Let me first ask the question for two-dimensional compact, connected manifolds and orbifolds. Then, if the answer is No, one can remove various conditions on the dimension, and allow non-compact examples and disconnected examples, to realize a (perhaps) wider range of rationals.

This came up after a class I'm teaching and I didn't know the answer.


  • 5
    Well, allowing non connected examples it seems to me that the answer is yes. In fact, for any $n$ there is an orbifold whose Euler characteristic is $1/n$, and it is well known that any rational number can be written as a finite sum of fractions of this form (egyptian fraction representation). – Francesco Polizzi Apr 12 '17 at 22:54
  • 5
    You don't even need Egyptian fractions, just $m/n=m*1/n$. Of course, for the case of negative Euler characteristic, you should start with an orbifold with Euler characteristic $-1/n$. – Victor Protsak Apr 13 '17 at 3:35
up vote 35 down vote accepted

Products of 2-orbifolds with manifolds will do the trick. There are 2-orbifolds of Euler characteristic $1/n$ (take a quotient of $S^2$ by a rotation of order $2n$). Then take a product with a manifold of Euler characteristic $m \in \mathbb{Z}$ to get all rationals.

The answer for connected 2-dimensional orbifolds is no. Euler characteristic is $$\chi(O)=\chi(M)-\sum\left(1-\frac{1}{q}\right)-\frac{1}{2}\sum\left(1-\frac{1}{p}\right),$$ where $p,q\geq 2$ are integers, and $\chi(M)$ is the characteristic of the surface of the orbifold. It immediately follows that $\chi(O)\leq 2$ for all 2D orbiforlds, and as $1-1/q\geq 1/2$, $1-1/p\geq 1/2$, most rational numbers will never occur (on any closed interval which does not contain half-integers, there are only finitely many of these numbers).

  • 1
    In fact, these numbers are well-ordered. – Ian Agol Apr 13 '17 at 23:01
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    @Ian Agol: Yes. Interesting whether this ordering of orbifolds has any meaning. – Alexandre Eremenko Apr 13 '17 at 23:46

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