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

As Kervaire and Milnor mentioned, an $n$-dim exotic sphere $\Sigma$ which bounds a parallelizable manifold $M$ is totally classified by the signature $\sigma(M)$ modulo the order of $bP_{n+1}$.

Let $n=2m$ be an even integer. Brieskorn had discovered that the singularity of complex hypersurfaces $V_k$, the zero set of $x_0^2+\cdots+x_{n-2}^2+x_{n-1}^3+x_n^{6k-1}=0$ has close relationship with exotic spheres. More precisely, if $\varepsilon>0$ is sufficiently small, let $S_\varepsilon$ be the $(2n+1)$-sphere with center $0$ and radius $\varepsilon$, then $\Sigma_k=S_\varepsilon\cap V_k$ is an exotic sphere, and actually every exotic sphere of dimension $(4m-1)$ which bounds a parallelizable manifold can be obtained in this way.

I want to know which element $\Sigma_k$ represents in $bP_{2n}$. In other words, I want to calculate the signature of the Milnor fibre. Since Brieskorn's original paper was written in German, I couldn't read it. Instead, I've read the papar 'Singularity and Exotic Sphere' written by Hirzebruch. In this paper, Hirzebruch gave the answer: actually $\Sigma_k$ represents $k$th multiple of the generator of $bP_{2n}$. However, he refered the proof to Brieskorn's paper.

Does anyone know a proof? Please tell me, thanks.

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.