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Meaning of isomorphism/diffeomorphism ## f: R^n\to M^m##

  1. Oct 1, 2015 #1
    Can one define an isomorphism/diffeomorphism map ## f: R^n\to M^m## when ##n>m##? ##M## is a non-compact Riemannian manifold..
     
  2. jcsd
  3. Oct 1, 2015 #2

    andrewkirk

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    Yes for diffeomorphism. The projection map is a diffeomorphism, eg ##f:\mathbb{R}^2\to\mathbb{R}## such that ##f((x,y))=x##.

    What do you mean by isomorphism though? What is the algebraic structure that you are wanting to preserve?

    If you define algebraic structures on domain and range, and find an isomorphism, it will not also be a diffeomorphism, because that would make it a homeomorphism, and there is a theorem that a n-dimensional manifold is not homeomorphic to a m-dimensional manifold if ##m\neq n##.
     
  4. Oct 1, 2015 #3

    WWGD

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    What do you mean, isn't a diffeomorphism invertible? ##p^{-1}(x_0)=(x_0, \mathbb R ) ##.
     
  5. Oct 1, 2015 #4

    WWGD

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    No, dimension is a diffeomorphism invariant. Take, e.g., the Jacobian to see why this is not possible. Dimension ( I think this is the Lebesgue dimension) is a topological, i.e., homeomorphism - invariant
     
  6. Oct 1, 2015 #5

    andrewkirk

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    Right you are. I forgot that part. :oops:

    In that case there cannot be a diffeomorphism.

    Edit: Ah, as I see you have noted in your next post.
     
  7. Oct 1, 2015 #6

    WWGD

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    No problem, Andrew, I have made a few of those -- just hit the delete button, and I will delete my post; let me know.
     
  8. Oct 4, 2015 #7

    lavinia

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    A key theorem that answers this question is Brouwer's Invariance of Domain which says that a continuous injective image of an open set of Rn into Rn is an open subset of Rn and that the mapping is a homeomorphism. It is not hard to show as a corollary that there is no continuous injection of Rn into Rn-m m>0. A differential map is continuous so the theorem applies to differential maps as well.

    BTW: This question is really a question about topology not differential geometry.
     
    Last edited: Oct 5, 2015
  9. Oct 4, 2015 #8

    WWGD

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    But if you're dealing with diffeomorphisms, you can use tools not available with topology alone: a diffeomorphism gives you a vector space isomorphism between tangent spaces at a point and at its image. By properties of linear maps, both dimensions are the same. OF course, a homeomorphism does not exist either, by, e.g.,as you stated, invariance of domain. You can twist this using the result that continuous bijection between compact and Hausdorff is a homeo. Consider the closed unit ball B(0,1).Then f: B(0,1)--> f(B(0,1)) is a homeo. Now consider a small open disk in B(0,1), which is sent to an open set in R^n , contradicting Invariance of Domain.
     
  10. Oct 5, 2015 #9

    lavinia

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    That is true.

    Not sure what you mean here.
     
    Last edited: Oct 7, 2015
  11. Oct 5, 2015 #10
    Actually, any two manifolds that are topologically equivalent* must have the same dimension, so the answer here is No.

    Possibly the simplest proof of this uses homology, which is a way of assigning one abelian group Hk(X) — which can at times be the trivial group {0} — to each topological space and choice of positive integer k, in such a way that topologically equivalent spaces have isomorphic groups. A mild generalization of this idea assigns an abelian group, likewise, to every pair of topological spaces (X,A) where A is a subspace of X, and every positive integer k. (This is called a relative homology group.)

    Then if M and N are topologically equivalent manifolds, there must be a homeomorphism h: M —> N. This implies that for any point p of M, the relative homology groups Hk(M,M-p) and Hk(N,N-h(p)) are isomorphic.

    If the dimensions dim(M) and dim(N) were not equal, let's assume, without loss of generality, that d = dim(M) > dim(N). It can be shown that for any manifold M of dimension d, the relative group Hd(M,M-p) is isomorphic to the group ℤ of integers. It can also be shown that for any manifold N with d greater than the dimension of N, the homology group Hd(N,N-q) is the trivial group {0}, where q is any point of N.

    But if M and N were homeomorphic they would have isomorphic relative groups. Since they do not, this shows that any two manifolds of unequal dimensions cannot be homeomorphic.
    ____________
    * Note: Two smooth manifolds might be "homeomorphic" (topologically equivalent) or even diffeomorphic (having a smooth homeomorphism between them, with a smooth inverse) but the word "isomorphic" is not used for topological spaces.
     
  12. Oct 5, 2015 #11

    WWGD

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    I was trying to avoid (co)homology; I thought if the OP was familiar with them, s/he would not have asked the question. And I guess you could also use (c0)homology of the whole space , so that if M>N , then , if M is orientable, its highest homology, say m is, as in the case of the relative homology you used ## \mathbb Z ## , while, again as you said, the m-th homology (and any one larger than Dim N) would be 0.
     
  13. Oct 5, 2015 #12

    andrewkirk

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    I would also tend to avoid purely topological explanations. While they have wider application, they can be much more abstract and conceptually demanding. If the context is Riemannian manifolds, there's no need for them.

    Indeed, my current feeling is that homology theory is the most abstract topic in mathematics I have studied. I love it, but to me it seems much less intuitive than diffeomorphisms between Riemannian manifolds, which are quite visual. One of the amazing things about homology theory is that one can achieve such practical, intuitive results, using such abstract mathematical structure, and that there is at present no easier known way to achieve such results. So it's not just abstruseness for the sake of abstruseness, as some branches of mathematics sometimes appear to be.
     
  14. Oct 5, 2015 #13

    mathwonk

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    actually I can define a diffeomorphism betwen any two manifolds, it is a differentiable map with a differentiable inverse. I just can't prove one exists.
     
  15. Oct 5, 2015 #14
    I recommend that you close your eyes when coming upon a topological explanation. Comparing diffeomorphisms with homology groups is like saying you like beautiful landscapes more than GPS devices. (Translation: You can't prove much with just a diffeomorphism. )

    And as for "If the context is Riemannian manifolds, there's no need for them," just maybe you don't know all the ways that topological explanations can provide useful information in that context?
     
  16. Oct 13, 2015 #15

    lavinia

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    The manifolds may not be compact so highest homology may be zero For instance the homology of Rn is zero in all positive dimensions.
    But Hn(M,M-p) = Z for any n manifold. This follows from excision.

    Let A be an open neighborhood of p that is homeomorphic to an open ball in Rn. By excision,
    Hq(M,M-p) = Hq(M-(M-A)),M-p-(M-A)) = Hq(A,A-p) for all q including n. A is contractible and A-p deforms onto a sphere so the exact sequence of the pair (A,A-p)

    ... Hq(A) -> Hq(A,A-p) -> Hq-1(A-p) -> Hq-1(A) ...

    is

    0 -> Hq(A,A-p) -> Hq-1(Sn-1) -> 0 except when q = 1.
     
    Last edited: Oct 14, 2015
  17. Oct 14, 2015 #16

    lavinia

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    To say that there is no need for topological proofs when one is dealing with smooth manifolds ignores much of the research in mathematics that deals with the interplay between algebraic topology, differential topology, and differential geometry.

    BTW: A Riemannian manifold refers to a smooth manifold together with a Riemannian metric. Your question does not involve a Riemannian metric and is not a question about Riemannian geometry. It is a question about topology. WWGD's proof uses calculus on manifolds not Riemannian geometry. It is a differential topology proof. Zinq's proof is an algebraic topology proof.
     
    Last edited: Oct 14, 2015
  18. Oct 14, 2015 #17

    WWGD

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    But if it is not possible to do it without a Riemannian metric, it is not possible to do it with one. So the Riemannian condition is somewhat superfluous. Is there some result that an n-dimensional Riemannian manifold cannot be equivalent to an m-dimensional one that only uses properties of the metric?
     
  19. Oct 14, 2015 #18

    lavinia

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    One point that should be made here is that smooth manifolds may be topologically equivalent but not diffeomorphic. For such manifolds, as mathwonk pointed out, it is possible to define smooth equivalence but not possible find one.

    The moral is that equivalence is relative to the category of mathematical objects. In the smooth category equivalence means diffeomorphic. In the topological category it mean homeomorphic. In the PL category it means piecewise linearly isomorphic.
     
  20. Oct 14, 2015 #19

    WWGD

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    I guess the standard example of homeo but not diffeo is that of ## \mathbb R ## and ## (x,x^3): x \in \mathbb R ##.
     
  21. Oct 14, 2015 #20

    lavinia

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    all connected closed smooth 1 manifolds are either diffeomorphic to R or to the circle.
     
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