I'm trying to show, using Frobenius’ theorem, that the integrability condition for the parallel transport equations $$ ∂_μX^a + Γ^a{}_{μb}X^b = 0 $$ is the vanishing of the Riemann curvature $$ R^a{}_{μνb} = ∂_μΓ^a{}_{νb} + Γ^a{}_{μc}Γ^c{}_{νb} - (μ↔︎ν) .$$ I want to do this using the formulation of Frobenius’ theorem in terms of tangent distributions and involutivity.
What I can do: It is easiest to use the differential forms formulation of Frobenius’ theorem: an exterior differential system $$ ω^a = \mathrm dX^a + Γ^a{}_bX^b $$ is integrable iff the ideal $I = ⟨ω^a⟩$ generated by the one-forms $ω^a$ is closed under exterior differentiation, $\mathrm dI ⊆ I$.
Observe: $$ \mathrm dω^a = \mathrm dΓ^a{}_bX^b - Γ^a{}_b ∧ \mathrm dX^b $$ Since $Γ^a{}_b ∧ ω^b ∈ I$ is in the ideal, we may add this to the two-form above to obtain $$ \mathrm dω^a \!\!\!\!\mod I ≡ (\mathrm dΓ^a{}_b + Γ^a{}_c∧Γ^c{}_b)X^b ,$$ and therefore $\mathrm dω^a \!\!\mod I ≡ 0 ⟺ \mathrm dω^a ∈ I$ exactly when $F^a{}_b ≔ \mathrm dΓ^a{}_b + Γ^a{}_c∧Γ^c{}_b$ — the ‘Riemann’ curvature — vanishes. (Is this general to arbitrary vector bundles?)
What I can’t do: The above is all well and good. However, I want to show the same thing using the “dual” Frobenius theorem formulated in terms of vectors: a tangent distribution $Δ$ (a.k.a., a tangent subbundle or plane field) admits an integral manifold at each point iff it is involutive, i.e., closed under the Lie bracket, $[Δ, Δ] ⊆ Δ$.
My attempt at formulating the parallel transport equations in this way goes like this:
Let Greek $μ$ denote coordinates of the base manifold and Latin $a,b$ denote linear coordinates of the vector bundle. A vector $X$ then has components $(X^μ, X^a)$, where $X^μ$ is the base point and $X^a$ are the vector components.
The tangent subbundle $Δ$ corresponding to the parallel transport equations may be described as $$ Δ|_X = \{(δX^μ, -Γ^a{}_{μb}X^bδX^μ) \mid ∀ \; δX^μ\} .$$ That is, if we think of $Δ|_X$ as encoding the “permitted motions” of a point $X$ in the bulk, then the vector components $X^a$ are constrained to move in response to movements of the base point $δX^μ$ according to the equations of parallel transport.
The thing to verify now is that, given two such sections $X, Y ∈ \operatorname{Γ}(Δ)$, we have $[X,Y] = Z$, where $Z ∈ \operatorname{Γ}(Δ)$ is also of the form above.
$$ [X, Y] = [δX^μ∂_μ -Γ^a{}_{μb}X^bδX^μ∂_a, \; δY^μ∂_μ -Γ^a{}_{μb}Y^bδY^μ∂_a] = \text{(mess)} \tag{1} $$
Question: This route seems much messier to take. Is computing the Lie bracket (1) the best way to show that $Δ$ is involutive? I can’t manage to put (mess) into a useful form. How can I show $[X, Y] ∈ \operatorname{Γ}(Δ)$?
Cheers,