# Is the generic deformation of a symplectic variety affine?

Kaledin and Verbitsky have shown that symplectic varities have a remarkably nice deformation theory as symplectic varieties.

Let $X$ be a symplectic variety (a smooth quasi-projective variety over $\mathbb{C}$ equipped with a nondegenerate closed algebraic 2-form $\Omega$); then there is a universal formal deformation $\tilde X$ over $H^2(X;\mathbb{C})$ completed at the class $[\Omega]$.

In many examples, it seems that this deformation is generically affine; that is, its fiber at the generic point is an affine scheme. For example, $T^*G/B$ deforms to a generic coadjoint orbit in $\mathfrak{g}^*$, and if $X$ is a hyperkähler quotient, like a quiver variety, then the deformation comes from varying the complex moment map, and the deformation is also generically affine (since a GIT quotient by a free action of a reductive group is always affine).

Of course, I don't think this always happens; the product of two elliptic curves is symplectic, but doing this deformation should just change the $j$-functions of the curves simultaneously. So, clearly one needs some kind of extra condition. I've opted for "resolution of singularities of its affinization."

So my question is:

Let $X$ be a symplectic variety which is a resolution of singularities of its affinization. Is $\tilde X$ generically affine?

EDIT: As was pointed out in Misha Verbitksy's answer below, the deformation is not canonically algebraic. I believe though that if you assume that $X$ has a $\mathbb{C}^*$-action which is dilating (Definition 1.7 of this paper), then $\tilde X$ will also have a $\mathbb{C}^*$-action which acts by dilation on $H^2(X;\mathbb{C})$ and there will be a unique algebraic structure for which weight vectors of the $\mathbb{C}^*$-action are algebraic functions. It is this algebraic structure I want.

EDIT: Since it's buried a little bit in comments, let me just put here that it's true and proven by Kaledin in this paper.

Being "affine" in this case does not make much sense, because the hyperkaehler deformation is a complex manifold, without a fixed algebraic structure. Simpson produced an example of a hyperkaehler deformation of a space of flat bundles admitting several algebraic structures, both inducing the same Stein complex structure; one of them is affine, another has no global algebraic functions. In fact, the space F of flat line bundles on elliptic curve (with an appropriate algebraic structure, defined by Simpson) is an example of such a manifold, it is biholomorphic to $C^*\times C^*$, but this biholomorphic equivalence is not algebraic, and F has no global algebraic functions.
However, you can show that a hyperkaehler deformation of a resolution of something affine has no non-trivial complex subvarieties (arXiv:math/0312520), except, possibly, some hyperkaehler subvarieties The latter don't exist, because the holomorphic symplectic form $\Omega$ on such a manifold is is lifted from the base, which is affine, hence $\Omega$ vanishes on all complex subvarieties.
• Though, I suppose it's worth asking: in the situations I'm interested in, the variety $X$ has a $\mathbb{C}^*$-action which is dilating on the affinization (as defined in arxiv.org/pdf/math/0608143.pdf), and thus I can try to put an algebraic structure $\tilde X$ by making weight vectors algebraic. Do you know if that will give the deformation an algebraic structure for which it is generically affine? Dec 12 '10 at 21:17