# Bound on the dimension of $L(D)$ for a compact Riemann Surface

this is homework, but after banging my head against the wall for a day, I feel justified asking for a hint.

Let $X$ a compact Riemann Surface, $D,E \in Div(X)$ divisors on $X$ with $degE\geq 0$.

Show that $dimL(D+E) \leq dimL(D) + deg(E)$.

I think if I can show that for $E = 1x$, $dim(D+1x) \leq DimL(D) + 1$, then this follows by induction, but I don't even know how to show this. Ideas?

Please note that this class is rather elementary, so I am seeking a rather elementary answer.

Edit: Based off the hint, let $f_1$ and $f_2$ linearly independent in $L(D+x)$ and let the degree of $x$ in $D$ be given by $k$. Then in local coordinates on a neighborhood of $x$ we can write

$$f_1(z) = a/(z-x)^{k+1} + h(z)$$ and $$f_2(z) = b/(z-x)^{k+1} + g(z)$$

for $h,g\in L(D)$ and $a,b\in \mathbb{C}$. Observe that for any values of $a$ and $b$ there are coefficients $\alpha, \beta \in \mathbb{C}$ both not zero such that $\alpha f_1 +\beta f_2 \in L(D)$ (this requires enumerating a couple cases and showing we can get rid of the order $k+1$ pole) so at least one of the two is in $L(D)$. But then the dimension of $L(D+x)$ is at most one more than $L(D)$ since we have shown there is no two lin. indep. functions in $L(D+x)$ that are both not in $L(D)$.

To show $dimL(D+E) \leq dimL(D) + deg(E)$ for any positive $E$ we simply use induction.

I can also prove this using the same rough argument about the local laurent expansion using the other hint, but I will add that later. It should be exact because we can map $f(z) = a/(z-x)^{k+1} + h(z)$ to $a$ and the kernel will be $h\in L(D)$.

• The ultimate goal is to show that $L(D)$ is finite dimensional, but I have not even gotten past this step. – AnonymousCoward Mar 22 '11 at 2:54

## 2 Answers

Suppose that if you have two functions $f_1$ and $f_2$ in $L(D+x)$, show that for some scalars $\alpha$ and $\beta$, not both zero, that $\alpha f_1 + \beta f_2$ belongs in $L(D)$. (For this you have to go back to the very definitions of $L(D)$ and $L(D+x)$, and think about what it means for a meromorphic function to belong to these spaces.) From this, show that $\dim L(D + x) \leq \dim L(D) + 1$.

• there appears to be a typo in your last sentence. – AnonymousCoward Mar 22 '11 at 5:57
• @GottfriedLeibniz: Indeed; thanks! Regards, – Matt E Mar 22 '11 at 11:41

Hint: construct an exact sequence $0 \to L(D) \to L(D+x) \to \mathbb{C}$ where the last map is the residue at $x$ (at least if $x \notin supp(D)$; the argument can be modified if $x \in supp(D)$ accordingly).

• In the algebraic category (i.e. for smooth projective curves), the above argument still works, though one often appeals to general coherence theorems for projective (or more generally proper) varieties. (Actually, one can prove such coherence theorems in the analytic category too, which imply the finite-dimensionality of these spaces.) – Akhil Mathew Mar 22 '11 at 4:12