Jan 29, 2025 6:16 AM

Exterior covariant derivative

The exterior derivative $d$ acts on a regular function or differential form to measure how it "changes" in space. For example:

If $f$ is a smooth function on a manifold $M$ , then:
$d f (v) = v (f),$
where $v$ is a vector field. This means $d f$ is the 1-form that takes a vector field $v$ and outputs the derivative of $f$ in the direction of $v$ .
For higher-order forms (like 1-forms, 2-forms, etc.), $d$ generalizes to build forms of one higher degree while encoding their "variation."
If we have a vector bundle $E$ over $M$ , a covariant derivative $D$ or connection extends this idea of "differentiating" functions to differentiating sections $s$ of $E$ . The exterior covariant derivative $d_{D}$ combines the ideas of $d$ and $D$ to define differentiation for $E$ -valued differential forms (forms with values in the bundle $E$ ).
Suppose $s$ is a section of $E$ (a "0-form" with values in $E$ ). Then $d_{D} s$ is an $E$ -valued 1-form defined by:

(d_{D} s) (v) = D_{v} s,

for any vector field $v$ . This is analogous to $d f (v) = v (f)$ . In a local chart, $d_{D} s = D_{μ} s \otimes d x^{μ}$ .

For higher forms, $d_{D}$ generalizes by using the wedge product $\land$ and the ordinary $d$ , as in:

d_{D} (s \otimes ω) = (d_{D} s) \land ω + s \otimes d ω,

where $ω$ is an ordinary differential form.

For example, given an $E$ -valued 1-form $ω$ , written in a local chart as $ω = s_{μ} \otimes d x^{μ}$ , we have

d_{D} ω = d_{D} (s_{μ}) \land d x^{μ} = D_{ν} s_{μ} d x^{ν} \land d x^{μ} .

Property

See @baez1994gauge page 251.
The exterior covariant derivative satisfies that $d_{D}^{2}$ is proportional to the curvature of $D$ .

Old stuff

I think what follows is wrong or it refers to another concept.

Schuller_2013 page 197
Olver_2014 page 31
Olver_1995 page 123-124

Idea: when we have a decomposition of the cotangent bundle in horizontal and vertical subspaces, we can use the usual exterior derivative operator $d$ and then project the result on the horizontal subspace. This leads to a new operator $d_{H}$ .

I think that it has to do with the typical situation of a function with parameters (like in high school). Consider $f (x) = a x^{3} + x^{2} - 3 x + 1$ . We have variables $x$ and $a$ , but they are not of the same "status". Students ask why we derive $x$ but not $a$ ...

Even in function of several variables

f (x, y) = a x^{2} + y^{2}

they don't have the same "status": $x$ and $y$ are more "variables" and $a$ is more fixed.
I think that the exterior covariant derivative is an "external device" that we introduce in order to specify this distinction. Indeed I guess that the connection itself has this goal...