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Let $Y$ be a cyclic cover of the projective line. Assume it is defined by the equation $y^d=h(x)$. It is obvious that the degree of the projection map $\pi: Y \to \mathbb{C}_\infty$ sending $(x,y) \to x$. Moreover let $\sigma: Y \to Y$ be the automorphism given by $\sigma(x,y)=(x,\zeta y)$, where $\zeta$ is a $d^{th}$ primitive root of unity. Clearly, every fiber of the projection map is an orbit of $\sigma$.

Now, assume that $x_0$ is a root of $h(x)$ having order $n$. I have to show that there are exactly $gcd(n,d)$ points of $Y$ above $x_0$, each having multiplicity equal to $\frac{d}{gcd(d,n)}$. To this regard, I consider such a root. Without loss of generality, I take $x_0=0$. Hence $y^d=x^nr(x)$, where $r$ has non-zero constant term. Taking a $n^{th}$ root of $r$, we may write $y^d=x^n$ (I make a reparametrization) and then I resolve the monomial singularity. Reasoning as Miranda (pag. 71), there are $k:=gcd(d,n)$ points of $Y$ on $x_0=0$ and, locally, the equation factorizes in $\prod\limits_{i=0}^{k-1} (z^a-\zeta^i w^b)$, where $d=ka$ and $n=kb$. Is my argument right?

Concerning the multiplicity, I want to study it in two ways: firstly, since each factor is of the form $(z^a-\zeta^i w^b)$, then there are $a=\frac{d}{gcd(d,n)}$ points above any point $z$ satisfying the relation $(z^a-\zeta^i w^b)$. Does this argument suffice in order to ensure that each point has the claimed multiplicity?

The other way is to set $G:=<\sigma>$ and to use the fact that $Y/G=\mathbb{C}_\infty$ (I think it is true, but I cannot prove it). Hence, $mult_P(\pi)$ equals the order of the stabilizer of each point. How may I show that the stabilizer has exactly order $\frac{d}{(d,n)}$?

Finally, given the degree of $h$ (say $d$) and the orders of the root (say $m_i$ for any $i=1,\dots,m$, I want to write down the genus of $Y$ through Hurwitz's formula. For, we have

$$ 2g(Y)-2=-2d+ \sum\limits_{i=1}^m gcd(d,m_i) (\frac{d}{gcd(d,m_i)}-1)+R $$ where the term $R$ must take account of what happen when $x=\infty$. What about the term $R$? And may I simplify the Hurwitz's formula to gain an appealing formula for the genus of $Y$?

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Your argument is correct except for $$\prod\limits_{i=0}^{k-1} (z^a-\zeta^{ai} w^b),$$ once you have fixed $\zeta$ as a $d$th root. After resolving the singularity, act with $\sigma(z,y) =(\zeta z, w)$ to see that the stabilizer of each point above $x=0$ is generated by $\sigma^{d/a}$.

The map $\pi$ is just the canonical projection $Y \longrightarrow Y/G$ as you observed. Now for the $R$, in page 73 of Miranda's book you have the equation for the curve around $x=\infty$. See how $\sigma$ is written in the new coordinates and you can calculate as in the other cases.

EDIT: Let $U$ be a small neighborhood of $(0,0) \in \mathbb{C}^2$ and let $\sigma \colon U \longrightarrow U$ be given by $\sigma(x,y)=(rx,sy)$ with $r,s \in \mathbb{C}^\ast$. Let $X \longrightarrow U$ be the blow-up at the origin. $$ X =\{ (x,y;u:t)\in U\times \mathbb{P}^1 \mid xt=uy \} $$ Then $\sigma$ lifts to a biholomorphism given by $\sigma(x,y;u:t) = (rx,sy;ru:st)$. In coordinates, $\sigma(x,t) = rx, st/r)$ and $\sigma(u,y)=(ru/s, sy)$. With this in hand you can see how the map transforms under an embedded resolution of the singularity. For example, let $$y^6-x^4 = (y^3-x^2)(y^3+x^2), \, \sigma(x,y)=(x,\zeta y), \, \zeta^6=1$$ (d=6, n=4, a=3, b=2 in the previous notation). This singularity is resolved after three blow-ups and you'll get a curve given, in coordinates $(z,w)$, by $(w-1)(w+1)$ and the map $\sigma(z,w) = (\zeta^{-1}z,\zeta^3 w)$. The points above $x=0$ are $(0,1)$ and $(0,-1)$ and their stabilizer is generated by $\sigma^2(z,w) = (\zeta^{-2}z,w)$.

In the general case, you'll get, after a finite number of blow-ups, $d/a$ branches $w=\zeta^{ai}$ and $\sigma$ permutes them cyclically. Hence the stabilizer is given by $\sigma^{d/a}$.

Let me know if you need to clarify it further.

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  • $\begingroup$ Which is the genus of $Y$. I obtain several summands, but I don't see how to simplify the formula in terms of the degree of $h$. $\endgroup$ – TheWanderer Jun 23 '18 at 15:14
  • $\begingroup$ Furthermore, may you write down explictly the proof of the fact that the stabilizer of each point above $x=0$ us generated by $\sigma^{\frac{d}{a}}$? $\endgroup$ – TheWanderer Jun 23 '18 at 15:23
  • $\begingroup$ I think you cannot simplify it further without any assumption. For instance, you can get a nice formula if you assume that $h(x)$ has only simple roots. $\endgroup$ – Alan Muniz Jun 23 '18 at 15:27
  • $\begingroup$ The computation of the stabilizer is just by looking at how $\sigma$ transforms under blow-ups. I'll edit to include that. $\endgroup$ – Alan Muniz Jun 23 '18 at 15:31

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