Commutation property of covariant derivative

In summary, the conversation discusses the definition of the covariant derivative of a tangent vector field and the proof of a property related to orthogonal patches. The assumption of x_u \cdot x_v = 0 is needed to show that the covariant derivative is equal for both vectors. The Lie bracket is also mentioned as being zero in case of derivatives of a coordinate system.
  • #1
demonelite123
219
0
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, [itex] x_u \cdot x_v = 0[/itex], then [itex] \nabla_{x_u}x_v = \nabla_{x_v}x_u [/itex].

It seems clear to me that by the definition of the directional derivative, [itex]x_v[x_u] = \frac{\partial}{\partial v}x_u = x_{uv} = x_{vu} = x_u[x_v] [/itex]. 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 [itex] x_u \cdot x_v = 0[/itex] was needed. can someone help explain?
 
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  • #2
demonelite123 said:
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, [itex] x_u \cdot x_v = 0[/itex], then [itex] \nabla_{x_u}x_v = \nabla_{x_v}x_u [/itex].

It seems clear to me that by the definition of the directional derivative, [itex]x_v[x_u] = \frac{\partial}{\partial v}x_u = x_{uv} = x_{vu} = x_u[x_v] [/itex]. 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 [itex] x_u \cdot x_v = 0[/itex] was needed. can someone help explain?

What you say is right.

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

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

What is the commutation property of covariant derivative?

The commutation property of covariant derivative is a mathematical property that describes how the order of taking covariant derivatives affects the final result. In other words, it determines whether the covariant derivative of a product of two vector fields is equal to the product of their covariant derivatives or not.

Why is the commutation property important in differential geometry?

The commutation property is important in differential geometry because it allows us to properly define and manipulate vector fields on curved manifolds. It also plays a crucial role in the study of curvature and the construction of geometric objects such as connections and curvature tensors.

How is the commutation property related to the curvature of a manifold?

The commutation property is closely related to the curvature of a manifold as it determines how the covariant derivative of a vector field changes as we move along a curve on the manifold. If the commutation property holds, then the manifold is said to have zero curvature.

What are some applications of the commutation property in physics?

The commutation property has several applications in physics, particularly in the field of general relativity. It is used to define the curvature tensor, which is essential in describing the gravitational field. It is also used in the study of gauge theories and in the formulation of the equations of motion for particles on curved manifolds.

Are there any limitations or exceptions to the commutation property?

Yes, there are limitations to the commutation property. It is not always applicable on non-symmetric manifolds or in non-Euclidean spaces. Additionally, it may not hold for certain types of vector fields, such as those that have singularities or discontinuities. It is important to carefully consider the requirements and assumptions when using the commutation property in mathematical and physical applications.

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