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Lie derivative versus covariant derivative

  1. May 27, 2010 #1
    When calculating the derivative of a vector field X at a point p of a smooth manifold M, one uses the Lie derivative, which gives the derivative of X in the direction of another vector field Y at the same point p of the manifold.

    If the manifold is a Riemannian manifold (that is, equipped with a metric tensor), then there is a natural connection called the Levi-Civita connection that also tells you the derivative of a vector field X at a point p of a smooth manifold.

    Do these two methods of calculating the derivative give the same result?

    And why is the Lie derivative so complicated? It seems the reason is that you need to define a diffeomorphism f: M --> M (an active transformation) because doing so will induce a transformation between tangent spaces: f*: T(M) --> T(M), and to compare 2 vectors at different points you first need to bring one vector to the other through f* so that you can subtract them. This diffeomorphism f is provided by a "flow", which is induced by Y. So what is so special about a diffeomorphism generated by a vector field Y? Why not just use any diffeomorphism to define the derivative, and not necessarily one generated by a vector field?

    Also, is there an easy way to see that a flow generated by a vector field is a diffeomorphism?
     
  2. jcsd
  3. May 27, 2010 #2
    The Lie derivative is not linear over functions in the first argument, thus it can't be a connection of any kind.

    For the flow question, it generates a local one-parameter group of diffeomorphisms (this is existence and uniqueness of linear ODEs). There's no reason to expect that it gives a global diffeomorphism (to see this, punch a hole in the real line and point your arrows at it).
     
  4. May 27, 2010 #3
    Maybe I'm reading my book wrong, but it claims that a flow is a group of global diffeomorphisms.

    The flow [tex]\sigma(t,x) [/tex], where t is the group parameter and x is a point on the manifold, is given by:

    [tex]\sigma^\mu(t,x)=e^{tX^\mu(x)}x^\mu [/tex]

    where [tex]X^\mu(x) [/tex] is the vector field at the point x in the [tex]\mu [/tex] direction.

    Is this equation a diffeomorphism? It seems to be just f(x)=eg(x)x, which seems like a diffeomorphism.
     
  5. May 27, 2010 #4
    If your book is using coordinates, it certainly seems local.
     
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