# 1-dimensional Linear System of Divisors of Degree 2 is Complete

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.

• 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. – Notone Apr 19 '18 at 11:32
• @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? – An Coileanach Apr 19 '18 at 18:00
• Sorry, I was referring to $\mathscr{L}$! – Notone Apr 19 '18 at 19:09
• @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? – An Coileanach Apr 19 '18 at 19:59
• "...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. – Georges Elencwajg Apr 24 '18 at 22:39

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.
• 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] – Georges Elencwajg Apr 24 '18 at 22:51