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Some basic problems in Riemannian Geometry

  1. Jun 7, 2004 #1
    Hi all! I just found this site today, and I am really hoping that I can get some useful advice here. That said, I have two problems--one easy, one not so easy.

    Easy problem:

    Basically, I was wondering if anybody out there knows of an algorithm to calculate [itex] g_{ij} [/itex], given only the 'shape' of the underlying manifold? For example, I have what can best be described as a dumbell, or [itex] S^2 [/itex] with the equator contracted in to a small radius--no corners anywhere, though. Keep in mind that I am not looking so much for a solution for the dumbell, but something that could be applied to a more general, finite dimensional, connected [itex] C^\infty [/itex] manifold.

    Harder problem:

    As a related problem, I am also trying to come up with some sort of meaningful geometric interpretation of the Laplacian when it is being applied to some function that lies on a Riemannian manifold, rather than Euclidean space. In particular, it seems that something like the heat equation:

    [itex]\frac{\partial}{\partial t}u=\Delta u [/itex]

    should lend itself to a, let's say, rewording as something like a Gauss curvature flow:

    [itex]\frac{\partial}{\partial t}x=-K \nu [/itex]

    where [itex] x [/itex] is an embedding. Of course, the manifold where this new surface would be embedded would require some additional structure, since the geometric heat equation is valid only for (smooth) compact Riemannian surfaces without boundary--like [itex] S^2 [/itex], or something diffeomorphic to it. That problem, however, has already been taken care of, so please don't waste your time even considering it.

    Anyway, I thank you all in advance for your time, and if you require clarification of the question at all, please don't hesitate to ask.

    Last edited: Jun 7, 2004
  2. jcsd
  3. Jun 7, 2004 #2
    Oops, seems I didn't read the howto correctly for inserting LaTeX code into a post. I will endeavor to get it right next time--at least I can see what the mistake is!

  4. Jun 7, 2004 #3


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    You should be able to edit your post to fix it, too.
  5. Jun 7, 2004 #4
    It would seem that I can! This is really a cool forum--very nice! So, changes made, any ideas?

  6. Jun 7, 2004 #5


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    As far as the easy problem goes...

    You may be way beyond me already, but I have read that typically a space(time) does not allow one single coordinate system to coordinatize the entire space. "Patches" of coordinate systems can always be built up into an "atlas" in order to get the whole space coordinatized. (I am going by old, faint recollections here, so my terminology may be in error.) The values of the components of your metric tensor at any point on your dumbbell will obviously depend on your choice of coordinate system(s).

    Whether a dumbbell space is simple enough to allow a single coordinate system to completely cover it, I don't know. But I'll bet some of the math types here will be able to tell you the answer to that. My own intuition, for what it's worth, is that you will need at least two or three coordinate patches on the dumbbell.
    Last edited: Jun 7, 2004
  7. Jun 8, 2004 #6


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    Oh, I guess I can comment on the easy question too.

    Shape is a far more general property than metric; in particular, the same shape can have many inequivalent metrics.

    For example, considur the torus.

    One way we can endow the torus with a metric is to take its embedding in Euclidean 3-space, and inherit that metric. In particular, this torus has nonzero curvature.

    Another way is to take the model where the torus is a square with opposite sides identified. With care, we can also endow this torus with the Euclidean metric, and in this case, the torus is flat.
  8. Jun 8, 2004 #7
    In your example of the torus above, you have stated that "the same shape can have many inequivalent metrics", but I don't think this is quite correct. For example, in the above, we could denote the metric that is inherited from Euclidean 3-space as [itex] g_{ij} [/itex], and the one corresponding to the "flat" torus as [itex] g'_{ij} [/itex]. These will be related by

    [itex] g'_{ij}=g_{kl}\frac{\partial u^k}{\partial u'^i} \frac{\partial u^l}{\partial u'^j} [/itex]

    Thus, the set of all [itex] g_{ij} [/itex] satisfying the above eqn. will define an equivalence class on [itex] T^2\mathcal{M} [/itex] (i.e., (2,0) tensors on the manifold).

    So, what I am after, then, is some means of actually operating the machinery to generate the [itex] g_{ij} [/itex]. That is, I have a manifold, I understand what the metric tensor tells us, and I know how to carry the fight forward. However, I cannot seem to get out of the starting gate, since I don't see how one actually determines any [itex] g_{ij} [/itex], except of course the Euclidean one...

    So, any ideas?
  9. Jun 10, 2004 #8
    An update:

    Well, the first problem has been taken care of. The key was to find a text that treats Riemannian geometry from a somewhat elementary standpoint--numerous examples were found in various continuum mechanics texts. If anybody is interested, I can post the result here.

    Now, the second problem... Any takers yet? I think it might be benificial to consider forms on the manifold (more general than just functions, or 0-forms), and in doing this we have results like the Weitzenbock formula,

    [itex] \Delta = \nabla^*\nabla + R [/itex]

    which should, I hope, have some geometric interpretation available. Anyhow, hoping to rouse some interest...

    Last edited: Jun 10, 2004
  10. Jun 10, 2004 #9


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    I'll take you up on that offer, though it may be over my head.
  11. Jun 11, 2004 #10
    Oh, I don't know. If I can figure this stuff out I'm pretty sure that it cannot be that difficult! Thank you, though, for the interest--it's a good opportunity for me to practice.

    That said, let us begin.

    Let [itex] \mathcal{M} [/itex] be a manifold, and let [itex] \mathcal{U}\subset\mathcal{M} [/itex]. Define a homeomorphism

    [itex] \varphi_{\mathcal{U}}:\mathcal{U}\rightarrow\bbm{R}^n[/itex]

    so that [itex] \left(\mathcal{U},\varphi_{\mathcal{U}}\right) [/itex] is a coordinate chart on [itex] \mathcal{M} [/itex]. Since [itex] \varphi_{\mathcal{U}} [/itex] is a homeomorphism, for any [itex] x\in\mathcal{U} [/itex], we can define the coordinates of [itex] x [/itex] to be the coordinates of [itex] u=\varphi_{\mathcal{U}} \left( x \right) [/itex], i.e.

    [itex] u^i = \left(\varphi_{\mathcal{U}}\left(x\right)\right)^i [/itex]

    where [itex] i=1,2,\ldots,n [/itex]. Thus, via Pythagoris, we know that the distance between any two points [itex] u [/itex] and [itex] u+du [/itex] is [itex] ds [/itex], where

    [itex] ds^2=du^kdu^k[/itex]; [itex] k [/itex] summed.

    However, we also know that

    [itex] du^k =\dfrac{\partial}{\partial x^i}u^kdx^i [/itex]

    So that

    ds^2 &= \left(\dfrac{\partial u^k}{\partial x^i}dx^i\right)\left(\dfrac{\partial u^k}{\partial x^j}dx^j\right)\\
    &= g_{ij}dx^idx^j

    Which tells us that

    [itex] g_{ij} = \dfrac{\partial u^k}{\partial x^i}\dfrac{\partial u^k}{\partial x^i} [/itex]

    Where, again, [itex] k [/itex] is summed, and each [itex] u^k [/itex] is the [itex] k^{th} [/itex] coordinate function [itex] \left(\varphi_{\mathcal{U}}\left(x\right)\right)^k [/itex] on [itex] \mathcal{U}\subset\mathcal{M} [/itex].

    So how was that? Better yet, does anybody have any thoughts on the other question I originally posted?
    Last edited: Jun 11, 2004
  12. Jun 12, 2004 #11


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    Do you mind dwelling on your first problem a bit longer?

    Maybe this would be too much work to expect you to go through, but to try to make your method concrete for me could you answer some specific questions for me?

    Take a plane with a circle in it of radius 10. Add another circle of radius 10 to the plane, such that the circles osculate. On each side of the osculation, add a circle of radius 1 that osculates with the two bigger circles. Trim away arcs such that you are left with a simple closed curve that looks something like an hourglass. It is nice and smooth all the way around. It has two lines of symmetry. Choose the line of symmetry to rotate the figure around such that it sweeps out a dumbbell surface. (Choosing the other line of symmetry would have caused the figure to sweep out a double-dimpled surface that would look kind of like a red blood cell.)


    1. Can a single two-dimensional coordinate chart cover the entire dumbbell? Regardless of the answer, go ahead and coordinatize the dumbbell with as many charts as you need, in any way you care to. If you are using more than one chart, describe the locus of points that form the boundaries between charts.

    2. What would be the coordinates of a point at either of the two poles (your choice) of the dumbbell? I.e., where the axis of rotational symmetry pierces the surface.

    3. What would the four metric components be at the point you chose? (Only three independent components, of course.)

    4. Pick some point on the circle of minimum radius at the waist of the dumbbell. What coordinates does this point have? What are the components of the metric at this point?

    5. For the point you chose in (2), use the machinery of the metric to show that the curvature is 1/10. For the point you chose in (4), use the machinery to calculate the curvature there. (We know that it will be the negative reciprocal of the radius of the waist, radius being measured in the three-dimensional embedding space.)
    Last edited: Jun 12, 2004
  13. Jun 13, 2004 #12
    Well, I would be willing to take stab at this, except that I cannot figure out what your construct is supposed to look like--I don't understand what you mean by "the circles osculate". It's simply a matter that I have not come accross this term before, and thus need a little elaboration... If you could post or email a scetch, then I think I would like to have a go at this--my curiosity is peaked!

  14. Jun 13, 2004 #13


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    He means tangent. The resulting plane curve will not be twice differentiable, I don't know if that will cause problems.

    It would require at least two charts to cover the surface; it is compact, so it cannot be homeomorphic to the plane.
    Last edited: Jun 13, 2004
  15. Jun 13, 2004 #14
    OK, so to make sure I have this picture correct, we can make this construct by drawing a circle of radius R, and then drawing another copy of the same circle, centered at a point at a distance of |2R| from the center of the the first. Then, at a point 'above' the cusp, we draw a smaller circle of radius r that is tangent to both of the two larger circles, repeat 'below' the cusp, throw away the excess line segments, and rotate the shape through \pi radians? Is this the same as what Janitor proposed?
  16. Jun 13, 2004 #15


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    You've got it. In fact, I like your idea of using R and r better than my idea of using 10 and 1. The curvatures that you calculate will then need to work out to be 1/R for the point out at the "tip" of the dumbbell, and some negative number involving both r and R at the waist (since it is a saddle point that looks concave of curvature 1/r in one direction and convex of curvature that is a function of both r and R in the direction orthogonal to that direction, if my intuition serves correctly).

    Hurkyl, I don't doubt you when you say at least two charts are needed. But how did you figure that out? Is it also true that a sphere is not homeomorphic to a plane, but a sphere can be charted by a single chart breaking it down into longitude and co-latitude?

    And the failure to be twice-differentiable I presume occurs where the waist meets the other two parts of the dumbbell?
    Last edited: Jun 13, 2004
  17. Jun 13, 2004 #16
    OK, so I got the picture correct, so I may as well get to work. Wish me luck!

    As for the question of two charts, consider that this dumbell is topologically equivalent to [itex] S^2 [/itex], so we need only consider that shape. On an intuitve level, to say that a manifold can be covered with one chart means that it can be bent / stretched / moved around in such a way that it becomes [itex] \bbm{R}^n [/itex], where the deformation process is defined by the chart [itex] \varphi [/itex]. So, in the case of the 2-sphere, we cannot deform it in any way to 'become' [itex] \bbm{R}^2 [/itex] without making at leat one cut, or hole--we'll call it [itex] \mathcal{W} [/itex]. Thus, since any chart will only be able to cover, at most, [itex] S^2\setminus \mathcal{W} [/itex], we will need at least one more chart to 'cover' the hole.

    Now, on to the dumbells...
  18. Jun 13, 2004 #17


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    Good luck!

    If I understand your last post, the sphere technically requires two charts, but one chart is almost enough to cover it, maybe just leaving one (or two antipodal?) points of the sphere uncovered.
  19. Jun 13, 2004 #18
    Correct, depending on how you define your charts. However 'almost' enough is not enough when in comes to math. Engineering on the other hand... :smile:
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