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Friday, May 11, 2012

Flat tori in 3D

Note: re-edited thanks to Steven Halter's wonderful input. Please check his comments below. 

It's been a while since I've read a pure math paper, but when I saw the picture I knew I had to pick this one up. For the pure mathematicians out there: I haven't done pure math since my grad years, so feel free to pitch in and correct me if I misunderstood any of the following!

"Torus" is mathematics for donut. Take a very flexible square -- imagine it's made of rubber -- roll it, then glue together the circles at the two ends. Congratulations. You've made a torus.

Now suppose you live on the torus and you need a map that takes you from A to B. Think of an atlas that shows you all the streets and cities on the torus. How do you map the torus onto a flat surface so that you can actually hold the map in your hands? Well, you go back to that square you used to make the torus, right? Cut out the donut vertically, then horizontally, and you've got your square back and now you can map all the streets you want.

Problem: the distances on your map will now be distorted, just like continent sizes are distorted on world maps (have you ever seen this image?). So now let's go back to the rubber square we've used to make the torus. Imagine you can move continuously from one edge to the opposite one both vertically and horizontally. That's what mathematicians call a square flat torus. The problem we want to solve is the following: we want to map this flat torus into the 3-D with the additional constraint that we all distances preserved.

Well, it turns out, there's a famous theorem, the Nash embedding theorem, that states that any Riemannian manifold (replace that with the torus we were talking about) can be isometrically embedded into a Euclidean space. Isometrically here means in such a way that it preserves the distances.

The above theorem tells us that there's a way to map it into a torus that will preserve the distances. Problem: how to visualize it? See, that's always been my issue with math. It's so beautiful at telling you what exists and what doesn't, but then you get to the practical side, as in, "Okay, now give me such map," and the mathematician shrugs and looks at you all weird: "I told you it exists, aren't you happy with that?"

Sorry, I'm joking, let's get serious again.

Here's the news: we now have a visualization of a flat torus (the square) in the three-dimensional space. In the latest issue of PNAS, Vincent Borrellia, Said Jabranea, Francis Lazarusb, and Boris Thibert present an isometric embedding of the flat torus in three-dimensional space. Forget the jargon and just look at the picture: how cool is that? And here is the best part: see all those corrugation in the figure? It's because it's a "smooth" fractal surface, a sort of hybrid between a fractal and smooth surface. The embedding is
"a continuously differentiable map that cannot be enhanced to be twice continuously differentiable. As a consequence, the image surface is smooth enough to have a tangent plane everywhere, but not sufficient to admit extrinsic curvatures."
Yes. I'm still a mathematician at heart, because I read this and got all excited. Of course, the rest of the paper went right past my head, but any of you willing to add a few more insights, you are more than welcome to do so in the comments below. Thanks! A few more details on the paper here.

Borrelli, V., Jabrane, S., Lazarus, F., & Thibert, B. (2012). From the Cover: Flat tori in three-dimensional space and convex integration Proceedings of the National Academy of Sciences, 109 (19), 7218-7223 DOI: 10.1073/pnas.1118478109


  1. That's very cool.
    The flat torus is actually the initial rubber square with the property that as an object moves across the surface of the flat torus, if it hits an edge, it will appear at the opposite edge.
    When you convert the flat torus to the "donut" you loose the distances as they stretch. The picture then shows the initial flat torus formed into a 3d structure with the distances (isomorphism) kept intact.
    Very neat, thanks for pointing this paper out!

    1. Thank you! Wait, I'm a little confused now, can you explain it to me: I thought that by "flat torus" they meant an isometric embedding, hence the square, though it is a flat embedding of the torus, isn't really a flat torus since it doesn't preserve the distances... so maybe one should specify "isometric flat torus"... right?

    2. Ooohh... you're saying it's the other way around...

  2. Here's a link to the project site that visually shows the flat torus:
    So (I think) the problem is going from the flat parallelogram with "identified" edges to a three d representation. Conversely, you could go from that 3d representation of the flat torus (in the picture above) to the flat parallelogram isomorphically.
    To put it another way, with this technique you could take a piece of paper (not stretchy), and glue the top and bottom together to get a cylinder. By carefully! corrugating the surface, you could then glue the ends of the paper cylinder together without tearing the paper. I'm not sure if paper would actually allow for the proper "corrugations" in reality, but it is a fun concept.
    You can get various packages that make your computer monitor work like a flat torus--your mouse will go to the bottom of the screen when you move it off the top instead of just stopping--just like the old video game Asteroids.

  3. Steve, thanks for the link and for the insight, I totally get it now and actually, now that you've pointed this out, I can totally see how it becomes a fractal... way cool! thanks!

  4. antisocialbutterflieMay 11, 2012 at 4:49 PM

    This is pretty awesome. Spatial puzzles are pretty much part and parcel with my field but we never get to play with cylindrical shapes. It's cool to see the math behind it.

  5. "Well, it turns out, there's a famous theorem, the Nash embedding theorem, that states that any Riemannian manifold (replace that with the torus we were talking about) can be isometrically embedded into ^^^^^the^^^^^ Euclidean space."

    Change the "the" to "a". It doesn't mean 3-d. The theorem says that for any riemannian manifold there exists a finite "n" such that this manifold can be isometrically embedded in Euclidean n-space. For the flat torus it is 4-d if you want to preserve curvature and C2 surface smoothness. Since surface curvature is not definable in this fractal embedding it seems somewhat presumptuous to call it a *flat* torus. It is a fascinating result though.


    1. Done, thank you. I thought the flat torus was the starting surface, then once you embed it it's probably no longer flat since the embedding is a fractal embedding...


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