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Let $X$ be the hyperelliptic surface defined by $y^2 = x^5-x.$ Note that $x$ and $y$ are meromorphic functions on $X.$ Compute the principal divisors div($x$) and div($y$).

We have the following definition:

The Divisor of a Meromorphic Function: Principal Divisors. Let $X$ be a Riemann surface and let $f$ be a meromorphic function on $X$ which is not identically zero. The divisor of $f$, denoted by div($f$), is the divisor defined by the order function: $$div(f)=\sum_{p}Ord_p(f) p$$ Any divisor of this form is called a principal divisor on $X$.

Using the following Lemma, I tried a naive approach:

Lemma 1.4. Let $f$ and $g$ be nonzero meromorphic functions on $X.$ Then we have:

  • div($fg$)=div($f$)+div($g$).
  • div($f/g$)=div($f$)-div($g$).
  • div($1/f$)=div(1)-div($f$).

Observe that div($y^2$)=div($x(x^4-1)$)=div(x)+div($x^4-1$). So, div($x$)=div($y^2$)-div($x^4-1$). Then I am not sure about div($y^2$) and div($x^4-1$). I think they are both zero.

Also, I have the following at disposal:

Lemma 3.12. Let $\omega$ be a meromorphic 1-form defined in a neighborhood of $p\in X.$ Let $\gamma$ be a small path on $X$ enclosing $p$ and not enclosing any other pole of $\omega.$ Then Res$_p$($\omega$)=$\frac{1}{2\pi i}\int_{\gamma}\omega.$ Lemma 3.14. Suppose $f$ is a meromorphic function at $p\in X.$ Then $df/f$ is a meromorphic 1-form at $p$, and Res$_p$($df/f$) = ord$_p$($f$).

Then $Ord_p(y^2)=Res_p(\frac{2}{y})=\frac{1}{2\pi i}\int_{\gamma}\frac{2}{y}.$ But how to solve this integral?

Any suggestions/hints? Thanks.

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  • $\begingroup$ I think that maybe you need to think a little bit more about what the definition of divisor in terms of orders really means. For instance, if $f$ has the value 0 at a point $p$, then certainly $p$ should appear in the divisor of $f$ with a positive coefficient. Now, what are the points on your hyperelliptic Riemann surface where $x$ takes the value 0? What are the points where $y$ takes the value $0$? (This is only a start: you also need to think about what the orders of $x$ and $y$ at those points are; then you need to worry about poles.) $\endgroup$ – user64687 Mar 17 '13 at 22:54
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"... and so we conclude that $div(x)=2P_0-2P_\infty$ and $div(y)=P_0+P_1+P_{-1}+P_i+P_{-i}-5P_\infty$ in the notation explained with exquisite precision above."

[Excerpt from a manuscript found in a Klein bottle]

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  • $\begingroup$ I couldn't decipher the beginning of the manuscript I found in that bottle $\endgroup$ – Georges Elencwajg Mar 18 '13 at 1:05
  • $\begingroup$ I don't understand the notation. Can you please explain what $P_0$ is supposed to represent? $\endgroup$ – Lyapunov Mar 21 '13 at 2:38
  • $\begingroup$ The notation $P_0$ represents the divisor corresponding to the number $0$ . In other words $div(x)=2P_0-2P_\infty$ is synonymous with $div(x)=2\cdot 0-2\cdot \infty$ $\endgroup$ – Georges Elencwajg Mar 21 '13 at 7:53
  • $\begingroup$ Thank you. I still don't understand where you are getting the coefficient $2$ in $div(x)$. Isn't $div(x)=div_0(x)-div_{\infty}(x)=1\cdot 0-1\cdot \infty $ for Riemann surfaces? $\endgroup$ – Lyapunov Mar 21 '13 at 16:34
  • $\begingroup$ Near the point $P_0\in X$, you can write $x=\frac {y^2}{x^4-1}$ and the crucial points are: i) that $y$ is a local coordinate at $P_0$ and ii) that $\frac {1}{x^4-1}$ is a unit $u\in \mathcal O_{X,P_0}$, so that indeed $ord_{P_0}(x)=ord_{P_0}(y^2u)=2$, as asserted. $\endgroup$ – Georges Elencwajg Mar 21 '13 at 17:52

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