# Covariant derivative and connection of a covector field

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• Vyrkk

#### 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?

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 ...

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

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.

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

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.

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!