As stacks are the weighted projective line $\mathbb{P}$(1,n-1) and $\mathbb{P}$(k,n-k) isomorphic? Is there any reference for this?
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2$\begingroup$ You should provide more details and give some idea of why you are interested in this problem. $\endgroup$– Chris RamseyCommented Feb 1, 2016 at 17:13
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1$\begingroup$ I think the question is pretty clear and that is all people here need to know, I guess. $\endgroup$– mathvaderCommented Feb 1, 2016 at 18:07
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$\begingroup$ Fair enough. I am glad that you got the answers you were seeking. $\endgroup$– Chris RamseyCommented Feb 1, 2016 at 18:48
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2 Answers
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Let $X$ be a weighted projective (stacky) line. At least if we work over a field $K$, the Picard group of $X$ is $\mathbb{Z}$. Only powers of one of the two generators have global sections. Call this generator $\mathcal{O}(1)$ and its powers $\mathcal{O}(l)$. Thus, we can attach to $X$ the graded ring $R_*(X) =\bigoplus_{l\in\mathbb{Z}}H^0(X; \mathcal{O}(l))$.
We have $R_*(\mathbb{P}(n,k)) = K[x,y]$ with $|x| = n$ and $|y| = k$. Thus, we can recover the degrees $n$ and $k$ from the graded ring $R_*(X)$. In particular, the weights $k$ and $n$ are uniquely determined.
answered Feb 1, 2016 at 18:17
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$\begingroup$ Is there a difference in terminology for weighted projective line or Picard group that explains why you say the Picard group is $\mathbb{Z}$ whereas e.g. section 2 of arxiv.org/abs/1409.7050v1 says that is a bit bigger than that? $\endgroup$– pbelmansCommented Feb 5, 2016 at 18:45
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3$\begingroup$ For me, a weighted projective (stacky) line is the stack quotient quotient $(Spec K[x,y]-0)/\mathbb{G}_m$, where $\mathbb{G}_m$ acts by $\lambda^a$ on $x$ and by $\lambda^b$ on $y$ for some weights $a,b\in\mathbb{Z}_{>0}$. We are in the world of the linked paper if $a$ and $b$ are coprime (no generic stabilizer). Now note that $\mathbb{Z}w\oplus \mathbb{Z}z/(aw = bz) \cong \mathbb{Z}$ with generator $cw+dz$, where $cb+ad = 1$. Thus, in the intersection of the two concepts, we get the same answer for $Pic$. $\endgroup$ Commented Feb 7, 2016 at 11:01
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This is not true, and you can verify that it is not so by checking the group of automorphisms of the $k$-points for $k$ a field. In the case of $\Bbb P(1,n-1)$, the inclusion of graded rings $k[x,y^{n-1}] \to k[x,y]$ (where $|y| = 1$ and $|x| = n-1$) gives a projection down to a coarse moduli space $\Bbb P^1$. The stabilizers of the points where $y \neq 0$ are all trivial (there is an open embedding $\Bbb A^1 \to \Bbb P(1,n-1)$ with this image), but the stabilizer of the point where $y=0$ is the cyclic group of $(n-1)$'st roots of unity.
By contrast, $\Bbb P(2,2)$ also has coarse moduli space $\Bbb P^1$ but the automorphism group of any point is $\{\pm 1\}$.
answered Feb 1, 2016 at 17:23
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$\begingroup$ Actually I do not want to consider cases like $\mathbb{P}(2,2)$. I have in mind cases like $\mathbb{P}(1,4)$ and $\mathbb{P}(2,3)$ $\endgroup$ Commented Feb 1, 2016 at 18:05
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1$\begingroup$ @mathvader Very well, the same type of analysis will be effective. In the case of $\Bbb P(1,4)$ there is only one $k$-point with a nontrivial automorphism group, and it is the group of fourth roots of unity. In the case of $\Bbb P(2,3)$ there are exactly two $k$-points with nontrivial automorphism groups--the second roots of unity and the third roots of unity respectively. $\endgroup$ Commented Feb 1, 2016 at 18:14