Explicit Computations of Dynkin Diagrams of Isolated Singularities - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T23:27:19Z http://mathoverflow.net/feeds/question/102400 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102400/explicit-computations-of-dynkin-diagrams-of-isolated-singularities Explicit Computations of Dynkin Diagrams of Isolated Singularities Umut Varolgunes 2012-07-16T22:36:02Z 2012-07-16T22:36:02Z <p>Let $f$ be a complex polynomial with an isolated singularity at the origin. Take a Morse deformation $\tilde{f}$, and consider the braid group action on the set of distinguished bases of vanishing cycles. This in turn induces an action on the matrix (or Dynkin diagram) representations of the (skew)-symmetric intersection form of the Milnor fibre. This is all well known, but note that I am only considering distinguished bases, in other words I am keeping track of the orders of the vanishing cycles.</p> <p>If we have two Dynkin diagrams corresponding to two different choices of distinguished bases of vanishing cycles, we should be able to get from one to the other by successively applying the standard generators (and inverses) of the braid group in the sense above (usually called Hurwitz moves, or mutations). But, in practice, I find it hard to actually find the right moves that does this. I was wondering if any of the following is available in the literature;</p> <ul> <li>A notion of a distance between two possible Dynkin diagrams</li> <li>Generalized moves that proved to be useful in some computations before</li> <li>Explicit computations</li> </ul> <p>A nice example would be a computation that transforms the standard form of the Dynkin diagram of $T(3,3,3)$ - three legs, with two dashed edges between the two vertices at the center - to the standard (Sebastiani-Thom, Gabrielov) form of $x^3+y^3+z^3$ - unit cube plus the positive diagonals. This must be possible since linearly adding the $xyz$ term is a $\mu$-constant deformation. </p> <p>Thank you.</p>