Graduate Solving Covariant Derivative Notation Confusion

[Markus Kahn]

TL;DR

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
$$\frac{\partial \mathbf{e}^{\alpha}}{\partial x^{\mu}}= \Gamma_{\mu \sigma}^{\alpha} \mathbf{e}_{\alpha}.$$

 

[PeterDonis]

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.

 

[Cem]

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 $\alpha$.