Torsion on the tangent frame bundle

The covariant derivative of the solder form defines the torsion on $${P}$$

\displaystyle \begin{aligned}\vec{T}_{P} & \equiv\mathrm{D}\vec{\theta}_{P}\\ & =\mathrm{d}\vec{\theta}_{P}+\check{\Gamma}_{P}\wedge\vec{\theta}_{P}. \end{aligned}

$${\vec{T}_{P}}$$ is a horizontal equivariant form since $${\vec{\theta}_{P}}$$ is. Examining the first few components, we have:

\displaystyle \begin{aligned}\vec{T}_{P}\left(v,w\right) & =\mathrm{d}\vec{\theta}_{P}\left(v^{⦵},w^{⦵}\right)\\ \Rightarrow\varepsilon^{2}\vec{T}_{P}\left(v,w\right) & =\vec{\theta}_{P}\left(\varepsilon w^{⦵}\left|_{p+\varepsilon v^{⦵}}\right.\right)-\vec{\theta}_{P}\left(\varepsilon w^{⦵}\left|_{p}\right.\right)-\ldots\\ & =\mathrm{d}\pi\left(\varepsilon w^{⦵}\left|_{p+\varepsilon v^{⦵}}\right.\right)_{_{p+\varepsilon v^{⦵}}}^{\mu}-\mathrm{d}\pi\left(\varepsilon w^{⦵}\left|_{p}\right.\right)_{p}^{\mu}-\ldots \end{aligned}

The first term projects the horizontal component of $${w}$$ at $${p+\varepsilon v^{⦵}}$$ down to $${M}$$, which is the same as projecting $${w}$$ itself down to $${M}$$ since the projection of the vertical part vanishes. Then we take its components in the basis at $${p+\varepsilon v^{⦵}}$$, which is the parallel transport of the basis at $${p}$$ in the direction $${v}$$. These are the same components as that of the projection of $${w}$$ at $${p+\varepsilon v^{⦵}}$$ parallel transported back to $${p}$$ in the basis at $${p}$$. Thus the torsion on $${P}$$ is the “sum of the boundary vectors of the surface defined by the projection of its arguments down to $${M}$$ after being parallel transported back to $${p}$$.”

This analysis makes it clear that the pullback of the torsion on $${P}$$ by the identity section

$$\displaystyle \vec{T}_{i}\equiv\sigma_{i}^{*}\vec{T}_{P},$$

which by our previous pullback results recovers the torsion on $${M}$$, just bounces the argument vectors to the identity section and back, thus yielding the same interpretation for torsion that we arrived at for manifolds.

It can also be shown that the analog of the first Bianchi identity on $${M}$$ holds on $${P}$$, with the original being recovered upon pulling back by the identity section:

\displaystyle \begin{aligned}\mathrm{D}\vec{T}_{P} & =\check{R}_{P}\wedge\vec{\theta}_{P}\\ \mathrm{D}\vec{T}_{i} & =\check{R}_{i}\wedge\vec{\theta}_{i} \end{aligned}