TITLE: Covariant Derivative of Basis Vectors
QUESTION [2 upvotes]: Consider the standard covariant derivative of Riemannian Geometry (torsion free with metric compatibility) in the $\frac{\partial}{\partial x^i}$ direction. Application to a vector field will be denoted $\nabla_i \vec{v} $ . For the purposes of this question, I will restrict myself to flat space (namely the plane).
Many introductory sources initially define the Christoffel Symbols by the relationship
$$\frac{\partial \vec{\mathbf{e_i}}}{\partial x^j}=\Gamma^k_{ij}\vec{\mathbf{e_k}}$$
where $\vec{\mathbf{e_i}} = \frac{\partial}{\partial x^i}$ . The covariant derivative is then derived quite simply for contravariant and covariant vector fields as being
$$\nabla_i \vec{v} =\bigg( \frac{\partial v^j}{ \partial x^i } + \Gamma^j_{ik} v^k\bigg) \frac{\partial}{\partial x^j}$$
$$\mbox{and}$$ $$\nabla_i \alpha =\bigg( \frac{\partial \alpha_j}{ \partial x^i } - \Gamma^k_{ij} \alpha_k\bigg) dx^j$$ respectively. Now let's consider the covariant derivative of the covariant basis vector. Observe
$$\nabla_i \vec{\mathbf{e_j}} = \frac{\partial \vec{\mathbf{e_j}}}{ \partial x^i } - \Gamma^k_{ij} \vec{\mathbf{e_k}}$$ $$\mbox{and by our definition of the Christoffel Symbols with symmetric lower indices}$$
$$ \nabla_i \vec{\mathbf{e_j}} = \Gamma^k_{ij} \vec{\mathbf{e_k}} - \Gamma^k_{ij} \vec{\mathbf{e_k}} = \vec{0} \mbox{ .}$$
When I was in my first course on tensors and Riemannian Geometry, we did not arrive at the same result in the plane. We did not use the above definition of the Christoffel Symbols but rather defined them by the geodesic equation (which we arrived at by using the G$\hat{\text{a}}$teaux Variation). I am aware that the intrinsic definition of the Christoffel Symbols
$$\Gamma^k_{ij} = \frac{1}{2}g^{k\ell}\bigg[\frac{\partial g_i\ell}{\partial x^j} + \frac{\partial g_j\ell}{\partial x^i} - \frac{\partial g_ij}{\partial x^\ell}\bigg]$$
is equivalent to the first definition I provided. In our class, we used the following argument for the derivative of the covariant basis.
Let all the components of $v^i $ be $0$ except for the $i^{\text{th}}$ component which is $1$. It is then clear that $\vec{v} = v^i \frac{\partial}{\partial x^i}$ is the invariant (rank $0$) form of the basis vector. The covariant derivative would then be
$$\nabla_i \vec{v} = \nabla_i\bigg( \frac{\partial}{\partial x^k} \bigg) = \bigg( \frac{\partial v^j}{ \partial x^i } + \Gamma^j_{ik} v^k\bigg) \frac{\partial}{\partial x^j} = \bigg( 0 + \Gamma^j_{ik} \bigg) \frac{\partial}{\partial x^j} = \Gamma^j_{ik} \frac{\partial}{\partial x^j}$$ which is clearly not identically $\vec{0}$ .
It is at this point where I turned to the physical/geometric interpretation of the covariant derivative: parallel transport. I walked myself through many examples including the following example of the polar coordinate system in the plane (a nice, flat space).
Consider a vector field $V$ in the polar coordinate system along with the two nearby points $p$ at $(r,\theta)$ and $p'$ at $(r, \theta + \Delta \theta)$. The covariant derivative (w.r.t. the theta covariant basis vector) is said to be the result of parallel transporting the vector $v' = V(p')$ along the direction of a short curve to point $p$ and then subtracting the vectors $v'_{||}-v$ where $v'_{||}$ is the transported vector $v'$ at point $p$. Note that I realize there is also a division by a pathlength parameter and a limit in the definition but this notion should work for arguments sake.
At this point I drew a circle and considered the covariant derivative $\nabla_\theta \bigg( \frac{\partial}{\partial \theta} \bigg)$. This derivative should (if I understand correctly) track the rate of change of the $\frac{\partial}{\partial \theta}$ basis vector using parallel transport along the circle upon which $p$ and $p'$ both reside. If one of the formulations above is correct, it will either come out to be $\vec0$ or $-r\frac{\partial}{\partial r}$ . Drawing this out, it is quite obvious that the vector $v'_{||}$ points slightly inward on this circle. The vectors should be the same length since they were both generated by the vector field $\frac{\partial}{\partial \theta}$ at the same radius and thus the vector subtraction $v'_{||} - v$ points directly inward in the $-\frac{\partial}{\partial r}$ direction. This is looking good for the second formulation! It is also intuitively easy to see in this example that as the radius grows, the length of the $\frac{\partial}{\partial \theta}$ vector also grows and thus the projection of $v'_{||}$ onto $-\frac{\partial}{\partial r}$ would also increase.
PRESTO! The physical intuition matches the second formulation. It may be helpful to note that I am aware that the first definition I provided for the Christoffel Symbols does not extend well to the intrinsic geometry of embedded surfaces. That being said, the plane is nice and flat and exactly what this definition seems to be made for (not to mention the fact that the definition appears in derivations of the covariant derivative for every differentiable object as far as I know).
How can I rectify these seemingly contradictory notions of differentiating basis vectors in flat space or in general?
If anyone is interested, this is where I first saw the covariant derivative sending basis vectors to zero in flat space.

REPLY [0 votes]: The covariant derivative formula you used is only to be use with vector or covector components.
You can not insert a basis vector into it and get the correct answer.
If you assume the components of a basis vector are constant and use an upper index you would have gotten the correct answer.  The partial derivative is zero and the last term with the Christoffel symbol is left over with a plus sign.
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