# Which isolated surface singularity comes from a -5 curve?

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 $$x^2+yz=0$$.

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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! – eventually 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

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.

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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 $[0:0:1]$.

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