For a smooth variety $X$ when $Pic^0(X)$ is trivial, we get an isomorphism between $N^1(X)$ and Picard group and life become easier.

My question is whether in general there are theorems, criteria ... which results in triviality of $Pic^0$ in some useful examples ?? Like for hyeprsurfaces in projective space, or in toric varieties, or for complete intersections, or in Fano varieties,...

Please tell, if you know any theorem or criteria like this, or if you know any method which helps one to answer this question in a specific example.

Don't get mad, I know it is vague question.

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    $\begingroup$ Lefschetz hyperplane theorem tells us that $H^1(X)=0$ for a hypersurface $X\subset P^n$ when $n>2$ which implies $Pic^0(X)$ is trivial. $\endgroup$ – Jim Bryan Apr 21 '11 at 21:03

Working over $\mathbb{C}$:

We have the short exact sequence of sheaves $0 \to \mathbb{Z} \to \mathcal{O} \to \mathcal{O}^* \to 0$ and we thus have $$H^1(X, \mathbb{Z}) \to H^1(X, \mathcal{O}) \to \mathrm{Pic}(X) \to H^2(X, \mathbb{Z}) \to H^2(X, \mathcal{O}).$$ Note that the kernel of the last map is discrete, while the cokernel of the first map is connected. So $\mathrm{Pic}^0(X) = \mathrm{CoKer}(H^1(X, \mathbb{Z}) \to H^1(X, \mathcal{O}))$ and $NS(X) = \mathrm{Ker}(H^2(X, \mathbb{Z}) \to H^2(X, \mathcal{O}))$.

Moreover, $H^1(X, \mathcal{O})$ is a $\mathbb{C}$-vector space, and $H^1(X, \mathbb{Z})$ is a finitely generated abelian group. So $\mathrm{Pic}^0(X)$ is trivial if and only if $H^1(X, \mathcal{O})=0$.

I know two main situations where you can conclude that $H^1(X, \mathcal{O})$ is $0$. If $X$ is Stein (for example, affine), then $H^1(X, \mathcal{O})$ vanishes because cohomology of any coherent sheaf on $X$ vanishes. Also, if $X$ is compact Kahler (for example, projective) then $H^1(X, \mathcal{O}) \cong H^{1,0}(X)$ and $\dim H^{1,0}(X) = (1/2) \dim H^1(X, \mathbb{C})$. So $H^1(X, \mathcal{O})$ vanishes if $X$ has no first cohomology. Most (I think all) of the examples you mention are of the second type.


By definition, for $X$ a compact Kähler manifold, $$ Pic^0(X)=H^1(X,\mathcal O_X)/H^1(X,\mathbb Z) $$ and $H^1(X,\mathbb Z)$ is a lattice of maximal rank. Thus the vanishing of $Pic^0(X)$ is equivalent to the vanishing of $H^1(X,\mathcal O_X)$, which is, by Hodge theory, purely topological: $H^1(X,\mathcal O_X)\simeq H^{0,1}(X;\mathbb C)$ and $h^{1,0}(X)=h^{0,1}(X)=b_1(X)/2$.

Hence you are asking for Kähler manifolds whose first Betti number vanishes.

Now, $H^1(X,\mathbb C)$ is the complexification of the abelianization of $\pi_1(X)$. So a compact Kähler manifold has zero Jacobian variety if and only if the abelianization of its fundamental group vanishes after complexification, i.e. it is completely torsion.

A first rich class of examples is given, as Jim said, by the Lefschetz hyperplane theorem. Take an $n$-dimensional smooth projective variety $X$ with zero $H^1(X,\mathcal O_X)$ (for instance a simply connected smooth projective manifold, e.g. $\mathbb P^n$), and $D\subset X$ any smooth ample divisor. Then one has isomorphisms $$ H^k(X,\mathbb Z)\to H^k(D,\mathbb Z),\quad k<n-1 $$ and it is injective when $k=n-1$. In particular $H^1(D,\mathbb Z)=0$ and all such hypersurfaces will have zero Jacobian variety, provided $n>2$. You can figure out a similar construction in the smooth complete intersection case.

Fano varieties are known to be rationally connected and hence simply connected, therefore provide other examples (to see it more elementary if $-K_X$ is ample then $H^q(X,O_X)=0$ if $q>0$ by Kodaira's vanishing. Thus $H^1(X,\mathcal O_X)=0$).

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    $\begingroup$ The abelianization of $\pi_1(X)$ is $H_1(X, \mathbb{Z})$, not $H^1(X, \mathbb{Z})$. For instance, if $X$ is an Enriques surface, one has $H^1(X, \mathbb{Z})=0$ but $H_1(X, \mathbb{Z})=\pi_1(X)=\mathbb{Z}_2$. $\endgroup$ – Francesco Polizzi Apr 21 '11 at 23:19
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    $\begingroup$ You are right. But since here I was looking just at the dimensions of the spaces this does not modify very much my answer. I'll correct slightly my answer, then. Thank you anyway! $\endgroup$ – diverietti Apr 22 '11 at 8:37

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