# “Arithmetic genus” of a plane curve singularity.

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

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In fact, the (smooth) topological genus of the link of the plane curve singularity is equal to the genus of a smoothing by the resolution of the Milnor conjecture. en.wikipedia.org/wiki/Milnor_conjecture_(topology) – Ian Agol Feb 23 '13 at 21:40
It seems to me that Milnor conjecture concerns only toric knots. – aglearner Feb 23 '13 at 22:26
If you find yourself asking more complicated questions of this sort, you may find (especially the first few chapters, and the appendix of) Zariski's book The moduli problem for plane branches'' to be useful. – Vivek Shende Feb 23 '13 at 22:31
Vivek, thank you this recommendation and as well for the explanation of what is the conductor. – aglearner Feb 24 '13 at 0:11
@aglearner: I think links of plane curve singularities are quasipositive, which implies that their Seifert genus is equal to their 4-ball genus, since they may be drawn as a separating curve on the Seifert surface of a torus knot. arxiv.org/abs/math/0411115 – Ian Agol Feb 24 '13 at 0:39

## 2 Answers

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 normalization, 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"): $2 \delta = \mu + b - 1$ 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, $\delta = 1$, $\mu =2$).

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Thank you! Do you know if there is as well some reasonable formula for Question 2? I am asking this since in the case that I want to consider I have an explicit parametrization of $C$ but I don't have the local equation of it – aglearner Feb 23 '13 at 17:57
won't macaulay transform a parametrization into an equation? Also I seem to recall that the defect is calculable from the multiplicity plus the multiplicities of all infinitely near points (points on proper transform after blowups). I.e. I believe it is (1/2)m(m-1) for each such multiplicity. So maybe your parametrization can be lifted to those blowups. More precisely, on p. 214 of Intro. to singularities and deformations, by Greuel Lossen and Shustin, is a formula (3.4.11) for computing the "delta invariant" in terms of the conductor ideal defined by a parametrization. That should do it. – roy smith Feb 23 '13 at 19:25
Dear Roy, thank you for the reference! I will have a look. (in fact when I said that I have an "explicit" parametrization, I was a bit sloppy, I just meant that in my problem the curve is given naturally by a parametrization $f(t),g(t)$ rather than by an equation $F=0$ (and $f(t), g(t)$ can be arbitrary), so I was hoping for some calculation directly applied to $f(t), g(t)$) – aglearner Feb 23 '13 at 21:13
Making Roy's comment explicit: in the analytically irreducible case, you have a subring $R = \mathbb{C}[[f(t), g(t)]] \subset \mathbb{C}[[t]]$. The $\delta$ invariant is the vector space dimension $\delta = \mathrm{dim} \mathbb{C}[[t]]/R$. One way to think about this is the following: let $\Gamma$ be the subset of $n \in 0, 1, \ldots$ where some element of $R$ is of the form $t^n + \cdots$. Then $\delta = \# \mathbb{N} \setminus \Gamma$. The conductor is the minimal $c$ such that $(t^c) \subset R$, and you always have $2\delta = c$ because plane curves are Gorenstein. – Vivek Shende Feb 23 '13 at 22:39
Only a minor correction to prevent confusion: The formula should be $2\delta = \mu + b - 1$. – Gregor Bruns Feb 9 '15 at 9:34

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.

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Thank you for this answer Jeremy! – aglearner Feb 24 '13 at 13:57