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Consider first the algebraic equation $$y^{2}=\prod_{k=1}^{2g+1}(x-a_k)$$ where $\{a_k\}_{k=1}^{2g+1}$ is a collection of $2g+1$ distinct complex numbers, and let $$ S^\circ =\{(x,y) \in \mathbb{C}^{2}: y^{2}=\prod_{k=1}^{2g+1}(x-a_k) \}. $$ We shall now define a chart $(U,\phi)$ around each given point $P_0=(x_0,y_0)$. As it is easier, we rather describe parametrizations:

1) If $x_0 \ne a_i$ for all $i$ (and so $y_0 \ne 0$), we take $$ \phi^{-1}(z)=\left (z+x_0,\sqrt{\prod_{k=1}^{2g+1}(x-a_k)}\right ), $$ defined in the disc $\{|z|<\epsilon\}$ with $\epsilon$ small enough for $z$ not to reach any of the values $a_i$. The branch of the square root is chosen so that its value at $z=x_0$ equals $y_0$ (and not $-y_0$).

2) For $P_0=(a_j,0)$, we take $$ \phi_j^{-1}(z)=\left( z^{2}+a_j,z\sqrt{\prod_{k \ne j}(z^{2}+a_j-a_k)} \right), \, \, \, |z|<\epsilon, $$ again with $\epsilon$ small enough to guarantee that $z^{2}+a_j$ does not reach $a_k$ for every $k \ne j$. Note that if we had taken the first coordinate to be $z+a_j$, then the second one would not to be a well-defined holomorphic function in $|z| < \epsilon$.

I don't understand why the second coordinate isn't a well-defined holomorphic function and why there's $z$ before the square root in the expression $\phi_j^{-1}(z)$.

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Except for the $a_j$, above each point of the plane there are two points in the surface (corresponding to the two values of the square root). When you consider a small neighbourhood $W$ of a point $w \notin \{a_j : 1 \leqslant j \leqslant 2g+1\}$, you have two branches of

$$\sqrt{\prod_{j=1}^{2g+1}(x-a_j)}$$

defined in $W$, and the two branches give a parametrisation of the surface above $W$. The projection $\pi_W \colon S^\circ\cap \pi^{-1}(W) \to W$ looks like the projection from $W\times \{-1,1\}$ to $W$.

But in a small neighbourhood $W = D_\varepsilon(a_j)$ of one of the $a_j$, the projection looks different, it looks like $z \mapsto z^2$, in particular $S^\circ \cap \pi^{-1}(W)$ is connected (even $S^\circ \cap \pi^{-1}(W\setminus \{a_j\})$ is connected).

Suppose you have a parametrisation of $S^\circ$ around $(a_j,0)$, that is, two hlomorphic functions $\alpha,\,\beta \colon D_\varepsilon(0) \to \mathbb{C}$ with $\alpha(0) = a_j$, $\beta(0) = 0$, such that $\alpha'$ and $\beta'$ don't ever vanish in the same point (so it's a regular parametrisation), and with

$$\beta(z)^2 = \prod_{k=1}^{2g+1} \bigl(\alpha(z) - a_k\bigr).$$

If $\varepsilon$ is small enough,

$$\prod_{k\neq j} \bigl(\alpha(z) - a_k\bigr)$$

has no zeros in $D_\varepsilon(0)$, hence, since the disk is simply connected, a holomorphic square root, say $\rho$ (then $-\rho$ is the other square root). $\rho$ vanishes nowhere, so we obtain

$$\left(\frac{\beta(z)}{\rho(z)}\right)^2 = \alpha(z) - a_j.$$

$\beta/\rho$ is holomorphic, so $\alpha(z) - a_j$ must have a holomorphic square root. The simplest holomorphic function $\gamma$ on $D_\varepsilon(0)$ with $\gamma(0) = 0$ that isn't constant and has a holomorphic square root is $\gamma(z) = z^2$, which leads to the choice $\alpha(z) = z^2 + a_j$. The two square roots of $\alpha(z) - a_j$ are $z \mapsto z$ and $z \mapsto -z$, so we obtain

$$\beta(z) = \pm z\rho(z).$$

The sign can be absorbed in the choice of $\rho$, so altogether, we obtain the paramerisation

$$\phi(z) = \left(\alpha(z), z\sqrt{\prod_{k\neq j}\bigl(\alpha(z) - a_k\bigr)}\right).$$

Since $z$ is a square root of $\alpha(z) - a_j$, we could also write the second component as

$$\sqrt{\prod_{k=1}^{2g+1} \bigl(\alpha(z) - a_k\bigr)},$$

but pulling the factor $z$ out of the square root makes the mapping behaviour more obvious.

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  • $\begingroup$ Thanks a lot... Can you write also the inverse function of $\phi^{-1}$ and of $\phi_j^{-1}$, that's to say $\phi$ and $phi_j$, please? $\endgroup$ – TheWanderer Sep 11 '13 at 8:01
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    $\begingroup$ For a point $(x_0,y_0) \in S^\circ$ with $y_0 \neq 0$, you simply have $\phi(x,y) = x-x_0$ (where the neighbourhood is small enough, it mustn't contain a point with $y = 0$), for the $(a_j,0)$, you have $\phi_j(x,y) = \sqrt{x-a_j}$, for one branch of the square root. Due to the way $S^\circ$ lies over the plane, a holomorphic square root of $x-a_j$ exists on $S^\circ$ near $a_j$. $\endgroup$ – Daniel Fischer Sep 11 '13 at 8:54

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