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

$$\DeclareMathOperator\Spec{Spec}$$Let $$L \rightarrow X$$ be an ample line bundle over $$X$$ which is 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 sense of Kodaira and Spencer? If one considers 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?

• 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. Oct 11, 2011 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? Oct 11, 2011 at 17:59
• I hope my question is more clear now?
– carl
Oct 12, 2011 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 ? Oct 12, 2011 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, 2011 at 13:54

What you are typically looking for is a "true" deformation over an algebraic or analytic 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 analytic versions 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.