Blowup of a (very) simple singularity

Question

Take the action of $\mathbb{Z}_2$ on $\mathbb{C}^2$ given by $(-1) \cdot (z,w) = (-z,-w)$ and of course, $(1)\cdot (z,w) = (z,w)$. If you look at the resulting quotient space $\mathbb{C}^2 / \mathbb{Z}_2$, there is a simple singularity (rational double point) at the origin. It is of type $A_2$, so you can resolve it by blowing up once. I have it written that the resulting space is equal to $T^* \mathbb{P}^1$. This last part I don't understand. So why is \begin{equation*} \text{Bl}_{(0,0)} \left( \mathbb{C}^2 / \mathbb{Z}_2 \right) = T^* \mathbb{P}^1\;? \end{equation*} Thanks for your help.

Answer 1

The most straightforward way to see this (in my opinion) is to view this variety as $V(XZ=Y^2) \subset \mathbb{C}^3$. This comes from the fact that the invariant functions under the $\pm 1$ action are $z^2, w^2, zw$, and the relationship is $(z^2)(x^2) = (zw)^2$. The blow-up at the singularity, which is the origin is given by $$ Bl_{(0,0)}(V) = \{((X,Y,Z), [t:u:v]) \in \mathbb{C}^3 \times \mathbb{P}^2 \vert \quad X u= Yt, \quad Xv = Zt, \quad Yv = Zu, \quad XZ=Y^2\} $$ If you look at the open neighborhoods of $\mathbb{C}^3 \times \mathbb{P}^2$ given by $t\neq 0$, $u\neq 0$ and $v \neq 0$, you can look at their overlaps and find how they glue together. This can be notationally heavy so I won't do it here, but you get that a) the second open neighborhood is redundant and b) the transformation of the first to the second and vice versa basically yield a square term. This tells us that the space is actually the total space of $\mathcal{O}_{\mathbb{P}^1}(-2)$.

So $Bl_{(0,0)} (\mathbb{C}^2 / \mathbb{Z}_2) = \mathcal{O}_{\mathbb{P}^1}(-2)$. Now, by Huybrechts' "Complex Geometry: An Introduction", Proposition 2.4.3, we know that the canonical bundle $K_{\mathbb{P}^1} = \mathcal{O}_{\mathbb{P}^1}(-2)$. But the canonical bundle is $det(\Omega_{\mathbb{P}^1}) = \Omega_{\mathbb{P}^1}^1 = T^*\mathbb{P^1}$.

Therefore $$ Bl_{(0,0)} (\mathbb{C}^2 / \mathbb{Z}_2) = T^*\mathbb{P}^1 $$