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?