## ICTP Day 9

So, today a few lecture series finished up, and I’ve posted notes from some of the older ones in pdfs on my personal website. And here’s the notes, which I’ll admit are getting spottier, as Griffiths is lecturing from his notes on a projector, and Kerr just goes VERY quickly. I’ve stopped taking notes for these two series.

1. Murre 5

There exist varieties ${X\subset\mathbb{P}^N}$ over ${\mathbb{C}}$ smooth and ${i>1}$ such that ${Z^i_{alg}(X)\neq Z^i_{hom}(X)}$.

In fact, there exists ${Z\in Z^i_{hom}(X)}$ such that ${Z\neq 0}$ in ${Griff^i(X)\otimes \mathbb{Q}}$.

The key theorem is suppose that ${V_{d+1}\subset \mathbb{P}^N}$ with ${d+1=2m}$ such that ${H^{2m-1}(V)=0}$. Suppose there exists a Lefschetz pencil ${\{W_t\}\subset V}$ such that ${H^{2m-1}(V_t,\mathbb{C})\neq H^{m-1,m}\oplus H^{m,m-1}}$. Assume that ${Z\in Z^m(W_t)}$ such that ${ZW_t=Z_t}$ in ${CH^m_{alg}(W_t)}$ for general ${t}$. Then ${Z\sim_{hom} 0}$ on ${V}$.

For instance, take ${V_4\subset\mathbb{P}^5}$ a smooth quartic in ${\mathbb{P}^5}$. Then ${H^3(V)=0}$, and we can take ${W_t=VF_t}$ where ${F_t}$ is a hypersurface of degree ${r}$. Now, if ${r\geq 5}$, then ${H^3(W_t)\neq H^{21}+H^{12}}$.

Because ${K_{W_t}>0}$ if ${r\geq 5}$, we have that ${H^{3,0}((W_t)=H^0(W_t,\Omega^3)\neq 0}$.

On ${V}$, there are two systems of planes ${\{L\}}$ and ${\{L'\}}$ and ${H^*(V)=\mathbb{Z}+\mathbb{Z}=\mathbb{Z}(L+L')\oplus \mathbb{Z}(L-L')}$. We can check that ${(L-L')(L-L')=-2}$ and so we take ${Z=L-L'}$, it’s not homologically equivalent to zero.

${Z_t=ZW_t=C-C'}$ where ${C=LW_t}$ and ${C'=L'W_t}$. Then ${Z_t}$ is not homologically trivial on ${W_t}$. But ${Z_t}$ is (and I didn’t follow what happened next, looked like a proof that ${Z_t}$ IS homologically trivial on ${W_t}$ or something)

For sufficiently general quintic hypersurfaces in ${\mathbb{P}^4}$, ${\dim Griff^2(X)\otimes\mathbb{Q}=\infty}$.

1.1. Albanese Kernel

Let ${X=X_d}$ be a smooth projective irreducible variety over an algebraically closed field ${k}$. For ${i=1}$, we get divisors and ${CH^1_{alg}(X)\cong \mathrm{Pic}^0(X)}$ is an isomorphism of abelian varieties.

Now, look at ${i=d}$. These are zero cycles. We have ${CH^{alg}_0(X)\rightarrow Alb(X)}$ a surjective map.

What’s the kernel? Let ${T(X)=\ker( CH^{(0)}_0(X)\rightarrow Alb(X))}$.

Let ${X=S}$ a surface over ${\mathbb{C}}$ with ${p_g(S)\neq 0}$, then ${T(S)\neq 0}$.

In fact, it is “infinite dimensional” that is, ${T(S)}$ cannot be parameterized by an algebraic variety.

Equivalently, ${\phi_m:\mathrm{Sym}^r S\times \mathrm{Sym}^r S\rightarrow CH^{(0)}_0(S)}$ by ${(z_1,z_2)\mapsto z_1-z_2}$ such that ${p_g(S)\neq 0}$, then there is no ${m}$ such that ${\phi_m}$ is surjective.

1.2. Generalization of Bloch 1979

Let ${X/k=\bar{k}}$ and define ${CH^i_{alg}(X)}$ is “weakly” representable if there exists ${(C,s\in C, T\in CH^i(C\times X))}$ such that ${T:CH^1_{alg}(C)=J(C)\rightarrow CH^i_{alg}(X)}$ surjective.

Fix a “good” cohomology theory. If ${CH^2_{alg}(S)}$ is weakly representable, then ${H^2(S)_{tr}=0}$.

Recall that ${H^2(S)=H^2(S)_{alg}\oplus H^2(S)_{tr}}$, with ${H^2(S)_{alg}}$ the image of ${CH^1(S)}$ and ${H^2(S)_{tr}}$ the orthogonal complement. Then the theorem implies the theorem of Mumford, because if ${X=S/\mathbb{C}}$, then ${H^{2,0}\subset H^2(S)_{tr}}$ is nonzero, so ${H^2(S)_{tr}}$ is nonzero.

${X_d/k=\bar{k}}$ Assume there exists ${Y\subset X}$ such that ${CH^{(0)}_{alg}(Y_L)\rightarrow CH^{(0)}_{alg}(X_L)}$ for all ${L=Z\supset k}$. Then there exists a divisor ${D\subset X}$ and two corredpondences ${\Gamma_1,\Gamma_2}$ in ${CH^d(X\times X)}$ with ${|F_1|\subset Y\times X}$ and ${|\Gamma_2|\subset X\times D}$ such that ${N\Delta (X)=\Gamma_1+\Gamma_2}$ in ${CH^d(X\times X)}$ for some ${N>0}$.

Note: If ${X=S}$, and ${CH^{(0)}_0(S)}$ is weakly representable, the assumption if then fulfilled, so this implies Mumford.

If ${\Delta(X)}$ is as before, then ${H^2(S)_{tr}=0}$.

The ${\Gamma_i}$ operate trivially on ${H^2(S)_{tr}}$ for ${i=1,2}$ and so ${\Gamma:H^2(X)\rightarrow H^2(S)}$ must restrict to zero on ${H^2(S)_{tr}}$.

Now, a sketch of the second Bloch theorem.

Take generaic point ${\eta}$ of ${X}$. ${k\subset k(y)\subset \overline{k(y)}}$, for ${y\in Y}$. Then ${\eta-y\in CH^0_0(X_{k(y)})\rightarrow CH^0(X_{\overline{k(y)}})}$ (couldn’t read the last bit)

2. Griffiths 2

From his notes.

3. Kerr 2

3.1. Hermitian Symmetric Domains

Let ${X=G(\mathbb{R})^+}$ a collection of homomorphisms ${\phi:\mathbb{U}\rightarrow G}$. ${G}$ a real adjoint algberaic group such that only ${z,1,z^{-1}}$ appear in the representation ${\mathrm{ad}\circ\phi}$, ${\theta=\mathrm{Ad}(\phi(-1))}$ Cartan and ${\phi(-1)}$ doesn’t project to the identity in any factor of ${G}$.

3.2. ID- Hodge Theoretic Interpretation

Let ${V}$ be a ${\mathbb{Q}}$ vector space.

A hodge structure on ${V}$ is a homomorphism defined over ${\mathbb{R}}$ ${\tilde{\phi}:\mathbb{S}\rightarrow\mathrm{GL}(V)}$ such that ${w_{\tilde{\phi}}:\mathbb{G}_m\rightarrow \mathbb{S}\rightarrow \mathrm{GL}(V)}$ is defined over ${\mathbb{Q}}$.

Associated to ${\tilde{\phi}}$ is ${\mu_{\tilde{\phi}}:\mathbb{G}_m\rightarrow \mathrm{GL}(V)}$ given by ${z\mapsto \tilde{\phi}_\mathbb{C}(z,1)}$. Then we get a decomposition into ${V^{p,q}}$ such that the eigenvalue of ${\tilde{\phi}_\mathbb{C}(z,w)}$ on ${V^{p,q}}$ is ${z^pw^q}$.

Now fix a weight ${n}$. The Hodge numbers ${\{h^{p,q}\}}$ and ${Q:V\times V\rightarrow \mathbb{Q}}$.

Let ${D}$ be the period domain for polarized HS’s of this type, ${t\in \oplus_{k,\ell} V^{\otimes k}\otimes \hat{V}^{\otimes \ell}}$

(I got lost here, and started reading Milne’s notes)

4. Griffiths 3

From his notes.

5. Green 1 – Applications to the Beilinson-Bloch Conjecture

California is like Italy without the art. – Oscar Wilde

Let ${X}$ be a smooth projective variety. There are two ways to look at it. One is to look at it as a compact Kähler manifold with a Hodge metric giving an projective embedding. The other is to look at ${X}$ as already a subset of ${\mathbb{P}^n(k)}$ given by explicit equations and work algebraically.

We use the Hodge metric to get a Hodge structure, which sits inside ${\Gamma\backslash D}$. A lot of what you get here comes from algebra, but aren’t Hodge structures of varieties, and even occasionally from analysis.

We can take the field of definition of ${X}$, ${k}$, to be finitely generated over ${\mathbb{Q}}$, by noting that ${X}$ is defined over ${\mathbb{Q}}$ adjoin the coefficients of the equations. Now, some things we can do with ${k}$ as an abstract field, but others, we can do only when ${k\subset\mathbb{C}}$.

Let’s look at ${\mathbb{Q}(\pi)}$. As ${\pi}$ is transcendental, we can represent elements of this field by ${p(\pi)/q(\pi)}$ where ${p,q\in\mathbb{Q}[x]}$. We can similarly describe ${\mathbb{Q}(e)}$. Abstractly, these fields are isomorphic. But they’re different subfields of ${\mathbb{C}}$. We can think of these as both being ${\mathbb{Q}(x)}$, which is distinct from ${\mathbb{Q}(\sqrt{7})}$.

So look at the elliptic curves ${y^2=x(x-1)(x-\pi)}$ and ${y^2=x(x-1)(x-e)}$. As abstract curves over ${\mathbb{Q}(x)}$, they’re isomorphic. However, over ${\mathbb{C}}$, the Hodge structures are not equivalent.

In mathematics, we’ve got a break between technology, which we need to prove theorems, and intuition, which we need to figure out theorems.

If we write ${k}$ as a field finitely generated over ${\mathbb{Q}}$, then ${k=\mathbb{Q}(\alpha_1,\ldots,\alpha_t)[\beta_1,\ldots,\beta_r]}$ where ${\alpha_1,\ldots,\alpha_t}$ are algebraically independent and ${\beta_1,\ldots,\beta_r}$ are algebraic over it.

Enter geometry. ${\mathbb{Q}(x)=\mathbb{Q}(\mathbb{P}^1)}$, so ${k=\mathbb{Q}(S)}$ for ${S}$ defined over ${\mathbb{Q}}$, and ${\mathbb{Q}(S_1)\cong \mathbb{Q}(S_2)}$ if ${S_1}$ and ${S_2}$ are birational over ${\mathbb{Q}}$.

A point ${s\in S(\mathbb{C})}$ is a geometric point, or a very general point of ${S}$ if ${s}$ does not lie on any proper ${\mathbb{Q}}$-subvariety of ${S}$ (that is, the Zariski closure of ${s}$ over ${\mathbb{Q}}$ is ${S}$)

For ${S=\mathbb{P}^1}$, ${s}$ is very general if and only if ${s}$ is transcendental.

Now, let ${k=\mathbb{Q}(\alpha)}$ for ${\alpha}$ transcendental, and look at ${y^2=x(x-1)(x-\alpha)}$. For every very general point of ${S}$, in fact, for ${s\neq 0,1,\infty}$, we get a smooth elliptic curve. So we have ${\mathcal{X}\rightarrow S\setminus E}$, and ${E}$, the discriminant locus, is defined over ${\mathbb{Q}}$.

Let ${k}$ be a number field. Then ${k=\mathbb{Q}(\alpha)}$ and ${\alpha}$ has minimal polynomial ${p(x)\in \mathbb{Q}[x]}$. Now let ${S}$ be the variety defined by ${p}$, that is, a finite number of points. ${S}$ is defined over ${\mathbb{Q}}$, but the points are not, individually, defined over ${\mathbb{Q}}$. ${X}$ defined over ${k}$ means that we get ${[k:\mathbb{Q}]}$ complex varieties, one for each embedding into ${\mathbb{C}}$.

Let ${X}$ be defined over ${\mathbb{Q}}$ and ${p\in X}$ a very general point. Then ${X\rightarrow \mathbb{P}^n(\mathbb{Q})}$ by ${p\mapsto (p_0,\ldots,p_n)}$ then ${k=\mathbb{Q}(p_1/p_0,\ldots,p_n/p_0)}$. We can take ${S=X}$, so transcendence degree is ${\dim X}$.

Let ${X}$ be defined over ${\mathbb{Q}}$. Take ${(p,q)\in X\times X}$ a very general point. Then we get an embedding into ${\mathbb{P}^n}$, and the field’s transcendence degree is ${2\dim X}$.

Look at hypersurfaces of degree ${d}$ in ${\mathbb{P}^N}$. If ${F=\sum_{|I|=d} a_I x^I}$, then if we look at ${k=\mathbb{Q}(a_I)}$ we get a much larger dimensional projective space for ${S}$, and we hafve ${\mathcal{X}\rightarrow S}$ the universal family of hypersurfaces.

So, which computations do we actually need the complex embeddings for? Grothendieck learned to compute cohomology groups using just ${k}$, but for the Hodge structures, really need ${\mathbb{C}}$.

Big Point: Hodge structures require a complex embedding.

Set ${y^2=x(x-1)(x-\alpha)=f(x)}$. If we differentiate, we get ${2ydy=f'(x)dx}$, so we have ${\frac{2dy}{f'(x)}=\frac{dx}{y}}$, and this lets us represent the holomorphic 1-form on ${E}$. Call it ${\omega}$.

Now, let ${\lambda}$ be a simple closed homotopically nontrivial curve in ${E}$. Look at ${\int_\lambda \omega}$. We can write ${\lambda=\sigma_1+\ldots+\sigma_n}$ a bunch of lines, and so we need to integrate ${\omega}$ along each of these, and ${\int_{\sigma_m} \frac{dx}{y}=\int_{\sigma_m} \frac{dx}{\sqrt{f(x)}}}$, and we need ${\int_C \frac{dx}{\sqrt{x(x-1)(x-\alpha)}}}$. We can expand as a power series and integrate.

Now, the point is that hte integral lattice ${H^r(X,\mathbb{Z})\subset H^r(X,\mathbb{C})}$ is going to depend transcendentally on ${s}$, on complex embedding ${k\rightarrow \mathbb{C}}$.

Now, take ${X}$ a smooth variety defined over ${k}$. We’ll be wanting ${\Omega^1_{X(k)/k}}$ to be the Kähler differentials over ${k}$. That is, the module ${d(f+g)=df+dg}$, ${da=0}$ for ${a\in k}$ and ${d(fg)=fdg+gdf}$. We also have ${\Omega^1_{X(k)/\mathbb{Q}}}$ where we only have ${da=0}$ for ${a\in\mathbb{Q}}$.

Now, ${da=0}$ for all ${a\in k}$ actually implies ${da=0}$ for all ${a\in\bar{k}}$. Why? Look at the minimal polynomial of ${\alpha}$, ${a_0 \alpha^m+\ldots a_m=0}$. Take ${d}$ of this, and the Liebniz rule implies that we have some element of ${k}$ times ${d\alpha=0}$, and so ${d\alpha=0}$.

So ${\Omega^1_{k/\mathbb{Q}}}$ is a ${k}$ vector space with basis ${d\alpha_1,\ldots,d\alpha_\ell}$.

These Kähler differentials give us a complex ${\Omega^*_{X(k)/\mathbb{Q}}}$, and need to be careful, this complex doesn’t end at ${\dim X}$.

6. Brosnan 5

The following are equivalent:

1. The Hodge Conjecture holds for all smooth projective varieties ${X}$
2. For all smooth projective varieties ${X\subset\mathbb{P}^M}$, of even dimension ${2n}$ and all Hodge classes ${\alpha\in H^{2m}(X,\mathbb{Z}(n))}$ there exist a ${k\in \mathbb{Z}_{>0}}$ and a ${D\in |\mathscr{O}(k)|}$ such that ${0\neq \alpha|_D\in H^{2n}(D,\mathbb{Z}(n))}$.

There is a similar result in Saito’s “Admissable Normal Functions”

Proof: ${\Rightarrow}$: If HC holds, then given a Hodge class ${\alpha\in H^{2n}(X,\mathbb{Z}(n))}$, there exists another Hodge class which is algebraic ${\beta=[Z]}$ such that ${\alpha\cup\beta\neq 0}$. “Clearly” ${Z\subset D}$ for some divisor ${D}$, therefore ${\alpha|_D\neq 0}$. $\Box$

Suppose that ${X}$ is even dimensional as above and ${\alpha\in Prim X}$. Then for ${k>>0}$, we have ${sing_s \nu(s)=0}$ if and only if ${\alpha|_{X_s}=0}$, where ${X}$ is the universal hyperplane over ${\mathbb{P}}$.

This is a result of Brosnan, Fong, Nie, and Pearlstein, and of de Cataldo and Migliorini, and uses the decomposition theorem for perverse sheaves.

${\nu}$ an admissable normal function implies that ${sing_s\nu\in IH^*(B\cap S,\mathscr{H})\subseteq H^*(B\cap S,\mathscr{H})}$ for ${B}$ a small contractible ball.

Construction: let ${\Delta}$ be the unit disc, and ${\Delta^*}$ the punctured disc. Suppose that ${\mathscr{V}}$ is a VMHS on ${(\Delta^*)^r}$. Pick ${s_0\in (\Delta^*)^r}$. Then the underlying local system ${\mathscr{V}_\mathbb{Z}}$ is a ${\mathbb{Z}^r=\pi_1((\Delta^*)^r,s_0)}$-module determined by the action of ${\gamma}$ commuting invertible operators ${T_1,\ldots,T_r}$ on ${\mathscr{V}_\mathbb{Z}}$.

The ${T_i}$ are quasi-unipotent operators. There exists positive integers ${a,b}$ such that ${(T^a-1)^b=0}$.

Remark: ${T_i}$ are quasi-unipotent on ${\mathrm{Gr}^W\mathscr{V}}$ implies that they are on ${\mathscr{V}}$, so the monodromy theorem holds for MHS.

If ${\mathscr{V}}$ is polarizable variation, then we can find ${\Delta^r\rightarrow \Delta^r}$ such that the monodromy is unipotent.

6.1. Monodromy Filtration

Let ${V}$ be a vector space and ${N}$ a nilpotent operator, ${\log T}$.

There exists a unique increasing filtration ${W}$ on ${V}$ satisfying ${N(W_*)=W_{*-2}}$, ${N^aGr_a^W(V)=\mathrm{Gr}^W_{-a}(V)}$ for ${a>0}$.

Relative Weight Filtration: Let ${N}$ be nilpotent operator on a finite dimensional vector space ${V}$ equipped with an increasing filtration ${W_*}$.

there exists at most 1 increasing filtration ${M=M(N,W)}$ of ${V}$ satisfying ${NM_*\subset M_{*-2}}$ and ${N^a\mathrm{Gr}^M_{k+a}\mathrm{Gr}^W_k V\rightarrow \mathrm{Gr}^M_{k-a}\mathrm{Gr}_k^W V}$ an isomorphism for all ${k}$ and all ${a\geq 0}$.

Deligne noticed in Weil 2 that ${\ell}$-adic sheaves coming from geometry always have this property.

Suppose ${\mathscr{V}\in VMHS(\Delta^*)}$ with unipotent monodromy. We say ${\mathscr{V}}$ is admissible relative to ${\Delta}$ if:

1. ${\mathrm{Gr}^W_k\mathscr{V}}$ is polarizable
2. The Hodge filtration extends to holomorphic subbundles of ${\mathscr{V}_{can}=}$Deligne canonical extension of ${\mathscr{V}\otimes_\mathbb{Z}\mathscr{O}_{\Delta^*}}$ to a vector bundle on ${\Delta}$ such that the connection has regular singular points.
3. If we pick ${s_0\in \Delta^*}$, and use ${V=\mathscr{V}_{s_0}}$ and ${N}$ the monodromy, then ${M(N,V)}$ exists.

Define ${VMHS(\Delta^*)^{ad}_{\Delta}}$ the category of variations which are admissible after pullback to make the monodromy unipotent.

Def (Kashiwara “A study of variations of mixed Hodge structure”) suppose that ${\bar{S}}$ a complex manifold, ${Z\subset \bar{S}}$ a closed algebraic set with complement ${S}$. Then ${\mathscr{V}}$ a VMHS on ${S}$ is admissible relative to ${\bar{S}}$ is whenever we map ${\Delta^*\rightarrow S}$ such that we can complete to ${f:\Delta\rightarrow \bar{S}}$, we have ${f^*\mathscr{V}}$ admissible relative to ${\Delta}$.

Then we set ${VMHS(S)^{ad}_{\bar{S}}}$ the abelian category of admissible variations.

If ${S}$ is quasi-projective, then ${VMHS(S)^{ad}_{\bar{S}}=VMHS(S)^{ad}_{\bar{S}'}}$ for any projective completion of ${S}$.

Define (Saito) ${NF(S,\mathscr{H})^{ad}_{\bar{S}}=\mathrm{Ext}^1_{VMHS(S)_{\bar{S}}^{ad}}(\mathbb{Z},\mathscr{H})}$ for ${\mathscr{H}}$ a variation of pure Hodge structure on ${S}$.

Note: polarizable VHS are always admissible.

Recall the big claim that if ${\nu}$ is an admissible normal funciton, then ${Z(\nu)}$ is algebraic.

We observe that ${NF(S,\mathbb{Z}(1))=\mathrm{Ext}^1_{VMHS}(\mathbb{Z},\mathbb{Z}(1))=\mathscr{O}^*_{S^{an}}}$, so we need an extension where the extension class is ${e^z}$ to see why we really need admissible.

As an example, take ${S=\mathbb{P}^1\setminus\{0\}=\mathbb{A}^1}$. We define an extension ${0\rightarrow \mathbb{Z}(1)\rightarrow E\rightarrow \mathbb{Z}(0)\rightarrow 0}$ which violates the algebraicity of the zero locus.

Take ${E_\mathbb{Z}=\mathbb{Z}(2\pi i)e_{-2}+\mathbb{Z} e_0}$. Checking carefully, the extension class is ${e^z}$, because the Hodge filtration doesn’t extend to a subbundle!

Check that the problem is with extension of ${F^*}$, the Hodge filtration.

Let ${\mathscr{V}_{\mathscr{O}}=\mathscr{O}_Sa\oplus\mathscr{O}_Sb\oplus\mathscr{O}_S c}$ and ${\mathscr{V}_\mathbb{Z}}$ generated by ${a+\frac{\log s}{2\pi i} c}$, ${b}$ and ${c}$. Then take ${F^0=\mathscr{O}a+\mathscr{O}(b+ic)}$ and ${W_{-1}=\mathscr{O}b+\mathscr{O}c}$, ${W_0=}$everything and ${W_{-2}=0}$.

Show that relaive weight filtration chosen exists here.

Point: ${NV\cap W_{-1}\subset NW_{-1}}$ and the problem in this case is the weight filtration.