# Affine manifolds

An affine manifold is a topological manifold which admits a system of charts such that the coordinate changes are (restrictions of) affine transformations. Let $M$ be a compact affine manifold. Let $G$ be the fundamental group of $M$ and $\tilde M$ be its universal cover. One can show that each $n$-dimensional affine manifold comes with a developing map $D\colon \tilde M \to \mathbb R^n$, and a homomorphism $\varphi \colon G \to {\rm Aff}(\mathbb R^n)$, such that $D$ is an immersion and equivariant with respect to $\varphi$.

An affine manifold is called complete if $D$ is a homeomorphism, in this case: $\varphi$ is injective, $G$ is a Bieberbach group, and $M$ is aspherical, i.e. $\tilde M$ is contractible. The non-complete case seems to be far more complicated.

Question 1: Is there an easy example, where $D$ is not surjective?

Question 2: Is there an easy example, where $\varphi$ is not injective?

Question 3: Is there an easy example, where $M$ is not aspherical?

EDIT: As André suggested, let's ask for examples for which $\varphi$ takes values in $SL(n,\mathbb R) \ltimes \mathbb R^n$ or even $SL(n,\mathbb Z) \ltimes \mathbb R^n$, seen as subgroups of ${\rm Aff}(\mathbb R^n)$.

• When a universal cover exists, the fundamental group acts faithfully on it, so there should be no examples for Question 2, unless I am confused by something. – Greg Muller Jul 18 '11 at 20:49
• You are right, $G$ acts faithfully on $\tilde M$; but not necessarily on $\mathbb R^n$. – Andreas Thom Jul 19 '11 at 6:00

There is a conjecture due to Markus which states that any compact affine manifold has parallel volume (i.e. the linear part of $\varphi$ lies in $\mathrm{SL}(n;\mathbb{R})$) if and only if it is complete. To the best of my knowledge, this conjecture is still open, which goes towards saying that there should be no easy examples to questions 1 and 3 for affine manifolds with parallel volume.

If the fundamental group $G$ of a compact affine manifold with parallel volume is nilpotent, the beautiful Affine manifolds with nilpotent holonomy by Fried, Goldman and Hirsch, Comm. Math. Helv. 56 (1981) proves that Markus' conjecture holds in this case and, thus, there are no examples to questions 1 and 3 with nilpotent fundamental group. The proof is a cunning mixture of representation theory and geometry, so I strongly recommend taking a look at it. The results in this paper also imply that the above examples (to questions 1 and 3) constructed by Andre Henriques cannot be adapted so that the resulting manifolds admit parallel volume (these are, nonetheless, very nice examples to the original question!).

An answer to questions 1 and 3 is prodvided by $(\mathbb R^n\setminus\{0\})\big/\lambda^{\mathbb Z}$, where $\lambda^{\mathbb Z}$ acts by scalings. That manifold is diffeomorphic to $S^{n-1}\times S^1$, and fails to be aspherical for $n\ge 3$.

The case $n = 2$ of the above construction can be used to produce an example of 2. Take the double cover of $\mathbb R^2\setminus\{0\}$, and mod it out by $\lambda^{\mathbb Z}$. That manifold is diffeomorphic $S^1\times S^1$.

All this being said, my answers probably aren't very interesting.

I therefore suggest modifying the question, and asking for examples of integral affine manifolds.
These are manifolds whose transition functions are taken from the group $SL(n,\mathbb Z)\ltimes \mathbb R^n$. I would be very interested to see examples of those.

• For 2 take an appropriate double cover of your example with n=2. – Dmitri Panov Jul 18 '11 at 20:57
• Thank you Dmitri. We had the same idea at the same time. – André Henriques Jul 18 '11 at 21:01
• Thanks for the answer. I modified the question a bit to make place for different answers. – Andreas Thom Jul 19 '11 at 5:58
• What is $\lambda^\mathbb{Z}$ in this context? – goblin Jan 2 at 6:23
• @goblin It's the set $\{..., \lambda^{-2}, \lambda^{-1}, 1, \lambda, \lambda^2, \lambda^3, ... \}$, viewed as a group under multiplication. – André Henriques Jan 3 at 0:46