I am working on the following questions:

Let $X$ be a compact Riemann surface of genus $3$ with two points $p\neq q$.

  • Find a non-constant meromorphic function on $X$ with at least a double zero at $p$ and holomorphic everywhere except possibly at $q$.

  • What is the smallest possible pole under at $q$ we need to accept in order to guarantee the existence of such a function?


Construct an example of a Riemann surface of genus $3$ that has a holomorphic $1$-form with a zero of order $1$ at a point $p$, and zero of order $3$ at a different point $q$.

My attempt:

The second part of the first problem is easy. I think it is just some simple application of Riemann-Roch theorem.

But I am desperate to construct something on Riemann surface. Could anyone show me how to do this?


2 Answers 2


a) Consider the divisor $D=-2\cdot p+N\cdot q$ on $X$. We have $L(D)=\Gamma (X,\mathcal O(D))\neq 0$ as soon as $N\geq 5$ (by Riemann-Roch). Choose such an $N$ and then choose a non-zero $f\in L(D)$.
That meromorphic function $f$ satisfies $\operatorname {div}f-2\cdot p +N\cdot q\geq 0 $ and thus has a pole of order at most $N$ at $q$ and is holomorphic everywhere else with a a zero of order $\geq2$ at $p$, just as required.

b) Take any smooth plane projective quartic $X\subset \mathbb P^2$ (for example $x^4+y^4+z^4=0$): it has genus $3$.
The curve $X$ always has an inflexion point $q\in X$ and the tangent line $T_q(X)\subset \mathbb P^2$ cuts $X$ in another point $p\in X$.
Since our tangent line $T_q(X)$ cuts $C$ in a canonical divisor (like any line in the plane!), we see that $3\cdot q+1\cdot p$ is a canonical divisor and that divisor is the divisor of a holomorphic differential form $\omega$ vanishing at $p$ and having a zero of order $3$ at $q$, just as required.


Take the Riemann surface $X$ as the Klein's surface, i.e. the Riemann surface associated to the algebraic function $$w^7=z(z-1)^2$$ Applying Riemann-Hurwitz formula, we have the genus of $X$ is $3$.

To construct the holomorphic $1$-form, you can see the section 7.2 of


  • 1
    $\begingroup$ Your link is dead and you have not answered your questions $\endgroup$ Feb 13, 2022 at 16:44
  • $\begingroup$ @GeorgesElencwajg I found the same article elsewhere on the web. so I updated the link. $\endgroup$
    – hbghlyj
    Mar 14 at 9:10
  • $\begingroup$ Thanks for updating. $\endgroup$ Mar 14 at 10:24

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