## ICTP Day 4

And today, Hodge theory is in full swing. There were some commutative diagrams today, and wordpress is famous for not supporting xy, so I’ll do what I can.

1. ElZein 2

Take a family of elliptic curves: ${\mathcal{C}_\lambda}$ given by ${y^2=x(x-1)(x-\lambda)}$ for ${\lambda\neq0,1}$ in ${\mathbb{C}^2}$, for each ${\lambda}$. We need to compactify:

Look at ${\bar{\mathcal{C}}_\lambda\subset \mathbb{P}^2}$ given by ${P_\lambda=Y^2Z-X(X-Z)(X-\lambda Z)=0}$. The fibers are then projective curves of genus 1. Each one is homeomorphic to ${\mathbb{R}^2/\mathbb{Z}^2}$.

For each ${\lambda}$, we could look at the fiber as ${R=\mathbb{C}[X,Y,Z]/P_\lambda}$, but we’re going to rely a lot on topology, so we’ll be better served by using the ${\mathbb{R}^2/\mathbb{Z}^2}$ model.

Thus, we have a map ${\bar{\mathcal{C}}\rightarrow \mathbb{C}\setminus\{0,1\}\ni\lambda_0}$ and by Ehresmann’s lemma, we have that for a small neighborhood of the point, the preimage is diffeomorphic to ${U\times \bar{C}_{\lambda_0}}$.

Calling the projection map ${f}$, this tells us that ${R^if_*\mathbb{Z}|_{f^{-1}(U)}\cong \mathbb{Z}_U}$. So for every ${\lambda}$ in ${U}$, we have ${\bar{C}_{\lambda_0}\stackrel{h}{\rightarrow} \bar{C}_\lambda}$ a homeomorphism, and if we take a different path and get ${h'}$, the two homeomorphisms are the same.

However, if we go around a hole, so that the two paths comprise a nontrivial loop, we can’t homotope one to the other, and so we get a nontrivial map when we go around the loop. This gives us maps ${H^i(\mathcal{C}_{\lambda_0})\rightarrow H^i(\mathcal{C}_\lambda)\rightarrow H^i(\mathcal{C}_{\lambda_0})}$, which is not necessarily the identity. We call this monodromy.

The monodromy transformation is not compatible with Hodge structure!

Lemma 1 (Deligne) Let ${I^{p,q}=(F^p\cap W_{p+q})\cap \left[(\bar{F}^q\cap W_{p+q})+(\bar{F}^{q-1}\cap W_{p+q-2})+\ldots\right]}$. Then under the projection ${W_m\rightarrow \mathrm{Gr}^W_m=W_m/W_{m-1}}$ we have ${I^{p,q}\rightarrow (\mathrm{Gr}^W_mH)^{p,q}}$ for ${p+q=m}$.

Now, ${W_m=\oplus_{p+q=m}I^{pq}}$ and ${F^p=\oplus_{i\geq p} I^{pq}}$. However, ${\bar{I}^{p,q}\neq I^{q,p}}$. However, it is mod ${W_{p+q-2}}$. If it’s exactly true, we call the mixed Hodge structure split.

A morphism of mixed Hodge structures is ${\mathscr{L}:(H,W,F)\rightarrow (H',W',F')}$ such that ${\mathscr{L}:H\rightarrow H'}$, ${\mathscr{L}(W_m)\subset W_m'}$ and ${\mathscr{L}(F_p)\subset F_p'}$.

Lemma 2 ${\mathscr{L}}$ is strict for ${W}$ and for ${F}$ and the kernel and cokernel are natural mixed Hodge structures.

The category of MHS’s is an abelian category.

Let ${\ldots\rightarrow H^{n-1}\rightarrow H^n\rightarrow H^{n+1}\rightarrow\ldots}$ then the cohomology of this sequence is a MHS.

Let ${X}$ be a compact Kähler manifold of dimension ${n}$ and ${\Omega_X^*}$ the holomorphic deRham complex. Then we can set ${F^p\Omega_X^*}$ to be ${0\rightarrow \Omega^p_X\rightarrow\ldots}$, and ${(\Omega_X^*,F)}$ is a filtered complex. We take a resolution of Filtered complexes ${\Omega_X^*\rightarrow \mathscr{E}_X^*=\mathcal{A}_X^*}$ the differential forms, a quasi-isomorphism. Then we can define ${F^p\mathscr{E}^*_X}$ by ${\ldots\rightarrow 0\rightarrow \mathscr{E}_X^{p,0}\rightarrow \mathscr{E}_X^{p+1,0}\rightarrow\ldots}$.

In degree ${k}$, ${(F^p\mathscr{E}_X^*)^k=\oplus_{p+q=k,p'\geq p}}$.

Now, ${F^p\Omega_X^*\rightarrow F^p\mathscr{E}_X^*}$ is a quasi-isomorphism, and ${\mathrm{Gr}_F^p\Omega_X^*=F^p\Omega^*_X/F^{p+q}\Omega_X^*\rightarrow F^p\mathscr{E}^*_X/F^{p+q}\mathscr{E}_X^*}$, and this gives a qis ${\Omega_X^p}$ to the Dolbeaut resolution.

${H^i(X,\mathbb{C})=H^i(\mathscr{E}^*_X)}$ and ${F^pH^i(X,\mathbb{C})}$ is ${\mathrm{im} H^i(F^p\mathscr{E}^*_X)\rightarrow H^i(\mathscr{E}^*_X)}$.

Define ${\mathrm{Gr}^p_F H^i(X,\mathbb{C})}$ ot be ${F^pH^i/F^{p+1}H^i}$. Then we have ${_FE_1^{pq}\stackrel{\gamma}{\rightarrow} _FE_1^{p+1,q}\rightarrow H^q(X,\Omega_X^p)\stackrel{d}{\rightarrow} H^q(X,\Omega_X^{p+1})}$.

The spectral sequence degenerates at rank 1. That is, ${d=0}$.

Thus, ${H^q(X,\Omega_X^p)\cong \mathrm{Gr}_F^p H^{p+q}(X,\mathbb{C})}$.

But even more, ${F^pH^m(X,\mathbb{C})\oplus \bar{F}^{m-p+1}H^m\cong H^m}$.

If ${X}$ is projective, then ${[\omega]\in H^2(X,\mathbb{Q})}$ is a hyperplane section ${[H]}$.

The fundamental class of a subvariety ${Z}$ of a compact complex manifold of codimension ${r}$ in ${X}$ where ${\dim X=n}$ is ${[Z]\in H_{2n-2r}(X,\mathbb{Z})}$ and then this sits inside ${H_{2n-2r}(X,\mathbb{C})}$, which is isomorphic to ${H^{2n-2r}(X)^*}$. This gives us a class ${\omega}$ and ${\int_{[Z]}\omega\in \mathbb{C}}$, and so ${\omega}$ is Poincaré dual to ${\eta_Z\in H^{2r}}$.

Lemma 3 ${[\eta_Z]\in H^{r,r}(X)\cap \mathrm{Im}(H^{2r}(X,\mathbb{Z})\rightarrow H^{2r}(X,\mathbb{C}))}$ and ${[\eta_Z]\neq 0}$

Define ${[\eta_Z]}$ for ${Z}$ a subvariety in ${X}$, including the possibility that it might be singular.

2. Migliorini 2

Today we’ll be following Chapter 4 of Volume II in C. Voisin’s book.

Let ${f:X\rightarrow Y}$ a smooth projective morphism. We want to prove that the Leray Spectral Sequence degenerates at ${E_2}$. So then morally, ${H^p(X)=\oplus_{a+b=p}H^q(Y,R^bf_*\mathbb{Q})}$.

The key fact we’ll need is the Hard Lefschetz theorem, and here it tells us that for ${y_0\in Y}$, we have ${H^k(f^{-1}(y_0))=\oplus L^r H_0^{k-2r}(f^{-1}(y_0))}$ where ${L}$ is the operator that takes the cup product with the Chern class of a relatively ample line bundle ${\mathscr{L}}$ on ${X}$, ${H_0}$ is the primitive cohomology, and ${H_0^{n-k}=\ker L^{k+1}}$.

Because ${\mathscr{L}}$ is globally define, this decomposition is monodromy invariant.

This decomposition is compatible with the differentials in the LSS. In fact, this decomposition, by Cattani’s first set of notes, holds already at the level of forms.

${L\in H^2(X)=\hom(\mathbb{Q},\mathbb{Q}[2])}$, cupping with ${L}$ gives a morphism ${Rf_*\mathbb{Q}\rightarrow Rf_*\mathbb{Q}[2]}$.

EG, prove ${d_2=0}$. Enough to show on the primitive part. There, we have $L^{k+1}\circ d_2=d_2\circ L^{k+1}$, but the $L^{k+1}$ on the right is zero, and on the left is $H.L$. The same argument gives that ${Rf_*\mathbb{Q}=\oplus R^kf_*\mathbb{Q}[-k]}$.

This splitting, however, is not canonical.

Now, degeneration at ${E_2}$ implies that ${H^p(X)\rightarrow H^p(Y,R^qf_*\mathbb{Q})=H^p(f^{-1}(y_0))^{\pi_1}\rightarrow 0}$

Now, if we have a projective map ${\bar{f}:\bar{X}\rightarrow \bar{Y}}$, ${Y\rightarrow \bar{Y}}$ the inclusion of an open sense subset such that ${f:X\rightarrow Y}$, the base change, is smooth, then there is a strong generalization of the above.

Let ${C}$ be a nodal cubic, given by ${Y^2Z=X^2(X-Z)}$. Then ${H^1(C)=\mathbb{Q}}$, and we can’t expect ${H^1(C,\mathbb{Z})=H^{1,0}\oplus \bar{H}^{1,0}}$.

This can be fixed by looking at MHS, as discussed by ElZein. Here we have a finite increasing filtration ${W}$ and a finite decreasing filtration ${F}$, where ${F}$ induces a pure Hodge structure on each ${W_k/W_{k-1}}$ over ${\mathbb{C}}$ of weight ${k}$.

A map of MHS is strict with respect to ${W}$ and ${F}$.

Let ${\phi:H\rightarrow H'}$ nad ${a\in \mathrm{Im}(\phi)\cap W_kH'}$ then ${a\in \mathrm{Im}W_kH}$.

The following theorem is actually true:

There is a functorial mixed Hodge structure on the cohomology groups of every open algebraic variety ${X}$.

In general, ${W_aH^\ell(X)=0}$ if ${a<0}$ and ${W_aH^\ell(X)=H^\ell(X)}$ if ${a\geq 2\ell}$.

So on the nodal cubic, the ${H^1}$ classes are of type ${(0,0)}$.

If ${X}$ is nonsingular, maybe no compact, then ${W_aH^\ell(X)=0}$ if ${a<\ell}$ and if ${X}$ is compact, possibly singular, then ${W_\ell H^\ell(X)=H^\ell(X)}$.

So this implies that the Hodge structure guaranteed by the theorem of Deligne above is pure if ${X}$ is smooth and compact.

Weight Trick: Let ${Y}$ compact inside ${X}$ smooth lie inside ${\bar{X}}$, smooth and compact. Then ${\mathrm{Im}(H(X)\rightarrow H(Y))=\mathrm{Im}(H(\bar{X})\rightarrow H(Y))}$.

${S^1\subset \mathbb{R}^2\setminus\{0\}\subset S^2}$, then ${H^1(\mathbb{R}^2\setminus\{0\})\rightarrow H^1(S^1)}$ is an isomorphism, but ${H^1(S^2)\rightarrow H^1(S^1)}$ is not, it’s the zero map. The problem here is that the spaces are not algebraic. The Weight trick is a very algebraic phenomenon.

In the algebraic setting, we have ${W_\ell H^\ell(X)=\mathrm{Im}(H^\ell(\bar{X})\rightarrow H^\ell(X))}$ for ${\bar{X}}$ is any smooth compactification of ${X}$.

Now, let’s prove the Weight trick:

We have ${H^\ell(\bar{X})\rightarrow H^\ell(X)\rightarrow H^\ell(Y)}$. Now, any class of weight ${\leq \ell}$ in ${H^\ell(Y)}$ comes from ${H^\ell(\bar{X})}$.

From ${E_2}$ degeneration and MHS theory, we have that ${H^\ell(f^{-1}(y_0))^{\pi_1}\subseteq H^\ell(f^{-1}(y_0))}$ is a subHodge structure, and the invariant classes are compatible with the ${(p,q)}$ decomposition.

The monodromy representation is completely reducible, when ${f:X\rightarrow Y}$ is a smooth projective morphism to a quasiprojective variety.

3. Cattani 2 – Variations of HS, Degenerations of HS

Let ${\phi:\mathcal{X}\rightarrow B}$ be a holomorphic proper submersion of complex manifolds. So by Ehresmann’s Theorem, we can assume that that around each ${b_0}$ we have a ${U}$ such that ${\phi^{-1}(U)\cong U\times X_{b_0}}$.

For each ${x\in X_{b_0}}$, we have ${U\rightarrow U\times X_{b_0}}$ by ${b\mapsto (b,x)}$, and then we can send it along the inverse of the diffeomorphism to ${\mathcal{X}}$, and this map is holomorphic.

Now, assume that ${U}$ is a polycylinder and that ${b_0=0}$.

The naive approach is that ${X=X_0}$ sits insider ${\{t\}\times X\stackrel{G}{\rightarrow} X_t}$, and this composition is the diffeomorphism ${g_t}$.

This gives ${J_t=g^*_t(J_{X_t})}$, with ${J_t^2=-I}$, and so we have a family ${\{J_t\}_{t\in\Delta}}$ which give decompositions ${[T_x(X)]_\mathbb{C}=T'_{x,t}\oplus T''_{x,t}}$ for all ${x\in X}$. For ${t=0}$, we get the usual decomposition for ${X}$

So in local coordinates, ${(U,z_1^U,\ldots,z_n^U)}$, we have ${T''_{x,t}}$ has a basis of the form ${\frac{\partial}{\partial \bar{z}_j}-\sum_{i=1}^n \omega_{ij}\frac{\partial}{\partial z_i}}$.

Thus, if ${v\in T_0^h(\Delta)}$, then we can write it out as ${\sum_{i=1}^n v(\omega_{ij}(z,t))\frac{\partial}{\partial z_i}}$.

But of course, this is not all well defined, it is not global. Given ${v\in T^h_0(\Delta)}$, we can consider the expression ${\sum_{i=1}^n \bar{\partial}v(\omega_{ij}(z,t))\otimes\frac{\partial}{\partial z_i}}$, and this will kill any holomorphic dependence. So what we get is a map ${T_0^h(\Delta)\rightarrow A^{0,1}(T^h(X))}$. But this is closed, it’s not just a form, so we actually have ${T_0^h(\Delta)\rightarrow H^{0,1}_{\bar{\partial}}(T^h(X))=H^1(X,\mathscr{O}(T^h(X)))}$. This is called the Kodaira-Spencer Map.

So for eahc ${p\in \mathcal{X}}$, we have ${0\rightarrow T_0(\phi^{-1}(\phi(p))\rightarrow T_p\mathcal{X}\rightarrow T_{\phi(p)}(B)\rightarrow 0}$, and these fit together into ${0\rightarrow T^hX\rightarrow T^h(\mathcal{X})|_X\rightarrow X\times T_0^h(B)\rightarrow 0}$

This gives us a long exact sequence, and in particular, a map ${H^0(X,\mathscr{O}(X\times T_0^h(B)))\rightarrow H^1(X,\mathscr{O}(T^hX))}$. However, the first is simply ${T_0^h(B)}$, and so this gives us again the Kodaira-Spencer map.

Now, assume that the ${X_t}$ are 1-dimensional. These are all diffeomorphic, so we can say that they are Riemann Surfaces of genus ${g}$. So then ${H^1(X_t,\mathbb{C})=H^{1,0}(X_t)\oplus H^{0,1}(X_t)}$. Now, ${H^1(X_t,\mathbb{C})}$ is constant with respect to ${t}$, but the decomposition may not be. We still have all the maps ${g_t:X\rightarrow X_t}$, and they give isomorphisms ${g_t^*:H^1(X_t,\mathbb{C})\rightarrow H^1(X,\mathbb{C})}$.

So we can view this as having ${H^1(X,\mathbb{C})=H^{1,0}_t\oplus H^{0,1}_t}$, a fixed vector space with a varying decomposition, satisfying ${H^{0,1}_t=\bar{H}^{1,0}_t}$. But we also have the polarization form ${Q=\int_X \cdot\wedge\cdot}$ and ${Q(H^{1,0}_t,H^{1,0}_t)=0}$ and ${iQ(H^{1,0}_t,\bar{H}^{1,0}_t)>0}$.

So now we have ${\Delta=\{W\in G(g,H^{-1}(X,\mathbb{C}))}$ satisfying these conditions${\}}$. So we can represent these by ${2g}$ by ${g}$ matrices, satisfying the conditions. So the conditions guarantee we can make half into ${I}$ and the other ${Z}$, and the condition then becomes ${^tZ=Z}$ and ${\mathrm{Im}(Z)>0}$ (the imaginary part).

For a Riemann surface, we have that ${H^{1,0}(X_t)=H^0(X_t,\Omega^1(X_t))}$, so we just need to look at the subspace of ${(1,0)}$ forms among ${C^\infty}$-forms.

So this whole approach is what we’ll be generalizing to higher dimensional fibers.

In general, given ${\phi:\mathcal{X}\rightarrow B}$, we get ${\pi_1(B,b_0)\stackrel{\rho}{\rightarrow} \mathrm{GL}(H^k(X,R))}$ where ${R=\mathbb{Z},\mathbb{Q},\mathbb{R}}$ or ${\mathbb{C}}$. We also have ${R^k\phi_*R}$ to work with.

So how do these two things relate? Can we go between them without going back to the fibration? We can, via flat vector bundles. (Note: Riemann-Hilbert Correspondence)

Let ${\tilde{B}\rightarrow B}$ be the universal cover, and let ${\pi_1}$ act on the right, and assume we have a rep ${\rho:\pi_1(B,b_0)\rightarrow \mathrm{GL}(V)}$. Then we define ${\tilde{B}\times_{\pi_1} V\rightarrow B}$ where this is just the product modulo ${\sim}$ where ${(\tilde{b},v)\sim (\tilde{b}\gamma,\rho^{-1}(\gamma) v)}$ for all ${\gamma\in\pi_1}$.

We say that a section of a bundle with connection is flat if ${D\sigma=0}$.

1. Representations of the fundamental group
2. Local systems
3. vector bundles with flat connection

4. ElZein 3 – Mixed Hodge Complex (MHC)

4.1. Desingularization

Let ${X}$ be a complex irreducible algebraic variety. Then there exists a Zariski open dense subset of smooth point ${U_{smooth}\subset X}$, and its complement is ${X_{sing}}$ the singular locus.

There exists a diagram $\begin{array}{ccccc}Y&\to& X' & \leftarrow & U'\\ \downarrow & & \downarrow & & \downarrow \\ Y_{sing} & \to & X & \leftarrow & U_{smooth}\end{array}$
such that ${X'}$ is smooth, ${Y}$ is a normal crossing divisor on ${X'}$ and ${U'=X\setminus Y}$ is isomorphic to ${U_{smooth}}$.

Consider ${X}$ projective and smooth complete variety, ${i:Z\rightarrow X}$ closed in ${X}$ irreducible of codimension ${r}$, then ${\pi:Z'\rightarrow Z}$ a desingularization. Then ${[Z']\in H_{2n-2r}(Z',\mathbb{Z})}$ gives ${i_*\pi_*[Z']:=[Z]\in H_{2n-2r}(X,\mathbb{Z})}$. By Poincaré duality, we get ${[\eta_Z]\in H^{2r}(X,\mathbb{Z})}$.

Lemma 4 ${[\eta_Z]\in H^{r,r}(X)\cap \mathrm{Im}(H^{2r}(X,\mathbb{Z})\rightarrow H^{2r}(X,\mathbb{C}))}$, and we will call this space ${H^{r,r}(X,\mathbb{Z})}$.

Proof: ${\int_{[Z]}\omega=0}$ if ${\omega\notin \mathscr{E}_X^{n-r,n-r}}$, and so we take ${\omega}$ to be the Kähler (1,1)-form on ${X}$, then we can deduce that ${\int_{[Z]}i_Z^*(\bigwedge^{n-r}\omega)\neq 0}$, by positivity. $\Box$

We define ${Z_r(X)}$ to be the set of all formal integer linear combinations of irreducible subvarieties of ${X}$ of codimension ${r}$. We have a map ${c\ell:Z_r(X)\rightarrow H^{r,r}(X,\mathbb{Z})}$.

Question: Is this map surjective?

Let ${X}$ be a smooth, open algebraic variety, ${X}$ singular.

Remark: If ${X}$ is smooth and proper complex algebraic variety ${H^i(X,\mathbb{C})}$ carries a pure Hodge structure of weight ${i}$.

There exists ${X'}$ projective and ${\pi:X'\rightarrow X}$ which induces an isomorphism on an open dense Zariski subset, and ${\pi^*:H^i(X,\mathbb{C})\rightarrow H^i(X',\mathbb{C})}$.

Now, the Hodge filtration ${F^pH^i}$ is algebraically defined. We deduce that the Hodge filtration ${F}$ on ${H^i(X,\mathbb{Z})}$ is a Hodge filtration of a Hodge structure.

4.2. The Hodge Complex

A Hodge complex of weight ${n}$ is defined as follows: a complex of groups ${K_\mathbb{Z}}$ bounded below, a filtered complex of complex vector spaces ${(K_\mathbb{C},F)}$, ${F}$ finite on each degree, and ${K_\mathbb{Z}\otimes\mathbb{C}\rightarrow K_\mathbb{C}}$, a quasi-isomorphism, and the differential on ${K_\mathbb{C}}$ is strict with respect to ${F}$.

So, for all ${k}$, ${H^k(K_\mathbb{C})\cong H^k(K_\mathbb{Z})\otimes\mathbb{C}}$, and ${H^k(K_\mathbb{C})}$ with induced filtration ${F}$ satisfies ${F^pH^\ell(K_\mathbb{C})=\mathrm{Im}(H^k(F^pK_\mathbb{C})\rightarrow H^\ell(K_\mathbb{C}))}$, and ${(H^k,F)}$ is a Hodge structure of weight ${n+k}$.

Let ${X}$ be a topological space, a CHC of weight ${n}$ ${K}$ on ${X}$ is a complex of sheaves ${K_\mathbb{Z}}$ bounded below, a filtered complex of sheaves ${(K_\mathbb{C},F)}$, a quasi-isomorphism ${\alpha:K_\mathbb{C}\rightarrow K_\mathbb{Z}\otimes_\mathbb{Z} \mathbb{C}}$ such that ${(\mathbb{R}\Gamma K_\mathbb{Z},\mathbb{R}\Gamma(K_\mathbb{C},F),\mathbb{R}\Gamma(\alpha))}$ is a Hodge complex of weight ${n}$

A mixed Hodge complex is ${K_\mathbb{Z}}$ a complex of groups bounded below, ${(K_\mathbb{Q},W)}$ a filtered complex of ${\mathbb{Q}}$ vector spaces with ${K_\mathbb{Q}\cong K_\mathbb{Z}\otimes_\mathbb{Z} \mathbb{Q}}$ and ${(K_\mathbb{C},W,F)}$ bifiltered, ${\alpha:(K_\mathbb{Q},W)\otimes\mathbb{C}\rightarrow (K_\mathbb{C},W)}$ an quasi-isomorphism such that for all ${n}$ we have ${(\mathrm{Gr}^W_n K_\mathbb{Q},(\mathrm{Gr}^W_nK_\mathbb{C},F))}$ along with ${\mathrm{Gr}^W_n(\alpha)}$ is a Hodge complex of weight ${n}$.

The definition of a cohomological mixed Hodge complex is similar.

Let ${K}$ be a MHC then the filtration ${W[n]}$ on ${H^n(K_\mathbb{Q})\cong H^n(K_\mathbb{Z})\otimes \mathbb{Q}}$ and the filtration ${F}$ on ${H^n(K_\mathbb{C})}$ define a MHS.

Proof: Take the spectral sequence ${_W E_r^{pq}}$. It has ${_W E_1^{pq}=H^{p+q}(\mathrm{Gr}_{-p}^W K)}$. These are HS of weight ${q}$. Look at ${d_1}$. We must prove that this map is compatible with ${F}$. Then ${E^{p,q}_2}$, the cohomology, is a Hodge structure of weight ${q}$. And then, ${d_2}$ can be shown to be zero, because it is a morphism of Hodge structures of different weights. $\Box$

5. Migliorini 3

Summary: ${f:X\rightarrow Y}$ a smooth projective morphism. Then ${Rf_*\mathbb{R}_X}$ is isomorphic in some sense to a complex wsith zero differentials and entries ${R^\ell f_*\mathbb{Q}_X}$, which are semisimple local systems and ${L^k:R^{n-k}f_*\mathbb{Q}_X\rightarrow R^{n+k} f_*\mathbb{R}_X}$ is an isomorphism, where ${n=\dim X-\dim Y}$.

What about these three facts for a general projective map ${f:X\rightarrow Y}$, ${X}$ nonsingular. Then ${E_2}$ degeneration fails! But yet, these three facts hold if we replace “local system” with “perverse sheaves”.

1. Let ${f:\tilde{X}\rightarrow X}$ and ${(X,x_0)}$ a germ of a normal surface singularity, with ${f}$ a resolution of singularities. Then ${f^{-1}(x_0)=\cup E_i}$ with ${E_i}$ curves, and the intersection form ${E_i\cdot E_j}$ is negative definite. (Grauert-Mumford Theorem)
2. ${X'\rightarrow C}$ with ${(C,y_0)}$ is a germ of a smooth curve, then ${X'\rightarrow C}$ is a family of projective curves. Assume ${f}$ is smooth ${f^{-1}(C\setminus y_0)\rightarrow C\setminus y_0}$, the inverse image is ${\cup E_i}$, and the intersection form is negative semidefinite and the radical is generated by the class of the fiber (Zariski Lemma)

1 gives as a consequence the fact that ${Rf_*\mathbb{Q}_{\tilde{X}}=\tau_{\leq i} R_{j*}\mathbb{Q}_{x_0}^{\oplus \# E_i}[-2]}$, which has cohomological version ${H^q(\tilde{X})=H^q(X-x_0)}$ if ${q\leq 1}$ and ${H^2(\tilde{X})=\mathbb{Q}^{\# E_i}}$

Number 2 gives ${Rf'_*\mathbb{Q}_{X'}=\mathbb{Q}_C\oplus \mathbb{Q}_C[-2]\oplus j_* R^1[-1]\oplus \mathbb{Q}_{y_0}^{\# E_i-1}[-2]}$ which is, cohomologically, ${H^0(X')=\mathbb{Q}}$, ${H^2(X')=\mathbb{Q}^{\#E_i}}$ and ${H^1(X')=H^0(j_*R^1)}$.

The first is the intersection cohomology complex of ${X}$. The second is the intersection cohomology complex of ${R^1f_*\mathbb{Q}}$.

Now, let ${X'\rightarrow \Delta\ni y_0}$, which is ${\Delta}$ is small enough retracts onto ${f^{-1}(y_0)}$. Then ${H^1(X')\cong H^1(f^{-1}(y_0))}$ which both map to ${H^1(X_\eta)}$ the generic fiber. The image lands in ${H^1(X_\eta)^{\pi_1(C\setminus y_0)}}$ l.i.c.t.

Does this hold in general?

${H^*(\mbox{central fibre})\rightarrow H^*(\mbox{generic fiber})^\pi\rightarrow 0}$, with the first a MHS.

On ${H^*(\mbox{generic fibre})}$ (smooth compact) it is possible to define a MHS such that the above map is a morphism of MHS.

We call this the limit MHS.

R. Friedman “On the Clemens Schmid exact sequence”

Suppose that ${T}$, the monodromy operator, is unipotent (that is, ${T-I}$ is nilpotent) then ${W}$ is defined out of the Jordan form of ${T-I}$.

6. Cattani 3

Look in the appendix of the first set of lecture notes, A.5, on the Weight filtration. Also Flat Bundles.

Let ${\phi:\mathcal{X}\rightarrow B}$ projective, ${\omega=R^2\phi_*\mathbb{Z}}$. We can look at ${R^k\phi_*\mathbb{C}\otimes \mathscr{O}_B}$. Over a point, we have ${\phi^{-1}(\Delta)\rightarrow \Delta}$, and ${\phi^{-1}(\Delta)}$ is diffeomorphic via ${F}$ to ${\Delta\times X}$, with inverse ${G}$, and these give ${g_t:X\rightarrow X_t}$ and ${f_t:X_t\rightarrow X}$.

If ${\alpha\in H^k(X,\mathbb{C})}$, we have ${t\mapsto t_t^*\alpha}$ the set of flat sections.

We look at the Gauss-Manin connection ${\nabla}$. Then ${\nabla (\\sum f_js_j)=\sum df_j\otimes s_j}$ for ${s_j}$ flat.

Look at ${H^k(X_t,\mathbb{C})=\oplus_{p+q=k}H^{p,q}(X_t)}$ then ${H^{p,q}_t=g_t^*(H^{p,q}(X_t))\subset H^k(X,\mathbb{C})}$. These terms are all upper semicontinuous, so they cannot decrease in dimension, only increase, and as the total is constant, each must be.

Now, set ${F^p_t=\oplus_{a\geq p} H^{k,k-a}_t}$. For each ${p}$, we get a map ${\Delta\rightarrow G(f^p,H^k(X,\mathbb{C})}$ where ${f^p=\dim F^p}$, taking ${t}$ to ${F^p_t=g_t^*\left(\oplus_{a\geq p} H^{a,k-a}(X_t,\mathbb{C})\right)}$. This map ${\mathcal{P}^p}$ is smooth.

So back on the bundle, we get ${C^\infty}$ subbundles ${\mathbb{F}^p}$.

Then the amazing fact is that ${\mathcal{P}^p:\Delta\rightarrow G(f^p,H^k(X,\mathbb{C}))}$ is holomorphic, so ${\mathbb{F}^p}$ is a holomorphic subbundle.

For notational convenience, we take ${\dim\Delta=1}$. Then ${\mathcal{P}^p_{*,0}\left(\frac{\partial}{\partial \bar{z}}\right)\in \hom(F^p(X),H^k(X,\mathbb{C})/F^p(X))}$ is zero.

Once we have holomorphicity, there’s Griffiths Transversality, which says that ${\mathcal{P}^P_{*,0}\left(\frac{\partial}{\partial z}\right)}$ belongs to ${\hom(F^p(X),F^{p-1}(X)/F^p(X))}$.

Now, we look at the weight 3 Hodge structure on ${H^3}$. We can break it up in two ways. If we take one half ot be ${H^{3,0}\oplus H^{1,2}}$, then we get the Weil filtration, and these are polarized, but Griffiths looked at ${H^{3,0}\oplus H^{2,1}}$, and lost the polarization in order to get things to vary holomorphically.

Sketch of Griffiths Transversality

Let ${\alpha\in F^p(X)=\oplus_{a\geq p} H^{a,k-a}(X)}$. Then we have ${\alpha(t)\in F^p(X_t)}$. We can look at ${g_t^*(\alpha(t))\in F^p_t}$. Then we have ${\left[\frac{\partial}{\partial \bar{t}}|_{t=0} g^*_t(\alpha(t))\right]\in H^k(X,\mathbb{C})/F^p(X)}$. Then there exists ${\Theta\in \oplus_{a\geq p} A^{a,k-a}(\mathcal{X})}$ such that ${d\left(\Theta|_{X_t}\right)=0}$ and ${\left[\Theta|_{X_t}\right]=\alpha(t)}$.

${G^*\Theta=dt\wedge\phi+\psi}$ and ${d(G^*\Theta)=dt\wedge d\phi+dt\wedge \frac{\partial\psi}{\partial t}+d\bar{t}\wedge\frac{\partial\psi}{\partial\bar{t}}}$ because the contractions of ${\phi}$ and ${\psi}$ with ${\frac{\partial}{\partial t}}$ are both zero.

Then, it’s a matter of counting the number of holomorphic and antiholomorphic terms.

We have a map ${\mathcal{P}^p_{*,0}:T_0(B)\rightarrow \hom(H^q(X,\Omega^p),H^{q+1}(X,\Omega^{p-1})}$ and a map ${T_0^h(B)\rightarrow H^1(X,T_X)}$. This gives us ${H^q(X,\Omega^p)\times H^1(X,T_X)\rightarrow H^{q+1}(X,\Omega^{p-1})}$ by contraction.

For the rest of these talks, we’ll forget geometry and talk about abstract variations of Hodge structures.

Let ${H\rightarrow B}$ be a local system of free ${\mathbb{Z}}$-modules, and let ${\nabla}$ be the Gauss-Manin connection on ${\mathbb{H}=H\otimes \mathscr{O}_B}$. A variation of Hodge structures of weight ${k}$ is an increasing flag of holomorphic subbundles ${0\subset \mathbb{F}^k\subset\mathbb{F}^{k-1}\subset\ldots}$ such that ${\mathbb{F}^p\oplus \bar{\mathbb{F}}^{k-p+1}=\mathbb{H}}$ and if ${\nabla_v \Gamma(U,\mathbb{F}^p)\subset \Gamma(U,\mathbb{F}^{p-1})}$ then ${v\in T^{1,0}(B)}$.

A polarization of VHS is a flat bilinear form ${\mathscr{Q}}$ on ${\mathbb{H}}$ of parity ${(-1)^k}$ such that for each ${x\in B}$, ${\mathscr{Q}_x}$ polarizes the Hodge structure on ${\mathbb{H}_x}$.

For every ${\gamma\in\pi_1(B,b_0)}$, ${\rho(\gamma)}$ is quasi-unipotent, that is, there exists ${r}$ such that ${\rho(\gamma)^r}$ is unipotent.

In the geometric case, this theorem goes back to Langlands.

Let’s call ${N}$ the nilpotent part. So the monodromy theorem actually also says that ${N^{k+1}=0}$.

In general, the first part goes back to Borel, and the second to Schmid.

The monodromy theorem requires that you have a polarization.

Tomorrow, we’ll take this definition and we’ll look how to map it into some nice space, a period domain, and look at the properties of this map, especially when ${B=\Delta^*}$, and look at limiting behavior as we go to ${0}$.

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
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### 6 Responses to ICTP Day 4

1. GS says:

Why not copy-paste it into a LaTeX editor and make a pdf and post it on your webpage? That way, you will be able to use xy.

2. I’ve also done that, though I still need to do it for the notes after the first week. See http://www.math.upenn.edu/~siegelch/notes.html at the bottom

3. GS says:

Thanks a lot. This was useful.

May I inquire how you are able to type and TeX so much? Any special tricks? I am somewhat overwhelmed by the comprehensiveness and would like to find out how this is possible.

4. I just sit in the lecture and type…I don’t really do anything special. I mostly make sure I get anything that’s on the board, and sometimes try to catch the important things the lecturer says, and then fill in with my own connectives, which seems to work out. I know that I’m rather above average in my typing speed, to the point where except for commutative diagrams (complex ones, I have macros for simple ones, like stacking SES’s) I can type faster than I can write things by hand, but that’s just a matter of diving in and getting used to doing it.

5. GS says:

Ah! Ah! I see. Thanks for the explanation.

I have got the “more than average typing speed” part down. But I suppose I would get bogged down with too many \$’s, and every sort of technical detail difficulty, if I were to try out what you are describing.

If I were doing biology, law or something like that, I suppose I could type faster than I write a lecture note. But math symbols would be too hard.

Another thing I am worried about is the annoyance to other people in the lecture hall, seeing someone type away at his laptop, and also from the (minor but audible and distracting) sound.

6. Actually, I find that latex is much easier with the math symbols. And at conferences and such, lots of people are on their laptops, anyway. I’d actually suggest asking a prof if they’d mind and seeing if you can keep up taking notes directly on latex for a week of some class. People seem to be impressed by it until about 20 minutes after they start trying, and realize it’s just not that hard to do.