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A map from an algebraic variety $X$ to a projective space is the same thing as a globally generated line bundle on $X$. What geometric object on $X$ corresponds to a map to a weighted projective space?

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A line bundle $L$ on $X$ and a set of sections $(s_1, s_2, \ldots, s_k)$ where $s_i \in H^0(X, \, L^{a_i})$? (If you want a morphism, you must also require that the common zero locus of these sections is empty). –  Francesco Polizzi Jul 12 at 8:33
    
@FrancescoPolizzi That makes sense! Maybe you can write it as an answer. I was thinking that it should be some kind of orbi-bundle, but the fact that $X$ has no orbifold structure was making me confused. –  Dan Petersen Jul 12 at 9:59

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up vote 11 down vote accepted

Let me expand my comment in a short answer.

One of the most common way to build a rational map $f \colon X \dashrightarrow \mathbb{P}(a_1, \ldots, a_n)$ is to consider a line bundle $\mathscr{L}$ on $X$ together with sections $$\sigma_1 \in H^0(X, \, \mathscr{L}^{a_1}), \quad \sigma_2 \in H^0(X, \, \mathscr{L}^{a_2}), \ldots, \sigma_n \in H^0(X, \, \mathscr{L}^{a_n}),$$ and to define $f(x) := [\sigma_1(x): \ldots : \sigma_n(x)].$

Clearly, $f$ is a morphism if and only if the intersection of the zero loci of the sections $\sigma_i$ is empty.

The basic example is given by hyperelliptic curves. Let $X$ be a smooth hyperelliptic curve of genus $g$ and take as $\mathscr{L}$ the hyperelliptic involution, i.e. the unique $g^1_2$ of $X$. Then it is no difficult to show that there exist $x_0, \, x_1 \in H^0(X, \, \mathscr{L})$ and $y \in H^0(X, \mathscr{L}^{g+1})$ that satisfy a polynomial relation of type $$y^2 = F_{2g+2}(x_0, \, x_1), \quad (\sharp)$$ where $F_{2g+2}$ is a homogeneous form of degree $2g+2$.

This precisely means that we can write $X$ as a double cover of $\mathbb{P}^1$, branched at $2g+2$ points. Trying to homogenize the equation $(\sharp)$ in the usual way introduces a complicate singularity at infinity. Instead, it is more convenient to see $(\sharp)$ as the global equation of $X$ into the weighted projective plane $\mathbb{P}(1, \, 1,\, g+1)$.

In fact, it is easy to see that the map $f \colon X \longrightarrow \mathbb{P}(1, \, 1, \, g+1)$ defined by the sections $x_0, \, x_1, \, y$ is an embedding (notice that the image does not pass through the singular point $[0: 0: 1]$ of the weighted projective plane).

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Thanks! (blank space) –  Dan Petersen Jul 12 at 15:04

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