I apologize if the question is a well-known theorem, but I'm just starting to learn about laminations, so I don't know much.

The question is roughly, if interval exchange maps have an underlying closed smooth surface, or if not, what is known about conditions on that.

Now I try to be more precise.

Usual interval exchange functions are bijective functions $\mathbb R/\mathbb Z=S^1\to S^1$ which are piecewise translations where "piecewise" is defined using a finite partition of $S^1$ in segments $s_i$.

Given a interval exchange function $\phi:S^1\to S^1$ one can consider its suspension $S_\phi=[0,1]\times S^1/\sim$ where the nontrivial identifications are $(0,x)\sim (1,\phi(x))$. Then one has a smooth 1-dimensional structure on $S_\phi$, given by the differentiable structure on $[0,1]$, which can be extended with continuity through the identifications. In the $S^1$-direction there is just some piecewise $C^1$-structure on the pieces $]0,1[\times s_i$, since $\phi$ is not even assumed to be continuous.

Then if I understood correctly one defines $\partial S_\phi=\cup_i [0,1]\times\partial s_i/\sim$, and this set is an union of (topological but not $C^1$, since there are cusps) copies of $S^1$. Then one obtains a surface by gluing some annuli to these circles.

The question is if there is some way of obtaining a smooth surface in this way. The case where the lenghts of $s_i$ are rational and $\phi$ is piecewise equal to translations by rational numbers is simple (one refined the segments and gets a longer, smooth, boundary), so I am interested if there is a result in the other case.