1 Lecture 1

Possible Topics:

1. Hodge Conjecture. Probably about 5 lectures.

2. Geometric Langlands. Perhaps 5 lectures, and might contain geometry ofderived categories

3. Geometry of Derived Categories

4. Grothendieck-Riemann-Roch? Maybe 2 lectures

5. Mirror Symmetry/Calabi-Yau’s. As many lectures as needed.

2 Lecture 2

Geometric Langlands Program.Rather than describe the program right out, we will discuss examples first.This is essentially a nonabelian phenomenon.Input: A curve (smooth, compact, genus g) C and G a reductive group.

Examples of reductive groups are GL(n), SL(n), SO(n), Sp(n), products, etc.Baby case: Abelian Geometric Langlands Conjecture. We have G = (C∗)r,

and will take r = 1. This will roughly correspond to Class Field Theory inNumber Theory, and the full GLC corresponds to nonabelian class field theory.

Roughly we have objects G-bundles versus G local systems. In the caseG = GL(n), we have that a G-bundle is a rank n vector bundle and a G-localsystem is a homomorphism π1(C) → G which is the same as a rank n vectorbundle with a flat connection. (All connections are flat, because curvature is a2-form, and we are on a 1-dimensional object)

The moduli of rank n vector bundles is H1(C,GLn(O)) = Bun and themoduli space of G-local systems is H1(C,GLn(C)) = Loc.

We know both that TVBun = H1(C,EndO(V )) and that T(V,∇)Loc =H1(C,EndC(V )).

We want to compute these dimensions, and we will use Riemann-Roch: h0−h1(V ⊗ V ∗) = deg−rank(g − 1).

If n = 1 then V = L . End(V ) = V ⊗ V ∗ = L ⊗L −1 = O, so deg is zero.If n = 2 and split, then V = L1 ⊕ L2. So EndV = L1 ⊗ L−1

1 ⊕ L1 ⊗ L−12 ⊕

L2 ⊗ L−11 ⊕ L2 ⊗ L−1

2 which is 0 + d1 − d2 + d2 − d1 + 0 = 0.And so, deg(End(V )) = 0 for all V . First we note that deg V = deg detV ,

and det(EndV ) =∧n2

(End(V )).And so dimV Bun = h1(C,End(V )) = n2(g − 1) + h0(C,End(V )).For a ”general” vector bundle, h0 = 1, and so generically dimV Bun =

n2(g − 1) + 1The dimension of T(V,∇)Loc is 2 dimV Bun.Rough Statement of the Geometric Langlands Program: Any G local system

E on C determines a local system cE on a Zariski open subset of BunC,G, the

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moduli space of bundles with structure group G on C. The key property isthat this is an automorphic sheaf. If we extend to the whole space, we get a”perverse sheaf.”

We will not be talking about perverse sheaves and D-modules, but will fo-cus on automorphic sheaves and restrict ourselves to open subsets whenevernecessary.

Example 2.1. Case where n = 1, so G = GL(1) = C∗. On this, BunC,GL(1) =Pic(C). Abelian Langlands tells us that every C∗ local system on C determinesa C∗ local system on Pic(C).

A local system on C means a map ρ : π1(C) → C∗. A C∗-local systemon Pic0(C) = Jac(C) is cρ : π1(Jac(C)) → C∗. We note that π1(Jac(C)) =H1(Jac(C),Z) = H1(C,Z), which is the abelianization of π1(C).

So a map ρ : π1(C)→ C∗ factors through π1(Jac(C)) because C∗ is abelian,and so gives us cρ : π1(Jac(C))→ C∗, a local system of Jac(C).

This naturally extends to all of Pic(C). Choose L ∈ Picd(C). Then ⊗L−1 :Picd(C) → Pic0(C), and we pull back along this map. This doesn’t depend onL, because the induced automorphism on π1(Jac(C)) given by translation by L0

is trivial.

Example 2.2. If we take a rank 2 local system on C, then we get a localsystem on and open subset of Bun whose rank is 23g−3. In general, rank n gives23g−335g−5 . . . n(2n−1)(g−1).

We need to discuss Hecke correspondences now.A correspondence between varieties X,Y is a subvariety H ⊂ X × Y .If H is proper over Y , then H induces a map on cohomology H∗(X) →

H∗(H)→ H∗(Y ) by first pulling back and then pushing forward.

Example 2.3. Let f : X → Y be any morphism and H be the graph of f . Thenthe operator induced by H is f∗.

Example 2.4. Let g : Y → X be a morphism and H it’s graph. Then theoperator induced by H is g∗.

Hecke Correspondence:Hecke

BunC,G BunC,G×C

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????

The Hecke Correspondence is (V, V ′, x)|V ⊂ V ′ ⊂ V (x) where V (x) =V ⊗ O(x), and the inclusions are inclusions of coherent sheaves.

An automorphic sheaf is an eigensheaf of Hecke.

3 Lecture 3

More on the Hecke Correspondence.

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Denote Bun = Bunn,C. Then Hecke ⊂ Bun × Bun × C is the locus(V ′, V, x)|V ⊂ V ′ ⊂ V (x).

We call p1 : Hecke → Bun the first projection and p2 : Hecke → Bun × Cthe second and third projections taken together.

What is p−12 (V, x)?

Example 3.1. x ∈ C and V = OC ⊕ OC . What is the set of all V ′ such thatOC ⊕ OC = V ⊂ C ′ ⊂ V (x) = OC(x)⊕ OC(x).

V and V ′ work, as do OC ⊕ OC(x) and OC(x) ⊕ OC . Is there anythingelse? Yes! You can take other coordinates. So in fact we get two points and aprojective line.

Choose any linear function f : Vx → C. Let V ′ = ker f , and this gives theanswer.

Back to the question: what is p−12 (V, x)?

It is disconnected. To see this we look at `(V ′/V ), the length of this sheaf.Because V ′/V is supported at x, and it’s fiber there is V ′x/Vx, then (V/V )x ⊂V ′x/Vx ⊂ (V (x)/V )x, so the first is zero and the last can be anything up to n.

So the answer is that p−12 (V, x) =

∐ni=0Gr(i, Vx), where n = dim(Vx).

We saw that the vector bundle V ′ determines a subspace of Vx. Conversely,given a subspace W ⊂ Vx, Γ(U, V ′) = x ∈ Γ(U, V (x))|pole(s) ⊂W.

Summary: Given a vector bundle V and a subspace W ⊂ Vx, we get the”elementary transform” V ′ =

V ′ is a line bundle, and V ⊂ V ′ ⊂ V (x), 0 → W → Vx → Q → 0 and0→ Q→ V ′x →W → 0 are short exact

Hecke =∐ni=0Hecke

i, where Heckeip2→ Bun × C is a fiber bundle with

fiber Gr(i, n).Hecke Operators: Hi := (pi2)∗(pi1)∗. Explicitly, Hi(F ) := pi2∗p

i∗1 F .

Let F be a sheaf on X and G a sheaf on Y , then F G is defined to be(p∗XF )⊗ (p∗Y G ).

Conjecture 3.1 (Baby Version of GLC). For each local system E on C, thereexists a ”local system” cE on Bun such that Hi(cE) = cE

∧iE.

Case n = 1, then i = 0, 1. Bun = Bun1,C = Pic(C) =∐∞d=−∞ Picd(C).

Then Hecke = (L′, L, x)|L,L′ ∈ Pic(C), L ⊂ L′ ⊂ L(x). So Hecke0 =(L′, L, x)|L′ = L and Hecke1 = (L′, L, x)|L′ = L(x). As abstract varieties,Hecke0 = Pic(C)× C and Hecke1 = Pic(C)× C.

What does it mean to be a Hecke0 eigensheaf? Nothing. The condition isp∗1cE = cE 1 = p∗1cE .

What does it mean to be a Hecke1 eigensheaf? Want cE on Pic(C) withthe condition A∂∗cE ∼= cE E where A∂ : Hecke1 → Pic(C) and takes (L, x)to L′.

It must be E on C, then on C × C we have E E. So in general EN onCN .

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4 Lecture 4

We have been looking at GLCC for the case G = GL(1), the abelian case.Then we have E a local system on C and it induces natural local systems onC×d = C × . . .× C and Symd C = C [d] = C×d/Sd.

We have E . . .E on C×d whose fiber at (x1, . . . , xd) = Ex1 ⊗ . . .⊗Exd.

We claim that there exists a natural local system on Symd(C) whose pull-back to C×d is Ed. The fiber at an unordered d-tuple x1, . . . , xd = D isBij(1, . . . , d, D).

We do have a map π : C×d → Symd(C), but we can’t take π∗(Ed), becauseit has rank d!. We have two actions of Sd, permute the sheets of π and get anaction on F and we can act directly on the fiber e1⊗. . .⊗ed 7→ eσ(1)⊗. . .⊗eσ(d).

So we look at (π∗Ed)Sd , the sheaf of the Sd-equivariant sections of Ed.The fiber at D is the assignment f : Bij → (π∗E)|D such that f(σx) = σf(x).

So this gives a sheaf on Symd C. It is a local system for n = 1 but nototherwise.

Topologically, ρ : π1(C) → GL(1), and π1(C×n) has maps to π1(C) andGL(1) such that they all commute, and π1(C×n) → GL(1) factors throughπ1(Symn(C)), because as GL(1) is abelian, the representation of π1(C×n) fac-tors through π1(Symn(C)).

Algebraically, we need to calculate the local invariants when two of the xicome together. eg, E = C < a > ⊕C < b > and C = A1. Then E2 = C⊗4

and E2(x,y) = C < ax ⊗ ay, bx ⊗ ay, ax ⊗ by, ay ⊗ by >. S2 invariants are ax ⊗ ay,

bx ⊗ by, ax ⊗ bY + ay ⊗ bx, but ax ⊗ by − ay ⊗ bx goes to minus itself.Summary: In rank ≥ 2, we get a local system of rank nd away from the

discriminant, its extension across the discriminant has local monodromy: it isnot a local system.

For rank n = 1, it does extend.Anyway, on SymdC we do get a rank 1 local system. We want a rank 1

linear system pE on Pic(C).We’ll use A∂d : SymdC → Picd C (Abel-Jacobi d). But for d >> 0, (d >

2g− 2 = degKC) thus A∂d is smooth, that is, it is a morphism, it is surjective,and its differential is everywhere surjective, so it has smooth fibers, in fact, thefibers are Pd−g. In particular, the fibers are simply-connected.

Conclusion: we get a natural rank 1 linear system on Picd(C) for d > 2g−2.Namely (A∂d)∗(Ed)Sd . This is a rank 1 linear system on Picd(C). Finally,this extends to cE on Pic(C) as there exists a natural isomorphism Picd(C)→Picd−(2g−2)(C) for all d. For d > 4g − 4, the pullback of our cE,d−2g+2 is cE,d.

5 Lecture 5

Today we’ll talk about the arithmetic/number theory version of the LanglandsConjecture.

Dictionary:

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”SpecF”, or more likely the collection of ”places” in the field, correspondsto our curve C.

A number field corresponds to a function field k(C).A=adeles.For all x ∈ C, OX ⊂ Fx with quotient field kx.Over k = C, kx = C and Ox = C[[t]] and Fx = C((t)).Define O =

∏x Ox ⊂ A :=

∏′x Fx ⊂

∏x Fx = F .

GL(F ) \GL0(A)/GLn(O) ∼ BunGLn,C .Above, O → A an inclusion and F → A is also an inclusion. The first is the

collection of things with no poles and the second are the global meromorphicfunctions. And so we can restrict GLn(A) to each subset, and quotient out.This is Weil’s approach to the moduli of vector bundles.

Case n = 1: GL1(A) = I is the set of ideles.This leads naturally to stacks, because for n > 1 we don’t have a moduli

SPACE, but rather need a STACK

5.1 Assigning a Vector Bundle to an Adele Matrix

What we need is an open cover of C and gluing matrices. As we want GLn(O)to be trivial, we should focus on the poles.

Try listing the poles x1, . . . , x`. Let U0 = C \ x1 and let Ui be a smalldisc containing xi. Then the vector bundle is trivial on each Ui and U0, and weneed gluing data on U0 ∩ Ui for 1 ≤ i ≤ `. There is no compatibility conditionnecessary.

This is given by the adelic matrix at xi.Back to the dictionary, we can see that Bunn,C corresponds to GLn(F ) \

GLn(A)/GLn(O).Local systems correspond to unramified `-adic representationsGeneral `-adic representations correspond to locally systems on curve CAutomorphic functions correspond to the trace of the Frobenius morphism

acting on Hecke eigensheafs.An `-adic representation: take ` to be prime and not divide q = pn, and

F = K(C). Then σ : Gal(F /F )→ GLn(Q`).(A local system is an `-adic representation where F = K(C), Q` = C,

k = Cand we consider only unramified representations)Reasonable conditions:

1. Nowhere ramified (ie, σ(Interiax) = 1 for all x)

2. σ is geometrically irreducible, that is, irreducible on Gal(F /Fk).

6 Lecture 6

The correspondence is best expressed in three columns, Number Fields, FunctionFields and Geometric Langlands.

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The base field will be Q or Fq(t) or k(P1), and we will look at F ⊃ Q,F = k(C) where C/Fq, and k(C) where C/k.

So we have G is any connected(?) reductive group, and then for functionfields, we look at G(F ) \G(A)/G(O) and corresponds in the geometric case toBunG,C , and G(A) corresponds to the Bundles with a meromorphic trivializa-tion and a local trivialization at each x. (In terms of schemes, we demand atrivialization at each point, including the generic point)

We note that G(A) =∏′xG(Fx), and this will give us gluing data for the

bundles, giving a bijection.Quotient by /G(O) corresponds to forgetting local trivializations and by

G(F )\ to forgetting the generic trivialization.`-adic representations correspond to local systems, that is, σ : Gal(F /F )→

G(Q`) correspond to σ : π1(V )→ G.An automorphic function f : G(A)→ Q` which is invariant under G(O and

G(F ) which satisfies

1. It is cuspidal

2. It is a Hecke eigenfunction.

corresponds to an automorphic sheaf which is irreducible and a Hecke eigen-sheaf.

Now take G = GLn. Then 0 ≤ i ≤ n, x ∈ C, we have πx a uniformizer at x,that is, a local coordinate on C at x.

then look at Hix = GLn(Ox)diag(πx, . . . , πx, 1, . . . , 1)GLn(Ox), a double

coset in G.We define a Hecke operator to be (T ixf)(g) =

∫Hi

Xf(ghx)dhx.

The picture is that Heckeix is a correspondence Bun → Bun, specificallythe set (V, V ′)|V ⊂ V ′ ⊂ V (x).

So given a function, we are pulling back to Heckeix and then integrating overthe fibers to get a function on Bun.

Conjecture 6.1. For all σ `-adic representations, there exists fσ an automor-phic function, unique up to Q∗` with T ixfσ = |k(x)|−i(i+1)/2 tr(∧iσ(Frobx))fσ.

This suggests a relation between sheaves and functions. tr(Frobx0), but thismakes sense only over Fq.

7 Lecture 7

7.1 Extensions and Connections

1. Connection with Shimura-Taniyama-Weil

2. GLC for other reductive grops

3. Ramified versions

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4. D-modules and Derived Categories

5. Stacks, Gerbes

6. Hitchen’s System and its quantization

7. Connection to Mirror Symmetry

We will do an ”impressionistic” look at the first, and it won’t be used later.Can we see geometric objects in the number field version of LC?Geometrically, we take x ∈ C → P1 and in number theory we have a prime

in the ring of integers in a finite extension F of Q.For a rank n bundle on C, we don’t know a good analogue in the number

theory, but if n is two, then we have ”tate modules” which are associated toelliptic curves.

If E is an elliptic curve over K and ` doesn’t divide the discriminant, itcorresponds to E elliptic over C(P1) with discriminant the set of points wherethe differential of E → P1 vanishes. The discriminant for the number theoreticelliptic curve can be thought of as a family of curves over the primes, and then` not dividing the discriminant is the same as E reduced modulo ` is smooth(this is the fiber over `).

Consider the `k-torsion points of E. Then the Tate module of the curve Eis lim←−k E[`k].

And TSW is a special case of Langlands.Back to geometric Langlands, what about other reductive groups?A reductive group is the complexification of a connected compact real group.

Equivalently (we think) every representation is a direct sum of irreducibles.

Example 7.1. S1 is a connected compact real group. C∗ = GL(1,C) is acomplexification, because it has an involution whose fixed points are S1. C∗ isa complexification of R+ with multiplication.

Similarly, Tn gets (C∗)n.

Example 7.2. U(n), viewed as a real group, has GL(n,C) as a complexification.

Example 7.3. SU(n) complexifies to SL(n,C). (possibly quotients by finitesubgroups of the center are also complexifications of SU(n))

Example 7.4. SO(n,R) complexifies to SO(n,C).

Example 7.5. Sp(n,C) ∩ U(n) complexifies to Sp(n,C).

Any complex simple group is reductive. There exists a complete classifi-cation: SL(n), SO(2n + 1), Sp(n) and SO(2n) and five exceptional groupsF4, G2, E6, E7 and E8 and quotients of the above by finite subgroups of thecenter. When talking about simple Lie groups, we mean no connected normalLie subgroup.

Two Liegroups G1, G2 are isogenous if they are quotients of a third group Gby finite subgrousp of its center, that is, G1

∼= G/S1 and G2 =∼= G/S2.

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We will give names to the groups mentioned before, SL(n) will be An−1,SO(2n+ 1) is Bn, Sp(n) is Cn, SO(2n) is Dn, and then there are E6, E7, E8, F4

and G2, the Cartan-Killing List, when An for n ≥ 1, Bn for n ≥ 2, Cn for n ≥ 3and Dn for n ≥ 4.

This classifies teh isogeny classes of simple complex groups, the simply-connected groups, and the centerless groups. (This last class is called groups ofadjoint type)

Classification of reductive groups, R = (A×S)/F where A is abelian, whichcan be abelian varieties, complex tori, Cn and (C∗)n, S is simple and f is afinite subgroup of the center.

Eg, GL(n,C) = C∗ × SL(n,C)/ZnThe Lie algebra of a Lie group G is g = TeG, the left invariant vector fields on

G. The operation [, ] is the commutator of vector fields. Lie algebra determinesthe Lie group up to isogeny.

8 Lecture 8

Today we’ll talk about GLC for other groups.Let G be reductive, that is, (A× S)/F where A is abelian, S is semisimple

and F is finite.A couple of key concepts:”Maximal Torus” T ⊂ G is a connected abelian subgroup isomorphic to

(C∗)r which is maximal among such subgroups consisting of elements that aresemisimple in the adjoint representation. (that is, diagonalizable in the adjointaction on the Lie Algebra)

Example 8.1. If G = GL(n), then T can be taken to be the diagonal matrices.The adjoint representation is a map ad : GL(n) → Aut(gl(n)) = GL(n2). Itmapes g 7→ (A 7→ gAg−1). For g = diag(g1, . . . , gn) ∈ T , ad(g) is the diagonaln2 × n2 matrix with eigenvalues gig−1

j .

Caveat: T is not the same as a maximal abelian subgroup.Basic Fact: All maximal tori in a group G are conjugate. In particaulr, r is

an invariant of G called rank.Rank(GL(n)) = n. Rank(SL(n)) = n − 1, and this is why we call this

An−1. GL(n) = An−1 × C∗/Zn. Rank(G × H) = Rank(G) + Rank(H).Rank((C∗)n)) = n.

Better, replace G by g and T by t, and call this the Cartan Subalgebra.Aut,Ad, ad. Let g ∈ G and A ∈ g.Then Autg ∈ Aut(G) takes h 7→ ghg−1, and Aut : G→ AutG is a map.Then Ad : G→ Aut(g), and ad : g→ End(g).Autg : G→ G induces a linear map TeG→ TeG, that is, g→ g, this is Adg.

Then ad = De(Ad), that is, the induced map ad : g → End(g), acuting on thetangent space of Aut(g).

Caveat: a maximal abelian subalgebra of g is not necessary a Cartan subal-gebra.

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charG = charT = hom(T,C∗) ∼= Zr. hom(T,C∗) ⊂ t∨

On g, there exists a natural pairding A,B 7→ Tr(adA adB). This is theKilling Form, it’s a natural bilinear pairing on g, and hence on t. It induces oneon t∨, and it takes Q values on char.

Example 8.2. Let G = SL(3), then char is the hexagonal lattice.

How about isogenous groups?Let G2 = G1/F where F is finite. Then T1 → T2 is surjective and the kernel

is F . Sp then we have charG2 → charG1 → F∨ and we get lattices of the samerank.

Picture of an isogeny class: Starts with Gsc the universal cover and goes toGad = Ad(G) ⊂ Aut(g) = G/Z(G).

Then π1(Gsc) = 0 and Z(Gsc) is the biggest possible. Then Gad has thebiggest π1 possible, and Z(Gad) = 0, and there are things in between.

Eg, SL(n). SL(n) is simply connected, and P(GL(n)) is centerless. We callchar(Gsc) the weights and char(Gad) the roots.

So then we have roots ⊂ chars ⊂ weights. Swapping these gives the Lang-lands Dual Group, which we will discuss next time.

9 Lecture 9

Let G be reductive, that means that G = (S ×A)/F as before.ASIDE: Complex Torus is a word used inconsistently. In complex geometry,

a complex torus is (S1)2g with a complex structure (this is the same as compactconnected abelian (abelian is implied, not a necessary hypothesis) complex Liegroup), and in group theory a complex torus is (C∗)r, this is also called analgebraic torus.

Let T ⊂ G be a maximal torus and t ⊂ g the Cartan subalgebra given by T .

Example 9.1. If G = GL(n) and T is the collection of invertible diagonalmatrices, then t is the set of arbitrary diagonal matrices.

To a reductive group G we associate six lattices as follows. roots ⊂ chars ⊂weights, coroots ⊂ cochars ⊂ coweights.

chars = hom(TG,C∗) ∼= Zr where r is the dimension of T and is defined tobe the rank of G.

The isogeny class of G includes Gsc and Gad. The roots are the charactersof Gad and the weights are teh characters of Gsc

Available structure: Lattice=(free abelian group with a Q valued pairing)The dual lattice means the dual group with the dual metric.So we can define the Langlands dual group LG is the group whose characters

are cocharacters of G. We have no justification yet for existence or uniquenessof such.

We will assume that such a group exists and is ”reasonably functorial” ingneeral.

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But explicitly define N to be the normalizer of T in G and W to be N/Tthe group of connected components of N .

Example 9.2. G = GL(n) and T the diagonals. Then N is the collection ofinvertible matrices with a single nonzero entry in each row and column. ThenW is Sn.

Example 9.3. Take G = SO(2n). Then if we take this as the group preservingthe form x1x2n + x2x2n−1 + . . .+ xnxn+1, then it is anti-skew-symmetric, andso has diagonal α1, . . . , αn,−αn, . . . ,−α1 as the maximal torus. Thus, SO(2n)has rank n.

To determine W , we see that it fits into the short exact sequence 1 →(Z/2)n−1 →W → Sn → 1 where the first is the group of even sign changes.

Facts

1. W depends on g, not G.

2. W (G) = W (LG).

3. W acts (by conjugation) on T . Therefore on characters hom(T,C∗). Wpreserves the pairing and so is a group of isometries of the lattice.

Example 9.4. G = GL(n), then the characters are Zn the square lattice, sowe really have (Z, standard)⊕n. So W = Sn. The obvious permutation actionon an orthonormal basis preserves the pairing only up to ±.

The maximal torus of LG is the dual of hte maximal torus of G. TLG =(TG)∨, that is, tLG = (tG)∨.

General yoga: The group G can be recponstructed from t, char,W action,metric. An, Dn, En are ”simply laced” and Bn corresponds to Cn, G2 to G2

and F4 to F4.

10 Lecture 10

Example 10.1. If G = GL(n) then LG = GL(n)If G = SL(n) then LG = PGL(n)

In general, with Lie Algebras, each type maps to itself except that B and Cswap.

At the group level, LGsc is the appropriate adjoint group.

Example 10.2. Z(SO(2n)) = Z/2 and π1(SO(2n)) = Z/2. So there is adouble cover of SO(2n) which we call Spin(2n) and is simply connected. So wehave Spin(2n) → SO(2n) → PSO(2n) and this has fundamental group eitherZ/4 (if n odd) and Z/2× Z/2 (if n even).

So now, let C be a smooth compact complex curve of genus g. Define BunG,Cto be the moduli space of semistable principal G-bundles on C and LocG,C tobe the moduli space of semistable principal G local systems on C.

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Theorem 10.1. TFAE

1. ρ : π1(C)→ G

2. (P,∇) where P is a principal holomorphic G-bundle over C and ∇ is aflat holomorphic connection.

3. Locally constant sheaf of groups on C with fiber isomorphic to G.

Roughly, what Geometric Langlands says is that points of LocG correspondto Hecke Eigensheaves on BunLG.

A better version is: introduce the notion of D-modules. Define D to be thesheaf of rings on BunLG given by the holomorphic differential operators. On anyX, we have OX the functions, TX the vector fields, and DX ⊂ HomC(OX ,OX).We define a filtration D0

X ⊂ D1X ⊂ . . . and the union is DX . Then D0

X = OX .We say that L ∈ Di if [L,OX ] ⊂ Di−1

X . That is, L(fg)− gL(f) ∈ Di−1.In local coordinates, a section of DX is a differential operator f0(z) +

f1(z) ∂∂z + f2(z) ∂2

∂z2 + . . ..

Conjecture 10.1 (Geometric Langlands Conecture). There exists a naturalequivalence of catagories c : Db(Loc,O)→ Db(LBun,D). c of a structure sheafof a pont is a Hecke eigensheaf on LBun with eigenvalues corresponding to theoriginal local system.

Next time: Hecke Operators.

11 Lecture 11

As a point x, we have Heckex = (V, V ′, β)|V, V ′ ∈ Bun and β : VC\x ' VC\xWe define Hecke to be (V, V ′, β, x)|x ∈ C, . . . a relation in Bun× (Bun×

C).Note: The fibers of p, q, px, qx are infinite dimensional. They are affine

grassmanians, that is, V is a principal G-bundle on C with trivialization onC \ x.

These are Ind-schemes.How do we get finite dimensional fibers?Let ρ : G → GLN be any representation. Let T ⊂ G be a maximal torus.

Then ρ|T can be simultaneously diagonalized: there exists w1, . . . , wN simulta-neous eigenvectors.

Then there exist functions λi : T → C∗ such that for all A ∈ T , ρ(A)wi =λi(A)wi. Clearly, λi is a group homomorphism, and so it is a character.

Upshot to an N -dimensional representation ρ of G, we associate an N -tupleof characters. This set of characters is invariant under the Weyl group W anduniquely characterizes an irreducible ρ.

So now we have finite dimensional Heckes, and we have

LHeckeµx = (V, V ′, β)|∀ finite dimensional representations ρ : LG→ GLn(C)

11

with ρ(β) : ρ(V )→ ρV ′)⊗ OC(〈µ, λρ〉, x)

where λρ is the highest weight of ρ.This has finite dimensional fibers, and µ1 ≤ µ2 implies that LHeckeµ1

x ⊂LHeckeµ2

x . So we can form lim−→µLHeckeµx = LHeckex.

Example 11.1. For G = SL(n), this runs over Zn−1/Sn. This is gneerated byn− 1 elements µ1, . . . , µn−1. The fiber of Heckeµi

SL(n),x = Gr(i, n).Unfortunately, for all other µ, the fibers of HeckeµSL(n),x are singular.

For general G and µ, the fibers are smooth iff the representation with highestweigth µ has only Wµ and 0 as weights, this is phrased as µ being miniscule.

Hecke OperatorsLµH : Db(LBun,D)→ Db(LBun,D) by m 7→ qµ! (pµ

∗m⊗ ICµ).

These operators generate an algebra. This algebra is commutative.

12 Lecture 12

Conjecture 12.1 (GLC, tentatively). There exists a natural equivalence c :Db(Loc,O) → Db(LBun,D) sending structure sheaves of a point V to auto-morphic D-modules M , ie, to the Hecke eigensheaf LHeckeµ(M ) = M V onLBun× C.

Define pµ : Heckeµ → LBun to be the first projection, and qµ : Heckeµ →LBn× C the second. Define H eckeµ(M ) = qµ! ((pµ)∗M ⊗ ICµ).

So LH echeµ(M ) = M pµ(V ) on LBun× C.Discrepancy: LBun is disconnected. π0(LBun) = H2(C, π1(LG)) = π1(LG) =

(Z(G))∨.

Example 12.1. G = GL(n) = LG. detGL(n) → C∗ gives π1(GL(n)) →π1(Z∗) = Z.

So topological types of rank n bundles on C are parametrized by their degreed = c1 ∈ Z.

There exist infinitely many components of n ”types”

Example 12.2. LG = SL(n), then π1(LG) = 0, and so BunSL(n) is connected,so we have BunSL(n) → Bun0

GL(n) →J , with the first map inclusion and thesecond the surjective determinant map.

Example 12.3. LG = PGL(n). A PGL(n) bundle on C is an equivalence classof GL(n) bundles modulo ⊗LB. Then we have C∗ → GL(n) → PGL(n), andso the number of components of BunPGL(n) = n.

So now, we note that Db(LBun,D) =∏d∈π1(LG)D

b(LBund,D). However,we get no such statement for Loc, due to the flat connection! So either LBunis too big or Loc is too small.

12

A solution is to replace the moduli space Log by the moduli stack L oc.(Actually, it’s a C∗ gerbe)

The new objects are ”twisted sheaves”:A ”sheaf” on X can be twisted by a cohomology class in H2(X,O∗). Fix a

Cech 2-cocycle Ui an open cover αijk ∈ Γ(Uijk,O∗) with αijkαjk`αk`iα`ij = 1on Uijk`.

An α-twisted sheaf F on X is an assignment of a sheaf of abelian groupsF (Ui) to each Ui. and transition functions gij : F (Ui)|Uij

→ F (Uj)|Uijsatis-

fying gijgjkgki = αijk.This is precisely the same thing as a sheaf, except the condition involving α.The case with αijk = 1 is an ordinary sheaf.If αijk and α′ijk are cohomologous, there is an obvious equivalence between

α-twisted sheaves and α′-twisted sheaves.So we can talk about sheaves twisted by classes in H2(X,O∗).Sheaves on L oc are twisted sheave on Loc, and under GLC, we haveSheaves on Loc (sheaves on L oc with α ≡ 1) correspond to the neutral

component of LBun and genuinely twisted sheaves correspond to the othercomponents.

It’s complicated...easier to consider only ”regularly stable” local systems. They form Zariski

open Locrs, and L ocrs is a gerbe over Locrs with structure group Z(G) actingtrivially.

So twists correspond to characters of Z(G) = π1(LG)Next time will be preempted by a physicist, and Monday we start Hodge

Conjecture.

13 Lecture 13 - Hodge Theory Begins Here

Let X be a smooth projective variety. Let Z ⊂ X a subvariety of complexcodimension p. Then we have [Z] ∈ H2p(X,Z)→ H2p(X,C).

Properties of [Z]

1. It is integral.

2. It is of Hodge type (p, p).

3. It is positive.

The set of classes in H2p(X,Z) that arise from Z’s is a cone. Characterizingit is very complicated. Instead, ask for its linear span:

Conjecture 13.1 (Hodge Conjecture). Any Hodge class (ie, (p, p) class inH∗(X,Q)) is a Q-linear combination of algebraic cycle classes.

The original version, stated by Hodge in 1952 was this over Z. It is false.There is a generalized Hodge Conjecture which was shown false by Grothendieck,

but is fixable. There is also a variant by replacing [Z] by ci(V ), the chern classes

13

of a vector bundle. Variants for compact Kahler. Variants for other cohomologytheories.

Why would we conjecture this? The conditions above are necessary, but noone has found any examples of sufficient conditions beyond them.

The main ”evidence” is the case that p = 1, due to Solomon Lefshetz.This case holds over Z and for compact Kahler X.The basic ideas is to look at the exponential sequence 1→ Z→ O → O∗ →

1. Taking cohomology, we get 0 → H0(X,Z) → H0(X,O) → H0(X,O∗) →H1(X,Z) → H1(X,O) → H1(X,O∗) → H2(X,Z) → H2(X,O). The mapH1(X,O∗)→ H2(X,Z) is the first chern class.

In the case whereX is a curve, c1(L) = deg(L). In general c1(L) : H2(X,Z)→Z takes c to degL/C.

Plan: Characterize the image of c1, and then identify with the image ofcodimension 1 cycles.

Part 1: The image of c1 is ker(H2(X,Z) → H2(X,O)) and this map isH2(X,C) → H20 ⊕H11 ⊕H02 to H02. So then H2(X,Z) ∩H11 is the Hodge(1, 1) classes.

Given an effective divisor Z ⊂ X, we get a line bundle L = OX(Z). Con-versely, any algebraic line bundle L has rational sections so it is O(Z) for someZ. For holomorphic bundles, use GAGA for PicalgX ∼= PicanX.

Summary: on a smooth projective X

1. Any Hodge (1,1) class is c1 of a line bundle

2. Algebraic line bundles are analytic line bundles

3. Any algebraic line bundle is O(Z) for some divisor Z.

4. c1(O(Z)) = [Z]

So the Hodge conjecture is true for p = 1 over Z.This turns out to be most of what is known about the Hodge conjecture.(Consider the subring of H∗(X,C) generated by H1,1 ∩H2(X,Z). It follows

that any class here is algebraic.)What would a counterexample be?Not a curve, not a surface, not a 3-fold...so problem starts on a 4-fold for

(2,2) classes.Start with a 4-fold, for (2, 2) classes. In a typical family of 4-folds, eg, all

smooth hypersurfaces of fied degree in P5 (or any other 5-fold) H2,2 is a vectorspace of fixed dimension H4(X,Z) is a lattice of fixed rank, but H4(X,Z) ∩H2,2(X) jumps (we call this Hodge2.)

For generic X, Hodge2(X) is as small as possible, but for special X, it jumpsup.

The Noether-Lefshetz or Jump locus is a countable union of algebraic sub-varieties (usually dense) of the parameter space.

Fix a class in H4(X,Z). What is the locus in the base where it becomesHodge? That is, in H2,2?

14

The Hodge Conjecture implies that this is algebraic.Cattani, Kaplan, and Deligne proved this corollary.The most likely counterexample would be this type.A Weil descrbed such classes on Abelian 4-folds: for special abelian 4-folds,

Hodge2 is not generated by Hodge1.

14 Lecture 14

NO CLASS FRIDAYLet X be a complex manifold (compact, often). Let Ak(X) be the C∞ k-

forms. Then these decompose into ⊕p+q=kAp,q(X), and elements of these arew =

∑|I|=p,|J|=q wIJdzIdzJ .

HkdR(X,C) is homology with respect to d = ∂ + ∂, with each being the

differential in the Ap,q’s.On a general compact complex manifold, there is no Hodge decomposition.

Example 14.1 (Hopf Surface). Look at T 2 → S → P1 surfaces S. That is,C∗ → C2 \ 0 → P1. Choose q ∈ C∗, |q| 6= 1. Then we have (C2 \ 0)/(q). Thisstill maps to P1, but it’s fiber isn’t C∗, it’s C∗/(q), which gives T 2.

Let h be a hermitian matrix,∑i,j hijdzi⊗dzj . Define ω = i

2

∑i,j hijdzi∧dzj .

Then ω is the Kahler form of h, it’s a (1, 1) form on X. So then X (or (X,h))is Kahler if dω = 0. Any projective manifold is compact and Kahler.

Basic results:

1. (X,h) is Kahler iff the complex structure is parallel with respect to Levi-Civita connection.

2. ∆d = 2∆∂ on a Kahler manifold.

3. A form on a Kahler manifold is harmonic iff all its (p, q) components areharmonic. Thus, on a compact Kahler X, we have the Hodge Decompo-sition.

So the Hodge conjecture is that any (p, p) rational cohomology class is ”al-gebraic”. Counterexamples on compact Kahler manifolds:

Example 14.2. Let X = T 4 = C2/Λ. Where Λ is any discrete lattice isomor-phic to Z4. Pick any constant hermitian form. It gives a (1, 1) form on C2.This may or may not descent to a rational form on X. It will be the class of adivisor (curve) in X iff h is positive.

Don’t impose this, take an indefinite h. Deformation tehory works: for somecomplex structures, this is a rational (1, 1)-form. So there exists a holomorphicLine bundle L with ω = c1(L), but h0(Ln) = 0 for all n ∈ Z nonzero. On ageneral such X, this is the only line bundle or the only hodge class.

15

so for Kahler manifolds, need to replaces algebraic by ”chern class of a vectorbundle”

notions of algebraic: Hodge classes 〈ci(F )|F is a coherent sheaf on X〉.This contains both the classes coming from subvarieties and those from vectorbundles.

15 Lecture 15

16 Chern Classes

Let X be a topological manifold and E a complex vector bundle over X.Then ci ∈ H2i(X,Z).If X is a complex manifold and E is a holomorphic bundle, then ci are Hodge

classes, in Hi,i(C,Z) ∩H2i(X,Z).Chern-Weil: ∇ any complex connection on E, ∇ : E → E ⊗ A1, then R∇

the curvature is in A2(End(E)).Then let σi be the ith symmetric polynomial of the eigenvalues of a matrix.

Then σ1(A) = tr(A). σ2(A) = 12 (− tr(A2) + (trA)2), etcetera.

Applied to R∇, we get ck =(i

2π

)kσk(R∇) ∈ A2k(X).

R∇ is an endomorphism, so defined only up to conjugation. But σk(R∇) isbasis independent. This depends on a choice, ∇.

Effect of change: ∇ 7→ ∇ + ω for ω ∈ A1(End(E)) is R∇ 7→ R∇ + Exact,and so ck 7→ ck + Exact.

The upshot, [ck(E,∇)] ∈ H2k(X,C). This is independent of ∇, and so wecall it ck(E). It’s actually in H2k(X,Z)→ H2k(X,C)’s image.

Another approach: Grk(n) the set of k-dimensional quotients of Cn. OverGrn(N), there is a universal rank n vector bundle Q (in fact, we have a shortexact sequence of vector bundles 0→ S → V → Q→ 0 over Grn(N)).

Let f : X → Grn(N) by any C∞ map. Then we get a pullback vector bundlef∗Q on X. (In a sense, grassmannians are universal...make this precise.)

Facts:

1. Any rank n vector bundle E on X is f∗Q for some f and some N >> 0.

2. Any two such (f,N), (f ′, N ′) are stably equivalent. That is, if we havethe same bundle arising from two such maps, there is another N ′′ largerthan N,N ′ and a map f ′′ : X → Grn(N ′′) such that the diagram withthese four objects and the natural maps commutes.

So then, for any class c ∈ H∗(Grn(∞)) the pullback f∗c is the characeristicclass of X,E.

Grn(∞) is a classifying space for vector bundles. A topological vector bundleE → X is equivalent to [f : X → Grn(∞)].

Example 16.1. For n = 1, Gr1(∞) = P∞. As H∗(PN−1,Z) = Z[H]/Hn, weobtain H∗(P∞,Z) = Z[H].

16

So if E = f∗O(H), then c1(E) = f∗H ∈ H2(X,Z) is the degree of E. If Xis a complex curve, then this is in H2(X,Z) = Z.

Splitting Principle: Formally, any E → X a vector budnle, there existsπ : Y → X smooth map such that π∗E is a direct sum of line bundles.

Informally, any fact that holds for direct sums of line bundles holds for allvector bundles.

ck(E) ∈ H2k(X,Z), because this is true for line bundles and the Whitneyformula c(E) = c(E1)c(E2) if E = E1 ⊕ E2 or even 0→ E1 → E → E2 → 0.

We want the Hodge classes to contain a type ∗, which contains as specialcases the chern classes of vector bundles and the cycle classes of algebraic sub-varieties. The possibilities for ∗ are the chern classes of coherent sheaves, thechern classes of complexes, and teh chern classes of Db(X,Coh).

There is a natural map c : Db(X,Coh)→ H∗(X,Q).We define the chern class of a complex to be c(complex) =

∏c(Fi)(−1)i

.Check that this is well defined up to quasi-isomorphism.

17 Lecture 16

c∗(coherent complexes) containes c∗(VB Complexes), which contains the classesof algebraic sycles and c∗(V B), whic is contained in c∗(coherent sheaves) whichis also contained in c∗(coherent complexes).

Why are classes of algebraic cycles related to chern classes of VBs or coherentsheaves?

Due to resolutions: to an algebraic subvariety Z ⊂ X we associate IZ ⊂ OX .Then [Z] = c(i∗OZ).

Use 0 → IZ → OZ → OZ → 0. This gives a quasi-isomorphism 0 → IZ →OZ → 0 with 0 → 0 → i∗OZ → 0, and so this chern class is c(IZ → OX). Sowe have algebraic cycles contained in coherent sheaves, contained in coherentcomplexes. But NOT contained in VB. However, it can be written as a VBcomplex:

If codimZ = 1, then IZ is a LB, and so IZ → OX is a complex of VB.On smooth n-dimensional X, any coherent sheaf has an n-step locally free

resolution.In fact, let O(1) be ample, then any coherent sheaf has resolution by vector

bundles of the form ⊕O(di).

Example 17.1. X = P2, and Z is a point. Then we have 0 → O(−2) →(O(−1)⊕2)→ OP2 → Opt → 0.

H0(OP2(1)) =< x, y, z >, and say the point is x = y = 0. Then H0(IZ) =<x, y >.

So what is c(Opt)? It is c(O(−2)→ O(−1)2 → O) = c(O)c(O(−2))/c(O(−1))2 =1(1−2H)(1−H)2 = 1−2H

1−2H+H2 . This is 1−2H+H2

1−2H+H2 − H2

1−2H+H2 , and so we have 1−H2 (byexpanding the denomenator and remembering that H3 = 0.)

17

Similarly for P3, except we have to go another step (and get(

3k

)instead of(

2k

)as the expondents). We get that the chern class of the complex is 1 + 2H3.

The class of Ipt is 1− 2H3. In general, c(IZ) = (−1)k/(k − 1)![Z].Cohomology of ToriLet Γ ∼= Z2g, and ΓQ = Γ⊗Q , and ΓC = Γ⊗ C2g.So then set X = Γ \ ΓC/W , with W = Cg, and W ∩ W = 0.Interpretation: Γ = H1(X,Z), ΓQ is cohomology with Q coefficients, etc.Hk(X,Z) =

∧k Γ.The Hodge Filtration Hp,q =

∧pW ⊗

∧qW ⊂

∧p+q ΓC.Are there Hodge classes? In general (this doesn’t make sense on a moduli

space, it makes sense before dividing by isomorphisms) there are none. (∧p

W⊗∧pW ) ∩

∧p+q Γ = 0. So the Hodge Conjecture is true, because it there are noHodge classes, then every hodge class is algebraic.

However, by moving W into special position, we can arrange to have Hodgeclasses.

The number of complex parameters is g(2g−g) = g2. For p = 1, the complextori that have a Hodge1 class depend on g(g+1)

2 complex parametersThere exist several components distinguished by teh signiture of the (1, 1)

form.Positive⇒abelian varietiesIndefinite⇒non-algebraic tori having Pic 6= 0.The interesting case is g = 4, p = 2. Generically, Hodge2 = 0. Always,

(Hodge1)⊗2 → Hodge2. So the Weil tori are the ones where Hodge2 is strictlybigger than the image of (Hodge1)⊗2. We will look at these more tomorrow.

They exist for both abelian varieties and for nonalgebraic tori. This willshow that the Hodge conjecture is false for Kahler manifolds

18 Lecture 17

Weil Tori:Look at the ring Z[i]. This acts on Γ, the lattice. Then the torus is ΓZ \

ΓC/H1,0, where γ ∼= Z2g with a bilinear form.

Let g = 2n.i acts on Γ, hence on ΓC, with eigenvalues ±i.Let C be a curve of genus g. Then Jac(C) = H1(C,O∗) = H1(C,O)/H1(C,Z) =

Γ.And so 0 → C → O → W → 0. Thus he have H1,0 = H0(C,W ) →

H1(C,C)→ H1(C,O) = H0,1, and take H1(C,Z) \H1(C,C)/H1,0.X = Γ \ ΓC/H

1,0, W = Wi ⊕W−i.Stability of Vector BundlesWe saw examples of jumping phenomenon for X a variety, 0 ∈ P a parameter

space and V a vector bundle over X × P , with p, q ∈ P , p 6= 0 6= q, thenV |X×p ∼= V |X×q, but V |X×p 6∼= V |X×0.

So in the natural topology, the set of isomorphism classes of vector bundlesis not separated.

18

How to construct a reasonable moduli space?Restrict to stable bundles, then we get a nice moduli space. Or we can

consider all vector bundles and then we have to work with a moduli stack. Wewill focus on the first.

In the case that X is a curve, C, then we define the slope of a vector bundleV → X to be µ(V ) = deg(V )/ rank(V ). If V = ⊕Li, then µ(V ) is the average ofthe degLi. A vector bundle V is stable if for all proper subbundles 0 ( V ′ ( Vthen µ(V ′) < µ(V ).

A better, equivalent definition: allow V ′ to be any coherent subsheaf, thenrankV ′ is defined to be additive in exact sequences, and so take a locally freeresolution. Equivalently, it is the rank at the generic point of X.

Example 18.1. On P1, and bundle is a sum of line bundles. So the only stablebundles are line bundles.

Because of this, we introduce the following:

Definition 18.1 (Semistable). We call V semistable if ∀V ′ 6= 0 ⊂ V , we haveµ(V ′) ≤ µ(V ).

So now on P1, O ⊕ O is semistable, but O(1)⊕ O(−1) is instable.For a curve, a projective moduli space BunC,n exists paramaterizing S-

equvialence classes of semistable bundles.It contains a Zariski open subset of isomorphism classes of stable bundles.

Definition 18.2 (S-equivalence). S-equivalence is the relation generated by0→ V1 → V → V2 → 0 where µ(V1) = µ(V2), then V ∼= V1 ⊕ V2.

This does extend to higher dimensions of X.The only thing we must do is define the slope of a vector bundle on X.

We need to fix a Kahler class or an Ample Line Bundle on X. That is, fixω ∈ H1,1(X) greater than 0. Then we define deg V = c1(V ) ∧ ωn−1/ωn, wheren = dimX.

If X is projective, then this is literally the projective degree of a divisorrepresenting c1(V ).

Typical Behavior:The Kahler cone is partitioned into subcones, and inside of each subcone,

we get the same moduli space, but as you cross between subcones, you get adifferent moduli space, and points on the boundary give something singularwhich is dominated by each.

We are going to a theorem of Donaldson-Uhlenbeck-Yau which gives a dif-ferential geometric interpretation of stability.

19 Lecture 18

Hodge Classes on 2n-ToriLet Γ ∼= Z4n, X = Γ \ ΓC/W . Note that Γ = H1(X,Z) = H1(X,Z)∗ and

ΓC = H1(X,C) = H1(X,C)∗.

19

To we have 0 → H1,0 → H1(X,C) → H0,1 → 0 and 0 → W = H0,1∗ →H1(X,C)→ H1,0∗ → 0.

How common are Hodge classes?There is no ”moduli space’ of complex tori of dim > 1, and so we have to

work in ”period space”. that is, we parameterize (X, i), i : Γ→ Z4n an iso.This period space is open (in the complex topology) in Gr(N, 2N) consisting

of W such that W ∩ ΓR = 0.the dimension is N2. So now we fix γ ∈ H2n(X,Z) =

∧2nH1(X,Z) =∧2n Γ∗ =

∧2nH1(X,Z) =

∧2n Γ (by Poincare Duality)When is it of type (n, n)?(Variants among N not necessarily even, γ ∈ H2k(X,Z), 0 ≤ k ≤ N)So γ ∈ Gr(N, 2N). TγGr ∩Gr ⊂ P(

∧N ΓC), the Plucker Space.This is a Schubert cycle σ1 = S ∈ Gr|S ∩ γ 6= 0.Claim: σ1 is a singular divisor in Gr. It’s singular locus σ22 is the locus of

W where γ is Hodge.General element of Plucker Space is linear combination of basic verieties

ei1 ∧ . . . ∧ einSay that γ = e1 ∧ . . . ∧ eN . Then TγGr = ei1 ∧ . . . ∧ eiN where at least one

ij is in 1, . . . , N.So Sing(TγGr∩Gr) is teh set where at least two of the ij ’s are in 1, . . . , N.

Thus, it is σ22 which has codimension 4.Explicitly, W ∩ σ should have vector space dimension 2, and so W |W∩σ has

N − 2.Gr(2, N)+Gr(N −2, 2N −2) = 2(N −2)+(N −2)N = N2−2n+2N −4 =

N2 − 4.Do the variant, then you will see that a generic torus on which a given

γ ∈ H4(X,Z) is of type (2, 2) has no Z classes of type (1, 1).Weil ToriX = Γ\ΓC/W . Add the assumtion that Γ is a Z[I]-module, where I2 = −1.

Then ΓC = W ⊕ W is the Hodge decomposition, and so is C2ni ⊕ C2n

−i.Z[i] acts on X iff W is Z[i]-stable iff W = Wi ⊕W−i and W±i = W ∩ C2n

i .Take dimW±i = n. Then we gave ΓC = W ⊕ W = Wi⊕W−i⊕ Wi⊕ W−i =

C2ni ⊕ C62n−i.

Our parameter space is Gr(n,C2ni ) ×Gr(n,C2n

−i). The only condition W =Wi ⊕W−i satisfies W ∩ W = 0 or W ∩ ΓC = 0, which is an open condition.

Claim; Hodge2 has rank ≥ 2 on a Weil Torus, and it is generically equal to2.

H2n(X,Q) ∼= H2n(X,Q) ∼=∧2n

Q ΓQ ⊃∧2n

Q[i] ΓQ ∼= Q[i].Claim: This is in Hn,n.∧2n C2n

i =∧2n(Wi⊕W−i) =

∧nWi⊕

∧nW−i ⊂

∧nW⊕

∧nW = Hn,n(X).

Now we note that the first part is the image of∧2n

Q[i] ΓQ[i],i →∧n ΓC over C.

20 Lecture 19

The analytic part can be read on our own in ”Voisin”

20

Obstructions to the Hodge Conjecture:There are three types

1. Torsion: H2k(X,Z)tor = kerH2k(X,Z)→ H2k(X,Q), which is also ker(H2k(X,Z)→H2k(X,C)). This is then contained in ker(H2k(X,Z) → F k−1) of theHodge filtration. And so, anything torsion is automatically Hodge. Sothen any torsion class should be algebraic, and there exists counterexam-ples: torsion classes on smooth projective varieties that cannot be alge-braic.

2. Integrality: Let X ⊂ Pn+1 a general hypersurface of deg d, eg n = 3, k = 2,then there exists a curve C ⊂ X such that [C] ∈ H4(X,Z) is aγ for a ∈ Z,a > 1 and γ ∈ H4(X,Z), but γ is not algebraic.

3. Generalized Hodge: Let X be projective, smooth, γ ∈ Hodgek(X) =Hk,k(X) ∩ H2k(X,Q). Then there exissts Z ⊂ X a subvariety of codi-mension k with γ ∈ Im(H∗(Z) → H∗(X)). So Generalized Hodge is,given γ ∈ H2k(X,Q) ∩ F k−i, then it is in the image of H∗(Z) → H∗(X)for some alg subvariety Z of codim k − i.

Note that if we have Z of codimension k, we have H2i(Z,Q)→ H2k(X,Z).Anything in H2i(X,Q) is within i steps of center, that is, H2k(X,Q) ∩ F k−i.The GHC is the converse.

Grothendieck shows that this is false, but a quick fix is to replace ”class γ”by ”Hodge substructure”

So the corrected version is Any Hodge Structure A ⊂ H2k(X,Q) ∩ F k−i isin the image of H∗(Z,Q) for some Z of codimension k − i.

A Hodge substructure is A = ⊕p+q=2k(A ∩Hp,q).Grothendieck’s example is γ ∈ H2k(X,Q) ∩ F k which is not contained in a

Hodge substructure.Kahler’s Example: X ⊆ P4 a general hypersurface of degree d. (IE, it holds

for X ∈ P(C[x0, . . . , x4]d) \ ∪countableVi with Vi a proper subvariety).Hk(P4,Z) has Z’s down the middle of the Hodge diamond and zeroes else-

where. For H∗(X,Z), we have (rotate 45 degrees clockwise)Z 0 0 ?0 ZH ? 00 ? ZH2/d 0? 0 0 ZH2/d

Claim: Assume p > 3 prime and p3|d, then the degree of any curve C ⊂ Xgeneral is divisible by p. (And so, HC over Z fails)

We’ll prove this claim next time.

21 Lecture 20

Today, we will prove a theorem which has the following corollary:

21

Corollary 21.1. The Hodge conjecture over Z is false for X as described inthe theorem.

Theorem 21.2. Let p ≥ 5 be prime, d = p3s ∈ N and X a general hypersurfacein P4 of degree d. If C ⊂ X is a curve, then p divides degC.

Proof. We will recall the construction, due to Grothendieck, of the Hilbertscheme parameterizing all subschemes of a given projective variety with a givenHilbert Polynomial.

We can also define the Hilbert scheme of pairs C ⊂ X ⊂ P4 to be the subsetof Hilb(X0)×Hilb(C0) with (X,C)|C ⊂ X = Hilb(C0, X0).

By looking at fiber products of the universal families over Hilb(C0) andHilb(X0), we can see that Hilb(C0, X0) is algebraic and parameterizes all pairsC ⊂ X. It also has a universal curve and hypersurface, with C ⊂X .

Now consider the projection Hilb(C0, X0) → Hilb(X0). We can break thisinto Hilbgen ∪Hilbspec, the former of which are the curves which exist over anyX, and the latter the ones that only exist over some X. By assumption, wehave X general, and so it is in Hilbgen. Thus, we can deform C ⊂ X to a cruvein any other X.

As X moves towards any particular X ′, C will move with it. That is, ifX → B is any appropriate family, we have a family C → B such that Cb ⊂Xb

for all b ∈ B.Now let Y ⊂ P4 have degree s, and look at H0(P4,O(p)). Choose generic

sections φ0, . . . , φ4. These give a map φ : P4 → P4 with φ∗O(1) = O(p). LetX0 = φ(Y ).

Then degX0 = #X0 ·H3 = #Y ·(φ∗H)3 = #Y ·p3H3 = p3 deg Y = p3s = d.By choosing φ sufficiently general, we can assume the following:

1. φ : Y → X0 is a birational map.

2. There is a surface in Y which maps 2-to-1 to the image

3. There is a curve in Y which maps 3-to-1 to the image

4. There are finitely many points which map 4-to-1

5. Nothing maps 5-to-1

So C ⊂ X specialies to C0 ⊂ X0 = φ(Y ) with degC = degC0. Now,φ−1(C0) ⊆ Y is a curve, and φ∗[φ−1(C0)] divides 6[C0], and so the degreedivides 6 degC0.

Define C = φ−1(C0). Then degC0 = #C0 ·H = #C · φ∗H = p deg C, andso as long as p is not 2 or 3, it must divide degC0 = degC.

Note: The generator of H4(X,Z) = Z is algebraic on a dense subset ofHilb(X0), though not open (it is fails to be so on a countable collection ofproper subvarieties).

Idea of how to see this: Look at surfaces, W ⊂ X of degree d on W . Anyintegral Hodge class is algebraic. CAn get a countable collection of classes

22

in H2(W,Z), each is Hodge somewhere and so is algebraic (as HC is true incodimension 1) which map to 1 ∈ H4(X,Z).

22 Lecture 21

Grothendieck’s counterexample: “Generalized Hodge is false”Consider X = (S1)8 a 4 complex dimensional torusH1(X,Z) ∼→ Λ = Z < α1, . . . , α4, β1, . . . , β4 > with polarization, which is

the same as c1(L) ∈ H2(X,Z) =∧2 Λ

So then fix α ∈ H4(X,Z) =∧4 Λ.

We want to describe:

1. all complex structures on X

2. which are principally polarized abelian varieties

3. on which α is in H2,2

4. on which α is in H3,1 ⊕H2,2 ⊕H1,3

1. Look at Gr(4, 8) = Gr(4,Λ⊗C). A complex structure with a trivializationof Λ gives a point of Gr. To recover X, we have Λ \ Λ⊗ C/spanW . Thisworks iff spanW ∩ ΛR = 0. The dimension is then 16.

2. Riemann’s bilinear relations: we describe W via the period matrix, whichis four by eight: choose a complex basis w1, . . . , w4 for W such that <αi, wj >= δij . But them < βi, wj >= Ωij , the period matrix. Ω issymmetric and has imaginary part positive definite. Thus, dim A4 =4 · 5/2 = 10. This has a Hodge interpretation: we get an abelian varietyiff the polarization is in H1,1. In Gr(4, 8), W is a Lagrangian subspacewith a positivity condition.

3. Now, for each α ∈∧4 Λ, (eg α = α1 ∧ α2 ∧ β1 ∧ β2). Sometimes α is

algebraic, for instance, X = S1 × S2 where Si are 2-dimensional com-plex tori. Then if H1(Si) = Λi, we have Λ = Λ1 ⊕ Λ2, and we can fixα1, α2, β1, β2 ∈ Λ1. Then α ∈ H4(S1)×H0(S2). So then α = [S2].

Now a geometric description, is that α ∈ H4(X,Z) =∧4 Λ ⊂ H4(X,C)∗.

And so we have the Plucker embedding of the grassmanian. So α = 0 is ahyperplan section of Gr(4, 8), call it Yα. Then α ∈ (H31 ⊕ H22 ⊕ H13) iffw ∈ Yα. And so α ∈ H22 iff w ∈ Sing(Yα)

In general, H40 is the wedge of four things, all of them in W , H31 is threefromW one from W , etc. So α ∈ H31⊕H22⊕H13 says that α has no componentsin H40 of H04, which is the same as α having no component in H40, so thenα ∧ W = 0, which is W ∈ Yα. Similarly, α ∈ H22 means that it has nocomponent in H40 or H31, which means that α = 0 is tangent to Gr(4, 8) atW , an so W ∈ SingYα.

23

So now we claim that there exists a principally polarized abelian variety(PPAV) on which there exists α ∈ H4(X,Z) ∩ (H31 ⊕ H22 ⊕ H13) which isunique up to scalar and not in H22. So then the span of α is not a Hodgesubstructure.

Dimension Count: To be in H31 ⊕H22 ⊕H13 is one condition, so we have10-1=9 parameters left. H22 is 6, and so having imposed the first one, we havea lot of space to choose from. Now, the second Riemann bilinear condition holdsalong H22, and is an open condition, so it holds.

23 Lecture 22

Two more Hodge topics:First we will talk about Hodge loci and absolute Hodge classes, and then

we will prove the integral Hodge conjecture for cubic fourfolds/intermediateJacobian and rationality.

Next: Calabi-Yaus and Mirror Symmetry. We will mostly follow notes withMark Gross from a book “Calabi-Yau Manifolds and their friends”

Maybe also Cox’s “Toric Varieties”, and Cox and Katz’s “Mirror Symmetry”So there is a notion of an absolute Hodge class, and algebraic implies absolute

Hodge, which implies Hodge, so we will ask if we can show that Algebraic impliesabsolute Hodge.

Corollaries:

Corollary 23.1. X a complex projective variety over C. Then there exists afinitely generated Q ⊂ K ⊂ C such that X is defined over K.

For all embeddings σ : K → C, we get a new complex variety Xσ

GAGA says that algebraically, on X we have C qis Ω∗dR, by H∗(X,Ω∗) andanalytically we have H∗(Xan,C) and H∗(X,Ω∗dR) ∼→ H∗(Xan,C).

Moreover, Hodge filtration can be defined algebraically.We define a filtration F ∗ of Ω∗ by subcomplexes F kΩ∗ with F kH∗(X,Ω∗) =

Image(H∗(X,F kΩ∗)→ H∗(X,Ω∗)). Intuitively, F k should mean the at least k

holomorphic differentials. So we define (F kΩ∗)` =

Ω` ` ≥ k0 ` < k

.

If Z is a codimension k subvariety ofX, then it has a class [Z] ∈ F kH2k(X,Ω∗),which tells us that [Z] is Hodge in H2k(Xan

σ ,C) for all C→ C.Thus, Algebraic implies Absolute Hodge.Note: There exist examples such that Xan

σ is not diffeomorphic to Xan.The reason is that except for the identity and complex conjugation, none of

these automorphisms are continuous.So now, let π : X → B a deformation of X. Look at Rnπ∗Z is a sheaf on B,

and Rnπ0∗Z is locally constant, and so is Rnπ0∗C.So it is a vector bundle over B0 with a flat connection ∇ : Rnπ0∗C →

Rnπ0∗C⊗ Ω1B the “Gauss-Manin Connection”

24

It also has the Hodge filtration F kRnπ0∗C ⊂ Rnπ0∗C. This is a vectorsubbundle, but it is not a FLAT subbundle. ∇F k ⊂ F k−1 ⊗ Ω1

B , “GriffithsTransversality”

Now, on graded pieces, we have ∇k : (F k/F k+1)→ (F k−1/F k)⊗ Ω1B . Now

∇ is C-linear, but ∇k is O-linear, and so commutes with multiplication by aholomorphic function. This is because the extra term in the Leibniz formulagoes to F k, which we mod out by.

Equivalently, we get TB⊗Hk,n−k → Hk−1,n−k+1. This is called the Kodaira-Spencer map. This was a key ingredient in studying the Torelli problem andproving injectivity of the period map last semester.

Let B be the universal cover ob B0 3 b. Then any integral class c ∈ Hn(X,Z)extends to a flat section of Rnπ∗Z. Over B0, we get a multivalued section. Isthis locus algebraic?

Assume that c is Hodge. Consider Hodgec = b ∈ B|cb ∈ Hk,k(Xb,Z). Isthis algebraic? The Hodge Conjecture implies yes. In fact, if c is absolutelyHodge, then yes.

Major known results:

Theorem 23.2 (Cattani-Deligne-Kaplan). The Hodge locus is always algebraic.

Theorem 23.3 (Deligne). Hodge implies Absolute Hodge for abelian varieties.

24 Lecture 23

CubicsFor cubic curves in P2, they are elliptic curves.For a cubic surface in P3, the Hodge diamond is zero except 1, 7, 1 along the

middle. This is a rational surface, which is a blowup of P2 at 6 points in generalposition (no three on a line and not all six on a conic). It contains 27 lines.Look in Lecture Notes 777.

We embed into P3 via Jp1+...+p6(3) where the pi are the points blown up.So then a line in X is a curve in X such that L · C = deg(L/C) = 1.

So we have a mapX → P2 from the blowup. L ∈ Pic(X), π∗L = Jp1+...+p6(3).So the lines will be the 6 exceptional divisors, the 15 lines between pairs of points,and there are 6 conics through five of the six points which become lines.

If C ⊂ X has image C in P2 which has degree d and multiplicity mi at pi,then deg(X ⊂ P3) = 3d−

∑mi. To see how this gives us 27, if we take d = 0,

we get the six exceptional curves which have multiplicities −1 at one and zeroelse. We get 15 lines containing 2 of them in degree 1, and in degree 2 we get 6which pass through five points. No singular conics work because they are twolines intersecting, and similarly for higher degrees.

So now we’ve shown that there are at least 27 lines. This is precise, but wewill prove the following easier result:’

Theorem 24.1. There are finitely many lines in a smooth cubic surface.

25

Proof. Let M be the space of all smooth cubic surfaces. This is an open subsetof P19. Take X ⊂ P19 × P3 the universal family of cubic surfaces in P3.

Let L be the space of cubics and line, and let XL = (x, `,X)|x ∈ ` ⊂X ⊂ P3 × Gr(2, 4) ×M . So we have the forgetful maps in a commutativesquare.

We also have a map XL → Gr(2, 4). Note that for a cubic to contain aline is a linear condition in the coefficients. In fact, four linear conditions. Andso we have 6 + 1 + (19 − 4) = 22, a so there are finitely many lines on a cubicsurface.

Now we move on to the cubic threefold in P4. X is unirational, that is, thereis a dominant rational map P3 → X.

In the case of curves, rational is equivalent to unirational. In the case ofsurfaces Luroth proved it. However, for cubic threefolds, many proofs wereclaimed, but Serre showed that they were all wrong. The irrationality of thecubic threefold was proved by Clement and Griffiths, and a second proof wasgiven by Mumford.

How about lines on X? There will be infinitely many, because a generichyperplane section is a smooth cubic surface. (Bertini?)

Define F (X) = ` ∈ Gr(2, 5)|` ⊂ X. Now, we can see that the codimensioncodim(F (X);Gr(2, 5)) = 4, and so we have dimF (X) = 2.

For the moment, we will look at (`, π,X)|` ⊂ π ∩X where ` is a line, π isa space and X is a cubic threefold in P4

We can map this to the locus of (`, π ∩X), lines in a cubic surface.To see unirationality, choose a line ` ⊂ X, then the projection X → P2 is a

conic bundle. The general fiber is a smooth conic, and also there exists a curve∆ ⊂ P2, p ∈ ∆ such that π−1(p) is a pair of lines, where ∆ is the discriminantcurve.

Exercise 24.1. Find a cover of X which is rational.

25 Lecture 24

Cubic surfaces: there exist 27 lines.Cubic threefolds in P4: there exists a “Fano” surface of lines. (for cubic

4folds, there is a “fano” 4fold of lines)Let X ⊂ P4 be cubic, smooth, and F ⊂ Gr(2, 5) the Fano surface of lines in

X.X is a conic bundle (where X is the blowup of X along a line). There exists

a P2 bundle P2 → P → B and an embedding X → P over B such that thefibers of π : X → B are conics in P2. Generically this is a smooth conic, butover some ∆ ⊂ B, we have line pairs. But no double lines.

For each line, we have a different conic bundle structure.

26

P4 = PP4

X`

E X

P2 = B

oo

OO//

?

???

OO//____

//

The fibers of X → P2: a point of P2 correspondes to a plane A ⊂ P4

containing `, and so A ∩X is `∪ a conic.Intermediate Jacobian: Given any Hodge structure H, of odd weight 2k− 1

(that means thatH is a free abelian group over Z withHC = H2k−1,0⊕H2k−2,1⊕. . .⊕H0,2k−1.) then F kHC is a “half”, that is F k ⊕ F k = HC.

Example 25.1. k = 1: HC = H10 ⊕H01 acts on H1 (curve) or in fact H1 ofanything smooth and projective, then F 1 = H10.

k = 2: HC = H30 ⊕H21 ⊕H12 ⊕H03, so F 2 = H30 ⊕H21.

So given any Hodge structure of odd weight, we form the intermediateJacobian J k(H) = H \ HC/F

kHC. This is a complex torus, isomorphic to(S1)dimC HC .

Etymology: For anyX smooth projective of dimension d, J 1 = J 1(H1(X)) =Pic(X). The Italians also studied J d = J (H2d−1(X)) = Alb(X) which is“dual” to the Picard. So when X is a 3fold, it has J 1 = Pic, J 3 = Alb andJ 2, the “intermediate” Jacobian.

Theorem 25.1 (Clemens and Griffiths). If X and X ′ are birational 3folds,then there exist curves C,C ′ such that J 2(X)×J (C) ∼= J 2(X ′)×J (C ′).

For general 3fold X, J 2(X) is a polarized complex torus, but no more.For cubic threefolds, however, J 2(X) is a ppav of dimension 5. It is NOTa Jacobian. (Let A5 be the moduli space of ppav’s of dimension 5 and M5

the moduli space of genus 5 curves, we have a map M5 → A5. Is this mapsurjective? No. dim M5 = 12 and dim A5 = 14, the Torelli question is whetherthe map is injective, and last semester we discussed the case of curves, where itis known to be injective.)

So the Clemens-Griffith result implies that J 2(X)×J (C) is J (C ′), andit follows that J 2(X) ∼= J (C ′ \ C), but J 2(X) is not the Jacobian of anycurve, so this fails.

To get J 2 of a cubic 3-fold, we compute the Hodge diamon, which is 1down the center and 0, 5, 5, 0 in the middle row (which we compute with lastsemester’s techniques)

Signs of the polarization are given by the Hodge Index theorem, which saysthat in the middle dimension, it alternates signs and so is +−+−.

We look at the first half of the cohomology, and the pairing is positive on h21

and negative on h30. In our case, h30 = 0, and so we have a definite polarization.General description of J 2(X) when π : X → B is a conic bundle and ∆ the

discriminant curve in B. Let ∆ be the Stein factorzation of π−1(∆)−Sing → ∆.J (∆)→J 2(X) is a amp H1

Z \H1C(∆)/F 1 → H2

Z \H3C(X)/F 3.

27

Now, a mapH1(∆)→ H3(X) is given by L a line bundle over ∆ and overX,we pullback and then push forward to get a map Hi(∆)→ Hi(L )→ Hi+2(X)(must go through homology via Poincare duality). This is the Gysin map.

Under mild conditions, this induces an isomorphism J (∆)/J (∆)→J 2(X).This is cally Prym(∆/∆), the Prym variety.

26 Lecture 25

Last time we defined the intermediate Jacobians, and we will be working withJ 2(X), which is constructed from the third cohomology byH3(X,C)/H3(X,Z)+F 2H3(X,C), which is F 2H3(X,C)∗/H3(X,Z). It can also be described as thekernel of the map Delignian → H4(X,Z), which is more algebraic. Thatis, there is an object called the Delignian of X such that 0 → J2(X) →Delignian(X)→ H2,2

Z (X)→ 0.Properties

1. Let X = BlCX where C is a curve in X. Then H3(X) = H3(X)⊕H1(C),and H30(X) = H30(X), H03(X) = H03(X), but H21(X) = H21(X) ⊕H10(C) and H12(X) = H12(X)⊕H01(C).

Thus, J 2(X) = J 2(X)×J (C), which is an isomorphism of ppav’s.

Corollary 26.1. The “non-Jacobian part” of J 2(X) is a birational in-variant.

The idea: Blp(X) doesn’t change H3 or J 2, but BlC(X) add a Jacobian.

Need: The category of ppav is semisimple.

2. Let C ⊂ B × X be a family of curves in X parameterized by B. Thechoice of base point b0 ∈ B gives a map AJ : B → J 2(X) which isAbel-Jacobi-like. A variant is to allow C to be a family of cycles in X(not necessarily effective) and assume that for all b ∈ B, Cb is homologousto zero, then AJ : B →J 2(X) is independent of base point.

Analytically, the former is a special case of the latter by taking Cb toCb − Cb0 .

So now each Cb is ∂Γb where Γb is a real 3-chain in X with homologyCb.

∫Γb

is a well defined linear function from harmonic 3-forms to C, andF 2H3(X) is the set of harmonic (3, 0) and (2, 1) forms, so we can restrictthe integration to this subset. Thus, for all b, we get

∫Γb∈ F 2H3(X,C)∗.

The choice of Γb changes only by H3(X,Z). So we get a well-defined imageindependent of the choice of Γb in J 2(X).

The Deligne group is defined to be H4(Z → Od→ Ω1). We call this

complex the Deligne complex D , and we have K → D → Z morphismsof complexes with K the complex O → Ω1 in degrees 1 and 2, and Z thecomplex Z→ 0→ 0 with Z in degree zero.

28

So we get a long exact sequence on hypercohomology, and so we haveH3(Z)→ H4(K)→ H4(D)→ H4(Z). We can also get that E → dR→ Kfor some complex E . We don’t want to spend too much time on this, butthe end result is H4(K) = F 2H3∗.

The upshot is that H4(K) = H3(C)/F 2 = F 2∗, and H4(K)/H3(Z) = J 2.So we have 0→J 2 → H4(D)→ H2,2

Z → 0.

A general result is that given any family C → B of curves there exists anatural AJ : B → Del which, when composed with c2 into H2,2

Z , sendseach point to the second chern class of the ideal sheaf of the curve.

The image of this AJ lands in shJ2(X), which is the kernel of c2, if andonly if Cb are homologous to zero.

For J = Pic, the analogue is 0→J 0 →J → H1,1Z → 0. So given a family

of codimension k subvarieties of X, S → B, we get a natural Abel-Jacobi mapB → Dk(X), which then has 0 → J k(X) → Dk(X) → Hk,k

Z (X) → 0. In thecase k = 1, there exist families such that this Abel Jacobi map is surjective. Fork > 1, the image is very thin. So Jacobi Inversion fails.

Third important property: General fact: if α : Pn → A, an Abelian variety,then α is constant.

Proof. H0(A,Ω1A) generates Ω1

A. But α∗(H0(Ω1A)) consists of projective forms

on Pn, and so dα = 0. This in fact holds for any rational projective variety.

Example 26.1. Let X be a conic bundle, so we have a map π : X → B wherethe general fiber is a smooth conic and there is a divisor ∆ ⊂ B over whichthe fibers are pairs of lines. We get a double cover ∆ → ∆ from the steinfactorization of the lines, and now we let C → ∆ be the P1-bundle. Apply theAbel-Jacobi map, and we get ∆→J 2(X) which factors through J (∆) (this isGLC for GL(1)) (any map to an abelian variety factors through the Jacobian)

So we have a map J (∆)/J (∆) = Prym(∆/∆) → J 2(X), which is anisomorphism in general.

Exercise 26.1. Describe ∆ and ∆ explicitly when X is a cubic 3-fold.

27 Lecture 26

Last time we defined the intermediate Jacobians and the Delignian Delk(X) =H2k(Z→ O → . . .→ Ωk−1).

For k = 1, we have the Picard group of X.If k is dimC X, then we get Alb(X), which has the property that any map

from X to an abelian variety factors through an inclusion X → Alb(X).More generally, if C ⊂ B ×X → B is a family of codimension k cycles for

any k, then there exists a natural map B → Delk(X). (It goes into J k(X) ifCb is always homologous to zero)

29

In the case k = dimX, then J k = Alb, and C = X as a family of dimen-sion 0 cycles over X. This gives a map on Alb(X), but depends on choice ofbasepoint.

Example 27.1. Let X be a conic bundle over B. then there is a subvariety of B,∆, which is the locus where the fiber is a pair of lines intersecting. Then L → ∆is a family of codimension two cycles in X. So we get a map B →J 2(X) whichis constant. Thus, ∆ →J 2(X) is constant, and so J →J 2 is zero. So weget a natural map Prym(∆/∆)→J 2(X), which is often an isomorphism.

Example 27.2. Let X be a Cubic three fold and F a Fano surface. ThenC → F × F a family of cycles with fibers ` − `′. Then we get a natural mapF × F →J 2(X).

It can be shown that ∆ is going to be a plane quintic and thus has arithmeticgenus 6, and ∆ → ∆ is an unramified double cover. Thus, g(∆) has genus 11.So the dimension of Prym is 11− 6 = 5.

So we have a smooth cubic surface S ⊂ P3 containing a line `0 an thequestion is: if the pencil of planes in P3 through `0, how many planes meet Sin a pair of lines?

One way to find out is by looking at the euler characteristic. We note thatS = Bl`0S = S, and so χ(S) = 9. On the other hand, it is also χ(P1)χ(Conic)+deg(∆)(χ(V (xy))− χ(conic)), and this is equal to 2 ∗ 2 + deg, so deg = 5.

Another way: say S = Blp1,...,p6P2, and `0 is exceptional over p1. It meetsCi iff i 6= 1, and it meets Cij iff 1 ∈ i, j, and additionally, it meets pi never.

More generally, S = dPn = BlnP2 for n ≤ 8 is a Del Pezzo surface. Thesecontain finite numbers of “lines”, that is, curves ` such that ` · (K−1

S ) is 1.The intersection configursation of these have symmetry groups W (En), with

the root system En.Cases: n = 6 is the cubic surface, for n = 8, we have a rational elliptic

surface. For n = 0 (that is, P2), there are no “lines”, and for n = 1, there is aunique line, the exceptional curve.

For n = 2, we have 3 lines, and the Weyl Group will interchange two of themwhich intersect the third (the two exceptional ones)

For n = 3, there are six lines in a hexagonl, etc.Look at H2

0 (dPn,Z), that is, the part perpendicular to the anticanonicalbundle. This is the lattice of characters of En.

So anyway, the imag of F × F →J 2(X) is the theta divisor of J 2(X).Next: discuss the singularities of Θ and deduce the Torelli Theorem and the

irrationality of cubic threefolds.Finally, we’ll consider cubic 4folds and improve this slightly better to get

the Hodge conjecture for them.

28 Lecture 27

Let X be a 3fold, C → B a family of curves in X, then we have a natural mapto Del2(X), which factors the natural map to H2,2

Z (X).

30

The map lands in J 2(X) iff the curves are homologous to zero.

Example 28.1. Conic bundles X → P2 gives Prym(∆/∆)→J 2(X).

Example 28.2. If F is a Fano surface in X, a cubic three fold, then we haveF × F →J 2(X) which is dimension 5, and the image is the Theta Divisor.

That is, the dimension of the image is 4 = 5− 1, the class of the image is inH1,1

Z (J 2(X)) =∧2(rank 10 lattice), and the elementary divisors equal 1.

This implies that L = OJ 2(X)(image) is a line bundle andb h0(L ) = 1.The divisor of a nonzero section is unique Θ.

We may as well describe all the fibers of F × F → J 2(X). Recall thatF ×X ⊂ L → F where L is the universal line. Then we have C → F ×F withC`1,`2 = `1 − `2. So this map is the Abel-Jacobi map for the family of cycles.

The dagonal F ⊂ F × F is sent to 0. All other fibers are finite. i(`1 − `2) =i(`′1 − `′2). The basi crelation is: take a cubic surface S ⊂ X and take all fourof these lines to be in S. Then we have Ei +Ci = Ej +Cj . Precisely, they willhave teh same index under the map F × F →J 2(X).

Exercise 28.1. Show that this relation generates the fibers of i (except on thediagonal)

Checking that this is the Θ-divisor is completed by a topological calculation.Applications

Theorem 28.1 (Torelli). The principlally polarized abelian variety (J 2(X),Θ)uniquely determines X.

Analize i : F × F → Θ. Conclude that Θ has a singuilar point at 0 andthat Θ is nonsingular elsewhere, so BlF (F × F )→ Bl0(Θ) is a finite map, anda nonsingular quotient of BlF (F × F ).

The projectivized tangent space at 0 to J 2(X) is isomorphic to the ambienP4, and so the projectivized cone at 0 to Θ is X.

Idea: ` ∈ F . What is the exceptional curve in BlF (F × F ) over `?Then we take P ⊂ Exceptional Divisor⊂ BlF (F × F ) to ` ⊂ F ⊂ F × F .

The exceptional divisor is the universal line, and the P1 is `.Proof: T`(F ×F ) = T`F ×T`F , and this contains T`(F ), So then N`(F/F ×

F ) = T`F , and projectivizing, we get `.The rest is an IOUThe map from the exceptional divisor to P4 is dominant onto X ⊂ P4 and

has image contained in it.A second application is teh irrationality of X: J 2(X) is not a Jacobian.Proof: C a curve of genus five, then J (C) ⊃ Θ ⊃ Θsing, which is a

curve of double points. Recall Riemann’s Singularity Theorem, which statesthat Picg−1(C) ⊃ Θ = L ∈ Picg−1(C)|h0(L) > 0, that is, the image ofAJ : Symg−1(C) → Picg−1(C). So Θsing consists of L|h0(L) > 1, and somultL(Θ) = h0(L).

31

On a curve of genus 5, there is a one parameter family of such L’s. In fact,

on C of any genus, dim(Θsing) =

g − 3 C hyperellipticg − 4 else

.

In our case, there exists a 1-parameter family of line bundles of degree 4 onC ∈M5.

In fact, for a curve C ∈ M5, ∆ = Θsing is a plane quintic, and we have∆→ ∆, so J (C) = Prym(∆/∆).

We have a map C → P4 of degree 2g−2 = 8, and C is a complete intersectionof 4 quadrics. So P3 =linear system of quadrics in P4 contains ∆, the singualrquadrics which is a quintic in P2. Looking at ∆ is the ruling in these quadrics.Each ruling cuts out a g1

4 on C, and induces an involution on Θsing : L 7→KC ⊗ L−1

29 Lecture 28

Question: Calculate T`F where X ⊂ P4 is a cubic three fold and ` ⊂ X is a linewith F the Fano surface of lines in X.

This is the same as deformations of ` ∈ X = H0(`,N`⊂X).

(NX/P)|` (NX/P)|`

T` TP|` N`/P

T` TX|` N`/X

// //

// //

Now we use the Euler sequence 0 → O → O(1)5 → TP4 → 0. Upon re-striction to `, we have 0 → O → O(1)5 →?? → 0. What is the missingterm? It has to be rank 4, and degree five, so it is O(a)⊕ O(b)⊕ O(c)⊕ O(d)with a + b + c + d = 5. Now the long exact sequence gives us 0 → 1 →10 →? → 0 so there should be 9 independent global sections. Now, how canwe find four numbers which have h0 summing to 9. We can see by compu-tation that we cannot have any -2 or lower, so some of the possibilities are(1, 1, 1, 2), (0, 1, 1, 3), (0, 0, 1, 4), (0, 0, 0, 5), (−1, 1, 1, 4), etcetera.

We claim that is it (1, 1, 1, 2). We tensor with O(−1) in the ses, and get0 → O(−1) → O5 → O(a − 1) ⊕ O(b − 1) ⊕ O(c − 1) ⊕ O(d − 1) → 0. Takinglong exact sequence and looking at dimensions of global sections, we get 0 →0 → 5 → h0 → 0, so we need to get a 5 after decreasing all the numbers. Wecan now exclue the sequences with a −1. Tensoring with O(−1) again rules outanything involving a 0.

With hindsight, we can tensor with the sheaf O(−2). This gives us 0 →O(−2)→ O(−1)5 → T ⊗O(−2)→ 0, which gives us 0→ 0→ 0→ h0 → 1→ 0in the long exact sequence. Thus, h0 = 0 after tensoring with O(−2). So1 = h0(a − 2) + . . . + h0(d − 2), which implies that the largest number is two,and that it can only appear once. Thus, a = b = c = 1 and d = 2.

So the map TP|` = O(1)3 ⊕ O(2) → (NX/P)|` = O(3) comes from TPX →

32

NX/P. The map O(1)5 → NX/P is given by the five partial derivatives of thedefining equation of X. So now we need to restrict this to the line.

On `, then, we have O(1)5 → O(3) down to O(1)3 ⊕ O(2)→ O(3).Conclusion so far: N`/X = ker(O(1)3 → O(3)) given by 3 of the five partials

Xi, namely the three transversal derivatives.It is of rank two, so we have 0 → O(a) ⊕ O(b) → O(1)3 → O(3) → 0,

with a + b = 0. Taking the long exact sequence and dimensions, we have

0 → h0 → 6 → 4 → h1 →, now h0 =

a+ 1 a > 02 a = 0

and h1 =

a− 1 a > 00 a = 0

.

So we have a = 0, 1, 2, 3, 4, 5 as possiblities.We will use a similar trick to rule things out. We will tensor with O(−2).

Then we have 0→ O(a− 2)⊕ O(−a− 2)→ O(−1)3 → O(1)→ 0, which givesus O(a − 2) ⊕ O(−a − 2) has no global sections. Thus a − 2 < 0, so a < 2,this gives a = 0, 1. So we have 0 → 0 → 0 → 2 →? → 0, so ? = 2. Thush1(O(a − 2) ⊕ O(−a − 2)) = 2. If a = 0, then we have h1(O(−2) ⊕ O(−2)) =1 + 1 = 2. If a = 1 we have h1(O(−1) ⊕ O(−3)) = 0 + 2, so both are stillpossibilities.

If a = 1, we get that T`F = H0(`,N`/X) = H0(`,O(1)) which gives thecoordinate ring of `, as we wanted.

We have that ` ⊂ S ⊂ X, and we know N`/S = O(−1), and so we have0→ N`/S → N`/X → O(1)→ 0, that is, 0→ O(−1)→ N`/X → O(1)→ 0, butstill, either a = 0 or a = 1 works. So we just need to determine if this ses splits.Still not done...

Example 29.1. Let X be the zero set of∑x3i = 0. Look at [t : −t : s : −s : 0],

a line in here. We’re stuck. Continued next time.

30 Lecture 29

We have ` ⊂ S ⊂ X ⊂ P4, and we can think of ` as a point of F , the Fanosurface of X.

We are looking at the short exact sequence 0→ N`/S → N`/X → NS/X → 0,which is 0 → O(−1) → N`/X → O(1) → 0. So it is an extension. NowExt1(O(1),O(−1)) = H1(P1,O(−2)) = H1(P1,O) = C, so if the extension is 0,we have O(1)⊕ O(−1), and otherwise it is O2.

We now look a TP4/` = O(2) ⊕ O(1)3, which gives Euler sequence 0 →O(2) → O(2)⊕ O(1)3 → O(1)3 → 0. Now we also have 0 → N`/X → N`/P4 →NX/P4 → 0, and the middle term we know is O(1)3, and the last is O(3), so wehave 0→ N`/X → O(1)3 → O(3)→ 0.

Now, which of the two candidates is this? We will try to calculate hom(O(1), N`/X).Whether or not is it zero tells us the answer. Note that the map O(1)3 → O(3)is given by the partials Xi which are nonzero along `.

So all we need to do is determine if there is a normal direction in P4 suchthat Xi = 0.

33

Example 30.1. Let X be ux2+vy2+u3+v3+w3 and ` given by u = v = w = 0.Then X is a nonsingular cubic, with partials ux, x2 + 3u2, vy, y2 + 3v2, w2.The partials x2, y2 are nonzero along the line, and so hom(O(1), N) 6= 0, andN ∼= O(1)⊕ O(−1), because the extension involved is zero.

Example 30.2. Define Xε = X + εxyw. Still contians `, and is still nonsin-gular.

So we originally wanted an identification P(T`F ) with `.Let p ∈ `, we send it to TpX∩X, a singular cubic surface with its singularity

at p. Behavior of the “27” lines on a singular surface is that the conic throughfive points and the exceptional divisor at the sixth point coincide, so there are15+6=21 distinct lines, 6 of which have multiplicity 2.

(Check it for 6 points lying on a conic)Our set up: F ∩Gr(2, Tp(X)) is 2`+ . . .. So it gives a well-defined tangent

vector to F at `.Conversely, consider a curve ` ∈ C ⊂ F . Each point `′ ∈ C determines a

hyperplane 〈`, `′〉 (there might be a finite number where it isn’t a hyperplane),so this gives us a map C → (P4)∗ defined away from `. This extends to `,however. So we get P3 through ` such that ` counts twice in F ∩Gr(2,P3).

So P3 ∩X must be singular, so it is the tangent space at X for some p ∈ X,which is in fact in `.

Back in context: We were looking at AJ : F × F → J 2(X), which hasimage the Θ divisor.

We get two main results:

1. Irrationality of X

2. Torelli for X

Both follow from geometry of Θ. The irrationality follows by Θ having aunique singular point (the image of the diagonal), and if this were the Jacobianof a curve, we’d have a curve of singularities.

To get Torelli, we look at P(T0Θ) ⊂ P(T0J 2(X)), where 0 is the singularpoint, and so the tangent cone is a cubic, and in fact the X we started with.

Less-Canonical Version of Abel-Jacobi: F → J 2(X), which depends on achoice of basepoint, but its derivative is intrinsic (the Gauss map G : F →Gr(2, T0J 2(X)) = Gr(2, 5))

Claim: G is the tautological map.This would imply that for AJ : F × F → J 2(X) that its Gauss map

G : F×F → Gr(4, 5) taking (`, `′) 7→ span〈`, `′〉. Inside there is F , the diagonal,and we need to blow this up in order to have the map defined everywhere.

F = P(TF ), the projectivization of the tangent bundle, and points of thisare sent to tangent space to X at a point of `.

The image of F under the Gauss map is the set of tangent hyperplanes toX, and so the tangent cone to the image will in fact be X.

The only thing that still needs to be proved is that G : F → Gr(2, 5) is thetautological map.

34

To see the identification of AJ with the Gauss map, we need to restrictattention to ∆. `→ ∆ = `′|`′ ∩ ` 6= ∅, and ∆ maps down to ∆ ⊂ P2.

∆ sits naturally in F .Next time: Prym Varieties and their Abel-Jacobi map.

31 Lecture 30

Prym VarietiesGeneral situation: if C → C is a double cover, and there are r ramification

points, then g(C) = g and g(C) = 2g − 1 + r/2.We define Nm(π) : J (C) → J (C) for π∗ : J (C) → J (C), then

Nm(π)(p1 ± . . .± pk) = π(p1)± . . .± π(pk)In general, we define O(Nm(π)(D)) = det(π∗O(D))

(12Branch

). How do

we take half of the branch locus?Well, we need a choice of L =

√O(B). That is, a line bundle L along with

an isomorphism L⊗2 → O(B). This gives a double cover C → C.Take the inverse image under ⊗2 of teh natural section of O(B), and we get

this double cover, branched along B.

Exercise 31.1 (in Harshorne). The converse is true. That is, the data of adouble cover is equivalent to a choice of square root of the branch divisor.

So π∗(OC) = OC ⊕ L−1.(Explicitly, we need to see that if a divisor D on C is linearly equivalent to

zero, then so is Nm(π) on C)There exists f rational function on C with (f) = D, we need a rational func-

tion g = Nm(π)(f) on C such that (Nm(f)) = π(D), Then g(p) = f(p1)f(p2)where p1, p2 are the points in π−1(p).

Relations: Nm(π) π∗ = 2 id. π∗ Nm(π) = 1 + i∗ where i : C → C is theinvolution over C.

ker0Nm(π) ⊂ J (C) maps down to cokerπ∗. We have abelian varieis ofdimension g − 1 + r/2. If r = 0, then ther kenreal has 2 components, for r > 0it is connected.

Back to cubic threefolds:∆` = ∆ = `′ ∈ Fano(X)|`′ ∩ ` 6= ∅ → ∆, a plane quintic. So we have

J (∆) → J 2(X). But also π∗ : J (∆) → J (∆) which sends this locus tozero in the composition. So J 2(X) is the Prym variety.

Example 31.1. If g = 0 and r = 0, then C → C is an unramified map ofelliptic curves, so the Jacobians are the curves. thus, Nm(π) = π : C → Cand π∗ : C → C is the dual map, so C → C → C is multiplication by 2, andC → C → C is also multiplication by 2. Finally, the Prym variety is zero.

Example 31.2. g = 0, r = 2b, then C = P1 adn C is a hyperelliptic curve ofgenus b− 1. Then Prym = J (C).

35

If r = 0, we can describe Prym as a map Rg → Ag−1, where Ag−1 consistsof abelian varieties of dimension g−1 and Rg is the moduli of pairs (C,L) whereC ∈Mg and L ∈J (C)[2].

Consequence:

Rg Ag−1

Mg Ag AJ //

Prym//Prym

OO

Prym has positive dimensional fibers for g ≤ 5.g dimRg dimAg−1

1 1 02 3 13 6 34 9 65 12 106 15 157 18 21

It is, in fact, never injective. (Though it is generically injective for g largeenough)

For g = 6, we have deg(Prym) = 27, and Gal(R6/A5) = W (E6), the sym-metry group of the lines on a cubic surface.

g = 5: fiber of Prym is a surface. In fact, it is a souble cover of a Fanosurface of lines in a cubic threefold.

A4 is the moduli space of abelian varieties of dimension 4, which is isomorphicto the moduli of cubic 3folds X and a point of order 2 in J 2(X).

Last time: we looked at F × F →J 2(X), F →J 2(X) projectivized.We have TF → F the tangent bundle, which is a rank two vector bundle,

and we projectivize it, which gives the Gauss map G : P(TF )→ P4.We used the fact that the image of P(T`F ) is ` ⊂ X ⊂ P4.Choose any `0 ∈ ∆`, and consider ∆`0 ⊂ F (X), and a point b on it.Then T`(∆`0) ⊂ T`(F (X)), and so we may as well describe P(T (∆`0)), which

is a curve in the threefold P(TF ). It is isomorphic to ∆`0 .So we have a map ∆`0 → P4.This is the Abel-Prym Map!Abel-Jacobi gives C → P(H0(ωC))∗, and we get a map J (C)→ P(H0(ωC))∗.The Abel-Prym map, then goes C →J (C)→ Prym(C/C)→ P(H0(C,ωC⊗

L)∗).So note: deg(L) = 0 (unramified case) and so deg(ωC⊗L) = degωC = 2g−2,

and so h0(ωC) = 2g− 2− g+ 1 +h0(0) and h0(Ωc⊗L) = 2g− 2− g+ 1 +h0(L).Now h0(0) = 1 and h0(L) = 0, so h0(ωC) = g and h0(ωC ⊗ L) = g − 1, and sothe Abel-Jacobi map goes to Pg−1 but Abel-Prym goes to Pg−2.

36

32 Lecture 31

We now look at X ⊂ P5 a cubic fourfold. It has hodge diamond

?0

00

1

0?

01

0

00

?0

0

01

0?

0

10

00

?

So now we need to determine the middle row. From last semester, we haveH40 ∼ R−3 = 0, H31 ∼ R0 has dimension 1, H22

0 ∼ R3, and so h22 = 21,because J 3 → S3 → R3 consists of a 36 dimensional space into a 56 dimensionalspace, and so R3 is 20 dimensional, but this is the primitive Hodge number, sowe add one. Thus the Hodge diamond is

00

00

1

01

01

0

00

210

0

01

01

0

10

00

0

The classes in Hp,p for p 6= 2 are clearly algebraic, so we only need to look atH2,2. (Why is it clear? H00 is the whole thing, H11 generated by a hyperplane,H22 we need to do work, H44 is generated by a point, and H33 is generated bythe class of a line, which we know exist because it is a cubic)

Let Σ be a cuibc surface, and let Y = BlΣX, and Yt = π−1(t).

X

Y

Yt

P1

t

τ

π //

//

Then choose α ∈ H2,2(X,Z). We want to show that α is algebraic.We define αt = (τ∗α)|Yt

= β[`] for β ∈ Z.

37

Now, set α′ = τ∗α − β[` × P1] This is a class on Y , and α′|Yt= 0. (Note

that the exceptional divisor E is a product Σ × P1, so the inverse image of aline ` ⊂ Σ is `× P1.

So we can restrict to worrying about the classes α′ being algebraic.Consider the intermediate Jacobian fibration J = J (Y/P1) → P1 which

has fibers intermediate Jacobians. We do need to be careful, however. As wehave J (Yt) = H3(Yt)/F 2H3 + H3(Yt,Z), we define J to be R3π∗C moduloR3π∗(0→ 0→ Ω2 → Ω3) +R3π∗Z.

We can also describe it as ker(H4Delignian(Yt,Z[2])→ H4(Yt,Z)).

If K = OYt → Ω1Yi

, then we have 0 → K → ZD[2] → Z → 0, this givesH3(Y,Z)→ H3(Y,K)→ H4

D(Y,Z[2])→ H4(4,Z).So the plan is to take α and get α′, which then gives us a section να of

J → P1, a “normal function”, which then gives us a cycle by Jacobi inversion.We pick a topological cycle A representing α′. Then A|Yt is ∂Γt for some

3-chain Γt ⊂ Yt. Then∫

Γt∈J (Yt), and this gives the normal function.

So we now have an element of H0(P1,J ) = H4D(Y,Z[2]). This lifts α.

So we now must convert a section ν of J to a cycle.Analogue for a surface: Z → P1 a surface and Zt the fibers. Then J =

Pic0(Z/P1)→ P1 and ν a section. Jacobi inversion tells us that Symg(Z/P1)→J is a surjective map over P1.

So a section ν lifts to a section of Symg(Z/P1) which IS a curve in Z (ofdegree g over P1)

In our case, we need a family of cycles in Y/P1 which dominate J /P1.Rationally, this is easy. We have F × F → Θ ⊂ J , and F × F × F → J

surjective, but we want it over Z.

Theorem 32.1 (Markushevich, Tikhmirov,. . . ). Let Mt be the moduli space ofsemistable rank 2 torsion free sheaves on Yt with c1 = 0 and c2 = 2`. Then Mt

is birational to J (Yt) (its Abel-Jacobi map, in fact, is a birational morphism).

The Abel-Jacobi map for a moduli space of bundles is given by looking athU →Mt×Yt the universal bundle, and L >> 0 a surfficiently ample line bundleon Yt. Then look at π∗(U ⊗ L) where π : Mt × Yt → Mt is the projection. Itis a sheaf on Mt, and for L >> 0, it is a vector bundle, V . We look at P(V ), aprojective bundle over Mt. Now, P(V ) parameterizes a family of codimension 2cycles in Yt. So we have AJ : P(V )→J (Yt), which factors through Mt, sincethe fibers are rational.

Thus, Jacobi inversion works over Z.

33 Lecture 32

We will be looking at X with c1(X) = 0. There exists a classification:There exists a finite cover Y → X with Y = Z ×

∏Si ×

∏Cj where Z =

Cn/Z2n is a complex torus, Si is holomorphic and symplectic, irreducible andh0(Si,Ω2) = 1.

38

(Any closed two-form α on a manifold X gives iα : TX → T ∗X, and α issymplectic if it is an isomorphism. Equivalently, α is everywhere of maximalrank.)

Also, Cj is Calabi-Yau, that is, c1(Cj) = 0, h1(Ci,O) = h2(Cj ,O) = 0. (ACY is a manifold X such that KX = O)

In dimension 1, the only possibilities are elliptic curvesIn dimension 2, we have complex tori and K3 Surfaces (like quartics in P3)In dimension 3, we can have a complex torus, an elliptic curves cross a K3

modulo a finite group action, or a Calabi-Yau threefold.The classification continues, and we will choose to stop at 3 (in physics,

sometimes 4 dimensional ones come up)Complex tori and K3-surfaces are classical objects, so we’re just going to

focus on studying Calabi-Yau 3folds.

The Hodge diamond of a CY3 is 10

01

0h21

h11

0

0h11

h21

0

10

01 . So there are two

parameters, h11, h21.Note that as h20 = 0 we have H11 = H2 is generated by Z-classes, and so

all classes in H2 are algebraic.Now H21 = H1(C,Ω2). As we have a nonvanishing 3-form, this gives us

an isomorphism T → Ω2, so we have H21 = H1(C, TC), the set of formaldeformations of C.

For Calabi-Yau’s, we have

Theorem 33.1 (Bohomolov-Tian-Toderov). Deformations are unobstructed.

That is, for all CY, C, there exists a moduli space M and a family C →M(proper, flat, smooth) and a point 0 ∈M with C the fiber over 0, such that theKodaira-Spencer map T0M → H1(C, TC) is an isomorphism.

So, topological, there are two invariants with very different behaviors.

Conjecture 33.1 (Mirror Symmetry version 1). For any CY3, there exists C ′

another CY3, such that h11(C) = h21(C ′) and h21(C) = h11(C ′).

Example 33.1 (Projective Hypersurfaces). Let H ⊂ Pn be a hypersurface.Then adjuction says that KX = 0 if and only if degX = n + 1. So for n = 4,we need quintic threefolds.

In this case, h11 = 1 and h21 = dimR5 =(

94

)− 5× 5 = 126− 25 = 101.

Example 33.2 (Complete Intersection of k Hypersurfaces). Look in P3+k, andtake the hypersurfaces of degree di with

∑di = 4 + k.

For k = 1, we get the previous. For k = 2, we have H ∩ H ′ ⊂ P5, we getdegree 6=5+1=4+2=3+3, the new cases are 4+2 and 3+3. For k = 3, we get

39

7, which partitions into 3+2+2. For k = 4, we get 2+2+2+2=8, and that’s it,it can’t be done anymore without 1s.

So this gives us 5 new classes of examples.

Example 33.3 (Weighted Projective Spaces). Consider complex intersectionsin a weighted projective space. We get more Calabi-Yau’s here, in fact, hundredsof them.

Better: we replace P3 by a toric variety, that is, Cn+k/(C∗)k. Now we havethousands or tens of thousands of topological types of Calabi-Yaus

This set of examples is known to be closed under mirror symmetry.

Example 33.4 (Elliptic Fibrations). Choose a base B and a line bundle L onB, look at P(L2⊕L3⊕O), which is a P2 bundle, write the Weierstrass equationhere, and we get an Elliptic Fibrartion over B. To make sure this is CY, thenL = −KB. Mostly in the case where B is a Del-Pezzo surface (that is, Fano).But also, where B is the blow up at 9 points, a rational elliptic surface. So thisX is B ×P1 B′ for two rational elliptic surfaces.

34 Lecture 33

A Calabi-Yau is a variety with KX∼= OX , and h1(OX) = h2(OX) = 0 (really,

only the first, but we can factor the variety by the theorem to get down to onesatisfying the latter)

We looked at the Hodge diamond of a CY3, and gave the first approximationof mirror symmetry

Conjecture 34.1 (Mirror Symmetry, version two). MComplex(X) ∼= MKahler(X)naturally

That is, the moduli space of complex structures on X is naturally isomorphicot the moduli space of Kahler structures on X.

By Bogomolov-Tian-Todorov unobstructedness, TXMComplex(X) = H1(TX) =H1(Ω2

X) = H21.MKahler is the set of complexified Kahler classes (B-fields added), that is

B + iω with ω a Kahler form and B ∈ H2(X,R)/H2(X,Z).We will discuss two techniques: degnerations and Quotients

35 Quotients

Proposition 35.1. Let G be a finite group acting freely on CY3 X. Then X/Gis a CY3.

Proof. Hpq(X/G) = Hpq(X)G.Thus, 0 ≤ hpq(X/G) ≤ hpq(X), and so h00(X/G) = 1, h10(X/G) = 0 =

h20(X/G), and h30(X/G) = 1, though this takes a bit of work: Look atχ(OX/G), as we have an etale cover, this is χ(OX)/|G| = 0, and so we musthave h30 = 1.

40

Example 35.1 (Fermat Quintic). Let X be the Fermat quintic, that is∑4i=0 x

5i =

0. What are the automorphisms? Well, there are the S5 permutations, andthere’s also (Z/5)5/(Z/5). Take a cyclic permutation (x0, . . . , x4) 7→ (x1, . . . , x4, x0).If ρ is a fifth root of unity, then (1, ρ, ρ2, ρ3, ρ4) are the fixed points in projectivespace, but none of these points lie on X, as if you plug them in, you get 5, notzero.

There’s also (x0, . . . , x4) 7→ (ρx0, . . . , x4), which has as fixed points (1, 0, 0, 0, 0)and (0, x, y, z, w). So this is not fixed point free. But take G = ker((Z/5)5/Z/5→Z/5) then the fixed points are (1, 1, 1, 1, 1). So we want the fixed points to beisolated.

So there are various possibilities, like Z/5×Z/5 with the first factor a cyclicpermutation and the second multiplication by (1, ρ, ρ2, ρ3, ρ4). This acts freelyon X (and also

∑x5i + ψπxi). So we look at the quotient.

h11 = 1 and h21 = 5, because the original 101 came from 1 + 4× 25 in termsof representations, and there is one invariant in each.

So X/G has Hodge numbers (1, 5). This is one of the smallest known Calabi-Yau’s.

Question from physics: There are three generations of particles, and a CYgives absolute value of h11 − h21 generations, which is 1

2χ(X), so we need onethat has three. This ones says 4.

This was used in a paper by Candelas, (other authors), Witten around 1985as the basic example of how to compactify string theory.

Example 35.2. Take G = (Z/5)3, that is, the kernel from above. This doesn’tact freely. Thus, the quotient is singular, and h11(X) = 1 and h21(X) = 101,but now h11(X/G) = 1 and h21(X/G) = 1. These need to be carefully definedusing the fact that X/G is an orbifold rather than a manifold.

The number of singular points is the number of orbits of fixed points, whichis 100. (we’ll do this next time)

The resolution has h11 = 101 and h21 = 1, and this is the mirror of thequintic.

36 Lecture 34

1. Our first example is the Quintic Threefold (1,101)

2. We looked at a quintic containing a place, which is singular at 16 points,we find a small resolution and get a new CY3 which is (2,86)

Local Models (Locally in the ANALYTIC topology)Let Xt be a family of 3folds and X0 has an ordinary double point:

∑4i=1 x

2i =

t. Then we have a diagram

41

ˆX0

X ′0

X0

X0

???

??

??

???

?

these are isomorphisms away from the inverse image of 0.Note, in C4 with coordinates xi, we have two places x1 = ix2, x3 = ix4, π,

and x1 = ix2, x3 = −ix4, π′. Blow up π and π′ in C4, and take the propertransform of X0. Get X, X ′. The effect on cohomology is that X0, X

′0 is χ+ 1

and ˆX0 is χ+ 3.

In a global model, if one exists, (Paper by Bob Friedman ?) then X0 → X0

a small resolution of n ordinary double points. Then χ(X0) = n+ χ(X0).Options: h11 goes up by n, because we are creating algebraic cycles, but

similarly, h21 goes down by n, and a third possibility is some combination.

1. Quintic with isolated triple point (1, 0, 0, 0, 0) ∈ P4, by x20 ⊂3 (x1, x2, x3, x4)+

x0F4 + . . .+G5. We resolve by blowing up the point. Replace 0 by a non-singular cubic surface. Exercise: This is a crepand reoltuion, and it is aCY3 with (2,90)

2. Let X be singular quintic along P1 given by x0 = x1 = x2 = 0, ThenX ∈ (x0, x1, x2)2. Blow up the line to get X → X. The exceptional divisorwhich maps to P1 will be a conic bundle over P1. Each ci = ci(x0, x1, x2)is a conic, the fiber. We have a map P1

(x3,x4) → P5 = P(Sym2 C3) =conics in P2, and it’s a twisted cubic.

Wilson’s results on the Kahler cone: The set of Kahler classes is an openconvex cone in H1,1

R . It is polyhedral away from the topological intersectiondivisor: ω3 = 0. The faces have three types of faces, and each face correpondsto a contraction of a curve to a point, surface to a point or surface to a curve.

37 Lecture 35

Bogomolov decomposition theroem: c1(X) = 0 implies that Y → X can bewritten Z ×

∏Si ×

∏Cj

Examples of CYsConstructing techniques - Quotients with free actions and resolutions of

quotients by nonfree actions.Examples: Degenerations of types I, II, and IIIShape of the Kahler Cone (results mostly due to P. Wilson)Let X be a projective CY3(NOTE: Here KX is the Kahler cone, not the canonical divisor)KX ⊂ KX ⊂ H1,1

R (X) = H2R() ⊃W = D ∈ H2|D3 = 0.

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Now, W ⊂ U open implies that in the complement of U , KX is a rationalpolyhedral cone, σ ⊂ KX a codimension 1 dace (or facet) and σ 6⊂ W , impliesthat there exists a primitive contraction f : X → Y with Y normal and f abirational map such that span(σ) = f∗(PicY )⊗ R.

Theorem 37.1 (Classification of Primitive Birational Contraction).

1. Finite number of curves collapse to points implies that the irreducible com-ponents are P1. (Example: 16 lines to 16 points)

2. Irreducible surface→ point, implies that the surface is a del Pezzo, possiblysingular (Example: Cubic surface)

3. Surface to a nonsingular curve with conic fibers

Behavior of a facet (iff contraction) under deformations of complex structure:Type I - Deforms to type 1Type II deforms to type IIType III deforms to type III if g = 0 and for g > 1,can deform to type I

or III. The contraction can disappear if E → C as a fibration has either allsmooth fibers or all nodal fibers.

Corollary 37.2. KX is deformation invariant unless X contains a surfaceE → C as above.

Let S be a monoid (associative with identity) with the finite partition prop-erty (that is, given any element z there are finitly many (x, y) such that x+y =z.) This gives a formal monoid ring (with coefficients in R) of R[[q;S]] =∑η∈S aηq

η|aη ∈ R.

Example 37.1. S = (Z≥0)n ⇒ R[[q1, . . . , qn]].

Example 37.2. S is integral Mori monad, which is ¯NEF (X,Z).

NEF is NOT Numerically Effective, it is Nef(X,Z) = η ∈ H2(X,Z)|ω ·η ≥0,∀ω ∈ KX.

Yukawa Coupling: input is three classesD1, D2, D3 inH2, and have 〈D1, D2, D3〉 =D1D2D3 +

∑η∈H2(X,Z)\0

∑∞m=1 q

mηφη(D1, D2, D3) where φη is the Gromov-Witten Invariant (which count rational curves P1 → X of class η meetingD1, D2, D3)

Imaigne that each Di is the class of an actual divisor (which we will identifywith Di). We want φη(D1, D2, D3) =the number of maps u : P1 → X such thatu(0) ∈ D1, u(1) ∈ D2 and u(∞) ∈ D3, that is, roughly, the numebr of points inthe quotient M (η,P1)/PGL(2) times (D1 · η)(D2 · η)(D3 · η).

Options:

1. The best option is to be lucky, and the above expression makes sense andis finite

2. If you aren’t lucky, then you can work hard - J -holomorphic curves whichis essentially analytic and nonalgebraic.

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3. An algebraic method due to Barbara Fentechi, Seibert and others is tointerpret M (η,P1) as an algebraic stack and it will be zero dimensionalas a stack, and then count the “points”, that is, we integrate a virtualfundamental class, and that gives the number.

38 Lecture 36

Last time we defined the (1, 1) Yukawa coupleing on H1,1, W +∑η 6=0, with the

instanton corrections counting rational curves of class η ∈ H2(X,Z) meetingD1, D2, D3, with D1 ·D2 ·D3 the classical intersection. This is taking place onthe “A-side”

Next: “B-model”, or (1,2)-Yukawa coupling on H1,2.VHS (Variation of Hodge Structure) is (HZ,∇, S, F ) where S is a complex

manifold (the base), HZ is a local system of abelian groups (generally taken tobe free, Z`), ∇ is a flat holomorphic connection on H = HZ ⊗Z OS , HZ flatwith respect to ∇ and F induces a Hodge structure on each fiber, that is, ∇ :H →H ⊗Ω1

S restricts to F p → F p−1⊗Ω1S . This is “Griffiths Transversality”.

A smooth proper map H → S with Xs the fiber over s ∈ S with projectivefibers (or at least Kahler) then we have VHS for all n ∈ Z given by HZ = Rnπ∗Z.

F ∗ is the family of Hodge filtrations on the fibers.GT : ∇S : Hn → Hn is given by α 7→ ∂α

∂s where α(s) is a local section of H ,and by differentiating, we lose at most one holomorphic differential.

Combine this with the Kodaira-Spencer map, the natural connection factorsthrough KS, and so we have H1(TX) × F p+1 → F p. We have a map down toH1(TX)×F p → F p−1, which maps to H1(TX)×Hp,n−p → Hp−1,n−p+1 by thecup product.

Now we assume that X is a CYn and X → S is a family. Then Ωs is aholomorphic n form on Xs, and Ωs 6= 0, so Ω is unique up to f ∈ Γ(S,O∗S). Itinduces TX

∼→ Ωn−1X .

This induces an isomorphism H1(TX) ∼→ H1(Ωn−1X ) from deformations of X

to Hn−1,1.So now we get the (1, n−1)-Yukawa couplingHn−1,1×Hp,n−p → Hp−1,n+1−p.

Example 38.1. For n = 3, we have H2,1×H3,0 → H2,1, H2,1×H2,1 → H1,2,H2,1×H1,2 → H0,3. A vector v ∈ H2,1 gives three maps, and each map dependslinearly on v.

Their composition H30 → H03 depends cubically on v, we se have a cubicmap H2,1 → hom(H3,0, H0,3) ∼= C. The upshot is that we get a cubic polynomialon H21 (or it is an element of ⊗3H12, and it is an exercise to check that it isin Sym3H12)

Situation so far:

classical quantumA−model intersection +

∑η gromov − witten

B −model V HS, Y ukawa 0

44

Conjecture 38.1 (Mirror Symmetry). Let f : X → (∆∗)s (the product of spunctured discs) a family of Calabi-Yau threefolds with 0 a maximally unipotentdegeneration (think as singular as possible, called a “large complex structure” byphyicists) then there exists a “mirror” Calabi-Yau threefold X and a choice offraming Σ ⊂ KX (ie, choose s linearly independent Kahler classes and considerthe simplex they space, this corresponds to canonical coordiantes q1, . . . , qs on(∆∗)s)

The ei’s give coordinates qi on MKah,Σ(X) iff m : (∆∗)s →MKah,Σ(X) isa mirrir map such that

〈 ∂∂qi

,∂

∂qj,∂

∂qk〉p = 〈 ∂

∂qi,∂

∂qj,∂

∂qk〉m(p)

39 Lecture 37

Today we’re going to talk about Borcea-Voisin Calabi-Yaus and Mirror Sym-metry

Let Y = Yd be a d-dimensional compact Kahler manifold and Xσd ⊂ Yd a

smooth divisor in class 2c1(Y ). (c1(Y ) is c1(TY ).)Then there exists Xd → Y is a double cover branched over Xσ

d where Xd isnonsingular and c1(Xd) = 0.

Example 39.1. If d = 1, we have Y = P1. Then Xσd is some set of 2k points.

Then X → Y branched at these 2k points is a curve of genus k − 1. To haveKX = 0, we need k = 2, because degKX = 2g − 2 = 4(k − 1). So X is CY iffthe branche locus is in 2c1(P1).

We call any Calabi-Yau defined this way to be a Calabi-Yau with Involution.Nikulun’s Classification of K3 surfaces with involution:Let X be a K3, and L = H2(X,Z). This is U3 ⊕ (−E8)2 where U is a

2 dimensional lattice with bilinear form(

0 11 0

)where −E8 is the weight

lattice for the Lie group of type E8 (the - is a matter of taste, choosing thenegative definite part). Define Lσ = ` ∈ L|σ(`) = `.

Deformation classes of K3s with σ correspond to pairs (L,Lσ). Classication:(r, a, δ) where r = rank(Lσ), Lσ ⊂ (Lσ)∗ ⊂ (Lσ) ⊗ Q. Now (Lσ)∗/Lσ ∼=(Z/2Z)a. (the reason is that 2(Lσ)∗ ⊂ Lσ), because π : X → Y gives π∗π∗ = 2)

δ =

0 if ∀x∗ ∈ (Lσ)∗, (x∗)2 ∈ Z1 else

=

0 if [Xσ

d ] is even1 else

This does give us a complete classification of K3’s with involution.GET PICTURE FROM RONReflection through the middle line is mirror symmetry for K3’s.

Example 39.2. (10, 10, 0) has the property that Xσ = ∅. This is called theEnriques involution, it is the only example of a fixed-point free involution on aK3. The quotient by σ is called an Enriques Surface, and all 11 dimensions ofthem arise in this way.

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Example 39.3. (10, 8, 0), Xσ = E1 ∪ E2 where Ei are elliptic curves.

These are the exceptional cases. In all other cases, Xσ = Cg∪⋃ki=1Ei where

this is a disjoint union, Cg is genus g, and Ei are rational.We have (g, k)↔ (r, a) with g = (22− r − a)/2 and k = (r − a)/2.Then Mirror Symmetry is (r, a, δ)↔ (20− r, a, δ) and g ↔ k + 1.Now, let (Xm, σm), (Xn, σn) be two CY’s with involution. Thne we can

construct (Xm+n, σm+n), a resolution of Xm ×Xn/(σm, σn). Namely, Yn+m →Yn×Ym containing Xσ

n+m → (Xσn×Xσ

m), where the first is the proper transformof Xσ

n × Ym ∪ Yn ×Xσm.

Application: n = 2,m = 1 takes a K3 with inv and an elltiptic curve withinv, and you get the BV 3folds.

Then we have h11 = 5 + 3r− 2a = 1 + r+ 4(k+ 1) (1 class from the ellipticcurve, r from the K3, and 4(k + 1) from the resoltuion)

h21 = 65 − 3r − 2a = 1 + (20 − γ) + 4g (1 modulus for the elliptic curve,20− r complex def in the K3, and 4g moduli for the curve in the K3)

40 Lecture 38

We’ve discussed BV CY3’s.Classification (Nikulin) of K3’s with involution as (r, a, δ).A BJ is K3×Elliptic/1,−1.Now, L = H2(X,Z) = U3 ⊕ (−E8)2 with r = Lσ, a is (Lσ)∗/Lσ ∼= (Z/2)a

and δ ∈ 0, 1.Then h11 = 5 + 3r − 2a and h21 = 65− 3r − 2a.

Example 40.1. (10, 8, 0) has (19, 19) Hodge numbers. In fact, these are thefiber products of rational elliptic surfaces.

RES → P1 in general 12 I1. Specialize to 2 × I∗0 . We need to resolve thesingularities in order to get a nonsingular space (see earlier)

So for BV’s, the mirror should be (20− r, a, δ).For K3’s, we have lattice polarizing K3’s:Let M ⊂ L be a fixed sublattice

Example 40.2. rankM = 1, then = Z and has generator squaring to 2d cor-responding to the moduli of K3’s of degree d

Consider the moduli space of lattice polarized K3’s. (S, α)|S a K3, α :M → Pic(S) = H11(S,Z).

Define T to be the transcendental lattice (the lattice of nonalgebraic classes),then set DM = [Ω] ∈ P(T ⊗Z C)|Ω2 = 0,ΩΩ > 0.

Then DM is the period space of the M -polarized K3’s.Choose an embedding U ⊕M → L and let M be (U ⊕M)⊥L . Then rk(M) =

20− rk(M) and we get U ⊕ M → L.The Complexified Kahler Cone is Tµ = B + iω ∈M ⊗Z C|ω2 > 0.

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Lemma 40.1. φ : TM → DM is an isomorphism, the mirror map. It dependson a generator E of U , such that U = E ⊕ E′.

Proof. φ(B + iω) = B + E′ + ω2−B2

2 E + i(ω − (ωB)E).

Context: SYZ (Strominger-Yau-Zader) Conjecture: Explicit construction ofthe mirror.

Tn → Xn → Sn where π is a fibration with fibers “Special lagrangian” toriin X.

Structure on X: ω is Kahler in H11 and Ω is holo volume in Hn0. So Z ⊂ Xis Lagrangian if dimR Z = n and ω|Z ≡ 0. It is special lagrangian if ImΩ|Z ≡ 0iff ReΩ|Z = vol|Z .

Maclean: deformation theory of smooth special lagrangians is unobstructed,so deformations are H1(Z,R).

Z ' Tn impliues that h1(Z,R) = n.SYZ: Whenever a mirror exists, there is a SLAG fibration π : X → Sn and

then the mirror X “is” the dual SLAG fibration.The case n = 2 has X = K3. So a SLAG fibration IS an elliptic fibration.

After a 90 degree roation in twistor space, we have ω ∈ H11, Ω ∈ H20 andΩ ∈ H02 so (X,ω) → S2 becomes (X,Ω) → CP1 with a section. The first isSLAG, the second is holo.

So an elliptic K3 with section corresponds to a K3 with U -structure.Back to BV’s:There is a short list of exceptions to this process: lattice Lσ which cannot

be embedded in U2 ⊕ (−E8)2.Gross-Wilson: For BV’s, we can fully justify the SYZ prediction: there exists

a recipe for completing the singular fibers so χ(X) = −χ(X).

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