## ICTP Day 6

Today we started the third part of the summer school: Algebraic Cycles, Arithmetic aspects of Hodge Structures-Mumford-Tate Groups. There was actually quite a bit of review today, which surprised me, and instead of taking notes on the definition of Hodge structures again, I just noted that that’s what we did, and moved on.

1. Carlson 1 – Period Domains

Period domains are parameter spaces for marked Hodge structures. We call ${\Gamma\backslash D}$ the period space, which is a parameter space of isomorphism classes of Hodge structures.

Let ${X\rightarrow S}$ be a morphism of varieties, ${X_s}$ the fiber over ${s}$ and ${\Delta}$ the subset of ${S}$ where the fibers are singular. Then the base change to ${S\setminus \Delta}$ is a smooth family. This induces a period map ${S\setminus \Delta\rightarrow \Gamma\backslash D}$ by ${s\mapsto H^n(X_s)}$ with its Hodge structure. Really we lift to the universal cover, which maps into ${D}$ and associates to ${X_{\tilde{s}}}$ its marked Hodge structure.

(Review of what a Hodge structure is, and then Primitive Cohomology)

So what is a marking on a Hodge structure? Take a Hodge structure ${H_\mathbb{Z}}$ and ${H_\mathbb{C}=\oplus_{p+q=k} H^{p,q}}$ with polarization ${Q}$. The lattice ${H_\mathbb{Z}}$ is isomorphic to ${\mathbb{Z}^r}$ for some ${r}$. A marking is an isomorphism ${\mathbb{Z}^r\stackrel{m}{\rightarrow} H_\mathbb{Z}}$ and ${Q_0}$ on ${\mathbb{Z}^r}$ such that ${m}$ is an isometry.

Now, we set for curves, that is, rank 1 Hodge structures, ${D=\{m^{-1}(H^{1,0})\subset\mathbb{C}^{2g}\}\subset \{S\subset \mathbb{C}^{2g}\}=G(g,2g)}$ and the more general definition is similar.

Now let ${\Gamma}$ be the isometry group of ${Q_0}$, and it is a discontinuous group, and ${\Gamma\backslash D}$ is an analytic space.

2. Murre 1 – Chow Groups

Conventions: ${k}$ is an algebraically closed field, ${X,Y,\ldots}$ are varieties over ${k}$, which are projetive (at worst, quasi-projective), irreducible and smooth.

2.1. Algebraic Cycles

Let ${X/k}$ be of dimension ${d}$. Let ${0\leq i\leq d}$ and ${q=d-i}$. Then ${Z_q(X)=Z^i(X)=\{Z|Z=\sum n_\mathbb{A}lpha W_\mathbb{A}lpha\mbox{ irreducible codimension }i\mbox{ varieties}\}}$ be the group of algebraic cycles on ${X}$ of dimension ${q}$ (codimension ${i}$).

Examples:

1. ${Z^1(X)}$ is the set of Weil divisors
2. ${Z^d(X)=Z_0(X)}$ is teh group of zero cycles, which are formal sums of points.
3. ${Z^2(X)=Z_1(X)}$ on a threefold is the free abelian group on the curves on ${X}$.

2.2. Operators on Cycles

There are three standard operations. The first is Cartesian product, if ${V\in Z_{q_1}(X_1)}$ and ${W\in Z_{q_2}(X_2)}$, then ${V\times W\in Z_{q_1+q_2}(X_1\times X_2)}$.

The second is pushdown of cycles. If ${f:X\rightarrow Y}$ is a morphism and ${Z\in Z_q(X)}$, then ${f_*(Z)\in Z_q(Y)}$ is defined as follows: take ${V\subset X}$. If ${f(V)}$ is the set theoretic image of an irreducible subvariety, then ${\dim f(V)\leq \dim V}$. If ${\dim f(V)<\dim V}$, then ${f_*(V)=0}$. If ${\dim f(V)=\dim V}$, then we have ${k(V)\supset k(f(V))}$ a finite extension of fields, and we set ${f_*(V)=[k(V):k(f(V))]f(V)}$, where ${f(V)}$ is always taken set-theoretically.

The third operation is the intersection product of two cycles, but it is not always defined. Let ${V,W}$ be subvarieties of ${X}$ smooth, of codimension ${i}$ and ${j}$. Then ${V\cap W=\cup A_\ell}$ where the ${A_\ell}$ are irreducible subvarieties of codimension at most ${i+j}$. The intersection is proper if the codimension is exactly ${i+j}$. We define the intersection multiplicity at ${A_\ell}$ to be ${i(V\cap W,A_\ell)=}$

If ${X=S}$ is a surface and ${V,W}$ are curves and ${P}$ a point on ${S}$, then we define the multiplicity at ${P}$ to be ${\dim \mathscr{O}_{P,S}/(f,g)}$ where ${f,g}$ are defining local equations for ${V}$ and ${W}$. But this isn’t the right definition if ${d>2}$.

In general, we take ${i(V\cap W,A_\ell)=\sum(-1)^r \mathrm{length}_{\mathscr{O}} \mathrm{Tor}_r^{\mathscr{O}} (\mathscr{O}/\mathscr{I}_V,\mathscr{O}/\mathscr{I}_W)}$ where ${\mathscr{O}=\mathscr{O}_{A_\ell,X}}$.

And so, we define ${V.W=\sum_\ell i(V\cap W,A_\ell)A_\ell}$ to be the intersection product if all the intersections are “good”. This then extends by linearity to ${Z^i(X)\times Z^j(X)\rightarrow Z^{i+j}(X)}$.

There are other operations: let ${f:X\rightarrow Y}$ and ${Z\in Z^i(Y)}$, then ${f^*(Z)\in Z^i(X)}$ the pullback.

We also have operation by correspondences, in particular if ${W\subset Y}$, then we have that ${f_*(W)}$ is the image under the projection of ${(X\times Z)\cdot \Gamma_f}$.

Let ${X}$ and ${Y}$ be varieties of dimension ${d}$ and ${\ell}$ and let ${T\in Z^m(X\times Y)}$ be a correspondence from ${X}$ to ${Y}$. If ${Z\in Z^i(X)}$, then ${T_*(Z)}$ is just the pushforward along the projection to ${Y}$ of ${T\cdot (X\times Y}$ in ${Z^{i+m-\ell}}$.

2.3. Good=adequate equivalence relations on cycles

Samuel in 1956 said that ${\sim}$ on the groups of algebraic cycles is an adequate relation if:

1. The set of cycles equivalent to zero should be a subgroup of ${Z^i(X)}$.
2. If ${Z,Z'}$ are equivalent and ${Z_1\in Z^i(X)}$, such that ${Z\cdot Z_1}$ and ${Z'\cdot Z_1}$ are defined, then they are equivalent.
3. If ${Z\in Z^i(X)}$ and ${Z_1\in Z^j(X)}$, then there exists ${Z'}$ equivalent to ${Z}$ such that ${Z'\cdot Z_1}$ is defined. (This is called the moving lemma, which motivated it)

If ${\sim}$ is good, then ${C^i_\sim (X)}$, the set of equivalence classes of this equivalence relation, are defined and ${C_\sim(X)=\oplus_{i=0}^d C^i_\sim(X)}$ is a commutative ring with respect to the intersection product, and ${f_*,f^*,T}$ are defined when ${f}$ is proper.

Some commonly used adequate equivalence relations are

1. Rational equivalence (Chow, Samuel 1956)
2. Algebraic equivalence (Weil 1952)
3. Homological equivalence
4. Numerical equivalence

2.4. Rational Equivalence

This is a generalization of linear equivalence.

Linear equivalence is a relation on ${\mathrm{Div}(X)=Z^1(X)}$. Here, for every ${\phi\in k(X)^*}$, we have ${\mathrm{div}(\phi)=sum_Y v_Y(\phi)Y}$ for ${X}$ smooth. If ${X}$ is not smooth, need to use ${\mathrm{ord}_Y(\phi)}$, which can be reviewed in Hartshorne.

For ${0\leq i\leq d}$, and ${Z\in Z^i(X)}$, we say that ${Z}$ is rationally equivalent to zero if ${Z\in \langle\mathrm{div}(\rho\rangle\subset Z^i(X)}$ where ${\langle\rangle}$ denotes the subgroup generated by cycles of the form ${\mathrm{div}\phi}$ where ${\phi\in k(Y)^*}$ for ${\mathrm{codim} Y=i-1}$.

Equivalently, there exists a finite collection ${\{Y_\mathbb{A}lpha,\phi_\mathbb{A}lpha\}}$ of codimension ${i-1}$ subvarieties of ${X}$.

3. Brosnan 1 – Normal Functions

The motivation is that Lefschetz used normal functions to prove the ${(1,1)}$ theorem, so maybe understanding normal functions will help prove the Hodge conjecture. The first person to explicitly define admissable normal functions seems to have been M. Saito in JAG and studied their zero loci and related them to the Hodge conjecture.

Let ${\bar{S}}$ be a complex manifold and ${\mathscr{H}}$ a variation of pure Hodge structure of negative weight on ${S}$, a Zariski open subset of ${\bar{S}}$. Let ${NF(S,\mathscr{H})^{ad}_{\bar{S}}}$ be the group of admissable normal functions on ${S}$. For ${\nu\in NF(S,\mathscr{H})^{ad}_{\bar{S}}}$, set ${Z(\nu)}$ be the zero locus of ${\nu}$. Then the closure of ${Z(\nu)}$ in ${\bar{S}}$ is a complex analytic subvariety of ${\bar{S}}$.

If ${S}$ is algebraic and ${\bar{S}}$ projective, then ${Z(\nu)}$ is an algebraic subvariety of ${S}$.

Remarks: There are at least two proofs of the above. One by Brosnan and Pearlstein, using information from the mixed ${SL_2}$ orbit theorem of Kato, Nakayama and Usui. The other is by Schnell. He introduces an extension of the family ${J(\mathscr{H})\rightarrow S}$ of Griffiths intermediate Jacobians. Uses that to extend the normal functions and the idea of Kato-Nakayama-Usui to compactify ${J(\mathscr{H})}$ is a more-or-less toroidal way.

There were some prior results by Saito in a JAG paper and by Brosnan-Pearlstein when ${\dim S=1}$. Also when the singularity of ${\nu}$ vanishes.

There are two basics: the first is Hodge structures. The category of pure Hodge structures of weight ${w\in\mathbb{Z}}$ is by definition the category of pairs ${(H,(H_{p,q})_{p,q\in\mathbb{Z}})}$ such that ${H}$ is finitely generated abelian group, ${H^{p,q}}$ are subspaces of ${H_\mathbb{C}}$ such that ${H}$ is their direct sum and ${\bar{H}^{pq}=H^{qp}}$, and morphisms are maps preserivng the ${H^{pq}}$‘s. We can replace finitely generated abelian group with ${A}$-modules for ${A}$ a subgroup of ${\mathbb{R}}$, though ${\mathbb{Z},\mathbb{Q},\mathbb{R}}$ are the only useful ones.

(defines MHS)

The category of MHS’s is an abelian category, and if ${X}$ is an algebraic variety over ${\mathbb{C}}$, then ${H^k(X,\mathbb{Z})}$ carries an MHS.

The second part is much more difficult than the first, adn involves geometry and a lot of homological algebra. The first part is just a masterpiece of linear algebra. The idea is to define ${I^{pq}=F^[\cap (\bar{F}^q\cap W_{p+q}+\bar{F}^{q-1}\cap W_{p+q-2}+\ldots)}$, as in El Zein’s notes. These give a bigrading, and a morphism of MHS’s is a morphism that preserves the ${I^{pq}}$, and that’s enough to show that it’s an abelian category.

The category of pure Hodge structures is essentially semi-simple. If ${X}$ is a smooth projective variety then ${H^k(X)}$ is a direct sum of irreducible Hodge structures. The category of MHS is not, there are nontrivial extension.

Take ${\lambda\in \mathbb{C}\setminus\{0,1\}}$ and set ${H_\lambda=H^1(\mathbb{P}\setminus \{0,\infty\},\{1,\lambda\})}$. Then ${H_\lambda}$ is a nontrivial extension of ${\mathrm{Gr}_2^W H_\lambda=\mathbb{Q}(-1)}$ by ${\mathrm{Gr}_0^W H_\lambda=\mathbb{Q}(0)}$.

In fact, we can explain the extension geometrically we have ${0\rightarrow H^1(\mathbb{P}^1,\{1,\lambda\})\rightarrow H^1(\mathbb{P}^1\setminus\{0,\infty\},\{1,\lambda\})\rightarrow H^1(\mathbb{P}^1\setminus\{0,\infty\})\rightarrow 0}$ which is naturally ${0\rightarrow \mathbb{Z}(0)\rightarrow H_\lambda\rightarrow \mathbb{Z}_{-1}\rightarrow 0}$.

To see that the extension is nontrivial, really need to calculate it, but to calculate it, we need to find where it goes.

Let ${H,K}$ be pure Hodge structures with ${K}$ of lower weight than ${H}$ and ${H_\mathbb{Z}}$ torsion free. Then ${\mathrm{Ext}^1_{MHS}(H,K)=\underline{\hom}(H,K)/F^0\hom(H,K)+\hom(H_\mathbb{Z},K_\mathbb{Z})}$.

There, ${\underline{\hom}(H,K)}$ is a pure Hodge structure of weight ${k-h}$ where ${F^p\underline{\hom}(H,K)}$ are the homomorphisms ${\phi(F^k)\subset F^{k+p}}$.

Proof: We’ll define a map by taking an extension ${0\rightarrow K\rightarrow E\stackrel{\pi}{\rightarrow}H\rightarrow 0}$. We can always find ${\phi_\mathbb{Z}:H_\mathbb{Z}\rightarrow E_\mathbb{Z}}$ of finitely generated abelian groups splitting the extension. On the other hand, maps ${\pi:E\rightarrow H}$ of mixed Hodge structure are strict with respect to the Hodge filtration, so ${\pi(F^p_E)=F^pH\cap \pi(E)}$, and from this it follows that we can find a map ${\phi:H_\mathbb{C}\rightarrow E_\mathbb{C}}$ such that ${\phi_F}$ is in the right place in the filtration. $\Box$

4. Carlson 2

So our set up is that ${D}$ is a period domain, it’s the set of ${m^{-1}(H^{p,q})}$‘s, and we have a marked polarized Hodge structure ${H}$. We really want to describe the period domain in terms of the Hodge filtration.

In the weight 1 case, ${D=\{m^{-1}(F^1)\subset \mathbb{C}^{2g}\}\subset G(g,2g)}$, and it needs to satisfy ${Q(F^1,F^1)=0}$ so it is contained in a proper subvariety, but then there’s also ${i^{p-q}Q(x,\bar{x})>0}$ where ${x\in H^{p,q}\setminus 0}$, so there’s some open conditions. So ${D\subset \check{D}\subset G(g,2g)}$.

In the weight 2 case, we first note that for polarized Hodge structures, then ${F^2}$ determines ${F^1}$. So we have ${D\subset \check{D}\subset G(p,2p+q)}$ where ${p=\dim H^{2,0}}$ and ${q=\dim H^{1,1}_0}$.

Let’s look at elliptic curves. Set ${\mathscr{E}=\{(x,y)|y^2=p(x)\}}$ where ${p(x)}$ is a cubic with distinct roots. For example, ${x(x-1)(x-t)}$, so long as ${t\neq 0,1}$ we get an elliptic curve, which is a nonsingular Riemann surface of genus 1.

The unique (up to scaling) abelian differential is ${\omega=\frac{dx}{\sqrt{p(x)}}=\frac{dx}{y}}$.

Show that ${\omega}$ is holomorphic at ${x=0,1,t,\infty}$.

So, we can write ${H^1(\mathscr{E},\mathbb{C})=\mathbb{C}\frac{dx}{y}\oplus \mathbb{C}\frac{\bar{dx}}{y}}$. We define a marking ${m:\mathbb{Z}^2\rightarrow H^1(\mathscr{E},\mathbb{Z})}$, picking out a symplectic basis for the cohomology. Take the basis to be ${\delta,\gamma}$, and the dual basis ${\delta^*,\gamma^*}$.

We know that ${\omega=A\delta^*+B\gamma^*}$, and that ${i\int_M \omega\wedge\bar{\omega}>0}$ and so ${i(\omega\cup\bar{\omega})[M]>0}$, which means that ${i(A\delta^*+B\gamma^*)\cup (\bar{A}\delta^*+\bar{B}\gamma^*)>0}$, and so we get that ${i(A\bar{B}-B\bar{A})>0}$.

Consequence of Riemann Bilinear relation is that ${A\neq 0,B\neq 0}$. So we can rescale ${\omega}$ so that the ${A}$-period is 1. So the period matrix ${[A B]}$ because ${[1;Z]}$. What can we say about ${Z}$? The relation is now ${i(\bar{B}-B)>0}$, so we have that ${\mathrm{Im}(Z)>0}$.

So we now know that ${D=\mathbb{H}_1=\{Z|\mathrm{Im}(Z)>0\}}$.

4.1. Period Map

Let ${f:X\rightarrow S}$ a morphism and ${f}$ of maximal rank on ${S\setminus \Delta}$. The smooth part of the map is locally differentiably trivial. The period map is the map ${p:\widetilde{S\setminus\Delta}\rightarrow D}$ and also the map it induces ${p:S\setminus\Delta\rightarrow \Gamma\backslash D}$.

Look at the family ${y^2=x(x-1)(x-t)}$ of elliptic curves. We claim that the period is defined and holomorphic on an open set away from ${0,1}$ in ${\mathbb{C}}$. Take our marking to pick the basis ${\delta,\gamma}$ with intersection ${+1}$. Then ${A(t)=\int_\gamma \frac{dx}{\sqrt{x(x-1)(x-t)}}}$ and ${B(t)=\int_\gamma \frac{dx}{\sqrt{x(x-1)(x-t)}}}$.

These are integrals with fixed domains of integration, so we can take the ${\bar{t}}$ derivative of the inside to check holomorphicisty, and it’s clear that the ${A}$-period is holomorphic, and so is the ${B}$-period for the same reasons.

So the period map is ${Z(t)=B(t)/A(t)}$. So we have ${\tilde{P}:U\rightarrow D}$ the period map. And then there’s ${\Gamma=\mathrm{Aut}(H^1(M,\mathbb{Z}))\cong \mathrm{Sp}(2,\mathbb{Z})}$.

Show that ${\mathrm{Sp}(2,\mathbb{Z})\cong \mathrm{SL}(2,\mathbb{Z})}$.

So, we get a map ${\mathbb{C}\setminus\{0,1\}\rightarrow \Gamma\backslash\mathbb{H}}$.

Let ${F}$ be a fundamental region for ${\Gamma}$. Almost al of ${\mathbb{H}}$ is given by ${\Gamma\cdot F}$. No two points of ${F}$ are equivalent under ${\Gamma}$, so if ${x,y\in F}$, then there is no ${g}$ such that ${gx=y}$.

The space ${\Gamma\backslash \mathbb{H}}$ is an orbifold, not quite a manifold, it has a few special points.

Reference for Modular Forms: Serre’s Course in Arithmetic.

Set ${g_2(Z)=60\sum_{(m,n)\neq 0} \frac{1}{(m+nZ)^4}}$ and say it has weight 4 and set ${g_3(Z)=140\sum_{(m,n)\neq 0} \frac{1}{(m+nZ)^6}}$ of weight 6. These are modular forms, they’re not invariant under ${\Gamma}$, but transform in a controlled way. Define a polynomial in them by ${\Delta=g_2^3-27g_3^2}$ and set ${j(Z)=\frac{g_2^3}{\Delta}}$. We call ${\Delta}$ the discriminant.

If the elliptic curve is ${y^2=4x^3-g_2 x-g_3}$ (and all complex elliptic curves can be put in this form by a change of variables) then ${\Delta=0}$ if and only if the curve is singular.

The function ${j}$ gives a biholomorphic map ${j:\Gamma\backslash \mathbb{H}\rightarrow \mathbb{C}}$.

5. Murre 2

5.1. Rational Equivalence

We have a cycle ${Z\in Z^i(X)=Z_q(X)}$ where ${q=d-i}$ and ${d=\dim X}$.

Lemma 1 The following conditions on ${Z\in Z^i(X)}$ are equivalent:

1. ${Z}$ is rationally equivalent to zero
2. there exists ${T\in Z^i(\mathbb{P}^1\times X)}$ and ${a,b\in\mathbb{P}^1}$ such that ${Z=T(a)-T(b)}$.

Proof: 1 implies 2 is easy. For the other direction, we use a theorem of Fulton, which says that if ${f:X\rightarrow Y}$ is proper and ${\phi\in k(X)^*}$ and ${\dim X=\dim Y}$, then ${f_*(\mathrm{div}(\phi))=\mathrm{div}(\mathrm{Nm}(\phi))}$.

Take the condition for ${\lambda\in\mathbb{P}^1}$. ${\phi=(\pi_{\mathbb{P}^1})^*(T)}$ and ${T:X\rightarrow Y}$. ${Z=T(0)-T(\infty)=\mathrm{div}\phi}$. $\Box$

Rational equivalence is a “good” equivalence relation. ${Z_{rat}(X)}$ is a subgroup, for the second condition, let ${Z,Z'\in Z^i(X)}$ such that ${Z\cdot W}$ and ${Z'\cdot W}$ are defined for any given ${W}$. Then take ${T\subset\mathbb{P}^1\times X}$ and ${T(0)-T(\infty)=Z-Z'}$. Because we can assume no horizontal components, it works.

The interesting part is the moving lemma. Let ${Z\in Z^i(X)}$ and choose finitely many ${W_j}$. We want to find ${Z'\sim Z}$ such that ${Z'\cdot W_j}$ are defined.

Case 1: ${X=\mathbb{P}^N}$, ${\tau}$ a general projective transforming on ${\mathbb{P}^N}$. ${\tau(V)}$ and ${V}$ are rationally equivalent, so we can make sure ${\tau(V)\cdot W}$ is defined.

Case 2: ${\mathrm{codim}(V\cap W)=i+j-e}$ where ${e>0}$ is called the excess. Take a general linear space ${L_{N-d-1}\subset \mathbb{P}^N}$. Take ${C_{L,V}}$ the cone spanned by ${L}$ and ${V}$. Then ${\dim C_{L,V}=N-i}$. ${X\cdot C=V+V_*}$, the excess intersection. Because ${L}$ is general, the excess is ${, and things work out.

5.2. Chow Groups

We define ${CH_q(X)=CH^i(X)=Z^i(X)/Z^i_{rat}(X)}$.

1. ${CH(X)=\oplus_{i=0}^d CH^i(X)}$ is a commutative ring with identity.
2. ${CH(X)}$ behaves functorially with respect to ${f:X\rightarrow Y}$. More precisely:
1. if ${f}$ is proper, then ${f_*}$ is
2. if ${f}$ arbitrary, ${f^*}$ is
3. ${T\in CH(X\times Y)}$ gives ${T_*:CH_q(X)\rightarrow CH_q(Y)}$ is a homomorphism.

Let ${Y\stackrel{i}{\rightarrow} X\stackrel{j}{\leftarrow}U=X-Y}$ inclusions. Then ${CH_q(Y)\stackrel{i_*}{\rightarrow} CH_q(X)\stackrel{j^*}{\rightarrow} CH_q(U)\rightarrow 1}$ is exact.

${X\times \mathbb{A}^n\rightarrow X}$ induces ${CH^i(X)\rightarrow CH^i(X\times \mathbb{A}^n)}$ is an isomorphism for ${0\leq i\leq \dim X=d}$.

5.3. Algebraic equivalence

Let ${Z\in Z^i(X)}$, we say that ${Z}$ is algebraically equivalent to zero if there exists ${T\subset C\times X}$ such that ${Z=T(a)-T(b)}$ where ${a,b\in C}$ and ${C}$ is a curve.

It’s easy to see that ${Z_{rat}\subset Z_{alg}}$, because ${\mathbb{P}^1}$ can be taken as ${C}$.

5.4. Homological Equivalence

Fixing a good cohomology theory, for instance, the usual ones, but also etale cohomology will work. Then we get the usual intersection theory in cohomology.

We require that there exist a cycle map ${Z^i(X)\rightarrow CH^i(X)\rightarrow H^{2i}(X)}$ such that the intersection product is compatible with the cohomological cup product. Set ${Z_{hom}(X)}$ to be the homologically trivial cycles. This is also a good equivalence relation.

${Z_{alg}\subset Z_{hom}}$, by Matsusaka in 1956, it was shown for divisors. It is not true in general, proved in 1969.

5.5. Numerical Equivalence

Let ${X}$ be a ${d}$-dimensional projective variety, ${Z\in Z^i(X)}$ and ${W\in Z^{d-i}(X)}$ such that ${Z\cdot W}$ is defined, and it’s just a number of points, a zero cycle. We say that ${Z}$ is numerically equivalent to zero if ${\deg(Z\cdot W)=0}$ for all ${W\in Z^{d-i}(X)}$ where it is defined.

Is ${Z_{hom}(X)\subseteq Z_{num}(X)}$? It is known for divisor, and it’s a conjecture for general ${i}$.

For ${X}$ a variety over ${\mathbb{C}}$, the Hodge conjecture implies this one.

6. Brosnan 2

We’ll finish proving the theorem of Carlson.

$\displaystyle \mathrm{Ext}^1_{MHS}(H,K)=\hom(H_\mathbb{C},K_\mathbb{C})/F^0\underline{\hom}(H,K)+\hom(H_\mathbb{Z},K_\mathbb{Z})$

Sketch continued:

So if we start with ${0\rightarrow K\rightarrow E\stackrel{\pi}{\rightarrow}H\rightarrow 0}$, because ${H_\mathbb{Z}}$ is torsion free, we can find a splitting ${\sigma_\mathbb{Z}:H_\mathbb{Z}\rightarrow E_\mathbb{Z}}$ because ${\pi:E\rightarrow H}$ is strict with respect to ${F}$, we can find a splitting ${\sigma_F:H_\mathbb{C}\rightarrow E_\mathbb{C}}$ preserving the Hodge filtration.

Using the fact that ${F^p=\oplus_{p'\geq p} I^{p',q}}$to decude that ${\pi}$ is strict with respect to teh Hodge filtration. Define ${cl(E)=\sigma_F-\sigma_\mathbb{Z}:H_\mathbb{C}\rightarrow K_\mathbb{C}}$. But ${cl(E)}$ depends on some choices: we could add any morphism ${\phi_\mathbb{C}\in F^0(\hom(H_\mathbb{C},K_\mathbb{C}))}$ to ${\sigma_F}$, and we could add any morphism ${\phi_\mathbb{Z}\in \hom(H_\mathbb{Z},K_\mathbb{Z})}$ to ${\sigma_\mathbb{Z}}$. Modding out by these choices determines the morphism we need.

The second step is to product a map backwards. Suppose that ${f\in \hom(H_\mathbb{C},K_\mathbb{C})}$. Define ${E_0=H\oplus K}$, and define ${T_f\in \mathrm{End} H_\mathbb{C}\oplus K_\mathbb{C}}$ by ${(h,k)\mapsto (h,k+f(h))}$. Define an extension ${E_f}$ by taking the underlying group to be ${E_0}$, ${W_nE_f=W_nE_0}$ and ${F^*E_f=T_f(F^*)}$. Then ${T_f}$ induces the identity on ${\mathrm{Gr}^W_0}$ and therefore ${F^*E_f=F^*E_0}$ on ${\mathrm{Gr}_2^W}$ and therefore ${E_f}$ is a Hodge structure. We need to check that ${cl(E_f)=f}$.

Most important case of the theorem is if ${H=\mathbb{Z}}$ and ${K}$ is of negative weight, we set ${J(K)=\mathrm{Ext}_{MHS}(\mathbb{Z},K)=K_\mathbb{C}/F^0K_\mathbb{C}+K_\mathbb{Z}}$, this is a complex torus. To check this we need to check that ${K_\mathbb{Z}}$ is discrete in ${F_\mathbb{C}/F^0}$. In fact, ${\mathrm{Ext}^1_{MHS}(K,H)=\mathrm{Ext}^1_{MHS}(\mathbb{Z}(0),\hom(H,K))}$.

In the exercise, we have ${0\rightarrow \mathbb{Z}(0)\rightarrow H_\lambda\rightarrow \mathbb{Z}(-1)\rightarrow 0}$. In that case, we have ${\mathrm{Ext}^1_{MHS}(\mathbb{Z}(-1),\mathbb{Z}(0))=\mathrm{Ext}^1_{MHS}(\mathbb{Z}(-1)\otimes \mathbb{Z}(1),\mathbb{Z}(0)\otimes \mathbb{Z}(1))}$ (all previous ${\mathbb{Z}}$‘s are ${\mathbb{Z}(0)}$)

So this is just ${\mathrm{Ext}^1(\mathbb{Z},\mathbb{Z}(1))}$, which is ${\mathbb{C}/\mathbb{Z}(1)=\mathbb{C}/2\pi i\mathbb{Z}}$ and by the exponential map, this is ${\mathbb{C}^*}$.

In general, ${J(H)}$ is called the Griffiths intermediate Jacobian, and is generally not algebraic. However, when ${H_\mathbb{C}=H^{-1,0}\oplus H^{0,-1}}$ is polarized, then ${J(H)}$ is an abelian variety: ${J(H_1(C))}$ is the Jacobian of the curve.

If the weight is ${-1}$, then ${J(H)}$ is a compact complex group. The reason is that we need to show that ${H_\mathbb{Z}}$ i sdiscrete. Look at ${H_\mathbb{R}\rightarrow H_\mathbb{C}/F^0}$. Ths is injective because ${H_\mathbb{R}\cap F^0}$ is zero, both have the same dimension as real vector spaces, and so it is an isomorphism of real vector spaces. Since ${H_\mathbb{Z}}$ is discrete in ${H_\mathbb{R}}$, it is in ${H_\mathbb{C}/F^0}$ as well.

Let ${C}$ be a smooth projective complex curve, and ${D=\sum_{p\in |D|}n_p[p]}$ a divisor of degree 0. We get a long exact sequence of MHS as follows:

$\displaystyle 0=H^1(C,C\setminus D)\rightarrow H^1(C)\rightarrow H^1(C\setminus D)\rightarrow H^2(C,C-D)\rightarrow H^2(C)$

and the last two terms are ${\oplus_{p\in D} \mathbb{Z}(-1)}$ and ${\mathbb{Z}(-1)}$. The divisor ${D}$ then gives a map ${\mathbb{Z}(-1)\rightarrow \mathbb{Z}(-1)^D}$ by sending ${\frac{1}{2\pi i}}$ to ${\frac{1}{2\pi i} \sum n_p}$.

So ${D}$ gives an extension ${[E_d]\in \mathrm{Ext}^1_{MHS}(\mathbb{Z}(-1),H^1(C))=\mathrm{Ext}_{MHS}(\mathbb{Z}(0),H^1(C)(1))=J(H^1(C))}$, which is just the Jacobian of ${C}$, and the map is the Abel-Jacobi map ${\mathrm{AJ}:\mathrm{Div}^0(C)\rightarrow \mathrm{Jac}(C)}$.

6.1. Normal Functions

Let ${D}$ be a divisor on a surface ${\sum n_iD_i}$ where the ${D_i}$ are irreducible curves not contained in fibers of a map ${f:X\rightarrow \mathbb{P}^1}$.

For each ${p\in \mathbb{P}^1}$ we get a curve as the fiber, and on a dense open ${U\subset \mathbb{P}^1}$, these curves are smooth.

Assume that for one and helce all ${p\in U}$, we have ${D_p=D\cdot X_p}$ has degree ${0}$. Then the above construction applied to the curves ${X_p}$ gives a section of the family whose fibers are the Jacobians of ${X_p}$. For eahc point, we’ve got the Abel-Jacobi map, and this section is called a normal function.