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 $\Omega(-D)$?

  • $\begingroup$ 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? $\endgroup$ Jun 7, 2020 at 15:31
  • $\begingroup$ What kind of meromorphic differentials do you know? Can you construct a differential with a double pole at $p_1$? A triple? A single? $\endgroup$ Jun 7, 2020 at 15:34
  • $\begingroup$ 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. $D + p_1$? $\endgroup$ Jun 7, 2020 at 16:41
  • $\begingroup$ @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? $\endgroup$ Jun 7, 2020 at 18:01
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    $\begingroup$ 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. $\endgroup$ Jun 7, 2020 at 18:25

1 Answer 1


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 $D = 3 p_1 + 1 p_2 + 1 p_3 + 2 p_4 + 4p_5$.

enter image description here

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.

enter image description here

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:

enter image description here

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

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    $\begingroup$ 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. $\endgroup$ Jun 7, 2020 at 19:46

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