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I am having some trouble understanding linear systems of divisors on Riemann Surfaces. I feel moderately comfortable with the notion of a complete linear system $|D|$, namely the set of effective divisors which are linearly equivalent to $D$, but I am struggling to grasp more general linear systems, even though they are (as I understand it) subspaces of $\mathbb{P}(L(D))$.

For example, I would like to show that given a linear system on a Riemann surface of genus $\geq 1$, $\mathscr{L}$ which is a $g^1_2$ (i.e. a 1-dimensional projective space of degree 2 divisors), that $\mathscr{L}$ must automatically be a complete linear system, i.e. $\exists D\in \mathrm{Div}(X)$ such that $\mathscr{L} = |D|$. [From this, I think we can argue that every $g^1_2$ is base-point free(?) which is a neat consequence.]

I believe that if there exists $D\in\mathscr{L}$ with $L(D)=\mathrm{span}\{1,f\}$ for some non-constant $f\in L(D)$, then we're done, since we can take $\mathscr{L} = \{D+\mathrm{div}(n\cdot f): n\in \mathbb{N}\}$. However, if we find $D\in \mathscr{L}$ with $l(D)> 2$, I don't understand how we can find an appropriate basis of $\mathscr{L}$. Can we just "subtract points" (i.e. consider divisors of the form $D - P_1 - \ldots -P_r$) until the resulting divisor $E$ satisfies $l(E) = 2$? This approach feels rather arbitrary to me, and I'm not totally convinced why the corresponding non-constant $f\in L(E)$ should generate the space $\mathscr{L}$.

As can be ascertained from the above, I don't have a good intuition of why we should consider more general linear systems than complete ones, and I would really appreciate any insights anyone could offer into the subject.

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  • $\begingroup$ If you take a rational section of the line bundle, i.e. just take a section which might have also poles, then this defines a (possibly non effective) Divisor $D$ by looking at the zeros and poles of the section. Your line bundle will then be isomorphic to $L(D)$ (here I mean $L(D)$ as the line bundle). This is basically just writing out the 1:1 correspondence between Divisors and Line bundles. $\endgroup$ – Notone Apr 19 '18 at 11:32
  • $\begingroup$ @Notone, thanks for your comment. Since I'm not entirely familiar with this correspondence between line bundles and divisors (I've seen the notion of a canonical divisor and from what I can see in Miranda's book, this correspondence seems to generalise that to arbitrary line bundles?), I was wondering if you could maybe clarify which line bundle you mean in the first sentence of your comment: do you mean that $\mathscr{L}$ is a line bundle, or are you talking about the cotangent bundle (so that $D$ is the canonical divisor) or something else entirely? $\endgroup$ – An Coileanach Apr 19 '18 at 18:00
  • $\begingroup$ Sorry, I was referring to $\mathscr{L}$! $\endgroup$ – Notone Apr 19 '18 at 19:09
  • $\begingroup$ @Notone, thanks for clarifying. Sorry to be dense here, but I'm not sure I fully understand why we wouldn't have a scenario where $\mathscr{L} \subsetneq L(D)$, i.e. why should $\mathscr{L}$ equal all of $L(D)$, and not just some proper linear subspace? $\endgroup$ – An Coileanach Apr 19 '18 at 19:59
  • $\begingroup$ "...a $g^1_2$ (i.e. a 1-dimensional vector space of degree 2 divisors)": No, it is a a 1-dimensional projective space of degree 2 divisors. $\endgroup$ – Georges Elencwajg Apr 24 '18 at 22:39
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Following a conversation with someone smarter than me, I will try to post an answer to my immediate question regarding completeness of $g^1_2$ linear systems (although I still lack good intuition about general linear systems).

Suppose $D \in \mathscr{L}$. Then $D$ is an effective divisor of degree 2, so it is of the form $P + Q$ for some $P,\ Q\in X$. By assumption, $\dim(\mathscr{L}) = 1$, so there is at least some meromorphic function contained in $L(P+Q)$, so $\ell(P + Q) \geq 2$; therefore, it only remains to show that $\ell(P+ Q) < 3$.

It is shown in Miranda's book that $\ell(P + Q) \in \{\ell(P),\ \ell(P) + 1\}$, $\forall Q\in X$. On the other hand, since the genus of $X$ is at least 1, it follows that $\ell(P) = 1$ (otherwise $X\cong \mathbb{P}^1$). Hence $\ell(P + Q) \leq 2$, so we find that $\ell(P + Q) = 2$, and $\mathscr{L}$ is complete.

In the comments, it was mentioned that there is a correspondence between line bundles and linear systems, but it isn't clear to me whether this correspondence only holds for complete linear systems, or whether it also works for more general systems.

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    $\begingroup$ It is a general fact that for an effective non-zero divisor $D$ of degree $d$ on a compact Riemann surface of positive genus we have $dim L(D) \leq deg(D)\;$ [Miranda Prop. 3.16 page 151+Problem I page 153] $\endgroup$ – Georges Elencwajg Apr 24 '18 at 22:51
  • $\begingroup$ Thanks for pointing this out. I hadn't gone through that exercise in Miranda yet, but it obviously yields a much more efficient solution than what I had written above. $\endgroup$ – An Coileanach Apr 25 '18 at 13:25

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