# Analytic and algebraic torsor of abelian scheme

Let $$M$$ be an affine complex manifold, let $$A$$ be an abelian scheme over $$M$$. Let $$\mathcal{A}$$ be the sheaf of local sections of $$A$$ over $$M$$. We can equip $$M$$ with etale topology $$M_{et}$$ or complex topology $$M_{an}$$. There is a natural comparison map $$\gamma\colon H^1(M_{et},\mathcal{A})\to H^1(M_{an},\mathcal{A})$$ between the corresponding sheaf cohomologies.

(a) Are there explicit examples such that $$\mathrm{ker}(\gamma)\neq 0$$?

(b) Are there explicit examples such that $$\mathrm{coker}(\gamma)\neq 0$$?

[In case (a) we need to find an algebraic $$A$$-torsor $$T$$, such that $$T/M$$ admits an analytic section but not an algebraic section. Suppose such $$T$$ exist, we can consider its relative albanese map $$a:T\to \mathrm{Alb}(T)$$, this is a family of algebraic maps over $$M$$, I think there should be plenty of non-algebraic family of algebraic maps, but I am not sure how to make such an example..]

[In case (b), I think we need to find a non-algebraic family of complex torus $$T'/M$$, whose albanese is algebraic, not sure how to find one...]

Here is an example when $$\gamma$$ is not injective.

In general, if $$A=A_0\times M$$ is a constant abelian scheme, choose a presentation for $$(A_0)_{an}$$ as $$\mathbb{C}^g/\mathbb{Z}^{2g}$$. This induces a short exact sequence of sheaves on $$M_{an}$$: $$0\to\underline{\mathbb{Z}}^{2g}\to\mathcal{O}_{M_{an}}^{\oplus g}\to\mathcal{A}\to 0$$

It induces a long exact sequence, part of which looks like $$H^1(M_{an},\mathcal{O})^{\oplus g}\to H^1(M_{an},\mathcal{A})\to H^2(M_{an},\mathbb{Z})^g$$. In particular, if $$M$$ is affine and $$H^2(M_{an},\mathbb{Z})=0$$, then $$H^1(M_{an},\mathcal{A})=0$$.

Let now $$A_0=E$$ be an elliptic curve and $$M=E'\setminus\{p\}$$ be the complement to a point in an elliptic curve $$E'$$ which is not isogenous to $$E$$. We have a long exact sequence of abelian groups $$Maps(M,E)\xrightarrow{[n]}Maps(M,E)\to H^1(M_{et},\mathcal{A}[n])\to H^1(M_{et},\mathcal{A})$$ Here $$\mathcal{A}[n]$$ is the sheaf of $$n$$-torsion in our abelian scheme, which is isomorphic to the constant sheaf $$\underline{(\mathbb{Z}/n)^2}$$. Any morphism $$M\to E$$ extends to a morphism $$E'\to E$$ (this fails for morphisms $$M_{an}\to E_{an}$$ of analytic spaces) which has to be constant by the assumption, so $$Maps(M,E)=0$$. Thus we get an injection $$(\mathbb{Z}/n)^4=H^1(M,(\mathbb{Z}/n)^2)\hookrightarrow H^1(M_{et},\mathcal{A})$$ which gives a non-zero element in $$H^1(M_{et},\mathcal{A})$$ while the source of the map $$\gamma$$ is zero.

(a) $$\gamma$$ will be injective if $$M$$ is proper, as then by GAGA every holomorphic section is automatically algebraic.

(b) The classical example of an element not in the image of $$\gamma$$ is provided by the Hopf surface as follows:

Let $$M = \mathbf{P}^1$$, and let $$X = (\mathbf{C}^2 \setminus \{(0,0)\})/q^{\mathbf{Z}}$$ for some complex number $$q$$ with $$0<|q|<1$$ acting on $$\mathbf{C}^2$$ by scaling. This is called the (a?) Hopf surface, and is the basic example of a non-Kahler compact complex manifold.

Let $$E = \mathbf{C}^*/q^{\mathbf{Z}}$$, this is an elliptic curve. We set $$\mathcal{A} = E \times M$$, the constant abelian scheme over $$M$$ with fiber $$E$$.

We endow $$X$$ with the structure of an $$\mathcal{A}$$-torsor over $$M$$ as follows. We have $$M = \mathbf{P}^1 = (\mathbf{C}^2 \setminus \{(0,0)\})/\mathbf{C}^*$$, and the inclusion $$q^{\mathbf{Z}} \subseteq \mathbf{C}^*$$ yields a natural map of the quotients $$\pi\colon X\to M$$ whose fibers can be identified with $$\mathbf{C}^*/q^{\mathbf{Z}} = E$$.

If the class $$[X] \in H^1(M_{\rm an}, \mathcal{A})$$ was in the image of $$\gamma$$, then $$X = Y_{\rm an}$$ for $$Y$$ the algebraic variety (or algebraic space) underlying the corresponding etale $$\mathcal{A}$$-torsor. But $$X$$ is not even Moishezon, so this can't happen.

• Thanks for the amazing example! (I had thought (b) was true when the base is proper...) – Qixiao Sep 19 '19 at 11:01
• @Qixiao "I had thought (b) was true when the base is proper..." - me too! I learned this example from Jason Starr somewhere on MO. – Piotr Achinger Sep 19 '19 at 12:25