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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...]

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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.

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This is a partial answer.

(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.

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

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