Why is the Christoffel symbol of the 2nd kind symmetric in lower indices? I have consulted multiple books on tensors for physicists, but they all take for granted this relation:
$\Gamma_{ij}^k = \Gamma_{ji}^k$
However, no proof is provided and I cannot find a single one online.
Can you prove this relation using basic tensor analysis ? (I have taken exactly one course on the subject, as a part of a physics undergraduate program.)
 A: Depends on what context did you study (old school, coordinate-based) differential geometry.
Sometimes in physics courses, you have something called tensor calculus, which often just deals with doing tensor calculus in general coordinates in euclidean space.
In that case, if you have a coordinate system $(u^1,...,u^n)$, and coordinate basis vectors $\mathbf{e}_i=\frac{\partial\mathbf{r}}{\partial u^i}$, and $\mathbf{X}$ and $\mathbf{Y}$ are vector fields, then the directional derivative is $$ d_\mathbf{X}\mathbf{Y}=d_\mathbf{X}(Y^i\mathbf{e}_i)=d_\mathbf{X}Y^i\mathbf{e}_i+Y^id_\mathbf{X}\mathbf{e}_i=X^j\partial_jY^i\mathbf{e}_i+Y^iX^jd_{\mathbf{e}_j}\mathbf{e}_i=\\=X^j\left(\partial_jY^i\mathbf{e}_i+Y^id_{\mathbf{e}_j}\mathbf{e}_i\right)=X^j\left(\partial_jY^i\mathbf{e}_i+Y^i\Gamma^k_{ji}\mathbf{e}_k\right)=X^j\left(\partial_jY^k+Y^i\Gamma^k_{ji}\right)\mathbf{e}_k, $$
so we see that in this case, the covariant derivative is the same thing as the usual directional derivative, just the components look different in curved coordinates. These components themselves are given by $$ \nabla_jY^i=\partial_jY^i+\Gamma^i_{jk}Y^k, $$ where by the previous derivation, $$ \Gamma^k_{ij}=\mathbf{e}^k\cdot d_{\mathbf{e_i}}\mathbf{e}_j .$$
Now, since $d_{\mathbf{e}_i}=\partial/\partial x^i$, and $\mathbf{e}_j=\partial\mathbf{r}/\partial x^j$, $d_{\mathbf{e}_i}\mathbf{e}_j=\partial^2\mathbf{r}/\partial x^i\partial x^j$, and the partials commute, we can see that $\Gamma$ is symmetric in the lower indices.
If you are working on a more general Riemannian manifold, the covariant derivative has to be defined by hand, and most of the time in physics, we wish to use a covariant derivative that behaves a lot like the flat space derivative, which, by the previous argument, is symmetric.
Also, if you have a metric tensor, then a covariant derivative is determined by the metric tensor uniquely if and only if it 1) annihilates the metric: $\nabla_ig_{jk}=0$, and 2) it is symmetric: $\Gamma^k_{ij}=\Gamma^k_{ji}$.
Most of the time we don't want to treat the covariant derivative as a dynamical variable, so we let it be determined uniquely by the metric this way, and, perhaps not coincidentally, the covariant derivative we get by this way is the covariant derivative that behaves the most like the derivative in flat space (annihilates the metric and is symmetric).
