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

1. Jul 26, 2010

### plasticfloor

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, ap,bp $$\in$$ TpM $$\subset$$ TpN

prove

$$\nabla$$Mba = pr($$\nabla$$Nba)

where pr is the projection funtion pr:TpN$$\rightarrow$$TpM
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 is taking a regular derivative and then projecting onto the tangent bundle. in a manifold the tangent bundle is just the manifold itself so the regular derivate is already the covariant derivative because the projection part is just identity. thus when I will project this vector on TpM ofcourse I wil get the covariant derivative on M because it's the projection onto Tp[\SUB]M of the regular derivative of vector fields on M.

is what I'm asked to prove actually just that the covariant derivative equal to the regular derivative when the vector fields in question belong to a sub manifold?

Last edited: Jul 26, 2010
2. Jul 27, 2010

### mehul ahir

please tell me what is meaning of dimension i knew only three dimension what about four ,five.............. ialso want to know that any point in the three dimension is a vector than why we writing vector as a special case suppose u(2,3,5) is any point in the three dimension than why we writing it as a u=2i+3j+5K please reply me via email indetai