Let $\mathcal{X} \to B$ be a flat family with some fibre $X_b \to b$. Suppose I have a coherent sheaf $F_b$ on $X_b$. When does it spread out to a sheaf $\mathcal{F}$ on $\mathcal{X}$ flat over $B$?

What about a subscheme $z \subset X_b$? Arbitrary diagrams of sheaves?

(I am only concerned with the case where everything is defined over $\mathbf{C}$, and moreover in the local case where $B$ is a disc and I am perfectly happy to take $\mathcal{X}$ to be affine. However $\mathcal{X}$ should not be assumed smooth, nor $X_b$ to even be reduced.)

  • $\begingroup$ Vivek, if $X$ is proper over $B$ then this is related to the local structure "near" $F_b$ (or the closed subscheme) viewed as a point in the corresponding moduli space/scheme for a Quot-functor or Hilbert functor. This doesn't help with non-proper $X$, nor for diagrams of sheaves. Doesn't really answer the question, but suggests what can go wrong (given the horrors one can find on Hilbert schemes). It also raises the question: are you happy to make a preliminary quasi-finite flat base change on $B$ near $b$? That may improve things. $\endgroup$
    – BCnrd
    Sep 23, 2010 at 7:30
  • $\begingroup$ Hmm, correction: for closed subschemes the link with Hilb is OK, but for "abstract" (coherent) sheaf $F_b$ we should assume projectivity to get an a-priori cohomological control of some Serre twist that will be generated by enough global sections even after coherent $B$-flat lifting, so one has an actual Quot-functor to grab onto. That's all quite abstract, so likely not going to really give a useful answer (but maybe some justified bad examples). $\endgroup$
    – BCnrd
    Sep 23, 2010 at 7:40
  • $\begingroup$ I am indeed happy to make a quasi-finite flat base change on B. $\endgroup$ Sep 23, 2010 at 7:41

2 Answers 2


Suppose that $B_n$ it the $n^{\rm th}$ infinitesimal neighborhood of $b$ in $B$; that is, if $\frak m$ is the maximal ideal of $b$ in $B$, we set $B_n := \mathop{\rm Spec} \mathcal O_B/{\frak m}^{n+1}$. If $\mathcal F_n$ is as extension of $\mathcal F_b$ to $B_n$, there is a canonically defined element of $({\frak m}^{n+2}/{\frak m}^{n+1})\otimes_{\mathbb C}\mathop{\rm Ext}_{\mathcal O_{X_b}}(\mathcal F, \mathcal F)$, called the obstruction; if this is zero, then the sheaf $\mathcal F_n$ extends to $B_{n+1}$. This depends on $\mathcal F_n$, not only on $\mathcal F_b$.

If this obstructions are always 0 (for example, if $\mathop{\rm Ext}_{\mathcal O_{X_b}}(\mathcal F, \mathcal F) = 0$), then $\mathcal F_b$ will extend to some étale neighborhood of $b$ in $B$. I don't think you can say much more in this generality.

There is a whole subject devoted to the study of this kind of problems (not only for sheaves, but for much more general objects), called deformation theory.

  • $\begingroup$ Does (how does) the obstruction depend on the family? How to compute it? I assume there's some sequence? $\endgroup$ Sep 23, 2010 at 7:42
  • 1
    $\begingroup$ My first answer was not very clear, nor correct, I edited it. The obstructions are described, for example, in Theorem 5.4 of <arxiv.org/abs/1006.0497>;, and (undoubtably) in lots of other references that I am too lazy to look up. $\endgroup$
    – Angelo
    Sep 23, 2010 at 9:09
  • $\begingroup$ A special case of what Angelo wrote: if $X\to S$ is smooth and $Z$ a smooth subscheme of $X_b$, the obstruction space is $H^1(Z,N_{Z/X})$, where $N$ is the normal sheaf. So, e.g., if $X_b$ is a surface and $Z$ a copy of $\mathbb P^1$ with $Z^2=-1$, then $Z$ always extends, but if $Z^2=-2$ then sometimes it does and sometimes it doesn't. $\endgroup$
    – inkspot
    Sep 23, 2010 at 17:16
  • $\begingroup$ @inkspot $H^1(Z, N_{Z/X})$ is the obstruction space to deform to the first order $Z$ in $X$ with $X$ fixed (embedded deformations). The correct obstruction space for the first-order extension problem is $H^2(X, T_X(- \log Z))$. However, if $H^1(Z, N_{Z/X})=0$ then there is a surjective map $H^1(X, T_X(- \log Z)) \to H^1(X, T_X)$, which means that no first-order deformation of $X$ makes $Z$ desappear. So your conclusion was correct, after all :-) $\endgroup$ Sep 23, 2010 at 20:12

If $\mathcal{F}_b$ is an invertible sheaf and $B=\textrm{Spec} \mathbb{C}[\epsilon]/(\epsilon^2)$ (first-order deformations) then the obstruction theory for deforming $\mathcal{F}_b$ described in Angelo's answer becomes very explicit. Indeed, there is the following result, whose proof can be found in Sernesi's book "Deformation of algebraic schemes", p. 147.

Set $\mathcal{L}:=\mathcal{F}_b$, $X:=X_b$.

THEOREM Given a first-order deformation $\xi$ of $X$, there is a first-order deformation of $\mathcal{L}$ along $\xi$ if and only if

$\kappa(\xi) \cdot c(\mathcal{L})=0$.

Here $\kappa$ is the Kodaira-spencer map, $c$ is the first Chern class and "$\cdot$" denotes the composition

$H^1(X, T_X) \times H^1(X, \Omega^1_X) \to H^2(X, T_X \otimes \Omega^1_X) \to H^2(X, \mathcal{O}_X)$,

where the first arrow is induced by the cup-product and the second one by the duality pairing $T_X \otimes \Omega^1_X \to \mathcal{O}_X$.

Using this, one can prove for instance that if $X$ is an Abelian variety of dimension $g$ and $\mathcal{L}$ is an ample line bundle, then $\mathcal{L}$ extends along a subspace of $H^1(X, T_X)$ of dimension $g(g+1)/2$. For $g \geq 2$, $\mathcal{L}$ does not extend to the whole of $H^1(X, T_X)$ (not even to the first-order!), since the general deformation of $X$ is not projective.


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