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Let $X$ be a complex normal projective variety.

Is there any sufficient condition to guarantee the torsion-freeness of Picard group of $X$?

One technique I sometimes use is following: If $X$ can be represented by GIT quotient $Y//G$ for some projective variety $Y$ with well-known Picard group, then by using Kempf's descent lemma, we can attack the computation of integral Picard group.

Of course, if we can make a sequence of smooth blow-ups/downs between $X$ and $X'$ with well known Picard group, then we can get the information of $\mathrm{Pic}(X)$ from $\mathrm{Pic}(X')$.

Is there any way to attack this problem?

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[EDIT: A previous version mistakenly argued that the fundamental group of X was responsible for torsion in the Picard group. I hope that this is correct now! Btw, there is probably a more direct way of arguing, but I cannot find one at the moment.]

The Picard group of X is torsion free if and only if the group ${\rm H_1}(X,\mathbf{Z})$ vanishes.

By the exponential sequence, the torsion in the Picard group of X comes from the torsion in ${\rm H^2}(X,\mathbf{Z})$ and from ${\rm H^1}(X,\mathbf{C})/{\rm H^1}(X,\mathbf{Z})$. Thus the vanishing of ${\rm H_1}(X,\mathbf{Z})$ is equivalent (by the Universal Coefficient Theorem) to the torsion-freeness of the Picard group of X.

ADDED (for explicitness) To make everything more explicit, assume that X is non-singular. The Picard group of X may contain torsion coming from two different sources. There might contain torsion in the connected component of the identity, and this is recorded by the torsion free part of the first homology group. Or there might be torsion in the component group of the Picard group, and this is recorded by the torsion in the first homology group. In terms of the exponential sequence, the first kind of torsion appears in the image of ${\rm H^1}(X,\mathbf{C})$, while the second one "appears" in torsion in ${\rm H^2}(X,\mathbf{Z})$. The Universal Coefficient Theorem implies that the "combination" of these two groups is the whole first integral homology group.

An example of torsion of the first kind is already present in the case of curves of genus at least one: the Jacobian of the curve contains plenty of torsion bundles. An example of torsion of the second kind is the case of Enriques surfaces: the canonical divisor on such a surface is a torsion line bundle that is non-trivial. If the characteristic of the ground-field is different from two, the corresponding cover of X is a K3 surface.

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    $\begingroup$ @Donu: thanks for the comment! The reason I find this argument not "direct" is that, by training, I tend to think of cohomology as the dual of homology, rather than the other way around. The reasoning above "retrieves the pieces of H_1" after they have been "scattered by taking the dual", reconstructing the homology of X as the dual of the cohomology. $\endgroup$
    – damiano
    Commented Jul 21, 2010 at 17:35
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    $\begingroup$ This is not quite correct, the proper condition is that $H_1(X,\mathbb Z)$ is torsion free not zero. $\endgroup$
    – Torsten Ekedahl
    Commented Jul 21, 2010 at 19:49
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    $\begingroup$ Torsten: I had the same thought initially, but then I remembered that $Pic^0(X)$ would be an abelian variety, so if it's nonzero... $\endgroup$
    – Donu Arapura
    Commented Jul 21, 2010 at 20:03
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    $\begingroup$ Another way to see this would be to use the Kummer sequence for multiplication by $n$ on $\mathcal{O}_X^*$. This shows that the $n$-torsion subgroup of $Pic(X)$ is (after fixing a primitive $n$-th root of unity) isomorphic to $H^1(X,\mathbb{Z}/n\mathbb{Z})$. So you need these groups to be all 0; or equivalently that $H_1(X,\mathbb{Z})=0$. $\endgroup$
    – Tony Scholl
    Commented Jul 21, 2010 at 21:42
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    $\begingroup$ Torsten -- the Picard group of a smooth complex projective curve has a lot of torsion. $\endgroup$
    – algori
    Commented Jul 21, 2010 at 21:54
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You have my sympathy, I have struggled with torsion-freeness of Picard and class groups too.

As Donu hinted in his comment on Damiano's answer, a torsion element of the Picard group gives you a cyclic cover of $X$, so triviality would follow from certain purity results for etale covers. A fact that is interesting (at least for me) is that one could use the purity of the local ring at the origin of the affine cone over $X$, see Lemma 9 of this paper.

(purity for the local ring here is in the sense of Grothendieck, that is to say, the restriction of etale covers over $\text{Spec} A$ to the punctured spectrum is an equivalence) .

This shows, for example, Grothendieck's classical result that if $X$ is a complete intersection of dimension at least $2$, then $\text{Pic}(X)$ is torsion-free. But Cutkosky's paper above gives quite more general results, see Theorem 19, the Cor after Theorem 26 and the Examples after Theorem 22 of his paper. Basically, his results say that if $X$ is locally a complete intersection in high enough codimension relative to the deviation of the local ring at the vertex of the cone, then $\text{Pic} (X)$ is torsion-free. These apply to a few Grasmanians and Pfaffians, and hopefully they can be helpful to what you want to do.

Also note that various results which give vanishing of $H_1$ or $\pi_1$ when $X$ is relatively small codimension subvariety of the projective space can be found in the paper of Lyubeznik and the references therein.

ADDED: I had a chance to look at Lyubeznik's paper and actually he has Theorem 11.2, which says that if $Y\subset \mathbb P^n_{\mathbb C}$ is an irreducible algebraic set of codimension $c< n/2$, then $\text{Pic}(Y)$ is torsion-free.

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