# from first-order deformation to complex deformation of a pair $(X,L)$

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

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 $(T, 0)$:
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.