## ICTP Day 10

And these posts are getting shorter, as the talks that I’m even attempting to take notes in are finishing. The other talks are good, they’re just moving very, very fast and given a choice between learning and notetaking, I’m choosing learning. Perhaps once this is done, I’ll make an attempt at posting blogified versions of some of these notes, making them a bit more interconnected, etc. We’ll see, if I do, it won’t be until the Fall, because right now, my big priorities are finishing writing the talk I’m giving on Thursday (my notes will be posted separately from the others) and also working on the lectures for the Math History for Liberal Arts course that I’m teaching (strictly, Ideas in Mathematics, but I’m doing something different with it). Anyway, here are today’s notes:

1. Green 2

Let ${X}$ be a smooth variety defined over ${k}$ where ${k}$ is finitely generated over ${\mathbb{Q}}$. Then we defined last time ${\Omega^1_{X(k)/k}}$ and ${\Omega^1_{X(k)/\mathbb{Q}}}$, and defined the complex ${\Omega^*_{X(k)/k}}$

Now, we need to define hypercohomology. Start with a complex ${A^*}$, then ${H^r(A^*)=\ker d/\mathrm{im}d}$ on ${A^r}$.

Now, look at ${A^{*,*}}$ a double complex, and take the total complex. This is a complex if ${d,\delta}$ the horizontal and vertical operators anticommute, and then we use ${D=d+\delta}$, and hypercohomology is defined by ${\mathbb{H}^r(A^{*,*})=H^r(T^*,D)}$.

Note: Hypercohomology is not on the list in “My Favorite Things”

Using Cech cohomology, we define ${A^{p,q}=C^q(\mathfrak{U},\Omega^p)}$, and add a sign to the Cech differential, then we can take Hypercohomology.

Now, there’s a theorem that tells us that for ${i_s:k\subset\mathbb{C}}$, ${\dim_k \mathbb{H}^n(\Omega^*_{X(k)/k})=\dim_\mathbb{C} H^n_{dR}(X_s,\mathbb{C})}$.

\mbox{}

1. ${\mathbb{H}^n(\Omega^*_{X(k)/k}\otimes_{i_s}\mathbb{C}\cong H^n(X_s,\mathbb{C})}$
2. ${H^q(\Omega^p_{X(k)/k})\otimes_{i_s}\mathbb{C}\cong H^{p,q}(X_s)}$
3. ${\mathbb{H}^n(\Omega^{\geq p}_{X(k)/k})\otimes_{i_s}\mathbb{C}\cong F^pH^n(X_s,\mathbb{C})}$

We can’t read off the integral lattice, though.

${\mathrm{Gr}^m\Omega^*_{X(k)/\mathbb{Q}}}$ is defined to by ${F^m\Omega^*/F^{m+1}\Omega^*}$ and that’s ${\Omega^m_{k/\mathbb{Q}}\otimes \Omega^{*-m}_{X(k)/k}}$.

He compares the derivation of the long exact sequence on cohomology to X-rays…not good for you to see too many times.

Looking at spreads: take ${\mathcal{X}\rightarrow S}$, this gives a variation of Hodge structure over ${S\setminus \Sigma}$, the smooth locus. Looking at the Gauss-Manin connection ${\nabla}$, we have a complex ${(\Omega^*_{k/\mathbb{Q}}\otimes \mathbb{H}^r(\Omega^*_{X(k)/k},\nabla)}$, and we can get Griffiths Transversality (also called the infinitesimal period relations)

Now, say that ${k=\mathbb{Q}(x_1,\ldots,x_T)[y_1,\ldots,y_A]/(p_1,\ldots,p_B)}$, then ${\Omega^1_{k/\mathbb{Q}}}$ is generated by ${dx_1,\ldots,dx_A}$.

Recall the example ${y^2=x(x-1)(x-\alpha)}$. If we differentiate, but don’t assume that ${d\alpha=0}$, we get ${2ydy=f'(x)dx-x(x-1)d\alpha}$. And now, ${\frac{2dy}{f'(x)}=\frac{dx}{y}-\frac{x(x-1)}{f'(x)y}d\alpha}$. And this last term gives us problems lifting. Fortunately, the coboundary map is essentially a measure of how much lifting fails, so the last term can be thought of as an element of ${\Omega^1_{k/\mathbb{Q}}\otimes H^1(\mathscr{O}_{X(k)})}$, so the whole thing is ${\nabla\omega}$ for some 1-form.

Now we introduct ${Z^p(X(k))}$, the codimension ${p}$ cycles defined over ${k}$ and ${CH^p(X(k))}$ the cycles mod rational equivalence defined over ${k}$.

So now, we define ${K_p(\mathscr{O}_{X(k)})}$ to be the Quillen ${p}$th K-group for ${\mathscr{O}_{X(k)}}$. Now, in the ${k}$ Zariski topology, we have ${CH^p(X(k))\cong H^p(K_p(\mathscr{O}_X))}$. Modulo torsion, it’s a theorem of Soulé that we can replace Quillen K-theory with Milnor K-theory.

${K_p^{Milnor}}$ is generated multiplicatively by ${f\in \mathscr{O}^*_{X(k)}}$ elements ${f_1\otimes\ldots\otimes f_p}$ modulo the Sternberg relations that if ${f_i=1-f_j}$ for some ${i\neq j}$, then ${f_1\otimes\ldots\otimes f_p=0}$.

(The lecture here became very difficult to typeset notes for, due to a sequence of complicated commutative diagrams. There are printed notes, and I will link to them as soon as they’re online)

The key point is that we have an arithmetic cycle class ${CH^p(X(k))\rightarrow H^p(\Omega^p_{X(k)/\mathbb{Q}})}$, and there is a criterion relating Hodge classes, and relating all of this to the Absolute Hodge conjecture.

2. Griffiths 4

No notes taken

3. Schnell 1 – Algebraic de Rham Cohomology and Betti Cohomology

We’re going to be talking about the arithmetic aspects of things. These are the “absolute Hodge classes” and fields of definition.

The basic insight is Grothendieck’s comparison theorem. Let ${X}$ be a smooth quasiprojective variety over ${k\supset \mathbb{Q}}$, and we have all of the various Kähler differentials.

We define ${H^i_{dR}(X/k)=\mathbb{H}^i(\Omega^*_{X/k})}$, which is a ${k}$-vector space.

Note that if ${L\supset k}$, then ${X_L=X\times_k L}$ then we have ${H^i_{dR}(X_L/L)=H^i_{dr}(X/k)\otimes_k L}$.

If ${X}$ is defined over ${\mathbb{C}}$ then ${H^i_{dR}(X/\mathbb{C})\cong H^i(X^{an},\mathbb{C})}$.

Under this isomorphism ${\mathbb{H}^i(\Omega^{\geq p})\cong F^pH^i(X^{an},\mathbb{C})}$.

Remark: This is true for any quasiprojective variety (only poles at infinity) and we have two structures on ${H^i(X^{an},\mathbb{C})}$. One is a ${\mathbb{Q}}$-structure, ${H^i(X^{an},\mathbb{Q})\otimes_\mathbb{Q} \mathbb{C}}$ the “Betti” structure, and the other is ${H^i_{dR}(X/k)\otimes_k \mathbb{C}}$, the deRham structure.

3.1. Families

Let ${f:X\rightarrow B}$ be a smooth projective variety over ${\mathbb{C}}$. By Katz-Oda, the Gauss-Manin connection is algebraically defined, and so we have our variations of Hodge structure are algebraic, and we have the SES

$\displaystyle 0\rightarrow f^*\Omega^1_{B/k}\otimes \Omega^{*-1}_{X/B}\rightarrow \Omega^*_{X/B}/L^2\Omega^*_{X/B}\rightarrow \Omega^*_{X/B}\rightarrow 0$

the connecting map gives the Gauss-Manin connection.

3.2. Cycle Classes and fields of definition

If ${X}$ is a smooth projective variety over ${\mathbb{C}}$ and ${Z}$ a subvariety of codimension ${p}$, then ${[Z^{an}]_k\in H^{2p}(X^{an},\mathbb{Q})}$ is always a Hodge class.

An important point is that we can also define an algebraic fundamental class ${[Z]_{dR}\in F^pH^{2p}_{dR}(X/k)}$.

Let ${X,Z}$ be defined over ${\mathbb{C}}$. Then ${(2\pi i)^p[Z^{an}]_B=[Z]_{dR}}$

In general, use the chen classes of vector bundles with sections vanishing along ${Z}$.

(Detailed definition of chern classes)

Now, we can extend Chern classes to coherent sheaves by resoltuing using locally free sheaves (because ${X}$ is smooth), and ${c_p(\mathscr{O}_Z)=(-1)^{p-1}(p-1)![Z]}$ for ${Z}$ codimension ${p}$. And so, finally, we set ${[Z]_{dR}=\frac{(-1)^{p-1}}{(p-1)!}c_p(\mathscr{O}_Z)}$.

4. Kerr 3

No notes taken

5. Charles 1 – Hodge Loci and Absolute Hodge Classes

Let ${X}$ be a compact Kähler manifold. We have Hodge classes ${H^{2p}(X,\mathbb{Z})\cap H^{p,p}(X)}$.

If ${X}$ is projective, Hodge classes are algebraic.

This is false for Kähler.

Now, let ${k}$ be a field, embeddable in ${\mathbb{C}}$. For ${X}$ smooth and projective over ${k}$, we have ${H^i_{dR}(X/k)}$. Now, take ${\sigma:k\rightarrow \mathbb{C}}$ an embedding.

We get Betti-deRham isomorphisms which are compatible with the cycle class map.

Let ${\alpha\in H^{2i}_{dR}(X/k)}$. We say that ${\alpha}$ is a Hodge class relative to ${\sigma}$ is ${\alpha\in F^iH^{2i}_{dR}(X/k)}$ and the image of ${\alpha}$ in ${H^{2i}(X_\sigma(\mathbb{C}),\mathbb{Q}(i))\otimes\mathbb{C}}$ lies in the reational subspace.

We say that ${\alpha}$ is absolute Hodge if it is a Hodge class relative to any ${\sigma}$.

Remarks: First, ${X/\mathbb{C}}$, we can define what it menas for a class to be an absolute Hodge class. Now, how dependent is this on ${k}$? The cohomology classes of algebraic cycles are absolute Hodge.

Proof: If ${Z}$ is an algebraic cycle in ${X}$, then for any ${\sigma:k\rightarrow \mathbb{C}}$, we have ${Z\times_\sigma\mathbb{C}}$ algebraic sycle in ${X\times_\sigma \mathbb{C}}$ So this gives a Hodge class relative to ${\sigma}$, and works for all ${\sigma}$.

Thus, we have two conjectures:

Hodge classes are absolute.

Absolute Hodge classes are algebraic.

These two conjectures imply the Hodge conjecture.

Let ${X/\mathbb{C}}$ be smooth and projective of dimension ${d}$. then we have ${H^d(X\times X)=\oplus_i H^i(X)\otimes H^{d-i}(X)}$ contains a class ${[\Delta]=\sum \pi_i}$ via the Künneth formula.

It’s a conjecture that the ${\pi_i}$ are algebraic.

The ${\pi_i}$ are absolute Hodge cycles.

Conjecture: The inverse of the Lefschetz map is algebraic.

This is absolute.

Hodge classes on abelian varieties are absolute.

Let ${X/\mathbb{C}}$ be smooth and projective, ${Z}$ an algebraic cycle of codimension ${i}$ in ${X}$, homologically equivalent to zero. We have ${AJ(Z)\in J^i(X)}$, which gives us an extension of mixed Hodge structures ${0\rightarrow H^{2n-1}(X,\mathbb{Z}(n))\rightarrow H\rightarrow \mathbb{Z}\rightarrow 0}$.

Now, ${AJ(Z)=0}$ if and onyl if the sequence splits which is if and only if there exists a Hodge class ${H}$ mapped to 1 in ${\mathbb{Z}}$.

This is related to the Bloch-Beilinson filtration on Chow Groups, ${F^2CH^i(X)=\ker AJ}$.

Now, we have out GM connection and a VHS structure.

Remark: If ${X/k}$ and ${k=\bar{k}}$, and ${K\supset k}$, then ${CH^i(X)K)\neq CH^i(X_k)}$. But, this does not for cohomological equivalence!

Indeed, take ${Z\subset X_K}$ be an algebraic cycle. Then the cohomology class ${[Z]_{dR}\in F^iH^{2i}(X/K)}$ actually comes from ${F^iH^{2i}(X/k)}$.

For ${\alpha\in F^iH^{2i}(X_K/K)}$ an absolute Hodge cycle, then ${\alpha}$ is defined over ${k}$.

Proof: We can assume that ${K}$ is finitely generated over ${k}$, and ${K}$ is the function field of some quasiprojective ${S/k}$. We can assume also that ${\alpha}$ extends to ${\bar{\alpha}}$, a section of ${F^1H^{2i}_{dR}(X/k)\otimes \mathscr{O}_S}$. We want to prove that ${\alpha}$ is constant.

Now, pull everything back to ${\mathbb{C}}$ by some ${k\subset\mathbb{C}}$. We want to check that ${\alpha_\mathbb{C}}$ is constant.

But, ${\alpha}$ being absolute, for every ${k\rightarrow \mathbb{C}}$, ${\alpha_\sigma}$ is going to lie the rational lattice of ${H^i_{dR}(X/\mathbb{C})\otimes \mathscr{O}_S}$, and so will be constant. $\Box$

${\pi:\mathcal{X}\rightarrow S}$ a smooth projective morphism with ${S}$ quasi-projective and connected. Let ${\alpha}$ be a global section of ${R^{2i}\pi_*\mathbb{Q}(i)}$, which is Hodge everywhere. Then if ${\alpha_s}$ is absolute Hodge for some ${s\in S(\mathbb{C})}$, it is for all ${s\in S}$.

Just conjugate ${\alpha}$. ${\nabla=\nabla^\sigma}$ for ${\sigma\in \mathrm{Aut}\mathbb{C}}$ Then ${\nabla \alpha^\sigma=\nabla^\sigma \alpha^\sigma=(\nabla\alpha)^\sigma=0}$. $\Box$

Now, we need ${\alpha}$ algebraic as a section of ${\mathscr{H}_{\mathcal{X}/S}}$. For this, we take the global invariant cycle theorem which says that ${\alpha}$ is the restriction of a Hodge class in ${H^{2i}(Y,\mathbb{Q}(-1))}$ where ${\mathcal{Y}\supset \mathcal{X}}$ is a smooth compactification.

Application: Let ${H}$ be a HS of weight 2 with ${h^{2,0}=1}$. Then there exists ${H'}$ a Hodge structure of weight 1, polarized, with ${H\rightarrow \mathrm{End} H'}$

Starting wtih ${\pi:\mathcal{X}\rightarrow S}$ a family of polarized K3’s, get ${\alpha:\mathscr{A}\rightarrow S}$ a polarized abelian scheme, then we have ${R^2\pi_*\mathbb{Q}(1)\rightarrow End(R^1\alpha_*\mathbb{Q})}$ for all ${S}$. connection.