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When I learned about the covariant derivative, it was motivated as a way of defining a good differentiation operation on tensors. To do this, we had to define a connection on the manifold, which was a substantial extra piece of structure.

However, the Lie derivative requires no connection at all; it just requires a vector field $V^\mu$ defined on the manifold. In particular, since we've already chosen coordinates, we can define the Lie derivative in any direction $n^\mu$ by using the vector field $V = n^\mu \partial_\mu$, which requires zero extra structure. Then $\mathcal{L}_V$ seems to be a perfectly good replacement for $n^\mu \nabla_\mu$. At the very least, it does everything that books say the covariant derivative was meant to do. Ignoring all the stuff the covariant derivative ends up getting used for, I don't know why we would have introduced it in the first place.

What good properties does $n^\mu \nabla_\mu$ have that $\mathcal{L}_{n^\mu \partial_\mu}$ does not?

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    $\begingroup$ (+1) related: Exterior Derivative vs. Covariant Derivative vs. Lie Derivative, Physicist's request for intuition on covariant derivatives and Lie derivatives. $\endgroup$ – AccidentalFourierTransform Mar 26 '16 at 18:51
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    $\begingroup$ Covariant derivative is scalar-field linear in one entry Lie derivative is not in both entries This feature strongly distinguishes between the two operators. The choice depends on which feature you need. $\endgroup$ – Valter Moretti Mar 26 '16 at 18:56
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    $\begingroup$ The Lie derivative does not give rise to a notion of parallel transport. It is how one vector field changes along another, it doesn't connect the tangent spaces at different points. Also, this is a pure differential geometry, not a physics question. $\endgroup$ – ACuriousMind♦ Mar 26 '16 at 18:56
  • $\begingroup$ @ACuriousMind Sure, but it gives some notion of transport, by pullback/push forward. Why is that not good enough? $\endgroup$ – knzhou Mar 26 '16 at 19:06
  • $\begingroup$ @knzhou But that transport is highly nonunique: it depends on the choice of $V$. The choice of a connection $\nabla$ does not depend on some arbitrary vector field, and in the Riemannian case, one can select a unique $\nabla$ with desirable properties. $\endgroup$ – Ryan Unger Mar 27 '16 at 16:14
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One distinctive feature is that the connection $\nabla$ has the tensor field property in its first entry $\nabla_{fX}=f\nabla_{X}$, while the Lie derivative doesn't. We have ${\cal L}_{fX}\neq f{\cal L}_{X}$ generically.

There are already manifestly covariant explanations on Math.SE and Mathoverflow.SE. Let's for simplicity consider a local coordinate patch $U$ with coordinates $x^{\mu}$. The tensor field property implies that the covariant derivative $\nabla_{X}=X^{\mu}\nabla_{\mu}$ is completely determine by some directional derivatives $\nabla_{\mu}$. Not so for the Lie derivative.

Say that we are given a metric tensor field $g$ and some tensor field $T$. Assume for simplicity that the tensor field components $g_{\mu\nu}$ and $T^{\mu_1 \ldots \mu_r}{}_{\nu_1 \ldots \nu_s}$ in the given local coordinate system are constant, i.e. $x$-independent. Let $\nabla$ be the Levi-Civita connection. Then $\nabla_Xg=0$ and $\nabla_XT=0$ as we would expect from a directional derivative (since after all, the tensor field components are constant in one coordinate system). But ${\cal L}_{X}g\neq 0$ and ${\cal L}_{X}T\neq 0$ generically.

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