# Most divisors on a curve aren't special?

I have a generic smooth curve $$C$$ of genus $$g$$ and fixed multiplicities $$a_1, \dots, a_n \geq 0$$ with $$\sum a_i = g+1$$.

Q1 : For generic marked points $$p_1, \dots, p_n \in C$$, must $$\sum a_i p_i$$ be a non-special divisor?

If $$n$$ is one, I just have to avoid the finitely many Weierstrass points, and I'm looking for an analogy of this for special divisors. If the union of the supports of all special divisors was finite, that'd be great, but the spaces $$G^r_d$$ can be greater dimension than the curve. Even though I can make the support of the divisor generic, the multiplicities are fixed.

Q2 : For generic marked points and any $$b_1, \dots, b_n$$, $$0 \leq b_i < a_i$$, must $$\sum b_i p_i$$ be non-special? What if $$\sum b_i = g$$?

I was reading Akhil Matthew's notes on a course by Joe Harris, and the proof that there are finitely many Weierstrass points is "something that eventually becomes obvious." I'm trying to generalize this, so I'd appreciate being made aware if there is a related result. A paper casually claims that $$H^1(\mathcal{O}(D)) = 0$$ for a generic divisor of degree $$g$$ and I'm trying to check it.

Apologies that I'm quite inexperienced with curves.

• The "fact" that there are only finitely many Weierstrass points is false for some curves (in positive characteristic) so you need to add some assumptions. Do you mean to take the curve as generic (in moduli) as well? Jan 1 at 8:37
• Happy New Year Leo. If your curve has generic moduli, then this should follow from the Harris-Morrison variant of the Brill-Noether theorem for linear series with specified ramification at a generic configuration of points on the curve. This variant is given in "Moduli of Curves" by Harris and Morrison (I do not have the book with me at present). Extensions of that theorem to "moving configurations of points" were proved by Rebecca Lehman, with a big generalization by Gavril Farkas later. Jan 1 at 11:40
• @JasonStarr What generic smoothness argument are you thinking of? Jan 10 at 17:19
• @WillSawin. "What generic smoothness argument are you thinking of?" On a smooth curve $X$, for a linear system $(L,V\subseteq H^0(X,L))$, the zero scheme of the "Wronskian" section of the associated bundle $\text{det}\mathbb{P}^r(L) \cong \Omega_X^{\otimes r(r+1)/2}\otimes L^{\otimes r}$ is nonzero, where $r=\text{dim}V$ and the characteristic is zero (or just "sufficiently large"). This can be proved by using "generic smoothness" on a suitable bundle over $X$. The proof in characteristic $0$, as well as counterexamples for small characteristics, are in a paper of Dan Laksov. Jan 10 at 19:18
• I'd think that if $\sum_i a_i p_i$ were special for generic $\vec p = (p_1,p_2,\ldots,p_n) \in C^n$ then it would remain special for any $\vec p$ on the diagonal $(p,p,\ldots,p)$. But we already know that the resulting divisor $(g+1)p$ is not special for generic $p$. Does anything more need to be done? (Meanwhile Weierstrass points come from special $g\, p$, not $(g+1)p$, though it's true that $(g+1)p$ can't be special unless $g\, p$ is.) Jan 18 at 4:43