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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.
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