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Let $R$ be a discrete valuation ring with residue field $k$, an algebraically closed field of characteristic zero and $\pi:X\to \mbox{spec}(R)$ a smooth, projective family of surfaces. Denote by $X_0$ the special fiber in the family. Suppose there exists a curve $C$ in $X_0$ which does not lift as a subcurve of $X$, meaning there does not exist a subcurve of $X$, flat over $R$, which restricts to $C$ on the special fiber.

Is it possible that for the infinitesimal, first order deformation $X_0'$ of $X_0$ along spec$(R)$, the curve $C$ lifts to a local complete intersection subcurve of $X_0'$? If so, is there any general condition when this does not happen?

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up vote 6 down vote accepted

That can happen. Here is one source of examples. Consider $\mathbb{P}^3_k = \text{Proj}\ k[s,t,u,v]$. Let $d\geq 4$ be an integer, and consider hypersurfaces $X\subset \mathbb{P}^3$ of degree $d$, i.e., $X$ is the zero locus of a degree $d$ homogeneous polynomial, $$F(s,t,u,v) = \sum_{\alpha+\beta+\gamma+\delta=d} c_{\alpha,\beta,\gamma,\delta} s^\alpha t^\beta u^\gamma v^\delta.$$ The parameter space for these hypersurfaces is the projective space $$P = \mathbb{P} H^0(\mathbb{P}^3,\mathcal{O}(d)) = \{ [c_{\alpha,\beta,\gamma,\delta}]_{\alpha+\beta+\gamma+\delta=d} \} .$$ For every line $L\subset \mathbb{P}^3$, e.g., for the line $C=\text{Zero}(u,v)$, the parameter space $P_L\subset P$ for hypersurfaces that contain $L$ is a codimension $d+1$ linear subspace, e.g., $$ P_C = \text{Zero}(c_{\alpha,\beta,0,0})_{\alpha+\beta=d}.$$ This parameter space contains points parameterizing smooth hypersurfaces, e.g., for $L=C$, one example is $$ X_0 = \text{Zero}(us^{d-1}+vt^{d-1}+u^d+v^d),$$ (this works when the characteristic is prime to $d$ and $d-1$, but there are other examples that work for the excluded characteristics).

In particular, since there is only a $4$-dimensional family of lines $L$ in $\mathbb{P}^3$, the union over all $L$ of $P_L$ is a closed subvariety of $P$ of codimension $d+1-4$. Since $d\geq 4$, this codimension is $\geq 1$. The complement of this proper closed subvariety is a Zariski dense open subset $U$. Thus, there exists a point $[b_{\alpha,\beta,\gamma,\delta}]$ in $U$ that parameterizes a smooth hypersurface $X_\infty\subset \mathbb{P}^3$ that contains no lines. (If need be, there are explicit equations for such a hypersurface $X_\infty$.)

Now choose $[c_{\alpha,\beta,\gamma,\delta}]\subset P_C$ parameterizing a smooth hypersurface, e.g., the hypersurface $X_0$ from above. For the DVR $R=k[z]_{\langle z \rangle}$, for every integer $r\geq 1$, form the homogeneous polynomial $$ F_{r,z}(s,t,u,v) = t^r(us^{d-1}+vt^{d-1}+u^d+v^d) + z^r\sum_{\alpha+\beta+\gamma+\delta = d} b_{\alpha,\beta,\gamma,\delta} s^\alpha t^\beta u^\gamma v^\delta. $$ The zero scheme of $F_{r,z}$ in $\mathbb{P}^3_R=\text{Proj}\ R[s,t,u,v]$ is a hypersurface $X$ in $\mathbb{P}^3_R$ such that $X_0$ contains $C$, and to order $r-1$ $X$ equals $X_0 \times_{\text{Spec}\ k}\text{Spec}\ R/\langle z^r \rangle$, so contains $C\times_{\text{Spec}\ k}\text{Spec}\ R/\langle z^r \rangle$. However, the generic fiber contains no line.

Note that $F_{r,z}$ is the base change of $F_{1,z}$ by the finite flat morphism of DVRs $\text{Spec}\ R \to \text{Spec}\ R$ obtained from the finite flat ring homomorphism $z\mapsto z^r$. Of course there are other examples that do not factor through such a base change. However, this shows a conceptual difficulty with your general question: you will need to rule out all examples that factor through a base change.

One way to do this would be to demand that the Kodaira-Spencer map $$\text{Hom}_k(\mathfrak{m}_R/\mathfrak{m}_R^2,k)\to H^1(X_0,T_{X_0/k}),$$ is nonzero. Then the image of the Kodaira-Spencer map is a $1$-dimensional subspace. When $C$ is a local complete intersection curve, the short exact sequence of sheaves of differentials, $$ 0 \to \mathcal{I}_{C/X_0}/\mathcal{I}^2_{C/X_0} \to \Omega_{X_0/k}|_{C} \to \Omega_{C/k} \to 0 $$ gives rise to a long exact sequence of groups $\text{Ext}^m_{\mathcal{O}_C}(\bullet,\mathcal{O}_C)$, one part of which is a connecting map $$ H^1(C,T_{X_0/k}|_C) \to \text{Ext}^1_{\mathcal{O}_C}(\mathcal{I}_{C/X_0}/\mathcal{I}_{C/X_0}^2,\mathcal{O}_C). $$ If the image of the Kodaira-Spencer map under this connecting map is nonzero, then there is no flat lift of $C$ to the first-order deformation $X_0'$.

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Thank you very much. The answer is very very helpful. +100 votes. – user43198 Mar 26 at 14:20

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