Day 1: Computing with coherent sheaves andsheaf cohomology with applications to geometry

and string theory

Mike Stillman (mike@math.cornell.edu)

Cornell University

1-4 July 2015 / Osaka

Plan for Lectures

Day 1 : July 1.

Why we care about sheaf cohomology ?Computing with graded modules over polynomial ringsCoherent sheaves on Pn : an algebraic definition.Key examples of coherent sheaves

Day 2 : July 2.

Global section modules of sheavesCech cohomologyComputing using exact sequencesLocal dualitySheaf of 1-forms, sheaf of p-formsApplication : Hodge diamond

Day 3 : July 3.

The exterior algebraTate resolution and the explicit Bernstein-Gelfand-GelfandcorrespondenceComputing sheaf cohomology via exterior algebraApplication : geometry on a surfaceA mystery surfaceCohomology and string theory

Macaulay2, working examples, and open problems

We will work through examples using Macaulay2 :

open source system for computing in algebraic geometry and relatedfields, (written with Dan Grayson, since the mid 1990’s)Download it, more information : www.macaulay2.comTry it out ! web.macaulay2.comUse whatever system you prefer, but Do examples !

We will present some exercises, some easy, some hard, some openproblems !

Open problems : we need better ways of computing with sheaves,computing sheaf cohomology, and related notions. We will mentionsome along the way.

Basic situation

Suppose :We are given X ⊂Pn by equations :

X =V(I ) := p ∈Pn | f1(p)= . . . = fr (p)= 0,

whereI = IX = ⟨f1, . . . , fr ⟩ ⊂ S = k[x0, . . . ,xn],

and

I is homogeneous : each fi is a homogeneous polynomial,

k is a field : specific field is not so important. Think k =C.

Example

Let φ :P1 −→P3 be the morphism defined by

(s ,t) 7→ (s4,s3t,st3,t4)= (a,b,c ,d)

Let X := imageφ, the rational quartic curve in P3.The ideal IX ⊂ S = k[a,b,c ,d ] of X :

IX = ⟨ bc −ad , c3−bd2, ac2−b2d , b3−a2c ⟩

Invariants from Grobner bases

Grobner bases help compute the following information about X :

dimX

degreeX

Hilbert polynomial : dimension, degree, genus, basic numericalinvariants of a variety.

dimSing(X )

Example (Numerical invariants of the rational quartic curve)

dimX = 1

degreeX = 4

Arithmetic genus of X is g = 0.

Hilbert polynomial is 4d +1.

sing(X )=;.

More invariants

More refined invariants : dimensions of sheaf cohomology vector spaces,for important sheaves on X :

sheaf of regular functions on X , OX ,

tangent sheaf TX ,

cotangent sheaf Ω1X , p-forms : Ωp

X ,

normal sheaf NX/Pn ,

and the canonical (=dualizing) sheaf ωX .

The Hodge diamond of a smooth variety is gold in the hands of abirational geometer :

hp,q(X ) := dimCHq(ΩpX ), for 0≤ p,q ≤ dimX

Geometry on a variety

Theorem (Castelnuovo’s theorem)

A smooth projective surface X is rational if and only ifH1(OX )=H0(ωX ⊗ωX )= 0.

In fact, one can use computational methods and the proof of thistheorem to construct a birational map to P2.

Intersections on a surface

Suppose X is a smooth projective surface, and D and E are curves on X ,possibly the same, or with many components.Can use sheaves and cohomology to

compute intersection numbers D ·Ework with divisors on X (linear equivalence, spaces of rationalfunctions with prescribed zeros and poles).

Application to String Theory in Physics

Cohomology is used all over the place in string theory.

One important example is the following (Anderson, Gray, He, Lukas, andothers) : the search for string theories that match our reality (e.g.number and types of particles)

Called : E8×E8 heterotic string theory models.

To construct a potential universe, the key data is the following :

Choose a (smooth) Calabi-Yau 3-fold X , e.g. a quintic hypersurfacein P4, such as x5

0 +x51 +x5

2 +x53 +x5

4 +2x0x1x2x3x4 = 0.

Choose two coherent sheaves V and V ′ on X (which are locally freeof rank n = 3, 4, or 5) the visible and hidden sector bundles

A number of constraints on which sheaves are useful. These are mostlyeasy to check.

Particle Content of the Theory

The “low energy Gauge group” is the commutant (subgroup generated byall commutators) of SU(n) in the Lie group E8. For

G = SU(3), SU(4), SU(5)

these giveH =E6, SO(10), SU(5).

Notation : hp(V ) := dimCHp(V )

After some more reductions, it turns out that the number of particles of agiven type (i.e. supermultiplets, i.e. representations) is given by thedimensions h1(V ⊗V ), h1(V ), h1(V ∗), h1(

∧2V ), and h1(∧2V ∗).

Example : Higgs content

In the case n = 4 (so H = SO(10)), n10 = h1(∧2V ) is the number of Higgs

super partners that exist in the theory.Alas, often the number, as computed, is 0. But : not always !

Back to Earth !Computing with S -modules.

Syzygies

Let R be a (commutative) ring. For these lectures : R = S := k[x0, . . . ,xn],or R will be a quotient R = S/I

Key computation

Given an a×b matrix

f : Rb −−−−−−−−−−−−−−→(f1 f2 . . . fb

) Ra,

Find a generating set for the syzygy module

ker f := v ∈Rb | f (v)= 0

Hilbert (1890) : Gave an algorithm, using elimination theory.

Schreyer (1982) : over a polynomial ring, this is a byproduct of theBuchberger algorithm for finding the Grobner basis of the submodule

image f = ⟨f1, . . . , fb⟩ ⊂Ra.

Example of Syzygies : the twisted cubic curve in P3

Example

Suppose that R = S = k[a,b,c ,d ], and

f1 = b2−ac , f2 = bc −ad , f3 = c2−bd .

Letf : S3 −−−−−−−−−−→(

f1 f2 f3) S1.

Use Buchberger’s algorithm to compute a Grobner basis ofI = image f = ⟨f1, f2, f3⟩ :

cf1−bf2+af3 = 0,

cf2−bf3−df1 = 0.

So

ker f =⟨ c

−ba

,

−dc−b

⟩

is a byproduct of Grobner basis construction.

Aside : Graded modules

For S = k[x0, . . . ,xn], set deg(xi )= 1, all i .Let Sd be the vector space of all forms of degree d .

Definition

A graded S-module is an S-module M with a k-vector spacedecomposition

M = ⊕d∈Z

Md ,

which satisfies SdMe ⊂Md+e (all d ,e).

For us : Md will always be finite dimensional as k vector space.

Definition

For e ∈Z, define M(e) to be the same S-module as M, but with thegrading

M(e)d :=Md+e .

Example

S(−3) has one generator. What degree is it in ? Answer :S(−3)3 = S0 = k. The generator is in degree 3.

Aside : Graded modules

For S = k[x0, . . . ,xn], set deg(xi )= 1, all i .Let Sd be the vector space of all forms of degree d .

Definition

A graded S-module is an S-module M with a k-vector spacedecomposition

M = ⊕d∈Z

Md ,

which satisfies SdMe ⊂Md+e (all d ,e).

For us : Md will always be finite dimensional as k vector space.

Definition

For e ∈Z, define M(e) to be the same S-module as M, but with thegrading

M(e)d :=Md+e .

Example

S(−3) has one generator. What degree is it in ? Answer :S(−3)3 = S0 = k. The generator is in degree 3.

Good exercise : Computing with modules

Exercise

Let R = S = k[x0, . . . ,xn], or R = S/I .Let M, N, P be finitely generated (graded) R-modules.

How to input a module ? Given an a×b matrix φ over R :

φ : Rb −→Ra,

there are two obvious ways to obtain an R-module :

M = cokerφ : φ is the presentation matrix of M.M = imageφ : φ is the generator matrix of M.

Show how to compute HomR(M ,R) (the R-dual of M).

Given f : M −→N, construct ker f , image f , and coker f .

Given M N Pf g

, such that gf = 0, construct thehomology R-module

kerg

image f.

Free resolutions

Definition

A free resolution of a graded S-module M is a complex

0←−F0d1←−F1

d2←− ·· · dr←−Fr ←− ·· ·

of graded free S-modules Fi , which is exact except that cokerd1 =M.The resolution is minimal if every entry of each matrix is either zero, orhas positive degree.

I often write the following loose statement : “Let

0←−M ←−F0d1←−F1

d2←− ·· · dr←−Fr ←− ·· ·

be a free resolution” but it is understood the M is not part of the freeresolution !

Free resolution of the twisted cubic

Example (twisted cubic)

The following is exact, for the twisted cubic example.

0 S/I S S3 S2 0d1(

f1 f2 f3

) d2c −d−b c

a −b

We make this into a graded free resolution by making sure the maps aredegree-preserving :

0 S/I S S(−2)3 S(−3)2 0d1 d2

d1 is the generator matrix for the ideal I .

d2 is the presentation matrix for the ideal I .

Free resolution of k = S/⟨x0, . . . ,xn⟩ (Koszul complex)

Example

Suppose S = k[x ,b,c]. A minimal graded free resolution of k is

0 k S S(−1)3 S(−2)3 S(−3) 0d1(

a b c) d2

−b −c 0

a 0 −c0 a b

d3c

−ba

Example (General case)

Let V = S1 = kn+1. Let

Fi := ∧iV ⊗S(−i) = S(n+1i )(−i).

The free resolution of k = S/⟨x0, . . . ,xn⟩ has the form

0 k S F1 . . . Fn Fn+1 0d1 d2 dn dn+1

Last example of a free resolution for now !

RecallIX = ⟨ bc −ad , b3−a2c , ac2−b2d , c3−bd2 ⟩.

Example (The minimal free resolution of the ring S/IX of the rationalquartic curve)

0←−S/IX ←−S ←−S(−2)

⊕S(−3)3

←−−−−−−−−−−−−−−−−−−−−−−−−−b2 −ac −bd −c2c d 0 0a b −c −d0 0 a b

S(−4)4 ←−−−−

d−c−ba

S(−5)←− 0

The Betti display presents the ranks and degrees in a compact, readableform.

0 1 2 3

total: 1 4 4 1

0: 1 . . .

1: . 1 . .

2: . 3 4 1

Regularity and projective dimension of a graded S-module

The Betti diagram again :

0 1 2 3

total: 1 4 4 1

0: 1 . . .

1: . 1 . .

2: . 3 4 1

Definition

Given the Betti display of a graded S-module M.

The index of the last row is reg(M), the (Castelnuovo-Mumford)regularity of M.

The index of the last column, projdim(M) is the projectivedimension of M.

For the rational quartic : reg(S/IX )= 2, and projdim(S/IX )= 3.

Small lemma of Hilbert in 1890

An amazing theorem from an amazing paper of Hilbert !

Theorem (Hilbert, 1890)

If S = k[x0, . . . ,xn], and if M is a finitely generated graded S-module, thenprojdim(M)≤ n+1, i.e. the minimal free resolution of M

0 M F0 F1 . . . Fr 0

with Fr 6= 0, has length r ≤ n+1.

Ext modules

One really important construction is Ext :

Example (Computing Exti (M ,S))

Compute a free resolution of M :

0 M F0 F1 . . . Fr 0d1 d2 dr

Apply HomS (−,S) (i.e. transpose everything)

0 F∗0 F∗

1 . . . F∗r 0

d∗1 d∗

2 d∗r

Exti (M ,S) := ker(d∗i+1)

image(d∗i )

We could define and compute Exti (M ,N) too.

Example : Rational quartic curve

Transpose the free resolution that we had before.

0−→S −→S(2)⊕

S(3)3−−−−−−−−−−−−−−−−−−→−b2 c a 0−ac d b 0−bd 0 −c a

−c2 0 −d b

S(4)4 −−−−−−−−−−−−−−−−→(

d −c −b a) S(5)−→ 0

Ext0(S/I ,S)=Ext1(S/I ,S)= 0.

Ext2(S/I ,S)= ker(dT3 )

image(dT2 )

.

Generated by 3 elements (of degree -3). The canonical module of I = IX .

Ext3(S/I ,S)=S/⟨a,b,c ,d⟩(5)= k(5), a copy of k, in degree -5.

Which Ext modules can be non-zero ?

Theorem/Exercise

Letc = codimM = codimV(ann(M))

andr = projdim(M).

Then c ≤ r and :

Extc(M ,S) 6= 0

Extc+1(M ,S)=??

. . .

Extr−1(M ,S)=??

Extr (M ,S) 6= 0

and all other Exti (M ,S) outside of this range are zero.

Definition

M is Cohen-Macaulay if codimM = projdim(M).In this case, there is only one non-zero Ext : Extc(M ,S).

Definition

ExtcS (S/I ,S) is called the canonical module of S/I , wherec = codimV(I ).

Get lots of information from Ext modules !

Example (Rational quartic)

S/I is not Cohen-Macaulay, as there are two non-zero Ext modules.codim(S/I )= 2.The canonical module Ext2(S/I ,S) has 3 generators, with presentationmatrix a c b 0 0

−b −d 0 a c0 0 c b d

Coherent sheaves on projectivevarieties

Coherent sheaves on projective space

First a definition : the truncation M≥r of the graded module M isdefined to be

M≥r =∞⊕d=r

Md .

Definition (Coherent sheaf on projective space)

A coherent sheaf on Pn is an equivalence class of finitely generatedgraded S-modules under the equivalence relation : if M and N areS-modules, then M ∼N if M≥r ∼=N≥r , for some r .

We let M denote the coherent sheaf associated to M.

Definition (Map of coherent sheaves)

A (degree-preserving) map of S-modules f : M≥r −→N gives a map ofcoherent sheaves f : M −→ N.

The operation˜defines an exact functor from the category of finitelygenerated graded S-modules to the category of coherent sheaves.