Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Define the surface $X$ to be the total space of $\mathcal{O}_{\mathbb{P}^1}(-5)$. By contracting the exceptional curve in $X$, we get a surface with an isolated singularity. I am looking for the equation (or the set of equations) that describes this singularity (as a surface in some $\mathbb{C}^n$, possibly just $\mathbb{C}^3$).

For example in the case of $X$ being the total space of $\mathcal{O}_{\mathbb{P}^1}(-2)$, the resulting singularity is the $A_1$ singularity given by $$x^2+yz=0$$.

share|improve this question
9  
This is the cone over the rational normal curve of degree $5$ in $\mathbb{P}^5$. So it has embedding dimension $6$. Its equations in $\mathbb{C}^6$ can be given in determinantal form as $$\textrm{rank} \,\begin{pmatrix} x_0 & x_1 & \ldots & x_4 \\ x_1 & x_2 & \ldots & x_5 \end{pmatrix} =1.$$ –  Francesco Polizzi Oct 28 '13 at 22:18
    
do you mind to tell me what is the meaning of the sentence "the surface X to be the total space of OP1(−5)"? I am not familiar with that expression! –  pmath Oct 30 '13 at 0:27
    
@ pmath: If $E\to X$ is a vector bundle, you can also look at $E$ as a manifold/variety. O_{P^1}(-5) is a line bundle on P^1. –  Mohammad F. Tehrani Oct 31 '13 at 20:03

2 Answers 2

up vote 10 down vote accepted

The weighted projective plane $\mathbb{P}(1,1,n)$ can be viewed as $\mathbb{P}(1,1,n)=\mathbb{C}^3\setminus \{0\}/(x,y,z)\sim (\lambda x,\lambda y,\lambda^n z)$. For $n=1$ we obtain the standard projective plane. For $n>1$, the point $(0,0,1)$ is the unique singular point, and the blow-up of this point is a Hirzebruch surface $\mathbb{F}_n$, with exceptional divisor $E\simeq \mathbb{P}^1$ of self-intersection $-n$. Hence, what you are looking for is just the quotient singularity of $\mathbb{P}(1,1,n)$.

If you want a local embedding into some affine space, take the local embedding $(x,y,z)\mapsto (x^n/z,x^{n-1}y/z,\dots,xy^{n-1}/z,y^n/z)$. The equations are probably easy to obtain from this explicit description. For $n=2$ you obtain what you already described.

share|improve this answer

Let $C\subset\mathbb{P}^n$ be a rational normal curve of degree $n$, and $X$ the cone over $C$ with vertex $p\in\mathbb{P}^n$. Blowing up $p$ in $X$ you get the Hirzebruch surface $\mathbb{F}_n = \mathbb{P}(\mathcal{O}\oplus\mathcal{O}(-n))$. The exceptional divisor $E\subset\mathbb{F}_n$ is a genus zero smooth curve with $E^2 = -n$.

Therefore blowing down $E$ you get the vertex of a cone over a rational normal curve of degree $n$, that is a Du Val singularity of type $A_n$. This singularity is given by the equation $$x^2+y^2+z^{n+1}= 0.$$

In your specific case $n = 5$ you have the vertex of a cone over a rational normal curve of degree $5$ in $\mathbb{P}^{5}$ given by $x^2+y^2+z^6 = 0$.

As Jérémy Blanc wrote these singualrities can be viewed as the rational quotient singularities of type $\frac{1}{n}(1,1)$ of the weighted projective plane $\mathbb{P}(1,1,n)$ at $[0:0:1]$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.