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I'm reading this paper with the following situation (Beginning of section 2 on page 3):

Let $X$ be a compact Riemann surface of genus $g$ with a finite group $G$ acting on that curve. We write the quotient curve $X/G$ as $X_G$ and the genus of the quotient as $g_0$. Let the cover $X\to X_G$ be branched at $r$ places, $q_1, \cdots, q_r$. The signature of the cover is an $(r+1)$-tuple $$[g_0; s_1, s_2,\cdots , s_r]$$ where the $s_i$ are the ramification indices of the covering at the branch points.

Denoting the map $X\to X_G$ by $f$, by the Riemann-Hurwitz formula $$2(g-1)=2\deg(f)(g_0-1)+\sum_{i=1}^r(s_i-1) $$

Since the left side is even and the degree term is even, it cannot be the case that $r$ is odd with all the $e_i$'s even. But in many tables in the paper, (Eg page 9 where $G$ is taken to be Aut$(X)$ ), there are instances like $$[0; 2,4,6] $$ which should be impossible by the parity argument above. Am I misreading something? If someone could clarify where I am off, I'd be really grateful. The only thing I can think of is that somehow the action of the group is bad so that the quotient map above is not complex analytic to make the Riemann-Hurwitz formula inapplicable. Is this the case?

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  • $\begingroup$ @reuns That isn't quite true. The first paragraph of page 7 says that they use a source of many curves up to a certain genus rather than just modular curves. Besides, how do you resolve the apparent contradiction I see? $\endgroup$
    – Arkady
    Sep 18, 2017 at 3:37
  • $\begingroup$ I don't, I'm trying to understand what they are talking about. They analyse Galois covers $Y \to X$ and its automorphism group $G$ (what is $X$ and $Y$ ?) from which they obtain a regular representation of $G$ decomposed in irreducible representations, they look at their degree and multiplicity to obtain $\mathbb{C}^g /\Lambda \simeq \prod_{j=1}^m A_j^{e_j}$ where $A_j$ are abelian varieties. $\endgroup$
    – reuns
    Sep 18, 2017 at 3:41
  • $\begingroup$ Where are you reading $X$ and $Y$? The usual situation is a complex curve $X$ with automorphism group $G$ and $X_G$ is the quotient of $X$ by the $G$ action and $X\to X_G$ the quotient map (as in my post). $\endgroup$
    – Arkady
    Sep 18, 2017 at 3:45
  • $\begingroup$ Maybe they work in magma with affine algebraic curves and they didn't mention the ramification at $\infty$ because it can be deduced from the others $\endgroup$
    – reuns
    Sep 18, 2017 at 4:00
  • $\begingroup$ I agree there is something strange I can't see. On the other hand I believe the quotient always exists, for $V = \text{Spec}(R)$ this is just $\text{Spec}(R^G)$ and, so if $X$ is projective one can cover it by two affines open (e.g $V_1 = X \backslash O_1, V_2 = X \backslash O_2$ where $O_1, O_2$ are orbits with no fixed points) and glue $V_1/G$ and $V_2/G$ along $(V_1 \cap V_2)/G$. $\endgroup$ Sep 18, 2017 at 6:09

1 Answer 1

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Whenever we have a map $X\to X_G$, we have to be careful to decide whether we are using ramification points (which live in $X$) or branch points (which live in $X_G$). The Riemann-Hurwitz formula above has a summation over ramification points. There is another version of it using branch points $y_1,\cdots y_k$ with multiplicity $r_i$, which when specialized to a group action as above can be written as

$$2(g(X)-2)=|G|(2g(X_G)-2)+|G|\sum_{i=1}^k(1-\frac{1}{r_i}) $$

which can essentially be derived from the above form by studying the stabilizers of the $G$ action on $X$.

So the example cited above with $|G|=240$ and $g=21$ is actually feasible with signature $[0;2,4,6]$. The mistake the OP (aka I) was making was conflating the notions of ramification points and branch points.

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