"Arithmetic genus" of a plane curve singularity. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T02:33:46Z http://mathoverflow.net/feeds/question/122725 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/122725/arithmetic-genus-of-a-plane-curve-singularity "Arithmetic genus" of a plane curve singularity. aglearner 2013-02-23T14:34:25Z 2013-02-24T13:39:21Z <p>I believe that the following questions are very basic, but I don't know how to get a reference. </p> <p>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. </p> <p><strong>Question 1.</strong> 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 $0$)?</p> <p><strong>Question 2.</strong> 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 $t=0$? </p> <p><em>Question 1 is settled in the answer of unknown and Question 2 in comments to it by Roy and Vivek</em></p> http://mathoverflow.net/questions/122725/arithmetic-genus-of-a-plane-curve-singularity/122738#122738 Answer by unknown (google) for "Arithmetic genus" of a plane curve singularity. unknown (google) 2013-02-23T17:43:47Z 2013-02-23T17:43:47Z <p>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.</p> <p>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, $\delta = 1$, $\mu =2$).</p> http://mathoverflow.net/questions/122725/arithmetic-genus-of-a-plane-curve-singularity/122796#122796 Answer by Jérémy Blanc for "Arithmetic genus" of a plane curve singularity. Jérémy Blanc 2013-02-24T13:39:21Z 2013-02-24T13:39:21Z <p>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.</p> <p>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. </p>