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What is the automorphisms group of the weighted projective space $\mathbb{P}(a_{0},...,a_{n})$ ? Consider the simplest case of a weighted projective plane, take for instance $\mathbb{P}(2,3,4)$; any automorphism has to fix the two singular points. Consider a smooth point $p\in\mathbb{P}(2,3,4)$. What is the subgroup of the automorphisms of $\mathbb{P}(2,3,4)$ fixing $p$ ?

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Will anyone answer in general what an automorphism group for a weighted projective space looks like? – Stanley Yao Xiao Sep 23 '12 at 20:48
It is a toric variety, so you could try Cox' paper on the homogenous coordinate ring of a toric variety.. – J.C. Ottem Sep 24 '12 at 16:19
up vote 16 down vote accepted

The automorphism group is the quotient of the automorphism group of the corresponding graded algebra by 1-dimensional torus acting by rescaling. In the particular case of $P(2,3,4)$ the graded algebra is $A = k[x_2,x_3,x_4]$ with $\deg x_i = i$. Note that any automorphism should take $x_2 \to a x_2$, $x_3 \to b x_3$ (since those are only elements of $A$ of degree 2 and 3) and $x_4 \mapsto c x_4 + d x_2^2$. So, the group can be written as $((k^*)^3 \ltimes k) / k^*$, where $\ltimes$ stands for the semidirect product.

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I am a bit confused about this. Consider $\mathbb P(1,2)$. This should be the same as $\mathbb P^1$. So the automorphism group is $\mathbb PGL(2)$. But with the method you describe, wouldn't you get a different answer? – mrw Jun 12 '11 at 19:08
I consider the weighted projective space as the quotient stack of a vector space (minus zero) by the torus action. This is not the same as a quotient in the category of schemes, and so the automorphism group is different. In your example, the quotient stack is $P^1$ but with an orbifold $Z/2Z$ point. The automorphism group should preserve the point. If you compare the subgroup of $PGL(2)$ preserving a point to the group analogous to the one in my answer you will see they are the same. – Sasha Jun 12 '11 at 19:53
ok. Thanks for the explanation. I also wanted to mention that the OP asked a slight different. He is asking what are the automorphisms that fix a smooth fixed point p. – mrw Jun 13 '11 at 13:52
The weighted projective plane $\mathbb{P}(2,3,4)$ is ismorphic to $\mathbb{P}(1,2,3)$. Following your method I find different automorphisms for $\mathbb{P}(1,2,3)$ and $\mathbb{P}(2,3,4)$. Where am I wrong ? – S_Z_S Jul 7 '11 at 15:03
They are not isomorphic. – Sasha Jul 7 '11 at 16:17

The group Aut(WPS) is studied in "Classes d'idéaux et Groupe de Picard des Fibrés projectifs tordus " §8.2.,8.3.

K-Theory 2 (1989),559-578 (A. AL-AMRANI)

A. Al-Amrani, May 16, 2013.

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Here is a short description of Aut(WPS), in English ! Aut(WPS) is completely computed in the article above.If the WPS is Proj(S), where S is the corresponding (weighted) A-graded algebra (here A is assumed a field), then consider G = Aut. of S as grad.-alg.. An explicit description of G is given, with an A* (units of A) weighted operation on. It is shown that the natural morphism from G to Aut(WPS) induces an isomorphism between G/A* and Aut(WPS). Remark : there is a weighted projective linear description of G/A* ( a generalization of PGL(n,A)) . A.Al-Amrani. – Al-Amrani May 17 '13 at 7:15
About last Sasha'comment above, P(2,3,4) and P(1,2,3) are isomorphic by reduction of the weights ( proved first by Charles Delorme in "Espaces projectifs anisotropes" ). The computation of Aut(PWS) I mentioned myself is done after reduction of the weights ( which is shown to be the best possible ). – Al-Amrani May 22 '13 at 14:22

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