$\DeclareMathOperator\Spec{Spec}$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?

  • 2
    $\begingroup$ 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. $\endgroup$ Oct 11, 2011 at 15:48
  • $\begingroup$ 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? $\endgroup$ Oct 11, 2011 at 17:59
  • $\begingroup$ I hope my question is more clear now? $\endgroup$
    – carl
    Oct 12, 2011 at 7:37
  • $\begingroup$ 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 ? $\endgroup$ Oct 12, 2011 at 10:10
  • $\begingroup$ "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? $\endgroup$
    – carl
    Oct 12, 2011 at 13:54

1 Answer 1


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 $(T, 0)$:

  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 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.


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