# Symmetric powers of curves and completion along the diagonal

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?

• For $d=2$ we have $\Delta^2=2-2g(C)$, so $\Delta$ is not ample if $g(C) \geq 1$. – 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 $$g(C)\geq 1$$.
• 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? – user127776 Feb 26 at 23:18
• 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. – abx Feb 27 at 5:03