I want to show that given a compact Riemann surface $X$ of genus 2, there is a map $\sigma:X\rightarrow \mathbb{CP}^2$, such that the followings are satisfied:

  • $\sigma(X)$ is a quartic curve with a double point $s$
  • $\sigma:X-\sigma^{-1}(s)\rightarrow \sigma(X)-s $ is biholomorphic.

What I have tried so far:

If we have found such map $\sigma$, then we can make a projection, $x = \pi \circ \sigma : X\rightarrow \mathbb{CP}^1$. And we can choose such projection that $\pi(s)$ is not a branch point of $x$.

Apply Riemann-Hurwitz formula, $\operatorname{deg}{R_x} = 6$. And we also have $\operatorname{deg}x = 4$.

I think there are more constraints on $x$, so that I can apply Riemann-Roch theorem to find such $x$ and then the quartic curve itself. But I don't know what to do next.

Any hints are welcome.

Thanks in advance!


Consider any divisor $D$ of degree 4 on $X$ (with $D \ne 2K_X$) and consider map $f \colon X \to \mathbb{P}^2$ given by the linear system $|D|$. It will have all the required properties.

Let me show that $f$ is almost injective. Assume $f(P) = f(Q)$ for a pair of points $P,Q \in X$. This means that $h^0(X,O_X(D-P-Q)) = h^0(D)-1 = 2$, hence $D - P - Q$ is a $g^1_2$, hence $D - P - Q = K_X$. This means that $P + Q \sim D - K_X$, and since $D - K_X \ne K_X$ (by the assumption $D \ne 2K_X$), the linear system $|D - K_X|$ consists of a unique divisor, hence $P$ and $Q$ are canonically defined up to permutation.

  • $\begingroup$ Would you please elaborate on it? $\endgroup$ – Y. Hu May 8 '18 at 9:09
  • $\begingroup$ @Y.Hu: What do you want to see? The definition of a morphism given by a linear system? Something else? Please, specify. $\endgroup$ – Sasha May 8 '18 at 10:28
  • $\begingroup$ I know the definition, but I can't see why it has all the required property. I know how to show that the image of the map lies in a quartic curve, for some certain $D$, but not the general case. Neither am I able to show that this map is injective with one exception. $\endgroup$ – Y. Hu May 8 '18 at 10:36
  • $\begingroup$ @Y.Hu: OK, I added an explanation. $\endgroup$ – Sasha May 8 '18 at 11:11
  • 1
    $\begingroup$ @Y.Hu if you follow Sasha's $h^0$ computations, then you have that $D - P - Q$ is a $g^1_2$ hence induces a degree $2$ map to $\mathbb P^1$. But every genus $2$ curve is hyperelliptic, and hyperelliptic curves are characterized by $K_X$ being the unique (up to linear equivalence) such divisor. $\endgroup$ – Tabes Bridges May 8 '18 at 19:44

Given the answers above, I think I can show that this is a surjection by considering the regularization $\mu:\overline{Y}\rightarrow Y$ , where $Y$ is the quartic curve. We claim that $Y$ is irreducible. (Though the regularization is defined for irreducible curves in the book, the process can still be carried out for a general curve, and if it is reducible, $\overline{Y}$ would be disjoint union of regularizations of each factor.) Now $\mu^{-1}\circ\sigma$ extends to be a isomorphism between $X$ and a connected component of $\overline{Y}$. If $Y$ is reducible, $X$ would be the regularization of some curve with lower degree, which is impossible by checking the genus. Then we must have $X$ as the regularization of $Y$, which implies surjectivity.

(I still don't konw why $\sigma(X)$ lies in a quartic curve actually. I wish you could explain it to me.)

  • $\begingroup$ Well, let $D = P + Q + K_X$, where $P + Q$ is not linear equivalent to $K_X$. If I'm right, Riemann-Roch gives that $L(D-P-Q) = L(D-P) \subset L(D)$. Let $x,y$ be a basis of $L(D - P) = L(D - P -Q)$, and $z\in L(D) - L(D-P)$. Consider a set $S = \{x^iy^jz^k:i + j + k = 4, k < 4\}$, which has 14 elements, all of them lying in $L(4D - P - Q)$, whose dimension is 13 by Riemann-Roch. Then we have a quartic equation. $\endgroup$ – Y. Hu May 30 '18 at 6:37
  • $\begingroup$ I think your argument makes sense, I'm not sure though. $\endgroup$ – Y. Hu May 30 '18 at 6:40
  • $\begingroup$ @y.hu I think your calculation is correct and, knowing the fact that $h^0(D-K_X)=1$, we can always assume that $D=P+Q+K_X$, so it applies to all such $D$ thx a lot ! $\endgroup$ – Bai Qingyuan May 30 '18 at 11:03

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