In differential geometry, there are several notions of differentiation, namely:

  • Exterior Derivative, $d$
  • Covariant Derivative/Connection, $\nabla$
  • Lie Derivative, $\mathcal{L}$.

I have listed them in order of appearance in my education/in descending order of my understanding of them. Note, there may be others that I am yet to encounter.

Conceptually, I am not sure how these three notions fit together. Looking at their definitions, I can see that there is even some overlap between the collection of objects they can each act on. I am trying to get my head around why there are (at least) three different notions of differentiation. I suppose my confusion can be summarised by the following question.

What does each one do that the other two can't?

I don't just mean which objects can they act on that the other two can't, I would like a deeper explanation (if it exists, which I believe it does). In terms of their geometric intuition/interpretation, does it make sense that we need these different notions?

Note, I have put the reference request tag on this question because I would be interested to find some resources which have a discussion of these notions concurrently, as opposed to being presented as individual concepts.

Michael Albanese
  • 90,765
  • 20
  • 171
  • 423
  • 3
    As a quick note, Lie derivative and exterior derivative are related, see https://en.wikipedia.org/wiki/Lie_coalgebra – Alexei Averchenko Oct 08 '12 at 15:32
  • @AlexeiAverchenko I would appreciate if you expanded your comment in a sort of answer, if you can. This could be interesting. – Yuri Vyatkin Oct 09 '12 at 07:56
  • 10
    Just wait until you start working in G bundles and need to deal with the [exterior covariant derivative](http://en.wikipedia.org/wiki/Exterior_covariant_derivative) `:-p`. – Willie Wong Oct 09 '12 at 08:47
  • 2
    @WillieWong I work with vector bundles and have to deal frequently with the exterior covariant derivatives. Secretly, they are in the question, just one needs to expand the notions appropriately – Yuri Vyatkin Oct 09 '12 at 09:05
  • 2
    @WillieWong: Isn't the extension of a connection to a map on $p$-forms with values in the vector bundle an example of an exterior covariant derivative? – Michael Albanese Dec 22 '12 at 00:43
  • 4
    An (maybe) interesting remark in Jost's *Riemannian Geometry and Geometric Analysis* on a relation between exterior derivative $\mathrm{d}$ and connection $\nabla$ (particularly curvature $R$). Namely, a connection $\nabla$ is called flat, if its curvature satisfies $R=\mathrm{d}^\nabla\circ\mathrm{d}^\nabla=0$. Then, Jost remarks: "The exterior derivative $\mathrm{d}$ thus yields a flat connection on the trivial bundle $M\times\mathbb{R}$." – gofvonx Aug 19 '13 at 10:36

10 Answers10


Short answer:

  • the exterior derivative acts on differential forms;
  • the Lie derivative acts on any tensors and some other geometric objects (they have to be natural, e.g. a connection, see the paper of P. Petersen below);
  • both the exterior and the Lie derivatives don't require any additional geometric structure: they rely on the differential structure of the manifold;
  • the covariant derivative needs a choice of connection which sometimes (e.g. in a presence of a semi-Riemannian metric) can be made canonically;
  • there are relationships between these derivatives.

For a longer answer I would suggest the following selection of papers

  1. T. J. Willmore, The definition of Lie derivative
  2. R. Palais, A definition of the exterior derivative in terms of Lie derivatives
  3. P. Petersen, The Ricci and Bianchi Identities

Of course, there is a lot more to say.

Edit. I decided to extend my answer as I believe that there are some essential points which have not been discussed yet.
  1. An encyclopedic reference that treats all these derivatives concurrently at a modern level of generality is
    I.Kolar, P.W. Michor, J. Slovak, Natural Operations in Differential Geometry (Springer 1993), freely available online here.
    I would not even dare to summarize this resource since it has an abysmal deepness and all-round completeness, and indeed covers all the parts of the original question.
    Moreover, I believe that the bibliography list of this book contains almost any further relevant reference.
  2. As it has been already mentioned by many in this discussion, these operations are intimately related. It cannot be overemphasized that the most important feature that they all share is naturality (they commute with pullback, and this, in particular, makes them coordinate-free).
    See KMS cited above and its bibliography, and specifically the following references may be useful:
    R. Palais, Natural Operations on Differential Forms, e.g. here or here.
    C.L. Terng, Natural Vector Bundles and Natural Differential Operators, e.g. here
  3. It turns out that their naturality forces them to be unique if we impose on them some basic properties, such as $d \circ d = 0$ for the exterior derivative. One way to prove that and further references could be found in:
    D. Krupka, V. Mikolasova, On the uniqueness of some differential invariants: $d$, $[,]$, $\nabla $, see here.
    Also it is interesting that the Bianchi identities for the connection follow from the naturality and the property $d \circ d = 0$ for the exterior derivative, see
    Ph. Delanoe, On Bianchi identities, e.g. here.
  4. The reference list that I produce here is too far from being complete in any sense. I only would add one classical treatment that I personally used to comprehend some of the fundamental notions related to Lie derivatives (in particular, the Lie derivative of a connection!):
    K. Yano, The Theory Of Lie Derivatives And Its Applications, freely available here

Indeed, my comments are speculative and sparse. I wish if this question were answered by someone like P. Michor, to be honest :-)

Yuri Vyatkin
  • 10,211
  • 2
  • 33
  • 58
  • 5
    Just to add: they all agree on scalar functions, and the reverse of Palais' construction (the definition of Lie derivatives in terms of exterior derivative) is known as [Cartan's formula](http://unapologetic.wordpress.com/2011/07/26/cartans-formula/). – Willie Wong Oct 09 '12 at 08:43
  • @WillieWong Thank you, this is really a good point. – Yuri Vyatkin Oct 09 '12 at 09:01
  • Thanks for all of these references. I wish I could have split the bounty between Jesse and yourself. Instead I awarded him the bounty and have accepted your answer. – Michael Albanese Dec 25 '12 at 05:27
  • Are you sure about (2)? If there are *two* distinct connections on a manifold $\nabla_1$ and $\nabla_2$, and if $i:(M, \nabla_1) \to (M, \nabla_2)$ is the identity map, would you say that $\nabla_1 i^* T = i^* \nabla_2 T$ for any covariant tensor $T$? This might be true under some uniqueness assumptions regarding the connection (such as it being the Levi-Civita connection of some metric), but I doubt that it holds in general. – Alex M. Mar 12 '16 at 22:19
  • @AlexM. No, I would not say that. As I mentioned in the beginning, connections involve choices (and are abundant). Of course, the naturality of the Levi-Civita connection (which is actually is treated in the reference to (3)) is the best illustration of what I was intending to allude. Another good [example](http://math.stackexchange.com/q/956086/2002) would be the naturality of the pullback connection. Giving precise statements was not what I was aiming at in this answer, but I tried to provide references for the curious to dig further. – Yuri Vyatkin Mar 13 '16 at 08:42

Since I don't have the time to give a super-detailed answer, allow me to just summarize some things that others have said, adding some additional points in the process. Hopefully this will be at least somewhat helpful.

Basic differences:

  • The exterior derivative and Lie derivative are defined in terms of the structure of a smooth manifold. By contrast, the choice of a connection is an additional structure.

  • All three agree on smooth functions. However, they generalize differently:

  • The exterior derivative takes differential forms as inputs.

  • Connections take sections of a vector bundle (such as tensor fields) as inputs, and differentiation is done with respect to a vector field.

  • The Lie derivative takes tensor fields as inputs, and differentiation is done with respect to a vector field.

Exterior derivative: The main feature here is $d^2 = 0$.

To me, the exterior derivative is the differentation operator we need for Stokes' Theorem to make sense. The $d^2 = 0$ property is dual to saying that "the boundary of a boundary is empty," and is the very thing that makes de Rham cohomology work. I sometimes think about $d^2 = 0$ as describing the commutativity of second derivatives.

Connections: The main feature here is differentiation along curves.

A choice of connection allows us to define the derivative of a vector field (more generally, a section of a vector bundle) with respect to another vector field. From there, we can define the notion of a "covariant derivative" along curves.

Connections generalize the case of $\mathbb{R}^n$, where

$$\nabla_XY := X^i\frac{\partial Y}{\partial x^i}$$

Connections also let us define the concepts of "parallel transport" and "torsion." When a Riemannian metric is given, there is a canonical choice of connection (the Levi-Civita connection) which gives lots of geometric information. In particular, many classical formulas from the differential geometry of curves and surfaces can be phrased in terms of connections.

Lie derivative: The main feature here is the relationship with integral curves and flows, and the fact that $\mathscr{L}_XY = XY - YX$.

Like connections, the Lie derivative also defines a derivative of a vector field (more generally, tensor fields) with respect to another vector field. Intuitively, the Lie derivative $\mathscr{L}_XY$ is the instantaneous change of $Y$ along the integral curves defined by $X$. This intuition comes directly from the definition:

$$\mathscr{L}_XY|_p := \lim_{t \to 0}\frac{D\phi_{-t}(Y_{\phi_t(p)}) - Y_p}{t},$$ where $\phi$ is the flow of $X$.

However, unlike connections, Lie derivatives do not give a well-defined directional derivative of vector fields "along curves." The following problem from Lee's Riemannian Manifolds book illustrates this:

Problem 4-3: b) There exists a vector fields $V,W$ on $\mathbb{R}^2$ such that $V=W=\partial_1$ on the $x^1$-axis, but the Lie derivatives $\mathcal L_V(\partial_2)$ and $\mathcal L_W(\partial_2)$ are not equal on the $x^1$-axis.

Pretty much all of this can be found in Lee's "Smooth Manifolds" and "Riemannian Manifolds" books.

Physical Mathematics
  • 4,151
  • 1
  • 7
  • 26
Jesse Madnick
  • 29,652
  • 7
  • 90
  • 147
  • 1
    I do not think that "differentiation is done with respect to a vector field" demonstrates a difference: one can take the exterior derivative of an appropriate tensor field (differential form) with respect to a vector field, or even with respect to an appropriate tensor field, if one wishes. And conversely, one does not need to take the covariant derivative (with connection) with respect to a vector field, because it is defined point-wise, so can be viewed as mapping one tensor filed to another tensor field. – Alexey Jun 07 '13 at 13:35
  • 1
    I have never heard of the concept of an exterior derivative of a tensor field with respect to an appropriate tensor field. That sounds fascinating. Could you provide a link, reference, or definition? Your "conversely" point is a good one, though. I may consider editing to include a perspective on Ehresmann connections. And of course, you're welcome to write your own answer. – Jesse Madnick Jun 07 '13 at 20:56
  • I didn't mean anything deep, i just wanted to say that the differential form can be "contracted" with a vector field, for example using its first argument. – Alexey Jun 07 '13 at 21:18
  • But maybe you are right, because after the contraction it wouldn't be a "differentiation" anymore. – Alexey Jun 08 '13 at 06:27
  • 2
    @JesseMadnick What is the most general relation between $d$ and $\nabla$? In one source, I notice that if $\nabla$ is torsion free, $(d\omega)_{\mu_0\mu_1\cdots\mu_l}=(l+1)\nabla_{[\mu_0}\omega_{\mu_1\cdots\mu_l]}$ in component form with $\omega_{\mu_1\cdots\mu_l}$ a $l$-form. What if $\nabla$ is not torsion free? Thank you! – Drake Marquis Jul 23 '15 at 23:28

Let me focus on the difference between Lie derivatives and covariant derivatives. Suppose I have a manifold with a connection $\nabla$ and a point $p$ in the manifold. Let $v$ be a vector field on $M$ and take $\xi \in T_pM$. The point to stress is that $\xi$ is not a vector field (although in practice it is often a vector field evaluated at $p$). We can then obtain $\nabla_{\xi}v \in T_pM$. Thus, covariant derivatives let you take directional derivatives of vector fields.

Lie derivatives of vector fields cannot quite be interpreted this way. The symbol "$\mathcal{L}_{\xi} v$" is not defined because $\xi$ is not a vector field. However, if we take a vector field $X$, we can think of $\mathcal{L}_{X} v$ (which is a new vector field) as the derivative of $v$ as we walk along the integral curves of $X$.

To understand the difference between these two ideas, consider working in coordinates. The Lie derivative will take the form $(\mathcal{L}_{X} v)^a=X^b \partial_b v^a-v^b \partial_b X^a$. The second term indicates that the first derivative of $X^\mu$ is involved when differentiating along flow lines. There is no sensible way to differentiate $v$ in the direction of a vector unless the vector is a vector field or you have additional structure that tells you how to connect nearby tangent spaces (a connection). (The only obvious guess would be $X^b \partial_b v^a$ which gives different vectors in different coordinate systems.)

  • 381
  • 3
  • 3

I think there is an important point that has been overlooked in the above answers: The exterior derivative is the only linear natural operator in the list. This is explained with several variations the book by Kolar, Michor and Slovak cited in Yuri Viatkin's answer.

The Lie derivative is also natural under general diffeomorphisms but only as a bilinear operator, which takes one vector field and one section of a general vector bundle (for example a tensor field) as it's entries. In particular it is a bi-differential operator, so both the vector field and the other section are differentiated.

The covariant derivative initially is natural in a bilinear sense and under the the much smaller (finite dimensional instead of infinite dimensional and generically trivial) group of affine transformations. However the advantage here is that it is tensorial in the vector field entry and only the additional section is differentiated. This allows one to view it as a linear operator mapping sections of a vector bundle $E$ to sections of $T^*M\otimes E$ and in this form (having also given a connection on $TM$) is can be iterated to define higher order operators, which is not possible with either the Lie derivative or the exterior derivative.

Michael Albanese
  • 90,765
  • 20
  • 171
  • 423
Andreas Cap
  • 17,427
  • 16
  • 31

A very short answer:

In finite dimensions and at least in characteristic 0, the equation $$\operatorname{d} \omega(x, y) = y(\omega(x)) - x(\omega(y)) - \omega([x, y])$$ allows you to define $[-,-]: V \wedge V \to V$ starting from $\operatorname{d}: V^* \to V^* \wedge V^*$, and vice versa.

Furthermore, you can prove that conditions $[[x, y], z] + [[y, z], x] + [[z, x], y] = 0$ and $d^2 = 0$ are equivalent under this correspondence.

Now, unfortunately this doesn't apply to infinite-dimensional case that is general Lie derivative and exterior derivative, and I don't know how relevant this is to general Lie derivative and exterior derivative, but it's very nice to know this relation at least in purely algebraic case.


  • 1,732
  • 1
  • 15
  • 22
Alexei Averchenko
  • 7,581
  • 2
  • 31
  • 62

Yet another, slightly different approach below :

The problem of course is that we can't compare vectors at different points ( different tangent spaces ) of a manifold when we don't know how to transport vectors from one tangent space to another.
If we want to generalize the notion of a directional derivative to vector fields (and other geometrical objects) on a manifold we have to be able to transport a vector from tangent space to another nearby tangent space in which we can take the difference of two vectors and divide by some notion of a distance.
In absence of some other structure on the manifold the only simple way to do this is by using the concept of Lie dragging a vector field along a congruence. This way we arrive at the Lie derivative which is most fundamental.
The other derivatives are just special cases of the more fundamental Lie derivative.

Lie derivative :

Let $M$ be a manifold with no extra structure ( in other words : no metric or parallel transport ) defined on it.
The notion of a vector field can be viewed as a congruence of curves ($\gamma(\lambda_{\sigma})$) defined on $M$.
The notation : $\gamma(\lambda_{\sigma})$ means that every value chosen for $\sigma$ defines an individual parameter $\lambda_{\sigma} \in \mathbb{R}$ to be used on one-and-only-one of the (space-filling) family of curves $\gamma()$ that define the congruence. So choosing a value for $\sigma$ is the same as choosing an individual curve on the congruence with its parameter $\lambda_{\sigma}$ , the values for $\lambda_{\sigma}$ then range from some $a \in \mathbb{R} $ to some other $b \in \mathbb{R}$. Choosing a value for $\lambda_{\sigma}$ is choosing a point on the selected curve.
Every point $p \in M$ is on exactly one of these curves $\gamma(\lambda_{\sigma})$ , then $\frac{d}{d \lambda_{\sigma}}$ defines a vector in the tangent space $T_pM$ of every point $p \in M$ . Therefore a congruence defines a vector field.

Since a congruence provides us with a way to correlate every point $p \in M$ with a unique curve on the congruence together with a specific parameter value on that curve, this means that a congruence also gives us a way to map all points of $M$ to different points by adding a value $\Delta \nu$ to their corresponding parameter values on their corresponding curves.
This way we can also project curves on a congruence ( and therefore on $M$ ) to other curves a parameter distance $\Delta \nu$ away. This is Lie dragging.

Now if we have two congruences : $\gamma(\lambda_{\sigma})$ and $\alpha(\nu_{\beta})$ we are able to Lie drag $\gamma$ a parameter distance $\Delta \nu$ along the $\alpha$ congruence. This gives us a way to compare the original $\gamma(\lambda_{\sigma})$ curves to the ones that were dragged along $\alpha$ by a parameter distance $\Delta \nu$ (the latter we can call $\gamma(\lambda_{\sigma}*)$ ) at the same point $p \in M$ .
So now in $T_pM$ we have two vectors : $\frac{d}{d \lambda_{\sigma}}$ and $\frac{d}{d \lambda_{\sigma}*}$.
The Lie derivative is now defined as : $\mathcal{L}_{\frac{d}{d \nu}} \space \frac{d}{d \lambda} = \lim_{\Delta \nu \to 0} \frac{\frac{d}{d \lambda*} - \frac{d}{d \lambda}}{\Delta \nu}$ ( we see that $\Delta \nu $ now serves as a measure of distance for lack of anything else to work with ).

It turns out that this is equivalent to : $\mathcal{L}_{V} U = [V, U]$ .

Lie dragging
Covariant derivative :

From the above we conclude that we can only define the Lie derivative if we have two congruences at our disposal.
But what if we only have one congruence $\gamma(\lambda_{\sigma})$ and a single curve , (say $\alpha(\nu_0)$) at our disposal? How can we define a derivative of a congruence along a curve that resembles the Lie derivative ?
The obvious way would be to envision some new congruence (say $\alpha'(\nu_{\sigma})$) that contains $\alpha(\nu_0)$ as one of its integral curves and then define the covariant derivative to be the Lie derivative with respect to the vector field of that new congruence $\alpha'$ : $\mathcal{L}_{\frac{d}{d \nu}} \space \frac{d}{d \lambda} $ . If we now restrict the outcome to the points of the original curve $\alpha(\nu_0)$ we have a generalization of the directional derivative along that curve.

First attempt might be to define $\alpha'$ such that it is Lie dragged along $\gamma(\lambda_{\sigma})$. But the resulting Lie derivative would then be zero.
The notion of a symmetric connection $\nabla$ (an operator that defines parallel transport of $\frac{d}{d \nu}$ along $\frac{d}{d \lambda}$ by way of : $\nabla_{\frac{d}{d \lambda}} \frac{d}{d \nu} = 0 $) helps us out:

We imagine constructing a new congruence $\alpha'$ from the curve $\alpha(\nu_0)$ such that it parallel transports its own tangent vectors along the congruence $\gamma(\lambda_{\sigma})$ and also agrees on its parameter $\nu_0$ with the curve $\alpha(\nu_0)$ :
So : $\alpha'(\nu_{\beta})$ , with $\nabla_{\frac{d}{d \lambda}} \frac{d}{d \nu} = 0 $ and $\forall \enspace \nu_0 \in \mathbb{R} : \alpha'(\nu_{0})=\alpha(\nu_{0})$ .
In general for a symmetric connection $\nabla$ and two vector fields $U$ and $V$ we have : $ \nabla_U V - \nabla_V U = [U , V] = \mathcal{L}_{U} V $, this implies that the Lie derivative in the above case with respect to $\alpha'(\nu)$ should be : $\nabla_{\frac{d}{d \nu}} \frac{d}{d \lambda} - \nabla_{\frac{d}{d \lambda}} \frac{d}{d \nu} = \nabla_{\frac{d}{d \nu}} \frac{d}{d \lambda} - 0 = \nabla_{\frac{d}{d \nu}} \frac{d}{d \lambda} $ . Indeed this is the usual definition for the covariant derivative. So the covariant derivative along curve $\alpha(\nu_0)$ is the Lie derivative along the 'parallel transported' vector field constructed from $\alpha(\nu_0)$.

$ \nabla_{V} U = \left ( v^ju^i\Gamma^k_{ij} + v^j \frac{\partial u^k}{\partial x^j}\right ) \vec{e}_k $
Parallel transport $V$ along $U$ :
$ 0= \nabla_{U} V = \left ( u^jv^i\Gamma^k_{ij} + u^j \frac{\partial v^k}{\partial x^j}\right ) \vec{e}_k \implies u^jv^i\Gamma^k_{ij} = -u^j \frac{\partial v^k}{\partial x^j} $
$ \mathcal{L}_{V} U = \left [ V, U \right ] = \left ( v^j \frac{\partial u^k}{\partial x^j} - u^j \frac{\partial v^k}{\partial x^j} \right ) \vec{e}_k \implies \mathcal{L}_{V} U = \left ( v^j \frac{\partial u^k}{\partial x^j} + u^jv^i\Gamma^k_{ij} \right ) \vec{e}_k = \nabla_{V} U $

The above shows why this definition is the only logical choice for a generalization of the Lie derivative in the absence of a second congruence (so the only obvious generalization for differentiation along a curve instead of a congruence).

Exterior derivative :

For similarities and differences between Exterior derivative and Lie derivative see this clear explanation : https://en.wikipedia.org/w/index.php?title=Lie_derivative#The_Lie_derivative_of_a_differential_form

Rutger Moody
  • 2,291
  • 1
  • 14
  • 28
  • 2
    I like this idea of starting with congruences, rather than vector fields, and going from there: it has a more intuitive "generalised coordinate" geometric feel to it than the usual rather algebraic definitions of the derivatives. Do you know a nice introductory text that goes into these ideas, and this exposition, in more detail? – Chappers Jul 24 '17 at 22:39
  • 2
    @Chappers : thx , this is a very nice introductory text : ' Schutz, Bernard - Geometrical methods of mathematical physics. Cambridge university press 1980. ' – Rutger Moody Jul 25 '17 at 07:15
  • Ah, thank you. I'll have a look at it. – Chappers Jul 25 '17 at 11:59
  • toward the end, the coefficient $\Gamma^k_{ij}$ suddenly appears. How do we connect this to what was said above? – Alex Jan 20 '22 at 05:10
  • @Alex The $\Gamma^k_{ij}$ part is just the coordinate dependent notation for the coordinate free formulation above. You can skip it if you want. $\Gamma^k_{ij}$ are the Christoffel symbols. For covariant derivative in terms of Christoffel symbols you can read here : https://en.wikipedia.org/wiki/Christoffel_symbols – Rutger Moody Jan 20 '22 at 09:22
  • Thank you. In vector bundle, this would be interpreted as a vector-valued one-form. Why is this a vector-valued one-form? – Alex Jan 20 '22 at 11:27
  • @Alex : I don't think the Christoffel symbols have anything to do with the $\Gamma$ that denotes smooth sections on a bundle. – Rutger Moody Jan 20 '22 at 12:07
  • I meant that in a vector bundle, the connection is a vector-valued one-form; for a principal bundle, "gauge potential" $A$ is a $\mathcal{g}$-valued one-form. Are they analogous to the Christoffel symbols? Christoffel symbols are not tensors; is $A$ a tensor? – Alex Jan 20 '22 at 12:12
  • @Alex: The A in gauge theory is a vector while Christoffel symbols are not. There is a similarity between the two connections. But as far as I understand the A in gauge theory is to be understood as residing in some extra dimension on top of our 4-dimensional space. Don't know much about this. Maybe good idea to post a new question about the similarities and differences between the two? – Rutger Moody Jan 20 '22 at 12:54
  • @Alex : Otherwise this : https://en.wikipedia.org/wiki/Gauge_covariant_derivative gives a rather good description it seems to me.. – Rutger Moody Jan 20 '22 at 13:00
  • @RutgerMoodyR Thank you for the explanations and the link! – Alex Jan 21 '22 at 02:03
  • @Alex PS: "Peskin and Schroder, An Introduction to Quantum Field Theory . Paragraph 15.1 The Geometry of Gauge Invariance". This also gives you a very good explanation regarding the differences between normal covariant derivative and gauge covariant derivative. – Rutger Moody Jan 21 '22 at 10:11
  • Thank you for the reference. I studied the book before. 15.1 gave a "physicists' view” on gauge theories, but did not mention much on the rigorous mathematical background. In the end, I realized that affine connections and gauge connections are two related but different concepts. – Alex Jan 31 '22 at 15:00

1) The vector derivative, $\partial$

In geometric calculus, one deals in not just vector fields but multivector fields--fields that associated oriented planes, volumes, or other types of primitives to each point. These multivector fields are differentiated by an operator denoted $\partial$. It can act on multivector fields in either of two ways. On a multivector field $A(r)$, it can act as $\partial \wedge A$, which is the familiar exterior derivative. This increases the grades of all components of the field by one--vectors become planes, planes become volumes, and so on.

But there is another derivative, denoted $\partial \cdot A$, which goes by various names: interior derivative, codifferential, and so on. Both these notions of differentiation arise from $\partial$, however. It is, in my opinion, foolish that differential forms treats the $\partial \cdot$ operation as somehow only expressible in terms of $\partial \wedge$, however. To me (and in GC) they are on equal footing with one another.

2) The covariant derivative, $\nabla$

Now, introduce a global rotation field called $\underline R(a; r)$, which acts linearly on the vector $a$ and is a function of position $r$. For brevity, we'll just call this $\underline R(a)$ in most cases. We can, at our discretion, use or set this rotation field to our liking, perhaps because it is convenient, perhaps because it is necessary--you can regard it as inherent to the space if you like.

We can then look at the transformation of $A \mapsto A' = \underline R(A)$. This naturally changes the way we must differentiate. See that

$$a \cdot \partial A' = \underline R(a \cdot \partial A) + (a \cdot \dot \partial) \dot{\underline{R}}(A)$$

This is just a fancy product rule, with the overdot saying we differentiate only the linear operator, not its argument.

We define the covariant derivative to get rid of the messy second term on the right-hand side. That is,

$$a \cdot \nabla A' = \underline R(a \cdot \nabla A)$$

Introducing or changing the rotation field changes the covariant derivative. This gives a way to talk about differentiation regardless of the current rotation field $\underline R$. Changing the rotation field can be beneficial to alter the geometry of the space in a way that is convenient. Thus, the rotation field represents generalized, position-dependent rotational degrees of freedom to rotate fields at all points in space by varying amounts and orientations at will. The covariant derivative allows us to do this and still recover results that are independent of the choice of rotation field--of the choice of gauge.

3) The Lie derivative

In GC, the Lie derivative has no special symbol. Rather, it can be built from covariant derivatives. Consider two vector fields $A, B$. The Lie derivative is simply

$$\mathcal L_A B = A \cdot \nabla B - B \cdot \nabla A$$

I'm not as familiar with Lie derivatives, but I'm given to understand that if $B$ were transported along a "flow" generated by $A$, this quantity would measure how much $B$ maintains its value during the process.

  • 18,790
  • 1
  • 23
  • 56

Here's an answer that's more abstract, and a bit different in spirit from the others. Not sure if you'll find it useful, but hopefully someone will. This approach is explained in chapter 5 of Kock's excellent text Synthetic Geometry of Manifolds, if you want more details.

In synthetic differential geometry (SDG), we work in a category of smooth spaces where infinitesimal objects exist. This is convenient because a lot of differential geometry (such as Lie's ideas) were originally conceived in terms of infinitesimals. But before you worry about how this applies to classical differential geometry, where we have (somewhat hastily) banished infinitesimals, note that the category of smooth manifolds has a fully faithful embedding into the Dubuc topos, which is a well-adapted model for SDG. So we can use SDG to reason about classical manifolds.

Given a manifold $M$ in SDG, we have the the 1-jet groupoid $\Pi^{(1)}(M) \rightrightarrows M$ of 1-jets of local diffeomorphisms. We also have the graph $M_{(1)} \rightrightarrows M$, which is the first order infinitesimal neighbourhood of the diagonal. That is, the elements of $M_{(1)}$ are pairs of infinitesimally close points (to first order). This is a reflexive symmetric graph: the symmetry is the involution $M_{(1)} \to M_{(1)}$ that interchanges the source and target of each edge, and reflexivity is the diagonal map $M \to M_{(1)}$. A morphism of reflexive symmetric graphs is a morphism which is compatible with the reflexivity and symmetry in the obvious way.

Given a vector bundle $E \to M$, we have the smooth groupoid $GL(E) \rightrightarrows M$ of linear isomorphisms between fibres in $E$.

From the point of view of representation theory and synthetic differential geometry, the Lie derivative of sections of a vector bundle $E \to M$ comes from a groupoid representation of the 1-jet groupoid $\Pi^{(1)}(M) \rightrightarrows M$ on the groupoid $GL(E) \rightrightarrows M $. Thus, we have a canonical Lie derivative of sections of any natural bundle.

An (infinitesimal) linear connection in $E$ is a choice of representation of the reflexive symmetric graph $M_{(1)} \rightrightarrows M$ on $GL(E) \rightrightarrows M$ (thought of as a reflexive symmetric graph in the obvious way). There is no canonical choice of such a representation on bundles over $M$, natural or not, unless they are trivial. In the case of a trivial bundle $M \times V \to M$ case we can just take the trivial representation of $M_{(1)} \rightrightarrows M$ which identifies all of the fibres, and we get the exterior derivative!

  • 9,712
  • 3
  • 25
  • 69

Here is my humble take on this very interesting question. I think of the exterior differentiation $d$ as some kind of infinitesimal Stoke's theorem, so to speak. Namely, if $\alpha$ is a $k$-form, then $d\alpha$ is of course a $k+1$-form, and it can be defined at some point $p$, in some directions determining a $k+1$ dimensional subspace of the tangent space $T_p$, by essentially taking a small $k+1$ dimensional closed ball, and associating to it the integral of $\alpha$ on the boundary $k$-dimensional sphere, and then normalizing this integral properly and then letting the radius of the ball go to $0$. This is similar to what one sees in some Mathematics for Engineering books actually, albeit in some special cases (divergence, curl etc.).

I think of a covariant derivative mathematically using parallel transport and holonomy, and physically, for a connection on some unitary vector bundle, as some kind of potential (similar to the electromagnetic potential, but of course, it could be more complicated).

I think of the Lie derivative along $X$ as some kind of derivative along the (local in "time") flow induced by $X$. I think of $[X,Y]$ at some point $p$ as the vector going from $p$ to the point obtained by flowing along $X$ for some time $t$, then $Y$ for some time $s$, then $-X$ for some time $t$, then $-Y$ for some time $s$. For this to be made rigorous, one has to divide by $st$, and then take the limit as $s$ and $t$ go to $0$. I learned about this intuitive picture while reading Sir Roger Penrose's "Road to Reality" (which I enjoyed reading).

Philosophically speaking, I like to think of these derivatives using the most intuitively appealing (for me) definitions, which are usually close to the historical ones. However, when it comes to computing them, of course, this is almost never the easiest approach: then one can resort to formulas such as Cartan's magic formula $L_X = d \circ \iota_X + \iota_X \circ d$, and so on.

  • 4,912
  • 1
  • 10
  • 22
  • how to view connections as vector-valued one-forms? – Alex Jan 20 '22 at 05:17
  • @Alex. Suppose you have a specific connection, say $D$, on a smooth vector bundle $E$ over a smooth manifold $B$. Then if $\alpha$ is a $1$-form on $B$ with values in $End(E)$ (so in some sense a matrix-valued $1$-form rather than a vector-valued $1$-form), then $D + \alpha$ is also a connection on $E$. Conversely, the difference of two connections on $E$ is a $1$-form on $B$ with values in $End(E)$. – Malkoun Jan 20 '22 at 12:27
  • There is a special case though which may be what you are interested in. If $(B, g)$ is a smooth Riemannian $3$-manifold, then a connection $D$ on $B$ (i.e. on $T(B)$) is said to be compatible with $g$ if $Dg = 0$. Then the difference of two connections on $B$ which are compatible with $g$ is a $1$-form on $B$ with values that are essentially skew-symmetric endomorphisms of the tangent bundle. But in this special case, any skew-symmetric endomorphism of $\mathbb{R}^3$ can be thought of as $v \times -$, for some $v \in \mathbb{R}^3$. – Malkoun Jan 20 '22 at 12:37
  • so in the special case in my previous comment, the difference of two connections on $(B, g)$ which are both compatible with $g$ can be thought of as a "vector"-valued $1$-form on $B$. Perhaps this is what you are interested in? – Malkoun Jan 20 '22 at 12:40
  • Thank you for the detailed explanations. I think I was confused on a different path: we say connections $\nabla$ on vector bundles are vector-valued one-forms. Then, on principal bundles there is the "gauge potential" $A$ which are Lie-algebra-valued one forms. In gauge theory $D\psi = \partial\psi + A\psi$. I thought $\nabla$ is the equivalent of $D$ but why we say $\nabla$ is vector-valued one-form and $A$ not $D$ is the vector-valued one form? – Alex Jan 21 '22 at 02:48
  • @Alex, some people use the mathematical language not very precisely. You have to be more careful. A connection on a smooth manifold is not itself an $End(T)$ valued $1$-form, but the difference of any two such connections is an $End(T)$ valued $1$-form. And in the case of connections on a bundle $E$, using the expression "vector-valued 1-form" is confusing in my opinion, since the relevant such object is the difference of two connections, and it is a $1$-form with values in $End(E)$, with $E$ being the vector bundle (if you are learning from physicists, they are not all always very careful). – Malkoun Jan 21 '22 at 03:55
  • Over flat space, an operator of the form $D = \partial + A$ where $A$ is a $1$-form with values in the corresponding Lie algebra is called a connection to a mathematician. To a physicist, they may write $D = \partial + ieA$ and call $A$ a gauge potential ($i$ is to make $A$ hermitian, for physicists, and $e$ is a coupling constant). What both mathematicians and physicists denote by $F$ is called curvature by mathematicians and field strength by physicists. For example, $F$ could be the curvature of a connection on a $U(1)$-bundle over Minkowski, which may be interpreted as an EM/Maxwell field. – Malkoun Jan 21 '22 at 04:06

I would like to make a remark. In a torsionless manifold, the link between these derivatives may be found in the (very good) reference mentionned by Yuri Vyatkin (book of Yano, 1955). Another point is very interesting for practical use of Lie derivative in the same reference : the index convention for the covariant derivative may lead to some errors when using Lie derivative of tensors in a manifold with torsion and curvature.