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Torsion, affine development and Levi-Civita connection

  1. Nov 16, 2008 #1


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    I quote http://en.wikipedia.org/wiki/Torsion_tensor#Affine_developments":

    I try to apply this to the natural connection on the tangent bundle of M = S2 (or more intuitively, of the surface of the Earth)
    I mean here natural connection the connection which defines the parallel transport so that

    1. Tangent vectors on the Equator pointing toward North are parallel transported along the Equator into each other.

    2. The longitude lines are geodesics, i.e. tangent vectors of a given longitude line are parallel transported along this longitude line into each other.

    Now I take xt a triangle on the sphere having three rectangles : x0 = (0,0), xπ/2 = (0, π/2), xπ = (π/2,π/2) = (π/2,0), x3π/2 = x0. (with (lat, long) coordinates)This is a loop.

    Applying the definition of Wikpedia, the affine development of this loop is the following open polygon in T(0,0)S2: C0 = (0,0), Cπ/2 = (0, π/2), Cπ = (π/2, π/2), C3π/2 = (0, π/2).

    If this is so, then according Wikipedia's definition, this connection has a non-vanishing torsion. But in this case, this connection can't be a Levi-Civita connection because the Levi-Civita connection is torsion-free by definition. Is this really so? The connection defined above is really not a Levi-Civita connection, or I have misunderstood or miscalculated something?
    Last edited by a moderator: Apr 23, 2017
  2. jcsd
  3. Nov 16, 2008 #2


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    Of course, I took the [tex] \tau_t^0[/tex] parallel transport along the xt curve, but Wikipedia doesn't tell that along what curve should it be understood. Perhaps along a geodesic from xt to x0?
  4. Nov 17, 2008 #3
    Your calculation seems correct. The usual connection on the sphere is of course torsion-less, but as you point out the development of a great triangle is not a closed curve. I wonder if it may indeed be the case that the development of a loop is closed iff the connection is flat (= curvaturefree and torsionfree). Maybe the idea of development is coming from Lie group theory. The connection of a Lie group has no curvature but possibly torsion, so that would explain the lack of considering connections with curvature in developments?

    The Wiki article is a bit misleading in assuming that torsion is only twisting around a curve. In fact torsion of a connection can manifest itself in several other ways too. A simple example is the group of orientation-preserving affine transformations of the line: [itex]x\mapsto bx+a[/itex]. An element of this group is a point [itex](a,b),\,b>0[/itex] in the upper plane. The torsion of the connection causes the vertical component of a vector to shrink and the geodesics are either vertical lines or curves that bend to the right or left. (Or something like that...) If I remember correctly Frankel's The Geometry of Physics has a good discussion of this example.

    Also, the idea that the image of closed curve in a manifold becomes non-closed in another space also shows up in the context of frame bundles. The lift of a closed curve to the bundle of Euclidean frames of a Riemannian manifold is not closed if the manifold has curvature.
  5. Nov 20, 2008 #4


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    Thank you for your answer! I try to analyze your example a bit (unfortunately, I don't have the book you mentioned)

    In the group you told, the product of (a1,b1) and (a,b) acts on x as (a1 + b1(a + bx)) = a1 + b1a + b1bx, i.e, the left action of (a1,b1) on (a,b) is (a1+b1a, b1b).

    Te identity element is (0,1).

    The inverse of (a,b) is (–a/b, 1/b)

    If g: [-1,1] -> G : t->(a(t), b(t)) is a curve in this G group, then this curve is transformed by (a(0), b(0))-1 to (-a(0)/b(0) + a(t)/b(0), b(t)/b(0)). The tangent vector of g at t=0 is (a'(t), b'(t)), while the tangent vector of (-a(0)/b(0) + a(t)/b(0), b(t)/b(0)) at t=0 is (a'(t)/b(0), b'(t)/b(0)).

    So, the Maurer-Cartan form at the point (a,b) is:

    (1/b, 0)
    ( 0, 1/b)

    That is, if we take a curve in G startig from the identity, then the tangent vector (a',b') in (a,b) is parallel transported to the tangent space at the identity as (a'/b, b'/b).

    Now take for example g(t) = [0,3] -> G:

    t ->
    (t , 1) if t is in [0,1],
    (2-t, t) if t is in [1,2]
    (0 ,4-t) if t is in [2,3]

    This is the line segments forming a triangle with vertices (0,1), (1,1) and (0,2).

    The tangent vectors of the three straight line segments are (1,0), (-1,1) and (0,-1).
    They are parallel transformed to the tangent space at identity as (1, 0), (-1/t, 1/t) and (0, -1/(4-t) respectively.

    So, the affine development of g(t) is C(t) : [0,3] -> T(0,1)G:

    (t , 0 ) if t is in [0,1],
    (1-log(t) , log(t) ) if t is in [1,2]
    (1-log(2) , log(4-t)) if t is in [2,3]

    This is an open polygon with vetices (0,0), (1,0), (1-log(2), log(2)), (1-log(2), 0).

    So this connection has really a nonvanishing torsion according to Wiki's definition.

    (I hope I didn't missed the calculations).
  6. Nov 20, 2008 #5
    That seems correct. Well done.
  7. Dec 1, 2008 #6


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    Perhaps I am now wrong, but I found that geodesics are straight lines. I show this in the following.

    Let (u,w) be an element of the Lie algebra of this group, and (A,B) an element of the group. Let's calculate the geodesic in the direction (u,w) through (A,B).
    If the tangent vector of this geodesic in the point (a,b) is (a', b'), then the Maurer-Cartan form must carry it into (u,v). That is, (a', b') = (ub, vb).
    So, the differential equations for the functions a(t) and b(t) are a' = ub and b' = vb. The solution is

    a(t) = A + (u/v)B(evt -1)
    b(t) = Bevt

    That is, a = A-(u/v)B + (u/v)b.

    This is a straight line through (A,B) on the (a,b) plane.
  8. Dec 1, 2008 #7
  9. Jun 15, 2010 #8
    I think the affine development of the closed path you're considering is closed; it is the boundary of the quarter of a circular disk.
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