$\mathbb{R}^n$ admits a tessellation by permutohedra. The corresponding identification of facets of a permutohedron therefore gives a well-defined space: call it $X_n$. For example, $X_2$, the hexagon with opposite sides identified, can be shown to be a 2-torus (see figure 2 in http://arxiv.org/pdf/cond-mat/0703326v2). Is $X_n \cong (S^1)^n$? Why (or why not)?
1 Answer
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The quotient $X_n$ is the same as the quotient of $\mathbb R^n$ by a subgroup of $\mathbb Z^n$ acting cocompactly through translations. Such a thing is always a torus.