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I have found, from this url, the link between Lie derivative and covariant derivative. It is said at the end of question that Lie derivative of of a vector field $\xi^{\alpha}$ with respect to a vector $v^{\nu}$, denoted, $\mathscr{L}_{\mathbf{v}}\xi^{\alpha}$ is equal to :

$$\mathscr{L}_{\mathbf{v}}\xi^{\alpha} = v^{\mu}\nabla_{\mu}\xi^{\alpha}$$

It seems this definition is false, (asked for the validity of this definition on other forums) and I would like to know why this is false ?

Maybe on this link, it is done confusions between (1) the general expression of directional derivative :

$\text{D}_{\mathbf{u}}\mathbf{v}=u^{\mu}\nabla_{\mu}\mathbf{v}$

and (2) Lie derivative (I think this one is correct) :

$\nabla_\mathbf{u} \mathbf{v} - \nabla_{\mathbf v} \mathbf{u} = [\mathbf{u},\mathbf{v}]=\mathscr L_u v$

Could anyone confirm the false information on this link above, i.e the following error of definition :

$$\mathscr{L}_{\mathbf{u}}v^{\alpha} = u^{\mu}\nabla_{\mu}v^{\alpha}$$

However, I see 663 view upvoters, it seems to me like fake informations or maybe I have forgotten something somewhere ...

Regards

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  • $\begingroup$ They should be equivalent, assuming $\nabla_\mu$ contains the Christoffel terms. For example, assume you're working in flat space. Then $\nabla_v u = 0$ since $u$ is effectively a constant vector field, so they coincide. $\endgroup$
    – Aaron
    Commented Jul 10, 2018 at 21:59
  • $\begingroup$ -@Aaron in this case, this works but in general case where $\nabla_{\mathbf v}\mathbf u$ is different from $0$, it seems they are not equivalent, doesn't it ? $\endgroup$
    – user87745
    Commented Jul 10, 2018 at 22:25

1 Answer 1

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From J. M. Lee, Introduction to Smooth Manifolds:

Theorem 9.38. If $M$ is a smooth manifold and $V, W \in \mathfrak{X}(M)$, then $\mathfrak{L}_V W = [V,W]$.

Now, turning to Problem 4-2 from the same author's Riemannian Manifolds:

(...) define a map [the torsion] by $\tau(X,Y) = \nabla_X Y - \nabla_Y X - [X,Y]$.
a) (...)
b) We say $\nabla$ is symmetric if its torsion vanishes identically. Show that $\nabla$ is symmetric if and only if its Christoffel symbols with respect to any coordinate frame are symmetric: $\Gamma^k_{ij} = \Gamma^k_{ji}$.

The Levi-Civita connection has symmetric Christoffel symbols. Therefore, the torsion is zero and $[X,Y] = \mathfrak{L}_X Y = \nabla_X Y - \nabla_Y X$.

So yes, the link is wrong.

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  • $\begingroup$ -@Javier thanks a lot ! So as a conclusion, there 3 things different : 1) the covariant derivative, 2) the generalized directionnal derivative of $v^\alpha$ vector field along $u$ vector $\text{D}_{\mathbf{u}}v^{\alpha}= u^{\mu}\nabla_{\mu}v^{\alpha}$ and 3) the Lie derivative : $\nabla_\mathbf{u} \mathbf{v} - \nabla_{\mathbf v} \mathbf{u} = [\mathbf{u},\mathbf{v}]=\mathscr L_u v$, is it right ? $\endgroup$
    – user87745
    Commented Jul 10, 2018 at 23:27
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    $\begingroup$ @youpilat13 Yes, though 2) is the contraction of 1) with the direction vector $u^\mu$. $\endgroup$
    – Javier
    Commented Jul 10, 2018 at 23:28
  • $\begingroup$ yes, the covariant derivative of a vector $\mathbf{v}$ along a vector $\mathbf{u}$ has a $j-th$ component under the form : $(\nabla_{\mathbf u}\mathbf v)^{j} = u^k\left(\frac{\partial v^j}{\partial x^k} + v^i\Gamma^j_{ki}\right)= u^{k} (\nabla_{k}\vec{v})^{j}$, is it good ? $\endgroup$
    – user87745
    Commented Jul 10, 2018 at 23:58
  • $\begingroup$ @Javier You are correct good sir, and I am mistaken. For a Lie derivative against a constant vector field, it becomes equivalent to the directional derivative with a partial derivative, not the covariant one. $\endgroup$
    – Aaron
    Commented Jul 11, 2018 at 0:16
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    $\begingroup$ @youpilat13 that's correct. $\endgroup$
    – Javier
    Commented Jul 11, 2018 at 11:05

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