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In the book "Degeneration of abelian varieties" by Faltings, Chai it reads (cf. p.81,82) as if the following holds:

take an abelian scheme $G$ over a base scheme $S$ with dual scheme $\hat{G}$ (for my applications you may always assume $S$ to be the spectrum of a field) and $P$ the Poincaré bundle on $G\times \hat{G}$.

They consider then the pullback of $P$ to the first infinitesimal neighborhood of $G\times {0}$ in $G\times \hat{G}$.

They say that so one gets a line bundle over $G\times Spec(\mathcal O_S \oplus \epsilon \Omega_{\hat{G}})$, which is a first order infinitesimal deformation of the trivial sheaf on $G$.

My questions:

(1) Most important: What do they mean with $Spec(\mathcal O_S \oplus \epsilon \Omega_{\hat{G}})$? It's not clear to me what this notation should mean and why the first inf. neighborhood of $0$ in $\hat{G}$ looks like this.

(2) Why is this a deformation?

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${\cal O}_S \oplus \epsilon \Omega_{\hat G}$ is a sheaf of rings on $S$ with $\epsilon^2 = 0$; you can locally take Spec and patch these together to produce a scheme over $S$. –  Tyler Lawson Oct 20 '11 at 19:33
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1 Answer

up vote 3 down vote accepted

Let $G$ be any group scheme over $S$. Let $I\subset\mathcal{O}_G$ be the augmentation ideal defining the identity section of $G$ over $S$. Then, by definition, the first infinitesimal neighborhood of the identity in $G$ is the closed sub-scheme of $G$ defined by the ideal $I^2$. Moreover, $I/I^2$ is a coherent sheaf on $S$: it is nothing but the sheaf of invariant differentials on $G$, namely $\Omega_{G}$. In the book 'Neron Models', for example, you will find in Proposition 4.2.1 that there is a canonical isomorphism from $e^*\Omega_{G/S}$ to $\Omega_G$; here, $\Omega_{G/S}$ is the full sheaf of differentials on $G$ and $e:S\to G$ is the identity section.

Now, $\Omega_{G/S}$ is the ideal of the diagonal embedding of $G$ in its first order neighborhood $G^{(1)}\subset G\times_S G$. Pull everything back by the map $g\mapsto (e,g)$ from $G$ to $G\times G$. Then the diagonal embedding pulls back to the identity section and its first order neighborhood pulls back to the first order neighborhood of $e$ in $G$. So we see that $e^*\Omega_{G/S}=I/I^2$.

So $\mathcal{O}_{G}/I^2$ is non-canonically isomorphic to $\mathcal{O}_S\oplus\epsilon\Omega_G$, where $\epsilon^2=0$. This is because we have a sequence of maps $$\mathcal{O}_S\to \mathcal{O}_G/I^2\to \mathcal{O}_G/I=\mathcal{O}_S$$ of sheaves of rings supported on the identity section. In particular, the short exact sequence $$0\to \Omega_G\to\mathcal{O}_G/I^2\to\mathcal{O}_S\to 0$$ admits a section. Which means that we can write $\mathcal{O}_G/I^2$ as a direct sum $\mathcal{O}_{S}\oplus\Omega_G$, with the multiplication given by $$(f,\omega)\cdot(g,\omega')=(fg,f\omega'+g\omega).$$ Writing it as $\mathcal{O}_S\oplus\epsilon\Omega_G$ is just a compressed way to indicate this.

That this pull-back is a deformation of the trivial sheaf follows from the fact that the restriction of the Poincare bundle to $G\times 0$ is canonically trivialized.

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Thanks for your answer, Keerthi. Just two things: why is it that $I/I^2$ is the sheaf of differentials? I thought one has to take the ideal of the diagonal? And how does one get the non-canonical iso you mention? Perhaps you know some convenient literature about group schemes and so on where these things are treated? –  Veen Oct 21 '11 at 6:47
    
I've edited the answer to add some more details. Chapter 4 of 'Neron models' is a good reference in my opinion for general facts about group schemes. –  Keerthi Madapusi Pera Oct 21 '11 at 12:31
    
@Keerthi: as this is really important for me, allow me one more question: the section of the second exact sequence you mention: is it a splitting of sheaves of $\mathcal O_S$-modules? And what is the morphism $\mathcal O_S \rightarrow \mathcal O_G /I^2?$ –  Veen Oct 24 '11 at 12:19
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The splitting is of $\mathcal{O}_S$-modules. The map $\mathcal{O}_S\to\mathcal{O}_G/I^2$ is just the one arising from the structure map from the first order neighborhood of the identity in $G$ to $S$. Note that the topological space underlying this first order neighborhood is just $S$! –  Keerthi Madapusi Pera Oct 24 '11 at 14:38
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