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.
Best Answer
I finally figured it out! The charts we want to use are those yielding the isomorphisms $U_{i}\cong\mathbb{A}^{2}/\sim$. In addition, we need to relabel our charts to obtain the given transition functions. More explicitly, we set \begin{align*} U_{1} &= \{ [z_0:z_1:1]|z_0,z_1\in\mathbb{C} \} \\ U_{2} &= \{ [1:z_1:z_2]|z_1,z_2\in\mathbb{C} \} \\ U_{3} &= \{ [z_0:1:z_2]|z_0,z_2\in\mathbb{C} \} \end{align*} and choose the following isomorphisms \begin{align*} & f_{1}:U_{1}\to\mathbb{A}^{2} \text{ , } [z_0:z_1:1] \mapsto (z_{0},z_{1}) \\ & f_{2}:U_{2}\to\mathbb{A}^{2}/\sim \text{ , } [1:z_1:z_2] \mapsto (z_{2}^3,z_{1}^3) \\ & f_{3}:U_{3}\to\mathbb{A}^{2}/\sim \text{ , } [z_0:1:z_2] \mapsto (z_{2}^2,z_{0}^2) \end{align*} where we have made a slightly non-canonical choixe in which we swap the ordsering of the factors in the second two maps. This avoids a relabelling at the end. We then find that \begin{align*} & (f_1\circ f_2^{-1})(x,y) = (x^{-1},x^{-2/3}y^{1/3}) \\ & (f_2\circ f_3^{-1})(x,y) = (y^{-1/2}x^{3/2},y^{-1}) \end{align*} and so \begin{align*} & u_{11} = u_{21}^{-1} \text{ , } u_{12}^{3} = u_{22}u_{21}^{-2} \\ & u_{22} = u_{32}^{-1} \text{ , } u_{21}^{2} = u_{31}^{3}u_{32}^{-1} \end{align*} as required. I think that the slightly funny choices of charts and maps are to make the similarity to $\mathbb{CP}^{2}$ as manifest as possible, although I am still not quite sure why they don;t just reverse the ordering and look at $\mathbb{P}_{\{1,2,3\}}$ in the first place!