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A Question about Holonomy of metric connecton

  1. Jul 16, 2016 #1
    I am trying to follow Nakahara's book about Holonomy.
    if parallel transporting a vector around a loop induces a linear map (an element of holonomy group)
    [tex]{P_c}:{T_p}M \to {T_p}M[/tex]

    the holonomy group should be a subgroup of

    then the book said for a metric connection, the property
    [tex]{g_p}({P_c}(X),{P_c}(Y)) = {g_p}(X,Y)[/tex]
    makes the holonomy group a subgroup of [tex]O(m)[/tex] if the manifold is Riemannian; and a subgroup of [tex]O(m-1)[/tex] if the manifold is Lorentzian.

    The author must think this is very straightforward and didn't explain why. Can anybody help?
  2. jcsd
  3. Jul 16, 2016 #2
    I assume the book you're referring to is Nakahara's Geometry, Topology and Physics, 2e.
    This isn't what the book says. Nakahara actually writes [itex]{g_p}({P_c}(X),{P_c}(X)) = {g_p}(X,X)[/itex]

    In any case, a linear transformation which preserves the inner product is the definition of an orthogonal transformation.

    Also, Nakahara doesn't say that the holonomy group is a subgroup of [itex]O(m-1)[/itex] in the Lorentzian case, he says it's a subgroup of [itex]SO(m-1,1)[/itex].
  4. Jul 17, 2016 #3


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    If the connection is compatible with the metric, then parallel translation preserves the metric..

    If the vector fields ##Y## and ##Z## are parallel along a curve with tangent vector ##X## then

    ##X⋅<Y,Z> = <∇_{X}Y,Z> + <Y,∇_{X}Z>## so the inner product of ##Y## with ##Z## is constant along the curve.

    If the metric is Riemannian, that is: it is positive definite on each tangent space, then parallel translation is an orthogonal linear map. If the metric is not positive definite then parallel translation is an element of ##O(m-p,p)##. In the case of a Lorentz metric ##p=1##.

    - If the manifold is not orientable then then a holonomy transformation may be orientation reversing.
  5. Jul 18, 2016 #4
    Thanks for replying, this is very helpful, and I will try to keep the book close next when I am typing.
  6. Jul 19, 2016 #5
    Now I seem to be able to understand the inner product preserving property make a orthogonal group, but can you explain where does the dimension (m-1) come from for Lorenzian case? Thanks
  7. Jul 19, 2016 #6
    The m-1 comes from the fact that the inner product in the Lorentzian case is usually written with signature (m-1,1). That is, (-1,1,1,1) or (1,-1,-1,-1,-1) for 4d spacetime, for example. In the genral case, this is the indefinite orthogonal group, and Nakahara is only considering the connected component which contains the identity. https://en.wikipedia.org/wiki/Indefinite_orthogonal_group
  8. Jul 19, 2016 #7
    Thanks very much, now I see the point.
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