I would like to discuss the following problem. Hopefully, you will suggest to me some ideas and bibliography.

At first I will provide some basic definitions to set up the notation.

Let us consider a line arrangement $\mathcal{A}=\left\lbrace l_{1},\ldots,l_{m}\right\rbrace$ in $\mathbb{CP}^{2}$ and let $M\left(\mathcal{A}\right)$ be its complement manifold, i.e., the space $M\left(\mathcal{A}\right)=\mathbb{CP}^{2}\setminus\bigcup_{k=1}^{m}l_{k}$ obtained taking the complement of the union of the lines.

Regarded as a differentiable manifold, $M\left(\mathcal{A}\right)$ is diffeomorphic to the interior of a connected compact manifold with boundary. As a consequence, by a standard result in low dimensional topology, there is a differentiable manifold $N$ which is homeomorphic but not diffeomorphic to $M\left(\mathcal{A}\right).$

Here is my question:

Let $\mathcal{A}$ and $\mathcal{B}$ be line arrangements in $\mathbb{CP}^{2}$ with complement manifolds $M\left(\mathcal{A}\right)$ and $M\left(\mathcal{B}\right).$ Is that true that if $M\left(\mathcal{A}\right)$ and $M\left(\mathcal{B}\right)$ are homeomorphic, then they are diffeomorphic, too? Otherwise, exhibit an explicit example of two line arrangements $\mathcal{A}$ and $\mathcal{B}$ in $\mathbb{CP}^{2}$ with complement manifolds $M\left(\mathcal{A}\right)$ and $M\left(\mathcal{B}\right)$ which are homeomorphic but not diffeomorphic.


1 Answer 1


As far as I know, that's still an open problem. The only work I'm aware of that addresses the rigidity of the topology is on the level of homotopy type: Randell's Isotopy Theorem and Rybnikov's example showing the matroid does not determine the homotopy type. But I see you know about these from your previous question (https://mathoverflow.net/users/53064/esaini582)


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