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I'm trying to find an algebraic curve that represents a specific Riemann surface and my question goes like this:

Given divisors $(\omega_1) = P_1 + 5 P_2 + 2 P_3,$ $(\omega_2) = 5 P_1 + P_2 + 2 P_3,$ $(\omega_3) = 2 P_1 + 2 P_2 + 4 P_3,$

is there a way one can come up with an algebraic equation $P(\omega_1, \omega_2, \omega_3) = 0$ with no further information?

What I'm trying to do here is to paste 8 sheets of 3-punctured spheres. At each point $P_i$(or slit), the $j$-th sheet gets glued to $(j+ a_i)(mod \, 8)$-th sheet given $(a_1, a_2, a_3) = (1, 5, 2).$ $\omega_i$s are the holomorphic 1-forms on this surface and assuming that $P_1$ is sent to $\infty,$ $P_2$ to 0, and $P_3$ to 1, I end up with $(\omega_1^2 - \omega_2^2) \omega_1 \omega_2 = \omega_3^4.$ Though I was wondering if there is a way to find this when only the divisors $(\omega_i)$ are given.

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If I understand correctly your question (?), you want a smooth curve $C$ of genus 5 and 3 holomorphic forms $\omega _i$ with the divisors you have written down, satisfying $\ (\omega_1^2 - \omega_2^2) \omega_1 \omega_2 = \omega_3^4$.

I do not think this is possible. It would mean that the map $C\rightarrow \mathbb{P}^2$ defined by $(\omega _1,\omega _2,\omega _3)$ maps onto the smooth quartic curve $\ \Gamma :\ XY(X^2-Y^2)=Z^4$. The only possibility is that $C$ is an étale double cover of $\Gamma $, but that would imply that the zero set of each $\omega _i$ is the pull back of a set on $\Gamma $, hence consists of an even number of points, contrary to your hypothesis.

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  • $\begingroup$ I'm not familiar with Etale covers yet, but thank you for the reference. Let me show you how I got to this equation. First of all, I think this surface is of genus 3 (using Hurwitz formula). Secondly, let me define $f:= \omega_1/\omega_3, g:= \omega_2/\omega_3.$ Then sending $P_i$s to $\infty, 0 ,1$ resp. and normalizing the quotients give me $\left((f/g)^2 - 1\right) f g^3 = 1.$ Please feel free to point out anything that I'm missing. $\endgroup$ – Dami Lee Aug 4 '14 at 23:12
  • $\begingroup$ If your surface has genus 3, the divisor of a nonzero holomorphic 1-form has degree 4. $\endgroup$ – abx Aug 5 '14 at 4:22
  • $\begingroup$ Dami Lee is indeed describing the smooth quartic curve $XY(X^2-Y^2)=Z^4$ pointed by abx. The $\omega_i$ have a nontrivial common divisors $\eta_3=P_1+P_2+2P_3$ and can be written $\omega_i=\eta_i +\eta_3$. Then $\eta_1, \eta_2, \eta_3$ are a basis of the canonical system, cut by the hyplerplane sections $X, Y, Z$. $\endgroup$ – Roberto Pignatelli Aug 5 '14 at 9:26
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abx has already pointed out that your divisors are not the canonical divisors. Still, your (not-so-clear) question was maybe slighlty different, on "how to find the equation"?

If I understand your question, you have three divisors in a linear system, you want to consider the induced map on ${\mathbb P}^2$ and find the equation of minimal degree vanishing on the image (that is the equation of the image, unless the map is constant). I run similar computations as follows, by induction on the degree.

First you have to remove the common part, $P_1+P_2+2P_3$. So your map is defined by $(\eta_1)=4P_2$, $(\eta_2)=4P_1$, $(\eta_3)=P_1+P_2+2P_3$ with $\omega_i=\eta_i\eta_3$.

Then look for a relation in degree $d$, for $d=1,2,\ldots$. To prove that there is no equation in small degree, look at the vanishing of your relation in a neighbourhood of each $P_i$.

For example, for $d=1$: $a_1\eta_1+a_2\eta_2+a_3\eta_3=0$ in a neighbourhood of $P_1$ forces $a_1=0$, in a neighbourhood of $P_2$ forces $a_2=0$. So we are left with $a_3\eta_3=0$ which implies $a_3=0$.

For $d=2$: $\sum_{i \leq j} a_{ij}\eta_i\eta_j=0$ in a neighbourhood of $P_1$ forces $a_{11}=0$, in a neighbourhood of $P_2$ forces $a_{22}=0$. So we are left with $a_{12}\eta_1\eta_2=-\eta_3\sum_i a_{i3}\eta_i$: a neighbourhood of $P_3$ gives $a_{12}=0$. Then an equation of degree $2$, if it exists, is divisible by $\eta_3$, a contradiction since the equation of minimal degree has to be irreducible.

For $d=3$: $\sum_{i \leq j \leq k} a_{ijk}\eta_i\eta_j\eta_k=0$ in a neighbourhood of $P_1$ forces $a_{111}=0$, in a neighbourhood of $P_2$ forces $a_{222}=0$. So we are left with $\eta_1\eta_2(a_{112}\eta_1+a_{122}\eta_2)=-\eta_3\sum_{i\leq j} a_{ij3}\eta_i\eta_j$. So $a_{112}\eta_1+a_{122}\eta_2$ vanishes twice on $P_3$, as $\eta_3$. Then, if the divisor of $a_{112}\eta_1+a_{122}\eta_2$ is different from the one of $\eta_3$, $P_1+P_2$ is movable and therefore the curve is hyperelliptic. On the other hand, since $4P_1$ and $4P_2$ are canonical divisors, $P_1$ and $P_2$ should be Weierstrass points, and then $P_1+P_2$ can't move, a contradiction. So up to a scalar $a_{112}\eta_1+a_{122}\eta_2$ equals $\eta_3$, again a relation in degree $1$, a contradiction.

Finally for $d=4$ we can directly notice that the divisor of $\eta_3^4$ contains the divisor of $\eta_1\eta_2$ so $\frac{\eta_3^4}{\eta_1\eta_2}$ equals a quadric in the $\eta_i$, with divisor $8P_3$. If your computations are correct, some more manipulations as above should give that this quadric can be written as a $\eta_1^2-\eta_2^2$, up to rescaling the $\eta_i$, of course.

Anyway, I see a problem here, maybe your computations are not correct? Because the section $Z=0$ gives $4$ distinct points, while you ask it to be $P_1+P_2+2P_3$, so the equation should be of the form, I think, $Z^4=XYQ(X,Y,Z)$ where $Q$ is a quadric such that $Q(X,Y,0)$ is a square, up to rescaling $Q=(X-Y)^2+Z(aX+bY+cZ)$.

So the condition you give (three divisors in the same linear system of the form above) seems not to determine the curve. We get a two-parameters family of not hyperelliptic curves of genus $4$, the family of plane quartics $$ Z^4=XY[(X-Y)^2+Z(aX+bY+cZ)] $$

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