Lie derivatives don't use the connection at all. They operate on the notion of evaluating a vector field along an integral curve of another vector field, this is inherently different to the notion of parallel transport.
Look at what happens when you take the commutator of integral curves, you get the Lie derivative. On the other hand if you take the commutator of parallel transport, you get the curvature tensor.
A smooth manifold $M$ may be equipped with many different geometric structures. For example:
- A Riemannian manifold is a pair $(M,g)$, where $g$ is a Riemannian metric.
- A manifold-with-connection is a pair $(M, \nabla)$, where $\nabla$ is an affine connection on $M$ (not necessarily torsion-free).
It turns out that on a Riemannian manifold $(M,g)$, there is a "best" connection to choose, the Levi-Civita connection. The Levi-Civita connection is characterized by being both (1) compatible with the metric, and (2) torsion-free.
To reiterate: once a metric $g$ is chosen, there are many connections $\nabla$ we can work with, but the Levi-Civita is the best one (in a certain precise sense).
Given an affine connection $\nabla$, any whatseover, we can define its associated torsion tensor by
$$T^\nabla(X,Y) = \nabla_XY - \nabla_YX - [X,Y].$$
We say that a connection is torsion-free iff $T^\nabla = 0$. Some connections (like the Levi-Civita connection) have this property, but others don't.
Finally, given an affine connection $\nabla$, any whatsoever, and a coordinate chart, we can talk about the associated Christoffel symbols, denoted $\Gamma^k_{ij}$. Again, they depend on both the choice of connection and the coordinate chart.
It is a fact (exercise) that $\Gamma^k_{ij} = \Gamma^k_{ji}$ in every coordinate chart if and only if $\nabla$ is torsion-free.
Best Answer
The Lie derivative $L_X T$ of any tensor $T$ along any vector field $X$ is defined directly and only from the underlying manifold structure. If, moreover, the manifold has a symmetric connection, (symmetric means zero torsion), then it is possible to express $\mathcal{L}_X T$ using this connection. Of course, the result is the same no matter which connection is used, (as long as it is symmetric).
Please note, accordingly, there is no meaning in writing $L_X$ and $\bar{L}_X$. This is wrong. The Lie derivative comes first, and then the expression in terms of a connection, not the other way around.
The Lie derivative has two (equivalent) definitions. A dynamical one and an algebraic one. It is very important to appreciate that these reflect two equally useful intuitions. I will not get into this here, (feel free to ask), but I will consider expressing the Lie derivative in terms of a symmetric connection.
If $\nabla$ is a symmetric connection, then for any vector fields $X,Y$
$$ L_XY = [X,Y] = \nabla_XY-\nabla_YX $$ where the first equality is by definition, but the second equality means $\nabla$ is symmetric.
For $(0,s)$-tensor $T$, (think of the metric if you would like), $L_XT$ is also a $(0,s)$-tensor, (one says that $L_X$ is type preserving), and by definition
$$ L_XT(Y_1,\ldots,Y_s) = X(T(Y_1,\ldots,Y_s)) - \sum_i T(\ldots,[X,Y_i],\ldots) $$
The first term is
$$ X(T(Y_1,\ldots,Y_s)) = (\nabla_X T)(Y_1,\ldots,Y_s) + \sum_i T(\ldots,\nabla_XY_i,\ldots) $$
By replacing in the definition of $L_XT$ and using the fact that $\nabla$ is symmetric, this yields the expression
$$ L_XT(Y_1,\ldots,Y_s) = \nabla_X T(Y_1,\ldots,Y_s) + \sum_i T(\ldots,\nabla_{Y_i}X,\ldots) $$
Please note this is not the definition of $L_XT$ but only a formula, using the connection $\nabla$. In other words, $L_XT$ does not depend on $\nabla$ but the righ-hand side does.
If $T$ is parallel, (this means $\nabla_X T = 0$ for any $X$),
$$ L_XT(Y_1,\ldots,Y_s) = \sum_i T(\ldots,\nabla_{Y_i}X,\ldots) $$
You may apply this last formula to the Lie derivative of the Riemannian metric $T=g$, and use the Levi-Civita connection as $\nabla$, to get the ''elasticity tensor'' $L_Xg$ and understand the definition of Killing vector fields, etc.
-- Salem