Cone over the Veronese surface

Question

Let $V\subset\mathbb{P}^5$ be the Veronese surface and let $X\subset\mathbb{P}^6$ be the cone over it. Since $X$ is $\mathbb{Q}$-factorial there are two integers $a,b$ such that $aK_X = \mathcal{O}_X(b)$.

Furthermore, if $f:Y\rightarrow X$ is the blow-up of the vertex then $Y$ is smooth and we may write $K_Y = f^{*}K_{X}+dE$, where $E$ is the exceptional divisor, and $d$ is a rational number.

How can I determine the numbers $a,b,d$ ?

Answer 1

Let me start being a little nitpicking with the formulation of the question. The fact that $X$ is $\mathbb Q$-factorial does not in itself imply that such $a$ and $b$ exists. One also needs the fact that the Picard number of $X$ is $1$. This is indeed true, but perhaps should be mentioned. In fact, the Picard group of $X$ is $\mathbb Z$ which was silently used in some answers. This is especially not obvious in cases where quotients are taken. Here it is (of course?) trivial, but in my opinion deserves mentioning. One possible way to see it is that $X\setminus \{P\}$ (where $P$ is the vertex) is an $\mathbb A^1$-bundle over $\mathbb P^2$, so its Picard group is $\mathbb Z$ and hence the group of Weil divisors on $X$ is also $\mathbb Z$ (this makes sense as $X$ is normal, because the Veronese embedding is projectively normal). This, by the way, is also a proof that $X$ is $\mathbb Q$-factorial (since it is projective), and that the Picard group is a subgroup which is also isomorphic to $\mathbb Z$.

So, let's get to answering the questions:

First, let us determine $a$ and $b$. These do not require blowing up. Let $H\subset X$ denote a general hyperplane section of $X$, so $H\simeq V\simeq \mathbb P^2$. Then since $V\subset \mathbb P^2$ is the $2$-uple embedding, $H\left|_H\right.$ is (linearly equivalent to the image of) a conic in $\mathbb P^2$. By adjunction $$-3=\deg K_H = \deg \left((K_X+H)\left|_H \right.\right)$$ (notice that $H$ avoids the singular point of $X$, so this is OK). It follows that $$\deg K_X\left|_H\right.= -5.$$

Since the Picard group of $X$ is $\mathbb Z$ (!!) it follows that $2K_X\sim -5H$, i.e., that $a=2$ and $b=-5$ (or any multiple of these values, e.g., $a=-2$ and $b=5$ works as well).

Next, let us determine $d$: From the equation $K_Y\sim f^*K_X + dE$, using adjunction, one obtains that $$(d+1)E\left|_E\right.\sim K_E.\tag{$*$}$$ Since $f$ was obtained as a blow-up of $\mathbb P^6$ at a point, $-E\left|_E\right.$ is linearly equivalent to the restriction of the hyperplane class of the exceptional $\mathbb P^5$ and hence it is a conic on $E$. Combined with $(*)$ this implies that $$-2(d+1)=-3,$$ that is, that $$d=\frac 12.$$

Answer 2

The answers are the following.

(1) It is well known that the singularity at the vertex of the cone over the Veronese surface is isomorphic to a quotient singularity of type $\frac{1}{2}(1, \, 1, \,1)$, that is the isolated singularity given by the quotient of $\mathbb{C}^3$ by the action of $\mathbb{Z}/2 \mathbb{Z}$ of the form $(x, \,y, \,z) \mapsto (-x, \, -y, \, -z)$. It is no difficult to check the claim directly by looking at the ring of invariants for the action. This is a terminal quotient singularity whose discrepancy is $1/2$ (I think that a detailed computation is carried out in [Reid, Young person guide to canonical singularities]).This means $$K_Y = f^*K_X + \frac{1}{2}E,$$ hence $d=\frac{1}{2}$.

(2) EDIT. This part was corrected by following Sandor Kovacs' observations.

We can identify $X$ with the weighted projective space $\mathbb{P}=\mathbb{P}(1,\, 1,\, 1, \, 2)$. Therefore, by using standard formulas for projective weighted complete intersections (see for instance Dolgachev's paper Weighted projective varieties) we compute the dualizing sheaf of $\mathbb{P}$, obtaining $$\omega_\mathbb{P} = \mathcal{O}_{\mathbb{P}}(-1-1-1-2) = \mathcal{O}_{\mathbb{P}}(-5).$$

Here the (non Cartier) reflexive coherent sheaf $\mathcal{O}_{\mathbb{P}}(1)$ is a generator for the class group $\textrm{Cl}(\mathbb{P})$, whereas the Cartier divisor $H= \mathcal{O}_\mathbb{P}(2)$ is a generator for the Picard group $\textrm{Pic}(\mathbb{P}).$ This shows that $\mathbb{P}$ is $2$-Gorenstein, in fact $$2K_\mathbb{P} = -5H.$$ Finally, the identification of $X$ with $\mathbb{P}$ induces an identification of $H$ with the hyperplane section $\mathcal{O}_X(1)$ of $X$, hence we obtain $$2K_X = \mathcal{O}_X(-5),$$ that is we can take $a=2$ and $b=-5$.

Answer 3

Here is a proof of $d = \frac{1}{2}$. Then you can argue as Francesco did in point two of his answer.

Consider the action: $$ \begin{array}{ccc} \mu_{2}\times\mathbb{A}^{3} & \longrightarrow & \mathbb{A}^{3}\\ (\epsilon,x_{1}, x_{2}, x_3) & \longmapsto & (\epsilon x_{1},\epsilon x_{2}, \epsilon x_3) \end{array} $$ The ring of invariants is given by: $$k[x_0^2,x_0x_1,x_0x_2,x_1^2,x_1x_2,x_2^2]\cong \frac{k[y_0,y_1,y_2,y_3,y_4,y_5]}{(y_0y_3-y_1^2,y_0y_4-y_1y_2,y_0y_5-y_2^2,y_1y_4-y_2y_3,y_1y_5-y_2y_4,y_3y_5-y_4^2)}$$ The singularity $X=\mathbb{A}^{3}/\mu_{2}$ corresponds to the vertex $v$ of the affine cone $X$ over a Veronese surface $V\subset\mathbb{P}^5$. The differential form $dx_0\wedge dx_1\wedge dx_2$ is a basis of $\bigwedge^3\Omega_{\mathbb{A}^3}$, and $(dx_0\wedge dx_1\wedge dx_2)^{\otimes 2}$ is invariant under the action. The form $$\omega = \frac{(dy_0\wedge dy_1\wedge dy_2)^{\otimes 2}}{y_0^3}\in \left(\bigwedge^3\Omega_{k(X)}\right)^{\otimes 2}$$ is a basis of $(\bigwedge^3\Omega_{X})^{\otimes 2}$ because the quotient map $\pi:\mathbb{A}^3\rightarrow X$ is \'etale on $X\setminus\{v\}$, and $$\pi^{*}\omega = \frac{(4x_0^6 (dx_0\wedge dx_1\wedge dx_2))^{\otimes 2}}{x_{0}^{6}} = 4(dx_0\wedge dx_1\wedge dx_2)^{\otimes 2}.$$ Blowing-up the vertex $v$ we get a resolution $f:Y\rightarrow X$, and we have an affine chart isomorphic to $\mathbb{A}^3$ with coordinates $(y_0,s,t)$ where the resolution is given by $(y_0,s,t)\mapsto (y_0,y_0s,y_0t,y_0s^2,y_0st,y_0t^2)$, and the exceptional divisor $E$ over $v$ is given by $\{y_0=0\}$. We have $$f^{*}\omega = y_0(dy_0\wedge ds\wedge dt)^{\otimes 2}.$$ Therefore, $f^{*}\omega$ has a zero along $E$, and we may write $K_Y = f^{*}K_X+\frac{1}{2}E$.

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Just to give you an idea of the computations which are involved here is the simpler example of a quadric cone.

Let us consider the action: $$ \begin{array}{ccc} \mu_{2}\times\mathbb{A}^{2} & \longrightarrow & \mathbb{A}^{2}\\ (\epsilon,x_{1}, x_{2}) & \longmapsto & (\epsilon x_{1},\epsilon x_{2}) \end{array} $$ The ring of invariants is given by: $$k[x_0^2,x_0x_1,x_1^2]\cong k[y_0,y_1,y_2]/(y_0y_2-y_1^2)$$ and we see that the singularity $X=\mathbb{A}^{2}/\mu_{2}$ corresponds to the vertex $v$ of the affine cone $$X=Spec(k[x_0^2,x_0x_1,x_1^2]\cong k[y_0,y_1,y_2]/(y_0y_2-y_1^2))$$ that is the vertex of a quadric cone $Q\subset\mathbb{P}^2$ or equivalently the singularity $\frac{1}{2}(1,1)$ of the weighted projective plane $\mathbb{P}(1,1,2)$. Now, $dx_0\wedge dx_1$ is a basis of $\bigwedge^2\Omega_{\mathbb{A}^2}$, and $(dx_0\wedge dx_1)^{\otimes 2}$ is invariant under the action. The form $$\omega = \frac{(dy_0\wedge dy_1)^{\otimes 2}}{y_0^2}\in \left(\bigwedge^2\Omega_{k(X)}\right)^{\otimes 2}$$ is a basis of $(\bigwedge^2\Omega_{X})^{\otimes 2}$ because the quotient map $\pi:\mathbb{A}^2\rightarrow X$ is \'etale on $X\setminus\{v\}$, and $\pi^{*}\omega = 4(dx_0\wedge dx_1)^{\otimes 2}$.\ Blowing-up the vertex $v$ we get a resolution $f:Y\rightarrow X$. If $[\lambda_0:\lambda_1:\lambda_2]$ are homogeneous coordinates on $\mathbb{P}^2$ then the equations of $Y$ in $\mathbb{A}^3\times\mathbb{P}^2$ are: $$ \left\lbrace\begin{array}{l} y_0\lambda_1-y_1\lambda_0=0, \\ y_0\lambda_2-y_2\lambda_0=0, \\ y_1\lambda_2-y_2\lambda_1=0, \\ y_0y_2-y_1^2. \end{array}\right. $$ Therefore, $y_1 = \frac{\lambda_1}{\lambda_0}y_0$, and $\frac{\lambda_2}{\lambda_1}=\frac{\lambda_1}{\lambda_0}$ yields $y_2 = \frac{\lambda_1}{\lambda_0}y_1 = (\frac{\lambda_1}{\lambda_0})^2y_0$. Then, in $Y$ we have an affine chart isomorphic to $\mathbb{A}^2$ with coordinates $(y_0,t)$ where the resolution is given by $(y_0,t)\mapsto (y_0,y_0t,y_0t^2)$, with $t = \frac{\lambda_1}{\lambda_0}$, and the exceptional divisor $E$ over $v$ is given by $\{y_0=0\}$. We have $$f^{*}\omega = (dy_0\wedge dt)^{\otimes 2}.$$ Therefore, $f^{*}\omega$ has neither a pole nor a zero along $E$, and we may write $K_Y = f^{*}K_X$.

Answer 4

Here's another way to get $d=1/2$, though it's probably good to do the computation in coordinates (once, anyway!). We have $K_Y = f^\ast K_X + dE$ and want to find $d$. Then $K_Y+E = f^\ast K_X + (d+1)E$, and $f^\ast K_X \vert_E = 0$, since it's a pullback. Restricting both sides to $E$ gives $K_E = (d+1) E \vert_E$, so if we knew the normal bundle of $E$ we'd be done.

But $Y = \mathbb P_{\mathbb P^2}(\mathcal O + \mathcal O(2))$, since it's the blow-up of the cone over the embedding of $\mathbb P^2$ by $\mathcal O(2)$. Moreover, $E$ is the section coming from $\mathcal O + \mathcal O(2) \to \mathcal O$, so $N_E = \mathcal O(-2)$; this is like the computation of the normal bundle to the negative section of a Hirzebruch surface. Since $K_E = \mathcal O(-3)$, we have $-3 = (d+1)(-2)$, which gives $d=1/2$.

I second Francesco's suggestion to read Reid's "Young Person's Guide", where there's lots of this sort of thing.