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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?

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    $\begingroup$ "co"dimension of the "Jacobian" ideal? $\endgroup$
    – quim
    Dec 17 '12 at 11:32
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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.

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  • $\begingroup$ Dear Liviu, thank you very much, this is a great answer! $\endgroup$
    – aglearner
    Dec 17 '12 at 22:49

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