1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Contravariant derivative

  1. Jan 3, 2006 #1
    So much has been talking about covariant derivative. Anyone knows about contravariant derivative? What is the precise definition and would that give rise to different [tex]\Gamma^{k}_{i,j}[/tex] and other concepts? :rolleyes:
     
  2. jcsd
  3. Jan 3, 2006 #2

    robphy

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    A "contravariant derivative operator" would probably be defined by [tex]\nabla^a=g^{ab}\nabla_b[/tex], where [tex]\nabla_b[/tex] is a torsion-free derivative operator that is compatible ([tex]\nabla_a g_{bc}=0[/tex]) with a nondegenerate metric [tex]g_{ab}[/tex].
     
  4. Jan 9, 2006 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    The connection needn't be torsion free, but metric compatibility is essential.

    Daniel.
     
  5. May 13, 2007 #4
    Let [tex](M,{\cal T})[/tex] be a sub-manifold of a Riemannian manifold [tex](N,{\cal R})[/tex] with metric tensor [tex]g[/tex], If we decompose the tangent space at the point [tex]p\in M\subseteq N[/tex] and accordingly decompose the tangent bundle [tex]T_pN=T_pM\circleplus {\tilde T}_pM[/tex] into tangential to [tex]M[/tex] and normal to [tex]M[/tex], could we say that the "converiant derivative" is the "tangential component" of the given connection [tex]\nabla_X: {\cal X}(N)\mapsto {\cal X}(N)[/tex] while the "contravariant derivative" is the "normal component" of [tex]\nabla_X[/tex] ?
    I mean the "convariant derivative along the vector fileld [tex]X[/tex]" is the projection of [tex]\nabla_X[/tex] onto the tangent space of the submanifold [tex]M[/tex], while the "contravariant derivative along the vector field [tex]X[/tex]" is the projection of [tex]X[/tex] onto the normal space of the submanifold [tex]M[/tex] in [tex]N[/tex]
    I would like to check if the above saying is correct
     
    Last edited: May 13, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Contravariant derivative
Loading...