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

Suppose $F$ is a holomorphic (or polynomial if you prefer) function on $\mathbb C^3$ and $0$ is an isolated singularity of the surface $F=0$. Then on the one hand we can define Milnor number of this singularity, which is equal to the co-dimension of the Jacobian ideal of $F$ (the ideal generated by derivatives of $F$ at zero). On the other hand we can consider minimal resolution of the singularity of the surface $F=0$.

Question. How can one calculate the Milnor number if one knows the exceptional divisor of the resolution?

share|improve this question
1  
"co"dimension of the "Jacobian" ideal? –  quim Dec 17 '12 at 11:32
    
Thank you for the correction quim –  aglearner Dec 17 '12 at 14:20

1 Answer 1

up vote 3 down vote accepted

In principle it is possible, but it you need to know a bit more than the exceptional divisor. Denote by $X_f$ the Milnor fiber of the singularity, and by $\mu_f$ the Milnor number. Then

$$ \mu_f= b_2(X_f)= \chi(X_f)-1. $$

So the computation boils down to computing the Euler characteristic of the Milnor fiber. Denote by $X_0$ the exceptional divisor of a good resolution, meaning that $X_0$ has only normal crossings.

There exists a natural map (Clemens map) $c: X_f\to X_0$, and one can use this to compute the Euler characteristic of $X_0$ in terms of the Euler characteristics of the irreducible components of $X_0$ and the orders of vanishing of $F$ along these components. Essentially, one performs an integration with respect to the Euler characteristic along the fibers of $c$, very similar in spirit with the classical proof of the Riemann-Hurwitz formula.

The fibers of $c$ over the singular points of $X_0$ are circles or tori so they do no contribute anything to the computation. If we denote by $X_0^*$ the smooth part of $X_0$ an we set $X_f^*:=c^{-1}(X_0^*)$, then over each component of $X_0^*$ $c$ is a finite cover so its fiber consists of finitely many points, as many as the multiplicity of $F$ along that component. If you put these things together you obtain the A'Campo formula that expresses the Euler characteristic of $X_f$ in terms of $X_0$ and the multiplicities of $F$ along $X_0$.

For more details see Chapter 14 of my course notes on singularities and the references therein.

share|improve this answer
    
Dear Liviu, thank you very much, this is a great answer! –  aglearner Dec 17 '12 at 22:49

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.