved by looking at the distinguished open subsetsof both spaces. The space C

2\{0} is covered by SpecC[x, 1x, y] and SpecC[x, y, 1

y],

while CP 1 is covered by SpecC[ yx] and SpecC[ x

y].

We have the following natural isomorphism:

(Mx)0 ⊗C[ y

x] C

[x,

1

x, y

]= (Mx)0 ⊗

C[ yx](C

[y

x

]⊗C C

[x,

1

x

])

= (Mx)0 ⊗C C

[x,

1

x

]= Mx.

Similarly,

(My)0 ⊗C[ xy] C

[x, y,

1

y

]= My.

These exactly say π∗F = H . �

As an immediate consequence, H is also locally free. By a theorem ofGrothendieck [3], F splits as a direct sum of line bundles (the corresponding state-ment fails for locally free sheaves over CP 2, and this is the key reason why graphcohomology is special in dimension two). So H is also a direct sum of line bun-dles, but the Picard group of C2\{0} is trivial, so H must be a trivial vector bundle.In particular, H (C2\{0}) must be a free module over OC2\{0}(C2\{0}) = C[x, y].

Now G (C2\{0}) = H (C2\{0}) and G (C2) = M , so the following lemma willenable us to conclude that M is a free C[x, y]-module.

J Algebr Comb (2014) 40:45–74 51

Lemma 3.4 The restriction map r : G (C2) → G (C2\{0}) is an isomorphism.

Proof of Lemma 3.4 We can think of G (C2\{0}) as⋂

a �=0 or b �=0 Max+by . The inter-section makes sense since M is torsion-free by definition, and thus Max+by can bethought of as a subset of M(0) = C(x, y)m, the localization of M at the zero primeideal.

Assume that (f1, . . . , fm) ∈ G (C2\{0}) = ⋂a �=0 or b �=0 Max+by ⊆ C(x, y)m. Then

we immediately see that fi ∈C[x, y] for all i. Since the graph is finite, we can pick anonzero linear polynomial px + qy such that it is not a multiple of any α(eij ). Then(f1, . . . , fm) ∈ Mpx+qy says that there exists n such that

α(eij )|(px + qy)n(fi − fj )

for all eij ∈ E. So

α(eij )|(fi − fj ),

which says exactly (f1, . . . , fm) ∈ M . �

Now we know that M is a free C[x, y]-module. To see its dimension, letf = ∏

eij ∈E α(eij ). The dimension of M as module over C[x, y] equals the di-mension of Mf as a module over C[x, y]f . But Mf is generated by (1,0, . . . ,0),(0,1, . . . ,0), . . . , (0,0, . . . ,1) as a C[x, y]f -module. So dimM = m. This completesthe proof of the theorem. �

The similar result does not hold in higher dimensions in general, even if we as-sume that Γ is a regular graph and α is induced from a map φ : V → R

l . An easycounterexample is the following.

Example 3.5 Consider Γ = (V ,E) given by V = {v1, v2, v3, v4} and E = {e12, e23,

e34, e14}. Note that Γ is a regular graph of degree 2. Define φ : V → R3 by

φ(v1) = (0,0,0), φ(v2) = (1,0,0),

φ(v3) = (1,1,0), φ(v4) = (1,1,1).

Then φ induces α : E →C[x, y, z]1 given by

α(e12) = x, α(e23) = y, α(e34) = z, α(e14) = x + y + z.

Then H ∗T3(Γ,α) as a module over C[x, y, z] is generated by

(1,1,1,1), (0, x, x + y, x + y + z), (0, xy,0,0),

(0,0, yz,0), and (0, xz, xz,0).

This is not free as a C[x, y, z]-module since the generators are related by

y(0, xz, xz,0) = z(0, xy,0,0) + x(0,0, yz,0).

This naturally leads to the following open question.

52 J Algebr Comb (2014) 40:45–74

Question 3.6 Given a regular graph Γ = (V ,E) together with an axial functionα : E → C[x1, . . . , xl]1 in the sense of Definition 1 in Sect. 2.1 in [5], is the graphcohomology necessarily a free module over C[x1, . . . , xl]? If not, what are the com-binatorial criteria on (Γ,α) that guarantee H ∗

Tl (Γ,α) to be free?

4 Properties of characteristic numbers

In this section we study some basic properties of characteristic numbers, defined inDefinition 1.5. These numbers do not satisfy the Poincaré duality in general, but wewill prove a weaker version of it. We will discuss how the “top characteristic num-ber” behaves for regular graphs. Then we discuss their relationship to the combina-torial Betti numbers. Next, we will compute the characteristic numbers for completegraphs, which are (graph theoretically) GKM graphs for complex projective spaces.Finally, we prove an analogue of the Künneth formula.

Notation 4.1 Continuing from Setup 1.3, the vertices of Γ are labeled as v1, v2,

. . . , vm. The edge connecting vi and vj will be denoted by eij . We do not distinguishbetween eij and eji , but most of the time we will use the smaller number as the firstindex. We will use λ(vi) to denote the degree of vi , i.e., the number of edges incidentto vi . We use λΓ (vi) to denote the degree of vi when we want to emphasize vi as avertex of Γ .

The characteristic number ci is not affected by composing φ : V → R2 with a

linear automorphism of R2, so without loss of generality, we may assume that φ(vi)

and φ(vj ) have different second components for any i �= j . Also, we notice that thedefinition of graph cohomology remains unchanged if we scale α, so if we assumethat φ(vi) = (pi, qi), φ(vj ) = (pj , qj ) and let aij = −pj −pi

qj −qi, we may redefine α as

α(eij ) = y − aij x.

For any positive integers p ≤ q , we will use bqp to denote the pth standard basis

vector in Cq , i.e., the pth entry of bq

p is 1, which is the only nonzero entry.For 1 ≤ i < j ≤ m, we let

v0ij = bm

i − bmj = (0, . . . ,1,0, . . . ,−1,0, . . . ,0) ∈C

m;v1ij = b2m

i − b2mj + aij b2m

i+m − aij b2mj+m

= (0, . . . ,1,0, . . . ,−1,0, . . . ,0,0, . . . , aij ,0, . . . ,−aij ,0, . . . ,0)

= (v0ij , aij v0

ij

) ∈C2m;

v2ij = b3m

i − b3mj + aij b3m

i+m − aij b3mj+m + a2

ij b3mi+2m − a2

ij b3mj+2m

= (v1ij , a2

ij v0ij

) ∈C3m;

and so forth, where we use (u,v) to denote concatenation of two vectors u and v.

J Algebr Comb (2014) 40:45–74 53

In general,

vkij = b(k+1)m

i − b(k+1)mj + aij b(k+1)m

i+m − aij b(k+1)mj+m + · · ·

+ akij b(k+1)m

i+km − akij b(k+1)m

j+km = (vk−1ij , ak

ij v0ij

) ∈C(k+1)m.

Denote by Mk(Γ ) the matrix of size |E|×(k+1)m, whose rows are indexed by E,and the row corresponding to eij is vk

ij . Let rk(Γ ) = rankMk(Γ ). This is the dimen-

sion of the vector space spanned by the vectors {vkij | eij ∈ E}. We set r−1(Γ ) = 0.

Let sk(Γ ) = |E|− rk(Γ ). This is the dimension of the vector space of linear relationsamong the vectors {vk

ij | eij ∈ E}. We set s−1(Γ ) = |E|.Throughout the rest of the paper, we will use the following notation as introduced

in Setup 1.3, Definition 1.5, and just above:

Γ, V, E, m = |V |;

v1, v2, . . . , vm, λ(vi), λΓ (vi), eij , for 1 ≤ i, j ≤ m;

φ, α, aij , vkij , for 1 ≤ i, j ≤ m and k ≥ 0;

Mk(·), ck(·) for k ≥ 0; rk(·), sk(·), for k ≥ −1,

where · in the last row can denote any graph in the sense of Setup 1.3.

Remark 4.2 Recall from Remark 1.4 the alternative definition of H ∗T(Γ ) as ker δ.

The matrix Mk defined above is the matrix representation of the degree k part of themap δ in a monomial basis. This will become more clear after the proof of Proposi-tion 4.5.

Remark 4.3 If we make Γ into a directed graph by letting each edge in E leave thevertex of smaller subscript and enter the vertex of larger subscript, then M0(Γ ) is thetranspose of the incidence matrix of this directed graph. The matrix M1(Γ ) is closelyrelated to the rigidity matrix of the graph Γ and the map φ, which is an importantconcept in rigidity theory. More information about the rigidity matrix and rigiditytheory can be found in [4]. The work in this paper is closely related to some work inrigidity theory by the author [7, 8]. For instance, if we let k = 1 in Lemma 4.14 andLemma 5.19 of this paper, they become Lemmas 2.3 and 2.7 in [7] respectively. If welet d = 4 in Proposition 6.8 of this paper, it becomes Proposition 4.7 in [7].

Example 4.4 Figure 2 shows a graph on four vertices. We have marked the coordi-nates of the image of φ and of the image of α. The edge set is E = {e12, e14, e23,

e24, e34}. So M0(Γ ) is a 5 × 4 matrix with rows indexed by E and columns indexed

54 J Algebr Comb (2014) 40:45–74

Fig. 2 An example of Γ , φ,and α

by V . In this example, we have

M0(Γ ) =

⎛⎜⎜⎜⎜⎝

v1 v2 v3 v4

e12 1 −1 0 0e14 1 0 0 −1e23 0 1 −1 0e24 0 1 0 −1e34 0 0 1 −1

⎞⎟⎟⎟⎟⎠.

We can compute Mk(Γ ) by recursively adding columns. As labeled in Fig. 2, we cansee that a12 = −3, a14 = 1, a23 = 1

2 , a24 = − 13 , and a34 = −2, so M2(Γ ) is

M2(Γ )

=

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

v01 v0

2 v03 v0

4 v11 v1

2 v13 v1

4 v21 v2

2 v23 v2

4

e12 1 −1 0 0 a12 −a12 0 0 a212 −a2

12 0 0

e14 1 0 0 −1 a14 0 0 −a14 a214 0 0 −a2

14

e23 0 1 −1 0 0 a23 −a23 0 0 a223 −a2

23 0

e24 0 1 0 −1 0 a24 0 −a24 0 a224 0 −a2

24

e34 0 0 1 −1 0 0 a34 −a34 0 0 a234 −a2

34

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

.

A direct computation shows that r0(Γ ) = 3 and rk(Γ ) = 5 for all k ≥ 1. It followsthat s0(Γ ) = 2 and sk(Γ ) = 0 for all k ≥ 1.

Proposition 4.5 The characteristic numbers may be computed as follows:

c0(Γ ) = π0(Γ ), the number of connected components of Γ, (4.6)

= m − r0(Γ ) = m − |E| + s0(Γ ); (4.7)

ck(Γ ) = 2rk−1(Γ ) − rk(Γ ) − rk−2(Γ ) (4.8)

= sk(Γ ) + sk−2(Γ ) − 2sk−1(Γ ), ∀k ≥ 1. (4.9)

J Algebr Comb (2014) 40:45–74 55

Proof Let f = (f1, f2, . . . , fm) ∈ Cm. Then it follows from the definition of graph

cohomology that f is a class in H 0T(Γ ) if and only if fi = fj ∈C whenever vi and vj

are in the same connected component of Γ . So c0 is equal to the number of connectedcomponents of Γ , which we denote by π0(Γ ). This proves (4.6).

For any k ≥ 0, assume that

f =(

k∑n=0

z1nxk−nyn,

k∑n=0

z2nxk−nyn, . . . ,

k∑n=0

zmnxk−nyn

)∈ Hk

T(Γ ),

where zst ∈ C for 1 ≤ s ≤ m, 0 ≤ t ≤ k. Then by the definition of graph cohomology,this is equivalent to

(y − aij x)

∣∣∣∣∣(

k∑n=0

zinxk−nyn

)−

(k∑

n=0

zjnxk−nyn

)

for each eij in E. This in turn means that when we substitute y with aij x in

(k∑

n=0

zinxk−nyn

)−

(k∑

n=0

zjnxk−nyn

),

we should get the zero polynomial:

0 =(

k∑n=0

anij zinx

k

)−

(k∑

n=0

anij zjnx

k

)

=(

k∑n=0

anij (zin − zjn)

)xk.

So∑k

n=0 anij (zin − zjn) = 0, and this expression can be rewritten as

vkij · (z10, z20, . . . , zm0, z11, z21, . . . , zm1, z12, . . . , z1k, z2k, . . . , zmk) = 0.

Therefore,

dimHkT(Γ ) = (k + 1)m − rk(Γ ).

By the definition of cis, we have

dimHkT(Γ ) =

k∑n=0

(k + 1 − n)cn,

and sok∑

n=0

(k + 1 − n)cn = (k + 1)m − rk(Γ ). (4.10)

56 J Algebr Comb (2014) 40:45–74

In particular, when k = 0, this immediately simplifies to

c0 = m − r0,

proving (4.7).The formula (4.8) for ck for k ≥ 1 can be proved inductively. For the base case

k = 1, we have, by Eq. (4.10),

c1 = 2m − r1 − 2c0

= 2m − r1 − 2(m − r0)

= 2r0 − r1

= 2r0 − r1 − r−1

= s1 + s−1 − 2s0.

Now for k > 1, we assume that formula (4.8) holds for smaller k. Then by Eq. (4.10)we have

ck = (k + 1)m − rk −k−1∑n=0

(k + 1 − n)cn

= (k + 1)m − rk −k−1∑n=1

(k + 1 − n)(2rn−1 − rn−2 − rn) − (k + 1)(m − r0).

We easily simplify this to conclude that

ck = 2rk−1 − rk − rk−2 = sk + sk−2 − 2sk−1,

as desired. �

Remark 4.11 Continuing with Remark 4.2, we have the following exact sequence:

0 −→ H ∗T(Γ ) −→

⊕v∈V

Sym•(C

2) δ−→⊕e∈E

Sym•(C

2)/α(e) −→ coker δ −→ 0.

Proposition 4.5 essentially relates the vector space dimension of H ∗T(Γ ) in each de-

gree to that of coker δ. The proposition itself may also be deduced by the fact that thealternating sum of the Hilbert series of an exact sequence equals zero.

Example 4.12 We continue with Example 4.4. Applying Proposition 4.5 to this ex-ample gives c0(Γ ) = 1, c1(Γ ) = 1, and c2(Γ ) = 2.

Notation 4.13 For any graph Γ = (V ,E) and v ∈ V , we write Ev for the set ofedges incident to v and write Γ − v for the subgraph of Γ obtained by deletingthe vertex v. We obviously have Γ − v = (V \{v},E\Ev). Similarly, for any subsetU ⊆ V , we write EU for the set of edges incident to some vertex in U and writeΓ − U for (V \U,E\EU).

J Algebr Comb (2014) 40:45–74 57

The following lemma will be used repeatedly throughout the paper. We call itthe Deletion Lemma. This concerns how the sks, hence the characteristic numbers,behave when we delete a vertex from a graph.

Lemma 4.14 (Deletion Lemma) Assume that a vertex vt ∈ V is of degree less thanor equal to k + 1, i.e., λ(vt ) ≤ k + 1. Then sk(Γ ) = sk(Γ − vt ).

Proof For simplicity, we write λ(vt ) = d . Without loss of generality, we may assumethat t = 1 and the d edges incident to vt are e12, e13, . . . , and e1,d+1. Now assumethat there is a linear relation among {vk

ij |eij ∈ E}:∑

eij ∈E

uij vkij = 0. (4.15)

We now consider the submatrix of Mk(Γ ) defined by the first d rows and columns 1,m + 1, . . . , km + 1:

N =

⎛⎜⎜⎜⎜⎝

1 a12 a212 . . . ak

12

1 a13 a213 . . . ak

13...

......

. . ....

1 a1,d+1 a21,d+1 . . . ak

1,d+1

⎞⎟⎟⎟⎟⎠ =

⎛⎜⎜⎜⎜⎝

xk12

xk13...

xk1,d+1

⎞⎟⎟⎟⎟⎠ ,

where xk1,i = (1, a1i , . . . , a

k1i ) is the vector consisting of the 1, m+ 1, . . . , (km+ 1)th

entries of vkij . This is a Vandermonde matrix. The assumption of φ in general position

guarantees that a1i �= a1j for i �= j , so the matrix N has full row rank. For the columns1, m + 1, . . . , (km + 1) of Mk(Γ ), the only nonzero entries are in the first d rows, soit follows from (4.15) that

d+1∑j=2

u1j xk1j = 0.

We conclude that u1j = 0 for all 2 ≤ j ≤ d + 1. Therefore, (4.15) reduces to

∑eij ∈E\Evt

uij vkij = 0.

This means that (4.15) is in fact a linear relation among {vkij |eij ∈ E\Evt }.

So sk(Γ ) = sk(Γ − vt ). �

Corollary 4.16 We may now bound the difference between sk(Γ ) and sk(Γ − vt ),

0 ≤ sk(Γ ) − sk(Γ − vt ) ≤ max(λ(vt ) − k − 1,0

).

Proof It immediately follows from the Deletion Lemma 4.14 and the definitionof sk . �

58 J Algebr Comb (2014) 40:45–74

Lemma 4.17 Let Γ = (V ,E), and let k ≥ maxi λ(vi) − 1. Then rk(Γ ) = |E| andsk(Γ ) = 0. Consequently, ck+2(Γ ) = 0.

Proof The repeated application of the Deletion Lemma will give sk(Γ ) = 0. Hence,rk(Γ ) = |E| − sk(Γ ) = |E|, and it follows from Proposition 4.5 that ck+2 = sk+2 −2sk+1 + sk = 0. �

Proposition 4.18 For any Γ = (V ,E), we have

∞∑i=0

ci(Γ ) = m = |V | (4.19)

and∞∑i=0

ici(Γ ) = |E|. (4.20)

Proof Equation (4.19) is already proved in Theorem 3.1. Turning to (4.20), accord-ing to Lemma 4.17, we can choose k ∈ N such that rk(Γ ) = |E| and ci = 0 for alli ≥ k + 1. Then, by formula (4.10),

k∑i=0

(k + 1 − i)ci = (k + 1)m − rk = (k + 1)m − |E|.

So∞∑i=0

ici =k∑

i=0

ici =k∑

i=0

(k + 1)ci − (k + 1)m + |E| = |E|.

This completes the proof. �

Remark 4.21 When Γ is a regular graph of degree d , then it follows from Proposi-tion 4.18 that the average degree of a set of generators of H ∗

T(Γ ) is d

2 . In this sense,the proposition provides a weak version of the Poincaré duality.

Proposition 4.22 Assume that Γ = (V ,E) is a connected regular graph of degree d .Then ci = 0 for i > d and cd = 0 or 1.

Proof The first part of the proposition is just a special case of Lemma 4.17.As for the second part, pick any edge eij and let Γ ′ = (V ,E\{eij }). Then we

have sd−2(Γ ) ≤ sd−2(Γ′)+ 1. By repeatedly applying the Deletion Lemma to Γ ′ we

get sd−2(Γ′) = 0. So sd−2(Γ ) ≤ 1. By Lemma 4.17 we have sd−1(Γ ) = sd(Γ ) = 0.

So we conclude

cd(Γ ) = sd(Γ ) + sd−2(Γ ) − 2sd−1(Γ ) ≤ 1,

as desired. �

J Algebr Comb (2014) 40:45–74 59

Remark 4.23 As the GKM graph of a GKM manifold is a regular graph, Propo-sition 4.22 can be viewed as a weaker version of the fact that the top geometricBetti number of a connected compact oriented manifold is 1. If the connected regulargraph Γ of degree d is also equipped with an axial function in the sense of [5], thenone can show by [5, Theorem 2.2] that cd(Γ ) = 1 and (

∏e1j ∈E α(e1j ),0,0, . . . ,0)

can be taken as the generator of H ∗T(Γ ) in degree d .

Example 4.24 Consider a regular graph of degree 2 whose vertices are in generalposition. By Proposition 4.22, ci = 0 for i ≥ 3. Also we know that c0 = 1 as thegraph is connected. Then the weaker version of the Poincaré duality, Proposition 4.18,forces c2 to be 1, hence c1 = 3.

In general, for regular graph of degree 2 and m vertices in general position, wehave c0 = c2 = 1 and c1 = m − 2.

Next we study the relations between the characteristic numbers and the combi-natorial Betti numbers. Proposition 4.25 provides a generalization of [5, inequality(2.28)] in dimension two.

Proposition 4.25 Given Γ = (V ,E), order the vertices as v1, . . . , vm. Define theindex of vi as

μi = ∣∣{j ∈ N|j > i and eij ∈ E}∣∣,and let

bk = ∣∣{i ∈ N|μi = k}∣∣.Then

k∑i=0

(k + 1 − i)ci ≤k∑

i=0

(k + 1 − i)bi

for any k ≥ 0.

Proof First, observe that∑∞

i=0 bi = |V | = m because each vertex is counted exactlyonce. Second, we claim that

∑∞i=0 ibi = |E|. To see this, we count the edge set in the

following way. First, we look at a vertex vi with μi = 0 and count how many edgesconnect vi to a vertex vj whose subscript j is larger than i. The answer must beμi = 0 by the definition of μi . Then we look at a vertex vi with μi = 1. In this case,there should be one edge connecting vi to some vertex of larger subscript. We repeatthis process until we have made the count for every vertex. Then altogether we willhave counted

∑∞i=0 ibi many edges. On the other hand, we see that every edge in E

is counted once and only once. So∑∞

i=0 ibi = |E|.Let Γ ′ = Γ − v1. Applying Corollary 4.16 to v1, we get the bound

sk(Γ ) ≤ sk(Γ ′) + max

(μ1 − (k + 1),0

).

60 J Algebr Comb (2014) 40:45–74

Applying Corollary 4.16 to v2, v3, . . . , vm in order, we get

sk(Γ ) ≤m∑

j=1

max(μj − (k + 1),0

)

=∞∑

i=k+2

(i − (k + 1)

)bi

=∞∑i=0

(i − (k + 1)

)bi −

k∑i=0

(i − (k + 1)

)bi

=∞∑i=0

ibi − (k + 1)

∞∑i=0

bi +k∑

i=0

(k + 1 − i)bi

= |E| − (k + 1)m +k∑

i=0

(k + 1 − i)bi .

So

k∑i=0

(k + 1 − i)ci = (k + 1)m − rk

= (k + 1)m − (|E| − sk)

≤ (k + 1)m − |E| + |E| − (k + 1)m +k∑

i=0

(k + 1 − i)bi

=k∑

i=0

(k + 1 − i)bi,

where the first equality is Eq. (4.10). �

We next compute the characteristic numbers of a complete graph. They turn out tobe the Betti numbers of complex projective space, as expected.

Definition 4.26 Define ω ∈ H 1T(Γ ) by ω = (x1x +y1y, x2x +y2y, . . . , xmx +ymy),

where (xi, yi) = φ(vi). We call this the equivariant symplectic form of Γ .Theorem 3.1 guarantees that H ∗

T(Γ ) is a free C[x, y]-module. If we denote by I

the ideal of H ∗T(Γ ) generated by (x, x, . . . , x) and (y, y, . . . , y), then the quotient

ring H ∗T(Γ )/I , which we will denote by H ∗(Γ ), is a vector space of dimension m.

A module basis of H ∗T(Γ ) descends to a vector space basis of H ∗(Γ ). The image

of ω in H ∗(Γ ) is denoted by ω and is called the ordinary symplectic form of Γ .As graded C[x, y]-modules, we have H ∗

T(Γ ) ∼= H ∗(Γ ) ⊗C C[x, y].

If Γ ′ = (V ′,E′) is a subgraph of Γ = (V ,E), then Γ ′ is equipped with φ′ and α′induced by φ and α. If we denote by i the natural inclusion map i : Γ ′ → Γ , then it

J Algebr Comb (2014) 40:45–74 61

induces a graded C[x, y]-algebra homomorphism i∗ : H ∗T(Γ ) → H ∗

T(Γ ′). This map

descends to a graded ring homomorphism i∗ : H ∗(Γ ) → H ∗(Γ ′).

Proposition 4.27 Given Γ = (V ,E) a complete graph on m vertices, we havec0 = c1 = cm−1 = 1 and ck = 0 for k ≥ m. Moreover, {ωi |0 ≤ i ≤ m − 1} forms abasis of the free module H ∗

T(Γ ) over C[x, y].

Proof Assume that φ(vi) = (xi, yi) for 1 ≤ i ≤ m. First, we observe that a trans-lation of φ to φ′ = φ + (p, q), where (p, q) ∈ R

2, will not affect the graph co-homology at all. The classes {ωi |0 ≤ i ≤ m − 1} form a module basis preciselywhen the classes {(ω + γ )i |0 ≤ i ≤ m − 1} form a module basis, where γ =(px + qy,px + qy, . . . ,px + qy). So, without loss of generality, we may assumethat xi �= 0 for all i and yi

xi�= yj

xjfor all i �= j .

We prove the statement by induction on m. When m = 1, the statement clearlyholds.

Now assume that we have proved that for m = k and Γ a complete graph on m

vertices, the classes {ωi |0 ≤ i ≤ m − 1} form a module basis of H ∗T(Γ ). We consider

the case m = k + 1.Denote by Γ ′ the complete graph on v1, v2, . . . , vk . This is a subgraph of Γ , and

we denote by i the inclusion map. Now ω is the equivariant symplectic form of Γ , andwe notice that i∗(ω) is the equivariant symplectic form of Γ ′. For any 0 ≤ i ≤ k − 1,by our induction hypothesis, we have i∗(ωi) �= 0 ∈ Hi(Γ ′), so ωi �= 0 ∈ Hi(Γ ), andhence ci(Γ ) ≥ 1.

Then the relationsk∑

i=0

ci(Γ ) = k + 1

andk∑

i=0

ici(Γ ) = |E| = k(k + 1)

2

force c0(Γ ) = c1(Γ ) = · · · = ck(Γ ) = 1 and ci(Γ ) = 0 for i ≥ k + 1.Now to complete the induction step, we only need to show that ωk �= 0 ∈ Hk(Γ ).

Assume that this is not the case. Then there must exist homogeneous polynomialsfi(x, y) of degree i for 1 ≤ i ≤ k such that

ωk = fk + fk−1ω + · · · + f1ωk−1 ∈ Hk

T(Γ ), (4.28)

where we have identified the polynomial fi with (fi(x, y), fi(x, y), . . . , fi(x, y)) ∈Hi

T(Γ ).At each vertex vi for 1 ≤ i ≤ k + 1, since φ(vi) = (xi, yi), we have ω|vi

= xix +yiy. It follows from Eq. (4.28) that

(xix + yiy)k ≡ fk(x, y) + fk−1(x, y)(xix + yiy) + · · · + f1(x, y)(xix + yiy)k−1,

(4.29)where ≡ means that the two sides are equal as polynomials in x and y.

62 J Algebr Comb (2014) 40:45–74

So (xix + yiy)|fk(x, y) for 1 ≤ i ≤ k + 1. Because yi

xi�= yj

xjfor all i �= j , we have∏k+1

i=1 (xix + yiy)|fk(x, y). This is only possible when fk(x, y) ≡ 0. Then Eq. (4.29)reduces to

(xix +yiy)k−1 ≡ fk−1(x, y)+fk−2(x, y)(xix +yiy)+· · ·+f1(x, y)(xix +yiy)k−2

for 1 ≤ i ≤ k + 1. In other words,

ωk−1 = fk−1 + fk−2ω + · · · + f1ωk−2 ∈ Hk−1

T(Γ ).

This contradicts the fact that ωk−1 �= 0 ∈ Hk−1(Γ ). �

We conclude this section by proving an analogue of the Künneth formula.

Definition 4.30 Given Γ1 = (V1,E1) and Γ2 = (V2,E2), assume that V1 = {v1,

v2, . . . , vm} and V2 = {u1, u2, . . . , un}. The (Cartesian) product graph Γ1�Γ2 =(V3,E3) is defined by V3 = V1 × V2, and two vertices (vi, us) and (vj , ut ) in V3

are adjacent if and only if

vi = vj ∈ V1, us and ut are adjacent in Γ2,

or

us = ut ∈ V2, vi and vj are adjacent in Γ1.

Proposition 4.31 Given Γ1 = (V1,E1) and Γ2 = (V2,E2), assume that V1 =(v1, . . . , vm), V2 = (u1, . . . , un), and that φ1 : V1 → R

2 and φ2 : V2 → R2 are two

moment maps in general position. Let Γ3 = Γ1�Γ2 = (V3,E3) and assume thatφ3 : V3 = V1 ×V2 → R

2 defined by φ3(vi, us) = aφ1(vi)+bφ2(us) is also in generalposition, where a, b ∈ R are both nonzero constants. Then we have an isomorphismof C[x, y]-algebras:

p : H ∗T(Γ1) ⊗C[x,y] H ∗

T(Γ2) −→ H ∗

T(Γ3)

(f1, f2, . . . , fm) ⊗ (g1, g2, . . . , gn) (4.32)

�→ (f1g1, f1g2, . . . , f1gn,f2g1, . . . , fmgn),

where the vertices in V3 are ordered lexicographically as

(v1, u1), (v1, u2), . . . , (v1, un), (v2, u1), . . . , (vm,un).

This immediately implies

ck(Γ3) =k∑

i=0

ci(Γ1)ck−i (Γ2). (4.33)

J Algebr Comb (2014) 40:45–74 63

Proof It is straightforward to verify that the map p defined by (4.32) is an algebrahomomorphism. We are going to prove that it is an isomorphism by induction on|E1| + |E2|. If |E1| = 0 or |E2| = 0, the conclusion obviously holds. Now we as-sume that |E1| > 0, |E2| > 0 and p defined by (4.32) is an isomorphism for smaller|E1| + |E2|.

Suppose that vi and vj are connected by an edge in E1, which we will denoteby e

vjvi

. Suppose that us and ut are connected by an edge in E2, which we will de-note by e

utus

. Assume that αk : Ek → C[x, y]1 is induced by φk for k = 1,2,3. Denoteα1(e

vjvi

) by f and α2(eutus

) by g. We first observe that f and g cannot be multiples ofeach other; otherwise, the three vertices (vi, us), (vj , us), (vj , ut ) in V3 would be onthe same line, contradicting the assumption that φ3 is in general position. We knowthat both H ∗

T(Γ1) ⊗C[x,y] H ∗

T(Γ2) and H ∗

T(Γ3) are free modules over C[x, y] of di-

mension mn, and we now fix bases for both. Then the map p can be represented bya matrix M(p), and p is an isomorphism if and only if M(p) is invertible as a ma-trix with coefficients in C[x, y]. This is equivalent to det(M(p)) being invertible inC[x, y].

Define E′1 = E1\{evj

vi} and Γ ′

1 = (V1,E′1). We may define

p′ : H ∗T

(Γ ′

1

) ⊗C[x,y] H ∗T(Γ2) → H ∗

T

(Γ ′

1�Γ2)

just as we have defined p. It follows from the induction hypothesis that p′ is an iso-morphism. For any C[x, y]-module N , we will use Nf to denote N ⊗C[x,y] C[x, y]f ,the localization at the polynomial f . It follows from the definition of graph cohomol-

ogy that for f = α1(evjvi

),

H ∗T(Γ1)f = H ∗

T

(Γ ′

1

)f

and

H ∗T(Γ1�Γ2)f = H ∗

T

(Γ ′

1�Γ2)f.

Then the two maps

pf : H ∗T(Γ1)f ⊗C[x,y]f H ∗

T(Γ2)f → H ∗

T(Γ1�Γ2)f

and

p′f : H ∗

T

(Γ ′

1

)f

⊗C[x,y]f H ∗T(Γ2)f → H ∗

T

(Γ ′

1�Γ2)f

can be identified. Because p′ is an isomorphism, so are p′f and pf . So det(M(p)) is

invertible in C[x, y]f . By the same argument we know that det(M(p)) is also invert-ible in C[x, y]g . Since f and g are not multiples of each other, this is only possiblewhen det(M(p)) ∈C

∗. So p is an isomorphism. This completes the induction step.Equation (4.33) now follows by counting dimensions. �

5 Connectivity properties of GKM graphs

In this section we prove two theorems about the connectivity of a regular d-valentgraph with cd = 1. As corollaries, we deduce connectivity properties of a GKMgraph.

64 J Algebr Comb (2014) 40:45–74

We first recall the standard graph theory notations of edge and vertex connectivity.

Definition 5.1 A graph Γ = (V ,E) is k-edge-connected for some k ∈ N if for anysubset F = {ei1j1, ei2j2, . . . , eit jt } ⊆ E with t < k, the subgraph Γ ′ = (V ,E\F) isconnected.

Definition 5.2 A graph Γ = (V ,E) is k-vertex-connected for some k ∈ N if for anysubset U = {vi1, vi2, . . . , vit } ⊆ V with t < k, the subgraph Γ − U = (V \U,E\EU)

is connected.

Now we are ready to state the main theorems in this section.

Theorem 5.3 Given Γ = (V ,E) a connected regular graph of degree d , ifcd(Γ ) = 1, then Γ is d-edge-connected.

Corollary 5.4 If a 2d-dimensional compact connected Hamiltonian GKM manifoldhas a moment map in general position, then its GKM graph is d-edge-connected.

Let � d2 � denote the least integer greater than or equal to d

2 .

Theorem 5.5 Given Γ = (V ,E) a connected regular graph of degree d , ifcd(Γ ) = 1, then Γ is (� d

2 � + 1)-vertex-connected.

Corollary 5.6 If a 2d-dimensional compact connected Hamiltonian GKM manifoldhas a moment map in general position, then its GKM graph is (� d

2 � + 1)-vertex-connected.

Remark 5.7 As we remarked earlier in Remark 4.23, cd = 1 is a weaker assumptionthan the existence of an axial function. So our theorems also apply to the graphsstudied by Guillemin and Zara [5].

Theorem 5.3 will play an important role in Sect. 6. We illustrate the use of it withan example.

Example 5.8 Figure 3 shows a regular graph of degree 3. According to Proposi-tion 4.22, c3 = 0 or 1. The graph will become disconnected upon removing edgese15 and e37, so the graph is not 3-edge-connected. By Theorem 5.3, we conclude thatc3 = 0.

Notation 5.9 Given Γ = (V ,E), we say a vector u = (u1, u2, . . . , u(k+1)m) ∈C

(k+1)m vanishes on vi , or u|vi= 0, if ui = um+i = · · · = ukm+i = 0. We say u

vanishes on U ⊆ V , or u|U = 0, if u vanishes on every point in U . We denote the setof vectors in C

(k+1)m that vanishes on V \U by WkU . We will use Wk

vifor Wk{vi } when

the set has only one point. It is easy to see that vkij ∈ Wk{vi ,vj }.

We denote by P kU : C(k+1)m → Wk

U the natural projection that sets all coordinatescorresponding to V \U to 0.

J Algebr Comb (2014) 40:45–74 65

Fig. 3 A 3-valent graph that isnot 3-edge-connected

For any U ⊆ V , we use K(U) to denote the edge set of the complete graph on U

and use KU = (U,K(U)) to denote the complete graph itself. Note that when U hasonly one element, K(U) = ∅, and KU is just a single vertex.

Lemma 5.10 Assume that Γ = (V ,E) and U ⊆ V is a nonempty subset. Then⟨vkij | eij ∈ E

⟩ ∩ WkU ⊆ ⟨

vkij | eij ∈ K(U)

⟩,

where 〈S〉 means the subspace of C(k+1)m spanned by S.

Proof Our vertex set is V = {v1, v2, . . . , vm}, and w.l.o.g. we assume that U = {v1,

v2, . . . , vn}. It is enough to show that⟨vkij | eij ∈ K(V )

⟩ ∩ WkU = ⟨

vkij | eij ∈ K(U)

⟩. (5.11)

It is clear that in the above equation, the RHS is contained in the LHS. It remains toshow the other containment. We divide the proof into three cases.

Case 1: n ≥ k + 1.

dim⟨vkij | eij ∈ K(V )

⟩ = rk(KV )

= (k + 1)m − ck(KV ) − 2ck−1(KV ) − · · · − (k + 1)c0(KV )

= (k + 1)m − (k + 1)(k + 2)

2. (5.12)

The first equality is the definition of rk , the second equality is by formula (4.10), andthe third equality is a consequence of Proposition 4.27.

It is immediate that

dimWkU = (k + 1)n. (5.13)

Now for any t ∈ N with n + 1 ≤ t ≤ m and any i ∈ N with 1 ≤ i ≤ k + 1 ≤ n, wehave vk

it ∈ Wk{vi ,vt } and P k{vt }(vkit ) ∈ 〈vk

it 〉 + Wkvi

. More explicitly, we have

P k{vt }(vkit

) = (0, . . . ,−1,0, . . . ,−ait ,0, . . . ,−ak

it ,0, . . . ,0),

where the only nonzero entries of the vector are the t , (m + t), . . ., (km + t)th en-tries, and they are −1, −ait , . . . , −ak

it , respectively. Then the elementary facts about

66 J Algebr Comb (2014) 40:45–74

Vandermonde matrices tell us that {P k{vt }(vkit ) | 1 ≤ i ≤ k + 1} forms a basis for Wk

vt.

So

Wkvt

⊆k+1∑i=1

(⟨vkit

⟩ + Wkvi

) ⊆ ⟨vkij | eij ∈ K(V )

⟩ + WkU

for n + 1 ≤ t ≤ m. So 〈vkij |eij ∈ K(V )〉 + Wk

U = WkV and

dim(⟨

vkij |eij ∈ K(V )

⟩ + WkU

) = m(k + 1). (5.14)

Now we put (5.12), (5.13), (5.14) together to get

dim(⟨

vkij |eij ∈ K(V )

⟩ ∩ WkU

)= dim

⟨vkij |eij ∈ K(V )

⟩ + dimWkU − dim

(⟨vkij |eij ∈ K(V )

⟩ + WkU

)= (k + 1)m − (k + 1)(k + 2)

2+ (k + 1)n − m(k + 1)

= (k + 1)n − (k + 1)(k + 2)

2

= dim⟨vkij |eij ∈ K(U)

⟩.

The last equality is obtained in exactly the same way as we obtained (5.12). Thus, weconclude that (5.11) holds in Case 1.

Case 2: k = 1, n = 1. We observe that each vector

u = (u1, u2, . . . , u2m) ∈ ⟨v1ij |eij ∈ K(V )

⟩must satisfy the linear equations

m∑i=1

ui = 0 and2m∑

i=m+1

ui = 0.

This forces the intersection of 〈v1ij |eij ∈ K(V )〉 with W 1

v1to be 0. So (5.11) holds

since K({v1}) = ∅.

Case 3: k ≥ 2, n < k + 1. For a fixed n ≥ 1, we prove (5.11) using induction on k.If n ≥ 2, we use k = n − 1 as the base case. If n = 1, we use k = 1 as the base case.Thus, these base cases have been verified in Case 1 and Case 2, respectively. So nowwe have k ≥ 2, k > n − 1, and we assume that (5.11) holds for smaller values of k.

Assume that u ∈ (〈vkij |eij ∈ K(V )〉 ∩ Wk

U), so we may write it as

u =∑

eij ∈K(V )

uij vkij ∈ Wk

U (5.15)

for some uij ∈C.

J Algebr Comb (2014) 40:45–74 67

For any vector w = (w1,w2, . . . ,w(k+1)m) ∈C(k+1)m, let

wini = (w1,w2, . . . ,wkm) ∈Ckm,

wend = (wm+1,wm+2, . . . ,w(k+1)m) ∈ Ckm, and

wmid = (wm+1,wm+2, . . . ,wkm) ∈C(k−1)m.

Then in particular we have

(vkij

)ini = vk−1ij ,

(vkij

)end = aij vk−1ij , and

(vkij

)mid = aij vk−2ij .

Then it follows from (5.15) that

uini =∑

eij ∈K(V )

uij vk−1ij ∈ Wk−1

U and

uend =∑

eij ∈K(V )

uij aij vk−1ij ∈ Wk−1

U .

Then by the induction hypothesis we have

uini =∑

eij ∈K(U)

xij vk−1ij and

uend =∑

eij ∈K(U)

yij vk−1ij

(5.16)

for some xij , yij ∈C. Hence,

umid =∑

eij ∈K(U)

xij aij vk−2ij =

∑eij ∈K(U)

yij vk−2ij . (5.17)

Since k > n − 1, we may apply Lemma 4.17 to KU to obtain sk−2(KU) = 0, whichmeans that the vectors {vk−2

ij |eij ∈ K(U)} are linearly independent. Thus, (5.17) im-plies that xij aij = yij . So the coefficients of the linear combinations in (5.16) are“compatible,” and we may write

u =∑

eij ∈K(U)

xij vkij .

This exactly implies (5.11), completing the induction. �We need the following technical lemma in the proof of Lemma 5.19.

Lemma 5.18 If 1 ≤ p ≤ n, 1 ≤ q ≤ n, then

p(p − 1)

2+ q(q − 1)

2+ n <

(p + q)(n + 1)

2.

68 J Algebr Comb (2014) 40:45–74

Proof

p(p − 1)

2+ q(q − 1)

2+ n − (p + q)(n + 1)

2

= 1

2

(p2 − (n + 2)p + q2 − (n + 2)q

) + n

= 1

2

((p − n + 2

2

)2

+(

q − n + 2

2

)2)−

(n + 2

2

)2

+ n

≤(

n

2

)2

−(

n + 2

2

)2

+ n

= −1 < 0. �

We now study the effect that disconnecting the graph by deleting edges has on thestatistic sk . We call the following lemma the Cut Lemma. It will be used to proveTheorem 5.3 and will also be used repeatedly in Sect. 6.

Lemma 5.19 Given a connected graph Γ = (V ,E) and a cut-set F = {ei1j1, ei2j2 ,

. . . , einjn} of size n, assume that (V ,E\F) = Γ1 � Γ2 is the disjoint union of twographs Γ1 and Γ2. Then

sk(Γ ) = sk(Γ1) + sk(Γ2)

for any k ≥ n − 1.

Proof Assume that Γ1 = (V1,E1) and Γ2 = (V2,E2). For any 1 ≤ t ≤ n, eithervit ∈ V1 and vjt ∈ V2, or vit ∈ V2 and vjt ∈ V1. Without loss of generality, we mayassume that vit ∈ V1 for all 1 ≤ t ≤ n. Let V3 = {vit |1 ≤ t ≤ n}, V4 = {vjt |1 ≤ t ≤ n}.Note that even though eisjs and eit jt are assumed to be distinct for s �= t , it couldstill happen that vis = vit or vjs = vjt . So we have |V3| ≤ n and |V4| ≤ n. DefineΓ ′ = (V ′,E′) by V ′ = V3 ∪ V4, E′ = K(V3) ∪ K(V4) ∪ F .

We claim that the graph Γ ′ has at least one vertex of degree less than n + 1. Oth-erwise, Γ ′ should have at least (|V3|+|V4|)(n+1)

2 edges. By Lemma 5.18, this is greater

than |V3|(|V3|+1)2 + |V4|(|V4|+1)

2 + n, which is the number of edges of Γ ′, a contradic-tion.

Now assume without loss of generality that vi1 is of degree less than n+ 1. DefineΓ ′′ = (V ′\{vi1},E′\Evi1

). Then we may apply the Deletion Lemma 4.14 to vi1 andΓ ′ to get sk(Γ

′) = sk(Γ′′).

By the same argument we can show that Γ ′′ also has at least one vertex of de-gree less than n + 1. Repeatedly applying the Deletion Lemma, we finally concludesk(Γ

′) = 0.Now assume that there is a linear relation among {vk

ij |eij ∈ E}:∑

eij ∈E

uij vkij = 0. (5.20)

J Algebr Comb (2014) 40:45–74 69

We may split the left-hand side and rewrite it as∑eij ∈E1

uij vkij +

∑eij ∈E2

uij vkij +

∑eij ∈F

uij vkij = 0. (5.21)

As both∑

eij ∈E2uij vk

ij and∑

eij ∈F uij vkij vanish on V1\V3, so does

∑eij ∈E1

uij vkij .

Then we may apply Lemma 5.10 to Γ1 and∑

eij ∈E1uij vk

ij to get

∑eij ∈E1

uij vkij =

∑eij ∈K(V3)

mij vkij

for some mij ∈ C. Similarly, we have∑eij ∈E2

uij vkij =

∑eij ∈K(V4)

nij vkij

for some nij ∈C. So (5.21) becomes∑eij ∈K(V3)

mij vkij +

∑eij ∈K(V4)

nij vkij +

∑eij ∈F

uij vkij = 0.

Because sk(Γ′) = 0, the above equation forces the coefficients mij = 0 for each

eij ∈ K(V3), nij = 0 for each eij ∈ K(V4), and uij = 0 for each eij ∈ F . So (5.21),hence (5.20), is in fact the sum of two linear relations:∑

eij ∈E1

uij vkij = 0, and

∑eij ∈E2

uij vkij = 0.

So sk(Γ ) = sk(Γ1) + sk(Γ2). �

Now we are ready to prove Theorem 5.3.

Proof of Theorem 5.3 We prove this by contradiction. Assume to the contrary thatΓ is not d-edge-connected. Then there exists a cut-set F = {ei1j1 , ei2j2, . . . , einjn} ofsize n < d .

Assume that (V ,E\F) = Γ1 � Γ2. Then, by Lemma 5.19, sd−2(Γ ) = sd−2(Γ1) +sd−2(Γ2). Repeated application of the Deletion Lemma to Γ1 will yield sd−2(Γ1) = 0and similarly sd−2(Γ2) = 0. Therefore, sd−2(Γ ) = 0.

We always have sd−1(Γ ) = sd(Γ ) = 0 by Lemma 4.17. So cd = sd + sd−2 −2sd−1 = 0. This contradicts our assumption that cd = 1. �

We prove Theorem 5.5 in a similar fashion.

Proof of Theorem 5.5 We prove this by contradiction. Assume to the contrary thatΓ is not (� d

2 � + 1)-vertex connected, so that there exists a smallest n ≤ � d2 � such

70 J Algebr Comb (2014) 40:45–74

that there are n vertices, which we may assume without loss of generality to beU = {v1, v2, . . . , vn} such that the graph Γ − U = (V \U,E\EU) is disconnected.

By the minimality of n we know that Γ − U consists of exactly two connectedcomponents, denoted by Γ1 = (V1,E1) and Γ2 = (V2,E2). For each vertex vi in U ,the edge set Evi

decomposes into three parts:

Evi= E1

vi� E2

vi� EU

vi,

where E1vi

, E2vi

, and EUvi

are the sets of edges connecting vi to V1, V2, and U , re-spectively. If E1

vi= ∅, then removing U\{vi} already disconnects Γ , contradicting

the minimality of n. So E1vi

�= ∅. For same reason, E2vi

�= ∅.Let E3 = ⋃

vi∈U E1vi

, E4 = ⋃vi∈U E2

vi, and E5 = ⋃

vi∈U EUvi

= E ∩ K(U). A lin-

ear relation among {vd−2ij |eij ∈ E}

∑eij ∈E

uij vd−2ij = 0 (5.22)

may be rewritten as∑eij ∈E1∪E3

uij vd−2ij +

∑eij ∈E2∪E4

uij vd−2ij +

∑eij ∈E5

uij vd−2ij = 0. (5.23)

Since |E1v1

| + |E2v1

| + |EUv1

| = d , either |E1v1

| ≤ d2 or |E2

v1| ≤ d

2 . We may as-

sume without loss of generality that |E1v1

| ≤ d2 . Since both

∑eij ∈E2∪E4

uij vd−2ij and∑

eij ∈E5uij vd−2

ij vanish on V1, it follows from (5.23) that∑

eij ∈E1∪E3uij vd−2

ij alsovanishes on V1. Define Γ3 = (V1 ∪U,E1 ∪E3 ∪K(U)). Then applying Lemma 5.10to Γ3 and

∑eij ∈E1∪E3

uij vd−2ij , we have

∑eij ∈E1∪E3

uij vd−2ij =

∑eij ∈K(U)

xij vd−2ij (5.24)

for some xij ∈C.

Claim sd−2(Γ3) = 0.

Proof of the claim The degree of v1 inside Γ3 is ≤ d2 +� d

2 �−1 ≤ d2 + d

2 + 12 −1 < d .

So we may apply the Deletion Lemma to Γ3 and v1 to remove v1 from Γ3 withoutchanging sd−2. If there is a vertex in the new graph of degree less than d , we mayrepeat this to form another new graph. We continue this process as long as there isa vertex in the new graph of degree less than d . This process will only stop whenwe are left with a subgraph of Γ3 whose vertices are all of degree greater than orequal to d . We will show that the only such subgraph is the empty graph. Assumeto the contrary that there is a nonempty subgraph Γ6 = (V6,E6) with V6 ⊆ V1 ∪ U ,E6 ⊆ E1 ∪ E3 ∪ K(U), such that all vertices of Γ6 are of degree ≥ d . First, weshow that V6 ∩ V1 = ∅. If not, assume that vt ∈ V6 ∩ V1. Since E1

v1�= ∅, we may

J Algebr Comb (2014) 40:45–74 71

assume that e1s ∈ E1v1

, and then vs ∈ V1. As Γ1 is connected, there is a path fromvt to vs in Γ1, which we denote by eti1 = ei0i1, ei1i2, . . . , eip−1ip , eipip+1 = eips . Letu = max{j |0 ≤ j ≤ p + 1, vj ∈ V6}. Then λΓ6(vu) < λΓ (vu) = d , contradicting ourassumption on Γ6. So V6 ⊆ U . But |U | ≤ � d

2 �, contradicting the assumption thatevery vertex in Γ6 has degree greater than or equal to d .

Therefore, sd−2(Γ3) = 0, and we have proved the claim.

The above claim and Eq. (5.24) now force uij = 0 for eij ∈ E1 ∪ E3. Then itfollows from (5.23) that ∑

eij ∈E2∪E4

uij vd−2ij +

∑eij ∈E5

uij vd−2ij = 0. (5.25)

Define Γ4 = (V2 ∪ U,E2 ∪ E4 ∪ E5). Applying the Deletion Lemma to Γ4 givessd−2(Γ4) = 0. So (5.25) forces uij = 0 for all eij ∈ E2 ∪ E4 ∪ E5.

So (5.23), and hence (5.22) is a trivial linear relation, which means sd−2(Γ ) = 0.So cd(Γ ) = sd(Γ ) + sd−2(Γ ) − 2sd−1(Γ ) = 0, a contradiction. �

6 An upper bound for the second Betti number of compact Hamiltonian GKMmanifolds

We now turn to the geometric consequences.

Theorem 6.1 Given Γ = (V ,E) a connected regular graph of degree d ≥ 2, ifcd(Γ ) = 1 and m = |V |, then cd−1(Γ ) ≤ m−2

d−1 . As a consequence, suppose thatM is a 2d-dimensional connected compact Hamiltonian GKM manifold whose mo-ment map is in general position. Let m be the number of vertices of its GKMgraph, which is equal to the sum of all the Betti numbers of M . Then we haveβ2(M) = β2d−2(M) ≤ m−2

d−1 , where βi(M) is the ith geometric Betti number of M .

Corollary 6.2 If M is an 8- or 10-dimensional connected compact HamiltonianGKM manifold whose moment map is in general position, then M has nondecreasingeven Betti numbers up to half dimension: β0(M) ≤ β2(M) ≤ β4(M).

Proof of Corollary This is a straightforward calculation using Theorem 6.1 and thePoincaré duality.

In the case of an 8-dimensional manifold, it follows from β2(M) ≤ m−23 that

β4(M) = m − 2 − 2β2(M) ≥ m − 2 − 2(m−2)3 = m−2

3 ≥ β2(M).In the case of a 10-dimensional manifold, it follows from β2(M) ≤ m−2

4 thatβ4(M) = 1

2 (m − 2 − 2β2(M)) ≥ m−24 ≥ β2(M).

In both cases, we have β0(M) = 1 ≤ β2(M) since the symplectic form representsa nontrivial cohomology class in H 2(M). �

Remark 6.3 Corollary 6.2 gives an affirmative answer to Question 1.8 in the caseof 8- and 10-dimensional Hamiltonian GKM manifolds whose moment maps are

72 J Algebr Comb (2014) 40:45–74

in general position. These are the first two nontrivial cases, as the inequalities areautomatic for 2-, 4-, and 6-dimensional manifolds.

Definition 6.4 A graph Γ = (V ,E) is called k-trimmed for some positive integer k

if

• each vertex of Γ is of degree at least k + 1,• each connected component of Γ is (k + 1)-edge-connected.

Lemma 6.5 Every graph Γ = (V ,E) has a unique maximal k-trimmed subgraph,which we will denote by Γ k . It can be obtained by the following algorithm:

(1) If Γ is k-trimmed, stop.(2) If Γ contains a vertex vi of degree less than or equal to k, we define Γ1 =

(V1,E1) by V1 = V \{vi} and E1 = E\Evi, where Evi

is the set of edges inci-dent to vi .

(3) If every vertex of Γ is of degree k + 1 or higher and there exists E′ ={ei1j1, . . . , eit jt } ⊆ E such that t ≤ k and removing these edges would increasethe number of connected components of Γ , but removing any t − 1 among themwould not, then we define Γ1 = (V1,E1) by V1 = V and E1 = E\E′.

(4) Repeat the algorithm on Γ1.

This process will finally stop. The resulting graph, which can be empty, is Γ k .

Proof The union of two k-trimmed subgraphs of Γ is also k-trimmed, so there is aunique maximal k-trimmed subgraph.

If Γ is itself k-trimmed, the lemma is trivial. Otherwise, following the algorithm,we get a sequence of graphs Γ1,Γ2, . . . ,Γp , where Γi+1 is a subgraph of Γi obtainedfrom Γi either as in Case (2) or as in Case (3) of the algorithm, and Γp is k-trimmed.This sequence is not necessarily unique, but we will show that in any case Γp = Γ k .

Assume that Γ k = (V k, Ek). First, Γp is a subgraph of Γ k since Γ k is the uniquemaximal k-trimmed subgraph of Γ . Secondly, if Γ is as in Case (2), then vi /∈ V k ,so Γ k is a subgraph of Γ1. If Γ is as in Case (3), then E′ ∩ Ek = ∅, so Γ k is also asubgraph of Γ1. Then we can show inductively that Γ k is a subgraph of Γp . So wemust have Γp = Γ k . �

Corollary 6.6 For a graph Γ and its unique maximal k-trimmed subgraph Γ k , wehave

sk−1(Γ k

) = sk−1(Γ ).

Proof By examining the algorithm in Lemma 6.5 we see that sk−1 remains un-changed in each step, either by the Deletion Lemma 4.14 or the Cut Lemma 5.19. �

We make the following definition so that we can make the statements and proofsin the rest of the section more concise.

Definition 6.7 For any d ≥ 2, we say that a graph Γ = (V ,E) is of type Td if

J Algebr Comb (2014) 40:45–74 73

• each vertex of Γ is of degree d or d − 1;• each connected component of Γ has at least one vertex of degree d − 1;• each connected component of Γ is (d − 1)-edge-connected.

Proposition 6.8 Assume that Γ = (V ,E) is a graph of type Td . Then

sd−3(Γ ) ≤ nd(Γ )

d − 1+ π0(Γ ), (6.9)

where nd(Γ ) = {vi ∈ V |λ(vi) = d} is the number of vertices of degree d , and π0(Γ )

is the number of connected components of Γ .

Proof We use induction on the size of |V |. The graph of type Td with the fewestvertices is the empty graph, and (6.9) holds as 0 ≤ 0 in this case.

Now we consider Γ = (V ,E) with |V | = m > 0 and assume that (6.9) holds forany graph of type Td and |V | < m. If π0(Γ ) > 1, then each connected component ofΓ is still of type Td and has fewer vertices. So the induction hypothesis implies that(6.9) holds for each connected component. We may add them up to show that (6.9)holds for Γ as well. Now we assume that Γ is connected.

Since Γ is of type Td , we may pick a vertex vt of degree d − 1. Define Γ ′ =(V ′,E′) by V ′ = V \{vt } and E′ = E\Evt , where Evt is the set of edges incident tovt . By Corollary 4.16 we have sd−3(Γ ) ≤ sd−3(Γ

′) + 1. Let Γ ′d−2 be the maximal(d − 2)-trimmed subgraph of Γ ′. Then we have sd−3(Γ

′d−2) = sd−3(Γ′). Assume

that π0(Γ′d−2) = p, Γ ′d−2 = ⊔p

i=1 Γi , and Γi = (Vi,Ei).Since Γ is (d − 1)-edge-connected, for each Γi , there must be at least d − 1

vertices in Vi whose degrees in Γ were d but now have degree d − 1 in Γi . So

p∑i=1

nd(Γi) ≤ nd(Γ ) − (d − 1)p.

Since Γ ′d−2 is also of type Td and with fewer vertices than Γ , by the inductionhypothesis we have

sd−3(Γ ′d−2) ≤ nd(Γ ′d−2)

d − 1+ π0

(Γ ′d−2).

So

sd−3(Γ ) ≤ sd−3(Γ ′d−2) + 1

≤ nd(Γ ′d−2)

d − 1+ π0

(Γ ′d−2) + 1

= 1

d − 1

p∑i=1

nd(Γi) + p + 1

≤ 1

d − 1

(nd(Γ ) − (d − 1)p

) + p + 1

74 J Algebr Comb (2014) 40:45–74

= nd(Γ )

d − 1+ 1

= nd(Γ )

d − 1+ π0(Γ ).

This completes the induction step. �

Now Theorem 6.1 follows easily.

Proof of Theorem 6.1 Since cd(Γ ) = 1, by Theorem 5.3 we know that Γ is d-edge-connected. Pick any edge eij ∈ E and form a new graph Γ ′ = (V ,E\{eij }). Then Γ ′is of type Td . By Proposition 6.8 we have

sd−3(Γ ′) ≤ m − 2

d − 1+ 1.

So sd−3(Γ ) ≤ sd−3(Γ′) + 1 ≤ m−2

d−1 + 2. Hence,

cd−1(Γ ) = sd−1(Γ ) + sd−3(Γ ) − 2sd−2(Γ ) ≤ 0 + m − 2

d − 1+ 2 − 2 = m − 2

d − 1,

where we have used the facts that sd−1(Γ ) = 0 by the Deletion Lemma and thatsd−2(Γ ) = cd(Γ ) − (sd(Γ ) − 2sd−1(Γ )) = 1. �

Acknowledgements I would like to thank Tara Holm, Bob Connelly, Victor Guillemin, Allen Knutson,Tomoo Matsumura, Edward Swartz, and Catalin Zara for many helpful discussions. Also, I would like tothank the referee for helpful feedback and suggestions.

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