Resolution of quadric cone in $\mathbb{C}^3$ is biholomorphic to $\text{T}^\ast\mathbb{CP}^1$

Question

I am learning about Calabi-Yau metrics on Kummer surfaces and I came across a question. The background is a s follows. We want to resolve the singularity of $\mathbb{C}^2/\{\text{id},-\text{id}\}$ at the origin. To do this observe that there is a bijection between $\mathbb{C}^2/\{\text{id},-\text{id}\}$ and the quadric cone $X=\{v^2=uw\}\subset\mathbb{C}^3$ (given by $(z_1,z_2)\mapsto(z_1^2,z_1z_2,z_2^2)$, where $(z_1,z_2)\in\mathbb{C}^2$). We now blow up the origin in $\mathbb{C}^3$ and take the proper transform of $X$, denoted by $\tilde X$. The book I am reading (page 293) now claims that $\tilde X$ is obviously biholomorphic to $\text{T}^\ast\mathbb{CP}^1$. However I do not really see the argument why this is true. If somebody could provide a hint or point me to a reference I would be very grateful.

Answer 1

We have $T^*P^1 = \{ f : T_pP^1 \to \mathbb C \text{ is linear}\}$. For describe $T_pP^1$ we can notice that $T_pP^1 = \rm Hom$$(p, \mathbb C^2/p) = \{ w \in \mathbb C^2 : \langle w, p \rangle = 0 \}$. The description $T_pP^k = \rm Hom$ $ (p, \mathbb C^{k+1}/p)$ comes from the exact sequence $ 0 \to p \to \mathbb C^{k+1} \to T_pP^k \to 0$ and holds more generally for Grassmanians.

Now, a linear map $f : T_pP^1 \to \mathbb C$ gives a nilpotent matrix $A$, defined by $Ap = 0$ and $A(w) = f(w)p$ for $w \in T_pP^1$. This is a 1-1 correspondance, so we can describe $T^*P^1$ as $\{(p,A) : p \in P^1, A \text{ is nilpotent}, A(\mathbb C^2) \subset p\}$. Nilpotent matrices are exactly matrices with zero trace and zero determinant : so they are matrices on the form $\pmatrix{a & b \\ c & - a}$ with $a^2 + bc = 0$, i.e we found again our quadratic cone $\mathscr N$!

Let's check that the projection $\pi : T^*P^1 \to \mathscr N$ is the blow-up. First, if $A = 0$, the condition $A(\mathbb C^2) \subset p$ is obviously true so the fiber over $A=0$ is $\mathbb P^1$ as required. Now, if $A \neq 0$ is nilpotent, there is a unique line $p \in P^1$ such that $A(\mathbb C^2) \subset p$ so the fiber over a non-zero matrix is a point : we deduce that $\pi$ is birational so we found that $T^*P^1$ is the total space of the blow-up of the quadratic cone in $\mathbb C^3$.

In fact, this is a very particular case of a more general story called Springer theory. Here is a reference talking a bit about the resolution of the quadratic cone if you are interested.

Edit : same result from toric geometry

I'll try to introduce quickly toric variety (say toric surfaces) and show how it works on your particular case. Of course you need probably more reference, I'll advice the book of Fulton which is great and the book of Cox/Little/Schenck which contains way more details. Probably now it looks complicate but once you are familar with the theory this is very easy to do such computations !

Let me give you an example of an affine toric surface without describing the general procedure. An affine toric surface is build from a combinatorial object, namely a cone $\sigma \subset \mathbb R^2$. Let's take $\sigma = \{\alpha e_1 + \beta (e_1 + e_2) : \alpha, \beta \geq 0 \}$. It is exactly the set of vectors in the plane with argument $\theta \in [\pi/4, \pi/2]$. I can look at the dual cone, defined as $\sigma^{\vee}= \{z \in (\mathbb R^2)^{\vee} : \langle z,u \rangle \geq 0 \forall u \in \sigma\}$.

Of course, $x \mapsto (y \mapsto \langle x,y \rangle)$ identify $\mathbb R^2$ with the dual $(\mathbb R^2)^{\vee}$. From this identification it is easy to see that $\sigma^{\vee}$ is defined by the set of vectors with argument $\theta \in [0, 3\pi/4]$. Now given such dual cone I can take its algebra, i.e the monomial in my algebra will be indexed by the vector of $\sigma^{\vee}$ with integer cordinates. I define the multiplication as $X^v \cdot X^w = X^{v + w}$. This is well defined since $\sigma^{\vee} \cap \mathbb Z^2$ is stable by sum. Now, additive generators for $\sigma^{\vee}$ are $e_1, e_2, e_2 - e_1$ so multiplicative generators for my algebra will be $x, y, yx^{-1}$. So really $\mathbb C[\sigma^{\vee} \cap \mathbb Z^2] = \mathbb C[x,y,yx^{-1}] = \mathbb C[a,b,c]/(ac = b)$ which is exactly the variety in $\mathbb C^3$ parametrized by $\psi((a,c)) = (a,ac, c)$. So we got a variety which is isomorphic to $\mathbb C^2$.

I'll take a different example now. Consider the cone $\tau = \{\alpha(e_2 - e_1) + \beta(e_1 + e_2), \alpha, \beta \geq 0\}$. Again from a picture it is clear that $\tau$ is the set of vectors of the plane with argument $\theta \in [\pi/4, 3\pi/4]$. The dual cone is $\tau = \tau^{\vee}$ (again a picture is helpful). In particular, $\tau^{\vee} \cap \mathbb Z^2$ is generated additively by $e_2, e_2 - e_1$ and $e_2 + e_1$. Now my algebra is $\mathbb C[x^{-1}y,y,xy] = \mathbb C[a,b,c]/(ac - b^2)$. So now the toric variety is the quadratic cone.

This is the way how to build affine surfaces : take two linearly independant vectors in $\mathbb R^2$, take the cone generated by these vectors, the dual cone and the spectrum of the algebra of the dual cone (intersected with $\mathbb Z^2$). Of course I didn't explain how to do it and didn't motivate this strange construction, but this is how to build affine toric varieties and they have pretty good properties. For example, my affine surface is smooth if and only if the two vectors which generate my cone also generate the lattice $\mathbb Z^2$.

Now we need to build projective varieties. We will build them as gluing toric affine variety. Again let's just restrict to surface, and let's try to glue two cones together. I'll take the self-dual cones $\sigma$ generated by $e_1,e_2$ and $\sigma'$ generated by $-e_1,e_2$. The algebra of the dual cones are $A = \mathbb C[x,y]$ and $A' = \mathbb C[x^{-1},y]$. Of course, the localisation agree, i.e $A_x = A'_{x^{-1}}$. This is exactly the gluing map between the two affine varieties, gluing two copies of $\mathbb C^* \times \mathbb C$. When we glue $Spec(A)$ and $Spec(A')$ together we will get $P^1 \times \mathbb C$.

Now let's finally return to your problem. The tangent bundle of $P^1$ is $O(2)$ as a vector field on the sphere has two zeroes. So the cotangent bundle $T^*P^1$ is given by $O(-2)$, i.e this is the gluing of two copies of $\mathbb C^2$ along $\mathbb C^{*} \times \mathbb C$, the coordinates are $(z, \lambda)$ on the first chart, and the gluing map is given by $(z,\lambda) \mapsto (z^{-1}, z^2\lambda)$. This is a standard computation but I can put more details if you want.

Another cool point of toric surface (but it works in all dimension) is that blow-up of a toric variety can be obtained from remplacing a cone by a fan, inserting an edge in the interior of the cone. Recall that the cone of your singular surface was $\{\alpha v_1 + \beta v_2 : \alpha, \beta \geq 0, v_1 = e_1 + e_2, v_2 = e_2 - e_1 \}$. The blow-up is simply putting one more edge in the middle, so we get a fan with $\sigma = \{\alpha(e_1 + e_2,) + \beta e_2\}$ and $\sigma' = \{\alpha(e_2 - e_1) + \beta e_2\}$. These two affine varieties are isomorphic to $\mathbb C^2$ as for example $\mathbb C[\sigma^{\vee}] = \mathbb C[x^{-1}y, y,x] = \mathbb C[x^{-1}y,x]$. So it is $\mathbb C^2$ with coordinates $a = x, b = x^{-1}y$. Similary $\mathbb C[\sigma'^{\vee}]$ is $\mathbb C^2$ with coordinate $c = x^{-1}, d = xy$. But up to localisation by $x/x^{-1}$, they agree ! What is the gluing map ? $c = a^{-1}, d = a^2b$. Sounds familiar ? This is exactly the cotangent bundle of $P^1$ !