an explicit, characteristic-free, equivariant homology equivalence...

AN EXPLICIT, CHARACTERISTIC-FREE, EQUIVARIANT

HOMOLOGY EQUIVALENCE BETWEEN KOSZUL COMPLEXES

Andrew R. Kustin

Abstract. Let E and G be free modules of rank e and g, respectively, over a

commutative noetherian ring R. The identity map on E∗ ⊗ G induces the Koszulcomplex

→ SymmE∗⊗SymnG⊗Vp(E∗⊗G) → Symm+1 E

∗⊗Symn+1G⊗Vp−1(E∗⊗G) →

and its dual

· · · → Dm+1E ⊗Dn+1G∗ ⊗

Vp−1(E ⊗G∗) → DmE ⊗DnG∗ ⊗

Vp(E ⊗G∗) → . . . .

Let HN (m,n, p) be the homology of the top complex at Symm E∗ ⊗ SymnG ⊗Vp(E∗ ⊗ G) and HM(m,n, p) the homology of the bottom complex at DmE ⊗DnG

∗ ⊗Vp(E ⊗ G∗). It is known that HN (m,n, p) ∼= HM(m′, n′, p′), provided

m+m′ = g − 1, n+ n′ = e− 1, p+ p′ = (e− 1)(g − 1), and 1 − e ≤ m− n ≤ g − 1.

In this paper we exhibit a complex Y and explicit quasi-isomorphisms from Y to twocomplexes, as described above, for the appropriate choice of parameters, which give

rise to this isomorphism. Our quasi-isomorphisms may be formed over the ring of

integers; they can be passed to an arbitrary ring or field by base change. All of ourwork is equivariant under the action of the group GL(E)×GL(G); that is, everything

we do is independent of the choice of basis.Knowledge of the homology of the top complex is equivalent to knowledge of the

modules in the resolution of the Segre module Segre(e, g, ℓ), for ℓ = m − n. The

modules {Segre(e, g, ℓ)|ℓ ∈ Z} are a set of representatives of the divisor class group ofthe determinantal ring defined by the 2×2 minors of an e×g matrix of indeterminates.

If R is the ring of integers, then the homology HN (m,n, p) is not always a free abelian

group. In other words, if R is a field, then the dimension of HN (m,n, p) dependson the characteristic of R. The module HN (m,n, p) is known when R is a field of

characteristic zero; however, this module is not yet known over arbitrary fields.The modules in the minimal resolution of the universal ring for finite length

modules of projective dimension two are equal to modules of the form HN (m,n, p).

2000 Mathematics Subject Classification. 13D25, 16E05, 18G35.Key words and phrases. Chessboard complex, Determinantal ring, Divisor class group, Equi-

variant quasi-isomorphism, Finite free resolution, Koszul complex, Segre ring, Universal resolution.

1

2 ANDREW R. KUSTIN

Introduction.

Let G be a free module rank g over the commutative noetherian ring R. Foreach integer n, the complex

Cn : · · · → Symn−2G⊗∧2G→ Symn−1G⊗

∧1G→ SymnG⊗

∧0G→ 0

of free R-modules is well understood: the complex C0 has homology H0(C0) = Rand, for any non-zero integer n, the complex Cn is split exact. Indeed, Cn isone homogeneous strand of a Koszul resolution. Let x1, . . . , xg be a basis for G,then

⊕n Cn is the minimal homogeneous resolution of R by free modules over the

polynomial ring R[x1, . . . xg].The situation is significantly different if two free modules E∗ and G are used. In

this case, the complexes look like

(0.1) · · · → Symm E∗⊗SymnG⊗Vp(E∗⊗G) → Symm+1 E

∗⊗Symn+1G⊗Vp−1(E∗⊗G) → . . .

and if V ∈ SymmE∗, X ∈ SymnG, vi ∈ E∗ and xi ∈ G, for 1 ≤ i ≤ p, then the

differential

SymmE∗ ⊗ SymnG⊗

∧p(E∗ ⊗G)→ Symm+1E

∗ ⊗ Symn+1G⊗∧p−1

(E∗ ⊗G)

sends

(0.2) V ⊗X ⊗ (v1 ⊗ x1) ∧ (v2 ⊗ x2) ∧ · · · ∧ (vp ⊗ xp)

top∑

i=1

(−1)i+1viV ⊗ xiX ⊗ (v1 ⊗ x1) ∧ · · · ∧ (vi ⊗ xi) ∧ · · · ∧ (vp ⊗ xp).

Let HN (m,n, p) equal the homology of (0.1) at SymmE∗⊗SymnG⊗

∧p(E∗⊗G).

Many of the complexes (0.1) have homology. Sometimes the homology of (0.1)occurs someplace in the middle of (0.1) rather than only at one of the endpoints.Some of the complexes (0.1) have homology in more than one position. The homol-ogy of (0.1) is not always free as an R-module. If R is a field, then the dimensionof the homology of (0.1) depends on the characteristic of R.

We listed 5 differences between the complexes Cn and the complexes (0.1). Thefirst three differences are not particularly surprising; however, the last two arestunning. These results were first established by Hashimoto [Ha]. Let S be theR-algebra Sym• E

∗ ⊗ Sym•G. If we fix bases v1, . . . , ve for E∗, and x1, . . . , xg forG, then one may think of S as the polynomial ring S = R[v1, . . . , ve, x1, . . . , xg].Let T be the subring

T =∑

m

SymmE∗ ⊗ SymmG

AN EXPLICIT HOMOLOGY EQUIVALENCE 3

of S. One may think of T as the subring R[vixj ] of S. Let P be the R-algebraSym•(E

∗⊗G). One may think of P as a polynomial ring over R in the eg indeter-minates {vi⊗xj}. It is convenient to let zij represent the element vi⊗xj of P. Theidentity map on E∗ ⊗G induces a surjective map ϕ : P → T . Let Z be the e × gmatrix whose entry in row i column j is the indeterminate zij . The kernel of ϕ isthe ideal I2(Z) generated by the 2× 2 minors of Z; and therefore, T is isomorphicto the determinantal ring P/I2(Z). Significant information about the homology

of (0.1), in the case that a = b, is contained in the graded module TorP•,•(T,R),where R is the graded P-module P/P+ concentrated in degree zero. Hashimotoproved that if R is equal to the ring of integers, and e and g are both at least five,then TorP3,5(T,R) is not a free R-module. On the other hand, the Koszul complex

P ⊗R∧•

(E∗ ⊗G) is a homogeneous resolution of the P-module R. It follows that

TorPp,n+p(T,R) = HN (n, n, p).

The assertion about the dependence of the dimension of the homology of (0.1) onthe characteristic of the base field follows immediately; see Roberts [R].

There is a determinantal interpretation of the complexes (0.1), even when m 6= n.For each integer ℓ, let Mℓ be the T -submodule

(0.3) Mℓ =∑

m−n=ℓ

SymmE∗ ⊗ SymnG

of S. View Mℓ as a graded T -module by giving Symn+ℓE∗⊗SymnG grade n. The

same reasoning we used before shows that

(0.4) TorPp,n+p(Mm−n, R) = HN (m,n, p).

The modules Mℓ arise in numerous situations. For example, take R = Z. Thedivisor class group of T is known to be Z and [BG] shows why ℓ 7→ [Mℓ] is anisomorphism from Z → Cℓ (T ). This numbering satisfies M0 = T , Mg−e is equalto the canonical class of T , and Mℓ is a Cohen-Macaulay T -module if and only if1− e ≤ ℓ ≤ g − 1. Furthermore, the modules {Mℓ | 1− e ≤ ℓ ≤ g − 1} satisfy

(0.5) Mg−e−ℓ∼= ExtαP(Mℓ,P);

one argument may be found on page 473 of [K05].Reiner and Roberts [RR] refer to T as the Segre ring Segre(e, g, 0) and they call

Mℓ the Segre module Segre(e, g, ℓ).If R =KKK is a field of characteristic zero, then the geometric method of Lascoux

[L] gives the modules in the resolution of T by free P-modules and hence dimension

of TorPp,q(T,KKK) for all p and q. The same technique can be used to resolve each

4 ANDREW R. KUSTIN

Mℓ over P; see [W, Section 6.5]. A very pretty description of TorPp,q(Mℓ,KKK), whenKKK is a field of characteristic zero, is recorded in [RR]. The complete description of

TorPp,q(Mℓ,KKK) for fields of arbitrary characteristic is not yet known.

The ring P = Sym•(E∗⊗G) has a natural Ne×Ng-grading; (once one picks bases

for E∗ and G). Each module Mℓ and each homology module TorPp (Mℓ,KKK) inherits

this grading. It is shown in [BH,RR] that, for each (γ, δ) ∈ Ne × Ng, the (γ, δ)th

graded piece of TorPp (Mℓ,KKK) is isomorphic to the p − 1 reduced homology of thechessboard simplicial complex with multiplicities ∆γ,δ. The vertex set of ∆γ,δ is theset of squares on an e× g chessboard; the simplices are the sets of squares havingat most γi squares from row i and at most δj squares from column j for all i and j.The complexes ∆γ,δ were introduced by Bruns and Herzog [BH]; these complexesgeneralize the original chessboard complexes ∆e,g of [BLVZ]. The complex ∆e,g

is equal to ∆γ,δ with (γ, δ) = ((1, . . . , 1), (1, . . . , 1)). It is shown in [BLVZ] that

H2(∆5,5,Z) has 3-torsion; thereby providing an independent proof of Hashimoto’s

Theorem. Bjorner, Lovasz, Vrecica, and Zivaljevic make conjectures about thehigher order connectivity of (or the higher order homotopy groups) of chessboardcomplexes. Friedman and Hanlon [FH] reformulate the conjectures of [BLVZ] intostatements about simplicial homology and they use characteristic zero techniques(Hodge Theory and eigenvalues of the Laplacian) to verify the conjectures in certaincases.

The matching complex of a graph Γ is the abstract simplicial complex whosevertex set is the set of edges of Γ and whose faces are sets of edges of Γ with no twoedges meeting at a vertex. If Γ is a complete bipartite graph, then the matchingcomplex of Γ is a chessboard complex. Karaguezian, Reiner, and Wachs [KRW]use representation theory over fields of characteristic zero to unite and extend allof the previously mentioned results.

Almost all of the results listed above require a field of characteristic zero. Wenotice; however, that [RR], where the homology of (0.1) is explicitly recorded incharacteristic zero, depends very heavily on duality. Duality continues to hold overZ – see (0.5). Our goal in the present paper is to splice (0.1) together with theappropriate resolution of the dual of the partner of (0.1) in order to produce afamily of split exact complexes {C(P,Q)}. This idea is motivated, for example,by the family of Eagon-Northcott like complexes (see, for example, [E, TheoremA2.10]) or the family of complexes which resolves the divisor class group of a residualintersection of a grade three Gorenstein ideal [KU]. Our approach, unlike that in[RR, W, FH, KRW], works over Z. In other words, the same complex works inall characteristics. Our approach, like that in [RR, W, FH, KRW], is completelyindependent of the choice of bases.

The complex (0.1) and its dual play prominent roles in the resolution of a univer-sal ring. For any triple of parameters e, f , and g, subject to the obvious constraints,Hochster [Ho] established the existence of a commutative noetherian ring R and a


universal resolution

(0.6) U : 0→Re →Rf →Rg → 0,

such that for any commutative noetherian ring S and any resolution

V : 0→ Se → Sf → Sg → 0,

there exists a unique ring homomorphism R → S with V = U⊗R S. In [K07], wefound a free resolution F of the universal ring R over an integral polynomial ring,P, in the border case f = e + g. The resolution F is not minimal; indeed, if eand g are both at least 5; then Hashimoto’s Theorem can be used to prove that Rdoes not posses a generic minimal resolution over the ring of integers. Nonetheless,for each field KKK, we are able to use the resolution F to express the modules inthe minimal homogeneous resolution of R⊗Z KKK by free P⊗Z KKK modules in termsof the homology of (0.1). The homology of (0.1) is known when KKK is a field ofcharacteristic zero. With the answer in hand, but without appealing to [K07], thegeometric method of calculating syzygies was directly applied in [KW] to find themodules in the minimal resolution of R ⊗Z KKK, when KKK is a field of characteristiczero.

Let E and G be free modules of rank e and g, respectively, over a commutativenoetherian ring R. The identity map on E∗ ⊗G induces the Koszul complex (0.1)and its dual

(0.7) · · · → Dm+1E ⊗Dn+1G∗ ⊗

Vp−1(E ⊗G∗) → DmE ⊗DnG∗ ⊗

Vp(E ⊗G∗) → . . . .

Let HM(m,n, p) be the homology of (0.7) at DmE ⊗ DnG∗ ⊗∧p

(E ⊗ G∗). It isshown in [BG], implicitly, and [K05], explicitly, that

(0.8) HN (m,n, p) ∼= HM(m′, n′, p′),

provided m+m′ = g− 1, n+n′ = e− 1, p+ p′ = α, and 1− e ≤ m−n ≤ g− 1, forα = (g − 1)(e− 1). The key step in establishing (0.8) is (0.5). Fix P = m+ p andQ = n + p. In (1.8) we define complexes N and M in such a way that N is (0.1),with

Np = SymmE∗ ⊗ SymnG⊗

∧p(E∗ ⊗G),

and M is like (0.7), with

Mp = Dg−1−mE ⊗De−1−nG∗ ⊗

∧α−p(E ⊗G∗),

for all p. In section 2 we exhibit a complex Y and quasi-isomorphisms

(0.9) ϕ : Y→M and ψ : Y→ N⊗ L,

6 ANDREW R. KUSTIN

where L is the rank one free R module of (1.1) which keeps are calculations equi-variant. The existence of quasi-isomorphisms ϕ and ψ yields an alternate proof of(0.8). The mapping cone of ψ is called C(P,Q). In sections 4 and 5 we prove thatψ is a quasi-isomorphism by showing that C(P,Q) is split exact. We create Y tobe the total complex of a double complex whose row homology is M; and therefore,there is no difficulty showing that ϕ is a quasi-isomorphism.

In the boundary cases, m − n equals 1 − e or g − 1, there exists a direct quasi-isomorphism ψ′ : M→ N⊗L; see section 3. However, in general, there does not exista characteristic-free, equivariant, quasi-isomorphism ψ′ : M→ N⊗ L, see Example3.15; and therefore, (0.9) is the best possible result in the general case. In section6 we apply the quasi-isomorphisms (0.9) in order to exhibit a generator of

HN (g − 1, e− 1, α) ∼= HM(0, 0, 0) = R.

In section 7 we use the technique of (0.9) to deduce a homogeneous version of (0.8).We also show that the homology modules HM(m,n, p) and HN (m,n, p) satisfy anextra duality when e is equal to three.

1. Preliminary results.

Throughout this paper, R is a commutative noetherian ring with one, E and Gare fixed finitely generated free R-modules of rank e and g, respectively, and

α = (e− 1)(g − 1).

We allow g to be zero, but e must be positive. Let G be the group GL(E)×GL(G)and L be the rank one R-free G-module

(1.1) L =∧e

E∗ ⊗∧g

G⊗∧eg

(E ⊗G∗).

We observe that the (R,G) bi-module L is isomorphic to Tg−1(∧e

E)⊗Te−1(∧g

G∗);however, we find the formulation in (1.1) to be most convenient for our purposes.

We always use ∗ to represent the functor HomR( , R), ⊗ to represent ⊗R, andTnE to represent the n-fold tensor product E ⊗ · · · ⊗ E. We make much use ofthe exterior algebra

∧•E, the symmetric algebra Sym• E, and the divided power

algebra D•E. In particular,∧•

E and∧•

E∗ are modules over one another, and

Sym•E and D•E∗ are modules over one another. Indeed, if V ∈

∧iE∗, U ∈

∧jE,

V ′ ∈ Symi(E∗), and U ′ ∈ Dj(E), then

V (U) ∈∧j−i

E, U(V ) ∈∧i−j

E∗, V ′(U ′) ∈ Dj−i(E), and U ′(V ′) ∈ Symi−j(E∗).

(We view∧i

E, SymiE, and DiE to be meaningful for every integer i; in particular,these modules are zero whenever i is negative.) The exterior, symmetric, and


divided power algebras A all come equipped with co-multiplication ∆: A→ A⊗A.More information about these algebras from multilinear algebra may be found in[BE75, section 1] and [BE77, Appendix]. In particular, if u ∈ E, U ∈

∧pE, and

V ∈∧q

E∗, then

(1.2) [u(V )](U) = u ∧ (V (U)) + (−1)1+qV (u ∧ U).

Throughout the paper we use the following conventions when naming elements: u,v, x, and y represent elements of E, E∗, G, and G∗, respectively; and U , V , X ,and Y represent elements of A(E), A(E∗), A(G), and A(G∗), respectively, for Aequal to

∧•, Sym•, or D•. Also, Z and W represent elements of

∧•(E ⊗G∗) and∧•

(E∗ ⊗G), respectively.

Definition 1.3. Let m, n, and p be integers. Define modules

N (m,n, p) = SymmE∗ ⊗ SymnG⊗

∧p(E∗ ⊗G) and

M(m,n, p) = DmE ⊗DnG∗ ⊗

∧p(E ⊗G∗),

and homomorphisms

nnn : N (m,n, p)→ N (m+ 1, n+ 1, p− 1) and

mmm : M(m,n, p)→M(m− 1, n− 1, p+ 1).

The homomorphism nnn is the composition

N (m,n, p)1⊗1⊗∆−−−−−→ SymmE

∗ ⊗ SymnG⊗ E∗ ⊗G⊗

∧p−1(E∗ ⊗G)

ρ−→

SymmE∗ ⊗ E∗ ⊗ SymnG⊗G⊗

∧p−1(E∗ ⊗G)

µ⊗µ⊗1−−−−→ N (m+ 1, n+ 1, p− 1),

where ∆ is co-multiplication in the exterior algebra, ρ rearranges terms, and µ ismultiplication in the symmetric algebra. The homomorphism mmm is the composition

M(m,n, p)∆⊗∆⊗1−−−−−→ Dm−1E ⊗E ⊗Dn−1G

∗ ⊗G∗ ⊗∧p

(E ⊗G∗)ρ−→

Dm−1E ⊗Dn−1G∗ ⊗ (E ⊗G∗)⊗

∧p(E ⊗G∗)

1⊗1⊗µ−−−−→M(m− 1, n− 1, p+ 1),

where ∆ is co-multiplication in the divided power algebra, ρ rearranges terms, andµ is multiplication in the exterior algebra. In practice, the action of nnn is given in(0.2) and the action of mmm is described by

mmm(u(m) ⊗ y(n) ⊗ Z) = u(m−1) ⊗ v(n−1) ⊗ (u⊗ y) ∧ Z,

for u ∈ E, y ∈ G∗, and Z ∈∧

(E ⊗G∗).

8 ANDREW R. KUSTIN

Definition 1.4. Fix integers P and Q. Let N(P,Q) be the complex

0→ N (P − eg,Q− eg, eg)nnn−→ . . .

nnn−→ N (P − 1, Q− 1, 1)

nnn−→ N (P,Q, 0)→ 0,

and M(P,Q) be

0→M(P,Q, 0)mmm−→M(P − 1, Q− 1, 1)

mmm−→ . . .

mmm−→M(P − eg,Q− eg, eg)→ 0.

The modules N (P,Q, 0) and M(P,Q, 0) are in position zero in the complexesN(P,Q) and M(P,Q), respectively; so

N(P,Q)c = N (P − c, Q− c, c) and M(P,Q)−c =M(P − c, Q− c, c).

The perfect paring

(1.5) M(m,n, p)⊗N (m,n, p)→ R,

which is given by

(U ⊗ Y ⊗ Z)⊗ (V ⊗X ⊗W ) 7→ U(V ) · Y (X) · Z(W ),

ensures that

(1.6) M(P,Q) ∼= N(P,Q)∗

as complexes. In particular, if t ∈ N (m− 1, n− 1, p+ 1) and t′ = u(m) ⊗ y(n) ⊗ ZinM(m,n, p), then

(1.7) [nnn(t)](t′) = t(u(m−1) ⊗ y(n−1) ⊗ Z ∧ (u⊗ y)

).

Indeed, if t = V ⊗X ⊗ (v1 ⊗ x1) ∧ · · · ∧ (vp+1 ⊗ xp+1), then both sides of (1.7) areequal to

p+1X

i=1

(−1)i+1(viV )(u(m)) · (xiX)(y(n)) ·h

(v1 ⊗ x1) ∧ · · · ∧ (vi ⊗ xi) ∧ · · · ∧ (vp+1 ⊗ xp+1)i

(Z).

Remark. The present complex M(P,Q) is a shift of the complex M(P,Q) from [K05,Definition 2.4]. The isomorphism (1.6) justifies our present indexing scheme.

Fix integers P and Q. Let

(1.8.) N = N(P,Q) and M = M(e(g − 1)− P, (e− 1)g −Q)[−α]

For each integer p, let m = P − p and n = Q− p. Observe that

(1.9) Np = N (m,n, p) and Mp =M(g − 1−m, e− 1− n, α− p);

and therefore, according to (0.8), if 1−e ≤ P −Q ≤ g−1, then Hp(N) ∼= Hp(M) forall integers p. In this paper we establish an explicit homology equivalence betweenN⊗ L and M; that is, we exhibit a complex Y and quasi-isomorphisms (0.9).

We collect some of the notation and conventions that are used throughout thepaper.


Notation 1.10. (a) Let m be an integer. Each pair of elements (U, Y ), withU ∈ DmE and Y ∈

∧mG∗, gives rise to an element of

∧m(E ⊗ G∗), which we

denote by U ⊲⊳ Y . We now give the definition of U ⊲⊳ Y . Consider the composition

DmE ⊗ TmG∗ ∆⊗1−−−→ TmE ⊗ TmG

∗ ξ−→∧m

(E ⊗G∗),

where ξ((U1 ⊗ · · · ⊗ Um)⊗ (Y1 ⊗ · · · ⊗ Ym)

)= (U1 ⊗ Y1) ∧ · · · ∧ (Um ⊗ Ym), for

Ui ∈ E and Yi ∈ G∗. It is easy to see that the above composition factors throughDmE ⊗

∧mG∗. Let U ⊗ Y 7→ U ⊲⊳ Y be the resulting map from DmE ⊗

∧mG∗ to∧m

(E ⊗G∗). The map

∧mE ⊗DmG

∗ →∧m

(E ⊗G∗),

which sends U ⊗ Y to U ⊲⊳ Y , for U ∈∧m

E and Y ∈ DmG∗, is defined in a

completely analogous manner. Akin, Buchsbaum, and Weyman [ABW,III.2] write<U, Y >, where we write U ⊲⊳ Y .

(b) For each statement “S”, let

χ(S) =

{1, if S is true, and

0, if S is false.

In particular, χ(i = j) has the same value as the Kronecker delta δij .

(c) The symbol λ always represents an ordered tuple of positive integers. If λ isthe ℓ-tuple (λ1, . . . , λℓ), then we write |λ| = λ1 + · · ·+ λℓ,

∧λE∗ =

∧λ1 E∗ ⊗ · · · ⊗∧λℓ E∗, and M(λ,m, n, p) =

∧λE∗ ⊗M(m,n, p).

(d) If A and B are free modules, then the adjoint isomorphism says that Hom(A,B)and Hom(A⊗B∗, R) are isomorphic. In order to be explicit, we introduce the follow-ing notation. If f : A→ B is a homomorphism, then we denote the correspondingelement of (A⊗B∗)∗ by Adj(f). In other words, if a ∈ A and β ∈ B∗, then

β(f(a)) = Adj(f)(a⊗ β)

in R.

(e) If F is a complex and a is an integer, then F[a] is the shift of F with F[a]n = Fa+n,for all n.

The complexes {W} constitute the main ingredient of the “rows” of the doublecomplex which gives rise to the complex Y of Definition 2.2.

10 ANDREW R. KUSTIN

Definition 1.11. Fix a free R-module E of rank e and integers a and ℓ with1 ≤ ℓ < a+ e. Define the complex WE(a, ℓ) = (W,www)

. . .www−→W1

www−→W0,

with Wp =⊕∧λ

E∗ ⊗Da+pE, where the sum is taken over all ℓ-tuples of positiveintegers λ with |λ| = ℓ+ p. The differential www : Wp →Wp−1 is the composition

Wp1⊗∆−−−→

∧λE∗ ⊗ E ⊗Da+p−1E

w#⊗1−−−−→Wp−1,

wherew# :

∧λE∗ ⊗E →

⊕

|λ′|=ℓ+p−1

∧λ′

E∗,

is given by

w#(V1 ⊗ · · · ⊗ Vℓ ⊗ u) =ℓ∑

i=1

(−1)λ1+···+λi−1χ(λi ≥ 2)V1 ⊗ · · · ⊗ u(Vi)⊗ · · · ⊗ Vℓ.

Define an augmentation h : W0 → Da−ℓE by

h(V1 ⊗ · · · ⊗ Vℓ ⊗ U) = (V1 · · ·Vℓ)(U).

(In W0, the ℓ-tuple λ satisfies |λ| = ℓ; hence, each λk = 1, each Vk is in E∗, and theelement V1 · · ·Vℓ of SymℓE

∗ acts on the element U ∈ DaE to produce an elementin Da−ℓE.)

Lemma 1.12. If 1 ≤ ℓ < a+ e, then the augmented complex

WE(a, ℓ)h−→ Da−ℓE → 0

is split exact.

Proof. The proof proceeds by induction on ℓ. We first consider ℓ = 1. In this case,the augmented complex

WE(a, 1)h−→ Da−1E → 0

is

(1.13) 0 →VeE∗ ⊗Da+e−1E −→ . . . −→

V2E∗ ⊗Da+1E −→V1E∗ ⊗DaE

h−→ Da−1E → 0,

with∧1E∗⊗DaE in position zero. Complex (1.13) is split exact provided a+e−1

is not equal to zero. Indeed, the Koszul complex

(1.14) · · · → Symb−1E∗ ⊗

∧1E∗ → SymbE

∗ ⊗∧0E∗ → 0


is split exact for all integers b 6= 0; hence, the dual of (1.14), which is equal to

(1.15) 0→∧0E ⊗DbE →

∧1E ⊗Db−1E → . . . ,

is also split exact, for all integers b 6= 0. Of course, (1.15) is isomorphic to(0→

∧eE∗ ⊗DbE →

∧e−1E∗ ⊗Db−1E → . . .

)⊗∧eE.

At any rate, the assertion holds when ℓ = 1. Fix ℓ and a with 1 ≤ ℓ < a + e − 1.The complex WE(a, ℓ+ 1) is the total complex of the double complex

⊕

λ0≥1

∧λ0E∗ ⊗W(a− 1 + λ0, ℓ) :

..

....

?

?

y

?

?

y

. . . −−−−−→V3E∗ ⊗ W(a+ 2, ℓ)1 −−−−−→

V3E∗ ⊗ W(a+ 2, ℓ)0?

?

y

?

?

y

. . . −−−−−→V2E∗ ⊗ W(a+ 1, ℓ)1 −−−−−→

V2E∗ ⊗ W(a+ 1, ℓ)0?

?

y

?

?

y

. . . −−−−−→V1E∗ ⊗ W(a, ℓ)1 −−−−−→

V1E∗ ⊗ W(a, ℓ)0.

The induction hypothesis ensures that each row is acyclic and that the zeroth

homology of row λ0 is∧λ0E∗⊗Da−1+λ0−ℓE. The induced map on the row homology

is the case ℓ = 1:

0→∧eE∗ ⊗Da−1+e−ℓE → · · · →

∧1E∗ ⊗Da−ℓE.

The critical number a − 1 + e − ℓ is positive; and therefore, the complex of rowhomology is acyclic and resolves Da−1−ℓE. �

We use the following well-known result (see, for example, [K07, Lemma 1.3]) inalmost all of our calculations. The result shows that one may prove a formula aboutthe elements of DmE by checking that the formula holds at each pure divided poweru(m), for u ∈ E, provided the formula can be obtained, by way of base change, froma corresponding formula over a polynomial ring over the ring of integers.

Lemma 1.16. Suppose R is a polynomial ring over the ring of integers, E and

G are free R-modules, and f : DmE → G is an R-module homomorphism. If

f(u(m)) = 0 for all u ∈ E, then f is identically zero.

The map η is the most important map in this paper. Three other maps arecreated using η as the main ingredient. The map η! is not particularly important,but is so similar to η that we define the two maps simultaneously.

12 ANDREW R. KUSTIN

Definition 1.17. Let ℓ be a positive integer, λ be an ℓ-tuple of positive integers,and m and n be integers. Set c = |λ| − ℓ. Define

η : DnE ⊗∧λ

E∗ ⊗DmE → Dn+c+1−ℓE ⊗∧c+1

E∗ ⊗Dm−cE and

η! : DnE ⊗∧λ

E∗ ⊗DmE → Dn+c−ℓE ⊗∧cE∗ ⊗Dm−cE

as follows. Both maps are identically zero unless λk ≤ 2 for all k. Henceforth, weassume that this condition holds. It follows that λk = 2 for exactly c values of kand λk = 1 for the remaining ℓ− c values of k. Identify the indices

(1.18)i1 < · · · < iℓ−c and j1 < · · · < jc with λik = 1 and λjk = 2for all k. Let J =

∑ck=1 jk.

Both maps begin with the composition

DnE ⊗∧λ

E∗ ⊗DmE1⊗∆−−−→ DnE ⊗

∧λE∗ ⊗ TcE ⊗Dm−cE

−−−→ DnE ⊗ Tℓ−cE∗ ⊗

∧cE∗ ⊗Dm−cE,

where the second map sends

U ′ ⊗ (V1 ⊗ · · · ⊗ Vℓ)⊗ (u1 ⊗ · · · ⊗ uc)⊗ U

to

(−1)JU ′ ⊗ (Vi1 ⊗ · · · ⊗ Viℓ−c)⊗ (u1(Vj1) ∧ · · · ∧ uc(Vjc))⊗ U.

The map η concludes with

U ′ ⊗ (v1 ⊗ · · · ⊗ vℓ−c)⊗ V ⊗ U 7→ χ(λℓ = 1)(v1 · · · vℓ−c−1)(U′)⊗ (V ∧ vℓ−c)⊗ U.

The map η! concludes with

U ′ ⊗ (v1 ⊗ · · · ⊗ vℓ−c)⊗ V ⊗ U 7→ (v1 · · · vℓ−c)(U′)⊗ V ⊗ U.

The product v1 · · · vℓ−c−1 takes place in the the symmetric algebra Sym•E∗. The

divided power algebra D•E is a module over Sym•E∗.

Remarks. 1. In practice “n” from the definition of η will be ℓ−c−1 and Dn+c+1−ℓEwill equal R.2. In the above notation, if

t = U ′ ⊗ (V1 ⊗ · · · ⊗ Vℓ)⊗ u(m) ∈ DnE ⊗

∧λE∗ ⊗DmE,


then

η!(t) =(−1)J(Vi1 · · ·Viℓ−c)(U ′)⊗ (u(Vj1) ∧ · · · ∧ u(Vjc))⊗ u

(m−c) and

η(t) =χ(λℓ = 1)(−1)J(Vi1 · · ·Viℓ−c−1)(U ′)⊗ (u(Vj1) ∧ · · · ∧ u(Vjc) ∧ Vℓ)⊗ u

(m−c).

The following result shows the interaction between the map www of Definition 1.11and the map η. Our first use of this Lemma occurs in the proof of Lemma 2.7 whenwe prove that ψ : Y → N ⊗ L is a map of complexes. Fix an ℓ-tuple of positiveintegers λ with |λ| = ℓ+ c. We have three maps

DnE ⊗∧λ

E∗ ⊗DmE −→ Dn+c−ℓE ⊗∧cE∗ ⊗Dm−cE;

namely,

(1.19) η ◦ (1⊗www), η!, and MA ◦η ◦∆,

where the first map is the composition,

DnE ⊗∧λ

E∗ ⊗DmE1⊗www−−−→

⊕

|λ′|=ℓ+c−1

DnE ⊗∧λ′

E∗ ⊗Dm−1E

η−→ Dn+c−ℓE ⊗


the third map is the composition

DnE ⊗∧λ

E∗ ⊗DmE∆⊗1−−−→ E ⊗Dn−1E ⊗

∧λE∗ ⊗DmE

1⊗η−−→ E ⊗Dn+c−ℓE ⊗

∧c+1E∗ ⊗Dm−cE

MA−−→ Dn+c−ℓE ⊗


and MA: E →∧•E∗ is the module action of

∧•E on

∧•E∗.

Lemma 1.20. Let ℓ be a positive integer, λ be an ℓ-tuple of positive integers, and

m and n be integers. Set c = |λ| − ℓ. The three maps of (1.19) are related by the

equation

η ◦ (1⊗www) + MA ◦η ◦∆ = (−1)cη!.

Proof. In light of Lemma 1.16, it suffices to verify the assertion using only pure

divided powers. Let t = u′(n) ⊗ V1 ⊗ · · · ⊗ Vℓ ⊗ u(m) ∈ DnE ⊗

∧λE∗ ⊗DmE. We

compute

A = η ◦ (1⊗www)(t), B = η!(t), and C = (MA ◦η ◦∆)(t).

14 ANDREW R. KUSTIN

We first observe that if λk ≥ 3, for any k, then all three terms are zero. (If λk = 3,for some k, then one of the factors of A is u(u(Vk)) = 0; and therefore, A = 0 in thiscase.) We adopt the notation of (1.18). Let V be the element u(Vj1) ∧ · · · ∧ u(Vjc)of∧cE∗. We see that

B =(−1)J (Vi1 · · ·Viℓ−c)(u′

(n))⊗ V ⊗ u(m−c),

C =χ(λℓ = 1)(−1)J(Vi1 · · ·Viℓ−c−1)(u′

(n−1))⊗ u′(V ∧ Vℓ)⊗ u

(m−c), and

A =η

(c∑

k=1

(−1)λ1+···+λjk−1u′(n)⊗ V1 ⊗ · · · ⊗ u(Vjk)⊗ · · · ⊗ Vℓ ⊗ u

(m−1)

).

We calculate A in two cases depending on the value of λℓ. Indeed,

A = χ(λℓ = 1)A+ χ(λℓ = 2)A.

We see that χ(λℓ = 1)A is equal to

c∑k=1

χ(λℓ = 1)(−1)λ1+···+λjk−1(−1)J−jk(Vi1 · · ·Viℓ−c−1u(Vjk))(u′

(n))⊗

(u(Vj1) ∧ · · · ∧ u(Vjk) ∧ · · · ∧ u(Vjc) ∧ Vℓ)⊗ u(m−c).

The set {1, . . . , jk} decomposes into the disjoint union of the sets

I ∪ {j1, . . . , jk},

where I is the set {i | i < jk and λi = 1}. Furthermore,

(−1)λ1+···+λjk−1 = (−1)card I ,

where card I is the cardinality of the set I. It follows that

(1.21) (−1)λ1+···+λjk−1(−1)jk = (−1)k.

The module action of∧•E on

∧•E∗ gives

(1.22) u′(V ) =c∑

k+1

(−1)k+1[u(Vjk)](u′) · (u(Vj1) ∧ · · · ∧ u(Vjk) ∧ · · · ∧ u(Vjc);

so, χ(λℓ = 1)A is equal to

(−1)J+1χ(λℓ = 1)(Vi1 · · ·Viℓ−c−1)(u′

(n−1))⊗ (u′(V ) ∧ Vℓ)⊗ u

(m−c).


The module action also gives

u′(V ) ∧ Vℓ = u′(V ∧ Vℓ) + (−1)c+1u′(Vℓ) · V.

The hypothesis λℓ = 1 ensures that iℓ−c = ℓ. We see that

χ(λℓ = 1)A = −C + χ(λℓ = 1)(−1)cB.

If λℓ = 2, then most values of k cause the corresponding term in A to be au-tomatically zero; the only value of k which for which the corresponding term isnon-zero is k = c. So, χ(λℓ = 2)A is equal to

χ(λℓ = 2)(−1)J+c(Vi1 · · ·Viℓ−c)(u′

(n))⊗ [u(Vj1) ∧ · · · ∧ u(Vjc−1

) ∧ u(Vℓ)]⊗ u(m−c).

Of course, the hypothesis λℓ = 2 ensures that jc = ℓ; hence, χ(λℓ = 2)A =(−1)cχ(λℓ = 2)B and A = −C + (−1)cB. �

We consider three very similar maps, ξ1, ξ2, ξ3, which involve η. The map ξ1 isthe main ingredient of the map of complexes ψ : Y → N⊗ L. The maps ξ2 and ξ3are used in the proof that ψ is a quasi-isomorphism; these maps do not come intoplay until we have selected a distinguished element xg of G, see (2.12).

Definition 1.23. Let ℓ be a positive integer and λ be an ℓ-tuple of positive integers.Set c = |λ| − ℓ. Let m, m2, n1, and n2 be integers and

M = M(λ,m, 0, n1)⊗M(m2, e− 1− c, n2)⊗ L∗.

For each index i, with 1 ≤ i ≤ 3, we define an R-module homomorphism ξi : M →R.

(1) The map ξ1 is defined provided m = g+ c, m2 = ℓ−1− c, and n1 +n2 = α+ c.The map ξ1 is the composition

Mρ1−→∧n1+n2(E ⊗G∗)⊗

(Dℓ−c−1E ⊗

∧λE∗ ⊗DmE

)⊗De−c−1G

∗ ⊗ L∗

1⊗η⊗1−−−−→

∧α+c(E∗ ⊗G)⊗

(∧c+1E∗ ⊗DgE

)⊗De−c−1G

∗ ⊗ L∗ ρ2−→ R,

where the map ρ1 rearranges the terms and performs exterior multiplication:

ρ1((V ⊗ U ⊗ 1⊗ Z)⊗ (U ′ ⊗ Y ′ ⊗ Z ′)⊗ L∗) = (Z ∧ Z ′)⊗ (U ′ ⊗ V ⊗ U)⊗ Y ′ ⊗ L∗,

and the map ρ2 involves the module action of∧c+1

E∗ on∧eE:

ρ2(Z ⊗ (V ⊗ U)⊗ Y ⊗ (U0 ⊗ Y0 ⊗W0)) = [Z ∧ (V (U0) ⊲⊳ Y ) ∧ (U ⊲⊳ Y0)](W0).

16 ANDREW R. KUSTIN

(2) The map ξ2 is defined provided m = g − 1 + c, m2 = ℓ− 1− c, and n1 + n2 =α+ c+ 1. The map ξ2 is the composition

Mρ1−→∧n1+n2(E ⊗G∗)⊗

(Dℓ−c−1E ⊗

∧λE∗ ⊗DmE

)⊗De−c−1G

∗ ⊗ L∗

1⊗η⊗1−−−−→

∧α+c+1(E∗ ⊗G)⊗

(∧c+1E∗ ⊗Dg−1E

)⊗De−c−1G

∗ ⊗ L∗ ρ3−→ R,

where ρ1 is unchanged from (1) and, in addition to the module action of∧c+1

E∗

on∧eE, ρ3 also involves the action of xg ∈ G on

∧gG∗:

ρ3(Z ⊗ (V ⊗ U)⊗ Y ⊗ (U0 ⊗ Y0 ⊗W0)) = [Z ∧ (V (U0) ⊲⊳ Y ) ∧ (U ⊲⊳ xg(Y0))](W0).

(3) The map ξ3 is defined providedm = g+c, m2 = ℓ−2−c, and n1+n2 = α+c+1.The map ξ3 is identically zero unless λℓ−1 = 1. Assume this condition holds. Letℓ = ℓ− 1, λ be the ℓ-tuple λ = (λ1, . . . , λℓ−2, λℓ). Notice that |λ| − ℓ remains equalto c. Let

ρ0 :∧λ

E∗ → E∗ ⊗∧λ

E∗

be the mapρ0(V1 ⊗ · · · ⊗ Vℓ) = Vℓ−1 ⊗ (V1 ⊗ · · · ⊗ Vℓ−2 ⊗ Vℓ).

The map ξ3 is the composition

Mρ0−→ E∗ ⊗M(λ,m, 0, n1)⊗M(ℓ− 1− c, e− 1− c, n2)⊗ L

∗

ρ1−→ E∗ ⊗∧n1+n2(E ⊗G∗)⊗

(Dℓ−c−1E ⊗

∧λE∗ ⊗DmE

)⊗De−c−1G

∗ ⊗ L∗

1⊗η⊗1−−−−→ E∗ ⊗

∧α+c+1(E∗ ⊗G)⊗

(∧c+1E∗ ⊗DgE

)⊗De−c−1G

∗ ⊗ L∗ ρ4−→ R,

where the map ρ1 is unchanged from part (1) and, in addition to the module action

of∧c+1

E∗ on∧eE, ρ4 also involves the module action of (E∗⊗G) on

∧•(E⊗G∗):

ρ4(v ⊗ Z ⊗ (V ⊗ U)⊗ Y ⊗ (U0 ⊗ Y0 ⊗W0))

=[Z ∧ (v ⊗ xg)

((V (U0) ⊲⊳ Y ) ∧ (U ⊲⊳ Y0)

)](W0).

Remark 1.24. For quick reference we record the value of the maps of Definition 1.23when they are applied to elements of the form

t = V1 ⊗ · · · ⊗ Vℓ ⊗ U ⊗ 1⊗ Z ⊗ U ′ ⊗ Y ′ ⊗ Z ′ ⊗ U0 ⊗ Y0 ⊗W0 ∈M,


when U is the pure divided power u(m). All three maps are identically zero unlessλk ≤ 2 for all k. Assume the notation of (1.18) is in effect and let

V = u(Vj1) ∧ · · · ∧ u(Vjc).

Then

ξ1(t) = (−1)Jχ(λℓ = 1) · (Vi1 . . . Viℓ−c−1)(U ′)·

[Z ∧ Z ′ ∧ ((V ∧ Vℓ)(U0) ⊲⊳ Y′) ∧ (u(m−c) ⊲⊳ Y0)](W0),

ξ2(t) = (−1)Jχ(λℓ = 1) · (Vi1 . . . Viℓ−c−1)(U ′)·

[Z ∧ Z ′ ∧ ((V ∧ Vℓ)(U0) ⊲⊳ Y′) ∧ (u(m−c) ⊲⊳ xg(Y0))](W0), and

ξ3(t) = (−1)Jχ(λℓ−1 = 1)χ(λℓ = 1) · (Vi1 . . . Viℓ−c−2)(U ′)·[

Z ∧ Z ′ ∧ (Vℓ−1 ⊗ xg)(((V ∧ Vℓ)(U0) ⊲⊳ Y

′) ∧ (u(m−c) ⊲⊳ Y0))]

(W0).

2. The complex Y.

The complex Y, which was promised in (0.9), is a shift of the complex X(P,Q, 0).The complex X(P,Q, 1) is used in our proof that ψ is a quasi-isomorphism. Ourstatements are about Y, but it we prefer to calculate using X. Recall the notationof 1.10. The differentials www and mmm are defined in 1.11 and 1.3.

Definition 2.1. Let E and G be free modules of rank e and g, respectively, andlet P , Q, and ǫ be integers with 1− e+ ǫ ≤ P −Q ≤ g− 1. We define the complexXE,G(P,Q, ǫ):

. . .xxx−→ X2

xxx−→ X1

xxx−→ X0 → 0.

Let ℓ = P −Q+ e− ǫ. For integers r and c, the module Xr,c is equal to

Xr,c =

{0, if r < 0 or c < 0,⊕

M(λ, g + r + c− ǫ, r, α+ e− 1−Q− r), if r ≥ 0 and c ≥ 0,

where the sum is taken over all ℓ-tuples of positive integers with |λ| = ℓ + c. Themodule Xp is equal to Xp =

⊕Xr,c, where the sum is taken over all pairs (r, c)

with r + c = p. If t ∈ Xr,c ⊂ Xp, then

xxx(t) =

{(www ⊗ 1⊗ 1)(t) ∈ Xr,c−1

+(−1)p(1⊗mmm)(t) ∈ Xr−1,c.

Remark. The hypothesis 1−e+ǫ ≤ P −Q ensures that ℓ ≥ 1. There is no difficultyin showing that (XE,G(P,Q, ǫ),xxx) is a complex.

18 ANDREW R. KUSTIN

Definition 2.2. Adopt the Data of 2.1. Define YE,G(P,Q) to be the complexX(P,Q, 0)[e− 1−Q]. In other words, for each integer p,

Yp = Xp+e−1−Q =⊕

r+c=p+e−1−Q

Xr,c.

Definition 2.3. Let Y = YE,G(P,Q) and M be the complexes of Definition 2.2and (1.8). We define a map of complexes ϕ : Y → M. Fix an integer p. Use (1.9)and the definition of ℓ to see that

Mp =M(g−1−P+p, e−1−Q+p, α−p) =M(g+p+e−1−Q−ℓ, p+e−1−Q,α−p).

Let Xr,c be a summand of Yp. If c > 0, then ϕ(Xr,c) = 0. If c = 0, then r =p+ e− 1−Q and

Xr,c = M(λ, g + p+ e− 1−Q, p+ e− 1−Q,α− p),

with λ equal to the ℓ-tuple (1, . . . , 1). We define ϕ : Xp+e−1−Q,0 → Mp to beh⊗ 1⊗ 1, where h is the map of Definition 1.11.

Proposition 2.4. The map of complexes ϕ : Y→M is a quasi-isomorphism.

Proof. We show that the mapping cone of ϕ : Y → M is split exact. The mappingcone of ϕ is the total complex of

......

......

yy

yy

. . . −−−−→ Xr,2 −−−−→ Xr,1 −−−−→ Xr,0 −−−−→ MQ+1−e+r → 0y

yy

y...

......

...y

yy

y

. . . −−−−→ X1,2 −−−−→ X1,1 −−−−→ X1,0 −−−−→ MQ+2−e → 0y

yy

y

. . . −−−−→ X0,2 −−−−→ X0,1 −−−−→ X0,0 −−−−→ MQ+1−e → 0.

Be sure to notice that all of M appears in the above double complex becauseMQ−e = 0. Observe that the augmented row

· · · → Xr,2 → Xr,1 → Xr,0 →M(g + r − ℓ, r, α+ e− 1−Q− r)→ 0


is isomorphic to the complex

(WE(g + r, ℓ)

h−→ Dg+r−ℓE

)⊗DrG

∗ ⊗∧α+e−1−Q−r

(E ⊗G∗),

which is split exact by Lemma 1.12, because the hypothesis P −Q ≤ g− 1 ensuresthat ℓ < g + e+ r for all non-negative r. �

Definition 2.5. Let X be the complex XE,G(P,Q, 0) of Definition 2.1, N be thecomplex N(P,Q) of Definition 1.4, and L be the rank one R-free G-module of (1.1).We define ψ : Y→ (N,−nnn)⊗ L by defining a module homomorphism

(2.6) ψ : X[+1]→ (N[Q+ 2− e],−nnn)⊗ L.

Fix p. Take (r, c) with r + c = p, and λ with |λ| = ℓ+ c. Let

M = M(λ, g + r + c, r, α+ e− 1−Q− r) ⊂ Xr,c ⊂ Xp.

We define

ψ : M→ NQ+1−e+p ⊗ L = N (ℓ− p− 1, e− p− 1, p− e+ 1 +Q)⊗ L

by setting Adj(ψ) = χ(r = 0)ξ1, where

ξ1 : M⊗M(ℓ− c− 1, e− c− 1, c− e+ 1 +Q)⊗ L∗ → R

is the map of Definition 1.23. (Our convention for “Adj” may be found in 1.10.d.)

Lemma 2.7. The module homomorphism ψ : Y → (N,−nnn) ⊗ L of Definition 2.5

is a map of complexes.

Proof. Our proof has three parts.

Part 1. If c is a fixed non-negative integer, then the diagram

X0,cxxx

−−−−−→ X0,c−1

ψ

?

?

yψ

?

?

y

N (ℓ− c− 1, e− c− 1, c+Q− e+ 1) ⊗ L−nnn

−−−−−→ N (ℓ− c, e− c, c+Q− e) ⊗ L

commutes. Fix λ with |λ| = ℓ+ c,

t = V1 ⊗ · · · ⊗ Vℓ ⊗ u(g+c) ⊗ 1⊗ Z ∈M(λ, g + c, 0, α+ e− 1−Q) ⊆ X0,c, and

t′ = u′(ℓ−c)

⊗ y′(e−c)

⊗ Z ′ ⊗ U0 ⊗ Y0 ⊗W0 ∈M(ℓ− c, e− c, c− e+Q)⊗ L∗.

20 ANDREW R. KUSTIN

We compute

[(ψ ◦ xxx)(t)](t′) = ξ1(xxx(t)⊗ t′) = ρ2(Z ∧ Z

′ ⊗ A⊗ y′(e−c)

⊗ U0 ⊗ Y0 ⊗W0),

where ρ2 is used in Definition 1.23 and

A = (η ◦ (1⊗www))(u′(ℓ−c)

⊗ V1 ⊗ · · · ⊗ Vℓ ⊗ u(g+c)).

Apply Lemma 1.20 to see that A is zero unless λk ≤ 2 for all k. Adopt the notationof (1.18) and let V = u(Vj1)∧· · ·∧u(Vjc). Lemma 1.20 shows that A = B+C with

B = (−1)J+c(Vi1 · · ·Viℓ−c)(u′

(ℓ−c)) · V ⊗ u(g) and

C = χ(λℓ = 1)(−1)J+1(Vi1 · · ·Viℓ−c−1)(u′

(ℓ−c−1)) · u′(V ∧ Vℓ)⊗ u

(g).

The term ρ2(Z ∧ Z ′ ⊗B ⊗ y′(e−c) ⊗ U0 ⊗ Y0 ⊗W0) contains a factor of

(2.8) (V (U0) ⊲⊳ y′(e−c)) ∧ (u(g) ⊲⊳ Y0) = 0.

Indeed, V = u(V ′) for V ′ = Vj1 ∧ u(Vj2) ∧ · · · ∧ u(Vjc); so by (1.2) the left side of(2.8) is

([u ∧ V ′(U0)] ⊲⊳ y′(e−c)) ∧ (u(g) ⊲⊳ Y0)

= (u⊗ y′) ∧ (V ′(U0) ⊲⊳ y′(e−c−1)

) ∧ (u(g) ⊲⊳ Y0)

= (−1)e−c−1(V ′(U0) ⊲⊳ y′(e−c−1)

) ∧ (u(g+1) ⊲⊳ (y′ ∧ Y0)) = 0.

It follows that

[(ψ ◦ xxx)(t)](t′) = ρ2(Z ∧ Z′ ⊗ C ⊗ y′

(e−c)⊗ U0 ⊗ Y0 ⊗W0)

=

{(−1)J+1χ(λℓ = 1)(Vi1 · · ·Viℓ−c−1

)(u′(ℓ−c−1)

)·[Z ∧ Z ′ ∧

((u′[V ∧ Vℓ])(U0) ⊲⊳ y

′(e−c))∧(u(g) ⊲⊳ Y0

)](W0)

=

{(−1)J+1χ(λℓ = 1)(Vi1 · · ·Viℓ−c−1

)(u′(ℓ−c−1)

)·[Z ∧ Z ′ ∧ (u′ ⊗ y′) ∧

((V ∧ Vℓ)(U0) ⊲⊳ y

′(e−c−1))∧(u(g) ⊲⊳ Y0

])(W0).

On the other hand, (1.7) shows that [(−nnn ◦ ψ)(t)](t′) is equal to

−[ψ(t)](u′

(ℓ−c−1)⊗ y′

(e−c−1)⊗ Z ′ ∧ (u′ ⊗ y′)⊗ U0 ⊗ Y0 ⊗W0

),

and this is equal to [(ψ ◦ xxx)(t)](t′), as given above.


Part 2. The diagram

X0,0 −−−−−→ 0

ψ

?

?

y

?

?

y

N (ℓ− 1, e− 1, Q− e+ 1)−nnn

−−−−−→ N (ℓ, e, Q− e)

commutes. Fix λ = (1, . . . , 1),

t = V1 ⊗ · · · ⊗ Vℓ ⊗ u(g) ⊗ 1⊗ Z ∈M(λ, g, 0, α+ e− 1−Q) ⊆ X0,0, and

t′ = u′(ℓ)⊗ y′

(e)⊗ Z ′ ⊗ U0 ⊗ Y0 ⊗W0 ∈M(ℓ, e, Q− e)⊗ L∗.

We see that [(nnn ◦ ψ)(t)](t′) is equal to{

(V1 · · ·Vℓ−1)(u′(ℓ−1)

)·[Z ∧ Z ′ ∧ (u′ ⊗ y′) ∧

(Vℓ(U0) ⊲⊳ y

′(e−1))∧(u(g) ⊲⊳ Y0

)](W0)

= (V1 · · ·Vℓ−1)(u′(ℓ−1)

) · u′(Vℓ) ·[Z ∧ Z ′ ∧

(U0 ⊲⊳ y

′(e))∧(u(g) ⊲⊳ Y0

)](W0),

and

(2.9) (U0 ⊲⊳ y′(e)) ∧ (u(g) ⊲⊳ Y0) = 0;

see, for example, [K05, Observation 2.11].

Part 3. If c is a fixed non-negative integer, then the composition

X1,c

xxx

y

X0,c

ψ

y

N (ℓ− c− 1, e− c− 1, c+Q− e+ 1)

is zero. Fix λ with |λ| = ℓ+ c,

t = V1 ⊗ · · · ⊗ Vℓ ⊗ u(g+c+1) ⊗ y ⊗ Z ∈M(λ, g + c+ 1, 1, α+ e− 2−Q) ⊆ X1,c,

and

t′ = U ′ ⊗ Y ′ ⊗ Z ′ ⊗ U0 ⊗ Y0 ⊗W0 ∈M(ℓ− c− 1, e− c− 1, c− e+ 1 +Q)⊗ L∗.

We see that

[(ψ ◦ xxx)(t)](t′) = (−1)c+1[ψ(V1 ⊗ · · · ⊗ Vℓ ⊗ u(g+c) ⊗ 1⊗ (u⊗ y) ∧ Z)](t′)

contains a factor of

(u⊗ y) ∧(u(g) ⊲⊳ Y0

)= u(g+1) ⊲⊳ (y ∧ Y0) = 0. �

22 ANDREW R. KUSTIN

Definition 2.10. Let E and G be free modules of rank e and g respectively andlet P and Q be non-negative integers with 1 − e ≤ P − Q ≤ g − 1. Define thecomplex (CE,G(P,Q), ccc). The module CE,G(P,Q)p is

XE,G(P,Q)p⊕

N (ℓ− 2− p, e− 2− p,Q− e+ 2 + p)⊗ L,

where ℓ = P −Q+ e. The differential cccp : Cp → Cp−1 is

[xxx 0ψ nnn

].

Our proof of Lemma 2.7 shows that

(2.6) ψ : (X,xxx)[+1]→ (N(P,Q),−nnn)[Q− e+ 2]⊗ L

is a map of complexes. Observe that (CE,G(P,Q), ccc) is the mapping cone of (2.6).

Example 2.11. If g = 0 and 1− e ≤ P −Q ≤ g− 1, then the complex CE,G(P,Q)is split exact.

Proof. The complex C = CE,G(P,Q) is identically zero unless Q is zero, and ifQ = 0, then C is the complex W(0, ℓ) of Definition 1.11, for ℓ = P − Q + e. Thehypothesis on P −Q ensures that 1 ≤ ℓ < e and therefore Lemma 1.12 applies. Weoffer a few more details. Let N be the submodule N(P,Q)c = N (P − c, Q − c, c)of C. We see that N is zero unless Q− c = c = 0. However, when Q = 0, then thehypothesis P −Q ≤ g− 1 forces P − c < 0 and N is zero in every event. Let M bethe summand M(λ, g+ r+ c, r, α+ e− 1−Q− r) of Xr,c ⊆ C. The parameter α isequal to −(e− 1); so M is zero unless r = −Q− r = 0. �

In Theorem 4.2 we prove that (CE,G(P,Q), ccc) is split exact. The proof is byinduction on the ℓ and the rank ofG. In order to reduce the rank ofG we decomposeG as

(2.12) G = G⊕Rxg,

where G is a free R-submodule of G of rank g − 1 and xg is an element of G. Thecorresponding decomposition of G∗ is

(2.12) G∗ = G∗⊕Ryg.

In particular,

yg(G) = 0, yg(xg) = 1, and xg(G∗) = 0.


Definition 2.13. The decompositions of (2.12) induce decompositions

SymnG =n⊕

a=0

Symn−aG⊗Rxag and DnG

∗ =n⊕

a=0

Dn−aG∗⊗Ry(a)

g ,

which in turn induces the decompositions

N (m,n, p) = N 0(m,n, p)⊕N+(m,n, p),

M(m,n, p) =M0(m,n, p)⊕M+(m,n, p), and

M(λ,m, n, p) = M0(λ,m, n, p)⊕M+(λ,m, n, p),

where

N+(m,n, p) =

n⊕

a=1

SymmE∗ ⊗ Symn−aG⊗Rx

ag ⊗

∧p(E∗ ⊗G),

N 0(m,n, p) = SymmE∗ ⊗ SymnG⊗

∧p(E∗ ⊗G),

M+(m,n, p) =n⊕

a=1

DmE ⊗Dn−aG∗⊗Ry(a)

g ⊗∧p

(E ⊗G∗),

M0(m,n, p) = DmE ⊗DnG∗⊗∧p

(E ⊗G∗),

M+(λ,m, n, p) =n⊕

a=1

∧λE∗ ⊕DmE ⊗Dn−aG

∗⊗Ry(a)

g ⊗∧p

(E ⊗G∗), and

M0(λ,m, n, p) =∧λ

E∗ ⊕DmE ⊗DnG∗⊗∧p

(E ⊗G∗).

Let N+(P,Q) be the subcomplex of N(P,Q) with N+(P,Q)c = N+(P − c, Q− c, c),and N0(P,Q) be the quotient complex N(P,Q)/N+(P,Q). The corresponding shortexact sequence for M(P,Q) is

0→M0(P,Q)→M(P,Q)→M

+(P,Q)→ 0.

3. The case ℓ = 1.

Fix a pair of integers (P,Q). The complexes M and N may be found in (1.8) and(1.9). Recall that ℓ is defined to be P −Q+ e. When ℓ is arbitrary, there does notexist an equivariant quasi-isomorphism M→ N, see Example 3.15; however, such aquasi-isomorphism does exist when ℓ = 1: we define ψ′ : M → N in Definition 3.3and we prove the following result.

24 ANDREW R. KUSTIN

Theorem 3.1. If ℓ = 1 and e+ g ≥ 2, then the complex (CE,G(P,Q), ccc) of Defini-

tion 2.10 is split exact and ψ′ : M→ N is a quasi-isomorphism.

Remark. The dual argument gives an equivariant quasi-isomorphism ψ′ : M → N

when ℓ = e+ g − 1. We suppress the details.

Proof. Let C1,P be the mapping cone of ψ′. We exhibit a quasi-isomorphismΦ: CE,G(P,Q) → C1,P [P + 1] in Proposition 3.5. The complexes which are anal-ogous to CE,G(P,Q) when ℓ = 0 were introduced and shown to be split exact in[K05]. In the present paper these complexes, C0,P , are defined in Definition 3.6. InObservation 3.9 we exhibit a map of complexes β : C1,P → C0,P . The mapping coneof β is split exact by Theorem 3.11. It follows that C1,P is split exact, CE,G(P,Q)is split exact, and ψ′ is a quasi-isomorphism. �

Observation 3.2. There exists a homomorphism

Ψ:∧eE ⊗De−1G

∗ ⊗Dg−1E ⊗∧gG∗ →

∧e+g−1(E ⊗G∗)

with

Ψ(U0 ⊗ x(Y′)⊗ U ⊗ Y0) = (U0 ⊲⊳ Y

′) ∧ (U ⊲⊳ x(Y0))

Ψ(U0 ⊗ Y ⊗ v(U′)⊗ Y0) = (−1)e+1(v(U0) ⊲⊳ Y ) ∧ (U ′ ⊲⊳ Y0)

for allU0 ∈

∧eE, U ∈ Dg−1E, v ∈ E∗, U ′ ∈ DgG∗,

Y0 ∈∧gG∗, Y ∈ De−1G

∗, x ∈ G, Y ′ ∈ DeG∗.

Proof. Consider the homomorphisms

Ψ1 :∧eE ⊗De−1G

∗ ⊗E∗ ⊗DgE ⊗∧gG∗ →

∧e+g−1(E ⊗G∗)

Ψ2 :∧eE ⊗G⊗DeG

∗ ⊗Dg−1E ⊗∧g

G∗ →∧e+g−1

(E ⊗G∗)

which are given by

Ψ1(U0 ⊗ Y ⊗ v ⊗ U′ ⊗ Y0) = (v(U0) ⊲⊳ Y ) ∧ (U ′ ⊲⊳ Y0),

Ψ2(U0 ⊗ x⊗ Y′ ⊗ U ⊗ Y0) = (U0 ⊲⊳ Y

′) ∧ (U ⊲⊳ x(Y0)).

We observe that the following diagram commutes

VeE ⊗G⊗DeG∗ ⊗E∗ ⊗DgE ⊗

VgG∗ MA−−−−−→

VeE ⊗De−1G∗ ⊗E∗ ⊗DgE ⊗

VgG∗

MA

?

?

yΨ1

?

?

y

VeE ⊗G⊗DeG∗ ⊗Dg−1E ⊗

VgG∗ (−1)e+1Ψ2−−−−−−−−→Ve+g−1(E ⊗G∗),


where the top map is the module action of G on D•G∗ and the map on the left is

the module action of of E∗ on D•E. Indeed, there is a third natural map

Ψ3 :∧eE ⊗G⊗DeG

∗ ⊗E∗ ⊗DgE ⊗∧g

G∗ →∧e+g−1

(E ⊗G∗),

given by

Ψ3(U0 ⊗ x⊗ Y′ ⊗ v ⊗ U ′ ⊗ Y0) = (v ⊗ x) [(U0 ⊲⊳ Y

′) ∧ (U ′ ⊲⊳ Y0)] .

On the one hand Ψ3 is identically zero by (2.9). On the other hand, the moduleaction of E∗ ⊗G on

∧•(E ⊗G∗) shows that Ψ3 = Ψ1 + (−1)eΨ2.

To finish the argument we show that Ψ2 factors through the kernel of∧eE ⊗G⊗DeG

∗ ⊗Dg−1E ⊗∧gG∗ MA−−→

∧eE ⊗De−1G

∗ ⊗Dg−1E ⊗∧g

G∗.

Every element of the kernel has the form t =∑i U0⊗Qi⊗Ui⊗Y0, where Ui ∈ Dg−1E

and Qi is in the kernel of G⊗DeG∗ → De−1G∗. For each i, lift Ui to an element

Ri of E∗⊗DgE. Consider t′ =∑i U0⊗Qi⊗Ri⊗Y0 in the upper left hand corner.

We see thatt′ −−−−→ 0y

y

t −−−−→ ±Ψ2(t). �

When ℓ = 1, then, in the notation of (1.8) and (1.9), the complex M is

. . .mmm−→M(g, 1, α− P − 1)

mmm−→M(g − 1, 0, α− P )→ 0

and the complex N is

0→ N (0, e− 1, P )nnn−→ N (1, e, P − 1)→ . . . ,

withM(g − 1, 0, α− P ) and N (0, e− 1, P ) both in position P .

Definition 3.3. Fix P and Q with ℓ = 1. Define ψ′ : M→ N⊗ L with

ψ′ : M(g − 1, 0, α− P )→ N (0, e− 1, P )⊗ L

given by

[ψ′(U ⊗ 1⊗Z)](1⊗ Y ′⊗Z ′⊗U0⊗ Y0⊗W0) = [Z ∧Z ′ ∧Ψ(U0⊗ Y′⊗U ⊗Y0)](W0)

for all U ⊗ 1⊗ Z ∈M(g − 1, 0, α− P ) and

1⊗ Y ′ ⊗ Z ′ ⊗ U0 ⊗ Y0 ⊗W0 ∈M(0, e− 1, P )⊗ L∗.

Remark. There is no difficulty in seeing that ψ′ : M→ N⊗L is a map of complexes.Indeed, the definition of Ψ shows that

v(u)·(u⊗y)∧Ψ(U0⊗Y′⊗u(g−1)⊗Y0) = (−1)e+1(u⊗y)∧(v(U0) ⊲⊳ Y )∧(u(g) ⊲⊳ Y0),

and this is zero for all v ∈ E∗ and u ∈ E.

26 ANDREW R. KUSTIN

Definition 3.4. Let C1,P be the mapping cone of ψ′ : M → N⊗ L. Thus, C1,P isthe complex

. . .mmm−→M(g, 1, α− P − 1)

mmm−→M(g − 1, 0, α− P )

ψ′

−→ N (0, e− 1, P )⊗ Lnnn−→ . . . ,

with N (0, e− 1, P )⊗ L in position P .

Proposition 3.5. If ℓ = 1, then the quasi-isomorphism ϕ : Y→ M of Proposition

2.4 induces a quasi-isomorphism Φ: (CE,G(P,Q), ccc)→ C1,P [P + 1].

Proof. We define Φ: (CE,G(P,Q), ccc)→ C1,P [P + 1] by

. . . −−−−→ X1xxx

−−−−→ X0ψ

−−−−→ NP ⊗ Lnnn

−−−−→ NP−1 ⊗ L −−−−→

ϕ

y ϕ

y (−1)e+1

y (−1)e+1

y

. . . −−−−→ MP+1mmm

−−−−→ MPψ′

−−−−→ NP ⊗ Lnnn

−−−−→ NP−1 ⊗ L −−−−→ ,

with the modules X0 and MP in position zero in their respective complexes. Observethat ψ′ ◦ φ = (−1)e+1ψ : X0 → N (0, e− 1, P )⊗ L. Indeed, if

t = V1 ⊗ U ⊗ 1⊗ Z ∈M(λ, g, 0, α− P ) = X0 and

t′ = 1⊗ Y ′ ⊗ Z ′ ⊗ U0 ⊗ Y0 ⊗W0 ∈M(0, e− 1, P )⊗ L∗,

then

[(ψ′ ◦ ϕ)(t)](t′) = [Z ∧ Z ′ ∧Ψ(U0 ⊗ Y′ ⊗ V1(U)⊗ Y0)](W0)

= (−1)e+1[Z ∧ Z ′ ∧ (V1(U0) ⊲⊳ Y′) ∧ (U ⊲⊳ Y0)](W0) = (−1)e+1[ψ(t)](t′).

It follows that Φ is a map of complexes. It is clear that Φ is a surjection. Thekernel of Φ is equal to the kernel of ϕ which is split exact since ϕ is a surjectivequasi-isomorphism. �

The complex C0,P is introduced in [K05].

Definition 3.6. Let E and G be free modules of rank e and g, respectively. LetP be an integer. Let P ′ = α− 1− P . Define

γ :M(g, 0, P ′) −→∧g+P ′

(E ⊗G∗)⊗∧gG by

(1⊗ Y0)[γ(U ⊗ 1⊗ Z)] = (U ⊲⊳ Y0) ∧ Z

for Y0 ∈∧gG∗, and U ⊗ 1⊗ Z ∈M(g, 0, P ′). Define

Γ:∧g+P ′

(E ⊗G∗)⊗∧gG −→ N (0, e, P )⊗ L by


[Γ(t)](1⊗ Y ′ ⊗ Z ′ ⊗ U0 ⊗ Y0 ⊗W0) = [(1⊗ Y0)(t) ∧ Z′ ∧ (U0 ⊲⊳ Y

′)](W0)

for t ∈∧g+P ′

(E ⊗G∗)⊗∧gG and 1⊗ Y ′ ⊗ Z ′ ⊗ U0 ⊗ Y0 ⊗W0 ∈M(0, e, P )⊗ L∗.

Let C0,P be the complex

0→M(P ′ + g, P ′, 0)mmm−→ . . .

mmm−→M(g, 0, P ′)

γ−→∧g+P ′

(E ⊗G∗)⊗∧gG

Γ−→ N (0, e, P )⊗ L

nnn−→ . . .

nnn−→ N (P, P + e, 0)⊗ L,

with N (P, P + e, 0)⊗ L in position 0.

Proposition 3.7. For all integers P , the complex C0,P is split exact.

Proof. See [K05, Corollary 2.26]. �

Recall the element xg of G from (2.12).

Definition 3.8. Let P and P ′ be integers with P + P ′ = α− 1. Define a modulehomomorphism β : C1,P → C0,P

→M(g, 1, P ′)mmm

−−−−→ M(g − 1, 0, α− P )ψ′

−−−−→ N (0, e− 1, P )⊗ L→ · · ·

xg

y β

y xg

y

→M(g, 0, P ′)γ

−−−−→∧P ′+g

(E ⊗G∗)⊗∧g

GΓ

−−−−→ N (0, e, P )⊗ L→ · · · .

The map β : N (a, b, c)⊗ L→ N (a, b+ 1, c)⊗ L is multiplication by xg in Sym•G.The map β : M(a, b, c)⊗ L →M(a, b− 1, c)⊗ L is given by the module action of

xg ∈ G on D•G∗. The map β : M(g − 1, 0, α − P ) →

∧P ′+g(E ⊗ G∗) ⊗

∧gG is

defined by(1⊗ Y0)[β(U ⊗ 1⊗ Z)] = (−1)g−1(U ⊲⊳ xg(Y0)) ∧ Z

for all U ⊗ 1⊗ Z ∈M(g − 1, 0, α− P ) and Y0 ∈∧gG∗.

Observation 3.9. For each integer P , β : C1,P → C0,P is a map of complexes.

Proof. Consider the diagram of Definition 3.8. Let t = U⊗1⊗Z ∈M(g−1, 0, α−P )and t′ = 1⊗ Y ′ ⊗ Z ′ ⊗ U0 ⊗ Y0 ⊗W0 ∈M(0, e, P )⊗ L∗. We see that

[xgψ′(t)](t′) = [Z ∧ Z ′ ∧Ψ(U0 ⊗ xg(Y

′)⊗ U ⊗ Y0)](W0)

= [Z ∧ Z ′ ∧ (U0 ⊲⊳ Y′)⊗ (U ⊲⊳ xg(Y0))](W0) and

[(Γ ◦ β)(t)](t′) = (−1)g−1[(U ⊲⊳ xg(Y0)) ∧ Z ∧ Z′ ∧ (U0 ⊲⊳ Y

′)](W0) = [xgψ′(t)](t′).

Let t = u(g) ⊗ y ⊗ Z ∈M(g, q, P ′) and Y0 ∈∧gG∗. We see

(1⊗ Y0)[(β ◦mmm)(t)] = (−1)g−1(ug−1) ⊲⊳ xg(Y0)) ∧ (u⊗ y) ∧ Z

= xg(y) · (u(g) ⊲⊳ Y0) ∧ Z = (1⊗ Y0)[γ(xg(t))]. �

28 ANDREW R. KUSTIN

Definition 3.10. Let (A, aaa) be the mapping cone of β : C1,P → C0,P . So, A is

→M(g − 1, 0, α− P )

⊕M(g, 0, P ′)

aaaP+2

−−−→N (0, e− 1, P )⊗ L

⊕∧P ′+g(E ⊗G∗)⊗

∧gG

aaaP+1

−−−→N (1, e, P − 1)⊗ L

⊕N (0, e, P )⊗ L

→

with aaai equal to

[mmm 0xg −mmm

]for i ≥ P + 3,

[ψ′ 0β −γ

]for i = P + 2,

[nnn 0xg −Γ

]for i = P + 1, and

[nnn 0xg −nnn

]for i ≤ P .

Theorem 3.11. The complex (A, aaa) of Definition 3.10 is split exact.

Proof. Let A′ be the submodule of A which consists of all modules of the form

M0(m,n, p) or N (m,n, p) from C1,P and∧P ′+g

(E ⊗G∗)⊗∧g

G together with allmodules of the form N (m,n, p) from C0,P . (The notation is explained in Definition2.13.) It is easy to see that aaa(A′) ⊆ A′; hence, (A′, aaa) is a subcomplex of (A, aaa).The quotient A/A′ consists of the modulesM+(m,n, p) from C1,P andM(m,n, p)from C0,P , with differential

. . . −→M+(m,n, p)

⊕M(m+ 1, n, p− 1)

"quot ◦mmm 0

xg −mmm

#

−−−−−−−−−−−−−−→M+(m− 1, n− 1, p+ 1)

⊕M(m,n− 1, p)

−→ . . . .

The map xg is an isomorphism; hence, A/A′ is split exact and Hi(A) = Hi(A′) for

all i.Let P be the submodule of A′ which consists of all modules of the form N (m,n, p)

from C1,P and all modules of the form N+(m,n, p) from C0,P . It is easy to see that(P, aaa) is a subcomplex of (A′, aaa); indeed, P looks like

. . . −→N (m,n, p)⊕

N+(m− 1, n, p+ 1)

"nnn 0xg −nnn

#

−−−−−−−−→N (m+ 1, n+ 1, p− 1)

⊕N+(m,n+ 1, p)

−→ . . . .

Each map xg is an isomorphism; thus, P is split exact and Hi(A) = Hi(A′/P) for

all i. The complex A′/P is

· · ·mmm−→M0(g, 1, α− P − 1)

mmm−→M0(g − 1, 0, α− P )

β−→∧P ′+g

(E ⊗G∗)⊗∧g

G

quot◦Γ−−−−→ N 0(0, e, P )⊗ L

quot◦nnn−−−−→ · · · .


Recall the free submodules G and G∗

be of G and G∗ from (2.12). LetM(m,n, p),

N (m,n, p), and C0,P

be the modules, maps, and complexes manufactured using thedata E and G.

Claim. The complex A′/P is isomorphic to the following direct sum of complexes:

(3.12)e⊕

w=0

C0,P−w

[−w]⊗ (∧e−w

E ⊲⊳ y(e−w)g ).

Once the claim is established, then the proof is complete, since each complex

C0,P−w

is split exact by Proposition 3.7.First we look at the modules involved in the claim. Let g = g − 1,

α = (e− 1)(g − 1) = α− e+ 1, and L =∧eE∗ ⊗

∧g−1G⊗

∧e(g−1)(E ⊗G

∗).

If 0 ≤ i ≤ P , then the module (A′/P)i = N 0(P − i, e+ P − i, i)⊗ L is equal to

e⊕

w=0

N (P − i, e+ P − i, i− w)⊗ L⊗ (∧w

E∗ ⊲⊳ x(w)g )⊗Rxg ⊗ (

∧eE ⊲⊳ y(e)

g )

= C0,P−w

[−w]i ⊗ (∧w

E∗ ⊲⊳ x(w)g )⊗Rxg ⊗ (

∧eE ⊲⊳ y(e)

g ).

The module (A′/P)P+1 =∧P ′+g

(E ⊗G∗)⊗∧gG is equal to

e⊕

w=0

∧P ′+g−e+w(E ⊗G

∗)⊗ (

∧e−wE ⊲⊳ y(e−w)

g )⊗∧g−1

G⊗Rxg.

We see that P ′ + g − e+ w = α+ g − 1− (P − w) and therefore

(A′/P)P+1 =e⊕

w=0

C0,P−w

[−w]P+1 ⊗ (∧e−w

E ⊲⊳ y(e−w)g )⊗Rxg.

For i ≥ 0, the module (A′/P)i+P+2 =M0(g − 1 + i, i, α− P − i) is equal to

e⊕

w=0

M(g − 1 + i, i, α− P − e+ w − i)⊗ (∧e−w

E ⊲⊳ y(e−w)g ).

We see that α− P − e+ w − i = α− 1− (P − w)− i; and therefore,

(A′/P)i+P+2 =

e⊕

w=0

C0,P−w

[−w]i+P+2 ⊗ (∧e−w

E ⊲⊳ y(e−w)g ).

30 ANDREW R. KUSTIN

It is clear that the differentials mmm and quot ◦ nnn of A′/P decompose to form thedifferentials of (3.12). There are natural isomorphisms

(3.13) (∧e−w

E ⊲⊳ y(e−w)g ) ∼= (

∧wE∗ ⊲⊳ x(w)

g )⊗ (∧eE ⊲⊳ y(e)

g ) and

(3.14)∧g−1

G→∧gG,

with (3.14) given by X 7→ X ∧ xg. It is not difficult to see that the diagrams

C0,P−wP−w+2 ⊗ (

∧e−wE ⊲⊳ y

(e−w)g )

γ⊗1−−−−→ C

0,P−wP−w+1 ⊗ (

∧e−wE ⊲⊳ y

(e−w)g )

incl

y (3.14)

y

M0(g − 1, 0, α− P )β

−−−−→∧P ′+g

(E ⊗G∗)⊗∧g

G

and

C0,P−wP−w+1 ⊗ (

∧e−wE ⊲⊳ y

(e−w)g )

Γ⊗(3.13)−−−−−→ C

0,P−wP−w ⊗ (

∧wE∗ ⊲⊳ x

(w)g )⊗ (

∧eE ⊲⊳ y

(e)g )

(3.14)

y (3.14)

y∧P ′+g

(E ⊗G∗)⊗∧g

Gquot◦Γ−−−−→ N 0(0, e, P )⊗ L

commute up to sign. �

Example 3.15. When e = g = P = Q = 2, then there does not exist a charac-teristic free, G-equivariant, quasi-isomorphism ψ′ : M ⊗ L∗ → N. The module of1-cycles, ZZZ1, in the complex

N : 0→∧2

(E∗ ⊗G)→ E∗ ⊗G⊗ (E∗ ⊗G)→ Sym2E∗ ⊗ Sym2G→ 0

is a free R-module of rank 7 with basis

A1 = v1 ⊗ x1 ⊗ (v1 ⊗ x2)− v1 ⊗ x2 ⊗ (v1 ⊗ x1)A2 = v1 ⊗ x1 ⊗ (v2 ⊗ x1)− v2 ⊗ x1 ⊗ (v1 ⊗ x1)A3 = v1 ⊗ x1 ⊗ (v2 ⊗ x2)− v2 ⊗ x2 ⊗ (v1 ⊗ x1)A4 = v1 ⊗ x2 ⊗ (v2 ⊗ x1)− v2 ⊗ x1 ⊗ (v1 ⊗ x2)A5 = v1 ⊗ x2 ⊗ (v2 ⊗ x2)− v2 ⊗ x2 ⊗ (v1 ⊗ x2)A6 = v2 ⊗ x1 ⊗ (v2 ⊗ x2)− v2 ⊗ x2 ⊗ (v2 ⊗ x1)A7 = v1 ⊗ x1 ⊗ (v2 ⊗ x2)− v1 ⊗ x2 ⊗ (v2 ⊗ x1).

The complex M⊗ L∗ is0→M(0, 0, 0)⊗ L∗ → 0,


concentrated in position 1. If

ψ′ : M(0, 0, 0)⊗ L∗ =∧eE∗ ⊗

∧gG→ ZZZ1

is a G-equivariant map, then every element in the image of ψ′ is invariant underthe action of SLE∗ × SLG. An easy calculation shows that the submodule of ZZZ1

which is left fixed by the action of SLE∗ × SLG is the free R-module of rank 1generated by A3 − A4 − 2A7 =

−v1⊗x1⊗ (v2⊗x2)−v2⊗x2⊗ (v1⊗x1)+v1⊗x2⊗ (v2⊗x1)+v2⊗x1⊗ (v1⊗x2).

On the other hand, the class of A3 − A4 − 2A7 in HN (1, 1, 1) corresponds to 2.

4. The map ψ is a quasi-isomorphism.

Theorem 4.1. The map of complexes ψ : Y → N ⊗ L of Lemma 2.7 is a quasi-

isomorphism.

Theorem 4.1 is an immediate consequence of Theorem 4.2.

Theorem 4.2. If 1 − e ≤ P − Q ≤ g − 1, then the complex (CE,G(P,Q), ccc) of

Definition 2.10 is split exact.

Proof. The proof proceeds by induction on ℓ and g. The complex CE,G(P,Q) issplit exact when ℓ = 1 by Theorem 3.1. The complex CE,G(P,Q) is split exactwhen g = 0 by Example 2.11. Assume 1 ≤ g and 2 ≤ ℓ ≤ e + g − 1. We define amodule homomorphism

(4.3) Θ: CE,G(P,Q)[+1]→ CE,G(P,Q+ 1)

in (4.5). In Theorem 4.9 we prove that Θ is a map of complexes. Let A be themapping cone of Θ. We exhibit a sequence of quasi-isomorphisms:

Aquot−−−→ A/P

H−→ B

incl←−− D ∼=

e⊕

w=0

CE,G(P − w,Q− w + 1)⊗ (∧e−w

E ⊲⊳ y(e−w)g ).

We see that1− e ≤ (P − w)− (Q− w + 1) ≤ g − 2;

so induction on g yields that CE,G(P −w,Q−w+ 1) is split exact. It follows that

A, which is the mapping cone of (4.3), is split exact. The complex CE,G(P,Q+ 1)is split exact by induction on ℓ. We conclude that CE,G(P,Q) is split exact.

The subcomplex P of A is introduced in Definition 5.1. The quasi-isomorphismA/P → B is Lemma 5.4. The subcomplex D of B is Definition 5.7. The decompo-sition

D ∼=

e⊕

w=0

CE,G(P − w,Q− w + 1)⊗ (∧e−w

E ⊲⊳ y(e−w)g )

is Lemma 5.8. �

32 ANDREW R. KUSTIN

Notation 4.4. Fix P and Q, with Q < P +e−2 (so 2 ≤ ℓ) and let (C, ccc) represent

(CE,G(P,Q), ccc). Let Q = Q+1, ℓ = ℓ−1, λ represent an ℓ-tuple of positive integers,

Xr,c =⊕

|eλ|=eℓ+c

M(λ, g + r + c, r, α+ e− 1− Q− r),

and (X, xxx) be XE,G(P, Q). Furthermore, we let (C, ccc) be the mapping cone of

ψ : (X, xxx)[+1]→ (N(P, Q),−nnn)[Q− e+ 2]⊗ L.

In particular, the module Cp is

Xp

⊕N (ℓ− 3− p, e− 2− p,Q− e+ 3 + p)⊗ L,

and the differential cccp : Cp → Cp−1 is

[xxx 0ψ nnn

].

Recall, from Definition 2.10, that

C[+1]p =

Xp+1

⊕N (ℓ− 3− p, e− 3− p,Q− e+ 3 + p)⊗ L.

The map Θ: C[+1]p → Cp is given by

(4.5) Θ =

[q 0θ xg

],

where the map xg : N (a, b, d) → N (a, b + 1, d) is multiplication in the symmetricalgebra:

V ⊗X ⊗W 7→ V ⊗ xgX ⊗W,

and θ is given in Definition 4.6. It is convenient to define the maps q : X[+1]→ X

and q : X[+1] → X simultaneously. We define X in Definition 4.7, and q and q inDefinition 4.8.


Definition 4.6. Define a module homomorphism

θ : X[+1]→ N(P, Q)[Q+ 2− e]⊗ L.


M = M(λ, g + r + c, r, α+ e− 1−Q− r) ⊂ Xr,c ⊂ Xp.

We defineθ : M→ N (ℓ− p− 2, e− p− 1, p− e+ 2 +Q)⊗ L

by setting Adj(θ) = χ(r = 0) · ξ3, where

ξ3 : M⊗M(ℓ− p− 2, e− p− 1, p− e+ 2 +Q)⊗ L∗ → R

is the homomorphism given in Definition 1.23.

Definition 4.7. Assume 2− e ≤ P −Q. Let (X, xxx) be the complex XE,G(P,Q, 1).

We write ℓ for ℓ− 1 and we always take λ to be an ℓ-tuple of positive integers. Themodules of X are Xp =

⊕r+c=p Xr,c, where

Xr,c =⊕

|λ|=ℓ+c

M(λ, g − 1 + r + c, r, α+ e− 1−Q− r).

Define h : X→ X, where X is the usual complex XE,G(P,Q, 0). If

t = V1 ⊗ · · · ⊗ Vℓ ⊗ U ⊗ Y ⊗ Z ∈M(λ, g + r + c, r, α+ e− 1−Q− r) ⊆ Xr,c,

then

h(t) = (−1)λℓχ(λℓ−1 = 1)V1 ⊗ · · · ⊗ Vℓ−1 ⊗ Vℓ ⊗ Vℓ−1(U)⊗ Y ⊗ Z

in M(λ, g − 1 + r + c, r, α+ e− 1−Q− r) ⊆ Xr,c, where λ = (λ1, . . . , λℓ−1, λℓ).

Remark. It is clear that h is a map of complexes. In Lemma 5.5 we use the techniqueof Proposition 2.4 to show that h is a quasi-isomorphism.

Up to sign, the map of complexes q : X[+1] → X is given by the module actionof xg on D•G

∗.

Definition 4.8. The map q : Xr,c → Xr−1,c is equal to (−1)α+r+1xg. In otherwords, if

t = V1 ⊗ . . . Vℓ ⊗ U ⊗ Y ⊗ Z ∈M(λ, g − 1 + r + c, r, α+ e− 1−Q− r) ⊆ Xr,c,

34 ANDREW R. KUSTIN

thenq(t) = (−1)α+r+1V1 ⊗ . . . Vℓ ⊗ U ⊗ xg(Y )⊗ Z

in M(λ, g − 1 + r + c, r − 1, α + e − 1 − Q − r) ⊆ Xr−1,c. The map of complexes

q : X[+1]→ X is the composition

X[+1]h−→ X[+1]

q−→ X.

In other words, if

t = V1 ⊗ . . . Vℓ ⊗ U ⊗ Y ⊗ Z ∈M(λ, g + r + c, r, α+ e− 1−Q− r) ⊆ Xr,c,

then

q(t) = (−1)α+r+1+λℓχ(λℓ−1 = 1)V1 ⊗ · · · ⊗ Vℓ−2 ⊗ Vℓ ⊗ Vℓ−1(U)⊗ xg(Y )⊗ Z

in M(λ, g + r − 1 + c, r − 1, α+ e− 1−Q− r) ⊆ Xr−1,c.

Theorem 4.9. The module homomorphism Θ: C[+1] → C of (4.5) is a map of

complexes.

Proof. Our proof has six parts.

Part 1. The diagram

X0,c

"xxx

ψ

#

−−−−−→X0,c−1

⊕N (ℓ− 1 − c, e− c− 1, Q− e+ 1 + c) ⊗ L

θ

?

?

y

hθ xg

i?

?

y

N (ℓ− 2 − c, e− 1 − c,Q− e+ 2 + c) ⊗ Lnnn

−−−−−→ N (ℓ− 1 − c, e− c,Q− e+ 1 + c) ⊗ L

commutes for all c ≥ 1. This calculation is parallel to the proof of Part 1 in Lemma2.7. Fix λ with |λ| = ℓ+ c. Take

t = V1 ⊗ · · · ⊗ Vℓ ⊗ u(g+c) ⊗ 1⊗ Z ∈M(λ, g + c, 0, α+ e− 1−Q) ⊂ X0,c and

t′ = u′(ℓ−1−c)

⊗ y′(e−c)

⊗ Z ′ ⊗ U0 ⊗ Y0 ⊗W0

inM(ℓ− 1− c, e− c, Q− e+ 1 + c)⊗ L∗. We compute

T1 = [(θ ◦ xxx)(t)](t′), T2 = [(xg ◦ ψ)(t)](t′), and T3 = [(nnn ◦ θ)(t)](t′).

We notice that T1, T2, and T3 are all zero, unless λk ≤ 2 for all k and λℓ = 1. (Acomplete proof that T1 6= 0 =⇒ λℓ = 1 involves the idea of (2.8).) We adopt the


notation of (1.18) and let V = u(Vj1)∧· · ·∧u(Vjc). There are two cases to considerdepending upon the value of λℓ−1.

Case 1. Assume λℓ−1 = 1. Definition 1.23 shows that

T1 = ρ4(Vℓ−1 ⊗ Z ∧ Z′ ⊗A⊗ y′

(e−c)⊗ U0 ⊗ Y0 ⊗W0),

whereA = (η ◦ (1⊗www))(u′

(ℓ−c−1)⊗ V1 ⊗ · · · ⊗ Vℓ−2 ⊗ Vℓ ⊗ u

(g+c)).

Apply Lemma 1.20 to see that A = B + C with

B = (−1)J+c(Vi1 · · ·Viℓ−1−c)(u′

(ℓ−c−1)) · V ⊗ u(g) and

C = χ(λℓ = 1)(−1)J+1(Vi1 · · ·Viℓ−c−2)(u′

(ℓ−c−2)) · u′(V ∧ Vℓ)⊗ u

(g).

The term B contributes 0 to T1; and therefore T1 is equal to8

<

:

(−1)1+J (Vi1 · · ·Viℓ−c−2)(u′(ℓ−2−c))·

h

Z ∧ Z′ ∧ (Vℓ−1 ⊗ xg)h

(u′ ⊗ y′) ∧“

[V ∧ Vℓ](U0) ⊲⊳ y′(e−c−1)”

∧ (u(g) ⊲⊳ Y0)ii

(W0),

and this is equal to T ′1 + T ′′

1 , for

T ′1 =

{xg(y

′) · (−1)1+J (Vi1 · · ·Viℓ−c−2· Vℓ−1)(u

′(ℓ−1−c))·[

Z ∧ Z ′ ∧([V ∧ Vℓ](U0) ⊲⊳ y

′(e−c−1))∧ (u(g) ⊲⊳ Y0)

](W0)

and T ′′1 is equal to

8

<

:

(−1)J (Vi1 · · ·Viℓ−c−2)(u′(ℓ−2−c))·

h

Z ∧ Z′ ∧ (u′ ⊗ y′) ∧ (Vℓ−1 ⊗ xg)h“

[V ∧ Vℓ](U0) ⊲⊳ y′(e−c−1)”

∧ (u(g) ⊲⊳ Y0)ii

(W0).

There is no difficulty in seeing that T ′1 + T2 = 0 and T ′′

1 = T3.

Case 2. Assume λℓ−1 = 2. Let V ′ = u(Vj1) ∧ · · · ∧ u(Vjc−1). We see that T3 = 0

because θ(t) = 0,

T1 = [θ((−1)λ1+···+λℓ−2V1 ⊗ · · · ⊗ Vℓ−2 ⊗ u(Vℓ−1)⊗ Vℓ ⊗ u(g+c−1) ⊗ 1⊗ Z)](t′)

=

{(−1)J+c(Vi1 · · ·Viℓ−c−1

)(u′(ℓ−1−c)

)·[Z ∧ Z ′ ∧ (u(Vℓ−1)⊗ xg)

[([V ′ ∧ Vℓ](U0) ⊲⊳ y

′(e−c))∧ (u(g) ⊲⊳ Y0)

]](W0).

The module action of∧•

(E∗⊗G) on∧•

(E ⊗G∗) gives rise to two terms; but oneof the terms involves u(u(Vℓ−1)) and is zero. It follows that

T1 =

{(−1)J+1xg(y

′) · (Vi1 · · ·Viℓ−c−1)(u′

(ℓ−1−c))·[

Z ∧ Z ′ ∧[(

[V ∧ Vℓ](U0) ⊲⊳ y′(e−c−1)

)∧ (u(g) ⊲⊳ Y0)

]](W0).

36 ANDREW R. KUSTIN

An easy calculation shows that T2 = −T1; thus, T1 + T2 = 0 = T3.

Part 2. The diagram of Part 1 continues to commute when c = 0. In this case,T1 = 0 and T2 = T3. One uses (2.9).

Part 3. The diagram

X1,cxxx

−−−−→ X0,c

q

y θ

y

X0,c

eψ−−−−→ N (ℓ− 2− c, e− 1− c, Q− e+ 2 + c)⊗ L

commutes for all c ≥ 0. Take λ, with |λ| = ℓ+ c,

t = V1 ⊗ · · · ⊗ Vℓ ⊗ u(g+c+1) ⊗ y ⊗ Z ∈M(λ, g + c+ 1, 1, α+ e− 2−Q) ⊆ X1,c

and

t′ = u′(ℓ−2−c)

⊗ y′(e−1−c)

⊗ Z ′ ⊗ U0 ⊗ Y0 ⊗W0

inM(ℓ− 2− c, e− 1− c, Q− e+ 2 + c)⊗ L∗. Observe that

A = [(θ ◦ xxx)(t)](t′) and B = [(ψ ◦ q)(t)](t′)

are both zero unless λk ≤ 2, for all k. We adopt the notation of (1.18) and letV = u(Vj1) ∧ · · · ∧ u(Vjc). We see that

A = (−1)c+1[θ(V1 ⊗ · · · ⊗ Vℓ ⊗ u(g+c) ⊗ 1⊗ (u⊗ y) ∧ Z)](t′)

=

8

<

:

(−1)c+1(−1)Jχ(jc < ℓ− 1)(Vi1 · · ·Viℓ−c−2)(u′(ℓ−2−c))·

h

(u⊗ y) ∧ Z ∧ Z′ ∧ (Vℓ−1 ⊗ xg)h“

[V ∧ Vℓ](U0) ⊲⊳ y′(e−1−c)

”

∧ (u(g) ⊲⊳ Y0)ii

(W0).

The element

(u⊗ y) ∧[(

[V ∧ Vℓ](U0) ⊲⊳ y′(e−1−c)

)∧ (u(g) ⊲⊳ Y0)

]

of∧•

(E ⊗G∗) is zero; and therefore, A is equal to

{(−1)1+α+Jχ(jc < ℓ− 1) · (Vℓ−1 ⊗ xg)(u⊗ y) · (Vi1 · · ·Viℓ−c−2

)(u′(ℓ−2−c)

)·[Z ∧ Z ′ ∧

([V ∧ Vℓ](U0) ⊲⊳ y

′(e−1−c))∧ (u(g) ⊲⊳ Y0)

](W0).

On the other hand, B is equal to

{(−1)λℓ+αχ(λℓ−1 = 1) · xg(y) · u(Vℓ−1)·[

ψ(V1 ⊗ · · · ⊗ Vℓ−2 ⊗ Vℓ ⊗ u(g+c) ⊗ 1⊗ Z

)](t′),


and this is equal to A.

Part 4. The diagram

Xr,cxxx

−−−−→ Xr−1,c

q

y q

y

Xr−1,cexxx

−−−−→ Xr−2,c

commutes for all r ≥ 2 and c ≥ 0. Take λ with |λ| = ℓ+ c and

t = V1⊗· · ·⊗Vℓ⊗u(g+c+r)⊗ y(r)⊗Z ∈M(λ, g+ c+ r, r, α+ e− 1−Q− r) ⊆ Xr,c.

We see that (q ◦ xxx)(t) and (xxx ◦ q)(t) are both equal to

{(−1)c+α+λℓχ(λℓ−1 = 1) · xg(y)·(

V1 ⊗ · · · ⊗ u(Vℓ−1)⊗ Vℓ ⊗ u(g+c+r−2) ⊗ y(r−2) ⊗ (u⊗ y) ∧ Z).

Part 5. The diagram

Xr,cxxx

−−−−→ Xr,c−1

q

y q

y

Xr−1,cexxx

−−−−→ Xr−1,c−1

commutes for all r, c ≥ 1. Keep t as in the proof of (4). We compare

A = (q ◦ xxx)(t) and B = (xxx ◦ q)(t).

The left side is

q(ℓ∑i=1

(−1)λ1+···+λi−1χ(λi ≥ 2)V1 ⊗ · · · ⊗ u(Vi)⊗ · · · ⊗ Vℓ ⊗ u(g+c+r−1) ⊗ y(r) ⊗ Z).

The value of this expression depends on i. Nothing tricky occurs when 1 ≤ i ≤ ℓ−2.When i = ℓ − 1, then u(u(Vℓ−1)) = 0 is a factor in the corresponding term. When

i = ℓ, we use the fact that u(Vℓ) ∈∧λℓ−1

E∗. At any rate, A is equal to

=

(−1)α+r+1+λℓχ(λℓ−1 = 1)xg(y)ℓ−2∑i=1

(−1)λ1+···+λi−1χ(λi ≥ 2)·

V1 ⊗ · · · ⊗ u(Vi)⊗ · · · ⊗ u(Vℓ−1)⊗ Vℓ ⊗ u(g+c+r−2) ⊗ y(r−1) ⊗ Z+(−1)α+r+λℓχ(λℓ−1 = 1)xg(y)(−1)λ1+···+λℓ−1χ(λℓ ≥ 2)·

V1 ⊗ · · · ⊗ u(Vℓ−1)⊗ u(Vℓ)⊗ u(g+c+r−2) ⊗ y(r−1) ⊗ Z.

38 ANDREW R. KUSTIN

On the other hand, B is equal to

(−1)α+r+1+λℓχ(λℓ−1 = 1)xg(y)xxx(V1⊗· · ·⊗u(Vℓ−1)⊗Vℓ⊗u(g+c+r−1)⊗y(r−1)⊗Z)

=

(−1)α+r+1+λℓχ(λℓ−1 = 1)xg(y)ℓ−2∑i=1

(−1)λ1+···+λi−1χ(λi ≥ 2)·

V1 ⊗ · · · ⊗ u(Vi)⊗ · · · ⊗ u(Vℓ−1)⊗ Vℓ ⊗ u(g+c+r−2) ⊗ y(r−1) ⊗ Z+(−1)α+r+1+λℓχ(λℓ−1 = 1)xg(y)(−1)λ1+···+λℓ−2χ(λℓ ≥ 2)·

V1 ⊗ · · · ⊗ u(Vℓ−1)⊗ u(Vℓ)⊗ u(g+c+r−2) ⊗ y(r−1) ⊗ Z.

Thus, A = B.

Part 6. It is clear that the diagram

N (a, b, c)nnn

−−−−→ N (a+ 1, b+ 1, c− 1)

xg

y xg

y

N (a, b+ 1, c)nnn

−−−−→ N (a+ 1, b+ 2, c− 1)

commutes for all integers a, b, c. �

Definition 4.10. Let (A, aaa) be the mapping cone of Θ: C[+1]→ C. So,

Ap = C[+1]p−1 ⊕ Cp = Cp ⊕ Cp =

Xp

⊕N (ℓ− 2− p, e− 2− p,Q− e+ 2 + p)⊗ L

⊕Xp

⊕N (ℓ− 3− p, e− 2− p,Q− e+ 3 + p)⊗ L,

and aaap : Ap → Ap−1 is given by

aaap =

[cccp 0Θ −cccp

]=

xxx 0 0 0ψ nnn 0 0q 0 −xxx 0

θ xg −ψ −nnn

.

5. Complexes homologically equivalent to A .We introduce a complex (B, bbb) which is homologically equivalent to, but signifi-

cantly smaller than, A. First, we identify a split exact subcomplex P of A, then wedefine B and exhibit a quasi-isomorphism H : A/P→ B. The idea of H is the sameas the idea of h in Lemma 1.12 and Proposition 2.4, except h removes all ℓ factors


of∧•E∗ from the complex W( , ℓ) and H only removes one factor of

∧•E∗. Of

course, the construction of B and H are somewhat delicate because B still capturesthe relevant data about a map from the split down version of X to the split downversion of N. We identify a subcomplex D of B with B/D split exact. It followsthat the complexes D and A are homologically equivalent. Finally, we show that D

is isomorphic to a direct sum of complexes built using the free module G, of rankg − 1, in place of the free module G.

We define a subcomplex (P, ppp) of the complex (A, aaa) of definition 4.10.

Definition 5.1. For each integer p, let Pp be the submodule

N (ℓ− 2− p, e− 2− p,Q− e+ 2 + p)⊗ L⊕

N+(ℓ− 3− p, e− 2− p,Q− e+ 3 + p)⊗ L

of Ap, (the top summand of Pp is a summand of Cp and the bottom summand of

Pp is a summand of Cp) and let pppp : Pp → Pp−1 be

pppp =

[nnn 0xg −nnn

].

It is clear that P is a subcomplex of A; furthermore, each differential in P hasthe form [

∗ 0∼= ∗

];

and therefore, P is split exact. The complex A is split exact if and only if A/P issplit exact. For future reference we record that (A/P)p is equal to

Xp

⊕Xp

⊕N 0(ℓ− 3− p, e− 2− p,Q− e+ 3 + p)⊗ L

and the differential (A/P)p → (A/P)p−1 is given by

xxx 0 0q −xxx 0

θ0 −ψ0 −nnn0

,

where nnn0 is the map N /N+ → N /N+, which is induced by nnn : N → N and θ0 and

ψ0 are θ and ψ, respectively, followed by the natural quotient map

N (ℓ− 2− p, e− 1− p,Q− e+ 2 + p)⊗ Lquot−−−→ N /N+ ⊗ L = N 0 ⊗ L.

40 ANDREW R. KUSTIN

In other words, if t ∈ Xp, then θ0(t) is the restriction of

θ(t) :M(ℓ− 2− p, e− 1− p,Q− e+ 2 + p)⊗ L∗ → R

toM0(ℓ− 2− p, e− 1− p,Q− e+ 2 + p)⊗ L∗ and the analogous statement holds

for ψ0.We next exhibit a quasi-isomorphism from A/P to a complex “B”. The advantage

of B is that all of the modules∧λ

E∗ will have ℓ − 1 factors rather than ℓ factors.Our first step is to define the homomorphism θ which is similar to θ from Definition4.6 and is built using the homomorphism ξ2 of Definition 1.23. Recall the complex(X, xxx) from Definition 4.7.

Definition 5.2. Define a module homomorphism

θ : X[+1]→ N0(P, Q)[Q+ 2− e]⊗ L.


M = M(λ, g − 1 + r + c, r, α+ e− 1−Q− r) ⊂ Xr,c ⊂ Xp.

We defineθ : M→ N 0(ℓ− p− 1, e− p− 1, p− e+ 2 +Q)⊗ L

by setting Adj(θ) = (−1)e−cχ(r = 0) · ξ02 , where ξ02 is the restriction of the homo-morphism

ξ2 : M⊗M(ℓ− p− 1, e− p− 1, p− e+ 2 +Q)⊗ L∗ → R

to M⊗M0(ℓ− p− 1, e− p− 1, p− e+ 2 +Q)⊗ L∗ .

Definition 5.3. For each integer p, let Bp be the module

Xp

⊕Xp

⊕N 0(ℓ− 3− p, e− 2− p,Q− e+ 3 + p)⊗ L

and bbbp : Bp → Bp−1 be the mapxxx 0 0q −xxx 0

θ −ψ0 −nnn0

.

Define Hp : (A/P)p → Bp to be the module homomorphismh 0 00 1 00 0 1

,

where the map of complexes h : X→ X is given in Definition 4.7.


Lemma 5.4. The module homomorphism H : A/P→ B is a quasi-isomorphism of

complexes.

Proof. We first show that (B, bbb) is a complex. Most of the composition bbbp−1 ◦ bbbpis seen to be zero without any difficulty. The only interesting calculation is thecomposition

Xp

2664

xxxqθ

3775

−−−→ Bp−1

hθ −ψ0 −nnn0

i

−−−−−−−−−−−−→ N 0(ℓ− 1− p, e− p,Q− e+ 1 + p)⊗ L.

Take (r, c) with r + c = p, λ a ℓ-tuple of positive integers with |λ| = ℓ+ c,

t = V1 ⊗ · · · ⊗ Vℓ ⊗ u(g−1+r+c) ⊗ y(r) ⊗ Z

∈M(λ; g − 1 + r + c; r;α+ e− 1−Q− r) ⊂ Xr,c ⊂ Xp, and

t′ = u′(ℓ−1−p)

⊗ y′(e−p)

⊗ Z ′ ⊗ U0 ⊗ Y0 ⊗W0

∈ M0(ℓ− 1− p, e− p,Q− e+ 1 + p)⊗ L∗.

We compute

T1 = [(θ ◦ xxx)(t)](t′), T2 = [(−ψ0 ◦ q)(t)](t′), and T3 = [(−nnn0 ◦ θ)(t)](t′).

All three terms are zero unless λk ≤ 2 for all k. (The assertion about T1 followsonce we apply Lemma 1.20.) Adopt the notation of (1.18) and let

V = u(Vj1) ∧ · · · ∧ u(Vjc).

We have

T3 = −[θ(t)](u′

(ℓ−2−p)⊗ y′

(e−1−p)⊗ Z ′ ∧ (u′ ⊗ y′)⊗ U0 ⊗ Y0 ⊗W0

)=

{χ(r = 0)χ(λℓ = 1)(−1)e−c+J+1(Vi1 · · ·Viℓ−c−1

)(u′(ℓ−2−c)

)·[Z ∧ Z ′ ∧ (u′ ⊗ y′) ∧

([V ∧ Vℓ](U0) ⊲⊳ y

′(e−1−c))∧ (u(g−1) ⊲⊳ xg(Y0))

](W0)

and

T2 = (−1)α+rxg(y)[ψ0(V1 ⊗ · · · ⊗ Vℓ ⊗ u

(g−1+r+c) ⊗ y(r−1) ⊗ Z)](t′)

=

{χ(r = 1)χ(λℓ = 1)(−1)α+r+Jxg(y)(Vi1 · · ·Viℓ−c−1

)(u′(ℓ−2−c)

)·[Z ∧ Z ′ ∧

([V ∧ Vℓ](U0) ⊲⊳ y

′(e−c−1))∧ (u(g) ⊲⊳ Y0)

](W0).

42 ANDREW R. KUSTIN

We see that T1 = T ′1 + T ′′

1 , where

T ′1 = [(θ ◦ (www ⊗ 1⊗ 1))(t)](t′) and T ′′

1 = (−1)r+c[(θ ◦ (1⊗mmm))(t)](t′);

furthermore, T ′′1 is equal to

{χ(r = 1)χ(λℓ = 1)(−1)r+J+e(Vi1 . . . Viℓ−c−1

)(u′(ℓ−1−p)

)[(u⊗ y) ∧ Z ∧ Z ′ ∧

([V ∧ Vℓ](U0) ⊲⊳ y

′(e−p))∧ (u(g−1) ⊲⊳ xg(Y0))

](W0),

and therefore, T ′′1 + T2 = 0. The element (www ⊗ 1⊗ 1)(t) is in Xr,c−1; thus,

T ′1 = (−1)e−(c−1)χ(r = 0) · ρ3(Z ∧ Z

′ ⊗ A⊗ y′(e−c)

⊗ U0 ⊗ Y0 ⊗W0),

for

A = η(u′(ℓ−c)

⊗www(V1 ⊗ · · · ⊗ Vℓ ⊗ u(g−1+c)))

and ρ3 given in Definition 1.23. Apply Lemma 1.20 to see that A = B + C for

B = (−1)J+c(Vi1 · · ·Viℓ−c)(u′

(ℓ−1−c)) · V ⊗ u(g−1) and

C = χ(λℓ = 1)(−1)J+1(Vi1 · · ·Viℓ−c−1)(u′

(ℓ−c−2)) · u′(V ∧ Vℓ)⊗ u

(g−1).

The idea of (2.8) shows that the contribution of B to T ′1 contains a factor of

(u⊗ y′) ∧ (u(g−1) ⊲⊳ xg(Y0)) = y′(xg) · (u(g) ⊲⊳ Y0),

and this is zero because y′ ∈ G∗. Thus,

T ′1 =

{χ(r = 0)χ(λℓ = 1)[(−1)e−c+J(Vi1 · · ·Viℓ−c−1

)(u′(ℓ−2−c)

)[Z ∧ Z ′ ∧

((u′[V ∧ Vℓ])(U0) ⊲⊳ y

′(e−c))∧(u(g−1) ⊲⊳ xg(Y0)

)](W0).

It follows that T ′1 + T3 = 0 and (B, bbb) is a complex.

We next show that H is a map of complexes. The map q of Definition 4.8 isdefined to be q ◦ h. We need only verify that θ0 = θ ◦ h. Take (r, c) with r+ c = p,λ with |λ| = ℓ+ c,

t = V1 ⊗ · · · ⊗ Vℓ ⊗ u(g+r+c) ⊗ y(r) ⊗ Z

∈M(λ, g + r + c, r, α+ e− 1−Q− r) ⊂ Xr,c ⊂ Xp, and

t′ = U ′ ⊗ Y ′ ⊗ Z ′ ⊗ U0 ⊗ Y0 ⊗W0

∈M0(ℓ− 2− r − c, e− 1− r − c, Q− e+ 2 + r + c)⊗ L∗.


Observe that (θ ◦ h)(t) and θ0(t) are both zero unless r = 0 and λk ≤ 2 for allk. Adopt the notation of (1.18) and let V = u(Vj1) ∧ · · · ∧ u(Vjc). We see that[θ0(t)](t′) = [θ(t)](t′)

=

{(−1)Jχ(r = 0)χ(λℓ−1 = λℓ = 1)(Vi1 · · ·Viℓ−c−2

)(U ′)·[Z ∧ Z ′ ∧ (Vℓ−1 ⊗ xg)

[([V ∧ Vℓ](U0) ⊲⊳ Y

′) ∧ (u(g) ⊲⊳ Y0)]]

(W0).

The element t′ is inM0 and therefore, xg(Y′) = 0; and [θ0(t)](t′) is equal to

{Vℓ−1(u) · (−1)J+e−1−cχ(r = 0)χ(λℓ−1 = λℓ = 1)(Vi1 · · ·Viℓ−c−2

)(U ′)·[Z ∧ Z ′ ∧ ([V ∧ Vℓ](U0) ⊲⊳ Y

′) ∧ (u(g−1) ⊲⊳ xg(Y0))](W0).

On the other hand, h(t) is equal to

Vℓ−1(u) · (−1)λℓχ(r = 0)χ(λℓ−1 = 1)V1 ⊗ · · · ⊗ Vℓ−1 ⊗ Vℓ ⊗ u(g+c−1) ⊗ 1⊗ Z

in M(λ, g − 1 + c, 0, α+ e− 1−Q) ⊆ X0,c and [(θ ◦ h)(t)](t′) is equal to

{Vℓ−1(u) · (−1)λℓχ(r = 0)χ(λℓ−1 = 1)(−1)J+e−cχ(λℓ = 1)(Vi1 · · ·Viℓ−c−2

)(U ′)·[Z ∧ Z ′ ∧ ([V ∧ Vℓ](U0) ⊲⊳ Y

′) ∧ (u(g−1) ⊲⊳ xg(Y0))](W0).

We have established that H is a map of complexes.

We conclude by proving that H is a quasi-isomorphism. Each map Hp is surjec-tive. One may prove that H is a quasi-isomorphism by showing that the kernel ofH is a split exact complex. The kernel of H is equal to the kernel of the map ofcomplexes h : X→ X. Each map hp is surjective and Lemma 5.5 shows that h is aquasi-isomorphism. �

Lemma 5.5. In the language of 4.7, the map of complexes h : X → X is a quasi-

isomorphism.

Proof. We show that the mapping cone of h is split exact. For each pair of integers(p, q), with 1 ≤ q and −1 ≤ p, let Tp,q be the following submodule of X:

Tp,q =⊕

r≥0

⊕

λ

M(λ, g + p+ q, r, α+ e− 1−Q− r),

where the inner sum is taken over all λ with

|λ| = p+ q − r + ℓ and λℓ−1 = q.

Fix a summand M = M(λ, a, b, d) of X. Let r = b and q = λℓ−1. It is not difficultto see that, if c = p+ q − b, then

M is a summand of Tp,q ⇐⇒ M is a summand of Xr,c.

44 ANDREW R. KUSTIN

It follows that if n is fixed, then

⊕

r+c=n

Xr,c =⊕

p+q=n

Tp,q .

We claim that the mapping cone of h : X→ X is equal to the total complex of

......

......

yy

y xxx

y

. . . −−−−→ T1,3 −−−−→ T1,2 −−−−→ T1,1h

−−−−→ X2yy

y xxx

y

. . . −−−−→ T0,3 −−−−→ T0,2 −−−−→ T0,1h

−−−−→ X1yy

y xxx

y

. . . −−−−→ T−1,3 −−−−→ T−1,2 −−−−→ T−1,1h

−−−−→ X0.

This is correct because the composition

Tp,qincl−−→ Xp+q

h−→ Xp+q

is zero unless q = 1 and the composition

Tp,qincl−−→ Xp+q

xxx−→ Xp+q−1

lands in Tp−1,q ⊕Tp,q−1. We also claim that each row in the above double complexis split exact. Indeed, the row

. . . −→ Tp,2 −→ Tp,1h−→ Xp+1 → 0

is isomorphic to the complex

(5.6)

(WE(g + p+ 1, 1)

h−→ Dg+pE → 0

)⊗⊕

r≥0

⊕

λ

M(λ, 0, r, α+ e− 1−Q− r),

where λ varies over all ℓ-tuples of positive integers with |λ| = p − r + ℓ + 1. Wehave identified the summand M(λ, g + p+ q, r, α+ e− 1−Q− r) of Tp,q with thesummand

∧λℓ−1E∗ ⊗Dg+p+qE ⊗M((λ1, . . . , λℓ−1, λℓ), 0, r, α+ e− 1−Q− r)


of (5.6). The complex on the left side of (5.6) is the split exact complex of (1.13),with g + p+ 1 playing the role of a. �

The module decompositions M(λ,m, n, p) = M0(λ,m, n, p) ⊕ M+(λ,m, n, p)from Definition 2.13 give rise to a module decomposition

X = X0 ⊕ X

+.

We observe that the map of complexes q : X[+1] → X of Definition 4.8 is the zeromap on X0 and is an isomorphism on X+.

Definition 5.7. For each integer p, let Dp be the module

X0p

⊕N 0(ℓ− 3− p, e− 2− p,Q− e+ 3 + p)⊗ L

and dddp : Dp → Dp−1 be the map

xxx 0

θ −nnn0

.

It is clear that (D, ddd) is a subcomplex of (B, bbb) and that the quotient complexB/D is a split exact complex of the form

· · · →X

+2

⊕X2

"∗ 0∼= ∗

#

−−−−−−→X

+1

⊕X1

[∼= ∗ ]−−−−−→ X0 → 0.

Lemma 5.8. The complex (D, ddd) is isomorphic to the following direct sum of com-

plexes:e⊕

w=0

CE,G(P − w,Q− w + 1)⊗ (∧e−w

E ⊲⊳ y(e−w)g ).

Proof. Let g = g − 1, ℓ = ℓ− 1, λ represent an ℓ-tuple of positive integers,

α = (e− 1)(g − 1) = α− e+ 1,

L =∧eE∗ ⊗

∧g−1G⊗

∧e(g−1)(E ⊗G

∗),

M(λ, a, b, d) be the submodule

∧λE∗ ⊗DaE ⊗DbG

∗⊗∧d

(E ⊗G∗)

46 ANDREW R. KUSTIN

of M(λ, a, b, d), and N (a, b, d) be the submodule

SymaE∗ ⊗ SymbG⊗

∧d(E∗ ⊗G)

of N (a, b, d). Fix (r, c) with r + c = p, and an ℓ-tuple of positive integers λ with|λ| = ℓ+ c. The summand

M = M(λ, g − 1 + r + c, r, α+ e− 1−Q− r)

of X0r,c ⊆ X0

p is isomorphic to

e⊕

w=0

M(w)⊗ (∧e−w

E ⊲⊳ y(e−w)),

whereM(w) = M(λ, g − 1 + r + c, r, α+ e− 1−Q− r − (e− w))

is equal to the summand

M(λ, g − 1 + r + c, r, α+ e− 1− (Q− w + 1)− r)

of XE,G(P − w,Q− w + 1)r,c. The module N 0(a, b, d) is equal to

e⊕

w=0

N (a, b, d− w)⊗ (∧w

E∗ ⊲⊳ x(w)g ).

The module L is equal to L⊗Rxg ⊗ (∧eE ⊲⊳ y

(e)g ). So the summand

N 0(ℓ− 3− p, e− 2− p,Q− e+ 3 + p)⊗ L

of Dp is equal to

e⊕

w=0

N (w)⊗ L⊗ (∧w

E∗ ⊲⊳ x(w)g )⊗Rxg ⊗ (

∧eE ⊲⊳ y(e)

g ),

where N (w)⊗ L is the summand

N (ℓ− 2− p, e− 2− p, (Q+ 1− w)− e+ 2 + p)⊗ L

of CE,G(P − w,Q− w + 1)p.


The differential of CE,G is

ccc =

[xxx 0ψ nnn

].

It is clear that nnn0 restricts to become nnn and xxx restricts to become xxx. There is nodifficulty in seeing that θ carries

A = M(w)⊗ (∧e−w

E ⊲⊳ y(e−w))

intoB = N (w)⊗ L⊗ (

∧wE∗ ⊲⊳ x(w)

g )⊗Rxg ⊗ (∧eE ⊲⊳ y(e)

g ).

Observe that the composition

A⊗B∗ nat−−→M(w)⊗

(N (w)⊗ L⊗Rxg

)∗ xg

−→M(w)⊗(N (w)⊗ L

)∗ ξ1−→ R

agrees, up to sign, with

A⊗B∗ incl−−→M⊗ (N ⊗ L)∗

ξ02−→ R.

Recall that Adj(ψ) = χ(r = 0)ξ1 and Adj(θ) = (−1)e−cχ(r = 0)ξ2. �

6. Homology generator.

We know from (0.8) that

HN (g − 1, e− 1, α) ∼= HM(0, 0, 0) = R.

We use the quasi-isomorphisms

Mφ←− Y

ψ−→ N⊗ L

to identify a cycle of N (g−1, e−1, α) which represents a generator of the homologyHN (g − 1, e − 1, α). Fix an element s ∈ Symg−1E

∗ ⊗ Dg−1E which is sent to 1under the evaluation map. We define ζs ∈ N (g − 1, e − 1, α) ⊗ L as follows. Foreach

(6.1) t′ = U ′ ⊗ Y ′ ⊗ Z ′ ⊗ U0 ⊗ Y0 ⊗W0 ∈M(g − 1, e− 1, α)⊗ L∗,

we haveζs(t

′) = [Z ′ ∧Ψ(U0 ⊗ Y′ ⊗ (U ′ ⊗ 1)(s)⊗ Y0)](W0),

where the map

Ψ:∧eE ⊗De−1G

∗ ⊗Dg−1E ⊗∧g

G∗ →∧e+g−1

(E ⊗G∗)

is defined in Observation 3.2. The cycle ζs in N (g− 1, e− 1, α)⊗L depends on thechoice of s; but the homology class of ζs in HN (g − 1, e− 1, α)⊗ L is independentof s.

48 ANDREW R. KUSTIN

Corollary 6.2. The element ζs is a cycle in N ⊗ L which represents a generator

of HN (g − 1, e− 1, α)⊗ L.

Proof. Set P = e(g − 1), Q = (e− 1)g, and ℓ = g. Let λ be the g-tuple (1, . . . , 1).In the notation of (1.9) we have:

M(λ, g, 0, 0) = X0,0ϕ

−−−−→ M(0, 0, 0) = Mα

ψ

y

N (g − 1, e− 1, α)⊗ L = Nα ⊗ L.

Let t be any element of TgE∗ ⊗DgE which is sent to s under the map

TgE∗ ⊗DgE → Symg−1E

∗ ⊗Dg−1E,

with v1 ⊗ · · · ⊗ vg ⊗ U 7→ v1 · · ·vg−1 ⊗ vg(U). In other words,

t =∑

i

vi,1 ⊗ · · · ⊗ vi,g ⊗ Ui

and s =∑

i vi,1 · · · vi,g−1⊗vi,g(Ui). We see that ϕ(t) =∑i(vi,1 · · · vi,g−1 · vi,g)(Ui),

which is 1, by the choice of s. It follows that ψ(t) is a cycle of N⊗ L and that thehomology class of ψ(t) generates HN (g − 1, e− 1, α)⊗ L. On the other hand, ψ(t)sends the element t′ of (6.1) to

∑

i

(vi,1 · · · vi,g−1)(U′) · [Z ′ ∧ (vi,g(U0) ⊲⊳ Y

′) ∧ (Ui ⊲⊳ Y0)](W0)

= (−1)e+1∑

i

(vi,1 · · · vi,g−1)(U′) · [Z ′ ∧Ψ(U0 ⊗ Y

′ ⊗ vi,g(Ui)⊗ Y0)](W0)

= (−1)e+1[Z ′ ∧Ψ(U0 ⊗ Y′ ⊗ (U ′ ⊗ 1)(s)⊗ Y0)](W0) = (−1)e+1ζs(t

′). �

7. Homogeneity.

The complexes M(P,Q) and N(P,Q) each have an enormous amount of ho-mogeneity. This homogeneity passes to the homology modules HM(m,n, p) andHN (m,n, p), and it even passes across the isomorphism (0.8); see Theorem 7.1.One consequence is Corollary 7.2 which says that the homology modules

HM(m,n, p) and HN (m,n, p)

satisfy an extra duality when e is equal to three. This duality is translated inCorollary 7.3 to give an extra symmetry in the graded betti numbers of the rankone reflexive modules of the determinantal ring defined by the 2 × 2 minors of ageneric 3× g matrix.


Notation. Fix bases

x1, . . . , xg for G; y1, . . . , yg for G∗; u1, . . . , ue for E; and v1, . . . , ve for E∗,

with {xi} dual to {yj}, and {ui} dual to {vj}. Let N = va1

1 · · · vaee be a monomial

of degree m + p and M = xc11 · · ·xcgg be a monomial of degree n + p. Define

N (m,n, p)|N,M to be the submodule of N (m,n, p) which consists of those elementswhich are homogeneous of degree ai in vi and of degree cj in xj for all i andall j. The submodules N (m,n, p)|N and N (m,n, p)|M , homogeneous in just the{vi} or just the {xj}, are defined in an analogous manner. The differential of N

is homogeneous in the u’s and y’s; and therefore, the complex N(P,Q) naturallydecomposes into a direct sum of subcomplexes N(P,Q)|N,M , where the sum is takenover all monomialsN andM of degree P andQ, respectively. Take HN (m,n, p)|N,Mto mean the homology of the complex N(m+ p, n+ p)|N,M at N (m,n, p)|N,M . Wesee that HN (m,n, p) is equal to

⊕N,M HN (m,n, p)|N,M . In a similar manner,

HM(m′, n′, p′) is equal to⊕

N ′,M ′ HM(m′, n′, p′)|N ′,M ′ , where the sum is taken

over all monomials N ′ and M ′ of degree m′ + p′ in the variables u1, . . . , ue, anddegree n′ + p′, in the variables y1, . . . , yg, respectively

Suppose that the triples (m,n, p) and (m′, n′, p′) satisfy

m+m′ = g − 1, n+ n′ = e− 1, p+ p′ = α, and 1− e ≤ m− n ≤ g − 1.

In this case, (0.8) assures us that

HN (m,n, p) ∼= HM(m′, n′, p′).

Furthermore, we know that

HN (m,n, p) =⊕

N,M

HN (m,n, p)|N,M and

HM(m′, n′, p′) =⊕

N ′,M ′

HM(m′, n′, p′)|N ′,M ′ ,

asN varies over all monomials of degreem+p in the v’s,M varies over all monomialsof degree n + p in the x’s, N ′ varies over all monomials of degree m′ + p′ in theu’s, and M ′ varies over all monomials of degree n′ + p′ in the y’s. It is natural towonder how a particular pair

HN (m,n, p)|N,M and HM(m′, n′, p′)|N ′,M ′

are related.

50 ANDREW R. KUSTIN

Theorem 7.1. Adopt the notation and hypotheses of (0.8). Let N0 = va1

1 · · · vaee ,

N ′0 = ub11 · · ·u

bee , M0 = xc11 · · ·x

cgg , and M ′

0 = yd11 . . . ydgg , with

∑ai = m + p,∑

bi = m′ + p′,∑ci = n+ p, and

∑di = n′ + p′. If

ai + bi = g − 1 and cj + dj = e− 1

for all i and j, then

HN (m,n, p)|N0,M0∼= HM(m′, n′, p′)|N ′

0,M′

0.

Proof. Let P = m + p, Q = n + p, P ′ = m′ + p′, and Q′ = n′ + p′. Observe thatP + P ′ = e(g − 1) and Q + Q′ = (e − 1)g. Let N = N(P,Q), Y = Y(P,Q), andM = M(P ′, Q′)[−α] as described in (1.8). We decompose Y into a direct sum ofsubcomplexes which interact nicely with the quasi-isomorphisms

N⊗ Lψ←− Y

ϕ−→M.

Fix a summand M = M(λ, g + r + c, r, α + e − 1 − Q − r) of Y and monomials

N ′ = uβ1

1 · · ·uβee and M ′ = yδ11 · · · y

δgg of degree P ′ and Q′, respectively. Observe

that each element of M has total degree |λ| in the v’s, total degree P ′ + |λ| in theu’s, and total degree Q′ in the y’s. Define M|N ′,M ′ to be the submodule of M whichconsists of all homogeneous elements for which the degree in vi minus the degree inui is βi and the degree in yj is γj, for all i and all j. It is clear that Y decomposesinto a direct sum of complexes

⊕Y|N ′,M ′ and that ϕ carries Y|N ′,M ′ onto M|N ′,M ′

for all pairs {N ′,M ′}.We complete the argument by showing that

ψ(M|N ′

0,M′

0) ⊆ N|N0,M0

⊗ L.

Taket = V1 ⊗ . . . Vℓ ⊗ U ⊗ Y ⊗ Z ∈M|N ′

0,M′

0.

We may replace t by a summand of t, if necessary, in order to assume that thereexist ǫ1, . . . , ǫe, so that t is homogeneous of degree ǫi in vi, for all i. It follows thatt is homogeneous of degree ǫi + bi in ui and t is homogeneous of degree dj in yj ,

for all i and j. Let N ′ = ub′11 · · ·u

b′ee and M ′ = y

d′11 · · · y

d′gg and take

t′ = U ′ ⊗ Y ′ ⊗ Z ′ ⊗ U0 ⊗ Y0 ⊗W0 ∈M(m,n, p)|N ′,M ′ ⊗ L∗.

So, U ′⊗Y ′⊗Z ′ is homogeneous of degree b′i in ui and of degree d′j in yj for all i and j.Of course, U0 is homogeneous of degree one in each ui; Y0 is homogeneous of degreeone in each yj ; and W0 is homogeneous of degree g in each vi and homogeneous ofdegree e in each xj . We see, from Definition 1.23 or Remark 1.24, that [ψ(t)](t′) ishomogeneous of degree

(ǫi + bi) + b′i + 1− ǫi − g = bi + b′i + 1− g

in each ui and is homogeneous of degree dj+d′j+1−e in yj , for all i and j. Thus, ψ(t)

sends M(m,n, p)|N ′,M ′ ⊗ L∗ to zero for all pairs {N ′,M ′}, except {N0,M0}. �


Corollary 7.2. (a) Assume that e = 3 and g is arbitrary. If P and Q are integers

with Q− 2 ≤ P ≤ 2Q− 1, then the homology of M(P,Q) satisfies

HM(m,n, p) ∼= HN (m′, n′, p′),

provided

m+m′ = Q− 1, n+ n′ = 2, and p+ p′ = 2Q− 2,

for m+ p = P and n+ p = Q.

(b) Assume that g = 3 and e is arbitrary. If P and Q are integers with P − 2 ≤Q ≤ 2P − 1, then the homology of M(P,Q) satisfies

HM(m,n, p) ∼= HN (m′, n′, p′),

provided

m+m′ = 2, n+ n′ = P − 1, and p+ p′ = 2P − 2,

for m+ p = P and n+ p = Q.

Proof. We prove (a). A symmetric argument establishes (b). The R-moduleHM(m,n, p) is isomorphic to the direct sum of the modules HM(m,n, p)|M asM varies over all of the monomials of degree Q in the variables y1, . . . , yg. IfM = yc11 · · · y

cgg , then Theorem 7.1 tells us that HM(m,n, p)|M is zero unless ci ≤ 2

for all i. For each pair of integers (a, b), with a+2b = Q, let Ma,b be the monomial

y1 · · · ya · y2a+1 · · ·y

2a+b.

If M is any other monomial which consists of a variables raised to the power oneand b variables raised to the power 2, then

HM(m,n, p)|M and HM(m,n, p)|Ma,b

are isomorphic as R-modules. Let εa,b equal the number of monomials in thevariables y1, . . . , yg which consist of a variables raised to the power one and bvariables raised to the power 2. The exact value of εa,b is not needed in this proof;on the other hand, it is obvious that the value is given by

εa,b =

{g!

a!b!(g−a−b)! if a+ b ≤ g

0 if g < a+ b.

We have shown that

HM(m,n, p) ∼=⊕

a+2b=Q

(HomM(m,n, p)|Ma,b

)εa,b .

52 ANDREW R. KUSTIN

The exact same analysis may be applied to HN (m′, n′, p′). We notice that thehypotheses ensure that n′ + p′ is also equal to Q. Let

M ′a,b = x1 · · ·xa · x

2a+1 · · ·x

2a+b.

It follows that

HN (m′, n′, p′) ∼=⊕

a+2b=Q

(HomN (m′, n′, p′)|M ′

a,b

)εa,b

.

To complete the proof, we apply Theorem 7.1 with G replaced by the free moduleof rank Q whose basis is x1, . . . , xQ. It does not matter whether (x1, . . . , xQ)is a submodule or an extension of the original G. The hypothesis ensures that−2 ≤ P −Q ≤ Q− 1. We conclude

HomM(m,n, p)|Ma,b∼= HomN (m′, n′, p′)|M ′ ,

for M ′ = x1 · · ·xa · x2a+b+1 · · ·x

2a+2b. Notice that for each i, with 1 ≤ i ≤ Q, the

exponent of yi in Ma,b plus the exponent of xi in M ′ is equal to 2. �

Take R to be a field (of arbitrary characteristic). We use (0.4) to translateCorollary 7.2 into the language of (0.3). Our base ring is a field, so HN (m,n, p) ∼=HM(m,n, p). If M is a graded P-module, X : · · · → X1 → X0 → M is a minimalhomogeneous P-free resolution of M , and Xp is equal to

⊕i P(−i)βp,i(M), then

the graded betti number βp,q(M) is equal to the dimension of the vector space

TorPp,q(M,R).

Corollary 7.3. Adopt the data of (0.3) with R equal to a field.

(a) Assume that e = 3. If ℓ and q are integers with −2 ≤ ℓ ≤ q − 1, then

βp,q(Mℓ) = βp′,q(Mℓ′)

for ℓ+ ℓ′ = q − 3 and p+ p′ = 2q − 2.(b) Assume that g = 3. If ℓ and q are integers with ℓ ≤ 2 and 1− 2ℓ ≤ q, then

βp,q(Mℓ) = βp′,q′(Mℓ′),

provided

ℓ+ ℓ′ = 3− ℓ− q, q + q′ = 3(ℓ+ q − 1), and p+ p′ = 2(ℓ+ q − 1).


Example. Retain the hypotheses of Corollary 7.3 with e = g = 3. The gradedbetti numbers of the P-modules Mℓ, for −5 ≤ ℓ ≤ 5 are

‡β0,5(M−5) = 21 ‡β0,4(M−4) = 15 ‡β0,3(M−3) = 10 β0,2(M−2) = 6‡β1,6(M−5) = 105 ‡β1,5(M−4) = 72 ‡β1,4(M−3) = 45 β1,3(M−2) = 24‡β2,7(M−5) = 216 ‡β2,6(M−4) = 141 ‡β2,5(M−3) = 81 β2,4(M−2) = 36‡β3,8(M−5) = 234 ‡β3,7(M−4) = 144 ‡β3,6(M−3) = 74 β3,5(M−2) = 24‡β4,9(M−5) = 141 ‡β4,8(M−4) = 81 β4,7(M−3) = 36 β4,6(M−2) = 6‡β5,10(M−5) = 45 β5,9(M−4) = 24 β5,8(M−3) = 9β6,11(M−5) = 6 β6,10(M−4) = 3 β6,9(M−3) = 1

β0,1(M−1) = 3 β0,0(M0) = 1 β0,0(M1) = 3 β0,0(M2) = 6β1,2(M−1) = 9 β1,2(M0) = 9 β1,1(M1) = 9 β1,1(M2) = 24β2,3(M−1) = 6 β2,3(M0) = 16 β2,2(M1) = 6 β2,2(M2) = 36β2,4(M−1) = 6 β3,4(M0) = 9 β2,3(M1) = 6 β3,3(M2) = 24β3,5(M−1) = 9 β4,6(M0) = 1 β3,4(M1) = 9 β4,4(M2) = 6β4,6(M−1) = 3 β4,5(M1) = 3

‡β0,0(M3) = 10 ‡β0,0(M4) = 15 ‡β0,0(M5) = 21‡β1,1(M3) = 45 ‡β1,1(M4) = 72 ‡β1,1(M5) = 105‡β2,2(M3) = 81 ‡β2,2(M4) = 141 ‡β2,2(M5) = 216‡β3,3(M3) = 74 ‡β3,3(M4) = 144 ‡β3,3(M5) = 234β4,4(M3) = 36 ‡β4,4(M4) = 81 ‡β4,4(M5) = 141β5,5(M3) = 9 β5,5(M4) = 24 ‡β5,5(M5) = 45β6,6(M3) = 1 β6,6(M4) = 3 β6,6(M5) = 6.

These numbers were calculated by the computer program Macaulay. There are foursymmetries running through this set of numbers. The R-modules E∗ and G areisomorphic because e = g; hence, the R-module automorphism of P, which sendsthe matrix of indeterminates Z to ZT, induces the relation

(7.4) βp,q(Mℓ) = βp,q+ℓ(M−ℓ),

for all integers p, q, and ℓ. If −2 ≤ ℓ ≤ 2, then

(7.5) TorPp,q(Mℓ,KKK) ∼= TorPp′,q′(Mℓ′ ,KKK),

whenever 1− e ≤ ℓ ≤ g− 1, ℓ+ ℓ′ = g− e, p+ p′ = (e− 1)(g− 1), q+ q′ = (e− 1)g.gives

βp,q(Mℓ) = βp′,q′(Mℓ′) for ℓ+ ℓ′ = 0, p+ p′ = 4, and q + q′ = 6.

The other two symmetries are listed in Corollary 7.3. We have marked (with ‡)the betti numbers which satisfy only (7.4). Each of the other betti numbers alsosatisfy at least one of the other symmetries. The module Mℓ is Cohen-Macaulay for−2 ≤ ℓ ≤ 2, and the symmetries of (7.5) apply only in this range. It is interestingto notice that the symmetries of Corollary 7.3 apply to some of the betti numbersof modules Mℓ which are not Cohen-Macaulay.

54 ANDREW R. KUSTIN

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