Basis of meromorphic $1$-forms on compact Riemann surface.

I am trying to solve the following exercise but I do not really know how to proceed.

For an integral divisor $$D$$ and any compact Riemann surface $$M$$, describe a basis of the space $$\Omega(-D)$$.

Where $$\Omega(-D) = \{\omega\in\mathcal{M}\Omega^1(M)\mid (\omega)\geq -D\}$$. Since $$D = \sum_v s_vp_v$$ is integral i.e. $$s_v>0$$ for at least one $$v$$, the elements of $$\Omega(-D)$$ are meromorphic $$1$$-forms which have a pole of order at most $$s_v$$ at $$p_v$$. This clearly contains all holomorphic differentials and I know how to generate those, but I have no idea about any classification of the strictly meromorphic differentials in terms of their dimension.

I have only seen proven the existence of such differentials with a single pole of higher order, or of differentials with two simple poles in the book of Wilhelm Schlag on A course in Complex Analysis and Riemann surfaces.

I also tried finding the dimension using the Riemann-Roch theorem, but for that I would need to find the dimension of $$L(-D)$$ the meromorphic functions on $$M$$ which only have poles at $$p_v$$ of order at most $$s_v$$. But I am not sure how to formally do this aswell.

Does anyone know how to find this basis $$\Omega(-D)$$?

• Note that you're using the Riemann-Roch theorem for the divisor $-D$, so you should consider $L(D)$ and not $L(-D)$. What can you say about $L(D)$ if $M$ is compact? Jun 7, 2020 at 15:31
• What kind of meromorphic differentials do you know? Can you construct a differential with a double pole at $p_1$? A triple? A single? Jun 7, 2020 at 15:34
• Suppose $D = p_1 + \cdots + p_5$. How many linearly independent non-holomorphic meromorphic differentials can you construct that lie in $\Omega(-D)$? (Hint: the answer is not $\binom{5}{2} = 10$). What happens when you repeat one point, e.g. $D + p_1$? Jun 7, 2020 at 16:41
• @GillesCastel Not sure what you can say about $L(D)$ if $M$ is compact. I suppose you would need meromorphic functions with zeros of multiplicity $s_v$ at $p_v$ to generate the whole space, but not sure how to make this rigorous. I am able to construct differentials with a pole of order $\geq 2$ at $p_v$ and nowhere else. But still I would not know whether these generate all the meromorphic differentials I want. Are meromorphic differentials completely characterised by their behaviour at poles or zeros? Jun 7, 2020 at 18:01
• In a previous comment, said that dimension of $L(D) = 1$, but that isn't true if $D$ is not trivial. We get that the dimension is $0$, because the only holomorphic functions having zeros at points in $D$ is the constant function $0$. So my formula for $\dim \Omega(-D)$ was off by one. Jun 7, 2020 at 18:25

Riemann-Roch tells us that $$L(D) = \deg(-D) - g + 1 + \dim \Omega(-D) .$$ If $$M$$ is compact and if $$D \ge 0$$ is not trivial, we have $$L(D) = \{0\}$$, so dimension is $$0$$. Indeed, $$L(D)$$ contains holomorphic functions, and the only holomorphic functions from a compact Riemann surfaces are constant functions. However if $$D$$ is not trivial, it forces us to have a zero somewhere. Hence the function is constant $$0$$.

Using $$\deg(-D) = - \deg (D)$$, we have $$\dim \Omega(-D) = g + \deg(D) - 1 .$$ As you've guessed, the $$g$$ comes from the dimension of holomorphic differentials on the surface. We are set out to find $$\deg D - 1$$ non-holomorphic meromorphic differentials which form a basis for $$\Omega(-D)$$.

Write $$D = \sum n_i p_i$$, where the $$p_i$$ are $$N$$ distinct points and $$n_i \ge 0$$. Then $$\Omega(-D)$$ contains meromorphic differentials which have poles $$p_i$$ of order at most $$n_i$$. There are two types of meromorphic differentials we can construct:

• Denote with $$\tau_{p_i, k}$$ a meromorphic differential with pole of order $$k\ge 2$$ at $$p_i$$
• Denote with $$\omega_{p_i, p_j}$$ a meromorphic differential with simple poles at $$p_i$$ and $$p_j$$ and residues $$1$$ and $$-1$$.

We also know there is a basis of $$g$$ holomorphic forms, so

• Denote with $$\alpha_i$$, $$i \in \{1, \ldots, g\}$$ a basis for holomorphic one forms.

Then we claim the following is a basis for $$\Omega(-D)$$:

$$\{ \tau_{p_i, k_{i,j}} \mid 2 \le k_{i,j} \le n_i \} \cup \{ \omega_{p_1, p_2}, \omega_{p_2, p_3}, \ldots, \omega_{p_{N-1}, p_N} \} \cup \{\alpha_i \mid 1 \le i \le g\} .$$

So in total, the dimension is indeed $$\deg(D) + g - 1$$.

As an example, consider $$D = 3 p_1 + 1 p_2 + 1 p_3 + 2 p_4 + 4p_5$$.

How many differentials of the $$\tau$$-type can we construct? Only at points which occur multiple times. So we get the following, where I omitted the reference to the point in the notation for $$\tau$$, an only included the degree of the pole.

Now, couldn't there be other meromorphic differentials we need to include in our basis which have the same singular behavior? Well, suppose $$\tau$$ and $$\tau'$$ have the same singular behaviour at a point. Then $$\tau - \tau'$$ is a holomorphic differential, which is already in our basis. So $$\tau'$$ is not independent.

What about the differentials of type $$\omega$$? You'd think we would need to include $$\omega_{p_i, p_j}$$ for all possible pairs. But this is not the case. For example $$\omega_{p_1, p_3}$$ is a linear combination of $$\omega_{p_1, p_2} + \omega_{p_2, p_3}$$ and some holomorphic differentials, by the same reasoning as above. So we only need to include adjacent pairs: $$\omega_{p_1, p_2}, ... \omega_{p_{N-1}, p_N}$$. Note that we don't even need to include the pair $$\omega_{p_N, p_1}$$. This way we end up with the following:

Here an $$\omega$$ on a line denotes the differential form with poles at the endpoints. This makes it clear that the dimension is $$g + \deg D - 1$$.

• Thank you! This is very clear. In the case that D is trivial. We are in the case of only holomorphic differentials, so the dimension is $g$ which corresponds to what we find from Riemann-Roch. Jun 7, 2020 at 19:46