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Let $C$ be a general curve, say of even genus $g=2s$. Then $C$ has finitely many pencils $|L|$ of degree $\deg L=[g+3]/2=s+1$. Choose one such. The residual series is of degree $\deg(K_C-L)=3s-3$.

Is it always the case that $K_C-L$ is projectively normal, i.e. for all $m>0$, $S^mH^0(K_C-L)\to H^0(mK_C-mL)$ is surjective? (for general $C$)

(Ex. C-4,C-5 in Arbarello et al. "geometry of algebraic curves, vol. I", says that on general $C$ any general linear series of degree $[3g/2]+2$ is projectively normal)

Edit: of course, we need to assume ${s+m \choose m}\geq 1-2s +3m(s-1)$. So, for fixed $s$, I actually want to consider only the values of $m$ satisfying this condition.

Edit: as Jason Starr points out in the comments below this already fails in genus 4. What if we only ask for quadratically normal (i.e. only $m=2$) ? (of course then assuming $g\geq10$)

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  • $\begingroup$ That fails for $s$ equal to $2$. Every non-hyperelliptic curve $C$ of genus $4$ is a complete intersection in $\mathbb{P}^3$ of a unique quadric surface $Q$ and a cubic surface. If $C$ is general, then $Q$ is smooth, i.e., $Q\cong \mathbb{P}^1\times \mathbb{P}^1$. The two pencils $|L_i|$ of degree $s+1=3$ on $C$ are the restrictions to $C$ of the two projections $\pi_i:Q\to \mathbb{P}^1$. The residual of $|L_1|$ is $|L_2|$ and vice versa. In particular, since the residual series does not separate points, that map fails to be surjective for all $m\geq 3$. $\endgroup$
    – Jason Starr
    Commented Mar 31, 2017 at 12:05
  • $\begingroup$ The same sort of thing happens for $s=3$. The residual linear system $|L'|$ maps the genus $6$ curve $C$ to a plane curve of degree $6$ with $4$ nodes. So the linear system $|L'|$ fails to separate pairs of points (the pairs mapping to nodes), while $|mL'|$ gives a projective embedding for every $m\geq 3$. Perhaps your question has a positive answer for $s\geq 4$, since then it is possible that the associated morphism $\phi_{|L'|}:C\to \mathbb{P}^{s-1}$ is an embedding. $\endgroup$
    – Jason Starr
    Commented Mar 31, 2017 at 12:27
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    $\begingroup$ If I recall correctly, Green and Lazarsfeld wrote a paper on the issue of failure of projective normality. Essentially failure of quadratic normality is the only issue. As I recall higher m 'come for free' via the bpf pencil trick. $\endgroup$
    – meh
    Commented Mar 31, 2017 at 16:28
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    $\begingroup$ I'm continuing this as a comment, because I don't have the time to double check my memory, and because my arithmetic may be correct modulo '1/2'. As I recall, the same paper shows that failure of quadratic normality is equivalent to there existing a 2e pointed e-1 plane in the projective space given by the embedding of the curve. So, since the original divisor D is a s+1 pointed (s-1) plane, the question becomes can there exist a $g^{1+e}_{s+1+2e}$ . Using B-N, and hoping I have my numbers right, I get we need $ (e+2)(g - (s+1+2e) +e+1) \leq g$ $\endgroup$
    – meh
    Commented Mar 31, 2017 at 17:11
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    $\begingroup$ Using $ g= 2s $ and playing around I get that we need to have the inequality $s-2 \leq e $. For $s=3$ I think (hope !) this recovers what Jason said. Since we also need $ 2g-2 \geq s+1 + 2e $, we also get the inequality $ 2e \leq 3s- 5 $ . If my numbers are correct, this means your question should always fail. Please double check my, ahem, math. $\endgroup$
    – meh
    Commented Mar 31, 2017 at 17:18

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