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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?

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    $\begingroup$ 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
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$\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 $g(C)\geq 1$.

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  • $\begingroup$ 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

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