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I am currently reading the paper "Holomorphic Differentials of Generalized Fermat Curves" by Rubén Hidalgo. The case I am interested in is that of a classical Fermat curve $F_k$, which in his terminology is a generalized Fermat curve of type $(k, 2)$. This is the curve in $\mathbb{P}^2(\mathbb{C})$ defined by the homogeneous equation $x_1^k + x_2^k + x_3^k = 0$.

With regard to this special case, Hidalgo defines meromorphic maps on $F_k$ via $y_2 = \dfrac{x_2}{x_1}$, $y_3 = \dfrac{x_3}{x_1}$ (note that Hidalgo writes $z:=y_2$). In section 3.2, he constructs a set of holomorphic differential forms given by the formula $\theta_{r;\alpha} = \dfrac{y_2^r dy_2}{y_3^{\alpha}}$, where $0 \le \alpha \le k-1$, and $0 \le r \le \alpha - 2$, for a total of $g(F_k) = \dfrac{(k-1)(k-2)}{2}$ distinct forms.

Hidalgo also defines symmetries of the curve by \begin{align} a_1([u:v:w] = [\omega_k u:v:w]) \\ a_2([u:v:w] = [u:\omega_k v:w]) \\ a_3([u:v:w] = [u:v:\omega_k w]) \end{align} where $\omega_k$ is a primitive $k$-th root of unity. He remarks that $\theta_{r;\alpha}$ pulls back under these symmetries according to \begin{equation} a_j^*(\theta_{r;\alpha}) = \begin{cases} \omega_k^{r + 1 - \alpha} \theta_{r;\alpha} & j = 1 \\ \omega_k^{-r-1} \theta_{r;\alpha} & j = 2 \\ \omega_k^{\alpha} \theta_{r;\alpha} & j = 3 \end{cases} \end{equation} He denotes the divisor with a value of 1 at each fixed point of $a_j$ by $\textrm{Fix}_{div}(a_j)$, so that the divisor of $\theta_{r;\alpha}$ is given by

\begin{equation} (\theta_{r;\alpha}) = (\alpha - 2 - r) \textrm{Fix}_{div}(a_1) + r \textrm{Fix}_{div}(a_2) + (k-1-\alpha) \textrm{Fix}_{div}(a_3) \end{equation}

His theorem 3.1 states that these $\theta_{r;\alpha}$ form a basis for the space of holomorphic differentials on $F_k$. He claims that linear independence follows immediately from the above pullback and divisor formulas for $\theta_{r;\alpha}$, and does not elaborate the details. I have as of yet been unable to see why this is true. I have been trying to see if there's a simple linear algebraic approach to the group generated by the pullbacks, to show that a single nontrivial linear dependence equation will transform into enough additional constraints under the pullbacks to be unsatisfiable and produce a contradiction. From what I have found so far it looks like I will get at most $k$ independent constraints using this method, so this alone does not seem promising. I do not see how the divisors come into it.

I understand that there is general classical theory behind constructing holomorphic differentials on projective curves, and have found similar-looking results in references such as Plane Algebraic Curves by Brieskorn and Knorrer, section 9.3, theorem 1. However, my knowledge of algebraic geometry is limited to Riemann surfaces. I think Hidalgo is indicating that the proof in this special case may be simpler. Can anyone please point me in the right direction?

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    $\begingroup$ I suspect what Hidalgo has in mind is the following representation-theoretic argument (which can also be rephrased in elementary terms): The automorphisms $a_1,a_2,a_3$ are part of a $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$ action. The pullback formula tells you that each $\theta_{r,\alpha}$ generates a character (1-dim representation) of the group, and the characters corresponding to different $(r,\alpha)$ are different. Therefore there can be no nontrivial relations between them. I don't really see why the divisor formula is necessary though. $\endgroup$
    – dhy
    May 31, 2021 at 7:19
  • $\begingroup$ @dhy: Thank you, that is very simple. I have been meaning to learn more about representation theory. If you want to turn it into an answer I will accept. $\endgroup$
    – yyc
    May 31, 2021 at 15:35

1 Answer 1

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The automorphisms $a_1,a_2,a_3$ are part of a $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$ action on $F_k$, which in turn induces an action on the space $H^0(\Omega^1_{F_k})$ of holomorphic differentials on $F_k$. Let me explain how the representation-theory of this group gives the desired independence.

The $\theta_{r,\alpha}$ define a linear map $\displaystyle\bigoplus_{r,\alpha}\mathbb{C}\rightarrow H^0(\Omega^1_{F_k}),$ and their linear independence is equivalent to this map being injective. The pullback rules give each $\mathbb{C}$-summand above a $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$-representation structure so that the map is equivariant. Call this representation (for a fixed $(r,\alpha)$) $\mathbb{C}_{r,\alpha}.$ Importantly, for different $(r,\alpha)$, the $C_{r,\alpha}$ are different representations.

Now assume that $\oplus_{r,\alpha}\mathbb{C}_{r,\alpha}\rightarrow H^0(\Omega^1_{F_k})$ is not injective. The kernel, which must be a $\mathbb{Z}/p\mathbb{Z}\times\mathbb{Z}/p\mathbb{Z}$-subrepresentation of $\displaystyle\bigoplus_{r,\alpha}\mathbb{C}_{r,\alpha},$ must in fact take the form $$\displaystyle\bigoplus_{(r,\alpha)\in S}\mathbb{C}_{r,\alpha}$$ for some set $S$.

But if $S$ is nontrivial, then that implies that some $\theta_{r,\alpha}$ is trivial, which is clearly not true.

If you wish to remove the representation theory from this argument, here is an elementary translation. (But I still recommend learning the representation theory, because it's broadly useful for this kind of study.)

Assume the $\theta_{r,\alpha}$ are not linearly independent. Then take a relation $$\displaystyle\sum_{(r,\alpha)\in S} c_{r,\alpha}\theta_{r,\alpha}=0$$ with $|S|$ minimal. Pulling back by $a_2$, we get another relation $$\displaystyle\sum_{(r,\alpha)\in S} \omega_k^{-r-1}c_{r,\alpha}\theta_{r,\alpha}=0.$$

Now, unless these two relations are multiplies of each other, we can take some linear combination of them to get a relation with less terms, a contradiction. So they must be multiples, i.e., all the $(r,\alpha)\in S$ must have the same value of $r$. Repeating this argument but for the pullback by $a_3$ we see that they must also all have the same value of $\alpha$, i.e., $S$ must be a one element set. But that would imply that $\theta_{r,\alpha}$ is trivial, a contradiction.

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  • $\begingroup$ Very interesting. I found the elementary proof online and was very happy to see how simple it was, but once I am done with my current project I will look over the details of the deeper setting you wrote out. Time to look for a good representation theory textbook. Thank you! $\endgroup$
    – yyc
    May 31, 2021 at 21:19

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