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LSpice
LSpice
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What you are typically looking for is a "true" deformation over an algebraic or analitycanalytic 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 versionanalytic 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.

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.

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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Miguel González
Miguel González
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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.