4
$\begingroup$

Are there two IRREDUCIBLE plane curve singularities having different equisingular type with the same monodromy (linear action on the first homology group of the (regular) Milnor fibre)?

$\endgroup$

1 Answer 1

7
$\begingroup$

No, it is a classical theorem of Zariski that the Alexander polynomial, i.e., the characteristic polynomial of the monodromy of the Milnor fibre, determines the equisingularity class in these cases.

In fact, from the theorem of Campillo, Delgado, and Gusein-Zade, one sees that the zeta function of the monodromy -- i.e., the Alexander polynomial divided by $(1-t)$ -- is the Poincaré series of the semigroup of the curve singularity.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .