Consider a polynomial map $f :\mathbb{C}^{n+1} \rightarrow \mathbb{C}$ with $f(0)=0$ (no constant term) and with isolated critical point at $0 \in \mathbb{C}^{n+1}$. We can choose a disc $D$ of some radius $\delta$ around $0 \in \mathbb{C}$ so that $0$ is the only critical value in $D$, and we can choose a ball $B$ of some radius $\epsilon$ around $0 \in \mathbb{C}^{n+1}$ so that $0$ is the only critical point in $B$.

It is known that restricting $f$ to $B^{*}=B \setminus f^{-1}(0) \cap B$ gives a fibration over $D^{*}=D \setminus 0$, the so-called Milnor fibration, and it is known furthermore that any fibre of $f:B^{*} \rightarrow D^{*}$ has the homotopy type of a bouquet of $\mu$ spheres $S^{n}$, so in particular the only interesting cohomology of the fibre is in degree $n$, where it has rank $\mu$. The number $\mu$ here is known as the Milnor number of singularity of $f^{-1}(0)$ at $0$.

Choosing a generator for $\pi_{1}(D^{*},p)$ (say a counterclockwise circle starting at a basepoint $p$), we get a monodromy automorphism $T : H^{n}(f^{-1}(p),Z) \rightarrow H^{n}(f^{-1}(p),Z)$. There are some cases when this monodromy automorphism is understood. For example, if the original polynomial is "quasi-homogeneous", $T$ is known to be semisimple and the eigenvalues can be read off from the quasi-homogeneous weights.

I would like a simple example of $f$ (and maybe small $\mu$) where $T$ is not semisimple.

  • 1
    $\begingroup$ If $n$ is odd and $f=\sum x_i^2$, $T$ is given by the Picard-Lefschetz formula : $T(x)=x+(x\cdot \delta )\delta $; it is unipotent $((T-I)^2=0$), hence very far from semi-simple. $\endgroup$
    – abx
    Apr 16, 2014 at 16:20
  • $\begingroup$ @Quid: are you sure? This is a homogeneous polynomial, and by general results the monodromy operator is diagonalizable, hence semisimple. In fact the Milnor number is $1$, so there is only one vanishing cycle, the middle cohomology group of the fibre is $1$-dimensional and the monodromy operator is either the identity or minus the identity (depending on the parity of $n$). $\endgroup$ Apr 16, 2014 at 16:31

1 Answer 1


Let us consider the polynomial $$f(x, \, y)= x^5+y^5+x^2y^2.$$ It defines a germ of isolated singularity at the origin of $\mathbb{C}^2$, whose Milnor number is $\mu(f) = 11$.

This computation with Singular shows that the corresponding monodromy matrix has a Jordan block of size $2$, hence it is not semi-simple.

Added. For the reader's convenience, the computation is as follows:

LIB "gaussman.lib";
ring R=0,(x,y),ds;
poly t=x5+x2y2+y5;

where the first line of output gives the Jordan data of the monodromy matrix, the second line gives the size of the Jordan blocks and the third line gives the corresponding multiplicities.


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