What are some ways of discovering the topology of the Fermat curve? Show that  $X=\{[x_0;x_1;x_2] \in \mathbb{C}P^2 | x_0^n + x_1^n +x_2^n = 0\}$ is an orientable surface of genus $\dfrac{(n-1)(n-2)}{2}$.
I want to know different ways of solving the problem.
Some ways I (along with my colleagues) have thought about are:

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*The standard Riemann-Hurwitz technique: This is a staple in compact Riemann surface theory. Consider the branched covering map $\pi:X \to \mathbb{C}P^1$ given by $\pi([x_0;x_1;x_2]) = [x_0;x_1]$. We triangulate the base and lift the triangulation to the branched cover. We relate the triangulation of a surface to its Euler characteristic and relate that to the genus.


*Classical Bernhard Riemann style: Consider the map $\pi:X \to \mathbb{C}P^1$ given by $\pi([x_0;x_1;x_2]) = [x_0;x_1]$. Now we see that it is a branched cover. We make slits joining the branch points and look at the restricted cover. So we have cover of polygons. Analysing the monodromy, we figure out how to glue the edges of the polygon to figure out topology of $X$.
Are there Kleinian methods to solve this problem? What about Morse theoretic methods? What about modern methods like TQFT?
 A: There are two more ways I know how prove this off the top of my head. Another thing to note is that this result say that the topology of your algebraic curve doesn't have anything to do with the actual polynomial equation chosen (here the degree $n$ Fermat curve) - it just matters what the degree of your polynomial is. Here are sketches of the two ways I had in mind:
(1) Start out with the union of $d$ lines in general position in $\mathbb{P}_{\mathbb{C}}^2$, which forms a singular curve that is topologically the wedge of a bunch of 2-dimensional spheres at $\binom{d}{2}$ points. Perturb the coefficients of the polynomial which defined the $d$ lines. The space of smooth degree $d$ curves in $\mathbb{P}_{\mathbb{C}}^2$ is dense in the space of all degree $d$ curves in $\mathbb{P}_{\mathbb{C}}^2$, so the perturbed polynomial will define a smooth algebraic curve. Its genus can be computed via the original singular picture, and is equal to $\binom{d}{2} -(d-1)$.
(2) Recall a simplified version of the adjunction formula states that, for a smooth degree $d$ algebraic curve $C \subseteq \mathbb{P}_{\mathbb{C}}^2$, we have the equation
$$\chi(C) = - [C]\cdot[C] + 3[L]\cdot [C].$$
Here $\chi$ is the topological Euler characteristic, $[L]$ is the hyperplane class in $\mathbb{P}_{\mathbb{C}}^2$, and $[A]\cdot[B]$ denotes the algebraic intersection number of $[A]$ and $[B]$. The left hand side measures the intrinsic topology of $C$, while the right hand side measures extrinsic information about how $C$ sits inside of $\mathbb{P}_{\mathbb{C}}^2$ (with an error term coming from the hyperplane class). By the topological classification of surfaces, $\chi(C) = 2 - 2g$, where $g$ is the genus of the curve, so our goal is to find $g$.
Every curve $C$ in the projective plane is a multiple of $[L]$ in homology (you can see this by computing the homology of $\mathbb{P}_{\mathbb{C}}^2$), and by the fundamental theorem of algebra (or Bezout's theorem, whatever you like), $[C] = d [L]$ in $H_2(\mathbb{P}_{\mathbb{C}}^2,\mathbb{Z})$. Each term on the right hand side can be computed using this, since $[L]\cdot[L] = 1$ (i.e. transverse lines intersect at one point). It follows that
$$2-2g = - d^2 + 3d,$$
so rearranging the equation gives us the degree-genus formula.
Hopefully those methods give you a few more ways to think about the degree-genus formula, and more generally appreciate the topology of algebraic varieties!
