# 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.

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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.

Therefore, a typical fiber of such a deformation is Stein.

Indeed, a hyperkaehler deformation of a resolution of something affine remains holomorphically convex. To see this if you produce a function which is strictly plurisubharmonic outside of a compact set (we have such a function, because we started from something affine), and apply the Remmert reduction.

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That's a very useful answer. Thanks a lot. –  Ben Webster Dec 12 '10 at 20:44
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? –  Ben Webster Dec 12 '10 at 21:17
In this case, it is affine, but I don't know a precise reference. I think it is due to Kaledin. I am surprised it's not stated in the paper you cited. For an easy proof, notice that your C^*-action gives a C^*-action with positive weights on a deformation of your original (singular) affine manifold. Therefore the total space of its deformation (which has the same generic fibers as the hyperkaehler deformation of its blow-up) is affine, hence its fibers are affine, too. –  Misha Verbitsky Dec 13 '10 at 0:24
It's not stated explicitly in the paper I referenced, but I see now that it is clear from the existence and properties of the twistor deformation. –  Ben Webster Dec 14 '10 at 6:58