# Contravariant derivative

1. Jan 3, 2006

### bchui

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 $$\Gamma^{k}_{i,j}$$ and other concepts?

2. Jan 3, 2006

### robphy

A "contravariant derivative operator" would probably be defined by $$\nabla^a=g^{ab}\nabla_b$$, where $$\nabla_b$$ is a torsion-free derivative operator that is compatible ($$\nabla_a g_{bc}=0$$) with a nondegenerate metric $$g_{ab}$$.

3. Jan 9, 2006

### dextercioby

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

Daniel.

4. May 13, 2007

### bchui

Let $$(M,{\cal T})$$ be a sub-manifold of a Riemannian manifold $$(N,{\cal R})$$ with metric tensor $$g$$, If we decompose the tangent space at the point $$p\in M\subseteq N$$ and accordingly decompose the tangent bundle $$T_pN=T_pM\circleplus {\tilde T}_pM$$ into tangential to $$M$$ and normal to $$M$$, could we say that the "converiant derivative" is the "tangential component" of the given connection $$\nabla_X: {\cal X}(N)\mapsto {\cal X}(N)$$ while the "contravariant derivative" is the "normal component" of $$\nabla_X$$ ?
I mean the "convariant derivative along the vector fileld $$X$$" is the projection of $$\nabla_X$$ onto the tangent space of the submanifold $$M$$, while the "contravariant derivative along the vector field $$X$$" is the projection of $$X$$ onto the normal space of the submanifold $$M$$ in $$N$$
I would like to check if the above saying is correct

Last edited: May 13, 2007