# Geometric interpretation of the second Bianchi identity?

Abstract index expressions can be difficult to interpret geometrically, and in general can be easily misinterpreted. For example if we drop the torsion-free assumption, the commutator of covariant derivatives applied to a vector yields curvature and torsion

$$[\nabla_{a},\nabla_{b}]w^{c}=R^{c}{}_{dab}w^{d}-T^{d}{}_{ab}\nabla_{d}w^{c},$$

while applied to a function we instead get just torsion

$$[\nabla_{a},\nabla_{b}]f=-T^{c}{}_{ab}\nabla_{c}f.$$

We can more easily geometrically interpret things if we express the second Bianchi identity as

$$\nabla_{u}\check{R}(v,w)\vec{a}+\nabla_{v}\check{R}(w,u)\vec{a}+\nabla_{w}\check{R}(u,v)\vec{a}=0,$$

where $$\check{R}$$ is a tensor-valued 2-form (of type $$(1,1)$$), and the covariant derivative acts on this tensor value before being applied to the vector $$\vec{a}$$.

To build a picture of what this means, we can take advantage of the fact that $$\check{R}(v,w)$$ is a tensor, and thus $$\check{R}(v,w)\vec{a}$$ only depends upon the local value of $$\vec{a}$$. We decide to construct its local vector field values to equal its parallel transport, e.g. $$\vec{a}\left|_{p+\varepsilon u}\right.=\parallel_{\varepsilon u}(\vec{a}\left|_{p}\right.)$$. Then using the definition of the covariant derivative in terms of the parallel transport, we have

\begin{aligned}\varepsilon\nabla_{u}\check{R}(v,w)\vec{a} & =\check{R}(v\left|_{p+\varepsilon u}\right.,w\left|_{p+\varepsilon u}\right.)\vec{a}\left|_{p+\varepsilon u}\right.-\parallel_{\varepsilon u}\check{R}(v,w)\parallel_{\varepsilon u}^{-1}\vec{a}\left|_{p+\varepsilon u}\right.\\ & =\check{R}(v\left|_{p+\varepsilon u}\right.,w\left|_{p+\varepsilon u}\right.)\parallel_{\varepsilon u}\vec{a}-\parallel_{\varepsilon u}\check{R}(v,w)\vec{a}. \end{aligned}

The first term parallel transports $$\vec{a}$$ along $$\varepsilon u$$ and then around the parallelogram defined by $$v$$ and $$w$$ at $$p+\varepsilon u$$, while the second parallel transports $$\vec{a}$$ around the parallelogram defined by $$v$$ and $$w$$ at $$p$$, then along $$\varepsilon u$$. Thus we construct a cube from the vector fields $$u$$, $$v$$, and $$w$$, and find that the second Bianchi identity reflects the fact that $$\nabla_{u}\check{R}(v,w)\vec{a}+\nabla_{v}\check{R}(w,u)\vec{a}+\nabla_{w}\check{R}(u,v)\vec{a}$$ parallel transports $$\vec{a}$$ along each edge of the cube an equal number of times in opposite directions, thus canceling out any changes.

Above, we see that $$\vec{a}$$ is parallel transported along each edge of the cube made of the three vector field arguments an equal number of times in opposite directions, thus canceling out any changes. Here $$\varepsilon\nabla_{u}\check{R}(v,w)\vec{a}=\check{R}(v\left|_{p+\varepsilon u}\right.,w\left|_{p+\varepsilon u}\right.)\parallel_{\varepsilon u}\vec{a}-\parallel_{\varepsilon u}\check{R}(v,w)\vec{a}$$ is highlighted by the bold arrows representing the path along which $$\vec{a}$$ is parallel transported in the first term, and by the remaining dark arrows representing the path along which $$\vec{a}$$ is parallel transported in the second term.

So geometrically, the second Bianchi identity can be seen as reflecting the same “boundary of a boundary” idea as that of $$d^2=0$$ when considering the exterior derivative of a 2-form.

A more detailed description of this approach and the somewhat unusual notation can be found here.

As an aside to Adam Marsh' answer, if one uses the Cartan-Chern approach with a connection form $$\omega$$, curvature form $$\Omega=d\omega+\omega\wedge^\cdot\omega=d\omega+\frac{1}{2}[\omega\wedge\omega]$$ and covariant exterior derivative $$d_\omega$$, the second Bianchi identity takes the form $$d_\omega\Omega=0$$. Writing this out explicit gives $$d_\omega\Omega=d\Omega+[\omega\wedge\Omega]=d\Omega+\omega\wedge^\cdot\Omega-\Omega\wedge^\cdot\omega \\ =d\Omega+\omega\wedge^\cdot d\omega+\omega\wedge^\cdot\omega\wedge^\cdot\omega-d\omega\wedge^\cdot\omega-\omega\wedge^\cdot\omega\wedge^\cdot\omega \\ =dd\omega+d\omega\wedge^\cdot\omega-\omega\wedge^\cdot d\omega+\omega\wedge^\cdot d\omega-d\omega\wedge^\cdot\omega=dd\omega=0,$$

so we find that the second Bianchi identity is literally the $$dd=0$$ identity for differential forms.