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I have a doubt about the real meaning of the derivative of a vector field. This question seems silly at first but the doubt came when I was studying the definition of tangent space.

If I understood well a vector is a directional derivative operator, i.e.: a vector is an operator that can produce derivatives of scalar fields. If that's the case then a vector acts on a scalar field and tells me how the field changes on that point.

However, if a vector is a derivative operator, a vector field defines a different derivative operator at each point. So differentiate a vector would be differentiate a derivate operator, and that seems strange to me at first. I thought for example that the total derivative of a vector field would produce rates of change of the field, but my studies led me to a different approach, where the total derivative produces rates of change only for scalar fields and for vector fields it produces the pushforward.

So, what's the real meaning of differentiating a vector field knowing all of this?

Thanks very much in advance.

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I can't really tell your background, the correct answer is that on an abstract manifold, your derivative is undefined, the item that defines these is called a connection, the best known is: I suppose m first specific question is, do you know anything about the Lie bracket? – Will Jagy Sep 13 '12 at 1:23
@WillJagy, unfortunately I haven't come yet to study the Lie bracket, but I've hear about that a lot of times. This Lie bracket has something to do with the meaning of differentiation of vector fields? Thanks and sorry if I've said something silly. – user1620696 Sep 13 '12 at 1:33
I see. Your best bet for general constructions for submanifolds of $\mathbb R^n$ is a book called Elementary Topics in Differential Geometry by John A. Thorpe. I don't know of any other book that goes so very far with the necessary constructions while staying in Euclidean space, all the time allowing arbitrary dimension (but codimension 1). – Will Jagy Sep 13 '12 at 1:36
@WillJagy, thanks for recommending the book, I think it'll be of great help. Thanks again. – user1620696 Sep 13 '12 at 1:39
A standard example is on the sphere. Take a vector field of constant length tangent to the equator. As you surely know from physics, its derivative should be a normal vector field, which, however, is not tangent to the sphere anymore. So this is something you cannot do thoughtlessly in an abstract manifold, where you cannot leave the intrinsic point of view. – Giuseppe Negro Sep 13 '12 at 1:52
up vote 7 down vote accepted

As I understand it, these are your questions:

  • How does one define the derivative of a vector field? Do we just take the "derivatives" of each vector in the field? If so, what does it mean to take the derivative of a differential operator, anyway?
  • Why does the total derivative of a scalar field give information about rates of change, while the "total derivative" of a vector field gives the pushforward (which doesn't seem to relate to rates of change)?

I think the best way to answer these questions is to provide a broader context:

In calculus, we ask how to find derivatives of functions $F\colon \mathbb{R}^m \to \mathbb{R}^n$. The typical answer is the total derivative $DF\colon \mathbb{R}^m \to L(\mathbb{R}^m, \mathbb{R}^n)$, which assigns to each point $p \in \mathbb{R}^m$ a linear map $D_pF \in L(\mathbb{R}^m, \mathbb{R}^n)$. With respect to to the standard bases, this linear map can be represented as a matrix: $$D_pF = \begin{pmatrix} \left.\frac{\partial F^1}{\partial x^1}\right|_p & \cdots & \left.\frac{\partial F^1}{\partial x^m}\right|_p \\ \vdots & & \vdots \\ \left.\frac{\partial F^n}{\partial x^1}\right|_p & \cdots & \left.\frac{\partial F^n}{\partial x^m}\right|_p \end{pmatrix}$$

Personally, I think this encodes the idea of "rate of change" very well. (Just look at all those partial derivatives!)

Let's now specialize to the case $m = n$. Psychologically, how does one intuit these functions $F\colon \mathbb{R}^n \to \mathbb{R}^n$? There are two usual answers:

(1) We intuit $F\colon \mathbb{R}^n \to \mathbb{R}^n$ as a map between two different spaces. Points from the domain space get sent to points in the codomain space.

(2) We intuit $F\colon \mathbb{R}^n \to \mathbb{R}^n$ as a vector field. Every point in $\mathbb{R}^n$ is assigned an arrow in $\mathbb{R}^n$.

This distinction is important. When we generalize from $\mathbb{R}^n$ to abstract manifolds, these two ideas will take on different forms. Consequently, this means that we will end up with different concepts of "derivative."

In case (1), the maps $F\colon \mathbb{R}^m \to \mathbb{R}^n$ generalize to smooth maps between manifolds $F \colon M \to N$. In this setting, the concept of "total derivative" generalizes nicely to "pushforward." That is, it makes sense to talk about the pushforward of a smooth map $F \colon M \to N$.

But you asked about vector fields, which brings us to case (2). In this case, we first have to be careful about what we mean by "vector" and "vector field."

A vector $v_p \in T_pM$ at a point $p$ is (as you say) a directional derivative operator at the point $p$. This means that $v_p$ inputs a scalar field $f\colon M \to \mathbb{R}$ and outputs a real number $v_p(f) \in \mathbb{R}$.

A vector field $v$ on $M$ is a map which associates to each point $p \in M$ a vector $v_p \in T_pM$. This means that a vector field defines a derivative operator at each point.

Therefore: a vector field $v$ can be regarded as an operator which inputs scalar fields $f\colon M \to \mathbb{R}$ and outputs scalar fields $v(f)\colon M \to \mathbb{R}$.

In this setting, it no longer makes sense to talk about the "total derivative" of a vector field. You've said it yourself: what would it even mean to talk about "derivatives" of vectors, anyway? This doesn't make sense, so we'll need to go a different route.

In differential geometry, there are two ways of talking about the derivative of a vector field with respect to another vector field:

  • Connections (usually denoted $\nabla_wv$ or $D_wv$)
  • Lie derivatives (usually denoted $\mathcal{L}_wv$ or $[w,v]$)

Intuitively, these notions capture the idea of "infinitesimal rate of change of a vector field $v$ in the direction of a vector field $w$."

Question: What do these constructions look like in $\mathbb{R}^n$?

Taking advantage of the fact that we're in $\mathbb{R}^n$, we can look at our vector fields in the calculus way: as functions $v\colon \mathbb{R}^n \to \mathbb{R}^n$. As such, we can write the components as $v = (v^1,\ldots, v^n)$.

The (Levi-Civita) connection of $v$ with respect to $w$ is defined as $$\nabla_wv = (w(v^1), \ldots, w(v^n)),$$ where $$w(v^i) := w^1\frac{\partial v^i}{\partial x^1} + \ldots + w^n\frac{\partial v^i}{\partial x^n}.$$

The Lie derivative of $v$ with respect to $w$ has a technical definition in terms of flows that I don't want to go into, but the bottom line is that it's similar to Rod Carvalho's answer.

Also, in $\mathbb{R}^n$ we have the pleasant formula

$$\mathcal{L}_wv = \nabla_wv - \nabla_vw,$$

which aids in computation.

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Note: The notation for "total derivatives" varies considerably. I used $D_pF$ above, but other people use $Df_p$ or $d_pf$ or $df_p$. Also, some people refer to "total derivatives" as "differentials" or "total differentials." These are all the same. – Jesse Madnick Sep 13 '12 at 3:50
Nice! Finally, an answer that addresses the question precisely. – Giuseppe Negro Sep 13 '12 at 8:34
@JesseMadnick, thanks a lot for your answer! You really gave a precise and complete answer to my question which I'm certain will help any others with this doubt who search here in the website. Thanks a lot again for your aid. – user1620696 Sep 13 '12 at 18:07

Let $\mathbb{v} : \mathbb{R}^n \to \mathbb{R}^n$ be a vector field, and let $\varphi : \mathbb{R}^n \to \mathbb{R}$ be a scalar field. Suppose that we would like to obtain the directional derivative of $\varphi$ at every $x$ in the direction of $\mathbb{v} (x)$, which is the following

$$(D_{\mathbb{v}} \varphi) (x) := \displaystyle\lim_{t \rightarrow 0^+} \frac{\varphi (x + t \mathbb{v} (x)) - \varphi (x)}{t} = \langle \nabla \varphi (x), \mathbb{v} (x) \rangle$$

This is the Lie derivative of $\varphi$ along $\mathbb{v}$. It's widely used in control theory, namely, in the study of Lyapunov stability of dynamical systems. If the vector field $\mathrm{v}$ is the gradient of a scalar field $\psi : \mathbb{R}^n \to \mathbb{R}$, then the Lie derivative of $\varphi$ along $\mathrm{v} (x) := \nabla \psi (x)$ is given by

$$(D_{\mathbb{v}} \varphi) (x) = \langle \nabla \varphi (x), \mathbb{v} (x) \rangle = \langle \nabla \varphi (x), \nabla \psi (x) \rangle$$

Has this answered, even if remotely, your question?

Update: Since my original post did not answer the OP's question, I will add this update. Let $\mathbb{u}, \mathbb{v} : \mathbb{R}^n \to \mathbb{R}^n$ be vector fields. Let $\mathbb{u}_i$ be the $i$-th component of $\mathbb{u}$, and note that $\mathbb{u}_i$ is a scalar field. We can compute the Lie derivative of $\mathbb{u}_i$ along $\mathbb{v}$, which is the scalar function

$$(D_{\mathbb{v}} \mathbb{u}_i) (x) = \langle \nabla \mathbb{u}_i (x), \mathbb{v} (x) \rangle$$

We could define the Lie derivative of vector field $\mathbb{u}$ along the vector field $\mathbb{v}$ as follows

$$(D_{\mathbb{v}} \mathbb{u}) (x) := \left[\begin{array}{c} (D_{\mathbb{v}} \mathbb{u}_1) (x)\\ (D_{\mathbb{v}} \mathbb{u}_2) (x)\\ \vdots \\ (D_{\mathbb{v}} \mathbb{u}_n) (x)\end{array}\right]$$

Finally, do note that $(D_{\mathbb{v}} \mathbb{u}) (x) = ((D \mathbb{u}) (x)) \, \mathbb{v} (x)$, where $(D \mathbb{u})$ is the Jacobian of $\mathbb{u}$.

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Rod Carvalho, thanks for your answer. From what you said I understood that I indeed can use a vector field as a derivative operator that varies from point to point, but my doubt has to do with the meaning of differentiating a vector field. For instance, if $v: \mathbb{R}^n \to \mathbb{R}^n$ is a vector field, then $v$ is indeed a derivative operator that varies from point to point. What is then the meaning of differentiating $v$ in the direction of another vector? I'm not sure I'm making myself clear. Sorry if I've said something very silly. – user1620696 Sep 13 '12 at 1:37
I'm afraid this is not an answer to the original question. The OP asks for the derivative of a vector field, not of a scalar field. – Giuseppe Negro Sep 13 '12 at 1:47
@user1620696: I added an update. Please let me know if it makes any sense. – Rod Carvalho Sep 13 '12 at 2:26

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