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Is there any example of a smooth, projective surface $S$ over $\mathbb{C}$, with Picard group $\mathbb{Z}$ and such that $H^1(S, L)$ is not zero for some ample line bundle $L$ ?

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Edit. My original exact sequences were wrong. I straightened them out now.

Let $X$ be a smooth, projective hyper-Kähler fourfold with $\text{Pic}(X)$ isomorphic to $\mathbb{Z}$, e.g., a sufficiently general deformation of the pair $(\text{Hilb}^2_{M/\mathbb{C}},\mathcal{O}(1))$ for $M$ a polarized K3 surface and $\mathcal{O}(1)$ the natural Plücker invertible sheaf. Let $Y\subset X$ be a smooth ample hypersurface such that $h^r(X,\mathcal{O}_X(\underline{Y}))$ vanishes for $r=1,2$. Let $Z\subset X$ be a very general hypersurface that is "sufficiently" ample. In particular, assume that $h^r(X,\mathcal{O}_X(\underline{Z}))$ vanishes for $r=1,2$ and $h^2(X,\mathcal{O}_X(\underline{Z}-\underline{Y}))$ vanishes. Let $i:S\hookrightarrow X$ be $Y\cap Z$, and denote by $\mathcal{I}$ the ideal sheaf of $S$ in $X$. Let $L$ be $\mathcal{O}_X(\underline{Y})|_S$.

There are short exact sequences, $$ 0 \to \mathcal{I}(\underline{Y}) \to \mathcal{O}_X(\underline{Y}) \to i_* L \to 0,$$ $$ 0 \to \mathcal{O}_X(-\underline{Z}) \to \mathcal{O}_X\oplus \mathcal{O}_X(\underline{Y}-\underline{Z}) \to \mathcal{I}(\underline{Y}) \to 0.$$ By hypothesis, $h^r(X,\mathcal{O}_X(\underline{Y})) = 0$ for $r= 1,2$. Thus, via the long exact sequence of cohomology, the following connecting map is an isomorphism, $$\delta_\Sigma^1:H^1(S,L)\xrightarrow{\cong} H^2(X,\mathcal{I}(\underline{Y})).$$ By Serre duality, $h^{2+r}(X,\mathcal{O}_X(-\underline{Z}))$ equals $h^{2-r}(X,\mathcal{O}_X(\underline{Z}))$. Thus, for $r=0,1$, both of these vanish. Finally, by Serre duality, $h^2(X,\mathcal{O}_X(\underline{Y}-\underline{Z})))$ equals $h^2(X,\mathcal{O}_X(\underline{Z}-\underline{Y}))$, so both of these vanish. Thus, the long exact sequence of the second short exact sequence gives an isomorphism, $$ H^2(X,\mathcal{O}_X)\to H^2(X,\mathcal{I}(\underline{Y})), $$ so also $H^1(S,L)$ is isomorphic to $H^2(X,\mathcal{O}_X)$. Of course for the hyper-Kähler fourfold $X$, $H^2(X,\mathcal{O}_X)$ is a $1$-dimensional vector space.

By the Lefschetz hyperplane theorem, the restriction map $\text{Pic}(X)\to \text{Pic}(Y)$ is an isomorphism (and the same holds for $Z$). By the generalized Noether-Lefschetz theorem (as in SGA7), for $Z$ sufficiently ample and very general, the restriction map $\text{Pic}(Y)\to \text{Pic}(S)$ is also an isomorphism. Thus $\text{Pic}(S)$ is isomorphic to $\mathbb{Z}$.

Edit. I was misremembering what is proved in SGA 7_2. The generalized Noether-Lefschetz theorem was not proved there. The generalized Noether-Lefschetz was proved by Kirti Joshi.

MR1299006 (96f:14005)
Joshi, Kirti(6-TIFR-SM)
A Noether-Lefschetz theorem and applications.
J. Algebraic Geom. 4 (1995), no. 1, 105–135.
https://arxiv.org/pdf/alg-geom/9305001v2.pdf

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  • $\begingroup$ Dear Jason, many many thanks for this :) $\endgroup$ – Sudarshan Gurjar Feb 12 '17 at 21:11
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For example, for an irregular surface of general type of Picard number one, $H^1(K_X)\neq0$. See Kovacs' answer here Smooth projective varieties of Picard number one.

Edit: As pointed out by @abx, in this case the picard group could not be $\mathbb{Z}$, as the surface is irregular.

Maybe the way of construct such example is to look at certain surface with $K_X=2L$ and try to show $H^1(L)\neq 0$.

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    $\begingroup$ For such a surface the Picard group is far from being $\mathbb{Z}$ — it contains a nontrivial abelian variety. $\endgroup$ – abx Feb 3 '17 at 7:06
  • $\begingroup$ @abx, you are right... $\endgroup$ – Chen Jiang Feb 3 '17 at 8:08
  • $\begingroup$ @ChenJiang. I made the same mistake that you made (and then I deleted my solution). I read "Picard group" as "Neron-Severi group". $\endgroup$ – Jason Starr Feb 3 '17 at 12:39

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