TITLE: Physicist's request for intuition on covariant derivatives and Lie derivatives
QUESTION [48 upvotes]: A friend of mine is studying physics, and asks the following question which, I am sure, others could respond to better:
What is the difference between  the covariant derivative of $X$ along the curve $(t)$ and a Lie derivative of $X$ along $y(t)?$  I know the technical stuff about not needing to define a connection with a Lie derivative, needing to define the fields $X$ and $Y$ over a greater neighborhood, etc.
I am looking for a more physical sense.  If a Lie derivative gives the sense of the change of a vector field along the direction of another field, how does the covariant derivative differ?

REPLY [5 votes]: I know I'm really late in the game here, but I'm teaching a differentiable manifolds class for the first time this semester and I have been thinking a bit about this question.  At least for the case of a torsion-free connection, there is a certain intuitive picture of the relation between the Lie and covariant derivatives that I find appealing.
Let $X$ be a smooth vector field and let $\nabla$ be a torsion-free affine connection.  Then $\nabla X$ defines a $C^\infty(M)$-linear endomorphism of vector fields: $Y \mapsto \nabla_Y X$.  Therefore functorially $\nabla X$ induces $C^\infty(M)$-linear endomorphisms of tensor fields, that is, sections of $TM^{\otimes s} \otimes T^*M^{\otimes r}$.
The intuitive picture is that the covariant derivative is pointwise a sort of affine transformation, with the Lie derivative $L_X$ playing the part of the translation and $\nabla X$ playing the part of the linear transformation.  For example, on vector fields,
$$
\nabla_X Y = L_X Y + (\nabla X)(Y) = [X,Y] + \nabla_Y X 
$$
which is the torsion-free condition.  This is more general than just vector fields, of course.  If $T$ is any tensor, then
$$
\nabla_X T = L_X T + (\nabla X)(T),
$$
where $\nabla X$ acts on $T$ in the natural way.