The Lie derivative for a covariant and contravariant vector is:

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

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

Therefore, $\mathcal{L}_V V^\mu=0$ using the above definition. My question is: what is the intuition behind this result?

Moreover, 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?

  • 1
    $\begingroup$ 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. $\endgroup$
    – Andrea
    Jul 13, 2018 at 14:00

3 Answers 3


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)

  • $\begingroup$ Thanks for the intuition it was very useful. What about the second part of the question? $\endgroup$
    – P. G. A.
    Jul 12, 2018 at 21:34
  • $\begingroup$ @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. $\endgroup$
    – Andrea
    Jul 13, 2018 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.


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 $U$):

$$ \mathcal{L}_U V^\mu=U^\nu \partial_\nu V^\mu- V^\nu \partial_\nu U^\mu \\ \mathcal{L}_U V_\mu=U^\nu\partial_\nu V_\mu+ V_\nu\partial_\mu U^\nu $$

However, it is sometimes useful to write it in terms of the covariant derivative, see Why can I use the Covariant Derivative in the Lie Derivative?.

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:

$$ \mathcal{L}_u u^\alpha = u^\nu \partial_\nu u^\alpha - u^\nu \partial_\nu u^\alpha = 0 $$


$$ \mathcal{L}_u u_\alpha = u^\nu\partial_\nu u_\alpha + u_\nu \partial_\alpha u^\nu = u^\nu \nabla_\nu u_\alpha + u_\nu \nabla_\alpha u^\nu = u^\nu\nabla_\nu u_\alpha $$

In the second passage of the above equation 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 is equivalent in computing a Lie derivative. The point is that, while it is intuitive that $\mathcal{L}_u u^\alpha=0$, this does not need to be the case for the corresponding 1-form because the Lie-derivative of the metric tensor does not need to be zero.


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