Do Christoffel symbols commute? Do Christoffel symbols commute? For example, does $\Gamma^{e}_{db}\Gamma^{c}_{ea} = \Gamma^{c}_{ea}\Gamma^{e}_{db}$?
 A: In classical theory, all observables commute. The components $\Gamma^a_{bc}$ are just real numbers so of course that they commute.
In quantum theory, they don't commute. It's probably a bit laborious to calculate the commutator.
A: No they don't commute in this sense, except you screwed up the commutator. It should be
$$\Gamma^{e}_{db}\Gamma^c_{ea} - \Gamma^c_{eb}\Gamma^{e}_{da}$$
Which is
$$ \Gamma_b \Gamma_a - \Gamma_a \Gamma_b $$
in a matrix form, where I have taken one upper and one lower index and suppressed them in the Christoffel symbol (which lower index doesn't matter, becuase of the symmetry on the lower indices). This has the interpretation I give below.
The Christoffel symbols are the coordinate form of the infinitesimal rotations associated with parallel transporting a vector a ways along a short distance. If all the frames were orthonormal, the parallel transport would be SO rotations, and the infinitesimal form would be a bunch of stuff in the Lie Algebra of SO, and these are antisymmetric matrices (or the upper index representation of antisymmetric forms for the Lorentzian case). Their commutators tell you when the rotations corresponding to moving in a certain direction is noncommutative with the rotation corresponding to moving in another direction.
For a coordinate basis for the tangent space, the basis vectors are not orthonormal, so the connection coefficients don't obey the Lie algebra conditions, but they still stay noncommutative. You can easily verify this in a generic example by direct computation, but it is also obvious from those cases where you happen to choose coordinates where the coordinate basis is orthogonal, like spherical coordinates, and think about the different coordinate rotations associated with parallel transport in different infinitesimal directions.
A: This might be overkill, but here it goes:
Let $\pi:E\rightarrow B$ be an arbitrary fiber bundle with typical fiber $F$ and $\Phi=\tau\times\varphi$ be a local trivialization over $U\subset B$
$$
\Phi:\pi^{-1}(U)\rightarrow U\times F
$$
Geometrically, the Christoffel symbol $\Gamma$ of a connection is an element
$$
\Gamma\in\mathrm{Hom}(\tau^*(\mathrm TU),\varphi^*(\mathrm TF))
$$
which is a fancy way of saying that for each $e\in E$, there is a linear map
$$
\Gamma(e):\mathrm T_{\tau(e)}U\rightarrow\mathrm T_{\varphi(e)}F
$$
In general, talking about the composition
$$
\Gamma(e)\circ\Gamma(e')
$$
makes no sense as domain and codomain don't agree.
In relativity however, $E=\mathrm TB$ which allows us to identify these spaces and we end up with
$$
\Gamma(V):\mathrm T_bU\rightarrow\mathrm T_bU
$$
where $V\in\mathrm T_bU$ is a tangent vector, eg a 4-velocity.
As $\Gamma(V)$ is linear, it can be expressed in local coordinates via matrix multiplication
$$
W^i\mapsto\Gamma(V)^i{}_j W^j
$$
In case of the Levi-Civita connection, the map
$$
V\mapsto\Gamma(V)
$$
is linear as well and we arrive at
$$
W^i\mapsto(\Gamma_k V^k)^i{}_j W^j = \Gamma_k{}^i{}_j V^k W^j
$$
The compositions read
$$
(\Gamma(V)\circ\Gamma(W))^i{}_j = \Gamma_k{}^i{}_a \Gamma_l{}^a{}_j V^k W^l
\\
(\Gamma(W)\circ\Gamma(V))^i{}_j = \Gamma_l{}^i{}_a \Gamma_k{}^a{}_j V^k W^l
$$
which in general do not commute.
