# Homology of Fermat curve

Let $$C(n):X^n+Y^n=Z^n$$ be the plane projective Fermat curve of degree $$n$$ over $$\mathbb{C}$$.

Shorter version of the question: How can I describe explicit representatives for a basis for the singular homology $$H_1(C(n),\mathbb{Q})$$?

Longer version of the question: Consider the following paths on $$C(n)$$:

$$\gamma_{r,s} : [0,1]\to C(n), \quad t \mapsto [\zeta^r(1-t)^{1/n}:\zeta^st^{1/n}:1],$$

where $$\zeta$$ is a primitive $$n-$$th root of unity and $$r,s \in \mathbb{Z}/n\mathbb{Z}$$. One can combine them to obtain cycles

$$\Delta_{r,s} := [\gamma_{0,s}-\gamma_{r,s}+\gamma_{r,0}-\gamma_{-r,0} + \gamma_{-r,-s}-\gamma_{0,-s}] \in H_1(C(n),\mathbb{Q})^-$$

that are anti-invariant under complex conjugation (that's what the $$^-$$ stands for), meaning that $$\overline{\Delta_{r,s}}=-\Delta_{r,s}$$. My aim is to show that the $$\Delta_{r,s}$$ span $$H_1(C(n),\mathbb{Q})^-$$, which has dimension equal to the genus

$$g=\dfrac{(n-1)(n-2)}{2}.$$

There are in principle enough elements. It's true that $$\Delta_{r,0}=\Delta_{0,s}=0$$ and that $$\Delta_{-r,-s}=-\Delta_{r,s}$$ and that $$\Delta_{\frac{n}{2},\frac{n}{2}}=0$$ for $$n$$ even, but one could in principle still find up to

$$\left\lfloor \frac{(n-1)^2}{2} \right\rfloor \ge g$$

cycles among the $$\Delta_{r,s}$$, and based on computer calculations I have reason to believe that they actually span $$H_1(C(n),\mathbb{Q})^-$$. But I would like to prove it. So I either need a direct argument or, if I can find a basis of $$H_1(C(n),\mathbb{Q})$$, I could try to express the $$\Delta_{r,s}$$ in terms of this basis and see what comes out.

The problem is that I do not have a nice way to visualize paths on the curve and describe them with explicit formulas.

Ideas:

• Of course one can take the standard basis on the compact Riemann surface with $$g$$ holes, but then to go from there to the concrete case of $$C(n)$$ probably requires complicated formulas (I honestly would have no idea how to do it).
• Find a basis of $$H_1(C(n),\mathbb{Q})$$ by induction. The cases with $$n=1,2$$ have trivial homology, so we just need the step. But it's difficult to do it without an idea for an explicit formula.
• Look at the Jacobian $$J(n)$$. It has the same homology as $$C(n)$$. It can be contructed as $$J(n) = \frac{\left(\Omega^{1}_{C(n)}\right)^*}{\Lambda},$$ i.e., as the dual of global holomorphic differential $$1-$$forms on $$C(n)$$ quotiented by the span $$\Lambda$$ of the functionals on $$\Omega^{1}_{C(n)}$$ of the form $$\lambda = \int_{[c]} \cdot$$ for some $$[c] \in H_1(C(n),\mathbb{Q})$$. $$\Omega^{1}_{C(n)}$$ is a complex vector space of dimension $$g$$, so if one finds a basis one can view $$J(n)$$ as $$\mathbb{C}^g$$ quotient a lattice, and cycles might be easier to describe there. And maybe, if one finds a basis of the homology of $$J(n)$$, its pullback via the Abel-Jacobi map $$C(n) \to J(n)$$ to $$C(n)$$ is again a basis and one can do something. But I am not sure about many points here.