"Arithmetic genus" of a plane curve singularity. - MathOverflow

"Arithmetic genus" of a plane curve singularity.

2013-02-23T14:34:25Z

I believe that the following questions are very basic, but I don't know how to get a reference.

Consider a curve in the plane $C\in \mathbb C^2$ with a singularity at $0$ and suppose it is unibranch at zero (i.e. analytically irreducible). Then I guess one should be able to define "arithmetic genus defect" of the curve at $0$. Namely if one smooths analytically $C$, its geometric genus will grow by a positive number (in case of the cusp $x^2=y^3$ it will grow by one), and let us call this number the defect.

Question 1. Is this defect well defined (independent of a smoothing)? How is it called and how one should calculate it (say it terms of the local ring of $C$ at $0$)?

Question 2. Suppose we have an explicit local parametrisation of $C$ at $0$, say by two holomorphic functions $f(t), g(t)$ (polynomials if you wish). Is it possible to find this "defect" as a certain invariant of this pair of functions at $t=0$?

Question 1 is settled in the answer of unknown and Question 2 in comments to it by Roy and Vivek

Answer by unknown (google) for "Arithmetic genus" of a plane curve singularity.

2013-02-23T17:43:47Z

The difference between the geometric genus of the singularity and the geometric genus of a smoothing (this one being called the arithmetic genus of the singularity) is often called the delta invariant. If A is the local ring of the singularity, B its normalisation then the delta invariant is the dimension of the complex vector space B/A.

It is rather easy to compute the delta invariant if one knows an equation f(x,y)=0 of the curve by a formula due to Milnor (see the book "Singularities of hypersurfaces") : $2 \delta = \mu + 1 - b $ where $\delta$ is the delta invariant, $\mu = dim_{\mathbb{C}} \mathbb{C}[[x,y]]/(\partial_{x}f, \partial_{y}f )$ and b is the number of branches. In the unibranch case, it is simply $2 \delta = \mu$ (example : for the cusp, $\delta = 1$, $\mu =2$).

Answer by Jérémy Blanc for "Arithmetic genus" of a plane curve singularity.

2013-02-24T13:39:21Z

Another way to compute the delta invariant in pratice: let $m_1$ be the multiplicity of the curve at the point $p_1$ you are looking for. You blow-up $p_1$, and look for singular points of the strict transform of the curve which lie on the exceptional curve obtained. Denote by $m_2,\dots,m_k$ the multiplicities obtained (which satisfy $m_2+\dots+m_k\le m_1$). Blow-up all these points. Repeat the process until the curve has no singular point infinitely near to $p_1$. You get a set of multiplcities $m_1,\dots,m_l$ (with $l\ge k$). The delta you want is exactly $\sum_{i=1}^l m_i(m_i-1)/2$. This is a direct consequence of adjunction formula.

It depends of the situation, but sometimes it is easier to compute than the formula of Milnor. The multiplicities are easy to get from either the equation of the curve or a parametrisation.