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 $S^2$.

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    $\begingroup$ Hi Bruno! What sort of restrictions do you put on the gluing maps between (not necessarily different?) Pairs-of-Pants? For example, how would you obtain $\mathbb{CP}^2$ in this picture? Is it explained in Mikhalkin's paper? In any case, it looks like a necessary condition is to have $\chi>0$, since your building blocks all have $\chi=1$, and both the submanifolds you glue them along and the corners have $\chi=0$. In particular, $\chi$ of your manifolds seems to determine how many PoPs you need. $\endgroup$ – Marco Golla Dec 13 '12 at 16:17
  • $\begingroup$ Hi Marco. I would put no restriction at all (I can't see any reasonable restriction...). You glue along diffeomorphisms of blocks of type $P\times S^1$: every such diffeomorphism must preserve the fiber, but there are infinitely many non-isotopic choices. You are absolutely right about $\chi$, I must admit I didn't think about it. It might be that you get only finitely many manifolds for each $\chi$, but since you might choose among infinitely many gluing maps between the $P\times S^1$ faces, this is not clear (to me). $\endgroup$ – Bruno Martelli Dec 13 '12 at 20:34
  • $\begingroup$ Thinking a bit more about your comment, I realized that Mikhalkin probably also admits the collapsing of the natural fibrations of the boundaries of the pants: the situation is slightly more complicated than I guessed, I have edited the text accordingly. $\endgroup$ – Bruno Martelli Dec 13 '12 at 20:57
  • $\begingroup$ I think that the following holds: let $X$ be obtained by gluing $n$ Pairs-of-Pants, and suppose that $X'$ is obtained by the same gluing rules, except that you add a Dehn twist in one of the gluing faces. Then, in analogy with the Lickorish-Wallace theorem, $X'$ should be obtained from $X$ by a single surgery along a (probably homologically essential, zero-square) torus. $\endgroup$ – Marco Golla Dec 13 '12 at 23:41

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