Let $X$ be a smooth complex algebraic variety with $H^0(X,\mathcal{O}_X) = \mathbb{C}$ and $V \subset X$ an open subvariety whose complement has codimension two. Now, let $L_{\varepsilon}$ be a line bundle on $V_{\varepsilon} = V \times Spec[\mathbb{C}[\varepsilon]/(\varepsilon ^2)$. If we denote by $j_{\varepsilon}: V_{\varepsilon} \hookrightarrow X_{\varepsilon} = X \times Spec(\mathbb{C}[\varepsilon]/(\varepsilon ^2))$ the natural inclusion, then is it true that $\tilde{L}_{\varepsilon} = (j_{\varepsilon})_* (L_{\varepsilon})$ defines a line bundle on $X_{\varepsilon}$ ? If not, is there any other approach to extend $L_{\varepsilon}$ to $X_{\varepsilon}$ ?
1 Answer
Under some conditions on $X,V$, your line bundle can be extended to $X_{\varepsilon}$. Indeed, let $\imath_X:X\hookrightarrow X_{\varepsilon}$ and $\imath_V:V\hookrightarrow V_{\varepsilon}$ be two closed immersions and $\mathcal{I}_X,\mathcal{I}_V$ be the ideal sheaves respectively. By the following exact sequence $$ 0\to \mathcal{I}_X \to \mathcal{O}^{\times}_{X_{\varepsilon}}\to \mathcal{O}^{\times}_X\to 0,$$ if we consider universal $\delta$functor theory and a natural transformation $H^0(X,)\to H^0(U,)$, we have a diagram of two exact sequences $$ \begin{aligned}H^1(X,\mathcal{I}_X)&\to \mathrm{Pic}(X_{\varepsilon})\to \mathrm{Pic}(X)\to H^2(X,\mathcal{I}_X) \\ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\downarrow \alpha&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\beta\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow \gamma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\delta \\ H^1(V,\mathcal{I}_V)&\to \mathrm{Pic}(V_{\varepsilon})\to \mathrm{Pic}(V)\to H^2(V,\mathcal{I}_V) \end{aligned}$$ (see Hartshorne Exercises 3.4.6). Note that in the case, $\mathcal{I}_X\cong \mathcal{O}_X$ as $\mathcal{O}_{X_{\varepsilon}}$modules and $\mathcal{I}_V\cong \mathcal{O}_V$ respectively.
Since $X$ is smooth, $\gamma$ is an isomorphism. Hence, if $\alpha$ is a surjection and $\delta$ is an injection, then $\beta$ is a surjection by five lemma and any line bundle on $V_{\varepsilon}$ can be extended to $X_{\varepsilon}$. Note that considering the local cohomology exact sequence $$ H^1_Z(X,\mathcal{O}_X)\to H^1(X,\mathcal{O}_X)\to H^1(V,\mathcal{O}_V)\to H^2_Z(X,\mathcal{O}_X)\to H^2(X,\mathcal{O}_X)\to H^2(V,\mathcal{O}_V),$$ where $Z:=X\setminus U$, if $H^2_Z(X,\mathcal{O}_X)=0$, then $\beta$ is a surjection (it is worth saying that if $\mathrm{codim}_X Z\ge 3$, then $H^2_Z(X,\mathcal{O}_X)=0$ by SGA2, III, Proposition 3.3).

$\begingroup$ Thanks for your answer @nariri. I think your construction is correct but it doesn't answer my first question at all. In my case, the complement of $V$ on $X$ has codimension 2. Do you have any counterexample in this case? Under such hypotheses, I strongly think that $(j_{\varepsilon})_*(L_{\varepsilon})$ is a line bundle and I'm trying to complete a proof. $\endgroup$ Sep 2 at 16:39

1$\begingroup$ Note that in the above diagram, $H^1(X,\mathcal{I}_X)\to \mathrm{Pic}(X_{\varepsilon})$ and $H^1(V,\mathcal{I}_V)\to \mathrm{Pic}(V_{\varepsilon})$ are injective because $H^0(X_{\varepsilon},\mathcal{O}^{\times}_{X_{\varepsilon}})\to H^0(X,\mathcal{O}^{\times}_X)$ and $H^0(V_{\varepsilon},\mathcal{O}^{\times}_{V_{\varepsilon}})\to H^0(V,\mathcal{O}^{\times}_V)$ are surjective. Now, assume $X$ is a smooth projective surface, and let $x\in X$ be any closed point on $X$ and $V:=X\setminus \{x\}$. Then local cohomology argument says us that $\endgroup$– naririSep 3 at 0:44

1$\begingroup$ if $H^2(X,\mathcal{O}_X)=0$, then $H^1(V,\mathcal{O}_V)\ne 0$ and $H^2(V,\mathcal{O}_V)=0$. Hence, we see that $\mathrm{Pic}(X_{\varepsilon})\to \mathrm{Pic}(V_{\varepsilon})$ is not a surjection and hence there is an example of line bundle $L_{\varepsilon}$ on $V_{\varepsilon}$ such that $(j_{\varepsilon})_*L_{\varepsilon}$ is not a line bundle on $X_{\varepsilon}$. $\endgroup$– naririSep 3 at 0:47

1$\begingroup$ It is worth saying that for any variety $X$ and any vector bundle $\mathcal{E}$ on $X$, the obstruction of $\mathcal{E}$ lies in $H^2(X,\mathcal{End}(\mathcal{E}))$ and if the obstruction is zero, then the group of isomorphism classes of the deformations of $\mathcal{E}$ is $H^1(X,\mathcal{End}(\mathcal{E}))$. See Hartshorne's Deformation Theory Theorem 2.6 and Theorem 19.1. $\endgroup$– naririSep 3 at 1:05

1$\begingroup$ Great, you are complete right. In terms of deformation theory it was clear that the question is equivalent to see that $H^1(X,\mathcal{O}_X) \rightarrow H^1(V,\mathcal{O}_V)$ is surjective. And the example of a smooth projective surface is correct. Thanks again @nariri. $\endgroup$ Sep 4 at 12:51