Commutation property of covariant derivative

[demonelite123]

My book defines the covariant derivative of a tangent vector field as the directional derivative of each component, and then we subtract out the normal component to the surface.

I am a little confused about proving some properties. One of them states:
If x(u, v) is an orthogonal patch, x_u \cdot x_v = 0, then \nabla_{x_u}x_v = \nabla_{x_v}x_u.

It seems clear to me that by the definition of the directional derivative, x_v[x_u] = \frac{\partial}{\partial v}x_u = x_{uv} = x_{vu} = x_u[x_v]. Therefore if it is true for the directional derivative, then it is clear that it is true for the directional derivative minus its normal component, or in other words the covariant derivative.

i can't figure out why the assumption x_u \cdot x_v = 0 was needed. can someone help explain?

 

[lavinia]

My book defines the covariant derivative of a tangent vector field as the directional derivative of each component, and then we subtract out the normal component to the surface.

I am a little confused about proving some properties. One of them states:
If x(u, v) is an orthogonal patch, x_u \cdot x_v = 0, then \nabla_{x_u}x_v = \nabla_{x_v}x_u.

It seems clear to me that by the definition of the directional derivative, x_v[x_u] = \frac{\partial}{\partial v}x_u = x_{uv} = x_{vu} = x_u[x_v]. Therefore if it is true for the directional derivative, then it is clear that it is true for the directional derivative minus its normal component, or in other words the covariant derivative.

i can't figure out why the assumption x_u \cdot x_v = 0 was needed. can someone help explain?

What you say is right.

\nabla_{x_u}x_v - \nabla_{x_v}x_u =[x_u, x_v]

If the vectors are the derivatives of a coordinate system the Lie bracket is always zero.