Lie derivative of a vector along itself

In Tensorial form, the definition of the Lie derivative for a covariant and contravariant vector are respectively:

$$\mathcal{L}_U V^\mu=U^\nu\nabla_\nu V^\mu- V^\nu\nabla_\nu U^\mu$$

$$\mathcal{L}_U V_\mu=U^\nu\nabla_\nu V_\mu+ n_\nu\nabla_\mu U^\nu$$

At some point of a calculation, I had to calculate $\mathcal{L}_V V^\mu$ that gives zero using that definition. My question is, What is the intuition behind this result?

When foliating a spacetime in Cauchy surfaces as

$$h_{\mu\nu}=g_{\mu\nu}-n_\mu n_\nu$$

where $n^\mu$ is a spacelike normal vector, we get an extra identity of the form

$$n^\alpha \nabla_\beta n_\alpha=0$$

derivating the definition $n^\alpha n_\alpha=1$.

With this in mind, the computation of $\mathcal{L}_n n_\alpha$ gives

$$\mathcal{L}_n n_\alpha=n^\beta\nabla_\beta n_\alpha$$

that is different from zero. Once again, what's the intuition behind this result?

• You should double check your second equation. It’s seems to me that the indices are inconsistent, and that you introduced a vector $n$ without defining it. – Andrea Jul 13 '18 at 14:00

When you take the Lie derivative of a vector, you are looking at how it changes as you move along integral curves. Now if you look at $L_UU$ you are asking how does $U$ change along its integral curves. But the point of an integral curve is that it’s tangent is always $U$. So $U$ does not change as you travel along the curve (its always pointing ahead)
• @P.G.A. Maybe I’m missing something, but if you raise the index in the LHS of your last equation, you get the Lie derivative of $n^\alpha$ by itself and thus 0, is the right hand side is also 0. – Andrea Jul 13 '18 at 13:59
The first kind of Lie derivative is anticommutative, which implies the desired result. In fact, identifying $V^\mu$ with $V^\mu\nabla_\mu$, the Lie derivative is a commutator.