A friend of mine is studying physics, and asks the following question which, I am sure, others could respond to better:

What is the difference between the covariant derivative of $X$ along the curve $(t)$ and a Lie derivative of $X$ along $y(t)?$ I know the technical stuff about not needing to define a connection with a Lie derivative, needing to define the fields $X$ and $Y$ over a greater neighborhood, etc.

I am looking for a more physical sense. If a Lie derivative gives the sense of the change of a vector field along the direction of another field, how does the covariant derivative differ?

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    $\begingroup$ I don't think the Lie derivative "gives the sense of the change of the vector field along the direction of another field". One primary difference is that $\nabla_Z$ is $C^\infty$-linear in $Z$, while $\mathcal{L}_Z$ is only $\mathbb{R}$-linear. Conceptually for the Lie derivative you flow the entire manifold/neighborhood along the vector field $Z$ (so in fact you cannot define the Lie derivative of a vector field $X$ along a curve $y(t)$; at the very least you need to have a congruence, instead of just a single curve). $\endgroup$ Commented Sep 12, 2011 at 14:01
  • 8
    $\begingroup$ you'll find an extensive discussion, with many pointers to the literature at physicsforums.com/showthread.php?t=150200 $\endgroup$ Commented Sep 12, 2011 at 14:12
  • 6
    $\begingroup$ @CarloBeenakker great link: why don't you quote it as an answer? I makes a lot more sense for poor physicists like me than any of the current answers. $\endgroup$
    – chris
    Commented Dec 31, 2014 at 11:01
  • $\begingroup$ "so in fact you cannot define the Lie derivative of a vector field XX along a curve y(t)y(t); at the very least you need to have a congruence, instead of just a single curve)." – @WillieWong That is completely incorrect. The Lie derivative of a vector field is defined on the level curve of another vector field. $\endgroup$
    – user98740
    Commented Sep 20, 2016 at 23:37
  • 1
    $\begingroup$ @guest I encourage you to learn some differential geometry, perhaps by doing the exercise Jack Lee mentioned in his comment. $\endgroup$ Commented Sep 21, 2016 at 3:29

8 Answers 8


The Lie derivative of a vector field $X$ with respect to another vector field $Y$ is just the Lie bracket of the two vector fields. It is well-defined given only the smooth structure and does not require any connection. In other words, it is independent of changes of co-ordinates and is preserved under any diffeomorphism. Given how flexible diffeomorphisms are, it can't be a pointwise or even curvewise concept, since you can basically map any pair of nonzero vectors to any other pair and even any nonvanishing transversal vector field along a curve to any other nonvanishing transversal vector field along another curve.

But we know what the Lie derivative tells us. It tells us how "coherent" or "independent" the two vector fields are with respect to each other locally (on an open set and not just at a point). It measures to what extent the generated flows commute, i.e. what happens if you first travel along an integral curve of one and then along one of the other versus the opposite order.

Another way to think about this is, discussed in control theory, to think about the set you get if you flow first along one vector field, then the other, then the first one again, etc. If the Lie bracket vanishes, then you stay inside a 2-dimensional surface. If it doesn't, then the value of the Lie bracket (and its iterates) tells you the dimension of the set that you stay inside.

A connection allows you to define the concept of a "constant" vector along a curve, i.e. parallel translation along a curve. It is important to understand that defining parallel translation is an extra assumption or geometric structure added to the smooth manifold.

  • 5
    $\begingroup$ Another natural question is whether the Lie derivative and connection are related somehow. The answer is that for an arbitrary connection, they don't have to be related at all. But it turns out to be useful (and therefore natural?) to assume that they are related. Again, however, this is an additional assumption and not forced on you. $\endgroup$
    – Deane Yang
    Commented Sep 12, 2011 at 15:19

First let me say that what is intuitive to a physicist may be not be so to a geometer and vice-versa. To many physicists a connection is the potential of a field satisfying a gauge invariance. For this point of view I refer to vol. 1, Chap. 6 sect 41 of the three volume book by Dubrovin-Fomenko-Novikov: Modern Geometry-Methods and applications.

I find this point of view less intuitive only because I was trained as a mathematician.

The notion of covariant derivative appears naturally when one tries to solve the following problem. Suppose that $E\to M$ is a smooth vector bundle over a smooth manifold $M$. For example, $E$ could be the tangent bundle of $M$. We seek a notion of parallel transport that will allow us to compare vectors situated in different fibers of the bundle. More precisely, this is a correspondence that associates to each smooth path

$$\gamma: [a,b]\to M$$

a linear map $T_\gamma$ from the fiber of $E$ at the initial point of $\gamma$ to the fiber of $E$ over the final point of $\gamma$

$$T_\gamma: E_{\gamma(a)}\to E_{\gamma(b)}.$$

The map $T_\gamma$ is called the parallel transport along the path $\gamma$.The assignment $\gamma\mapsto T_\gamma$ should satisfy two natural conditions.

(a) $T_\gamma$ should depend smoothly on $\gamma$. (The precise meaning of this smoothness is a bit technical to formulate, but in the end it means what your intuition tells you it should mean.)

(b) If $\gamma_0: [a,b]\to M$ and $\gamma_1:[b,c]\to M$ are two smooth paths such that the initial point of $\gamma_1$ coincides with the final point, then we obtain by concatenation a path $\gamma:[a,c]\to M$ and we require that

$$T_\gamma= T_{\gamma_1}\circ T_{\gamma_0}. $$

Suppose we have a concept of parallel transport. Given a smooth path $\gamma:[0,1]\to M$ and a section $\boldsymbol{u}(t)\in E_{\gamma(t)}$, $t\in [0,1]$ of $E$ over $\gamma$, then we can define a concept of derivative of $\boldsymbol{u}$ along $\gamma$. More precisely

$$ \nabla_{\dot{\gamma}} \boldsymbol{u}|_{t=t_0}=\lim_{\varepsilon \to 0} \frac{1}{\varepsilon} \left( T^{t_0,t_0+\varepsilon}_\gamma \boldsymbol{u}(t_0+\varepsilon)- \boldsymbol{u}(t_0)\right), $$

where $ T^{t_0,t_0+\varepsilon}_\gamma$ denotes the parallel transport along $\gamma$ from the fiber of $E$ over $\gamma(t_0+\varepsilon)$ to the fiber of $E$ over $\gamma(t_0)$. The left-hand-side of the above equality is called the covariant derivative of $\boldsymbol{u}$ along the vector field $\dot{\gamma}$ determined by the parallel transport. Thus, a choice of parallel transport leads to a concept of covariant derivative.

Conversely, a covariant derivative $\nabla$ leads to a parallel transport. Given a smooth path $\gamma:[0,1]\to M$ the parallel transport

$$T_{\gamma}: E_{\gamma(0)}\to E_{\gamma(1)} $$

is defined as follows. Fix $u_0\in E_{\gamma(0)}$. Then there exists a unique section $\boldsymbol{u}(t)$ of $E$ over $\gamma$ satisfying

$$ \boldsymbol{u}(0)=u_0,\;\;\nabla_{\dot{\gamma}}\boldsymbol{u}(t)=0,\;\;\forall t\in [0,1].$$

We then set $\newcommand{\bu}{\boldsymbol{u}}$

$$T_\gamma \bu_0:= \boldsymbol{u}(1).$$

This construction allows us to define the covariant derivative $\nabla_X\bu$ of a section $\bu$ of $E$ along a vector field $X$ of $M$. It satisfies the rescaling property

$$ \nabla_{fX}\bu=f\big(\nabla_X\bu\big),\;\;\forall f\in C^\infty(M). $$

A connection on $TM$ will then satisfy

$$\nabla_{fX} Y=f\big(\nabla_X Y),$$

for any vector fields $X,$ and any smooth function $f$. On the other hand the Lie derivative satisfies $$ L_{fX} Y= fL_XY-(Xf) Y, $$ so it cannot be a covariant derivative.

  • $\begingroup$ This is the best answer I've found on the internet. I am trained as a physicist and I follow your intuition very well. I am with you till the end part - you suddenly say this construction allows us to define the covariant derivative $\nabla_X u$ over a vector field, which is $C^\infty$-linear. I was lost here. Up till here you were talking about derivative along a curve, so 1. where is a vector field $X$ from 2. why is it (naturally or can be made) $C^\infty$-linear? $\endgroup$
    – Alex
    Commented Jan 19, 2022 at 12:47
  • $\begingroup$ Or did you mean that with a vector field $X$ we can generate its flow - a bunch of curves. Then, with the extra structure of "parallel transport" defined over the entire manifold we can take the covariant derivative along these curves. But the covariant derivative practically has nothing to do with $X$ (other than that $\nabla_X Y$ is just on a curve defined by $X$), rather, covariant derivative is all specified by the parallel transport. Unlike Lie derivative which is about $X$ just as much it is about $Y$? $\endgroup$
    – Alex
    Commented Jan 19, 2022 at 12:53
  • $\begingroup$ The $C^\infty$-linearity is very intuitive. Think you are traveling by car with velocity $v$ and you measure the rate of change in temperature per unit of time. This rate of change is the rate of change per unit of distance $\times$ velocity. Different cars will report different temporal rates of change at the same location $x$ if they have different velocities at that location. The proportionality factor depends on $x$, and is a function. $\endgroup$ Commented Jan 19, 2022 at 13:18
  • $\begingroup$ I think I got it now - at every point on $M$, $f$ is just a scaling number. The notion of parallel transport (or to say connection) is defined on $M$ independent of $X$, hence $f$ can be pulled to the front at each point independently. Is my understanding in the 2nd comment above correct? $\endgroup$
    – Alex
    Commented Jan 19, 2022 at 13:54
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    $\begingroup$ Here is how my adviser taught me. Regular one-forms are beasts that eat a vector and return a number. An operator valued 1-form eats a vector and returns an operator. If you feed a connection $\nabla$ you get an operator $\nabla_X$. This operator takes a vector $Y$ and returns the vector $\nabla_XY$. $\endgroup$ Commented Jan 20, 2022 at 11:30

I like to think about the Lie derivative in the following way. You are standing on a bridge over a river. Open a box of matches and throw them into the river. At time $t=0$ the matches define a vector field $X$, the velocity field of the river is a vector field $Y$. Fix your eyes at a point $p$ (immobile with respect to the bridge) and watch how the directions of the matches flowing through point $p$ are changing. The speed of this change is the Lie derivative $\mathcal{L}_Y(X)(p)$. This picture is not very exact, because the matches do not change their lengths. A truly elastic match can be stretched or shrunk by the flow.

To understand the Levi-Civita covariant derivative one has to understand geodesics. If you are driving on uneven terrain, then your car will move along the geodesic if the left and the right wheel rotate with the same speed. If $X$ is a vector field on the terrain, and your speed at the moment is $Y$, then $\nabla_YX$ is the rate of change of the vector field $X$ in the coordinate system bound to your car.

In particular, one sees that $\nabla_YX(p)$ depends only on the value of $Y$ at the point $p$, while $\mathcal{L}_Y(X)(p)$ depends on the values of $Y$ in a neighborhood of $p$.


Lie derivative is based on a Lie group (or Lie algebra) which acts on the manifold. This derivative cannot be defined just at one point because the action cannot be defined at a point even if you give explicitly the direction at that point. On the other hand, using connection, covariant derivative can be defined pointwise. I think this is the main technical difference between them.

  • $\begingroup$ I'm not sure what you mean by "defined pointwise". With either type of derivative, you need to know something about the vectors involved at more than one point. So you need to be more precise about what the distinction is. $\endgroup$
    – Deane Yang
    Commented Sep 12, 2011 at 19:53
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    $\begingroup$ Covariant derivative is the analogue of directional derivative in R^n case. So if we fix a connection and assign a direction to a point, the covariant derivative at that point is well-defined. But for Lie derivative, one direction is not enough. We have to point out the vector field. L_X(f) might not equal to L_Y(f) even if X(p)=Y(p). $\endgroup$
    – Xiao Xinli
    Commented Sep 12, 2011 at 20:20
  • $\begingroup$ Yes, that's a good clarification of what you wrote. $\endgroup$
    – Deane Yang
    Commented Sep 12, 2011 at 20:30
  • $\begingroup$ I suppose you mean the action of the Lie algebra of the diffeomorphism group, which is the Lie algebra $\mathfrak{X}(M)$ of vector fields on $M$, right? How does it clarify the picture about the Lie derivative? $\endgroup$
    – Qfwfq
    Commented Sep 12, 2011 at 21:50

Here's an example from Lee's Riemannian Geometry

Problem 4-3: b) There exists a vector field on $\mathbb R^2$ that vanishes along the x-axis, but whose Lie derivative with respect to $\partial_1$ does not vanish on the x-axis. [This shows that Lie differentiation does not give a well-defined way to take directional derivatives of vector fields along curves.]

We can for instance take the vector field to be $V = \exp(-\frac{1}{x_2} + x_1)$ for $x_2 > 0$ and 0 otherwise.

  • 9
    $\begingroup$ Note that this problem was incorrectly stated in my book. In my correction list, it's restated correctly as follows: "Show that there are vector fields $V$ and $W$ on $\mathbb R^2$ such that $V = W = \partial_1$ along the $x^1$-axis, but the Lie derivatives $\mathscr L_V (\partial_2)$ and $\mathscr L_W (\partial_2)$ are not equal on the $x^1$-axis.” $\endgroup$
    – Jack Lee
    Commented Sep 15, 2015 at 16:48

I know I'm really late in the game here, but I'm teaching a differentiable manifolds class for the first time this semester and I have been thinking a bit about this question. At least for the case of a torsion-free connection, there is a certain intuitive picture of the relation between the Lie and covariant derivatives that I find appealing.

Let $X$ be a smooth vector field and let $\nabla$ be a torsion-free affine connection. Then $\nabla X$ defines a $C^\infty(M)$-linear endomorphism of vector fields: $Y \mapsto \nabla_Y X$. Therefore functorially $\nabla X$ induces $C^\infty(M)$-linear endomorphisms of tensor fields, that is, sections of $TM^{\otimes s} \otimes T^*M^{\otimes r}$.

The intuitive picture is that the covariant derivative is pointwise a sort of affine transformation, with the Lie derivative $L_X$ playing the part of the translation and $\nabla X$ playing the part of the linear transformation. For example, on vector fields, $$ \nabla_X Y = L_X Y + (\nabla X)(Y) = [X,Y] + \nabla_Y X $$ which is the torsion-free condition. This is more general than just vector fields, of course. If $T$ is any tensor, then $$ \nabla_X T = L_X T + (\nabla X)(T), $$ where $\nabla X$ acts on $T$ in the natural way.


Let $T$ be a tensor field on the manifold $M$, $\nabla$ a connection, $v$ a tangent vector at $x\in M$, and $V$ a vector field such that $V(x)=v$.

Then the intuition is as follows:

The covariant derivative $\nabla_v T$ is the derivative of $T$ along a geodesic arc $\gamma$ for $\nabla$ which has direction $v$ at $x=\gamma(0)$. The derivation is computed in the finite dimensional tangent space $T_xM$, as nearby values $T(y)$, $y\in M$, are compared via parallel transport.

(Remark: here "geodesic arc" should be made more precise, as geodesics emanating from $x$ are determined as parametrized curves and it may happen that the geodesic in the direction $v$ doesn't have velocity $v$)

The (value at the point $x$ of the) Lie derivative $\mathcal{L}_VT$ is the derivative of $T$ along the flowline of $V$ (passing through $x$). The derivation is computed in the finite dimensional tangent space $T_xM$, as nearby values $T(y)$, $y\in M$, are compared via pullback along the local flow of $V$.

Edit. I happen to have re-read this old answer of mine, and I find that it was indeed misleading as indicated by Dean Yang in the comments. Let's see if it can be phrased better:

In both cases, we want to understand the derivatives as the velocity of curves in the same finite dimensional vector space $T_x M$ (or its tensor powers $T^{p,q}_x M$). How to do this?

  • In the case of $\mathcal{L}_V T$, we use the flow $\varphi_V^t$ of $V$.
    So the curve is $t\mapsto \varphi_{V,\star}^{-t}(T(\varphi_V^t (x)))$ (for contravariant tensors; for covariant ones or mixed ones, we use pullback along $\varphi_V^t$ instead, where needed).
  • In the case of $\nabla_v T$, we use the parallel transport $\Pi_{\eta,t}$ along $\eta$ with respect to $\nabla$. So the curve is $t \mapsto \Pi_{\eta,t}^{-1}(T(\eta(t)))$. (Here $\eta$ denotes any smooth curve passing through $x$ at $t=0$ with velocity $v$, and $\Pi_{\eta,t}:T_x M\to T_{\eta(t)}M$)
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    $\begingroup$ I don't see any reason why a covariant derivative has to be computed using a geodesic. You get the same answer using any curve with tangent vector $V$. $\endgroup$
    – Deane Yang
    Commented Sep 12, 2011 at 16:18
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    $\begingroup$ And the explanation of Lie derivative is quite misleading, because it makes it seem like the Lie derivative depends only on $T$ along the flow line of $V$. In fact, it depends on how both $T$ and $V$ behave in a neighborhood of the flowline. If you really want to use only data along the flow line, then you need to know their first order jets along the curve. $\endgroup$
    – Deane Yang
    Commented Sep 12, 2011 at 17:27
  • $\begingroup$ @DeanYang: of course I agree the connection only depends on $v$, but he asked for an intuitive explanation, and I think the most intuitive meaning I can attach to the covariant derivative along a direction is: "directional derivative along the stright line (with respect to the connection, or metric if the connection is metric)". I agree you don't have to compute it along a geodesic, but it's what is it morally supposed to mean. $\endgroup$
    – Qfwfq
    Commented Sep 12, 2011 at 21:38
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    $\begingroup$ "I agree you don't have to compute it along a geodesic, but it's what is it morally supposed to mean": I don't really agree with the second half. It's just a directional derivative associated with the tangent vector. A geodesic works, but in this case plays no special role. So mentioning it is misleading. It is true that when you first learn about directional derivatives on $R^n$, you tend to define them in terms of straight lines. However, it is rather important in differential geometry to understand that straight lines are not special when computing or defining a directional derivative. $\endgroup$
    – Deane Yang
    Commented Sep 13, 2011 at 9:49
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    $\begingroup$ Looks better to me. $\endgroup$
    – Deane Yang
    Commented Oct 16, 2020 at 18:59

Lie derivate is the variation of the a tensor field at a point after deforming the space by the integral curve of a vector field.

Covariant derivative is a generalization of the directional derivative applied to the tensors of the any range in such a way that its result is a tensor and is expressed in the same form for any arbitrary coordinate system, for which it is necessary to define additional functions of connection, demanding that they be transformed in a certain way with respect to coordinate changes.

Exterior derivative is applied for antisymmetric covariant tensor or diferencial forms.


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