Given a smooth curve $C$, denote by $\text{Sym}^d(C)$ its $d$-th symmetric power. Let $\Delta$ be the diagonal subvariety which is defined as the codimension $1$ subvariety that at least two of the points coincide. Let $\text{Sym}^{d-1}(C)$ be the closed variety that its embedding is given by adding some extra fixed point. Let $Z$ be either $\Delta$ or $\text{Sym}^{d-1}(C)$.

Question 1. Is $Z$ an ample divisor?

Question 2. Is there any interesting interpretation of coherent modules/vector bundles on the formal completion of $Z$ in $\text{Sym}^d(C)$?

Edit: Is it true that for $Z=\text{Sym}^{d-1}(C)$, the formal completion is going to be a bundle on $Z$? Is it like completing a zero section of a vector bundle?

  • 1
    $\begingroup$ For $d=2$ we have $\Delta^2=2-2g(C)$, so $\Delta$ is not ample if $g(C) \geq 1$. $\endgroup$ – Francesco Polizzi Feb 26 at 5:46

$\operatorname{Sym}^{d-1}(C) $ is ample, and $\Delta $ is not unless $C\cong \mathbb{P}^1$. To see this, consider the finite map $\pi :C^d\rightarrow \operatorname{Sym}^{d}(C) $, and the projections $\pi_i:C^d\rightarrow C$. Your divisor $Z$ is ample if and only if $\pi ^*Z$ is ample. Now $\pi ^*\operatorname{Sym}^{d-1}(C) =\sum p_i^*[p]$, where $p$ is your fixed point. Since $[p]$ is an ample divisor on $C$, $\operatorname{Sym}^{d-1}(C) $ is ample.

On the other hand, take $d=2$; then $\pi ^*\Delta =2 \Delta '$ where $\Delta '$ is the diagonal in $C^2$, and $\Delta '^2=2-2g$ is $\leq 0$ if $g(C)\geq 1$.

  • $\begingroup$ I discovered this paper :arxiv.org/pdf/1707.09484.pdf. On page 7 they claim the ampleness of $\text{Sym}^{d-1}$ holds in char $0$ and not necessarily in char $p$. That has made me confused, does your argument depend on characteristics? $\endgroup$ – user127776 Feb 26 at 23:18
  • $\begingroup$ No, the argument works in any characteristic. I think the authors of the preprint you mention add char. 0 because they quote [ACGH] which is written over $\Bbb{C}$; they say nothing about char. $p$. Note that [ACGH] gives the same argument as mine, and another one based on Nakai's criterion; both work in arbitrary characteristic. $\endgroup$ – abx Feb 27 at 5:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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