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Let $L \rightarrow X$ be an ample line bundle over $X$ a compact complex manifold. Suppose that I have a first-order deformation of the pair $(X,L)$. When does this first-order deformation gives rise to a complex deformation of the pair $(X,L)$ in the sence of Kodaira and Spencer? If one consider the extension $$ 0 \rightarrow \mathcal{O}_X \rightarrow \mathcal{E}_L \rightarrow T_X \rightarrow 0$$ defined by the first Chern class of $L$, then $H^2(X, \mathcal{E}_L)$ is an obstruction space for the functor of infinitesimal deformations of $(X,L)$. Is $H^2(X, \mathcal{E}_L)$=0 sufficient to ensure the existence of an associated complex deformation?

I took the definitions from Sernesi's book. I call a deformation of $X$ a flat surjective morphism $$\mathcal{X} \rightarrow \Delta$$ with $X \rightarrow Spec(\mathbb{C})$ the central fiber. Then the deformation is : - "first-order" if $\Delta=Spec(\mathbb{C}[\epsilon])$ - "infinitesimal" if $\Delta=Spec(A)$ with $A$ a local artinian $\mathbb{C}$-algebra. - "complex" if $\Delta$ is a complex manifold. In that case the definition of the morphism $$\mathcal{X} \rightarrow \Delta$$ is to be a proper submersion.

The point that I do not understand is how do I go from infinitesimal to complex?

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Could you given the definitions of "a first-order deformation" and of "a complex deformation" ? I have a definition in mind for the first concept but I don't see what the second one is. –  Damian Rössler Oct 11 '11 at 15:48
    
I'm with Damian, I don't quite understand what you're asking. Are you asking about the difference between formal and unobstructed deformations? I.e. suppose that an element of $H^1$ gives an infinitesimal direction into which we can deform our manifold, but that the associated obstruction in $H^2$ is non-zero. Would that, for you, be an "infinitesimal deformation" and not a "complex deformation", while one where the obstruction in $H^2$ is zero would be both? –  Gunnar Þór Magnússon Oct 11 '11 at 17:59
    
I hope my question is more clear now? –  carl Oct 12 '11 at 7:37
    
An infinitesimal deformation contains much less information than a complex deformation. Indeed a complex deformation will provide you with infinitesimal deformations (as complex analytic spaces, not as projective schemes) of any order. How could one go from infinitesimal to complex ? But maybe I am missing your point ? –  Damian Rössler Oct 12 '11 at 10:10
    
"How could one go from infinitesimal to complex ?". This is the question I am asking. Do you know if there exists a criterion that ensures that an infinitesimal deformation gives rise to a complex one? –  carl Oct 12 '11 at 13:54

1 Answer 1

up vote 3 down vote accepted

What you are typically looking for is a "true" deformation over an algebraic or analityc pointed curve $(T, 0)$ such that the tangent vector to $T$ at $0$ corresponds to the infinitesimal deformation you have.

The functor of infinitesimal deformations of a pair $(X, L)$ admits a semiuniversal formal deformation (see e.g. Sernesi's book page 146). Assuming $(X, L)$ is unobstructed (which is equivalent to saying that the base of the semiuniversal formal deformation is the Spec of a power series ring) then $H^1(X, \mathcal E_L)$ is the tangent space to the base of this deformation.

Now in order to construct $(T, 0)$:

1) you can sometimes make an ad--hoc argument using some suitable Hilbert space, or

2) you can use the extremely powerful Artin's approximation and algebraization theorems (see Sernesi's book pages 87, 88). Artin's theorems ensures the existence of a "true" algebraic (I think there are also analityc version of Artin's theorems) deformation provided the semiuniversal formal deformation is effective (see Sernesi's book page 82, for a Grothendieck's criterion for effectiveness). This way you get an algebraic deformation over a smooth base, whose tangent space at a point $0$ is $H^1(X, \mathcal E_L)$, so you can get a path through $0$ pointing at your infinitesimal deformation.

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Thanks Miguel, that is what I was looking for! –  carl Oct 13 '11 at 7:16

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