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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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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). –  Willie Wong Sep 12 '11 at 14:01
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you'll find an extensive discussion, with many pointers to the literature at physicsforums.com/showthread.php?t=150200 –  Carlo Beenakker Sep 12 '11 at 14:12
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4 Answers 4

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.

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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. –  Deane Yang Sep 12 '11 at 15:19
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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.

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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. –  Deane Yang Sep 12 '11 at 19:53
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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). –  Xiao Xinli Sep 12 '11 at 20:20
    
Yes, that's a good clarification of what you wrote. –  Deane Yang Sep 12 '11 at 20:30
    
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? –  Qfwfq Sep 12 '11 at 21:50
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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. 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

$$T_\gamma u_0:= \boldsymbol{u}(1).$$

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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 $V$.

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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 $V$. –  Deane Yang Sep 12 '11 at 16:18
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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. –  Deane Yang Sep 12 '11 at 17:27
    
@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. –  Qfwfq Sep 12 '11 at 21:38
    
@DeanYang: your second remark make me think my interpretation of the Lie derivative is not quite correct as stated. Is there a way (given $T$ and $V$ in a neighborhood of $x$) to compute the Lie derivative $\mathcal{L}_VT$ within a fixed finite dimensional vector space (depending on $x$)? –  Qfwfq Sep 12 '11 at 21:45
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"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. –  Deane Yang Sep 13 '11 at 9:49
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