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Hello,

I want to take a unit $n$-ball $B_n$ and quotient it by some subgroup $G \subset \mathbb{Z}_2^n$.

Here is a sketch of the construction: take the obvious inscribed $n$-hypercube of $B_n$; we can associate the hypercube's vertices with the $2^n$ bitstrings of length $n$, and take some sub-vector space (thought of as a subgroup) $G$ of $\mathbb{Z}_2^n$. We can then think of $G$ as identifying the vertices of the hypercube and it is fairly obvious that we can then quotient $B_n$ by $G$ to get some quotient space $B_n/G$.

Now, I'm interested in constructing a "tangent bundle" for this quotient which is not quite a manifold and then calculate / construct the Stiefel-Whitney classes for this bundle. I use quotes since I'm not sure about how to do this, or if it is well-defined.

  • are the quotients $B_n/G$ for these "well-behaved looking" $G$ well-understood as topological spaces? In particular, do we know their cohomologies? I can see $H^1(B_n/G) = G$, but that's about it for now =) It looks like equivariant cohomology would come into play, but my topology skills are elementary and I don't have a feeling of what is "easy" or "hard" in equivariant cohomology.
  • what is the correct notion of "tangent bundle" for this kind of orbifold, if any? $B_n$ itself is a nice $n$-manifold with boundary, so it seems like I want to look at something like $TB_n/G \rightarrow B_n/G$, but I am in alien waters and I feel I should be careful about singularities.
  • A natural followup question: if so, we can formally define Stiefel-Whitney classes (and spin structures). Are its Stiefel-Whitney classes well-understood?

Some googling has led me to two papers:

and I'm trying to read them now, but advice from experts would be great. =)

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I think this question partially has answers for yours: mathoverflow.net/questions/19530/… –  Anton Fetisov Jul 15 '13 at 18:43
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1 Answer

As far as (co)homology is concerned, you may replace the orbifold quotient of a manifold $M$ by a finite group $G$ by the so-called Borel construction $(M\times EG)/G$. The "tangent bundle" of the orbifold is then a usual vector bundle over the space $(M\times EG)/G$, and you can define all the usual characteristic classes.

I should say:
- $EG$ is a contractible $G$-space with free action.
- The total space of the "tangent bundle" is $(TM\times EG)/G$.

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