6
$\begingroup$

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?

$\endgroup$
6
  • 1
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.