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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 $D$.

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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

2 Answers 2

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.

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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.

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