Let $X$ be a smooth, projective variety, $E$ a rank $r$ locally free sheaf on $X$. Fix a closed embedding $i:X \hookrightarrow \mathbb{P}^N$ and denote by $\mathcal{O}_X(m)=i^*\mathcal{O}_{\mathbb{P}^N}(m)$. For $d \gg 0$, take $r$ general global sections $s_1,...,s_r \in H^0(E(d))$. Denote by $Z$ the locus of points on $X$ at which the sections $s_1,...,s_r$ are linearly dependent. Denote by $D$ the singular locus of $Z$, $\tilde{X}$ the blow up of $X$ along $D$ and $\tilde{Z}$ the strict transform of $Z$ in $\tilde{X}$. Denote by $E$ the exceptional divisor in $\tilde{X}$, $F:=\tilde{Z} \cap E$ and $\pi:F \to D$ the natural morphism.

I am looking for conditions under which the morphism $\pi$ is smooth and projective. The vector bundle $E$ I am interested in, is the dual of the normal bundle $\mathcal{N}_{X|\mathbb{P}^N}$. Any idea/reference in this direction will be very helpful.

  • $\begingroup$ If $\text{dim}(X)\geq 9$, and if the sections are sufficiently generic, then $F\to D$ is smooth away from a closed subset that has codimension $\geq 9$ in $X$. However, the singular locus of $F\to D$ will be nonempty. $\endgroup$ – Jason Starr Apr 24 '17 at 12:16
  • $\begingroup$ @JasonStarr I found a Bertini type theorem (Proposition 7.4 of Eisenbud and Harris's intersection theory) which states that if the determinant $L$ of $E$ is very ample then the zero locus $Z$ mentioned above is smooth. Do I understand this correctly or I am missing something? $\endgroup$ – user43198 Apr 25 '17 at 10:25
  • $\begingroup$ My copy of '3264 and all that' does not say that for Proposition 7.4, but I have an early draft, and things may have changed in the published version. Certainly a "sufficiently generic" section of a vector bundle has smooth zero locus. But $Z$ is not the zero locus of a sufficiently generic section of a vector bundle. It is the zero locus of the section $s_1\wedge \dots \wedge s_r$ of the invertible sheaf $\bigwedge^r (E(d))$. That is not a general section. For yourself, compute the singular locus of the determinant $s_{1,1}s_{2,2}-s_{1,2}s_{2,1}$ of a general $2\times 2$ matrix. $\endgroup$ – Jason Starr Apr 25 '17 at 12:42
  • $\begingroup$ @JasonStarr Thanks a lot. That clears my doubt. $\endgroup$ – user43198 Apr 25 '17 at 12:53

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