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I've posted this question few days ago on math.stackexchange because it seems quite superficial. However, since I've got no responses at all, I'm posting it here. If the question is not suitable, please delete it.

In Deligne's article in Séminaire Bourbaki "Formes modulaires et représentation $\ell$-adiques" (http://archive.numdam.org/ARCHIVE/SB/SB_1968-1969__11_/SB_1968-1969__11__139_0/SB_1968-1969__11__139_0.pdf) some passages were not clear to me.

Let $E \xrightarrow{f} S$ be an elliptic curve over an analytic space $S$ and $T_{\mathbb{Z}} (E) = R^1 f_{*} \mathbb{Z}^{\vee} $

Why the map $\alpha$ in $R^1 f_{*} \mathbb{Z}^{\vee} \xrightarrow{\alpha} \text{Lie}_S (E) \xrightarrow{\text{exp}_E} E$ fits into the exact sequence?

Is somehow $T_\mathbb{Z} (E)$ a Tate module for the archimedean place? If yes, how can the analogy be seen?

Why $R^2 f_{*} \mathbb{Z} \cong \mathbb{Z}^{2}$?

In page 143, why $E_X$ exists as a scheme and not just as a stack?

In page 143, how he got the sequence $\omega_{E/S} \cong f^{*}\Omega^{1}_{E/S} \rightarrow R^{1}f_{*}{\mathbb{R}} \otimes_{\mathbb{R}} \mathscr{O}_X \rightarrow R^{1}f_{*} \mathscr{O}_X$? The expected sequence would be $\omega_{E/S} \rightarrow R^{1}f_{*} \Omega^{\bullet}_{dR, E/S} \rightarrow R^1f_{*} \mathscr{O}_X$. So why would $R^{1}f_{*} \Omega^{\bullet}_{dR, E/S} \cong R^{1}f_{*}{\mathbb{R}} \otimes_{\mathbb{R}} \mathscr{O}_X$?

Furthermore,(as in page 143) how to prove that under a suitable trivialization the map $q$ is just $q(ax + by) = a \frac{f(x)}{f(y)} + b$ where $f$?

Thanks in advance.

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That's too many questions at the same time. Here are some answers.

The exact sequence $0\rightarrow R_1f_*\mathbb{Z}\rightarrow \mathrm{Lie}_S(E)\rightarrow E\rightarrow 0$ is just the globalization of the exact sequence $0\rightarrow H_1(E_0,\mathbb{Z})\rightarrow \mathrm{Lie}(E_0)\rightarrow E_0\rightarrow 0$ for a complex elliptic curve $E_0$.

The group $H^2(E_0,\mathbb{Z})$ is canonically isomorphic to $\mathbb{Z}$, hence $R^2f_*\mathbb{Z}\cong \mathbb{Z}$.

$E_X$ is certainly not a scheme! It is a complex manifold, or if you like a relative group scheme over the Poincaré upper half-space.

Since $X$ is a complex manifold, $R^1f_*\mathbb{R}\otimes _{\mathbb{R}}\mathscr{O}_X$ is the same thing as $R^1f_*\mathbb{C}\otimes _{\mathbb{C}}\mathscr{O}_X$, which is isomorphic to $R^1f_*\Omega ^{\bullet}_{DR}$ .

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  • $\begingroup$ Thanks for the answer. Could you explain why $R^1 f_{*} \mathbb{C} \otimes_\mathbb{C} \mathscr{O}_X \cong R^{1}f_{*} \Omega^{\bullet}_{dR}$? $\endgroup$
    – user40276
    Feb 16, 2015 at 18:31
  • $\begingroup$ Furthermore, how could you conclude that $E_X$ exists as a group scheme (over $S$)? It's not clear that the moduli problem is solvable and a fine moduli space exists. $\endgroup$
    – user40276
    Feb 16, 2015 at 18:37
  • $\begingroup$ Well, that is the assertion 2.2 (ii) of Deligne -- which is quite standard. You have to understand why elliptic curves $\mathbb{C}/\Gamma $ together with a basis of the lattice $\Gamma $ are classified by the Poincaré upper half-space. As for your other question, this is again the globalization of the canonical isomorphism $H^1(E,\mathbb{C})\cong H^1_{\mathrm{DR}}(E)$ for a complex elliptic curve $E$. $\endgroup$
    – abx
    Feb 16, 2015 at 21:13
  • $\begingroup$ The problem is not $X$ (I know that it parametrizes the lattices). It's just that he does not prove the existence of $E_X$ which has fibers with non-trivial automorphisms apparently (so a fine moduli space would not exist, maybe a coarser one). And I could not understand why there's a tensor with $\mathscr{O}_X$ in what you said that is a globalization of this canonical isomorphism. $\endgroup$
    – user40276
    Feb 17, 2015 at 1:21
  • $\begingroup$ No, there are no nontrivial automorphisms fixing a basis of the lattice. And you need to tensor with $\mathscr{O}_X$ because $R^1f_*\mathbb{C}$ is a local system, not a vector bundle. $\endgroup$
    – abx
    Feb 17, 2015 at 5:47

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