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It seems that there is no chance to explain the Hodge theory (to students) in an hour or so. Yet do there exist any cases when the Hodge filtration (or the Hodge decomposition) of the cohomology of a complex (projective) variety can be more or less clearly explicitly described in terms of periods?

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What do the students know? If they know manifolds and de Rham cohomology, then the statement of Hodge decomposition can be reached in an hour. If not, then you're going to have to think very carefully about what you are telling them it is a decomposition of. –  David Speyer Oct 2 '12 at 10:47
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This isn't precisely an answer to your question, but two years ago I taught a course on Hodge theory and I tried to design a lecture for the opening day which presented the key ideas of the course in as elementary a way as possible. You might be interesting in seeing my notes math.lsa.umich.edu/~speyer/632/jan-6.pdf –  David Speyer Oct 2 '12 at 13:34
    
The students are of different levels of knowledge, and some of them know little about de Rham cohomology. Yet I would like them to obtain some impression on the Hodge filtration/decomposition. Thank you for your notes! –  Mikhail Bondarko Oct 2 '12 at 19:43
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3 Answers

Building on Donu Arapura's answer (although it's again not clear you can do it in an hour), I would suggest the computing the Hodge filtration in $H^2$ of a product $E_1\times E_2$ of two elliptic curves. The beautiful thing about this example is that (1) you can see how the Neron-Severi group changes rank with the periods, and explicitly relate this back to the properties of the elliptic curves, so it also serves as an example of why the Hodge conjecture is so mysterious and significant, and (2) you can see how the filtration varies with the periods, and so see Griffiths transversality in action.

Here is an outline of this calculation. Choose a basis $\delta_1$, $\delta_2$ for $H^1(E_1,\mathbf{Z})$, oriented so that $\delta_1\cap\delta_2 =1$, and normalize the holomorphic differential $\omega$ so that its integral along (the Poincare duals of) $\delta_1$ and $\delta_2$ are $\tau_1$ (the period of $E_1$), and $1$ respectively. Then this tells us that in $H^1(E_1,\mathbf{C})$ we have the relations

$$\delta_1=\frac{1}{\overline{\tau}_1 - \tau_1}(\overline{\tau}_1\omega - \tau_1\overline{\omega})\quad\mbox{and}\quad \delta_2=\frac{1}{\overline{\tau}_1-\tau_1}(-\omega-\overline{\omega}).$$

Similarly, choose an oriented basis $\epsilon_1$, $\epsilon_2$ for $H^{1}(E_2,\mathbf{Z})$ and normalize the holomorphic differential $\eta$ of $E_2$ to have integrals (along the Poincare duals of $\epsilon_1$ and $\epsilon_2$) of $\tau_2$ (the period of $E_2$) and $1$ respectively. Then in $H^1(E_2,\mathbf{C})$ we get the similar formulas

$$\epsilon_1=\frac{1}{\overline{\tau}_2 - \tau_2}(\overline{\tau}_2\eta - \tau_2\overline{\eta})\quad\mbox{and}\quad \epsilon_2=\frac{1}{\overline{\tau}_2-\tau_2}(-\eta-\overline{\eta}).$$

Wedging $\delta_1$, $\delta_2$, $\epsilon_1$, and $\epsilon_2$ together gives integral bases for $H^2(E_1\times E_2,\mathbf{Z})$, and from the formulas above see their expressions in terms of wedges of $\omega$, $\overline{\omega}$, $\eta$, and $\overline{\eta}$, and thus see the Hodge filtration. Explicitly, the change of basis matrix (up to a factor of $(\overline{\tau}_1-\tau_1)(\overline{\tau_2}-\tau_2)$) is

$$ \begin{array}{cc} & \begin{array}{cccc} \omega\wedge\eta & \overline{\omega}\wedge\eta & \omega\wedge\overline{\eta} & \overline{\omega}\wedge\overline{\eta} \\ \end{array} \\\\ \begin{array}{c} \delta_1\wedge\epsilon_1 \\\\ \delta_1\wedge\epsilon_2 \\\\ \delta_2\wedge\epsilon_1 \\\\ \delta_2\wedge\epsilon_2\\ \end{array} &\left[{ \begin{array}{cccc} \overline{\tau}_1\tau_2 & -\tau_1\overline{\tau}_2 & -\overline{\tau}_1\tau_2 & \tau_1\tau_2 \\\\ -\overline{\tau}_1 & \tau_1 & \overline{\tau}_1 & -\tau_1 \\\\ -\overline{\tau}_2 & \overline{\tau}_2 & \overline{\tau}_2 & -\tau_2 \\\\ 1 & -1 & -1 & 1 \\ \\ \end{array}}\right] \end{array}$$ (So the first row expresses $\delta_1\wedge\epsilon_1$ in terms of the wedges of $\omega$, $\eta$ and their conjugates.) Here $\delta_1\wedge\delta_2$ and $\epsilon_1\wedge\epsilon_2$ have been omitted since they are clearly $(1,1)$-forms.

The change of basis matrix allows us to see how to express the $(2,0)$ forms (spanned by $\omega\wedge\eta$), the (1,1)-forms (spanned by $\overline{\omega}\wedge\eta$, $\omega\wedge\overline{\eta}$ and the (1,1)-forms above), and the $(0,2)$ forms (spanned by $\overline{\omega}\wedge\overline{\eta}$) in terms of the integral classes.

To explain (1) above: Since the Hodge conjecture holds in codimension 1, the rank of the Neron-Severi group of $E_1\times E_2$ is $2$ (for the fibre classes $\delta_1\wedge\delta_2$ and $\epsilon_1\wedge\epsilon_2$) plus the integral or rational $(1,1)$ classes coming from the forms in the matrix. To get a $(1,1)$ form what we need to do is make sure that the coefficients of $\omega\wedge\eta$ and $\overline{\omega}\wedge\overline{\eta}$ are zero and so we end up with the formula:

$$\mbox{Rank of $NS(E_1\times E_2)$} = 2 + \mbox{dimension of space of relations of $\tau_1\tau_2$, $\tau_1$, $\tau_2$, $1$ over $\mathbf{Q}$ or $\mathbf{Z}$.} $$

If you feed this back into standard facts about when two elliptic curves are isogenous (in terms of their periods) and when an elliptic curve has CM (again in terms of the period) the formula that results is that

$$\mbox{Rank of $NS(E_1\times E_2)$} = \begin{cases} 2 & \mbox{if $E_1$ and $E_2$ are not isogenous} \\\\ 3 & \mbox{if $E_1$ and $E_2$ are isogenous and neither has CM} \\\\ 4 & \mbox{if $E_1$ and $E_2$ are isogenous and both have CM} \\\\ \end{cases}. $$ Which is a beautiful demonstration of how arithmetic relations among the periods imply geometric results (which is what the Hodge conjecture is saying).

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In an hour? Hmmm... you could do the classical example of an elliptic curve $X$ given by $y^2=f(x)$, where $f(x)$ is a cubic with distinct real roots $a_1<a_2<a_3$. Then lines $\gamma=[-\infty, a_1]$ and $[a_3,\infty]$ correspond to a basis of $H_1(X,\mathbb{Z})$ projected to the plane. The differential $dx/y$ spans $H^0(X,\Omega^1)$. The periods representing the inclusion of $$H^0(X,\Omega^1)\subset H^1(X,\mathbb{C})\cong Hom(H_1(X,\mathbb{Z}),\mathbb{C})$$ are precisely the elliptic integrals $$c_\gamma\int_\gamma \frac{dx}{\sqrt{f(x)}}$$ over the curves given earlier, and $c_\gamma$ are suitable normalizations. I don't have time to dig up precise references now, but see http://www.math.purdue.edu/~dvb/graph/elliptic.html

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A very "down to earth" explicit example where Hodge decompositions can be computed comes from the so-called square-tiled cyclic covers. In a nutshell, square-tiled cyclic covers are "essentially" branched covers of the Riemann sphere branched at 4 points $z_1$, ..., $z_4$, or equivalently, curves determined by algebraic equations of the form $w^N=(z-z_1)^{a_1}...(z-z_4)^{a_4}$. In this particular case, as it is nicely explained in Section 2 (see e.g. their Lemma 2) of this paper of Eskin-Kontsevich-Zorich (http://arxiv.org/pdf/1007.5330.pdf), one obtains variations of Hodge structures that can be explicitly computed in terms of the data $z_1$, ..., $z_4$, $N$, $a_1$, ..., $a_4$.

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