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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18Would [math.se] be a better home for this question? – rob Feb 06 '22 at 15:58
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4The 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. – Moishe Kohan Feb 08 '22 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.
Steven Gubkin
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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! – Pathfinder Feb 08 '22 at 03:47
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2+1 and crucially, PacMan gets back the same vector even when parallel transporting it through a big loop... – giobrach Nov 28 '23 at 08:28
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Nice answer. Can you refer any resource where I can get the full detour of torus like compute metric to curvature? @StevenGubkin. – N00BMaster Jan 23 '24 at 18:15
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@N00BMaster Any book on the geometry of curves and surfaces should have this stuff. I have not taught from this book so I don't know it well, but I do love everything I have ever read by Shifrin so it is probably a solid bet. – Steven Gubkin Jan 23 '24 at 19:09
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I know it's super late but I was confused with the metric. Like, If we consider the parametrization of the torus $\mathbf{x}(u,v)=((a+b\cos u)\cos u,(a+b\cos u)\sin v,b\sin u)$ then we can have the metric. But what if we want to find the metric when we consider it $\mathbb T^2=\mathbb R^2/\ \mathbb Z^2$ as quotient space (I guess it's your PacMan universe description)? I didn't find any information related to this metric. It will be a great help if you suggest anything @StevenGubkin – falamiw Jun 16 '24 at 10:35
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@falamiw If you think about the quotient map $\pi: \mathbb{R}^2 \to \mathbb{R}^2/\mathbb{Z}^2$ there is only one way to put a Riemannian metric on the quotient space which turns $\pi$ into a local isometry. Basically just "lift" your tangent vectors to any pre-image at the same basepoint and compute your inner product there. It might be useful to see the argument in full generality. The key words to google would be "Riemannian quotient metric" and "discrete group acting freely and properly discontinuously". – Steven Gubkin Jun 16 '24 at 11:08
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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.
Mozibur Ullah
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@Steve Gubkin: ... into 3-space. This metric is not flat. Hence there is a good reason to say that this metric is not intrinsic (even though it is). – Mozibur Ullah Feb 08 '22 at 03:09
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@Steve Gubkin: So to a degree, we were talking past each other. Though also to a degree, my answer wasn't as clear as it should have been. And to a degree, nor was your comment - though it pushed me to be clearer. – Mozibur Ullah Feb 08 '22 at 03:10
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@Matthew Drury: You are correct in that a smooth torus is not flat because it doesn't have a metric. But this isn't sufficient, because to say something is not flat usually means that it has a metric that isn't flat and not that it has no metric at all, in which case the assertion becomes vacuous. Still, thanks for the clarification. – Mozibur Ullah Feb 08 '22 at 03:14
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3@MoziburUllah The torus doesn't just have a family of metric classified by $\mathbb{R}^2/SL(2,\mathbb{R})$. You can perturb the metric locally pretty much however you want (you can make a patch as weird as you want using partititions of unity). Globally the average will still be zero by Gauss-Bonnet though. You are not using the word "intrinsic" in the way mathematicians do. You might say that the "canonical" metric on the torus (as a product of two circles) is flat. "Intrinsic" means that it can be calculated from only the metric, not the embedding. – Steven Gubkin Feb 08 '22 at 03:54
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Actually you don't even need partitions of unity: take any parallelogram P, take any positive P periodic function f, and define a Riemannian metric on the torus $\mathbb{R}^2/P$. A constant f will give a flat torus, but when f is non-constant you get curvature. – Steven Gubkin Feb 08 '22 at 04:00
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@Steve Gubkin: You said earlier, that the sectional curvature isn't intrinsic. However, since the sectional curvature can be determined from the Riemannian curvature and the latter is intrinsic, then so must also be the sectional curvature. Or do you disagree? – Mozibur Ullah Feb 08 '22 at 04:19
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It's the zero curvature that is an artifact of the embedding in 4d. A spherical shell has positive curvature, just parallel transport a vector on a closed line. The same holds for the outside of the torus. A cylinder has zero curvature. Not a torus. – Pathfinder Feb 08 '22 at 17:39
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@Steve Gubkin: I said that every intrinsic Riemannian torus is flat. However, this is not quite correct. The torus supports a family of metrics and these are classified by R^2/SL(2,\mathhbb{R}). It is true, however, that the product Riemannian torus which is constructed from $S^1 \times S^1$ is flat. This is because each factor is flat. This is what I called the intrinsic torus. This can be embedded isometrically in $\mathbb{R}^4$. However, it cannot be isometrically embedded into $\mathbb{R}^3$. However, the smooth such torus can be and this has an induced metric coming from its embedding ... – Mozibur Ullah Feb 08 '22 at 03:07
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@Steve Gubkin: I meant upto isomorphism. Tori with a metric are in bijection with $\mathbb{R}^2$ quotiented by lattices and these are also in bijection with complex elliptic curves. Its this isomorphism class of metrics that I'm referring to. I'm using intrinsic in the traditional sense and in another sense which I have explained. – Mozibur Ullah Feb 08 '22 at 04:12
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2@MatthewDrury I don't think we are talking past each other. I think Mozibur has a misunderstanding about curvature. He says above that he understands that we are discussing Riemannian manifolds. Then later he says that every torus is flat. – Steven Gubkin Feb 07 '22 at 20:13
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@Steve Gubkin: I mean, were you using a non-traditional sense of intrinsic there or was it a misunderstanding of what sectuonal curvature is and entails? – Mozibur Ullah Feb 08 '22 at 04:56
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2I think y'all are talking past one another a bit. A differential manifold has no "intrinsic" Riemannian metric (defined here as a Riemannian metric uniquely defined by the differentiable structure), it can carry many very different ones, up to some global obstructions that are determined by the differential structure (coming from stuff like charecteristic classes of the tangent bundle). So in the sense that the torus is a differential manifold, it does not have an "intrinsic" Riemannian metric. In this sense, it is not (viewed as a smooth manifold) intrinsically flat. – Matthew Drury Feb 07 '22 at 17:43
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Isn't the mouth of the torus negatively curved? Aren't light rays traveling from the innermost circle to the outermost diverging? – Pathfinder Feb 06 '22 at 17:52
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3@felicia: Thats an artifact of embedding the torus in 3d. It can be embedded in 4d with vanishing extrinsic curvature. – Mozibur Ullah Feb 06 '22 at 18:06
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2This 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. – Steven Gubkin Feb 07 '22 at 10:57
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3You 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. – Matthew Drury Feb 07 '22 at 00:54
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1Are you saying that every torus is "intrinsically" flat? Your answer seems to be saying this. – Steven Gubkin Feb 07 '22 at 14:00
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7This 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. – Steven Gubkin Feb 07 '22 at 14:59
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1The Gaussian curvature is "intrinsic": it can be calculated just from the metric. More intuitively, an ant living on a donut could tell that space 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.
The "extrinsic" curvatures here are the "sectional curvatures", which the ant cannot say anything about.
– Steven Gubkin Feb 07 '22 at 15:45 -
1@steve Gubkin: Err, since we're talking about curvature then its implicit that we require a manifold with a metric, ie a Riemannian space. Your criticism is irrelevant. If all you are talking about is 'extra structure'. Then a topology is extra structure on top of the set of points. Without it, of course, you could not say a torus was continuous. – Mozibur Ullah Feb 07 '22 at 13:38
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@JoelKeene I like the final fantasy example because you can fly along all the geodesics! – Matthew Drury Feb 07 '22 at 17:45
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The 2-dimensional torus having the familiar doughnut shape has positive Gaussian curvature on the outside, and negative on the inside. As simple as that.
Oscar Lanzi
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2There 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'. – Moishe Kohan Feb 08 '22 at 22:42
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@MosheKohan A circle rotated around a straight line? Isn't it a circle rotated around a circle? – Pathfinder Feb 08 '22 at 23:41
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1That 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. – Moishe Kohan Feb 08 '22 at 23:52
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@MoisheKohan Felicia is using the phrase "rotate around a circle" to mean the same thing as what you mean when you say "rotate around a line". As you rotate the circle around a line, the centers of these circles sweep out another circle. This is the circle Felicia is saying the original circle is being "rotated around". – Steven Gubkin Feb 09 '22 at 01:00
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2@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. – Steven Gubkin Feb 09 '22 at 01:06