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Fix two positive integers $a$ and $b$. Consider a weighted projective line $\mathbb{P}(a,b)$ as a quotient stack $$[(\mathbb{C}^2-\{0\})/\mathbb{C}^*]$$ where $\mathbb{C}^*$ acts on $\mathbb{C}^2-\{0\}$ by $$\lambda \cdot (x,y) = (\lambda^a x,\lambda^b y).$$ Then $\mathbb{P}(a,b)$ has its underlying quotient space $|\mathbb{P}(a,b)| = (\mathbb{C}^2-\{0\})/\mathbb{C}^*$ as a coarse moduli space. It's known that $\mathbb{P}(a,b)$ also has the usual projective line $\mathbb{P}^1$ as its coarse moduli space (which I believe is not true in higher dimensions).

What's the morphism $\mathbb{P}(a,b) \to \mathbb{P}^1$ corresponding to the coarse moduli space $\mathbb{P}^1$? In other words, how to show that $|\mathbb{P}(a,b)|$ is homeomorphic to $\mathbb{P}^1$?

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  • $\begingroup$ Isn't the coarse moduli space of any weighted projective space just projective space? $\endgroup$
    – Harry Gindi
    Aug 2, 2020 at 3:08
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    $\begingroup$ I think you send $(x,y) \to (x^b,y^a)$? I believe the obvious generalization to higher dimensions works too. $\endgroup$
    – Asvin
    Aug 2, 2020 at 3:42
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    $\begingroup$ @HarryGindi: this is not true in general (but ok in dimension $1$). They are all toric varieties, and so can be described by their fans. I believe that $\mathbf P(d_0,\ldots,d_n)$ is isomorphic to $\mathbf P^n$ if and only if it is smooth; but already things like $\mathbf P(1,1,2)$ are singular! See for example Dolgachev's Weighted projective varieties (available on his website), 1.2.3 and Prop. 1.3.3(iii), complemented with Fulton's Toric varieties, §2.2. $\endgroup$
    – R. van Dobben de Bruyn
    Aug 2, 2020 at 4:48
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    $\begingroup$ I think @Asvin's suggestion is right, except you need to divide by their $\gcd$ (which is often assumed $1$). The proof is by computing the (GIT) invariants, or computing the $\operatorname{Proj}$ of the obvious graded ring. There is no "obvious generalisation to higher dimension". $\endgroup$
    – R. van Dobben de Bruyn
    Aug 2, 2020 at 5:06
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    $\begingroup$ It seems to be a special case of Prop. 1.3.1 in Dolgachev's paper (linked above). (Typo: I think it should be $q_0/a_0$, not $q_0/a$. The latter is not an integer.) Specifically, see the second corollary of 1.3.1. $\endgroup$
    – R. van Dobben de Bruyn
    Aug 2, 2020 at 5:19

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