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In the course of learning about the reolution of singularities in toric geometry, I have come across something which doesn't make much sense to me. Consider the weighted projective space $$ \mathbb{P}(1,2,2) := \mathbb{C}^{3}/(z_0,z_1,z_2)\sim(\lambda z_0,\lambda^2 z_1,\lambda^2 z_2) $$ I want to understand the singularities of this space and their resolution. Working directly, we note that the points $(0,z_1,z_2)$ have non-trivial stabilisers $\{\pm 1\}$, and so I would expect a curve of $\mathbb{Z}_{2}$ singularities. Indeed, if we introduce the standard charts $U_{i}\subseteq\mathbb{P}(1,2,2)$ such that $z_{i}\neq 0$, I find that $$ U_{0} \cong \mathbb{C}^{2} \text{ , } U_{1} \cong \mathbb{C}^{2}/(z_0,z_2)\sim(-z_0,z_2) \text{ , } U_{2} \cong \mathbb{C}^{2}/(z_0,z_1)\sim(-z_0,z_1) $$ which seems to support this idea. On the other hand, explicitly calculating the transition functions gives those of the unweighted projective space $\mathbb{CP}^{2}$. It thus seems to me that $\mathbb{P}(1,2,2)$ "looks like", $\mathbb{CP}^{2}$ with a curve of $\mathbb{Z}_{2}$ singularities.

This is all well and good so far, but my original motivation for looking at this example came from toric geometry, and for some reason I can't see these singularities reflected in the toric data.

I have taken the fan $\Delta$ consisting of the three large cones given by $$ \sigma_1=\mathbb{R}_{\geq 0}e_1+\mathbb{R}_{\geq 0}e_2 \text{ , } \sigma_2=\mathbb{R}_{\geq 0}e_1+\mathbb{R}_{\geq 0}(-2e_1-2e_2) \text{ , } \sigma_3=\mathbb{R}_{\geq 0}e_2+\mathbb{R}_{\geq 0}(-2e_1-2e_2) $$ The dual cones can then be given by $$ \check{\sigma}_1=\mathbb{R}_{\geq 0}e_1+\mathbb{R}_{\geq 0}e_2 \text{ , } \check{\sigma}_2=\mathbb{R}_{\geq 0}(-e_1)+\mathbb{R}_{\geq 0}(-e_1+e_2) \text{ , } \check{\sigma}_3=\mathbb{R}_{\geq 0}(-e_2)+\mathbb{R}_{\geq 0}(e_1-e_2) $$ Continuing with the caclulation to find the transition functions gives those of $\mathbb{CP}^{2}$, as found in the previous case.

My issues are as follow. Firstly, there are no relations between the vectors defining the dual cones, and thus each toric coordinate chart seems to be $\mathbb{C}^{2}$, rather than some $\mathbb{Z}_{2}$-quotient as I would have expected.

Secondly, each cone can be written as $\sigma = \mathbb{R}_{\geq 0}\bf{n}_{1}+\mathbb{R}_{\geq 0}\bf{n}_{2}$ where $\bf{n}_{1}$ and $\bf{n}_{2}$ form an integral basis of the $N$-lattice, $N\cong\mathbb{Z}^{2}$. As far as I understand, this is the condition for smoothness of the toric variety defined by $\Delta$.

In summary, the toric approach seems to suggest that $\mathbb{P}(1,2,2)$ is smooth and isomorphic to $\mathbb{CP}^{2}$, whereas the elementary definition of $\mathbb{P}(1,2,2)$ seems to imply a curve of $\mathbb{Z}_{2}$ singularities. Can anybody explain what is going wrong here?

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  • $\begingroup$ You should probably search a bit before posting. Perhaps this question will be helpful. $\endgroup$ Commented Dec 9, 2023 at 1:58

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A weighted projective space $\mathbb{P}(w_0,w_1,\dots,w_n)$ is called well formed if for any $i$ one has $$ \gcd\{ w_j \mid j\ne i\} = 1. $$ If $\mathbb{P}(w_0,w_1,\dots,w_n)$ is not well formed, and $d := \gcd\{ w_j \mid j\ne i\} > 1$ then one has an isomorphism $$ \mathbb{P}(w_0,w_1,\dots,w_n) \cong \mathbb{P}(w'_0,w'_1,\dots,w'_n), $$ where $w'_j = w_j/d$ for all $j \ne i$ and $w'_i = w_i$. This applies to your example and gives an isomorphism $$ \mathbb{P}(1,2,2) \cong \mathbb{P}(1,1,1) = \mathbb{P}^2. $$

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  • $\begingroup$ This makes a lot of sense. So both the toric argument and this well-formedness argument correctly predict that $\mathbb{P}(1,2,2)\cong\mathbb{P}^{2}$, but how does this resolve the apparent contradiction with the $\mathbb{Z}_{2}$ singularitites? $\mathbb{P}^{2}$ is obviously smooth, and I thought that smoothness would be preserved by isomorphism. $\endgroup$
    – CoffeeCrow
    Commented Dec 9, 2023 at 12:55
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    $\begingroup$ If a group acts on a smooth variety by complex reflections then the quotient is also smooth. For instance, the quotient of an affine line by the $\pm1$ action is again an affine line. $\endgroup$
    – Sasha
    Commented Dec 9, 2023 at 14:15
  • $\begingroup$ Oh, I see! That was surprisingly counter-intuitive for me, but I get it now. Thanks a lot for your help. $\endgroup$
    – CoffeeCrow
    Commented Dec 11, 2023 at 3:20

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