MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a smooth variety and $D \subset X$ be an effective, reduced, irreducible divisor. My question is the following.

1.What is the first order deformation and obstrution for the pair $(X, D)$?

2.In particular, if $D$ is a singular divisor and has a global smoothing such that the line bundle $\mathcal{O}(E)\vert_E$ also extends to a general smoothing, then under what conditions can we deform the pair so that the induced deformation of the pair gives a smoothing of $D$.

share|cite|improve this question
The answer to point 1. is $H^1(X, T_X(-\log D))$ and $H^2(X, T_X(-\log D))$, respectively. – Francesco Polizzi Nov 8 '10 at 20:02
up vote 4 down vote accepted

From another point of view, the isomorphism classes of first order deformations of the pair (X,D) are isomorphic to the 1st hypercohomology group of the 2 step complex from the tangent sheaf of X to the normal sheaf of D in X.

The 1st order deformations of just D are given by the 1st hypercohomology group of the 2 step complex from the restriction of the tangent sheaf of X to D, to the same normal sheaf of D in X.

By the long exact sequence induced by the natural map of these complexes, the forgetful map from T^1(X,D) to T^1(D) is an isomorphism whenever the first 2 hypercohomology groups vanish for the 2 step complex from the ideal sheaf of D in X times the tangent sheaf of X, to zero.

This happens for instance if X is a principally polarized abelian variety of dimension 3 or more, and D is the theta divisor, by Kodaira (or abstractly Mumford) vanishing.

ref: Compositio Math 76(1990) pp. 367-398.

share|cite|improve this answer
Thank you for the answer. So you are saying that if $H^1(T_X(-D))=H^2(T_X(-D))=0$, then the forgetful map from the deformation of the pair $(X,D)$ to the deformation of $D$ is isomorphic ( since the two term complex is quasi-isomorphic to $T_X(-D)$.) – Zhiyu Nov 9 '10 at 15:33
Yes, this is true, and it also follows from the short exact sequence $0 \to T_X(-D) \to T_X(- \log D) \to T_D \to 0$. – Francesco Polizzi Nov 9 '10 at 15:58
...which is nothing but Roy's answer, after all :-) – Francesco Polizzi Nov 9 '10 at 16:05
But does the exact sequence [ 0 \to T_X(-D) \to T_X(-\logD) \to T_D \to 0 ] also hold when $D$ is singular? – Zhiyu Nov 9 '10 at 21:12
Yes, see Sernesi's book, p. 177. But notice that in the general case $H^1(D,T_D)$ parametrizes only equisingular first-order deformations. The space parametrizing all first-order deformations is actually $\textrm{Ext}^1(\Omega^1_D,\mathcal{O}_D)$, which obviously coincides with $H^1(D,T_D)$ when $D$ is smooth. – Francesco Polizzi Nov 10 '10 at 8:39

Let me expand a little bit my previous comment. For the sake of simplicity, I will assume first that $D$ is smooth.

Then one has a short exact sequence

$0 \to T_X(- \log D) \to T_X \to N_{D|X} \to 0 \quad (*)$,

where $T_X(- \log D)$ is the sheaf of tangent vectors on $X$ which are tangent to $D$. The group $H^1(X, T_X(\log D)$ classifies the first-order deformations of the pair $(X, D)$, and the obstructions lie in $H^2(X, T_X(-\log D))$.

Taking cohomology in $(*)$, we obtain

$H^0(D, N_{X|D}) \stackrel{\alpha}{\to} H^1(X, T_X(-\log D)) \to H^1(X, T_X) \stackrel{\beta}{\to} H^1(D, N_{D|X}) \to H^2(X, T_X(-\log D)).$

It is well known that the group $H^0(D, N_{D|X})$ classifies first-order embedded deformations of $D$ in $X$, with $X$ fixed. Therefore the map $\alpha$ naturally associates to every such a deformation the corresponding deformation of the pair $(X, D)$.

Moreover, we can interpret the map $\beta$ as the obstruction to lifting an abstract first-order deformation of $X$ to a deformation of the pair $(X, D)$.

For instance, let $X$ be a smooth projective surface and $D$ a nonsingular rational curve with $D^2=-1$. Then $H^1(D, N_{D|X})=0$, so every abstract deformation of $X$ lifts to a deformation of the pair $(X, D)$.

If instead $D^2=-2$, there are in general abstract deformations of $X$ which do not come from deformations of the pair. For example, the general deformation of a Kummer surface (which contain 16 $(-2)$-curves) is a $K3$ surface without $(-2)$-curves.

When $D$ is singular, the same arguments hold with $N_{D|X}$ replaced by the so-called equisingular normal sheaf $N'_{D|X}$. See Sernesi's book [Deformation of algebraic schemes, Chapter 3] for further details.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.