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How does changing the metric on a manifold affect the shape of the manifold?

  1. Oct 4, 2012 #1
    Hi all,

    I am trying to understand geometric flows, and in particular the Ricci flow. I understand how to calculate the metric tensor from the parametrization of a surface, but I am facing a problems in the concept phase.

    A metric tensor's purpose is to provide a coordinate invariant expression for the shape (in the form of distance) of an object, i.e. a surface. A geometric flow changes the metric tensor. Therefore, the shape of the manifold will change, obviously for any coordinate system.

    So, what exactly is the product of a geometric flow? Let's stick to surfaces for now, so is it the original surface endowed with a new measure of distance? The way I understand it is that since the shape of the original surface is invariant, changing how we measure distance on the surface shouldn't yield a different shape. So does this mean that by defining a new metric for a surface we get a different surface?

    If so, I am missing something here because I can't understand how defining a new measure of distance for an object creates a new, different object.

    I would be grateful if someone could showcase or guide me through applying a geometric flow for a simple shape, i.e. an ellipsoid.
     
  2. jcsd
  3. Oct 4, 2012 #2

    quasar987

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    My knowledge of the Ricci flow is extremely minimal, and basically consists of this one example: You can put a metric on a 2-sphere by embedding it has the standard sphere of radius r in R³. In spherical coordinate (θ,[itex]\phi[/itex]), the metric is then r²(dθdθ + sin² d[itex]\phi[/itex]d[itex]\phi[/itex]). So we see that varying the r parameter in there corresponds to varying the radius of the embedded sphere. As r-->0, the sphere degenerates into a point.

    Conversely, any riemannian manifold can be embedded isometrically into some R^n (Nash's embedding theorem), so given any riemannian manifold (M,g), varying g can be seen as varying the embedding of M in R^n.
     
  4. Oct 4, 2012 #3
    a topological manifold may not have a metric yet defined on it. but you can construct a metric tensor field on the manifold if the manifold satisfies some (uncertain to me) geometrisation conditions (i think hausdorff compact or normalisable or something like that).

    the main point of the ricci flow is that if the manifold is locally geometrisable, it is globally geometrisable. that is, from a compatible family of local metrics, there is a single globally defined metric equivalent to it. this is under a suitable limit on a family of metrics that describe the diffeomorphism equivalence class. so by way of this equivalence, you see the diffeomorphism dually either as deforming the manifold and its attached tensor field, or as a fixed manifold conformally transforming the metric itself.

    the main problem is that you want some well defined way of defining geometrisation locally and them some way of bulding this geometrisation globally so that we can identify it with a product of canonical (Lie group) manifolds under the ricci flow. the obstruction to this is failing to be geometrisable somewhere - hense the known categorisation of type singularities.

    sorry i'm not sure if that helps or is even completely accurate.
     
  5. Oct 5, 2012 #4
    Maybe you need to get a better understanding of what deforming a metric in a Riemannian manifold implies as opposed to what you mean by just "defining a new measure of distance".
    There are some actually well versed people around here, hopefully they can help you with this.
     
    Last edited: Oct 5, 2012
  6. Oct 5, 2012 #5
    This helps a lot, thank you.


    You are correct, this is essentially the core of my question, what deforming a metric in a Riemannian manifold actually means. If I understand quasar's explanation correctly, it produces a different embedding of M in R^n, therefore a shape that looks different but can at any time be reverted to the original manifold using the 'current' metric?

    Also a practical issue, the Ricci flow equation is:

    [tex]\partial_t g_{ij}=-2R_{ij}[/tex]

    I have calculated the Ricci tensor, but I am having trouble with the 'time' parameter. The parametrization of a 3D surface needs two parameters, let's say u,v. How do I apply the time differentiation? Do I need to express u,v as functions of 't'? As far as I know it is possible to express a 3D curve using just one parameter, but I do not know how to express a surface in that form.
     
    Last edited: Oct 5, 2012
  7. Oct 5, 2012 #6

    Ben Niehoff

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    In purely intrinsic geometry, the "shape" of a manifold is meaningless. What we mean by "shape" has a lot more to do with how a manifold is embedded into some R^n. For example, a flat sheet of paper has many different isometric embeddings into R^3, which you can achieve by all the ways you might imagine bending, crumpling, etc., a sheet of paper. Hence the metric information alone is not enough to constrain the shape information; the "shape" is really the collection of all the extrinsic curvatures.

    You introduce "time" as a third parameter, i.e.

    [tex]g_{ij} = g_{ij}(u, v; t)[/tex]
    For example, you might write a sphere as

    [tex]ds^2 = a(t)^2 (d\theta^2 + \sin^2 \theta \; d\phi^2)[/tex]
     
  8. Oct 5, 2012 #7
    Thank you, this helps me immensely! I'll try it out tomorrow and see if I can get it to work :biggrin:
     
  9. Oct 10, 2012 #8
    Another, slightly different, way of writing the Ricci flow equation would be:

    \begin{equation}
    \partial_\beta \, g_{i j} (\beta) = - 2 R_{i j} (\beta) \, ,
    \end{equation}

    with the following property:

    \begin{equation}
    g_{i j} (0) = g_{i j} \, .
    \end{equation}

    gij would stand for the `unaffected' metric in the sense that, when you consider your additional parameter β equal to zero, you recover your `initial' metric (or initial geometry).

    So β should not in general be related with the time parameter of your coordinate system. In other words, for any given β, you can choose a different coordinate system that won't affect the geometry of your manifold. However, a different β gives you a different geometry.

    As far as I know/understand, the Ricci flow provides a way of studying smooth deformations of manifolds.
     
  10. Oct 10, 2012 #9
    Hmmmm. I am struggling quite a bit to wrap my mind around how to properly insert 'time' into the equation. Let's assume that I have a metric:

    [tex]g_{ij}=\left(\begin{array}{cc}g_{11}(u,v)&g_{12}(u,v)\\g_{21}(u,v)&g_{22}(u,v)\end{array}\right)[/tex]

    I could parametrize it for time like:

    [tex]g_{ij}(t)=(1-2t) \left(\begin{array}{cc}g_{11}(u,v)&g_{12}(u,v)\\g_{21}(u,v)&g_{22}(u,v)\end{array}\right)[/tex]

    so that the time integrations and differentiations (for the Ricci tensor as a function of [itex]g_{ij}(t)[/itex]) become trivial. I've seen this parameterization in a few papers. However, by doing this aren't I imposing a pre-determined way that the manifold will deform? It seems in this sense that whether the manifold becomes a 3-sphere or not depends intensely on choosing the proper parametrization. Perelman (I can't really keep up with the details of the proof yet) however proved that this is always possible for a closed smooth 3-manifold, so what am I missing? Does the proof imply/state that convergence depends on proper parametrization? What would the parameterization be for an arbitrary manifold then?

    Sorry if I'm confusing a few concepts, but differential topology is still quite new to me :tongue:
     
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