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Let $M$ be a connected compact manifold without boundary, $\pi:\widetilde{M}\to M$ be the universal covering map. A fundamental domain of $(\pi,\widetilde{M}, M)$ is a compact subset $D\subset \widetilde{M}$ such that
1. the union of $\gamma D$ over all $\gamma\in \pi_1(M)$ covers $\widetilde{M}$,
2. the collection $\gamma D^o$ are mutually disjoint,
3. $\pi(D)=M$ and the restriction $\pi|_{D^o}:D^o\to M$ is diffeomorphic onto its image.

My question is:
Does there always exist some simply connected fundamental domain?
Is every fundamental domain simply connected?

I saw the following statement in several papers about dynamical systems: let $B^d(0,1)$ be the unit ball in $\mathbb{R}^d$ and $M$ be a $d$-dimensional compact connected manifold without boundary. Then $M\simeq B^d(0,1)/\sim$ where $\sim $ represents some gluing along $S^{d-1}=\partial B^d(0,1)$.
I think the statement might be related to above question.


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The definition of fundamental domain may not be accurate. I just listed the properties that came to my mind. – Pengfei May 18 '11 at 13:25
I think Tom answered my question. What he left there is beyond my problem~ <br> Thanks! – Pengfei May 26 '11 at 0:00
up vote 5 down vote accepted

The first part of 3 follows from 1. The second part of 3 follows from 2.

There is always a contractible (in particular simply connected) fundamental domain: Triangulate $M$. Take the union of all the codimension zero open simplices, together with just enough codimension one open simplices to make it connected. This will be $\pi(D^{o})$. It is an open subset of $M$, dense and contractible (homotopy type of a maximal tree in the dual cell structure). Choose a lifting to $\tilde M$ and let $D$ be its closure.

For an example of a non simply connected fundamental domain in dimension $3$, take $M=S^2\times S^1$, $\tilde M=S^2\times \mathbb R$. One fundamental domain is $S^2\times I$, but you can make another by first cutting $S^2\times I$ into two non simply connected pieces meeting along a closed surface and then letting $D$ be the union of the left piece and a translate of the right piece.

Edit: This wrong, as pointed out in the comments. My error ws in imagining that the inclusion of the lifted open thing into its closure was necessarily a homotopy equivalence. I wonder if this approach can be fixed.

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Here's a counterexample to your claim. Take the manifold $S^2$. Its fundamental group is trivial, and so its universal cover is just itself. The only way of satisfying the requirement that the union of all $\gamma D$ be $S^2$ is to take the fundamental domain to be the manifold itself... now you may note that $S^2$ is not contractible. – André Henriques May 18 '11 at 16:43
Of course it is too much to hope that the fundamental domain can be contractible, as I now see. But can it be made simply connected? It's interior can, as my argument showed. – Tom Goodwillie May 25 '11 at 15:22

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