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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 (where 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.

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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.

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.

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    $\begingroup$ I'd just point out that if there is torsion, then $[\nabla_a,\nabla_b]w^c=R^c_{\ dab}w^d$ is not true, and there is an additional term involving torsion. $\endgroup$ – Bence Racskó Mar 2 at 13:15
  • $\begingroup$ Thank you Bence, corrected to make both consistent with non-zero torsion assumption. $\endgroup$ – Adam Marsh Mar 4 at 0:23
  • $\begingroup$ It is not clear that the opposite direction arrows will lie on top of one another if the vectors v and w are functions of the co-ordinates (or the co-ordinates themselves). I do not think that going epsilon distance in the u direction from point p followed by going epsilon distance in the v direction, will bring you to the same point as if you start at p and then travel first in the v direction and then the u direction? $\endgroup$ – Luke Mar 19 at 16:01
  • $\begingroup$ Does the torsion measure how much the opposite pointing arrows do or do not overlap? $\endgroup$ – Luke Mar 19 at 19:14
  • $\begingroup$ The figure above assumes vanishing Lie brackets, which results in the two infinitesimal transports you mention being equivalent. Since zero torsion is assumed, this coincides with assuming the vector fields equal their parallel transports. This level of detail, along with a picture describing torsion (and one depicting what the second Bianchi identity looks like with non-vanishing torsion), is available in this section of the above referenced book. $\endgroup$ – Adam Marsh Mar 20 at 17:20
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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.

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  • $\begingroup$ Here it is good to keep in mind that $\omega$ is a frame-dependent matrix-valued 1-form, and the differential acts on each matrix component as a 1-form. To me this is an excellent way to see how the second Bianchi is true algebraically, but hard to use to draw a picture and see the geometric meaning. $\endgroup$ – Adam Marsh Mar 4 at 0:48
  • $\begingroup$ An alternative: $\mathrm{D}\check{R}=0$. At $p$ the exterior covariant derivative is "the sum of $\check{R}$ on the boundary defined by its arguments after being parallel transported to $p$," while $\check{R}$ is "the difference between a vector and its parallel transport around the boundary defined by its arguments" -- see here and here. $\endgroup$ – Adam Marsh Mar 4 at 0:56
  • $\begingroup$ @AdamMarsh Yeah your answer is much better in this regard, I just wanted to illustrate how close the Bianchi identity is to $dd=0$. It is an interesting question to see if some finite holonomy formula could be used to illustrate this, as for Abelian connections, parallel transport can be described by the integral of the curvature over a surface being bounded by the loop, but for non-Abelian connections, this is more involved. But there is actually one such formula here deaneyang.com/papers/holonomy.pdf . $\endgroup$ – Bence Racskó Mar 4 at 10:31

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