On Mixed Brieskorn Variety

TL;DR

This paper proves that certain mixed Brieskorn varieties and their associated complex links are not only topologically equivalent but also diffeomorphic, establishing a smooth equivalence between their Milnor fibrations and links.

Contribution

It demonstrates that mixed Brieskorn links are diffeomorphic to their complex counterparts, extending previous topological equivalences to smooth equivalences.

Findings

Mixed links are $C^$ equivalent to complex links.

Milnor fibrations of mixed and complex polynomials are smoothly equivalent.

The links are shown to be diffeomorphic, not just homeomorphic.

Abstract

Let $f_{{\bf a},\{bf b}}({\bf z},\bar{\bf z})=z_1^{a_1+b_1}\bar z_1^{b_1}+...+z_n^{a_n+b_n}\bar z_n^{b_n}$ be a polar weighted homogeneous mixed polynomial with $a_j>0,b_j\ge 0$, $j=1,..., n$ and let $f_{{\bf a}}({\bf z})=z_1^{a_1}+...+z_n^{a_n}$ be the associated weighted homogeneous polynomial. Consider the corresponding link variety $K_{{\bf a},{\bf b}}=f_{{\bf a},{\bf b}}\inv(0)\cap S^{2n-1}$ and $K_{{\bf a}}=f_{{\bf a}}\inv(0)\cap S^{2n-1}$. Ruas-Seade-Verjovsky \cite{R-S-V} proved that the Milnor fibrations of $f_{{\bf a},{\bf b}}$ and $f_{{\bf a}}$ are topologically equivalent and the mixed link $K_{{\bf a},{\bf b}}$ is homeomorphic to the complex link $K_{{\bf a}}$. We will prove that they are $C^\infty$ equivalent and two links are diffeomorphic. We show the same assertion for $ f({\bf z},\bar{\bf z})=z_1^{a_1+b_1}\bar z_1^{b_1}z_2+...+z_{n-1}^{a_{n-1}+b_{n-1}}\bar z_{n-1}^{b_{n-1}}z_n+z_n^{a_n+b_n}\bar z_n^{b_n}$ and its associated polynomial $ g({\bf z})=z_1^{a_1}z_2+...+ z_{n-1}^{a_{n-1}}z_n+z_n^{a_n}$.