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Why is the surface of a torus is said to be flat? If you consider the geometry of the torus, its surface has locally positive (spherical), negative (hyperbolic) and flat curvature.

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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – rob
    Feb 6 at 15:58
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    $\begingroup$ The origin of confusion and disagreement between the answers is the implicit and wrong assumption in the question that there is such a thing as "the geometry of the torus." The torus (as a smooth manifold) has many different geometries (Riemannian metrics). Some of these will be flat, while others will be non-flat. This is akin to the fact that there is no such thing as "the price of the car:" Different cars will have different price. $\endgroup$ Feb 8 at 18:46

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Not every torus is flat.

A 2-dimensional torus is any topological space which is homeomorphic (topologically equivalent) to a product of two circles. However, knowing the topology is not enough to give you curvature information: you need to specify a Riemannian metric on the torus if you want geometric data like this. A metric is additional data on top of the topological structure.

The torus you are most familiar with can be obtained by rotating a circle around a line. This torus (a "doughnut") inherits a Riemannian metric from the ambient space it is embedded in. With this metric, this torus is not flat: as you observe it has both regions of both positive and negative curvature.

Some tori are flat though! The easiest example is the PacMan universe: a square where if you exit through one side, you appear on the other side. In other words, the left edge has been "identified" with the right edge, and the top edge has been "identified" with the bottom edge. However, as far as PacMan is concerned his universe is not curved. To be precise, when he parallel transports a vector around a tiny loop, he gets back exactly the same vector he started with. This flat torus cannot be embedded in $\mathbb{R}^3$, but it can be embedded in $\mathbb{R}^4$ as Mozibur describes in their answer.

The Gaussian curvature is "intrinsic": it can be calculated just from the metric. More intuitively, an ant living on a donut embedded in 3D space could tell that his home was curved: they could verify that the angle sum theorem is not true to within a first order approximation, or that parallel transport of vectors changes the vectors. None of this depends on awareness of the embedding. The "extrinsic" curvatures here are the "principle curvatures", which the ant cannot say anything about.

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  • $\begingroup$ Now this is a good answer! What is the other answerer talking about? Negative curvature around the mouth of the doughnut is maybe an artifact and maybe irrelevant but it's still a negative curvature. +1! $\endgroup$
    – Felicia
    Feb 8 at 3:47
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The metric on a circle $S^1$ is unique, it is always flat and hence its Riemannian curvature always vanishes.

However, there is a family of metrics on the torus $T^2 := S^1 \times S^1$. One of these is a flat metric. All these metrics are intrinsic

However, the flat metric in a sense has a claim to being an intrinsic metric and the usual metric in a sense is an extrinsic metric.

The torus can be defined as the product of two circles:

$T^2 := S^1 \times S^1$

Since the Riemann curvatures of both factors vanishes (since the curvature of any circle vanishes), the Riemann curvature of the torus vanishes too. And this gives the flat metric on the torus. Thus the flat metric turns up in a natural way without any embedding in any space being considered, either for the circular factors or for the torus itself

It is also possible to embed the smooth torus smoothly into 3-space and this latter space has a metric which we can induce onto the 3-torus. Thus this metric is naturally induced extrinsically, even though it is again an intrinsic metric.

In fact, the flat torus can be embedded into 4-space in such a way that its metric is induced from the metric of 4-space. This embedding is called the Clifford torus. Furthernore, it is the metric of the rectangular torus. This is the torus constructed from a rectangle by identifying opposite sides and we can see directly here the metric is flat because the rectangle is flat.

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    $\begingroup$ @felicia: Thats an artifact of embedding the torus in 3d. It can be embedded in 4d with vanishing extrinsic curvature. $\endgroup$ Feb 6 at 18:06
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    $\begingroup$ You can experience the flat torus yourself by getting a copy of Final Fantasy 4, 5, or 6 on the Super Nintendo, then playing far enough to obtain an airship. $\endgroup$ Feb 7 at 0:54
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    $\begingroup$ @Matthew Drury also Pacman $\endgroup$
    – Joel Keene
    Feb 7 at 1:44
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    $\begingroup$ This answer seems to be saying that the curvature of a torus is an artifact of the embedding. This is not true. A Riemannian metric is additional structure on top of topological structure. If I hand you a topological torus there is no canonical metric to put on it. The choice of metric determines the curvature. By Gauss Bonnet the average curvature must be zero, and so the curvature must vanish at some points, but that is all we can say. The "intrinsic" curvature can be vary. OP is correct in their understanding of the "intrinsic" curvature of the embedded torus. $\endgroup$ Feb 7 at 10:57
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    $\begingroup$ This is incorrect. The intrinsic curvature of many Tori are not flat. For example, the standard torus embedded in 3 space inherits a Riemannian metric from this embedding. The curvature is not zero. $\endgroup$ Feb 7 at 14:59
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The usual 2-dimensional torus has positive Gaussian curvature on the outside, and negative on the inside. As simple as that.

https://en.wikipedia.org/wiki/Gaussian_curvature

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    $\begingroup$ There is no such thing as "the usual torus", what you meant to say is the torus of revolution in the 3d space obtained by rotating a circle around a straight line. For many mathematicians (such as myself, or Mozibur, given his answer) your torus is 'unusual'. $\endgroup$ Feb 8 at 22:42
  • $\begingroup$ @MosheKohan A circle rotated around a straight line? Isn't it a circle rotated around a circle? $\endgroup$
    – Felicia
    Feb 8 at 23:41
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    $\begingroup$ That would be another one, if you are working inside the 3d sphere. In that case though, the natural thing to do would be to induce the metric from the sphere, not from the Euclidean space. But than you obtain a flat metric on the torus! In any case, there is no 'usual' metric, as you can see. $\endgroup$ Feb 8 at 23:52
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    $\begingroup$ @StevenGubkin I see. This is another nonstandard terminology.... $\endgroup$ Feb 9 at 1:05
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    $\begingroup$ @MoisheKohan Yes, I think geometry is particular rife with nonstandard terminology because people just try to describe their mental pictures with phrases they invent, but these often conflict with established usages. $\endgroup$ Feb 9 at 1:06

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