So in my relativity course, we recently learned about the covariant derivative. it is defined as:
\begin{equation}
\nabla_{\mu}V^{\nu} = \partial_{\mu}V^{\nu}+\Gamma^{\nu}_{\mu,\lambda}V^{\lambda}
\end{equation}
where $V^{\mu}$ is a vector, and $\Gamma^{\nu}_{\mu,\lambda}$ is the connection, or Christoffel symbol. So recently I came accross:
\begin{equation}
\nabla_{\mathbf{e}'_{\beta}}\mathbf{e}'_{\alpha} = \Gamma^{\tau'}_{\alpha'\beta'}\mathbf{e}'_{\tau'}
\end{equation}
So I interpret this as the covariant derivative of $\mathbf{e}'_{\alpha}$ along $\mathbf{e}'_{\beta}$. However, because of the definition of this derivative, doesn't that mean $\mathbf{e}'_{\alpha}$ is a vector field? It is hard for me to understand the meaning of "a vector field of basis vectors." Also, given the index placement, wouldn't these be forms? Taking the covariant derivative of something with an index downstairs is confusing to me.
So my main question is, I cannot seem to calculate this result. So if we calculate $\nabla_{\mathbf{e}'_{\beta}}\mathbf{e}'_{\alpha}$ using the definition of the covariant derivative shown above, we get:
\begin{equation}
\nabla_{\mathbf{e}'_{\beta}}\mathbf{e}'_{\alpha} = \left(\partial_{\mu}\mathbf{e}'_{\alpha}+\Gamma_{\mu\nu}^{\lambda}\mathbf{e}'_{\lambda}\right)\mathbf{e}'_{\beta}
\end{equation}
So how the heck am I supposed to get the formula shown above? I can see that if I made $\mu = \alpha,\nu=\beta$ and $\lambda = \tau$, I would get the right form (as in how it looks) in the second term. But what about the first term? Again the index placement is weird. Any help would be appreciated!
Best Answer
In your first equation you gave the expression for the components of the covariant derivative of a contravariant vector field $V^\nu$. Your second equation is a bit different there you have the covariant derivative of a basis vector along a basis vector: we are dealing with vectors there.
Vectors and components (of vectors) are very very different objects: components are always related to a fixed (chosen) set of basis vectors, where vectors on the other hand are basis independent. That being said you can not just plug in vectors into your first equation. Your first equation does not hold for vectors it only holds for components. For the covariant derivative of a vector along a vector you would need: $$\nabla_\mathbf{v}\mathbf{u}=\nabla_{v^i\mathbf{e_i}}u^j\mathbf{e_j}=v^i\nabla_\mathbf{e_i}u^j\mathbf{e_j}=v^iu^j\nabla_\mathbf{e_i}\mathbf{e_j}+v^i\mathbf{e_j}\nabla_\mathbf{e_i}u^j=(v^iu^j \Gamma^k_{~ij}+v^i\frac{\partial u^k}{\partial x^i})\mathbf{e_k} \tag{1}.$$ In the last step we actually used that $$\nabla_{\mathbf{e_i}}\mathbf{e_j}=\Gamma^k_{~ij}\mathbf{e_k}\tag{2},$$ which is one definition of the affine connection: The basis vectors change if one moves from one point to another one in infinitesimal distance and the transformation that describes this change is the affine connection.
If one accepts the left hand side and the right hand side of (1) as a definition one can easily verity that (2) would be a consequence (set $\mathbf{v}=\mathbf{e_i} \Leftrightarrow v^i=1$ and $\mathbf{u}=\mathbf{e_j} \Leftrightarrow u^j=1$ but note that now $i$ and $j$ are not longer summation indices.) But equation (2) holds deeper meaning than just a trivial consequence of (1) since (2) is actually used to define (1). Equation (2) really describes the curvature of the space: in terms of the change of basis vectors.
So to your question about the basis vectors being vector fields: Yes of course they are. Differential-geometry or General relativity for that matter only become non trivial if this is the case: if the basis vectors where constant in space, the metric would be constant and all higher objects would vanish. All objects of GR are from a formal point of view fields: vector-, scalar- and even tensor-fields of the four dimensional space time. And even in "classical" differential-geometry this would be the case. The prime example: the basis vectors on the surface of a (2D) sphere change with position/angles, this is why from a mathematical point of the surface of a sphere has a non-vanishing curvature.
On your question of the "nature"/kind of equations (1) and (2): the covariant derivative of a component/vector is a tensor of rank 2.