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? 
 A: 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)
A: 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.
A: The Lie derivative is a more primitive notion than the covariant derivative $\nabla$, since it does not require specification of a connection (although it does require a vector field, of course):
$$
\mathcal{L}_U V^\mu=U^\nu \partial_\nu V^\mu- V^\nu \partial_\nu U^\mu
\\
\mathcal{L}_U V_\mu=U^\nu\partial_\nu V_\mu+ V_\nu\partial_\mu U^\nu
$$
However, it is sometimes useful to write it in terms of the covariant derivative, see this.
Consider the case where $U=u$, $V^\mu =u^\mu$ and $u_\mu u^\mu = 1$. Hence, since $u_\mu u^\mu$ is a scalar (that is also a constant field),
$$  \partial_\beta ( u_\mu u^\mu) =  \nabla_\beta ( u_\mu u^\mu) = u_\mu \nabla_\beta  u^\mu = 0 , $$
where we used the fact that the covariant derivative is defined to obey the Leibnitz rule.
Now, you have:
$$
\mathcal{L}_u u^\alpha = u^\nu \partial_\nu u^\alpha - u^\nu \partial_\nu u^\alpha = 0
$$
and
$$
\mathcal{L}_u u_\alpha = u^\nu\partial_\nu u_\alpha +  u_\nu \partial_\alpha u^\nu
= u^\nu \nabla_\nu u_\alpha +  u_\nu \nabla_\alpha u^\nu
= u^\nu\nabla_\nu u_\alpha
$$
In the second passage we used the fact that  for scalar fields, vector fields and 1-forms (i.e. covectors), the use of standard $\partial$ or covariant $\nabla$ derivative are equivalent in computing a Lie derivative.
