TITLE: Exterior Derivative vs. Covariant Derivative vs. Lie Derivative
QUESTION [213 upvotes]: 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.

REPLY [12 votes]: 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 : $\mathcal{L}_{V} U = [V, U]$ . 



 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