Differential Geometry – Meaning of Derivatives of Vector Fields

Question

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?

Answer 1

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:

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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."

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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.

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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.