In this nice paper Mikhalkin uses certain (more geometrical than algebraic) aspects of tropical geometry to prove that every complex projective hypersurface in $\mathbb C \mathbb P ^n$ decomposes as a smooth manifold into some *$(n-1)$-dimensional pair-of-pants*.

Following Mikhalkin, a $k$-dimensional pair-of-pants is a particular real $(2k)$-manifold with corners. It is obtained by removing $k+2$ generic hyperplanes from $\mathbb C \mathbb P^k$. For instance, a $1$-dimensional pair-of-pants is a sphere minus 3 points (as everybody knows), while a $2$-dimensional pair-of-pants is $\mathbb C \mathbb P^2$ minus 4 generic lines.

Boundaries and corners arise when you take the compact version of these pair-of-pants, i.e. you remove an open regular neighborhood of the $k+2$ generic hyperplanes:

- When $k=1$ you get the genuine compact pair-of-pants $P$ with 3 boundary components.
- When $k=2$ you get a real 4-manifold with boundary and corners: the boundary consists of four compact 3-manifolds homeomorphic to $P\times S^1$ (corresponding to the four lines), glued together along six tori (corresponding to the six intersections between the lines). The tori are the corners.

The 4-dimensional manifold with corners we get with $k=2$ is a nice object, which may look intriguing to a low-dimensional topologist.

As everybody knows, not only complex hypersurfaces (i.e. curves) in $\mathbb C \mathbb P^2$ decompose into pair-of-pants: every oriented surface of negative Euler characteristic does! It would be then natural to ask the following:

Which compact 4-manifolds decompose into pair-of-pants?

The set of course includes all complex hypersurfaces in $\mathbb C \mathbb P^3$.

** Edit: ** It is not true that every curve in $\mathbb C \mathbb P^2$ decomposes into pants: a line or a cubic give $S^2$ and $T^2$ and they of course do not decompose into pants. Reading more carefully Mikhalkin's paper, it seems to me that he also allows to collapse the natural fibering of the boundaries of the blocks: for instance, by collapsing the boundaries of a two-dimensional pair-of-pants we get $S^2$.