# For an abelian scheme, $R^pf_* \Omega^q$ is locally free and its formation is compatible with any base change

Let $$k$$ be a field, $$\bar{R} \to R$$ a local homomorphism of artinian local rings with the residue fields $$k$$, $$I$$ its kernel, $$A/R$$ an abelian scheme, and $$\mathscr{T}$$ its tangent sheaf. Let $$A_0 = A \times_R k$$. Assume that $$\mathfrak{m}_\bar{R} I = 0$$. Then $$H^2(A, \mathscr{T}_{A/R} \otimes_R I) \cong H^2(A_0, \mathscr{T}_{A_0/k}) \otimes_k I$$?

This is a part of the proof of (2.2.4.1) of Kai-Wen Lan's "Arithmetic Compactifications of PEL-Type Shimura Varieties". To show it, I need the following proposition:

Let $$S$$ be a scheme, $$f : A \to S$$ an abelian scheme of relative dimension $$g$$. Then the sheaf $$R^pf_* \Omega^q$$ is locally free. And this formation commutes with any base change.

Are there "elementary" proof of this?

I know this is (2.5.2) of Berthelot, Breen, Messing's Théorie de Dieudonné Cristalline. But its proof is too hard for me, since it heavily relies on the theory which I don't know.

And I know that this post shows it elementary. But it uses the formally smoothness and the pro-representability of the deformation of "abelian schemes + polarization", which is what I want to show using this highlighted statement. So it is a circular reasoning for me.

• Isn't this just semi-continuity theorem? – Mohan May 18 at 17:00
• I think this is supposed to be a little tricky. As Daniel Litt points out in the comments of the linked post, over a reduced base this is an easy consequence of "cohomology and base change" (plus the fact that the dimension of the cohomology of the fibres is constant). In characteristic $0$ you might be able to argue by Hodge theory. But the case of an Artinian ring in positive characteristic is the hard one. The two methods that I'm aware of are the ones you cited: Dieudonné theory, or trying to find a reduced moduli space to work with. (But I'm hoping someone else knows a nice argument!) – R. van Dobben de Bruyn May 18 at 17:20
• @Mohan To use the base change theorem, we need that for every $s \in S$, $R^p f_* \Omega^q \otimes k(s) \to H^p(X_s, \Omega^q)$ is surjective. – k.j. May 19 at 0:30
• What do you consider "elementary"? You can prove this quickly for Jacobians of smooth curves (since there are parameter spaces that are smooth over Spec $\mathbb{Z}$). Every Abelian scheme over a local Artin ring with positive characteristic residue field is projective (since obstructions to deforming invertible sheaves have $p$-power torsion). Thus your Abelian scheme is a smooth quotient of a Jacobian -- now apply Leray. – Jason Starr May 19 at 19:24

Let $$A/R$$ be an abelian scheme of relative dimension $$g$$ over an Artinian local ring $$(R, \mathfrak m, \kappa)$$. I am going to give you a proof that works if the characteristic of $$\kappa$$ is not $$2$$.
Denote $$f : A \to \text{Spec}(R)$$ the structure morphism. By the usual trick (see for example here) we have $$\Omega_{A/R} \cong \mathcal{O}_A^{\oplus g}$$. Thus $$\Omega_{A/R}^q$$ is isomorphic to the free $$\mathcal{O}_A$$-module of rank $${g \choose q}$$. Hence it suffices to prove that $$H^i(A, \mathcal{O}_A)$$ is a free $$R$$-module of rank $${g \choose i}$$. Namely, we already know that formation of $$K = Rf_*\mathcal{O}_A$$ in the derived category $$D(R)$$ commutes with base change (by very general cohomology and base change results, see for example the exposition in Mumford's book on Abelian varieties) and freeness of its cohomology will imply it is the direct sum of its cohomology sheaves.
Denote $$[2] : K \to K$$ the pullback by multiplication by $$2$$ on $$A$$. By cohomology and base change (see above) we know that $$K \otimes_R^\mathbf{L} \kappa$$ is isomorphic to $$\wedge^*(\kappa^{\oplus g})$$. It follows that $$K$$ can be represented in $$D(R)$$ by a complex of the shape $$K^\bullet : R \to R^{\oplus g} \to \ldots \to R^{\oplus g} \to R$$ See for example here. Moreover, the map $$[2] : K \to K$$ in $$D(R)$$ can be represented by a map of complexes $$t^\bullet : K^\bullet \to K^\bullet$$ by usual homological algebra. Calculating on the special fibre we see that $$t^i \bmod \mathfrak m$$ is multiplication by $$2^i$$ on $$\wedge^i(\kappa)$$. A bit of elementary algebra then shows that the differentials of $$K^\bullet$$ have to be zero (look at what happens to the leading terms'').
• Is that representative of $K^\bullet$ the natural generalisation of a 'minimal resolution'? – R. van Dobben de Bruyn May 24 at 16:23