Assuming a torsion free Christoffel symbol, the covariant derivative can be shown to satisfy the second (differential) Bianchi identity:

\begin{equation} [[\nabla_a,\nabla_b],\nabla_c]+[[\nabla_c,\nabla_a],\nabla_b]+[[\nabla_b,\nabla_c],\nabla_a]=0 \end{equation}

Question: Is there a nice geometric interpretation of this identity?

One of my motivations for this question is that this condition can be interpreted as stating that the co-variant derivatives form a Lie algebra (with the algebra product given by the commutator). Thus a geometric interpretation of the second Bianchi identity may motivate why the Jacobi identity is natural/fundamental.

  • 1
    $\begingroup$ The second Bianchi identity can also be viewed as a consequence of the diffeomorphism invariance of the Riemann curvature tensor, i.e. the fact that $\operatorname{Rm}_{\varphi^*g} = \varphi^* \operatorname{Rm}_g$ for any metric $g$ and diffeomorphism $\varphi$. $\endgroup$ – Dietrich Burde Feb 28 at 20:26

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


while applied to a function we instead get just torsion


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


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.

Second Bianchi identity

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.