Let me start by saying that I understand the definitions of the Lie and covariant derivatives, and their fundamental differences (at least I think I do). However, when learning about Killing vectors I discovered I don't really have an intuitive understanding of the situations in which each one applies, and when to use one over the other.
An important property of a Killing vector $\xi$ (which can even be considered the definition) is that $\mathcal{L}_\xi\, g = 0$, where $g$ is the metric tensor and $\mathcal{L}$ is the lie derivative. This says, in a way, that the metric doesn't change in the direction of $\xi$, which is a notion that makes sense. However, if you had asked me how to represent the idea that the metric doesn't change in the direction of $\xi$, I would have gone with $\nabla_\xi g = 0$ (where $\nabla$ is the covariant derivative), since as far as I know the covariant derivative is, in general relativity, the way to generalize ordinary derivatives to curved spaces.
But of course that cannot be it, since in general relativity we use the Levi-Civita connection and so $\nabla g = 0$. It would seem that $\mathcal{L}_\xi\, g = 0$ is be the only way to say that the directional derivative of $g$ vanishes. Why is this? If I didn't know that $\nabla g = 0$, would there be any way for me to intuitively guess that "$g$ doesn't change in the direction of $\xi$" should be expressed with the Lie derivative? Also, the Lie derivative is not just a directional derivative since the vector $\xi$ gets differentiated too. Is this of any consequence here?
Best Answer
Nice question. One way to think about it is that given a metric $g$, the statement $\mathcal L_Xg = 0$ says something about the metric, whereas $\nabla_Xg = 0$ says something about the connection. Now what $\mathcal L_Xg = 0$ says, is that the flow of $X$, where defined, is an isometry for the metric, while $\nabla_Xg = 0$ says that $\nabla$ transports a pair of tangent vectors along the integral curves of $X$ in such a way that their inner product remains the same.
As an example, consider the upper half plane model of the hyperbolic plane. Its metric is $y^{-2}(dx^2 + dy^2)$, so clearly $\partial_x$ is a Killing vector field; its flow, horizontal translation, is an isometry. The fact that $\nabla_{\partial_x}g = 0$ doesn't say anything about $g$, but it does say that Euclidean parallel transport is compatible with this directional derivative of the connection.
Now consider $\partial_y$. This of course is not a Killing vector field, since vertical translation is not an isometry. The connection however can be made such (by the theorem of Levi-Civita) that a pair of tangent vectors can be parallel transported in such a way that the inner product is preserved.
EDIT
I think I have a more illustrative example: consider the sphere embedded in $\Bbb R^3$. Pick an axis and take the velocity vector field $\xi$ associated to rotation around the axis at some constant angular velocity. Also consider a second vector field $\zeta$ that is everywhere (in a neighbourhood of the equator, extend in any smooth way toward the poles) proportional to $\xi$, but that has constant velocity everywhere, something like in this image
(downloaded from this page).
Obviously $\xi$ is a Killing field, as it integrates to an isometry. An immediate way to see that $\zeta$ is not, is by noting that curves parallel to the equator remain parallel to the equator under the flow of $\zeta$, hence so do their tangent vectors. What happens to a curve whose tangent vector at the equator points toward a pole, is that the flow of $\zeta$ moves the point at the equator over a smaller angle than a point above the equator, so these two vectors don't remain perpendicular. For parallel transport on the other hand, two perpendicular tangent vectors to a point at the equator will remain perpendicular both under $\xi$ and in $\zeta$, since they only depend on the restriction to the vector fields to the equator, where they are equal. This doesn't say anything about the vector field generating an isometry, i.e. being a Killing vector field.