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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

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    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

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    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
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