# when is a quotient manifold of an affine manifold affine?

So, my question is this: given an affine manifold $X$, and a quotient manifold $Y$ of $X$, is $Y$ necessarily an affine manifold? If it helps, I'm especially interested in the case that $X$ is not any affine manifold but the affine space of dimension $n$ and I also know that $Y$ is orientable, compact, complete and flat.

edit: By affine manifold I mean a manifold that has an atlas such that the transition maps of the charts are locally restrictions of affine maps. Furthermore I ment the real affine space. And lastly, are there any conditions under which $Y$ admits an affine atlas?

edit2: As pointed out in the comments, the answer to my original question was trivially anwserable, therefor I modify my question. Given an quotient manifold $Y$ of an affine manifold $X$. Under which further assumptions is $Y$ an affine manifold? Is especially flattness sufficent, and if so, can someone tell my where I can find a proof for this?

edit3: Okay, I try one more time - this time as specific as I can. I have a group $\Gamma < Aff(\mathbb{E})$, $\mathbb{E}$ being the affine real space of dimension $n$, and $\Gamma$ acts on $\mathbb{E}$ in the natural way, freely and without accumulation points. Now I want to show that $\Gamma$ is an affine crystallographic group and I try to do this by showing that $\Gamma \backslash \mathbb{E}$ is a compact, complete manifold with an affine structure and flat affine connection. So far, all I could show, was that the quotient $\Gamma \backslash \mathbb{E}$ is a compact, orientable manifold and I don't know how to proceed, while apparently for Auslander* this yields already the claim ($\Gamma$ is affine crystallographic). So i was wondering, which properties, like being orientable for example, give the manifold all the properties I want it to have. I hope I made myself clear this time, as to what I want.

• "Examples of Locally Affine Spaces", The Annals of Mathematics, Second Series, Vol. 64, No. 2, 255-259 (1965).
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A quotient of $\mathbb{C}^n$ by a lattice is a complex torus, which is surely not affine (it is a compact complex manifold). So the answer is no. – Francesco Polizzi Jul 11 '12 at 12:57
@Francesco -- It is possible by "affine manifold" the OP means "manifold with affine structure" (not "affine algebraic variety"). A complex torus is a manifold with affine structure. So perhaps a more relevant examples is $\mathbb{C}P^n$ (non-affine), which is the quotient of $\mathbb{C}^{n+1} \setminus \\{0\\}$ (affine). – Jason Starr Jul 11 '12 at 13:06
I thought that all flat manifolds are affine (?) – Anton Petrunin Jul 11 '12 at 14:12
Take a closed non-affine manifold (such as a hyperbolic manifold) with universal cover $R^n$, then the universal cover is affine. I think this answers your question trivially, so I voted to close, although you may want to modify your question to ask for quotients of closed manifolds, which might be interesting. – Ian Agol Jul 11 '12 at 16:56
Amaq: Even after two edits, your question is very poorly phrased to the point where it is impossible to understand what is it that are you asking about: 1. What kind of quotients do you wish to consider (arbitrary equivalence relations, or quotients by groups of homeomorphisms, or by proper actions by connected Lie groups preserving affine structure, or by properly discontinuous groups preserving affine structure, or by properly discontinuous groups acting freely and preserving affine structure, etc.)? 2. What do you mean by flatness (flat affine connection, flat Riemannian metric, ...)? – Misha Jul 11 '12 at 21:41

With the 3rd edit, your question become a triviality: The affine space $A^n$ has canonical flat (i.e., zero curvature, zero torsion) linear connection $\nabla$: $\nabla_{X_i} X_j=0$ for all coordinate vector fields. This is a calculus exercise to check that $\nabla$ is invariant under all affine transformations. Therefore, $\nabla$ will descend to a flat affine connection on every quotient $M=A^n/\Gamma$, where $\Gamma$ is a discrete torsion-free group of affine transformations. Geodesics in $(A^n, \nabla)$ are points and Euclidean straight lines with linear parameterization (another calculus exercise), therefore $(A^n, \nabla)$ is geodesically complete. Geodesics in the quotient manifold $M$ are exactly the projections of geodesics from $(A^n, \nabla)$. Therefore, $M$ is also geodesically complete. Compactness or/and orientability are irrelevant here.
Thank you very much, this answers my question. I had a lot of pieces, actually too many, but was missing the one about $\nabla$ inducing a flat affine connection, if $\Gamma$ is discrete as well as torsion free. – amaq Jul 12 '12 at 20:39