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I keep running into the statement that "the generic k3 surface has Picard rank 1".

For instance the answer of this question (end) and this paper (following Example 1.1) or this paper (proof of Theorem 3.5) all state this fact without reference.

I simply cannot track down a reference for this (I tried e.g. Huybrechts) -- perhaps because it is very simple.

I would greatly appreciate a reference, or if very simple an argument.

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    $\begingroup$ For quartic K3s in $\mathbb{P}^3$, this follows from the Noether-Lefschetz theorem (which you can google). The general case can be proved using similar ideas. $\endgroup$ Nov 30 '21 at 16:00
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    $\begingroup$ Welcome new contributor. The generic K3 surface has Picard rank zero. A K3 surface is a simply connected, compact, Kaehler manifold of dimension $2$ whose canonical divisor class is trivial. There is a $20$-dimensional, connected, smooth deformation space of such objects, and the set of those with a nonzero Picard group is a countable union of connected, closed, complex analytic subspaces that are smooth of complex codimension one. Inside any of those smooth, closed, connected, complex analytic subspaces of codimension one, the locus of those with Picard rank at least $2$ is $F_\sigma$. $\endgroup$ Nov 30 '21 at 18:01
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Welcome new contributor. I am just writing my comment as an answer, and expanding on the observation of Prof. Arapura. For a smooth, projective scheme $X$ over a field $k$, the space of first order deformations of $X$ as a $k$-scheme is naturally isomorphic to the $k$-vector space $H^1(X,T_X)$.

For a locally free sheaf $E$ on $X$ of rank $r>0$, the "Atiyah extension" is an element in $\text{Ext}^1_{\mathcal{O}_X}(E,E\otimes_{\mathcal{O}_X}\Omega^1_{X/k})$ whose "characteristic polynomial" has coefficients in $\text{Ext}^r_{\mathcal{O}_X}(\mathcal{O}_X,\Omega^r_{X/k}) = H^r(X,\Omega^r_{X/k})$ that are equal to the images of the degree $r$ Chern classes of $E$ under the cycle class map from $\text{CH}^r(X)$ to the "de Rham cohomology groups" $H^r(X,\Omega^r_{X/k})$. In particular, the "pairing" of the Atiyah extension with an element of $H^1(X,T_X)=\text{Ext}^1_{\mathcal{O}_X}(\Omega^1_{X/k},\mathcal{O}_X)$ gives an element in $\text{Ext}^2_{\mathcal{O}_X}(E,E)$. As proved in references on deformation theory (e.g., Illusie's book on the cotangent comples or Grothendieck's earlier LNM on the good 2-term truncation of the cotangent complex), this element is the obstruction to lifting $E$ to a locally free sheaf on the corresponding first-order deformation of $X$. Altogether, this defines a $k$-linear map, $$ H^1(X,T_X) \to \text{Hom}_k(\text{Ext}^1_{\mathcal{O}_X}(E,E\otimes_{\mathcal{O}_X}\Omega^1_{X/k}), \text{Ext}^2_{\mathcal{O}_X}(E,E)). $$

Now consider the case when $E$ is an invertible sheaf. The $k$-linear map above reduces to the form, $$ H^1(X,T_X) \to \text{Hom}_k(H^1(X,\Omega^1_{X/k}),H^2(X,\mathcal{O}_X)). $$ For a polarized K3 surface $X$ over $\mathbb{C}$, using the trivialization of $\Omega^2_{X/\mathbb{C}}$, this map is "equivalent to" the usual cup-product pairing, $$ H^1(X,\Omega^1_{X/\mathbb{C}}) \to \text{Hom}_{\mathbb{C}}(H^1(X,\Omega^1_{X/\mathbb{C}}),H^2(X,\Omega^2_{X/\mathbb{C}}). $$ The cup-product pairing is a perfect pairing. Thus, this $\mathbb{C}$-linear map is an isomorphism of $\mathbb{C}$-vector spaces.

Finally, since $H^1(X,\mathcal{O}_X)$ vanishes for a K3 surface, the cycle class map from the Picard group to $H^1(X,\Omega^1_{X/\mathbb{C}})$ is injective (in characteristic $p$, the kernel of this cycle class map obviously contains the $p$-power image of the Picard group, so this is one place characteristic $0$ is crucial). Thus the induced $\mathbb{C}$-linear map, $$ H^1(X,T_X) \to \text{Hom}_{\mathbb{Z}}(\text{Pic}(X),H^2(X,\mathcal{O}_X)), $$ is surjective. In particular, if the Picard rank is at least $1$, if we fix a saturated rank $1$ sublattice, there are first order deformations such that the corresponding "obstruction map" on $\text{Pic}(X)$ has kernel precisely equal to this saturated rank $1$ sublattice. Since also $H^2(X,T_X)$ vanishes, these first order deformations extend. Therefore, there are deformations of the K3 surface over a $1$-dimensional base such that the Picard lattice of a very general member of the family equals the saturated rank $1$ sublattice, i.e., the Picard rank is $1$. In applications, typically we choose the saturated rank $1$ sublattice to be generated by an ample divisor class, so that the deformation is a family of projective K3 surfaces (rather than just Kaehler K3 surfaces that are not projective).

Some part of this is included in the proof of Claim 3.5 of my note about Artin's axioms.

Artin's axioms, composition and moduli spaces http://www.math.stonybrook.edu/~jstarr/papers/moduli4.pdf

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    $\begingroup$ Thank you so much for writing out the details @Jason! Now, if I understand correctly, after having obtained the last map, you show there is a (first order) deformation of $X$ to which this invertible sheaf actually extends. Following the argument in your notes (and notation) picking $\theta$ outside of these countably many $\mathcal{V}_L$, yields only a trivial extension to the deformation. The fact that sublattice is saturated implies now that no other invertible sheaf on $X$ can be extended to this particular deformation, and thus it has Picard rank $1$. Am I correct? $\endgroup$
    – user147163
    Dec 1 '21 at 11:28

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