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?

  • 1
    $\begingroup$ Good question, for reference there are several related threads on other forums: mathse, mathoverflow $\endgroup$
    – zzz
    Commented Jun 3, 2015 at 3:32

4 Answers 4


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.


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.

  • 2
    $\begingroup$ +1 "Nice question" - and nice answer. LOVED the first paragraph. $\endgroup$ Commented Jun 9, 2015 at 23:02
  • $\begingroup$ I will have to reflect on this for some time to make sure I understand it, but I think it's the best answer. Thank you! $\endgroup$
    – Javier
    Commented Jun 10, 2015 at 0:12
  • $\begingroup$ I realize now that the vector field in the second example only changes the curve to higher order at the equator. The field should be changed in magnitude to being strictly monotonous from North to South. $\endgroup$
    – doetoe
    Commented Jun 13, 2015 at 6:42
  • 2
    $\begingroup$ "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." These seem like the same to me. How is "flow" different from moving "along the integral curves", and how is "isometry" different from "inner product remains the same"? $\endgroup$
    – Keshav
    Commented Dec 2, 2020 at 19:46
  • $\begingroup$ @Keshav What you are asking is a bit the point of the question (so in that sense my answer may have failed). The flow is indeed "moving along the integral curves", but not of tangent vectors, but of points: it is an actual mapping from the manifold onto itself (i.e. its points). Of course, this allows to move any tensor from one point to another along the flow as well (not along a single integral curve though, we need the vector field), preserving inner products when we deal with a Killing vector field. $\endgroup$
    – doetoe
    Commented Dec 4, 2020 at 18:00

As derivatives, the Lie and covariant derivatives involve comparing tensors at different points on the manifold. They differ in the prescription given for comparing the tensors at two different points.

The key concept with a covariant derivative $\nabla_\xi = \xi^a\nabla_a$ is parallel transport. It is defined so that as you move along a geodesic in the direction of $\xi^a$, the inner products between parallel transported vectors are preserved. This inner product necessarily involves the metric (it is basically what the metric is for), so the covariant derivative is also necessarily metric-dependent. In order for this definition to make sense, it is also important that the metric is parallel transported in all directions, which leads to the defining condition which you are probably well-aware of, $\nabla_a g_{bc}=0$.

By contrast, the Lie derivative $£_\xi$ gives the change in a tensor due to a diffeomorphism. It tells you how a tensor changes due to a one-parameter family of diffeomorphisms following flows of the vector $\xi^a$. Since diffeomorphisms make no reference to a metric, one of the key properties of Lie derivatives is that they do not depend at all on what the metric is!

Another very nice way to think of Lie derivatives is to use a coordinate system adapted to the vector $\xi^a$, so that it has components $\xi^\alpha = \delta^\alpha_0$. Then in this coordinate system, the Lie derivative is simply the partial derivative $\partial/\partial x^0$. Killing's equation in this coordinate system is $£_\xi g_{\alpha\beta} = \frac{\partial}{\partial x^0}g_{\alpha\beta} = 0$, i.e. the metric does not depend on the coordinate $x^0$. This should make great intuitive sense now: the metric is the same even as you flow to different values of $x^0$.

Hopefully this illustrates the big differences between the two derivatives: the covariant derivative should be used to measure whether a tensor is parallel transported, while the Lie derivative measures whether a tensor is invariant under diffeomorphisms in the direction of the vector $\xi^a$.


Let $T$ be some tensor field, $V$ a vector field, intuitively:

  • Covariant derivative $\triangledown_V T$ measures how far away a tensor is from being parallel transported along a vector field $V$.

  • Lie derivative $\mathcal L _V T$ measures how much a tensor changes under the one-parameter group of transformations generated by vector field $V$.

Parallel transport is a very specific notion that depends on the connection you choose. In fact one can conceivably choose a connection in which parallel transport doesn't have the geometrical interpretation as 'geometry transported along a curve'. In that case, all that the covariant derivative measures is 'how different $T$ is from some other arbitrary tensor $S$, where $S$ is a tensor that is parallel transported.

On the other hand, as you well know, the Lie derivative does not have such silly dependence on your connection. By definition it measures change of the tensor along a integral curve of $V$, always.

  • $\begingroup$ But what consequences does this difference have? Why is the fact that the metric is "independent" of some direction expressed with a Lie derivative? $\endgroup$
    – Javier
    Commented Jun 7, 2015 at 2:09
  • 1
    $\begingroup$ You want to express the notion "the metric doesn't change along some direction specified by a vector field". This change is exactly what the lie derivative measures, by definition. In contrast this change is NOT what the covariant derivative measures - covariant derivative measures 'deviation' from a parallel transported curve, which can be a rather arbitrary notion as I argued above. $\endgroup$
    – zzz
    Commented Jun 7, 2015 at 2:14
  • $\begingroup$ I'm sorry, but the difference is not so clear to me. You say that the covariant derivative is a (somewhat) arbitrary notion, but it's the one used almost everywhere in GR whenever we want to differentiate things. Why is this case special? $\endgroup$
    – Javier
    Commented Jun 7, 2015 at 2:23
  • $\begingroup$ When the covariant derivative is used, what it measures is how different the object is from a parallel transported object. For intuition, think about the case where you use the covariant derivative on a vector field along itself, this measures how different the vector field is from a geodesic. So to specify a covariant derivative you specify 2 things: a vector to differentiate with respect to, and a notion of parallel transport. In the case of defining a killing vector, there is no natural notion of parallel transporting a metric tensor and hence we don't want to measure this difference. $\endgroup$
    – zzz
    Commented Jun 7, 2015 at 2:38

The notion of derivative requires a notion of comparison. In a general manifold, tangent vectors at different points belong to totally different vector spaces (see footnote 1), so we must define a way of mapping one tangent vector to another tangent space that we shall take, by definition to be the the "invariant image" of the vector in the new tangent space so that we can compare its image under other transformations with this invariant image for the purposes of calculating a "derivative" through the appropriate limiting process(see footnote 2). This is not exactly like the elementary - and co-ordinate dependent - notion of a derivative, so we make our definition so that our derivative is as much like the elementary one in its algebraic properties. The Liebnitz product rule is foremost amongst these, and both the Lie and covariant derivatives are derivations on the algebra of smooth vector fields on a manifold: they both fulfil Leibnitz's rule (see footnote 3) and both are co-ordinate free definitions.

So I think I'd summarize an answer to you thus: in the sense that there is more than one way of defining the "invariant image" of a tangent vector for calculating a "derivative" with, there is no one notion of a "directional derivative". They are both directional derivatives.

A Lie derivative is the derivative of a vector field along the flow of a "benchmark" field, $\xi$ in your notation. It is as though a pioneering surveyor has mapped the manifold for you in advance by laying down a field which we use to compare all other fields to. Everything is measured by its relationship with $\xi$.

I say a bit more about Lie derivatives on my website: the discussion is about a quarter way through this page, around Figure 11.1:

"Lie Groups as Manifolds: The Conventional Lie Group Definition 2"

In contrast, the covariant derivative does not need a "benchmark" field. Everything is defined in terms of the metric, which in physics is the "physical" thing - it defines what length measurements we as small creatures living in the manifold will make. It can be visualized thus: embed the manifold in a higher dimensional Euclidean (or Minkowskian) space through an isometric embedding (by Nash's theorem, this can always be done). Then calculate the tangent vectors in the higher usual derivative. The Levi-Civita covariant derivative (others are possible) of some vector along the direction of a tangent vector is the component of the "elementary" directional derivative (in the higher dimensional embedding space) of the vector that is tangential to the manifold. We throw away the normal component as being owing to the bending of the manifold itself, rather than owing to the "intrinsic" variation of the geometrical object we are trying to measure.

We can also define the covariant derivative without the higher dimensional space abstractly as a connection without the metric. There then two tensors which one defines abstracty to measure the deviation of a manifold from its fulfilling the Euclid parallel postulate, the curvature and the torsion. These are both nought in a neighborhood if and only if Euclid's parallel postulate holds throughout the neigborhood. It can be shown, on a Riemannian manifold, where one can define a metric, that there is a unique connection defined in this way that has a vanishing torsion, and all the noneuclidean behavior is thus encoded in the curvature.


  1. Think of tangent planes to the 2-sphere embedded in 3 dimensional Euclidean space: two tangent planes are in general totally different vector spaces.

  2. Historically, Élie Cartan and the late 19th and early 20th century geometers first thought in terms of tangent planes rolling without slipping over a two dimensional manifold to make this mapping.

  3. If you've not looked into the notion of a derivation before, do so. It is surprising how much of elementary differential calculus you can derive from the Leibnitz product rule alone.

  • 3
    $\begingroup$ I appreciate your answer, but I'm not sure it directly addresses my question. I more or less understand the main differences between the Lie and covariant derivatives; at least I understand the definitions. But I don't know the differences in the notions of change encoded by each one, or when one might use one over the other. $\endgroup$
    – Javier
    Commented May 1, 2015 at 1:51
  • 2
    $\begingroup$ The link to the answerer's website appears to be broken and leads to some highly questionable websites. $\endgroup$
    – Kai
    Commented Apr 5, 2021 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.