Do Quartic Curves give Genus 2 surfaces when mapped in the complex plane?

Question

Elliptic Curves (cubic equations) give toruses when mapped in a complex plane, that is equations of the form: $y^2=x(x-a)(x-b)$. Also, equations of form $y^2=x(x-a)$ give a sphere or homeomorphic equivalents when mapped in the complex plane. I think the natural next question is about $y^2= x(x-a)(x-b)(x-c)$. Do these give surfaces of genus $2?$ Thanks in advance!

Answer 1

The equation $y^2 = x(x-a)(x-b)$ gives a torus in the projective plane $\mathbb{CP}^2$, not in the affine plane. Moreover you have to ask that $a,b,0$ are distinct else you don't have topologically a torus but a singular torus, for example the curve $y^2 = x(x-1)^2$ is a "pinched torus" :

Now if you consider an equation in the projective plane with differents roots, you get curves which are called hyperelliptic curves, and the equation for such a curve is $$ y^2 = \prod_{k=1}^{n} (x- a_i)$$

where the $a_i$ are distincts complex numbers. For $n = 2g +1$ or $n=2g+2$ the curve is a curve of genus $g$.

So here for $g=1$ you get curve of degree $3$ or $4$, i.e your curve is again a torus.

More informations about hyperelliptic curves (such that their structure as Riemann surfaces) can be found in the first chapter of this book.

Edit : You can also verify it directly with Riemann-Hurwitz formula. Let $X$ be your curve and $\pi_x$ the projection to the $x$ coordinate. There are four ramifications points for $\pi$ (namely 0,a,b,c) so $\chi(X) = 4 - 4 = 0$ and it follows that $X$ is a complex torus.