I am trying to find the transition functions of the weighted projective space $\mathbb{P}^{2}_{\{3,2,1\}}$. Bear with, I am very new to algebraic geometry.
To my understanding, $\mathbb{P}^{2}_{\{3,2,1\}}$ is defined as the quotient of $\mathbb{C}^{3}\setminus{\{0\}}$ by the equivalence relation: $$ (z_{0},z_{1},z_{2})\sim(\lambda^{3}z_{0},\lambda^{2}z_{1},\lambda^{1}z_{2}) \text{ , with }\lambda\in\mathbb{C}^{*}\text{.} $$
As far as I understand, instead of having charts as in differential geometry, in algebraic geometry we look at "coordinate patches". The obvious guess (which clearly covers $\mathbb{P}_{\{3,2,1\}}$), is:
\begin{align*} & U_{1} = \{ [1:z_{1}:z_{2}] | z_{1},z_{2}\in\mathbb{C} \} \\ & U_{2} = \{ [z_{0}:1:z_{2}] | z_{0},z_{2}\in\mathbb{C} \} \\ & U_{3} = \{ [z_{0}:z_{1}:1] | z_{0},z_{1}\in\mathbb{C} \} \\ \end{align*} where we have used square brackets to denote equivalence classes.
Then as far as I understand, these patches are supposed to be isomorphic to (some quotient of?), affine $2$-space, in the same way we would expect homeomorphisms with $\mathbb{R}^{2}$ for manifolds.
The obvious isomorphisms are: \begin{align*} & \phi_{1}:U_{1}\to\mathbb{A}^{2} \text{ , } [1:z_{1}:z_{2}] \mapsto (z_{1},z_{2}) \\ & \phi_{2}:U_{2}\to\mathbb{A}^{2} \text{ , } [z_{0}:1:z_{2}] \mapsto (z_{0},z_{2}) \\ & \phi_{3}:U_{3}\to\mathbb{A}^{2} \text{ , } [z_{0}:z_{1}:1] \mapsto (z_{0},z_{1}) \end{align*} with inverses: \begin{align*} & \phi_{1}^{-1}:\mathbb{A}^{2}\to U_{1} \text{ , } (x,y)\mapsto[1:x:y] \\ & \phi_{2}^{-1}:\mathbb{A}^{2}\to U_{2} \text{ , } (x,y)\mapsto[x:1:y] \\ & \phi_{2}^{-1}:\mathbb{A}^{2}\to U_{2} \text{ , } (x,y)\mapsto[x:y:1] \\ \end{align*} which are also locally given by polynomials.
The transition functions should then be given by $\tau_{ij}=\phi_{j}\circ\phi_{i}^{-1}$.
This means that: $$ \tau_{12}(x,y) = (x^{-3/2},x^{-1/2}y) \text{ , and } \tau_{23}(x,y) = (y^{-3}x,y^{-2}) \text{.} $$
If we now write the components as $\tau_{ij}(u_{i,1},u_{i,2})=(u_{j,1},u_{j,2})$, this says that: \begin{align*} & u_{1,1}=u_{2,1}^{-2/3} \text{ , } u_{1,2}=u_{2,2}u_{2,1}^{-1/3} \\ & u_{2,2}=u_{3,2}^{-1/2} \text{ , } u_{2,1}=u_{3,1}u_{3,2}^{-3/2} \end{align*}
This seems fine to me, but the answer given in the paper I'm reading is: \begin{align*} & u_{1,1}=u_{2,1}^{-1} \text{ , } u_{1,2}^{3}=u_{2,2}u_{2,1}^{-2} \\ & u_{2,2}=u_{3,2}^{-1} \text{ , } u_{2,1}^{2}=u_{3,1}^{3}u_{3,2}^{-1} \end{align*}
I have gone over my workings several times and can't find any issues, so I am probably misunderstanding something fundamental. Any help would be much appreciated.
EDIT:
Thanks to reuns' comment, I have realised that $U_{1}$ and $U_{2}$ are not simply affine space. To see this, note that for $\zeta_{n}$ an n'th root of unity, $[1:z_{1}:z_{2}]=[1:\zeta_{3}^{2}z_{1}:\zeta_{3}z_{2}]$. This means that for the map $\phi_{1}$ to be well defined we must take the quotient by $\mathbb{Z}_{3}$ with this action. This gives:
\begin{align*} & U_{1}\cong\mathbb{A}^{2}/(z_{1},z_{2})\sim(\zeta_{3}^{2}z_{1},\zeta_{3}z_{2}) \\ & U_{2}\cong\mathbb{A}^{2}/(z_{0},z_{2})\sim(\zeta_{2}z_{0},\zeta_{2}z_{2}) \\ & U_{3}\cong\mathbb{A}^{2} \end{align*}
Since my transition functions are invariant under these actions, I am still unsure about how to obtain the form given in my reference.
UPDATE:
Thinking about this question again sometimes later, I have made a little progress. In particular, I have been able to identify the affine charts \begin{align*} & U_{1} \cong \mathbb{A}^{2}/(z_{1},z_{2})\sim(\zeta_{3}^{2}z_{1},\zeta_{3}z_{2}) \cong V(b^3-ac) \\ & U_{2} \cong \mathbb{A}^{2}/(z_{0},z_{2})\sim(\zeta_{2}z_{0},\zeta_{2}z_{2}) \cong V(b^2-ac) \\ & U_{3} \cong \mathbb{A}^{2} \cong V(b-ac) \end{align*} under the maps \begin{align*} & \phi_{1}:U_{1}\to V(b^3-ac) \text{ , } [1:z_{1}:z_{2}] \mapsto (z_{1}^3,z_{1}z_{2},z_{2}^3) \\ & \phi_{2}:U_{2}\to V(b^2-ac) \text{ , } [z_{0}:1:z_{2}] \mapsto (z_{0}^{2},z_{0}z_{2},z_{2}^{2}) \\ & \phi_{3}:U_{3}\to V(b-ac) \text{ , } [z_{0}:z_{1}:1] \mapsto (z_{0},z_{0}z_{1},z_{1}) \end{align*} We then find that \begin{align*} & (\phi_{1}\circ\phi_{2}^{-1})(a^2,ac,c^2) = \left( \frac{1}{a^{2}},\frac{c}{a},\frac{c^{3}}{a} \right) \\ & (\phi_{2}\circ\phi_{3}^{-1}) (a,ac,c) = \left( \frac{a^2}{c^3},\frac{a}{c^2},\frac{1}{c} \right) \end{align*} And so we see that indeed the transition functions are rational! However, I have still been unable to rephrase these transition functions into the form given above. In particular, I am struggling with now having three constrained variables to work with instead of two free variables.