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

I am having a hard time in finding an upper bound in terms of the degree and the dimension for the Milnor number of an isolated hypersurface singularity. I am mostly interested in surfaces on the projective space. Can some one please give me a hint on this?

Thanks!

share|improve this question
    
An upper bound in terms of the degree and dimension? –  Will Sawin Apr 8 '13 at 22:33
2  
I think the sum of all the Milnor numbers of a hypersurface with only isolated singularities, is the difference between the euler characteristic of the singular hypersurface, and the euler charac. of a smooth hypersurface of the same degree. does that help. It should probably give as upper bound the euler charac. of a smooth one, at least in odd dimensional projective space. e.g. in P^3 a normal cubic hypersurface might have say 9 (or even 7) as an upper bound on the Milnor number of an isolated singularity. does this seem right? just tossing this out for comment since no responses yet. –  roy smith Apr 9 '13 at 23:41
    
well you can use the Lefschetz hyperplane theorem to compute cohomology away from middle degree. You can use the top chern class of the tangent bundle to compute the Euler characteristic of the smooth one, and thereby compute its cohomology in the middle degree. The difference will be the Milnor number. As far as I can tell, you don't have any reason to suspect a lower bound for the cohomology of the singular one much better than $0$ in odd dimension and $1$ in even dimension. –  Will Sawin Apr 11 '13 at 15:44
1  
I do not think the method suggested by Will and Roy works, as shown by the answer of Dmitry below. Indeed, the Euler characteristic of a smooth hypersurface of degree d in $\mathbb{P}^n$ is $((1−d)^{n+1}−1)/d+ n + 1$, which is quite a bit less than $(d−1)^n$. The point is that the vanishing cycles can come either from the disappearance of $H^{mid}$ or the appearance of $H^{mid+1}$. Of course, $(d−1)^n$ really is an upper bound, since $\mu= \mathrm{dim} \,\mathbb{C}[x_1, \ldots, x_n]/(\partial f/ \partial x_i) \le (d-1)^n$ if finite by Bezout. –  Vivek Shende Jan 14 at 9:34
add comment

2 Answers

I also agree with roy smith's method.

1) There is a relation between Milnor number and the total number of vanishing cycles, you can find this in Deligne's paper: La formula de Milnor, SGA 7,Expose XVI. In Proposition 2.1, he gaves a identity between Euler-Poincare characteristic and summation of milnor numbers.

You can find a relationship between milnor number and degree of local Chern class in the following paper(and its' ref). 2)F.Orgogozo's paper: Conjecture de Bloch et nombres de Milnor.

I hope this two papers can help you in some sense.

share|improve this answer
add comment

Well, for the hypersurface $X_d\subset\Bbb{P}^n$ the "most degenerate" isolated singularity is of the type: $\{x^d_1+\cdots+x^d_n=0\}$. Thus, $\mu_{max}=(d-1)^n$. Is this what was meant?

share|improve this answer
add comment

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.