# Solving Covariant Derivative Notation Confusion

• A
• Markus Kahn
In summary, the conversation discusses a statement involving a vector field and a covariant derivative. One of the questions raised is why the index α is shifted up, which is due to sloppy notation in the article. The second question pertains to the relationship between the covariant derivative and the vector field, which can be shown through the definition of the (1,1) tensor and the contraction of the vector field with it.
Markus Kahn
TL;DR Summary
I'm having a hard time relating two things that should be the same, but don't appear to be the same...
I've stumbled over this article and while reading it I saw the following statement (##\xi## a vectorfield and ##d/d\tau## presumably a covariant derivative***):
\begin{align*}\frac{d \xi}{d \tau}&=\frac{d}{d \tau}\left(\xi^{\alpha} \mathbf{e}_{\alpha}\right)=\frac{d \xi^{\alpha}}{d \tau} \mathbf{e}_{\alpha}+\xi^{\alpha} \frac{d \mathbf{e}^{\alpha}}{d \tau} \\& =\frac{d \xi^{\alpha}}{d \tau} \mathbf{e}_{\alpha}+\xi^{\alpha} \frac{d x^{\mu}}{d \tau} \frac{\partial \mathbf{e}^{\alpha}}{\partial x^{\mu}} \\& =\frac{d \xi^{\alpha}}{d \tau} \mathbf{e}_{\alpha}+ \xi^{\sigma} u^{\mu} \Gamma_{\mu \sigma}^{\alpha} \mathbf{e}_{\alpha}.\end{align*}

Question:
This gives me some headaches... especially the
$$\frac{d \mathbf{e}^{\alpha}}{d \tau}$$
part. First of all, why does ##\alpha## gets shifted up? I don't see any metric or the like to compensate for that... Second, how exactly does one come to
$$\frac{d \mathbf{e}^{\alpha}}{d \tau} =\frac{d x^{\mu}}{d \tau} \frac{\partial \mathbf{e}^{\alpha}}{\partial x^{\mu}}= u^{\mu} \Gamma_{\mu \sigma}^{\alpha} \mathbf{e}_{\alpha}.$$
(just to make it clear, I don't mean using the chain rule, this much would be obvious, see below for a clearer explanation...)

I know that for any function ##f## and vector field ##X## we have ##\nabla_{X}f = Xf## so in particular ##\nabla_{\frac{\partial}{\partial x^\mu}}f = \frac{\partial}{\partial x^\mu}f##. I therefore just assumed that ##\frac{d}{d\tau} \mathbf{e}_\alpha = \nabla_{\frac{d}{d\tau}} \mathbf{e}_\alpha##. Now if ##\mathbf{e}_\alpha## is a function of ##x##, which is parametrized over ##\tau## , then how can you show that
$$\nabla_{\frac{d}{d\tau}} \mathbf{e}_\alpha(x(\tau)) = u^{\mu} \Gamma_{\mu \sigma}^{\alpha} \mathbf{e}_{\alpha}$$

---------------------
*** I'm pretty sure it is b.c. we are trying to prove that
$$\frac{D^{2}}{D \tau^{2}} \delta x^{\mu}=R_{\nu \sigma \rho}^{\mu} \frac{d x^{\nu}}{d \tau} \frac{d x^{\sigma}}{d \tau} \delta x^{\rho}$$
and apparently
$$\frac{\partial \mathbf{e}^{\alpha}}{\partial x^{\mu}}= \Gamma_{\mu \sigma}^{\alpha} \mathbf{e}_{\alpha}.$$

Markus Kahn said:
why does ##\alpha## gets shifted up?

Because the article is being sloppy with notation. It should be down, since each term should have no free indexes.

vanhees71
For the second part of the question, for some vector ##\mathbf Y##, if we define ##\nabla \mathbf Y## to be the (1,1) tensor that gives ##\nabla_{\mathbf X} \mathbf Y## when contracted with the vector ##\mathbf X##, then from the definition ##\Gamma^a_{bc} = \left < \mathbf e^a, \mathbf \nabla_{\mathbf e_b} \mathbf e_c \right >##, we can write ##\nabla \mathbf e_c =\Gamma^a_{bc} \mathbf e^b \otimes \mathbf e_a##. Contracting with ##\mathbf X = \frac{d}{d\tau} = X^a \mathbf e_a## we get ##\nabla_{\mathbf X} \mathbf e_c =\Gamma^a_{bc} X^b \mathbf e_a##. Then ##\xi^c \nabla_{\mathbf X} \mathbf e_c = \xi^c \Gamma^a_{bc} X^b \mathbf e_a##, and since ##X^b = \frac{dx^b}{d\tau} = u^b## this is the expression you originally have when you account for the mistakenly shifted up ##\alpha##.

## 1. What is covariant derivative notation?

Covariant derivative notation is a mathematical notation used in differential geometry and tensor calculus to represent the derivative of a tensor field with respect to another tensor field. It is used to describe how a tensor field changes as it moves along a curved manifold.

## 2. Why is covariant derivative notation confusing?

Covariant derivative notation can be confusing because it involves a combination of symbols and indices that may not be familiar to those who are not well-versed in differential geometry. It also requires a deep understanding of tensor calculus and the concept of covariant differentiation.

## 3. How can I solve covariant derivative notation confusion?

To solve covariant derivative notation confusion, it is important to have a solid understanding of differential geometry and tensor calculus. It may also be helpful to practice with specific examples and work through problems to become more comfortable with the notation.

## 4. What are some common mistakes when using covariant derivative notation?

Some common mistakes when using covariant derivative notation include confusing the placement of indices, forgetting to include the metric tensor, and mixing up the order of operations. It is important to pay close attention to the details and double check your work to avoid these mistakes.

## 5. How can I improve my understanding of covariant derivative notation?

To improve your understanding of covariant derivative notation, it is recommended to study the fundamentals of differential geometry and tensor calculus. Additionally, practicing with specific examples and seeking help from a mentor or tutor can also be beneficial. It may also be helpful to read textbooks or articles on the topic to gain a deeper understanding.

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