# Understanding the "shape" of a singular Riemann surface

Consider the singular Riemann surface given by the following expression: $$z^d w^d-z^d-w^d+t=0\ ,$$ where $t$ is a parameter in $(-1,1)$ and $d$ is a positive integer greater than 2.

For $t\neq0$ the genus is $(d-1)^2$ and the surface has 2 singularities $[1:0:0]$ and $[0:1:0]$, while for $t=0$ the genus drops to $\frac{1}{2}(d-1)(d-2)$ and a third singularity $[0:0:1]$ appears.

I would like to understand how the singularities look like locally (possibly in a pictorial way at the level of the underlying topological surface) and how the appearance of the third singularity changes the genus from $(d-1)^2$ to $\frac{1}{2}(d-1)(d-2)$.

I haven't found a way to imagine what happen at the topological level, and I would like to be able to draw some pictures (say for $d=3$ or $d=4$) to gain some intuition.

• What do you mean by genus? Anyway, you should complete your curves: your question is ambiguous as it stands since you can take the closure of your curves in $\mathbb P^2$ or take the normalization of that closure. Jul 28, 2015 at 9:28
• I'm considering the closure in $\mathbb{P}^2$, but I would also understand what happens when one desingularize... Jul 28, 2015 at 9:35