Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Does the Mystery Manifold theorem exist?

  1. Apr 17, 2012 #1

    Greylorn

    User Avatar
    Gold Member

    The theorem supposedly states:

    A space cannot be bent unless it is a manifold which is embedded in a space of at least one higher dimension.

    Does anyone know if this theorem actually exists? If it does, I would appreciate a reference to its name and proof.
     
  2. jcsd
  3. Apr 19, 2012 #2

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    This is not a proper theorem. First of all, any smooth manifold can be embedded in a Euclidean space of high enough dimension (strong Whitney embedding theorem http://en.wikipedia.org/wiki/Whitney_embedding_theorem). So the latter part of the statement is not a restriction at all.

    Secondly, there are multiple notions of curvature, but they can be roughly divided into classes. First, extrinsic curvature http://en.wikipedia.org/wiki/Extrinsic_curvature, which uses the nature of the way a manifold embeds into a higher-dimensional space. The value of this curvature can depend on the nature of the embedding space. Second, for so-called Riemannian manifolds, we have a notion of intrinsic curvature http://en.wikipedia.org/wiki/Curvature_of_Riemannian_manifolds, which can be defined independently of any embedding.
     
  4. Apr 19, 2012 #3

    Greylorn

    User Avatar
    Gold Member

    Thank you! I'll study those links and see what comes of it. The theorem appeared on the blackboard in a math course taken several decades ago, long since forgotten from disuse. It made sense then, and now seems relevant. I appreciate your help.
     
  5. Apr 22, 2012 #4
    Thank you, Fzero. Since I am the person who first raised this question with Greylorn, it is appropriate that I take his suggestion and "jump into the fray" here. Thank you for giving this question some thought and for taking the time to respond.

    In your first sentence you seem to dismiss the question out of hand. I must admit that I don't quite understand your statement since I don't know the difference between a "proper theorem" and an "improper theorem". 'Improper theorem' seems like an oxymoron.

    In any case, you seem to be pretty certain that this is not a proper theorem, which either means that it is not a theorem at all, or that it is an improper theorem – whatever that is. So my first question to you is, How do you know it is not a proper theorem? Do you know the complete list of proper theorems and that this one is not on the list?

    Next you cite the Whitney embedding theorem evidently to make the point that "the latter part of the statement [I assume "the statement" refers to my "mystery theorem"] is not a restriction at all."

    By "the latter part" I assume you mean "a space which is a manifold embedded in a space of at least one higher dimension." If your claim is true, that this condition (of being an embedded submanifold) does not restrict a space from being bent (my term), then it is possible for a space to be bent and not be an embedded manifold in a higher-dimensional space.

    But if the bent space is a smooth manifold, which is the type we are interested in, then the Whitney theorem implies that it can be embedded in a "high enough" dimensional Euclidean space. So it seems that if "at least one greater" is "high enough", then the "mystery theorem" holds. If not, then the mystery theorem is even stronger, possibly requiring more than one additional dimension.

    The rest of your response consistently uses the term 'curvature' and avoids all mention of my term 'bent'. Since I am not knowledgeable enough to be able to distinguish between 'bent', 'curved', 'warped', 'stretched', 'distorted', or other such terms, I would prefer to stay within vernacular and intuitive usage, if that's possible. Otherwise I would appreciate your teaching me the proper usage in this context. It might be that it is the mis-use of these terms that disqualifies the "mystery theorem" and relegates it to the set of "improper theorems". If so, let's switch to the terms 'curve' and 'curvature' and consider the following statements:

    Which of these statements is false or nonsensical?

    1. In a Euclidean plane, there can exist straight lines of zero curvature and curved lines with non-zero curvature.

    2. Both the straight lines and the curved lines in (1) are 1D manifolds embedded in the 2D Euclidean plane.

    3. The straight lines are each also embedded in a 1D Euclidean space defined by the points of the straight lines.

    4. The curved lines, on the other hand, cannot be similarly embedded in a 1D Euclidean space because of their curvature.

    5. Since the curved lines exist in the original 2D Euclidean space and cannot exist in a 1D Euclidean space, and since the dimension of the manifold of the curved line is 1, at least one higher dimension, viz. 2, is necessary for the existence of the curved line in Euclidean space. (This seems to be a special case of the "mystery theorem".)

    6. An analogous argument can be made for flat and curved 2D surfaces in 3D Euclidean spaces with the conclusion being that 3 dimensions are required for existence of curved 2D manifolds. (This would seem to be another special case of the "mystery theorem".)

    7. It is reasonable to expect that in general, n+1 dimensions are required for the existence of curved n-dimensional manifolds.

    Since (7) is an intuitive rendition of the "mystery theorem", it seems reasonable to expect that it is either a theorem, or a conjecture that might be provable.
     
  6. Apr 24, 2012 #5

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    First off, you've written quite a bit, so let me apologize if I don't get to address every single point in your post.

    A "proper" theorem must be a statement regarding well-defined objects. The statement must logically follow from the definitions and conditions on the objects, but should not obviously hold true in the absence of some of the conditions (otherwise we would state the stronger version of the theorem).

    Let me quote the "theorem" once again:

    A space cannot be bent unless it is a manifold which is embedded in a space of at least one higher dimension.

    To be a "proper theorem," at the very least we must have all terms used be well-defined. Unfortunately, they have not been, so we are left grasping among standard mathematical usage for context. We can go term by term:

    Space: A space is a set with some added structure, but what structure are we adding here? A topological space adds the notions of open sets. Mappings on and between topological spaces follow naturally, from which we get to the concept of a manifold. If we require that the charts are smooth functions, we get differentiable manifolds.

    Bent: This is apparently an archaic term that means "curved." You admit to some confusion over this terminology yourself. I was also confused until I googled it. If we can agree that it is synonymous with "curved," we still have not completely specified what we mean, hence why I linked to the wikipedia page on curvature. If you go there, you'll see that there are distinct notions of curvature in low dimensions. For curves in the plane, all curvature is extrinsic, which means it is solely due to the nature of the embedding of the curve. In 2d, we can have intrinsic curvature as well, such as the Gaussian curvature. In any case, the "theorem" does not specify.

    In any case, to define curvature properly, we need a suitable notion of differentiability. So are we to understand that the spaces we are talking about are differentiable manifolds? Well, maybe not, because the next part of the theorem introduces the term manifold in a way that forces us to conclude that some of the spaces under consideration were not manifolds. If true, this is confusing, because it is unlikely that we would get very far in defining the notion of curvature without already imposing the structure of a manifold and differentiability.

    So this was my logic in saying that the statement was not a proper theorem. For a "space" to be "bent," we are already in the setting of differentiable manifolds, since we cannot sufficiently develop the notion of curvature in the absence of that structure. So the part of the theorem which says that the space must be a manifold ought to be at the beginning, since it is part of the structure that we need to provide the setting for the rest of the theorem. The statement becomes

    A differentiable manifold cannot be curved unless it can be embedded in a space of at least one higher dimension.

    This is a theorem that has almost no value. By the Whitney theorem, all smooth manifolds can be embedded in a higher dimensional space. The curvature of the manifold is completely irrelevant. One could make other objections: for example, there are flat manifolds that can be embedded, so the theorem is purporting to only tell us when a manifold cannot be curved. It's not a complete test of whether or not the manifold can be. Obviously to say more along these lines, we'd need to know what type of curvature we're talking about.

    Note, however, that a possible loophole to some of what I just said is that the Whitney theorem requires a smooth manifold. Smooth means that an infinite number of derivatives of the chart functions exist. We can have differentiable manifolds where only k derivatives exist. If you look at the wiki page on curvature, you'll see that you only need 2 derivatives to define the curvature of a curve in a plane. In any case, the theorem does not specify this and if we're to understand this came from a reasonably standard university course, it's 99% likely that differentiable = smooth for all intents and purposes. (I just noted that you also suggested that we were referring to smooth manifolds in your post.)

    So to sum up, when I say that the theorem is "improper," I mean that it is seriously deficient in the context in which it was presented, for the reasons above.

    No it's not stronger. If we're in agreement that space = smooth manifold for the point of the theorem, then the space can be always be embeded in some higher-dimensional space, i.e., whether or not it is curved. The "mystery theorem" is essentially the Whitney theorem with an irrelevant piece of information tacked on.

    Unfortunately the vernacular and intuitive use of terms doesn't always match up with the mathematical usage. Since the rigor and logic of mathematics relies crucially on the precision used to define structure, we must use the mathematical notions. "Bent" is an archaic term that has been replaced by "curved." However, as I've elaborated on above, "curved" is a term that still refers to many different concepts and in order to formulate a theorem, we need to specify what we mean by curvature more precisely.

    All of these seem to be fine, but they are statements about curves in the plane. As I mentioned, the only curvature we have here is extrinsic (tied to the nature of the embedding). Here we have that straight lines have zero curvature by definition. We don't really gain much by attempting to formulate the theorem, because it is precisely the straight lines (or segments) that are isomorphic to (subsets of) the Euclidean line.

    In 2 and higher dimensions, we can have intrinsic curvature. Depending on the context, that Euclidean space is flat is either a definition or a small lemma. Then you seem to be arguing that the "mystery theorem" could be restated as

    A curved manifold is not Euclidean.

    That's fine, but it's not a particularly strong or illuminating result.
     
  7. Apr 24, 2012 #6

    lavinia

    User Avatar
    Science Advisor

    There is another definition of bending. A bending of a manifold is a deformation of it that preserves lengths and angles. For instance a flat piece of paper can be bent into a cylinder.

    It is possible for a manifold to be bent without being curved. Curved means that the Riemann curvature tensor is not identically zero but some manifolds do have a zero curvature tensor. In fact, there are manifolds in all dimensions that have no curvature. These manifolds are flat, not curved ,and can all be constructed from a flat chunk of Euclidean space by bending, just as the cylinder - or for that matter the Mobius band - can be obtained from a flat sheet of paper by bending. (Paper when deformed by bending does not strectch or shrink. All angles and lengths are preserved. So any bending of paper produces a flat rather than curved surface.)

    Sometimes a particular bending can not take place in a low dimension but can take place in a higher dimension. For instance, a flat rectangle of paper can not be bent into a torus in three dimensions. But it can in 4.

    Not all surfaces can be bent. For instance, the sphere in 3 space with constant Gauss curvature can not be bent. This is because there are no other surfaces without boundary that have constant Gauss curvature. I believe the same is true for hypersurfaces of constant mean curvature in any dimension.

    Bending is an idea that refers to a way in which the shape of a manifold can change when it is deformed inside another manifold. So it relates to extrinsic curvature.

    I suppose similar ideas can be applied to immersed manifolds rather than embedded manifolds but I am just guessing.
     
    Last edited: Apr 25, 2012
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook