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I Tangent spaces at different points

  1. Dec 8, 2018 at 1:26 PM #1
    How do you know if two given points on a manifold have the same tangent space? Checking if a vector does not change when transported from one point to the other is enough?
     
  2. jcsd
  3. Dec 8, 2018 at 1:29 PM #2

    fresh_42

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    What do you mean by the same? They cannot be the same at different points. They are isomorphic, as e.g. on a n-manifold they both are isomorphic to ##\mathbb{R}^n##. So again, what is supposed to mean "equal"?
     
  4. Dec 8, 2018 at 1:47 PM #3
    oh that makes sense
    I mean, if we consider the m-dimensional manifold to be ##\mathbb{R}^m## itself, we can find a (global) coordinate system where the basis don't change, namely a Cartesian coordinate system. In that coordinate system, a vector (not a vector field, so constant components) will not change at all if we move from one point to another. That's what we usually do geometrically when we e.g. drag the arrows around in the plane, right?

    Now for me it seems unecessary to continue saying that the points in ##\mathbb{R}^m## have different tangent spaces, as we can do what I just described above. From this follows my question in post #1, if checking the constancy of a vector is a sufficient condition.
     
  5. Dec 8, 2018 at 1:49 PM #4

    Orodruin

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    They are never the same so that is easy ...

    ##\mathbb R^n## has the special property of being an affine space.
     
  6. Dec 8, 2018 at 4:27 PM #5

    WWGD

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    Just to add that this motivates the concept of connections, which, well, connect tangent spaces at different points. In euclidean n-space, the isomorphism is natural, but not so in general manifolds.
     
  7. Dec 9, 2018 at 11:15 AM #6

    mathwonk

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    the tangent bundle is locally trivial, i.e. locally has the form MxV, where V is a vector space. In the rare case where the tangent bundle is actually a product globally, the manifold is called "parallelizable". One example of a parallelizable manifold is a manifold that is also a group, such as R^n or a torus, i.e. a manifold isomorphic to R^n/lattice. In the parallelizable case one can consider all tangent spaces as being the same, since there is a natural isomorphism from any one of them to any other, namely translation by an element of the group. This occurs of course in R^n.

    Even if the manifold is parallelizable, one must choose a global trivialization, and until this is done there is no natural isomorphism from one tangent space to another. The case of a group however is special in this regard, since the group operation gives the isomorphism. However even in this case one must be given the group structure to define the isomorphisms. I.e. just because a manifold is smoothly isomorphic to a group does not yet specify the group operation, since not every smooth isomorphism of a group manifold is a group isomorphism. Thus even if one knows ones manifold is smoothly isomorphic to R^n, one does not have a global trivialization of the tangent bundle until one chooses a global coordinate system. I.e. it is not enough to know such a global coordinate system exists, since many different ones exist, and they usually define different trivializations.

    https://en.wikipedia.org/wiki/Parallelizable_manifold

    This link mentions the interesting fact that all orientable 3 manifolds are parallelizable!

    Oh yes, and one must be careful about the meaning of a "trivialization". If one means it in the sense of the wikipedia link, i.e. an isomorphism of the tangent bundle of M with the product MxR^n, even a group structure is not quite enough. I.e. the group structure gives an isomorphism of the tangent bundle with MxV where V is the tangent space at the origin. To get an isomorphism with MxR^n, one must still choose an isomorphism of the tangent space at the origin with R^n, i.e. one must choose a basis for V. But just the group structure is already enough to identify any two tangent spaces with each other, although not with R^n.

    As already pointed out by Orodruin, actually a bit less than a group structure is sufficient to identify any two tangent spaces. I.e. in the case of R^n one only needs its structure as an affine space, i.e. one only needs the ("faithful", i.e. without fixed points) action of a group on the space by translation, one does not need to know where the origin is in the space itself. But one does need two points to be joined by a unique translation.

    The case of complex manifolds is somewhat more special, since I believe that any complex analytic isomorphism of a complex manifold with a group manifold of form C^n/lattice, actually defines the group structure uniquely up to translation; i.e. any complex analytic isomorphism of C^n/lattice with itself is a group automorphism plus a (possibly trivial) translation.
     
    Last edited: Dec 9, 2018 at 4:12 PM
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