# projection of the co-derivative == co-derivative of the projection ?

Hey,

here is the formal question.

M is a riemannian sub-manifold in N. a,b are vector fields such that for each p$\in$M, $a_p$,$b_p$ in $T_p$M $\subset$ $T_p$N

prove

$\nabla^M_b$a = pr($\nabla^N_b$a)

where pr is the projection funtion pr:$T_p$N$\rightarrow T_p$M and $\nabla^N$ and $\nabla^M$ are the covariant derivative operators (by riemannian connection) in N and M respectively.

I don't really understand why is this not immediate from definitions. the covariant derivative in a manifold is just the regular derivate since there's no need to late project onto the manifold since the derivative of a vector field in a sub-manufold will surely we already completley in the manifold thus the projection will just be Identity. thus when I will project this vector on $T_p$M ofcourse I wil get the covariant derivative on M with is also just the derivative then projected onto M. maybe what I'm asked to prove is that the derivative of the vector field a with respect to b is equal to the covariant derivative in N since the derived vector is 'fully' in N?

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This appears to be a homework problem. I suggest consulting either a professor or a classmate. –  Deane Yang Jul 27 '10 at 2:40

Homework or not, it is not true that your covariant derivative in N is already parallel to the submanifold M; indeed the normal component eliminated by the projection is the second fundamental form of M in N. Consider the simple example of the 2-sphere in $N=\mathbb{R}^3$; if X is a normal (=radial) vector field on the sphere then $(\nabla^N_b a)\cdot X=-a\cdot(\nabla^N_b X)$ is proportional to the inner product $a\cdot b$. To prove the projection formula, it suffices to (i) observe that the projected connection respects the induced metric on M, and (ii) prove that it has zero torsion, which is related to the fact that the second fundamental form is symmetric in a,b.