Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a non-hyperelliptic Riemann surface of genus $g = 4$. Let $p,q,r$ be three points on $X$, not necessarily distinct. Is there any obstruction to the existence of a meromorphic function $f: X\to \mathbb{C}$, such that $f$ has simple poles at exactly $p,q,r$, and no other poles?

In other words, if $D = p + q + r$ is a divisor, what is the dimension of $L(D)$? (I believe this is standard terminology, but $L(D)$ is the space of meromorphic functions on $X$ which have poles bounded by the divisor $D$)

The dimension of $L(D)$ can only be 1 or 2 (by the non-hyperellipticity hypotesis). I was wondering if, in general, a function like the one described above does exist (and then $L(D)$ has always dimension 2) or not.

share|cite|improve this question
up vote 3 down vote accepted

Let $K$ denote an effective divisor linearly equivalent to the canonical divisor. By Riemann-Roch we have $\ell(D) = \ell(K - D)$ (where $\ell(D) = \dim \mathcal{L}(D)$). The elements of $\ell(K - D)$ are the functions $f$ such that $\text{div}(f) + K - D \ge 0$. Now, since $X$ is non-hyperelliptic, the canonical map $\phi_K : X \to \mathbb{P}^3$ is an embedding, and we can choose coordinates $(X_0 : X_1 : X_2 : X_3)$ such that the hyperplane $X_0 = 0$ at infinity has hyperplane divisor $K$.

It follows that if $a_0 X_0 + ... + a_3 X_3 = 0$ is a hyperplane containing $p, q, r$, then $f = \frac{a_0 X_0 + ... + a_3 X_3}{X_0} \in \mathcal{L}(K - D)$. But three points generically determine a hyperplane in $\mathbb{P}^3$, so we ought to have $\ell(K - D) = 1$ in general (hence $\ell(D) = 1$ in general). If $p, q, r$ are distinct then $\ell(K - D) = 1$ if and only if the points $p, q, r$ are not collinear. If two of the points are identical, say $p = q$, then $\ell(K - D) = 1$ if and only if the tangent line to $p$ does not intersect $r$. If all three points are identical, then I think $\ell(K - D) = 1$ if and only if $p$ is not a point of inflection.

share|cite|improve this answer
Can you please argumentate a bit more on the second part of your answer? Specifically, the fact that you found that there always exists a function $f\in \mathcal{L}(K-D)$ implies only that $\mathcal{L}(K-D) \geq 1$ (but, of course $\leq 2$). How can you exclude the case $\mathcal{L}(K-D) = 2$? I must be missing something.. – Raziel May 1 '11 at 17:09
@Luca: $\mathcal{L}(K - D)$ is a subspace of $\mathcal{L}(K) = \text{span}(1, \frac{X_1}{X_0}, \frac{X_2}{X_0}, \frac{X_3}{X_0})$. The argument shows (if I haven't made a mistake) that it consists precisely of the equations of hyperplanes containing all of $p, q, r$, and in the generic case this hyperplane is unique, so $\ell(K - D) = 1$ in general. – Qiaochu Yuan May 1 '11 at 17:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.