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##x+y## on a Riemannian manifold

  1. Sep 29, 2015 #1
    Can one define a vector space structure on a Riemannian manifold ##(M,g)##?! By this I mean, does it make a sense to write ##x+y## where ##x,y## are arbitrary points on ##M##?
     
  2. jcsd
  3. Sep 29, 2015 #2

    WWGD

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    I don't think (at least I don't see how) the Riemannian metric here plays a role, since most reasonable spaces can be given a Riemannian metric. Maybe you can pullback locally the vector space structure of ## \mathbb R^n ##, I would say, given ##x:= \phi^{-1}(x'), y:= \phi^{-1}(y'); x',y' \in \mathbb R^n ## by defining ## x+y## to be ##\phi^{-1} (x'+y') ##then patch it up into a global one, I havent tried it . And, given that a Riemannian manifold has a global inner-product space, maybe you can find a way of making the two structures compatible.
     
  4. Sep 29, 2015 #3
    No. But do check out the exponential map for something close to what you're attempting.
     
  5. Sep 29, 2015 #4

    WWGD

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    Why can't you pullback the vector space structure from ## \mathbb R^n?## : for ##x,y \in M## , and
    ##x,y \in (U, \phi) : \phi(x)=x', \phi(y)=y'## , define ## x+y := \phi^{-1}(x'+y') ##
     
  6. Sep 29, 2015 #5
    Locally sure, but I guess he means a global vector space structure.
     
  7. Sep 29, 2015 #6

    WWGD

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    Yes, that is a harder question, I dont know if it can be done or not.
     
  8. Sep 29, 2015 #7
    It can't. There are topological obstructions.
     
  9. Sep 29, 2015 #8

    WWGD

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    But isnt vector space structure independent of topology ( if you ignore the inner-product generating a norm and "inducing" a normed space/vector space structure)?
     
  10. Sep 29, 2015 #9
    In that case, every manifold has the same cardinality of ##\mathbb{R}##, so you can find a bijection.
     
  11. Sep 29, 2015 #10

    WWGD

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    Sorry, I dont get your point. Do you mean there is no compatibility as _normed spaces_ , alone, tho there is as pure vector spaces? Could you cite the topological obstruction?
     
  12. Sep 29, 2015 #11
    The topological construction is contractibility.
    If you don't care for topology, then you can find a bijection and then just transport the structure.
     
  13. Sep 29, 2015 #12

    WWGD

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    I see, so it can be done for contractible manifolds then? Or is/are there some other obstruction(s)?
     
  14. Sep 29, 2015 #13
  15. Sep 29, 2015 #14

    WWGD

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    I see, makes sense, good points.
     
  16. Sep 29, 2015 #15

    lavinia

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    Here is an exercise. Prove that a manifold that is a vector space over the real numbers is diffeomorphic to Rn.

    - If you only require addition and not scalar multiplication, then the manifold is called a Lie group. If the addition is commutative, then the manifold can be an n dimensional torus.

    Your question will not work if you replace vector space by Lie group. The condition of being a Lie group is very restrictive. For instance the only closed 2 dimensional surface that is a Lie group is the torus.
     
    Last edited: Sep 29, 2015
  17. Sep 30, 2015 #16
    I couldn't get your point on this quote..how can one define the distance on M?!
     
  18. Sep 30, 2015 #17
    Yeah.. I think things are starting to be a bit clearer .. however, I couldn't understand what does really a Riemannian vector space look like?!
     
    Last edited: Sep 30, 2015
  19. Sep 30, 2015 #18

    lavinia

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    Not sure what you mean by Riemannian vector space - but Rn together with an inner product is a Riemannian manifold that is also a vector space. Algebraically, one sees the connection between geometry and vector algebra through the bilinearity of the inner product.

    Geometrically, the inner product assigns lengths to vectors and angles to pairs of vectors. Without the inner product there is no geometry.

    - Technically, a Riemannian metric is defined on the tangent bundle of the manifold. But in Rn the tangent space and the vector space are canonically identified.
     
    Last edited: Sep 30, 2015
  20. Sep 30, 2015 #19
    I'm interested in a Riemannian manifold in which ##x+x'## and ##\alpha.x## where ##\alpha## is a scalar from some field have a meaningful definition..so I thought that this kind of riemannian manifold should has an additional vector space structure..if I understand it correctly, one can locally consider it..
     
  21. Sep 30, 2015 #20

    lavinia

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    In any local coordinate system that does not map surjectively onto all of Rn e.g. normal coordinates for almost all manifolds, the scalars that keep you inside the coordinate system are limited(really big scalars throw you outside) and some vectors when added will be outside of the coordinate system as well. The exponential mapping is only a local diffeomorphism. That means that it fails to be 1-1 after a while (for sufficiently large radius in the tangent space) and it may have singularities( called conjugate points). In these cases the addition on the manifold is not well-defined. Try defining addition of the south pole of a sphere to any other point using the exponential map at the north pole.
     
    Last edited: Oct 1, 2015
  22. Sep 30, 2015 #21

    WWGD

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    The distance is dealt with by treating ##(M,g)## as a length space: the distance between two points equals the length of the shortest path between the points.

    @micromass; sorry for my thickness, obviously the topology generated by the inner-product has to agree with that of the ambient manifold, which agrees with the metric topology given by ##d(x,y)##:= length of shortest path between two points.
     
  23. Oct 1, 2015 #22

    WWGD

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    What it comes down to is this: You want your n-manifold to be a finite-dimensional (n -dimensional) inner-product space, with the inner-product given by the Riemannian metric. But there is a result that every finite-dimensional normed space (over a fixed field) of dimension n is isomorphic to ##\mathbb R^n ## . This means your manifold, if given an inner-product normed space condition, will have to be homeomorphic to ## \mathbb R^n ##, since the (metric) topology generated by the norm will be homeomorphic to that of ##\mathbb R^n ##.
     
  24. Oct 5, 2015 #23
    Depending on what you really mean by this question, there are various answers. (Though it is not clear how you might want the Riemannian structure to relate, if at all, to the vector space structure.)

    If you really mean a "vector space" structure, then this is possible on any Euclidean space Rn, just treating it as the vector space it is originally defined to be, over the field ℝ of real numbers.

    In case the dimension n is even, say n = 2k, then you can also identify Rn with ℂk, the vector space of dimension k over the field of complex numbers.

    If you really mean just a way to combine two elements of the manifold to get a third element of the manifold (and we will assume you want that operation to be continuous), as well as having an identity element and continuously defined inverse elements, then we are talking about what is called a "Lie group" ("Lie" rhymes with SEE). Lie groups are fascinating objects and of great important in physics, but can get quite complicated. I'll mention just a few: The circle group S1 := {z ∈ ℂ | |z| = 1} (which is also the group of rotations of the plane), the torus group T := S1×S1, and the group SO(3) of rotations of 3-space. Of these, the circle S1 and the torus T are abelian (commutative) Lie groups, but the rotation group SO(3) is not.
     
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