# Covariant derivative and connection of a covector field

• A
• Vyrkk
In summary: And now I understand why this requirement is needed, despite Caroll not stating its use anywhere in the derivation. Thank you!
Vyrkk
TL;DR Summary
Deriving the expression for the covariant derivative of a covector field in components
I am trying to derive the expression in components for the covariant derivative of a covector (a 1-form), i.e the Connection symbols for covectors.
What people usually do is
1. take the covariant derivative of the covector acting on a vector, the result being a scalar
2. Invoke a product rule to develop in "(covariant derivative of vector)(covector) + (covariant derivative of covector)(vector) = Covariant derivative of scalar"
3. substract the covariant derivative of the vector which is already known, leaving the expression for the covariant derivative for the covector on the other side.
E.g:
- https://www.physicsforums.com/threa...-derivative-for-covectors-lower-index.689141/
- https://math.stackexchange.com/questions/1069916/covariant-derivative-for-a-covector-field
- https://math.stackexchange.com/questions/1499513/covariant-derivative-of-a-covariant-vector
Well, I find something disturbing: the covariant derivative doesn't have such a product rule between a covector field and a vector field, or even between two vector fields. The only product rule it has is the Leibniz rule (which is one of the defining property of the axiomatic definition of the covariant derivative) between a vector field and a scalar field (a function on the manifold), but definitely no such thing between a vector and a covector.
What is an explanation for this, and a more rigorous derivation?

What makes you think the product rule when acting on tensors is not defined? What book are you using?

Orodruin said:
What makes you think the product rule when acting on tensors is not defined? What book are you using?

It's not that I think it's not defined, it's that I haven't seen it defined anywhere.

As far as I know, the covariant derivative has a Leibniz rule for a scalar field times a vector field, and for the tensor product of any two tensors.

I am using my professor's Lecture notes and complement them with Sean Caroll notes.

It is in the very first definition of the connection on page 55 of Carroll’s lecture notes ...

Orodruin said:
It is in the very first definition of the connection on page 55 of Carroll’s lecture notes ...
Well, the product rule given p. 55 as part of the definition is a product rule for tensor product, not for a one form acting on a vector field or for two vector fields

Vyrkk said:
Well, the product rule given p. 55 as part of the definition is a product rule for tensor product, not for a one form acting on a vector field or for two vector fields
A one form acting on a vector is nothing but a contraction of a tensor product.

Orodruin said:
A one form acting on a vector is nothing but a contraction of a tensor product.
Mmh finally we're getting somewhere.
So if I take the covariant derivative of a tensor product of a one-form and a vector, I can develop using the Leibniz product rule for the tensor product. Taking the contraction on each side, I get my result.

But under one condition: the contraction has to go under the covariant derivative of the tensor product, otherwise I wouldn't know what to do with this term. In other words, the covariant derivative has to commute with the tensor contraction operator. Is this right?

In that case, the commutation would have to be a requirement and an important step in the derivation, which I haven't seen stated/done in any of the aforementionned derivations

Vyrkk said:
In that case, the commutation would have to be a requirement and an important step in the derivation, which I haven't seen stated/done in any of the aforementionned derivations
This is stated two pages later in Carroll's lecture notes as requirement 3, right before requirement 4, which is that it should reduce to the partial derivative on scalar fields.

Orodruin said:
This is stated two pages later in Carroll's lecture notes as requirement 3, right before requirement 4, which is that it should reduce to the partial derivative on scalar fields.
And now I understand why this requirement is needed, despite Caroll not stating its use anywhere in the derivation. Thank you!

## 1. What is a covariant derivative?

A covariant derivative is a mathematical operation that describes how a vector field changes as one moves along a curve on a curved surface. It takes into account the curvature of the surface and allows for the differentiation of vector fields in a way that is independent of the coordinate system being used.

## 2. How is a covariant derivative different from an ordinary derivative?

An ordinary derivative only takes into account the change of a function with respect to one variable, while a covariant derivative takes into account the change of a vector field with respect to multiple variables on a curved surface. It also takes into account the curvature of the surface, making it a more general and powerful tool.

## 3. What is a connection of a covector field?

A connection of a covector field is a mathematical object that describes how a covector field changes as one moves along a curve on a curved surface. It is closely related to the covariant derivative and allows for the differentiation of covector fields in a way that is independent of the coordinate system being used.

## 4. How is a connection of a covector field related to a covariant derivative?

A connection of a covector field is closely related to a covariant derivative, as it is used to define the covariant derivative of a vector field. It describes how a covector field changes along a curve, while the covariant derivative describes how a vector field changes along a curve. Together, they allow for the differentiation of both vector and covector fields on a curved surface.

## 5. Why are covariant derivatives and connections important in mathematics and physics?

Covariant derivatives and connections are important tools in mathematics and physics because they allow for the differentiation of vector and covector fields on curved surfaces, which are common in many real-world applications. They also play a crucial role in the formulation of theories such as general relativity, which describes the curvature of spacetime. They are also used in various mathematical fields such as differential geometry and differential equations.

Replies
18
Views
405
Replies
16
Views
3K
Replies
10
Views
1K
Replies
6
Views
2K
Replies
5
Views
2K
Replies
15
Views
4K
Replies
7
Views
3K
Replies
6
Views
1K
Replies
7
Views
3K
Replies
5
Views
2K