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Let $D$ be an effective $ \mathbb{Z}$-divisor on $ \mathbb{P}^1$. Is there a form to associate a curve $C$, and morphism $C \to P^1$ to the divisor $D$ ? For example, let $Y$ be a singular plane curve, sometimes, we can built a cover of $ \mathbb{P}^2$ that branches along $Y$. Is there a similar construction for covers of $\mathbb{P}^1$ ? I will appreciate any reference in this direction, or anything related? Thanks

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Yes, this is the famous Riemann Existence Theorem.

In its general form, it can be stated as follows.

Theorem. Let $Y$ be a compact Riemann surface, and $D \subset Y$ be an effective reduced divisor. Then there is a $1$-$1$ correspondence between the following sets:

$\mathbf{1)}$ finite covers $f \colon X \to Y$ of degree $d$ whose branch locus lies in $D$, up to isomorphism;

$\mathbf{2)}$ group homomorphism $\rho \colon \pi_1(Y - D) \to S_d$ with transitive image, up to conjugacy in $S_d$.

Now, the fundamental group of $\mathbb{P}^1 -\{b_1, \ldots, b_k\}$ is the group generated by $k$ generators $\gamma_1, \cdots, \gamma_k$ with the unique relation $\gamma_1 \gamma_2 \cdots \gamma_k=1$, where each $\gamma_i$ is the homotopy class of a small loop on $\mathbb{P}^1$ around $b_i$. Then we obtain the following corollary:

Corollary. Let $D=\{b_1, \ldots, b_k\} \subset \mathbb{P}^1$ be a finite set of points.Then there is a $1$-$1$ correspondence between the following sets:

$\mathbf{1)}$ finite covers $f \colon X \to \mathbb{P}^1$ of degree $d$ whose branch locus lies in $D$, up to isomorphism;

$\mathbf{2)}$ conjugacy classes of $k$-tuples $(\sigma_1, \ldots, \sigma_k)$ of permutations in $S_d$ such that $\sigma_1 \sigma_2 \cdots \sigma_k=1$ and the subgroup generated by the $\sigma_i$ is transitive.

Notice that neither $X$ nor the degree $d$ of the cover are uniquely determined by $D$.

For more details, see Miranda's book [Algebraic curves and Riemann surfaces, p. 90-92].

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Thanks a lot for answer and your reference. However, the main point is the case of a non reduce divisor. For example, if I have a cover $X \to P^1$ with two branching points, may I "collapse" those points for having a branch covering over my double point? Or it makes non sense the use of fat points in this context? –  pmath Aug 8 '11 at 7:08
    
Yes you can do this since the flat limit exists. For instance, if you have a double cover and exactly two points in the branch divisor "collapse" together, you obtain a family of smooth covers whose flat limit will be a curve with a double point (liyng over the non-reduced point of the brach). If the degree of the cover is higher and more points collapse together, the situation will be more complicated. –  Francesco Polizzi Aug 8 '11 at 7:30
    
It can also happen that $X$ becomes reducible. For instance, take a double cover $\mathbb{P}^1 \to \mathbb{P}^1$ branched in exactly two points. When the two points "collapse together" the $\mathbb{P}^1$ upstairs splits into a sum of two $\mathbb{P}^1$'s intersecting in a single point. Needless to say, the intersection point lies over the branch locus (that consists of a unique non-reduced point). –  Francesco Polizzi Aug 8 '11 at 7:39

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