The exterior covariant derivative of vector-valued forms

A vector field $${w}$$ on $${M}$$ can be viewed as a vector-valued 0-form. As noted previously, the covariant derivative $${\nabla_{v}w}$$ is linear in $${v}$$ and depends only on its local value, and so can be viewed as a vector-valued 1-form $${\mathrm{D}\vec{w}(v)\equiv\nabla_{v}w}$$. $${\mathrm{D}\vec{w}}$$ is called the exterior covariant derivative of the vector-valued 0-form $${\vec{w}}$$. This definition is then extended to vector-valued $${k}$$-forms $${\vec{\varphi}}$$ by following the example of the exterior derivative $${\mathrm{d}}$$:

\begin{aligned} & \mathrm{D}\vec{\varphi}\left(v_{0},\dotsc,v_{k}\right)\\ & \equiv\underset{j=0}{\overset{k}{\sum}}\left(-1\right)^{j}\nabla_{v_{j}}\left(\vec{\varphi}\left(v_{0},\dotsc,v_{j-1},v_{j+1},\dotsc,v_{k}\right)\right)\\ & \phantom{{}=}+\underset{i<j}{\sum}\left(-1\right)^{i+j}\vec{\varphi}\left(\left[v_{i},v_{j}\right],v_{0},\dotsc,v_{i-1},v_{i+1},\dotsc,v_{j-1},v_{j+1},\dotsc,v_{k}\right) \end{aligned}

For example, if $${\vec{\varphi}}$$ is a vector-valued 1-form, we have

$$\displaystyle \mathrm{D}\vec{\varphi}\left(v,w\right)\equiv\nabla_{v}\vec{\varphi}\left(w\right)-\nabla_{w}\vec{\varphi}\left(v\right)-\vec{\varphi}\left(\left[v,w\right]\right).$$

So while the first term of $${\mathrm{d}\varphi}$$ takes the difference between the scalar values of $${\varphi(w)}$$ along $${v}$$, the first term of $${\mathrm{D}\vec{\varphi}}$$ takes the difference between the vector values of $${\vec{\varphi}(w)}$$ along $${v}$$ after parallel transporting them to the same point (which is required to compare them). At a point $${p}$$, $${\mathrm{D}\vec{\varphi}\left(v,w\right)}$$ can thus be viewed as the “sum of $${\vec{\varphi}}$$ on the boundary of the surface defined by its arguments after being parallel transported back to $${p}$$,” and if we use $${\Vert_{\varepsilon v}}$$ to denote parallel transport along an infinitesimal curve with tangent $${v}$$, we can write

\begin{aligned}\varepsilon^{2}\mathrm{D}\vec{\varphi}\left(v,w\right) & =\Vert_{-\varepsilon v}\vec{\varphi}\left(\varepsilon w\left|_{p+\varepsilon v}\right.\right)-\vec{\varphi}\left(\varepsilon w\left|_{p}\right.\right)\\ & -\Vert_{-\varepsilon w}\vec{\varphi}\left(\varepsilon v\left|_{p+\varepsilon w}\right.\right)+\vec{\varphi}\left(\varepsilon v\left|_{p}\right.\right)\\ & -\vec{\varphi}\left(\varepsilon^{2}\left[v,w\right]\right).\end{aligned}

The above depicts how the exterior covariant derivative $${\mathrm{D}\vec{\varphi}\left(v,w\right)}$$ sums the vectors $${\vec{\varphi}}$$ along the boundary of the surface defined by $${v}$$ and $${w}$$ by parallel transporting them to the same point. Note that the “completion of the parallelogram” $${[v,w]}$$ is already of order $${\varepsilon^{2}}$$, so its parallel transport has no effect to this order.

From its definition, it is clear that $${\mathrm{D}\vec{\varphi}}$$ is a frame-independent quantity. In terms of the connection, we must consider $${\vec{w}}$$ as a frame-dependent $${\mathbb{R}^{n}}$$-valued 0-form, so that

$$\displaystyle \mathrm{D}\vec{w}\left(v\right)=\nabla_{v}w=\mathrm{d}\vec{w}\left(v\right)+\check{\Gamma}\left(v\right)\vec{w}.$$

For a $${\mathbb{R}^{n}}$$-valued $${k}$$-form $${\vec{\varphi}}$$ we find that

$$\displaystyle \mathrm{D}\vec{\varphi}=\mathrm{d}\vec{\varphi}+\check{\Gamma}\wedge\vec{\varphi},$$

where the exterior derivative is defined to apply to the frame-dependent components, i.e. $${\mathrm{d}\vec{\varphi}(v_{0}\ldots v_{k})\equiv\mathrm{d}\varphi^{\mu}(v_{0}\ldots v_{k})e_{\mu}}$$. Recall that $${\check{\Gamma}}$$ is a $${gl(n,\mathbb{R})}$$-valued 1-form, so that for example if $${\vec{\varphi}}$$ is a $${\mathbb{R}^{n}}$$-valued 1-form then $${(\check{\Gamma}\wedge\vec{\varphi})\left(v,w\right)\equiv\check{\Gamma}\left(v\right)\vec{\varphi}\left(w\right)-\check{\Gamma}\left(w\right)\vec{\varphi}\left(v\right)=\Gamma^{\lambda}{}_{\mu}\left(v\right)\varphi^{\mu}\left(w\right)-\Gamma^{\lambda}{}_{\mu}\left(w\right)\varphi^{\mu}\left(v\right)}$$.

 Δ As with the covariant derivative, it is important to remember that $${\mathrm{D}\vec{\varphi}}$$ is frame-independent while $${\mathrm{d}\vec{\varphi}}$$ and $${\check{\Gamma}}$$ are not.

The set of vector-valued forms can be viewed as an infinite-dimensional algebra by defining multiplication via the vector field commutator; it turns out that $${\mathrm{D}}$$ does not satisfy the Leibniz rule in this algebra and so is not a derivation. However, following the above reasoning one can extend the definition of $${\mathrm{D}}$$ to the algebra of tensor-valued forms, or the subset of anti-symmetric tensor-valued forms; $${\mathrm{D}}$$ then is a derivation with respect to the tensor product in the former case and a graded derivation with respect to the exterior product in the latter case. We will not pursue either of these two generalizations.