As stacks are the weighted projective line $\mathbb{P}$(1,n1) and $\mathbb{P}$(k,nk) isomorphic? Is there any reference for this?

2$\begingroup$ You should provide more details and give some idea of why you are interested in this problem. $\endgroup$– Chris RamseyFeb 1, 2016 at 17:13

1$\begingroup$ I think the question is pretty clear and that is all people here need to know, I guess. $\endgroup$– mathvaderFeb 1, 2016 at 18:07

$\begingroup$ Fair enough. I am glad that you got the answers you were seeking. $\endgroup$– Chris RamseyFeb 1, 2016 at 18:48
2 Answers
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.

$\begingroup$ 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? $\endgroup$– pbelmansFeb 5, 2016 at 18:45

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 $Pic$. $\endgroup$ Feb 7, 2016 at 11:01
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 $\Bbb P(1,n1)$, the inclusion of graded rings $k[x,y^{n1}] \to k[x,y]$ (where $y = 1$ and $x = n1$) 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 $\Bbb A^1 \to \Bbb P(1,n1)$ with this image), but the stabilizer of the point where $y=0$ is the cyclic group of $(n1)$'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 $\{\pm 1\}$.

$\begingroup$ Actually I do not want to consider cases like $\mathbb{P}(2,2)$. I have in mind cases like $\mathbb{P}(1,4)$ and $\mathbb{P}(2,3)$ $\endgroup$ Feb 1, 2016 at 18:05

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 groupsthe second roots of unity and the third roots of unity respectively. $\endgroup$ Feb 1, 2016 at 18:14