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Let $C$ be a complex algebraic curve. It is well known that if $L$ is a special divisor on $C$, i.e., $h^0(L) > 0$ and $h^1(L) > 0$, then $$ h^0 (L) \le \frac{1}{2} \deg L + 1. $$ Assume that $L$ is not trivial and $L \ne K_C$. The equality holds if and only if $C$ is hyperelliptic.

My question is that: if there is a nontrivial special divisor $L$ on $C$ such that $L \ne K_C$ $$ h^0 (L) = \frac{1}{2} \deg L $$ holds, then what conditions $C$ should satisfy? For example, $C$ could be hyperelliptic. Is there a complete classification?

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This questions seems unclear at the moment. First of all, every curve has a special line bundle satisfying the equality $h^0(L) = \frac{1}{2}deg(L)+1$; namely the canonical bundle. Hyperelliptic curves are unique because they also possess a line bundle which satisfies the equality which is neither the trivial bundle nor the canonical bundle. So, is your question asking which curves possess some special line bundle satisfying $h^0(L) = \frac{1}{2}deg(L)$? Or, are you asking for a list of all line bundles on all curves which satisfy the equality, akin to Clifford's theorem? –  Mike Skirvin Nov 19 '11 at 17:29
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One more comment: If you are simply looking for curves which possess a line bundle satisfying the equality, then any curve possessing a $g^1_4$ obviously works. This includes all hyperelliptic curves, but also all curves of genus at most $6$. More generally, you might try checking when the Brill-Noether number associated to $d=deg(L)$ and $r=\frac{1}{2}deg(L)-1$ is non-negative. I haven't done the calculation, but it seems that using this strategy you might find that every curve possesses a line bundle satisfying your equality? –  Mike Skirvin Nov 19 '11 at 17:37
    
Right. I should assume that $L$ is not canonical bundle. Thank you! –  Michael Zhang Nov 19 '11 at 22:41
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Have you looked at Arbarello-Cornalba-Griffiths-Harris book? I don't have it with me right now, but is the first place I'd look. –  Felipe Voloch Nov 20 '11 at 0:03

2 Answers 2

Notice first of all that if $|L|$ has a fixed part, then applying Clifford's theorem to the moving part of $|L|$ one obtains immediately that the fixed part is a point, $C$ is hyperelliptic and $|L|$ is a multiple of the $g^1_2$. So we can assume that $|L|$ is free.
Another remark is that $K_C-L$ also satisfies $h^0(K_C-L)=\deg(K_C-L)/2$, so we can also assume $\deg L\le g(C)-1$.

There exists a more precise form of Clifford's theorem, sometimes called "Clifford+" that can be useful for answering your question. The statement is the following (cf. [Beauville, "L'application canonique pour les surfaces de type g\'en\eral", Invent. Math. 55, 121-140, (1979], Lemma 5.1):

Let $C$ be a smooth curve and $L$ a divisor of $C$ with $\deg L\le g(C)-1$. Then one of the following holds:

(i) $h^0(L)\le \deg L/3+4$;

(ii) there exists a a degree 2 map $f\colon C\to \Gamma$ such that $\Gamma$ is a smooth curve and the moving part of $|L|$ is equal to $f^*|\Delta|$ where $\Delta$ is a divisor of $\Gamma$. In this case $h^0(L)\le \deg L/2+1-g(\Gamma)$.

In your case you get immediately that for $h^0(L)\ge 13$ the map given by $|L|$ factors through a double cover $C\to \Gamma$, where $\Gamma$ is either rational or elliptic. If $\Gamma$ is rational, then $C$ is hyperelliptic and $L$ is multiple of the $g^1_2$, but this is not possible since we are assuming that $2h^0(L)=\deg L$. So $C$ is bielliptic and $|L|$ is pulled back from an elliptic curve.

The cases with $r:=h^0(L)-1\le 11$ I think should be done by hand. For $r=1$ any tetragonal curve will do, as pointed out in the comments. For $r>1$, the image $C'$ of $C$ via $|L|$ has degree $\ge r$, hence if $r>2$ then the map given by $|L|$ has degree $\le 2$. If the degree is 2 then $C'$ has degree $r+1$ hence $C'$ is elliptic (it cannot be rational, otherwise the system $|L|$ would not be complete) and we are in the previous situation. If the map given by $|L|$ is birational, then you can use Castelnuovo's bound to give an upper bound for $g(C)$ in terms of $r$. In case $r=2$ there's one more possibility: $C'$ is a conic and $|L|$ gives a map of degree 3. This corresponds to the case in which $C$ is a trigonal curve and $L$ is twice a $g^1_3$.

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I'm going to assume $L$ is base point free. I think it is clear how to change what I have written in the case where there are base points (numbers goes down by the degree of the base locus). In general, if $L$ is a line bundle of degree $d$ and $h^0(C,L)= r+1$, then the Clifford index of $L$, written Cliff(L) $= d-2r$. Cliffords theorem is Cliff(L) >= 0. Your line bundle satisfies Cliff(L) = 2. Two ways to achieve that if for the curve $C$ to have a $g^1_4$ or be a plane sextic. The Clifford index of a curve is the min{cliff(L)| $h^0(L)$ & $h^1(L)$ >=2}. For a general curve $C$, Cliff(C) is the floor of (g-1)/2. So for a general curve this can't be done. It is absolutely true that A-C-G-H will have more details and I believe a careful perusal will give the classification of all curves with Clifford Index 2. I think it is only the cases I mention.

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