# Branched coverings over orbifolds with reflector lines

It is well known that if $F\to B$ is a $n$-finite branched covering over an orbifold with cone-points then the orbifold Euler's characteristics are related via $\chi(F)=n(\chi(B)-\sum_i^r\frac{a_i-1}{a_i})$, where $r$ is the number of cones with stabilizer of orders $a_1,...,a_r$ respectively.

Now, I don't know if everyone would like to know what is the corresponding relation when $B$ has reflector intervals or reflector circles, as I'd rather... so I dare to question:

Is there a generalization in this direction?

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Rather than just accepting the "Well-known" identity, why not prove it. Then you can ask, "what happens if I use the same technique in the other situation?" and you should have your answer. –  Ryan Budney Dec 17 '09 at 4:01
Could you explain why you find Greg Kuperberg's answer insufficiently detailed? Does the following observation help? The identity really just says $\chi_o(F)=n\chi_o(B)$, where $\chi_o$ is orbifold Euler characteristic. Now just count vertices, edges and faces, as in the 'usual' proof. –  HJRW Jan 12 '14 at 17:27
there, I found street babbling difficult to follow and retain –  janmarqz Jan 13 '14 at 5:15
@janmarqz, I fear that something has been lost in translation. Calling an honest attempt to answer your question 'street babbling' will not encourage anyone else to help you. –  HJRW Jan 13 '14 at 10:10
for "street babbling" I was referring to the style there on the Greg's link... but now i understand over this two word the meaning that your culture carries... excuse my clumsy english and for being to rude –  janmarqz Jan 13 '14 at 15:59

A detailed write up of what you are looking for is also available in Chapter 13 of Thurston's notes. Specifically, formula 13.3.4 on page 331, has the desired formula $$\chi(O) = \chi(X_O) -\frac{1}{2} \sum_{i=1} ^N (1-\frac{1}{n_i}) -\sum_{j=1}^M(1-\frac{1}{m_j}),$$ where $O$ is the orbifold, $X_O$ is it's underlying space, there are N points fixed locally by dihedral groups of orders $2n_1,...,2n_N$ and $M$ points fixed locally by rotations of orders $m_1,...,m_M$.

However, I believe the question addressed (at least in part) by Hurwitz in the proof of Hurwitz's theorem, since the bound the order of an automorphism group H acting on surface of genus g is 168(g − 1) if H includes orientation reversing symmetries and 84(g-1) in the orientable case. The key observation is that the Gauss-Bonnet extends to orbifolds as well.

Computing the Euler characteristic for the quotient of $\mathbb{H}^2$ by the (full) (2,3,7) triangle group is a good way to see this in action.

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what about in the presence of reflector arcs and reflector circles? and say: $N=0$ and $M=0$ –  janmarqz Jan 14 '14 at 21:55
where reflector circle is taken à la Scott, in "Geometries of 3-manifolds" –  janmarqz Jan 14 '14 at 23:02
@janmarqz It turns out those don't really contribute unless they are part of a corner point in the language of Thurston. For example, take an annulus $A = S^1\times[-1,1]$. The core of the annulus, $S^1 \times \{0\}$ is the fixed point set of a reflection. Call this quotient $Q$. Note, $\chi(Q)=0$. Also, $X_Q$ is an annulus and so $\chi(X_Q)=0$. Thus, the orbifold locus, a reflector circle, does not contribute to the euler characteristic directly (however it does affect the topology of $X_Q$, and so it implicitly affects $\chi(X_Q)$). There are perhaps other ways to make this observation. –  Neil Hoffman Jan 15 '14 at 1:54
For an annulus where only one of the two boundary circles is a reflector one is the same to $M\ddot{o}/\sim$. Here we have $M\ddot{o}/\sim$ where $(p,t)\sim(p,-t)$ for $0\le t\le 1$ over the Möbius strip $M\ddot{o}$... So, is the same fate? –  janmarqz Jan 15 '14 at 2:10
@janmarqz I suppose I buried some point in the comment. It is a subtle, but reflector circles do not contribute to orbifold Euler characteristic. For the triangulation of an orbifold $Q$, any reflector circle would decompose into reflection arcs and vertices fixed by the dihedral group of order 2. For these vertices, $n_i =1$ and so looking at the formula in the answer above, these points contribute a $0$ to the sum. Thus, we can just ignore them. Again, I would recommend reading Thurston's notes, especially Chapter 13. It is rich with examples and many people have cut their teeth here. –  Neil Hoffman Jan 17 '14 at 11:21

Yes.

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