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Differential manifold without connection: Is it possible?

  1. Jul 8, 2011 #1
    We all are familiar with the kind of differential geometry where some affine connection always exists to relate various tangent spaces distributed over the manifold, and from this connection two fundamental tensors, namely the Cartan's torsion and the Riemann-Christoffel curvature, arise.
    Is it possible to have a differential manifold, where due to some topological anomaly, a connection cannot exist?
    Of course there exists symplectic manifolds where no connection property is required.
    But my question is related to the existence of non-metricity --- as in non-metric case, metric property id there but the connection is not metric. Similarly, can we have some property called non-connectivity where the affine connection is there but somehow it is lacking some fundamental connection requirements?
     
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  3. Jul 8, 2011 #2

    lavinia

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    Every smooth manifold can be given a Riemannian metric. The proof glues local metrics together using a partition of unity. I suppose you need to show that there is an open cover by coordinante charts such that each point of the manifold is contained in only finitely many of the charts.
     
  4. Jul 8, 2011 #3

    quasar987

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    ...hence, if by manifold you mean, as differential geometers do, a space which is paracompact (or something stronger like second countability), then every (smooth) manifold admits a Riemannian metric and a connection compatible with it.
     
    Last edited: Jul 8, 2011
  5. Jul 8, 2011 #4

    lavinia

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    there are manifolds that do no admit a differentiable structure. These can not have a connection.
     
  6. Jul 10, 2011 #5
    can you give me some examples where we cannot define a connection?
    I'm an engineering student. I asked these question regarding non-Riemannian description of defects in solids. In this field, there arises a manifold which is composed of disjoint non-compact parts and you cannot form a compact Euclidean subset from these parts by an unique global diffeomorphism.
     
  7. Jul 22, 2011 #6

    lavinia

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    examples are difficult to produce and high dimensional - I think.
     
  8. Jul 22, 2011 #7
    Could you give more information about this manifold? If it's a manifold, every point has a compact neighborhood about it. The proof that every manifold has a connection is based on the assumption that you can form a partition of unity. If your manifold is not paracompact, this fails. However, most definitions of manifold assume this to begin with.
     
  9. Jul 22, 2011 #8

    lavinia

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    I would be glad to research it. These manifolds do not have differentiable structures so can not have a connection. They are paracompact though - in fact compact.
     
  10. Jul 23, 2011 #9

    lavinia

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    http://www.math.rochester.edu/u/faculty/doug/otherpapers/kervaire_AMSRev.pdf

    I am told that this is the first example ever discovered. It is a ten dimensional manifold. I do not understand the paper but would be willing to read it through with you.
     
  11. Mar 17, 2012 #10
    many thanks for the discussion.
     
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