ICTP Day 11

Today was the penultimate day of the summer school. Once it finishes, there will be a conference, and things go through Friday, so all-in-all, there will be 15 days in this series. And here’s today’s notes:

1. Griffiths 5

No notes taken

2. Charles 2

Let ${X/k}$ be a smooth projective (maybe quasi-projective) variety and ${k}$ of characteristic 0 and embeddable into ${\mathbb{C}}$ .

Fix ${\sigma:k\rightarrow \mathbb{C}}$ and ${\alpha\in H^{2i}_{dR}(X/k)}$ , say that ${\alpha}$ is Hodge with respect to ${\sigma}$ if with the isomorphism ${H^{2i}_{dR}(X/k)\otimes \mathbb{C}\cong H_B(X_\mathbb{C}(\mathbb{C}),\mathbb{Q}(i))\otimes\mathbb{C}}$ then ${\alpha}$ lies in the rational subspace and ${\alpha\in F^iH^{2i}_{dR}(X/k)}$ .

We say that ${\alpha}$ is absolute Hodge if ${\alpha}$ is Hodge with respect to every ${\sigma}$ .

Now, if ${X/\mathbb{C}}$ and ${\alpha\in H_B^{2i}(X(\mathbb{C}),\mathbb{Q}(i))}$ is absolute Hodge if is is a hodge class and its image in ${H^{2i}_{dR}(X/\mathbb{C})}$ is absolutely Hodge, so Hodge with respect to all ${\sigma\in \mathrm{Aut}(\mathbb{C})}$ .

Let ${\pi:\mathcal{X}\rightarrow S}$ be smooth, projective over ${\mathbb{C}}$ with ${S}$ quasiprojective, smooth and connected. Let ${\alpha}$ be a global section of ${R^{2i}\pi_*\mathbb{Q}(i)}$ such that ${\alpha}$ is of type ${(i,i)}$ at any point of ${S}$ and for some ${s_0\in S(\mathbb{C})}$ , ${\alpha_{s_0}}$ is absolutely Hodge. Then ${\forall s\in S(\mathbb{C})}$ , we have that ${S(\mathbb{C})}$ is absolutely Hodge.

Remark: If we know ${\alpha_{s_0}}$ is algebraic, then we can prove ${\alpha_s}$ is.

We prove this before.

Let ${\pi:\mathcal{X}\rightarrow S}$ as above, ${\rho:\pi_1(S(\mathbb{C}),s)\rightarrow \mathrm{Aut}(H^{2i}(X_s,\mathbb{Q}(i))}$ the monodromy representation. Then the fixed points of ${\rho}$ are those in the image of ${H^{2i}(\mathcal{Y},\mathbb{Q}(i))\rightarrow H^{2i}(X_s,\mathbb{Q}(i))}$ where ${\mathcal{Y}\supset\mathcal{X}}$ is a smooth compactification.

In our proof, ${\alpha}$ is a global section of ${R^{2i}\pi_*\mathbb{Q}(i)}$ , and so it is ${\pi_1}$ -invariant and thus comes from ${\mathcal{Y}}$ , so is algebraic. Then ${\forall s\in S(\mathbb{C})}$ , we have that ${\alpha_s}$ is absolute Hodge.

Remark: if ${X}$ is defined over ${\mathbb{Q}}$ , then Hodge classes on ${X}$ are always absolute Hodge.

Question: How can we reduce questions in Hodge theory (Hodge cycles are absolute, Hodge conjecture) to questions over ${\mathbb{Q}}$ and number fields?

Let ${X/\mathbb{C}}$ be smooth projective. We can always find ${\mathcal{X}\stackrel{\pi}{\rightarrow} S/\mathbb{Q}}$ with ${\pi}$ smooth and projective, ${S}$ smooth and quasiprojective and ${s\in S(\mathbb{C})}$ such that ${X\cong X_s}$ ( ${X\subset\mathbb{P}^n_\mathbb{C}}$ corresponds to a point of some Hilbert-scheme of subschemes of ${\mathbb{P}^n_\mathbb{C}}$ )

What does it mean for Hodge classes fibers to be absolute?

Remark: ${\mathcal{X}\rightarrow S/\mathbb{Q}}$ , so the Hodge bundle ${\mathscr{H}^i_{dR}(\mathcal{X}/S)}$ are defined over ${\mathbb{Q}}$ , and so is ${\nabla}$ .

If ${\sigma\in \mathrm{Aut}(\mathbb{C})}$ , and ${\sigma}$ permutes the complex points of the Hodge bundle, then

${\alpha\in F^i H^{2i}_{dR}(X_s/\mathbb{C})}$ is an absolute Hodge class if and only if ${\alpha}$ is rational and ${\sigma(\alpha)}$ is an absolute Hodge class for all ${\sigma\in \mathrm{Aut}(\mathbb{C})}$ .

The locus of Hodge classes for ${\pi}$ is the set of ${\alpha_t\in H^{2i}_{dR}(X_t/\mathbb{C})}$ such that ${\alpha_t}$ is a hodge class.

Hodge classes for the fibers of ${\pi}$ are absolute iff the locus of Hodge classes is invariant under ${\mathrm{Aut}(\mathbb{C})}$ .

The locus of Hodge clases is actually a countable union of analytic subvarieties of ${\mathscr{H}^{2i}(\mathcal{X}/S)}$ .

Start with ${\alpha_t\in H^{2i}_{dR}(X_s/\mathbb{C})}$ . We get a component of the locus of Hodge classes by taking parallel transport of ${\alpha_s}$ over some small open subset and looking at the vanishing on ${\mathscr{H}^{2i}/F^i}$ .

Lemma 1 Let ${B}$ be defined over ${\mathbb{Q}}$ and ${Z\subset B(\mathbb{C})}$ be a countable union of analytic subvarieties such that ${Z(\mathbb{C})}$ is stable under the action of ${\mathrm{Aut}(\mathbb{C})}$ . Then ${Z}$ is a countable union of algebraic varieties over ${\mathbb{Q}}$ .

The idea is to take a very general point of ${Z}$ and look at the orbit of it under ${\mathrm{Aut}(\mathbb{C})}$ .

We know the geometric part of the conclusion:

Let ${\pi:\mathcal{X}\rightarrow S}$ as before. Then the locus of Hodge classes in ${\mathscr{H}^{2i}(\mathcal{X}/S)}$ is a countable union of algebraic subvarieties.

We don’t get information on the field of definition.

Let ${\mathcal{X}\rightarrow S/\mathbb{Q}}$ as before. The Hodge locus of ${\pi}$ is the image in ${S}$ of the component of the locus of Hodge classes passing through ${\alpha}$ . By DCK, this locus is algebraic. Let ${Z_\alpha}$ be in this locus. Assume it is defined over ${\bar{\mathbb{Q}}}$ . Then the Hodge conjecture for ${(X,Z_\alpha)}$ can be reduced to the HC for some ${\mathcal{X}/\bar{\mathbb{Q}}}$ .

Proof follows from DCK and the global invariant cycle theorem.

3. Green 3

Remember, we have ${k/\mathbb{Q}}$ finitely generated, and this gives us ${S}$ with ${\mathbb{Q}(S)\cong k}$ . We also start with ${X}$ defined over ${k}$ and get ${\mathcal{X}\rightarrow S}$ .

We set ${Z^p(X(k))}$ the cycles on ${X}$ defined over ${k}$ , take it mod rational equivalence on ${X}$ defined over ${k}$ , and we get ${CH^p(X(k))_\mathbb{Q}=CH^p(X(k))\otimes_\mathbb{Z} \mathbb{Q}}$ .

There is a conjectural filtration ${CH^p(X(k))_\mathbb{Q}=F^0\supset F^1\supset\ldots\supset F^{p+1}=0}$ with ${F^1}$ the homologically (but not rationally) trivial cycles.

For any ${\Gamma\in Z^n(X\times Y(k))}$ we get a map ${CH^p(X(k))\rightarrow CH^{p+n-\dim X}(Y(k))}$ which should take ${F^{n_1}}$ to ${F^{n_1+n_2}}$ .

${F^1CH^p(X(k))_\mathbb{Q}}$ . Let ${Z\in Z^p(X(k))}$ . Then ${[Z]\in \mathbb{H}^{2p}(\Omega^*_{X(k)/k})}$ , and then ${Z\in F^1}$ if and only if ${[Z]=0}$ .

Now look at ${F^2}$ . We expect that ${F^2}$ is the kernel of ${F^1\rightarrow J^p(X_s)\otimes\mathbb{Q}}$ , the Abel-Jacobi map for ${X_s}$ . So we need that this kernel doesn’t depend on which ${s}$ is chosen for very general ${s}$ .

Beilinson’s Conjectural Formula: ${\mathrm{Gr}^m CH^p(X(k))_\mathbb{Q}\cong \mathrm{Ext}^m_{MM_k}(\mathbb{Q},H^{2p-m}(X)(p))}$ where ${MM_k}$ is the category of mixed motives over ${k}$ . So, in particular, we have no idea how to compute this ${\mathrm{Ext}}$ group in general.

Now, we’re going to talk about ${\mathrm{Ext}^1}$ and the Abel-Jacobi map.

Let ${X}$ be a smooth projective variety defined over ${k}$ . Take ${Z_i}$ smooth cycles defined over ${k}$ , ${f_i:Z_i\rightarrow X}$ the inclusion, defined over ${k}$ . Set ${Z=\sum n_i {f_i}_*Z_i}$ . Then we have ${\Omega^*_{X(k)/k}\oplus \bigoplus_i \Omega^{*-1}_{Z_i(k)/k}}$ and we have differential ${\alpha\oplus \bigoplus \beta_i\mapsto d\alpha\oplus\bigoplus (d\beta_i-f_i^*\alpha)}$ .

Then ${d^2=0}$ , so we have a complex.

We can get an exact sequence ${0\rightarrow k(-(m-p))\rightarrow V\rightarrow H^{2m-2p+1}(\Omega^*_{X(k)/k})\rightarrow 0}$ , and we can construct splittings that respect the Hodge filtration, and splittings that respect the integral lattice, but not one that does both! So, working out the set of extension classes, we get ${J^p(X_s)}$ , so it is ${\mathrm{Ext}^1_{MHS}(\mathbb{Z},H^{2p-1}(X_s)(p))}$ .

4. Kerr 4

No notes taken

5. Schnell – Deligne’s Theorem on Abelian Varieties, Part I

On an abelian variety, all Hodge classes are absolutely Hodge.

The proof breaks up into two parts:

reduce to the case of CM abelian varieties
Deal with CM case.

We’ll deal with step 1 today.

Recall, in the case of weight 1:

A CM field is a number field ${E}$ of the form ${E=F[t]/(t^2-f)}$ where ${f\in F}$ and ${F}$ is totally real and under all embeddings ${F\subset \mathbb{R}}$ , ${f}$ is negative.

An abelian variety is CM if there exists a CM field ${E\subset \mathrm{End}(A)\otimes\mathbb{Q}}$ such that ${\dim_E H^1(A,\mathbb{Q})=1}$

This implies that ${[E:\mathbb{Q}]=\dim_\mathbb{C} H^1(A,\mathbb{Q})=2\dim A}$ .

There is a nice criterion ${MT(A)=MT(H^1(A,\mathbb{Q}))}$ (MT means Mumford-Tate group)

If ${A}$ is simple, then ${A}$ is CM if and only if ${MT(A)}$ is abelian.

Given any abelian variety ${A}$ and a Hodge class ${\alpha}$ on ${A}$ , there exists a family ${\mathcal{A}\stackrel{\pi}{\rightarrow}B}$ of abelian varieties with ${B}$ irreducible and quasi-projective such that there exists ${0\in B}$ with ${\mathcal{A}_0\cong A}$ and the Hodge locus of ${\alpha}$ is ${B}$ , and there is ${t\in B}$ where ${\mathcal{A}_t}$ is CM.

Proof: Choose a polarization ${Q}$ and let ${G=\mathrm{Aut}(H^1(A,\mathbb{Q}),Q)}$ and ${M=MT(A)}$ the smallest ${\mathbb{Q}}$ -subgroup whose ${\mathbb{R}}$ -points contain the image of ${\phi:\mathbb{S}^1\rightarrow G(\mathbb{R})}$ .

Abelian varieties of the same kind, along with a choice of basis for ${H^1(A,\mathbb{Z})}$ , are parameterized by the period domain ${D=G(\mathbb{R})/K}$ .

Note: points of ${D}$ are classes of ${gH}$ in terms of ${\phi}$ . ${\phi_{gH}=g\phi g^{-1}}$ .

Main idea: family comes from the Mumford-Tate domain: ${D_\phi=M(\mathbb{R})/M(\mathbb{R})\cap K\subset D}$ .

This should have the properties that for all Hodge structures ${H'\in D_\phi}$ ,

${MT(H')\subset M}$
any Hodge tensor for ${A}$ is a Hodge tensor for ${H'}$
${\phi_{H'}=g\phi g^{-1}}$ for ${g\in M(\mathbb{R})}$ .

Finding CM points corresponds to finding points with abelian MT. ${\phi(\mathbb{S}^1)\subset M(\mathbb{R})}$ contained in some maximal ${\mathbb{R}}$ -toruc ${T_0}$ , and we can show that for ${\xi_0\in m_\mathbb{R}}$ generic, ${T_0}$ is the stabilizer of ${\xi_0}$ .

Nearby, there exists ${\xi\in m_{\mathbb{R}}}$ close to ${\xi_0}$ , then if ${T}$ is the stabilizedr of ${\xi}$ , it is a ${\mathbb{Q}}$ -torus. There exists ${g\in M(\mathbb{R})}$ such that ${\xi=g\xi_0 g^{-1}}$ , and ${g\phi g^{-1}}$ has image in ${T}$ . Then ${MT(H_{g\phi g^{-1}})\subseteq T}$ is abelian.

Problem: family over quasi-proj base, not ${D_\phi}$ . Solution: Fix an ${N>>0}$ and use a level ${N}$ structure.

Define ${\mathcal{M}_{g,Q,N}}$ to be the moduli space of abelian varieties of dimension ${g}$ with polarization ${Q}$ and level ${N}$ structure (a basis of the ${N}$ -torsion points) and let ${\mathcal{A}_{g,Q,N}\rightarrow \mathcal{M}_{g,Q,N}}$ the universal family. OUr replacement for ${D_\phi}$ is to let ${B\subset \mathcal{M}_{g,Q,N}}$ be the Hodge locus of the Hodge tensors for ${H^1(A,\mathbb{C})}$ defining ${MT(A)}$ .

${B}$ is algebraic by CDK, and finite etale over ${\Gamma\backslash D_\phi}$ . In this case, things are ok. $\Box$

Proof that (for ${A}$ simple), ${MT(A)}$ abelian implies ${A}$ is CM.

We start with the fac tthat ${E=\mathrm{End}(A)\otimes \mathbb{Q}}$ is a division algebra, since ${A}$ is simple. It is also the sety of ${\mathbb{Q}}$ -endomorphisms that commute with ${MT(A)=M}$ . So we know that ${M}$ is abelian, and thus it acts on ${H^1(A,\mathbb{Q})}$ , and we can write ${H^1(A,\mathbb{C})=\oplus_\chi H^1(A,\mathbb{C})_\chi}$ for characters, and thus ${E\otimes\mathbb{C}=\oplus_i \mathrm{End} H^1(A,\mathbb{C})_\chi}$ .

And so ${\dim_\mathbb{Q} E\geq \dim H^1(A,\mathbb{Q})=2\dim A}$ is bounded above by ${2\dim A}$ . So ${2\dim A=\dim\mathbb{Q} E}$ and thus ${E}$ is a commutative field, so ${\dim_E H^1(A,\mathbb{Q})=1}$ .

Now, use teh Rosati involution ${\phi\mapsto \phi^t}$ on ${E}$ , and ${Q(\phi h_1,h_2)=Q(h_1,\phi^t h_2)}$ , and ${F}$ the fixed field. We claim that ${[E:F]=2}$ and ${F}$ is totally real.

We have that ${F=\mathbb{Q}(\phi)}$ , with ${\phi=\phi^t}$ and take the minimal polynomial. Then ${\lambda_j}$ , the roots, are the eigenvalues of the action of ${\phi}$ on ${H^1(A,\mathbb{Q})}$ , and if we set ${\lambda=\lambda_j}$ , and ${\phi}$ acts on ${H^1(A,\mathbb{C})}$ preserving ${H^{1,0}\oplus H^{0,1}}$ , there exists ${h\in H^{1,0}}$ with ${\phi(h)=\lambda h}$ , ${\phi(\bar{h})=\bar{\lambda}\bar{h}}$ . Look at ${Q(\phi h, \bar{h})=Q(h,\phi \bar{h})}$ , this is ${\lambda Q(h,\bar{h})=\bar{\lambda}Q(h,\bar{h})}$ and so ${\lambda=\bar{\lambda}}$ , so ${\lambda\in\mathbb{R}}$ .

6. Griffiths – Colloquium – Hodge Theory and Representation Theory

Joint work with Mark Green and Matt Kerr

Lecture went faster than I could type. No notes.

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	Proof of Hurwitz… on Hurwitz’s Theorem on…
	Monodromy Representa… on Monodromy Representations
	Problem underestandi… on Quadratic Differentials
	Set of branch points… on Hurwitz’s Theorem
	Simon on Monodromy Representations

ICTP Day 11

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