An explicit, characteristic-free, equivariant homology equivalence between Koszul complexes

(aka: Divisors over determinantal rings defined by two by two minors, II)

Andy Kustin Purdue University September, 2007

Where to find it:

I have posted this talk on my website. Also, a relevant paper and pre-print are available on my website.

The Set up:

Let R be a ring (probably ℤ or a field K) and E and G be free R-modules of rank e and g, respectively.

We study: the (Koszul) complex

···→ N (m − 1,n − 1,p + 1) → \underbrace{Sym_{m}E^{∗}⊗ Sym_{n}G ⊗ ⋀ ^{p}(E^{∗}⊗ G)}_{N (m,n,p)}  (1)
→ N (m + 1,n + 1,p − 1) → ...,

its homology (which I’ll call H_{N} (m,n,p)), its dual

···→ M (m + 1,n + 1,p − 1) → \underbrace{D_{m}E ⊗ D_{n}G^{∗}⊗ ⋀ ^{p}(E ⊗ G^{∗})}_{M (m,n,p)}  (2)
→ M (m − 1,n − 1,p + 1) → ...,

and the homology of its dual (which I’ll call H_{M} (m,n,p)).

The plan for this talk:

Topic 1. The complex (1) can have interesting homology. Topic 2. The connection with divisors of determinantal rings.

Topic 2.a. The homology of (1) depends on characteristic. Topic 2.b. Duality

Topic 3. Does there exist a quasi-isomorphism from (2) to (1), for the appropriate choice of parameters?

Topic 3.a. Yes, but this quasi-isomorphism depends (heavily) on the

choice of a basis. Topic 3.b. Independent of basis? NO. (This is new.) Topic 3.c. What can be done in an equivariant manner? (This also is

new.)

Topic 4. Show an example of 3.b. Topic 5. Fix 3.b. Topic 6. Show the answer to 3.c Topic 7. Ask a question, maybe? Topic 8. Application 1. We exhibit a generator for H_{N} (g − 1,e − 1,α − 1) ≃ R. Topic 9. Application 2. The betti numbers of the divisors of the determinantal rings defined by the 2 × 2 minors of a generic e × g matrix exhibit an extra symmetry when e = 3 or g = 3. Topic 10. The proof of 6.

Topic 1. The complex (1) can have interesting homology.

Consider

0 → \underbrace{N (0,0,2)}_{⋀ ^{2}(E^{∗}⊗G)} → \underbrace{N (1,1,1)}_{E^{∗}⊗G⊗(E^{∗}⊗G)} → \underbrace{N (2,2,0)}_{Sym_{2}E^{∗}⊗Sym_{2}G} → 0

The homology is concentrated in spot (1,1,1). If v_{1},v_{2},... is part of a basis for E^{∗} and x_{1},x_{2},... is part of a basis for G, then

v_{1}⊗ x_{1}⊗ (v_{2}⊗ x_{2}) − v_{1}⊗ x_{2}⊗ (v_{2}⊗ x_{1})

is a cycle which represents a non-zero element of homology.

In light of duality:

H_{N} (m,n,p) ≃ H_{M} (m′,n′,p′),

provided

m + m′= g − 1, n + n′= e − 1, p + p′= \underbrace{(e − 1)(g − 1)}_{α}, and

1 − e ≤ m − n ≤ g − 1,

the only homology of → M (g−1,e−1,α−2) → M (g−2,e−2,α−1) → M (g−3,e−3,α) → occurs at (g − 2,e − 2,α − 1).

Of course, Topic 3 asks if there is a map of complexes [figure: ...; M (g−1,e−1,α−2); M (g−2,e−2,α−1); M (g−3,e−3,α); ...; 0; N (0,0,2); N (1,1,1); N (2,2,0); 0,] so that the mapping cone of the resulting picture is exact.

Topic 2. The connection with divisors of determinantal rings.

• Let P = Sym_{•}(E^{∗}⊗ G), a polynomial ring in the eg variables v_{i}⊗ x_{j},

• S = Sym_{•}(E^{∗}⊕ G), a polynomial ring in e + g variables v_{1},...,v_{e},x_{1},...,x_{g}, and

• T be the subring ∑_{m}Sym_{m}E^{∗}⊗ Sym_{m}G of S. (So, T is the subring R[{x_{i}v_{j}}] of S = R[v_{1},...,v_{e},x_{1},...,x_{g}]).

Notice that v_{i}⊗ x_{j} ↦ v_{i}x_{j} gives a ring homomorphism P ↠ T whose kernel is I_{2} of the matrix (v_{i}⊗ x_{j}). Thus, T is the determinantal ring defined by the 2 × 2 minors of a generic e × g matrix.

The class group of T is ℤ with the integer s associated to the divisor M_{s} = ∑_{m−n=s} Sym_{m}E^{∗}⊗ Sym_{n}G. (Notice that M_{s} is a T -submodule of S.)

One can resolve R as a P -module. The resolution is the Koszul complex

P ⊗ ⋀ ^{•}(E^{∗}⊗ G).

Apply M_{s}⊗_{P} :

···→ ∑_{m−n=s} Sym_{m}E^{∗}⊗ Sym_{n}G ⊗ ⋀ ^{p}(E^{∗}⊗ G) → ...

Take one graded strand:

···→ Sym_{m}E^{∗}⊗ Sym_{n}G ⊗ ⋀ ^{p}(E^{∗}⊗ G) → ...

We conclude that

H_{N} (m,n,p) = Tor_{p,?}^{P}(M_{m−n},R).

(The grading on M_{s} is defined so that ? = n + p.) Thus,

H_{N} (m,n,p) = Tor_{p,n+p}^{P} (M_{m−n},R).

Topic 2.a. The homology of (1) depends on characteristic.

• If R = ℤ, then H_{N} (m,n,p) is not always free! (If e,g ≥ 5, then H_{N} (2,2,3) has 3-torsion.) [Hashimoto 1990]

• If R is a field, then the dimension of H_{N} (m,n,p) depends on the characteristic of R. In particular, H_{N} (2,2,3) has larger dimension in characteristic 3, when e,g ≥ 5, than it has in other characteristics.

Topic 2.b. Duality.

Theorem. Assume 1 − e ≤ m − n ≤ g − 1. If • m + m′= g − 1,

• n + n′= e − 1, and

• p + p′= α, then

H_{N} (m,n,p) ≃ H_{M} (m′,n′,p′).

Proof. The key step in the proof is: If 1 − e ≤ s ≤ g − 1, then M_{s} is Cohen-Macaulay and

M_{s} ≃ Ext_{P}^{α}(M_{g−e−s} ,P).

Once one has this, then the rest is booking keeping (which I will hide).

Show M_{s} ≃ Ext_{P}^{α}(M_{g−e−s} ,P).

The canonical class is M_{g−e} . The class group arithmetic tells us that

M_{s} ≃ Hom_{T}(M_{g−e−s} ,M_{g−e} ) ≃ Hom_{T}(M_{g−e−s} ,Ext_{P}^{α}(T,P))

= Ext_{P}^{0}(M_{g−e−s} ,Ext_{P}^{α}(T,P)) ≃_{⋆} Ext_{P}^{0}(T,Ext_{P}^{α}(M_{g−e−s} ,P))

= Hom_{P}(T,Ext_{P}^{α}(M_{g−e−s} ,P)) = Ext_{P}^{α}(M_{g−e−s} ,P).

I justify the isomorphism ⋆. Let M and N be perfect P modules of projective dimension c and let 𝔽 and 𝔾 be free resolutions of M and N, respectively. The complex

Hom(𝔽,𝔾^{∗}) = Hom(𝔽 ⊗ 𝔾,P) = Hom(𝔾,𝔽^{∗})

shows that

Ext_{P}^{j}(M,Ext_{P}^{c}(N,P)) ≃ Ext_{P}^{j}(N,Ext_{P}^{c}(M,P))

for all j.

Topic 3. Does there exist a quasi-isomorphism from (2) to (1), for the appropriate choice of parameters?

Main Question. Suppose 1 − e ≤ m − n ≤ g − 1. Choose m′,n′,p′to satisfy m + m′= g − 1, n + n′= e − 1, and p + p′= α. Does there exist a quasi-isomorphism

[figure: ...; M (m,n,p); M (m − 1,n − 1,p + 1); ...; ...; N (m′,n′,p′); N (m′+ 1,n′+ 1,p′− 1); ...?] Answer (a). Yes, even when R = ℤ.

Answer (b). The quasi-isomorphism of Answer (a), depends on the choice of basis. Does there exist a coordinate free quasi-isomorphism for the Main Question when R = ℤ? NO! (new)

Answer (c). Does there exist a sequence of coordinate-free quasi-isomorphisms

[figure: ℕ; ψ; 𝕐; φ; 𝕄]

when R = ℤ? YES! (new)

Topic 4. We show an example for Answer 3.b.

Take e = g = 2 and R = ℤ. I demonstrate that there does not exist a coordinate-free quasi-isomorphism

[figure: 0; M (0,0,0); 0; 0; N (0,0,2); N (1,1,1); N (0,0,2); 0]

I will demonstrate that it is impossible to select a cycle z in N (1,1,1), such that z is invariant under change of basis and the homology class of z generates all of H_{N} (1,1,1).

It is easy to calculate the cycles of N (1,1,1) which are invariant under under change of basis form the free group generated by the cycle

\{ \begin{matrix}+v_{1}⊗ x_{1}⊗ (v_{2}⊗ x_{2}) − v_{1}⊗ x_{2}⊗ (v_{2}⊗ x_{1}) \\ +v_{2}⊗ x_{2}⊗ (v_{1}⊗ x_{1}) − v_{2}⊗ x_{1}⊗ (v_{1}⊗ x_{2}) \end{matrix}

The given cycle corresponds to 2 in N (1,1,1). The green summand and the red summand each represent generators of H_{N} (1,1,1).

(Note for the purposes of this talk, I have pretended that all free modules are oriented, that is I have chosen a generator ω_{E} for ⋀ ^{e}E. This simplifies the exposition, allows us to consider only change of bases which have determinant one, and has no effect on any important idea.)

Topic 5. Fix 3.b.

It is easy to “fix” 3.b. We merely look for a complex 𝕐 with H_{0}(𝕐) = M (0,0,0) for which the mapping cone of

[figure: 0; 𝕐_{1}; 𝕐_{0}; 0; N (0,0,2); N (1,1,1); N (2,2,0); 0] is exact. One such 𝕐 is

0 → ⋀ ^{2}E^{∗}⊗ D_{2}E → E^{∗}⊗ E

Topic 6. We create equivariant quasi-isomorphisms:

[figure: ℕ; ψ; 𝕐; φ; 𝕄.]

• Fix P,Q with 1 ≤ \underbrace{P − Q + e}_{ℓ} ≤ e + g − 1.

• Define P + P′= e(g − 1) and Q + Q′= (e − 1)g.

• Let ℕ = ℕ(P,Q) : ···→ N (P −1,Q−1,1) → \underbrace{N (P,Q,0)}_{position 0}→ 0, and

• 𝕄 = 𝕄(P′,Q′)[−α] : 0 → \underbrace{M (P′,Q′,0)}_{position α}→ M (P′−1,Q′−1,1) →...

We create 𝕐 to be (a shift of) the total complex of

⋮  ⋮

[figure: ...; 𝕏_{1,1}; 𝕏_{1,0}]

[figure: ...; 𝕏_{0,1}; 𝕏_{0,0}.]

We make 𝕐 so that each row of

⋮  ⋮  ⋮
[figure: ...; 𝕏_{1,1}; 𝕏_{1,0}; φ; M (P′− Q′+ 1,1,Q′− 1); 0]
[figure: ...; 𝕏_{0,1}; 𝕏_{0,0}; φ; M (P′− Q′,0,Q′); 0]

is split exact; so,

φ: 𝕐 → 𝕄

is automatically a quasi-isomorphism.

We define ψ so that the total complex of

⋮  ⋮
[figure: ...; 𝕏_{1,1}; 𝕏_{1,0}]
[figure: ...; 𝕏_{0,1}; 𝕏_{0,0}]
[figure: ...; N (g−2+Q′−P′,e−2,α−Q′+1); ψ; N (g−1+Q′−P′,e−1,α−Q′); ψ; ...]

is split exact.

The module 𝕏_{r,c} is

⨁ ⋀ ^{λ_{1}}E^{∗}⊗ ··· ⊗ ⋀ ^{λ_{ℓ}}E^{∗}⊗ \underbrace{D_{g+r+c} E ⊗ D_{r}G^{∗}⊗ ⋀ ^{α+e−1−Q−r} (E ⊗ G^{∗})}_{M (g+r+c,r,α+e−1−Q−r)},

where the sum is taken over all ℓ-tuples (λ_{1},...,λ_{ℓ}) with λ_{i} ≥ 1 for all i and λ_{1}+ ··· + λ_{ℓ} = ℓ + c.

The horizontal map 𝕏_{r,c} → 𝕏_{r,c−1} is

V_{1}⊗ ··· ⊗ V_{ℓ}⊗ u^{(a)}⊗ Y ⊗ Z ↦

∑_{i} sign_{i}χ(λ_{i} ≥ 2)V_{1}⊗ ··· ⊗ u(V_{i}) ⊗ ··· ⊗ V_{ℓ}⊗ u^{(a−1)} ⊗ Y ⊗ Z.

The vertical map

\begin{matrix}𝕏_{r,c} \\ ↓ \\ 𝕏_{r−1,c} \end{matrix}  is  \begin{matrix}V ⊗ u^{(a)}⊗ y^{(b)}⊗ Z \\ ↓ \\ V ⊗ u^{(a−1)} ⊗ y^{(b−1)} ⊗ (u ⊗ y) ∧ Z. \end{matrix}

The horizontal augmentation

X_{r,0} = \underbrace{E^{∗}⊗ ··· ⊗ E^{∗}}_{ℓ}⊗D_{a}E ⊗ D_{b}G^{∗}⊗ ⋀ ^{d}(E ⊗ G^{∗}) → M (P′+ r,r,Q′− r)

is

v_{1}⊗ ··· ⊗ v_{ℓ}⊗ U ⊗ Y ⊗ Z ↦ v_{1}··· v_{ℓ}(U) ⊗ Y ⊗ Z.

I’ll tell you the vertical augmentation

\begin{matrix}𝕏_{0,c} \\ N (g − 1 + Q′− P′− c,e − 1 − c,α − Q′+ c) & \begin{aligned}ψ \\ & \end{aligned} \end{matrix}

by telling you [ψ(t)](t′) for each

t = V_{1}⊗ ··· ⊗ V_{ℓ}⊗ u^{(g+c)} ⊗ Y ⊗ Z

in ⋀ ^{λ_{1}}E^{∗}⊗ ··· ⊗ ⋀ ^{λ_{ℓ}}E^{∗}⊗ M (g + c,r,α + e − 1 − Q) ⊂ 𝕏_{0,c} and

t′= U′⊗ Y′⊗ Z′∈ M (\underbrace{g − 1 + Q′− P′− c}_{ℓ−1−c},e − 1 − c,α − Q′+ c).

(a) The value of [ψ(t)](t′) is zero unless every λ_{i} ≤ 2 and λ_{ℓ} = 1.

(b) Under hypothesis (a), identify i_{1} < ···< i_{ℓ−c} and j_{1} < ···< j_{c} with λ_{i_{k}} = 1 and λ_{j_{k}} = 2 for all k.

(c) The value of [ψ(t)](t′) is

\begin{matrix}(V_{i_{1}} ··· V_{i_{ℓ−c−1}} )(U′) \\ ·[Z ∧ Z′∧ ( ( V ∧ V_{ℓ})(ω_{E}) ▷◁ Y′) ∧ ( u^{(g)} ▷◁ ω_{G^{∗}}) ] (ω_{E^{∗}⊗G}) \end{matrix}

for V = u(V_{j_{1}}) ∧ ··· ∧ u(V_{j_{c}}). Recall t = V_{1}⊗ ··· ⊗ V_{ℓ}⊗ u^{(g+c)} ⊗ Y ⊗ Z in ⋀ ^{λ_{1}}E^{∗}⊗ ··· ⊗ ⋀ ^{λ_{ℓ}}E^{∗}⊗ M (g + c,r,α + e − 1 − Q) ⊂ 𝕏_{0,c} and

t′= U′⊗ Y′⊗ Z′∈ M (ℓ − 1 − c ,e − 1 − c,α − Q′+ c).

The homomorphism ▷◁: D_{a}E ⊗ ⋀ ^{a}G^{∗} → ⋀ ^{a}(E ⊗ G^{∗}) is on the next page.

The homomorphism

▷◁: D_{a}E ⊗ ⋀ ^{a}G^{∗} → ⋀ ^{a}(E ⊗ G^{∗})

satisfies:

u^{(a)} ▷◁ (y_{1}∧ ··· ∧ y_{a}) = (u ⊗ y_{1}) ∧ ··· ∧ (u ⊗ y_{a}).

Topic 7. Ask a question, maybe?

Let R be a ring – at this point, “ring” should be interpreted in a liberal enough manner to include the possibility R = ℤ[GL(E) × GL(G)].

Definition. The complexes 𝔽 and 𝔽′are homologically equivalent if there exists a sequence of quasi-isomorphisms

𝔽 ← Q_{1}, Q_{1} → Q_{2}, Q_{2} ← Q_{3}, ..., Q_{t} → 𝔽′.  (3)

Question. Let 𝔽 and 𝔽′be complexes of free R-modules. Find necessary and sufficient conditions so that 𝔽 and 𝔽′are homologically equivalent. One necessary condition is that

H_{i}(𝔽) ≃ H_{i}(𝔽′)  (4)

be isomorphic, for all i.

(a) Is condition (4) sufficient? Find a counterexample.

(b) Is there some significance to the least number t in (3)?

Topic 8. Application 1. We exhibit a generator for H_{N} (g − 1,e − 1,α − 1) ≃ R.

When P = e(g − 1) and Q = (e − 1)g, then we can use our homology equivalence:

[figure: \underbrace{E^{∗}⊗ ··· ⊗ E^{∗}}_{g}⊗D_{g}E ⊗ D_{0}G^{∗}⊗ ⋀ ^{0}(E ⊗ G^{∗}) = 𝕏_{0,0}; φ; M (,0,0,0); N (g − 1,e − 1,α); ψ]

to pick out a generator for H_{N} (g − 1,e − 1,α − 1) ≃ R.

There exists a homomorphism

Φ : D_{e−1} G^{∗}⊗ D_{g−1} E → ⋀ ^{e+g−1} (E ⊗ G^{∗})

which satisfies:

(a) Φ(x(Y′) ⊗ U) = (ω_{E} ▷◁ Y′) ∧ (U ▷◁ x(ω_{G^{∗}})) for Y′∈ D_{e}G^{∗} and x ∈ G, and

(b) Φ(Y ⊗ v(U′)) = (−1)^{e+1} (v(ω_{E}) ▷◁ Y) ∧ (U′▷◁ ω_{G^{∗}}) for U′∈ D_{g}E and v ∈ E^{∗}.

Exhibit a generator for H_{N} (g − 1,e − 1,α − 1) ≃ R.

• Take any s ∈ Sym_{g−1} E^{∗}⊗ D_{g−1} E, which is sent to 1 under the evaluation map.

• Define ζ_{s} ∈ N (g − 1,e − 1,α) by

ζ_{s}(U′⊗ Y′⊗ Z′) = [Z′∧ Φ(Y′⊗ (U′⊗ 1)(s))](ω_{E^{∗}⊗G}) for all U′⊗ Y′⊗ Z′∈ M (g − 1,e − 1,α).

• The cycle ζ_{s} ∈ N (g − 1,e − 1,α) depends on the choice of s.

• The homology class of ζ_{s} ∈ H_{N} (g − 1,e − 1,α) is independent of s and generates H_{N} (g − 1,e − 1,α).

Topic 9. Application 2. The betti numbers of the divisors of the determinantal rings defined by the 2 × 2 minors of a generic e × g matrix

exhibit an extra symmetry when e = 3 or g = 3.

Return to the determinantal ring set up. Let K be a field, P be a polynomial ring over K in eg variables, T be the determinantal ring defined by the 2 × 2 minors of a generic e × g matrix, and {M_{s}} be the previously identified set of representatives of the class group of T.

One can translate

H_{N} (m,n,p) ≃ H_{M}(m′,n′,p′)
for m + m′= g − 1, n + n′= e − 1 and p + p′= α

into:

\begin{matrix}β_{p,q}(M_{s}) = β_{p′,q′}(M_{s′})for \\ 1 − e ≤ s ≤ g − 1, s + s′= g − e, p + p′= α, q + q′= (e − 1)g. \end{matrix}  (5)

One also gets Corollary.

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

β_{p,q}(M_{s}) = β_{p′,q}(M_{s′})  (6)

for s + s′= q − 3 and p + p′= 2q − 2.

(b) Assume that g = 3. If s and q are integers with s ≤ 2 and 1 − 2s ≤ q, then

β_{p,q}(M_{s}) = β_{p′,q′}(M_{s′}),  (6)

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

Example. Take e = g = 3. The graded betti numbers of the P -modules M_{s}, for −5 ≤ s ≤ 5 are

‡β_{0,5}(M_{−5} )=21  ^{‡}β_{0,4}(M_{−4} )=15 ^{‡}β_{0,3}(M_{−3} )=10 β_{0,2}(M_{−2} )=6
‡β_{1,6}(M_{−5} )=105 ^{‡}β_{1,5}(M_{−4} )=72 ^{‡}β_{1,4}(M_{−3} )=45 β_{1,3}(M_{−2} )=24
‡β_{2,7}(M_{−5} )=216 ^{‡}β_{2,6}(M_{−4} )=141 ^{‡}β_{2,5}(M_{−3} )=81 β_{2,4}(M_{−2} )=36
‡β_{3,8}(M_{−5} )=234 ^{‡}β_{3,7}(M_{−4} )=144 ^{‡}β_{3,6}(M_{−3} )=74 β_{3,5}(M_{−2} )=24
‡β_{4,9}(M_{−5} )=141 ^{‡}β_{4,8}(M_{−4} )=81  β_{4,7}(M_{−3} )=36  β_{4,6}(M_{−2} )=6
‡β_{5,10}(M_{−5} )=45  β_{5,9}(M_{−4} )=24  β_{5,8}(M_{−3} )=9
β_{6,11}(M_{−5} )=6  β_{6,10}(M_{−4} )=3  β_{6,9}(M_{−3} )=1

\begin{matrix}β_{0,1}(M_{−1} )=3 & β_{0,0}(M_{0})=1 & β_{0,0}(M_{1})=3 & β_{0,0}(M_{2})=6 \\ β_{1,2}(M_{−1} )=9 & β_{1,2}(M_{0})=9 & β_{1,1}(M_{1})=9 & β_{1,1}(M_{2})=24 \\ β_{2,3}(M_{−1} )=6 & β_{2,3}(M_{0})=16 & β_{2,2}(M_{1})=6 & β_{2,2}(M_{2})=36 \\ β_{2,4}(M_{−1} )=6 & β_{3,4}(M_{0})=9 & β_{2,3}(M_{1})=6 & β_{3,3}(M_{2})=24 \\ β_{3,5}(M_{−1} )=9 & β_{4,6}(M_{0})=1 & β_{3,4}(M_{1})=9 & β_{4,4}(M_{2})=6 \\ β_{4,6}(M_{−1} )=3 & & β_{4,5}(M_{1})=3 \end{matrix}

38

\begin{matrix}^{‡}β_{0,0}(M_{3})=10 & ^{‡}β_{0,0}(M_{4})=15 & ^{‡}β_{0,0}(M_{5})=21 \\ ^{‡}β_{1,1}(M_{3})=45 & ^{‡}β_{1,1}(M_{4})=72 & ^{‡}β_{1,1}(M_{5})=105 \\ ^{‡}β_{2,2}(M_{3})=81 & ^{‡}β_{2,2}(M_{4})=141 & ^{‡}β_{2,2}(M_{5})=216 \\ ^{‡}β_{3,3}(M_{3})=74 & ^{‡}β_{3,3}(M_{4})=144 & ^{‡}β_{3,3}(M_{5})=234 \\ β_{4,4}(M_{3})=36 & ^{‡}β_{4,4}(M_{4})=81 & ^{‡}β_{4,4}(M_{5})=141 \\ β_{5,5}(M_{3})=9 & β_{5,5}(M_{4})=24 & ^{‡}β_{5,5}(M_{5})=45 \\ β_{6,6}(M_{3})=1 & β_{6,6}(M_{4})=3 & β_{6,6}(M_{5})=6. \end{matrix}

These numbers were calculated by the computer program Macaulay. There are four symmetries running through this set of numbers. In addition to the three mentioned above, the R-modules E^{∗} and G are isomorphic because e = g; hence, the R-module automorphism of P, which sends the matrix of indeterminates Z to Z^{T}, induces the relation

β_{p,q}(M_{s}) = β_{p,q+s} (M_{−s} ),  (7)

for all integers p, q, and s. We have marked (with^{‡}) the betti numbers which satisfy only (7). Each of the other betti numbers also satisfy at least one of the other symmetries. The module M_{s} is Cohen-Macaulay for −2 ≤ s ≤ 2, and the symmetries of (5) apply only in this range. It is interesting to notice that the symmetries of (6) apply to some of the betti numbers of modules M_{s} which are not Cohen-Macaulay.

Topic 10. The proof that ψ: 𝕐 → ℕ is a quasi-isomorphism.

Let ℂ(P,Q) be the mapping cone of ψ. We want to prove that all relevant ℂ(P,Q) are split exact. We proceed by induction on ℓ and g.

• The case g = 0 makes sense, is easy, and is exact.

• In today’s talk 1 ≤ ℓ ≤ e + g − 1. (Recall that ℓ = s + e and M_{s} is Cohen-Macaulay and the critical duality

M_{s} ≃ Ext_{P}^{α}(M_{g−e−s} ,P)

holds precisely when 1 − e ≤ s ≤ g − 1.) Nonetheless, I studied the case “ℓ = 0” (the module M_{−e} is just outside the Cohen-Macaulay range) in some previous work and one can use “Φ” to connect the cases “ℓ = 0” and ℓ = 1 and prove that ℂ(P,Q) is exact whenever ℓ = 1. (Recall that ℓ = P − Q + e.)

Now one decomposes G as Ḡ⊕ Rx_{g}. Use

\begin{matrix}x_{g} : N (a,b,d) → N (a,b + 1,d) and \\ x_{g} : M (a,b,d) → M (a,b − 1,d) \end{matrix}  (8)

to create a map

Θ : ℂ(P,Q) → ℂ(P,Q + 1)[−1].

The complex ℂ(P,Q+1)[−1] is split exact by induction on ℓ; so it suffices to show that the mapping cone, 𝔸, of Θ is split exact. Use (8) to split as much from 𝔸 as possible. What remains is a direct sum of complexes

⨁_{w=0}^{e} ℂ_{E,Ḡ}(P − w,Q − w + 1)

formed using the free module Ḡof rank g − 1.