1
$\begingroup$

What is the difference, if any, between a flat manifold (in which the Riemann tensor vanished identically) and an affine space?

$\endgroup$
1
$\begingroup$

A Riemmanian manifold is called flat if its curvature vanishes everywhere. However, this does not mean that this is is an affine space. It merely means (roughly) that locally it "is like an Euclidean space."

Examples of flat manifolds include circles (1-dim), cyclinders (2-dim), the Möbius strip (2-dim) and various other things.

Yet, let me add into the direction of your idea that the universal cover of a complete flat manifold is indeed an Euclidean space.

$\endgroup$
1
  • 2
    $\begingroup$ OK, that makes sense. Riemann tensor can tell us whether the manifold is intrinsically flat, but it can still be embeded so that it's extrinsically curved. $\endgroup$ – user54031 Aug 5 '14 at 12:08
0
$\begingroup$

If $G < Isom (\mathbb{R}^n) \cong O(n) \rtimes \mathbb{R}^n$ is a subgroup of isometries of some Euclidean space $\mathbb{R}^n$ acting freely and properly discontinuously on it, then the quotient will be a flat Riemannian manifold. Conversely, any flat Riemannian manifold arises in this way. In the realm of surfaces, this gives you planes, disks, annuli(=cylinders), Mobius bands, tori and Klein bottles. As you see you can have non trivial topology (the main constraint in this case is that the Euler characteristic must vanish, by Gauss-Bonnet).

On the other hand, an affine space is always a contractible space: fix any point $P$, then retract the whole space onto $P$ by straight line homotopy, i.e. $(x,t)\mapsto tP + (1-t)x$. Thus the only surface admitting the structure of an affine space is the plane.

Moreover, on a flat manifold you have well defined geo-metric notions: for example it makes sense to measure angles or lengths on a flat manifold (actually any Riemannian manifold). But it makes no sense to measure such entities on an affine space (such measurements do not belong to affine geometry). From this point of view, the structure of an affine space is more algebraic than geometric: here you have a way to add vectors to points or to perform affine combinations (something which is not available on a general Riemannian manifold). So I guess the point here is that the two structures belong to two different realms of "geometry" and are not really comparable to each other.

$\endgroup$
9
  • $\begingroup$ Very interesting perspective! So, you wouldn't consider affine spaces to be a precursor to Riemann spaces? Also, what is the use of affine spaces then? $\endgroup$ – user54031 Aug 6 '14 at 0:13
  • 1
    $\begingroup$ Of course you are right: Riemannian manifolds don't have a preferred point. [By Euclidean space I usually mean a real vector space endowed with a scalar product, so it would have an origin; but this is just a matter of terminology]. Anyway I guess the main feature of a structure is what it has, not what it lacks: I wouldn't consider an affine space similar to a Riemannian manifold "because" both of them lack the choice of a special point, since this property is shared by a lot of structures; for example a general topological space comes without the choice of a base point. And what you have... $\endgroup$ – Lor Aug 6 '14 at 11:20
  • 1
    $\begingroup$ ..on an affine space is the underlying vector space, which gives you the ability to add vectors to points and to perform affine combinations; this is something not available on a general Riemannian manifold. I do agree that you have a way to turn an affine space into a Riemannian manifold (by means of non canonical choices). The examples above show that, conversely, you cannot turn every Riemannian manifold into an affine space. $\endgroup$ – Lor Aug 6 '14 at 11:22
  • 1
    $\begingroup$ If we agree that the main feature of an affine space is the availability of affine combinations, then I would say no. I agree that on a Riemannian manifold you can pick a chart around a point and perform affine combinations, but what if you pick a different chart around the same point? then the affine combination will not be preserved, because the general change of variable is not going to be an affine map. Tu put it in another way, affine combinations are not intrinsecally/well defined on a general Riemannian manifold. $\endgroup$ – Lor Aug 7 '14 at 8:56
  • 2
    $\begingroup$ If you want to do affine combinations on differentiable manifold, then you have to require that the local charts are such that the change of charts are affine maps. This is what is called an affine manifold (see the wiki page for detailed discussion). Without this additional structure, the only "affine" aspects of a differentiable manifold is the lack of a preferred point, I guess. $\endgroup$ – Lor Aug 7 '14 at 9:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy