# Problem in Rick Miranda: finding genus of a projective curve

I have just started learning Riemann surfaces and I am using the book by Rick Miranda: Algebraic curves and Riemann Surfaces. #F in section 1.3 asks to determine the genus of the curve in $\mathbb{P}^3$ defined by the two equations $x_0x_3=2x_1x_2$ and $x_0^2 + x_1^2 +x_2^2 +x_3^2 = 0$. #G also has a similar question in which he asks to determine the genus of the twisted cubic. Please explain how to approach this type of question.

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I posted the same question in mathoverflow. You may check the various answers given there. mathoverflow.net/questions/55312/… –  Divakaran Feb 16 '11 at 8:03
See my answer link. It is a formula that you can prove with sheaf cohomology. It appears also in Exercise I.7.2(b) of Hartshorne, but I don't know an elementary proof. –  Andrea Aug 8 '11 at 14:03
I remember Miranda's book is not difficult. So maybe you can read the sections again and try to do it for yourself. –  Kerry Nov 20 '11 at 8:26
In my copy of Miranda, the equation is $x_0 x_3 = x_1 x_2$, not $x_0 x_3 = 2 x_1 x_2$. I have therefore changed the statement of the problem, under the assumption that the $2$ is a typo. –  Jim Belk Nov 21 '11 at 16:53
Having considered the problem a bit more, it seems that the 2 is required for the equations to define a smooth complete intersection curve. For example, without the 2, there is a critical point at $[1:i:-i:1]$. Presumably this is an error in my copy of Miranda, which was fixed in a later printing. I have therefore restored the 2 in the statement of the problem. –  Jim Belk Nov 21 '11 at 17:13

Here's a map from the curve $X$ in $\mathbb{P}^3$ defined by $x_1 x_4 = 2 x_2 x_3$ and $x_1^2+x_2^2+x_3^2+x_4^2 = 0$, to the elliptic curve $E$ in $\mathbb{P}^2$ given by the Weierstrass equation $$Y^2 Z = X^3 - \frac{17822265625}{62208} X Z^2 + \frac{2269744873046875}{40310784} Z^3,$$ which is topologically a torus:

$$[x_1:x_2:x_3:x_4 ] \mapsto [y_1:y_2:y_3] \mapsto [a:b:c] \mapsto [X:Y:Z]$$ with \begin{align*} y_1 &= x_2 x_3^2 - \frac{i}{2}x_2 x_3 x_4 - \frac{i}{2}x_3^2x_4 - \frac{1}{4} x_3 x_4^2 - \frac{i}{4} x_4^3 \\ y_2 &= x_2 x_3 x_4 - \frac{i}{2} x_2 x_4^2 \\ y_3 &= x_3 x_4^2 - i x_4^3 \end{align*}, (note that $y_2$ has a simple pole at $[1:i:0:0]$, $y_1$ has a simple pole at $[1:0:i:1]$ and $y_3$ has a simple pole at $[0:0:i:1]$), \begin{align*} a &= \frac{625}{36} i y_1 - \frac{625}{72} i y_2 \\ b &= \frac{-390625}{1296} i y_1 \\ c &= \frac{-4}{5} i y_1 - \frac{8}{5} i y_2 - i y_3 \end{align*}, and $$[a:b:1] \mapsto \left [\frac{1}{36}a - \frac{71875}{15552} : \frac{1}{216}b - \frac{125}{288} a + \frac{13671875}{124416} : 1 \right ].$$

The above maps then give you an explicit homeomorphism from your curve $X$ to the elliptic curve $E$ which is topologically a torus (these in fact form an algebraic isomorphism between the two curves).

(This was obtained from embedding $X$ in $\mathbb{P}^2$ through the divisor $([1:i:0:0]) + ([1:0:i:0]) + ([0:0:i:1])$ which is very ample, and then messing around to get a nice Weierstrass equation for the resulting elliptic curve.)

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Not sure if the proof below is totally correct as I read this book two year ago. Write this in local coordinates $(\frac{x_{1}}{x_{3}}, \frac{x_{2}}{x_{3}})=(y_{1},y_{2})$ you should get a curve as $$y_{1}^{2}+y_{2}^{2}+4y_{1}^{2}y_{2}^{2}+1=0$$ defined on the affine chart $(x_{1},x_{2},x_{3}), x_{3}\not=0$.

Locally this surface is singular if and only if $\frac{\partial}{\partial y_{1}}$ and $\frac{\partial}{\partial y_{2}}$ are both 0. Hence all the singularity points are $( \frac{1}{2}i, \frac{1}{2}i),(-\frac{1}{2}i, -\frac{1}{2}i)$ . Further both of them are of multiplicity 1. Therefore by the formula $g=\frac{(d-1)(d-2)}{2}-\sum_{p}\delta_{p}$ we should have $g=3*2/2-2=1$. (but I am not familiar with Plucker's formula myself so I might be wrong).

Another way of looking at this is to define $\displaystyle(y_{1}^{2}+y_{2}^{2}+4y_{1}^{2}y_{2}^{2})^{1/4}=y_{3}$. Then we are essentially working with a ramified map $y_{3}^{4}-1=0$. From Riemann-Hurwtiz we should have g=1 by $2-2g=4*(2-0)-4*2$.

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