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I have a simple question. Let $C$ be a compact Riemann surface of genus, say $g >= 2$, to avoid silly cases.

I think it should be true, but I want to prove the following concretely:

"there exists a divisor $D$ on $C$ of degree $g-1$, that is non-special."

(For those who do not know what special divisors are: a divisor is called special if it has $h^0 (D) >0$ and $h^1 (D) >0$.)

Notice that by the Riemann-Roch, for this degree $g-1$ case we immediately have $h^0 (D) = h^1 (D) = 0$. This is, in fact, equivalent to $D$ being non-special, when $\deg D = g-1$.

Is there an interesting (or any) way to prove this? I believe it should be fairly easy, and maybe I am very dumb so that I can't immediately produce a proof.

More generally, if this is possible, if the degree is a given $d$, when do we see that there exists a non-special or special divisor of given degree $d$ on a given compact Riemann surface?

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    $\begingroup$ Divisors of degree d modulo rational equivalence form a g dimensional algebraic variety, Pic^d(C), and there is a natural morphism C^d to Pic^d(C) given by sending an n-tuple (p_1,...p_d) to the divisor $\sum_i [p_i]$. The image consists exactly of the effective divisors. The morphism C^d to Pic^0(C) is generically finite (for d $\leq$ g) so there always exist non-special divisors for d < g. (The canonical reference for special divisors is the book of Arbarello, Cornalba, Griffiths, Harris.) $\endgroup$
    – naf
    Commented Apr 19, 2010 at 9:22
  • $\begingroup$ @unknown: For answering the original question, it is not even important that the morphism is generically finite, since we just need an upper bound on the dimension of image. $\endgroup$
    – t3suji
    Commented Apr 19, 2010 at 12:57

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Take $g+1$ general points $p_1, \dots, p_{g+1}$ on your curve. The divisor $p_1+ \cdots +p_g - p_{g+1}$ is non-special. The proof is easy from the following lemma: if $D$ is a divisor such that $\mathrm{h}^0(D) > 0$, then $\mathrm{h}^0(D - p) = \mathrm{h}^0(D) - 1$ for all but finitely many points $p$. First you use Riemann-Roch to deduce that $\mathrm{h}^0(p_1+ \cdots +p_g) = 1$, then you apply the lemma once again to $p_1+ \cdots +p_g$. This also works for $g = 0$ and $g = 1$.

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  • $\begingroup$ Ah, thank you very much! There was a very easy way! $\endgroup$
    – Jinhyun Park
    Commented Apr 19, 2010 at 11:36

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