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Once upon a time I was told that the torus is flat. This was supposed to be surprising, since the ordinary picture of a torus we have in our heads looks inherently curved. However, thinking instead of a torus as a square in the plane with opposite points identified, it becomes 'clear' that the torus at least admits a flat metric, because the plane admits a flat metric.

However, a two-holed torus can also be obtained in this way: it is an octagon in the plane with appropriate pairs of edges identified. However, by the Gauss-Bonnet theorem, this surface does not admit a flat metric.

Thus, something about the way we make the identifications for the 1-torus is compatible with the flat metric structure on the plane, and this is not so for the 2-torus. I am hence lead to ask:

Given a smooth manifold $M$ obtained from $\mathbb{R}^n$ by appropriate identifications, is there some general criterion for determining whether the flat metric on $\mathbb{R}^n$ descends to $M$? Or is the $n$-dimensional torus particularly special in its ability to inherit a metric from the plane? If so, what is special about it?

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    $\begingroup$ Metric tensors pullback. So if you have a manifold as a quotient, there is a map $X\to X/\sim$. Given a metric on the quotient, you may always pull it back to the starting space without further assumptions. Pushing the metric down from the parent space requires more work, which I guess is the point of the question. $\endgroup$ – ziggurism Nov 15 '17 at 15:08
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Let $M$ be a manifold and $G$ a discrete group acting freely and proper discontinuously, then $M/G$ is a manifold. If $g$ is a Riemannian metric on $M$, then it descends to a Riemannian metric on $M/G$ if and only if the group $G$ acts by isometries; to be precise, by 'descends to a Riemannian metric on $M/G$', I mean there is a Riemannian metric $g'$ on $M/G$ such that $g = \pi^*g'$ where $\pi : M \to M/G$ is the quotient map. To see how to construct $g'$ from $g$, see this answer where I show that the standard metric on $\mathbb{R}^n$ descends to the torus $\mathbb{R}^n/\mathbb{Z}^n$; here the group $G = \mathbb{Z}^n$ is acting by translations which are isometries of the usual metric on $\mathbb{R}^n$.

If $(M, g)$ is flat, and $g$ is complete, then its universal cover is isometric to $\mathbb{R}^n$ with its standard metric, so every complete flat manifold is a quotient of $\mathbb{R}^n$ with its standard metric by a discrete group of isometries which acts freely; note, every discrete group of isometries of $\mathbb{R}^n$ acts properly discontinuously. If in addition $M$ is compact, then the group action is cocompact; a discrete group of isometries which acts cocompactly is called a crystallographic group, if it also acts freely, it is called a Bieberbach group; note, a Bieberbach group can equivalently be defined as a torsion-free crystallographic group (this is the usual definition). Bieberbach showed that an $n$-dimensional crystallographic group has a finite index subgroup isomorphic to $\mathbb{Z}^n$, from which it follows that every compact flat manifold is finitely covered by a torus.

Note that not every subgroup of isometries of $\mathbb{R}^n$ gives rise to a flat manifold as it may not act freely on $\mathbb{R}^n$. For example, every isometry of $\mathbb{R}^2$ is a composition of reflections, rotations, and translations; the first two transformations have fixed points, so they don't act freely on $\mathbb{R}^2$. If the subgroup does not act freely, the quotient will be a flat orbifold instead of a flat manifold. Moreover, every flat orbifold arises this way, see Theorem $13.3.10$ of Foundations of Hyperbolic Manifolds (second edition) by Ratcliffe. In particular, a quotient of $\mathbb{R}^n$ by a crystallographic group which is not a Bieberbach group gives a compact flat orbifold which is not an manifold.

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  • $\begingroup$ This is a great answer, thank you. Could you comment about how this generalises to the orbifold case? It appears that there is a 1-to-1 correspondence between crystallographic groups and (2D flat) orbifolds: is this to say that some of the 17 wallpaper groups don't act freely on the plane? $\endgroup$ – gj255 Nov 15 '17 at 16:32
  • $\begingroup$ @gj255: I don't know much about orbifolds. I will try to look into it and update my answer later. $\endgroup$ – Michael Albanese Nov 15 '17 at 17:16
  • $\begingroup$ Yes, most of the 17 wallpaper groups do not act freely on the plane. In fact, the only ones which do act freely on the plane are the actions of $\mathbb{Z} \oplus \mathbb{Z}$ whose quotient is a torus, and the actions of the non-abelian semi direct product $\mathbb{Z} \rtimes \mathbb{Z}$ whose quotient is the Klein bottle. Every other wallpaper group has torsion elements, each of which is either a rotation (which fixes a point) or a reflection (which fixes a line). $\endgroup$ – Lee Mosher Nov 15 '17 at 18:55
  • $\begingroup$ I believe it's also true that every compact Euclidean orbifold is covered by the plane and hence is a quotient of a crystallographic group. $\endgroup$ – Lee Mosher Nov 15 '17 at 18:58
  • $\begingroup$ @MichaelAlbanese Thanks for expanding your answer. One final question: what is your working definition of a Bieberbach group and how does it differ from a crystallographic group? $\endgroup$ – gj255 Nov 16 '17 at 13:28
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The octagon with edges identified appears to be different because you cannot tile the entire flat (Euclidean) plane with octagons in such a way that everything is OK when you cross the "border" and move to the adjacent octagon.

Instead, you can tile the hyperbolic plane with octagons in the correct way. So the genus-2 surface can inherit the negative-curvature metric of the hyperbolic plane. Pictures: Wikipedia: Order-8 octagonal tiling

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You can use this idea to put a flat metric (in a sense) on any surface obtained by identifying the edges of a polygon in pairs. But the metric won't be smooth unless whenever you have a set of vertices being identified, the angles at those vertices add up to exactly $360^\circ$. It works for the square because you have four $90^\circ$ angles meeting at a point. To see what goes wrong, cut a regular octagon out of a piece of paper, then tear off eight pieces containing the eight vertices and try to tape them all together so that all the vertices meet at a point. You can do it, but you'll get a shape that can't be flattened out without a lot of creasing.

Here's an example of a hexagon that works:

enter image description here

The vertices labeled A are all identified with each other, as are the vertices labeled B. Since the angles all measure $120^\circ$, each set of three angles fits together to make a smooth flat surface. (The resulting surface is homeomorphic to the torus, as it must be because it has a flat metric. It's an interesting exercise to prove this directly by cutting and pasting.)

The reason you can put a smooth hyperbolic metric on a surface of genus $n=2$ or more is that as long as $n\ge 2$, it's possible to find a regular $4n$-sided geodesic polygon in the hyperbolic plane whose angles are all exactly $360^\circ/4n$, so they all fit together to make a smooth surface with a hyperbolic metric.

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    $\begingroup$ Thanks for the answer. Are you sure the hexagon example doesn't tile the plane? $\endgroup$ – gj255 Nov 16 '17 at 1:03
  • $\begingroup$ @gj255: Oops -- yes, of course it does. Forget that statement. $\endgroup$ – Jack Lee Nov 16 '17 at 1:09
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If the manifold is closed, it is finitely covered by the $n$-flat torus.

https://en.m.wikipedia.org/wiki/Flat_manifold

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