Geometric interpretation of the second covariant derivative

Question

I'm having some doubts about the geometric representation of the second covariant derivative.

I know that $\triangledown^{}_a(\triangledown^{}_bv)=(\triangledown^{}_a\triangledown^{}_b)v+\triangledown^{}_{\triangledown^{}_ab}v$

So the Riemann tensor can be defined in two ways :

$R(a,b)v=\triangledown^{}_{a}(\triangledown^{}_{b}v)-\triangledown^{}_{b}(\triangledown^{}_{a}v)-\triangledown^{}_{[a,b]}v\quad$ or $\quad R(a,b)v=(\triangledown^{}_{a}\triangledown^{}_{b})v-(\triangledown^{}_{b}\triangledown^{}_{a})v$

So far so good (correct me if I'm wrong).

But the problem arises when I read that the Riemann tensor represents the difference between a parallel transported vector along two parts of a closed path.

Because if $\triangledown^{}_{b}v$ is the parallel transported vector along b, then it stands to reason that $\triangledown^{}_{a}(\triangledown^{}_{b}v)$ is the transported vector along b and then a. But there would be a correction term, so the representation would be inexact.

And conversely, if $(\triangledown^{}_{a}\triangledown^{}_{b})v$ is the transported vector along the path b and then a, the representation would be correct, but I don't understand why...

So, which is correct, A or B ?

Edit : would the difference be linked with the torsion ? Are $\triangledown^{}_{a}(\triangledown^{}_{b}v)$ and $(\triangledown^{}_{a}\triangledown^{}_{b})v$ equal if the connection is torsion free ?

PS : why is the image not integrated in the post ? too large ?

Answer 1

I think you are mixing up (abstract) index notion with an index-free notation. If you take $\xi$, $\eta$ and $v$ to be vector fields, then $(\nabla_\xi\nabla_\eta)v$ does not make sense, since you can apply a covariant derivative only to a vector field and not to the operator $\nabla_\eta$. The equation you are looking for is that if you extend the covariant derivative to an operation on tensor fields, then given $v$, you can form the $\binom11$-tensor field $\nabla v$ defined by $\eta\mapsto \nabla_\eta v$. Applying the covariant derivative to that, one obtains a $\binom12$-tensor field which is usually denoted by $\nabla^2v$. This is indeed given by $(\nabla^2v)(\xi,\eta)=\nabla_\xi\nabla_\eta v-\nabla_{\nabla_\xi\eta}v$. Torsion-freness of the connection then implies that $\nabla_\xi\eta-\nabla_\eta\xi=[\xi,\eta]$ and using this, you see that the curvature is given by $$ R(\xi,\eta)(v)=(\nabla^2v)(\xi,\eta)-(\nabla^2v)(\eta,\xi)=\nabla_\xi\nabla_\eta v-\nabla_\eta\nabla_\xi v-\nabla_{[\xi,\eta]}v. $$ Your confusion probably comes from the fact that in (abstract) index notation, one would use $\nabla_av^b$ as a symbol for the $\binom11$-tensor field $\nabla v$. But here $a$ and $b$ are not vector fields but (abstract) indices (and you cannot leave out the "$b$", otherwise $v$ would be a function rather than a vector field). Correspondingly, $\nabla^2v$ will be denoted by $\nabla_a\nabla_bv^c$ (and again $a,b,c$ are indices and not vector fields). In this notation the definition of curvature via $\nabla^2v$ as above then indeed reads as $R_{ab}{}^c{}_dv^d=\nabla_a\nabla_bv^c-\nabla_b\nabla_av^c$.

Answer 2

After putting all this aside for a while, here's a proposition for a visual answer to my question. Please tell me if something doesn't look right ^^ (note : the vectors are of course random, it's a schematic representation)

Here are the second covariant derivatives and the (contracted) Riemann tensor in the simplest case where $\triangledown^{}_{b}a=\triangledown^{}_{a}b=0$ and $\triangledown^{}_{[a,b]}=\triangledown^{}_{\triangledown^{}_{b}a}-\triangledown^{}_{\triangledown^{}_{a}b}=0$ :

Now in the general case (still torsion-free) with $\triangledown^{}_{[a,b]}v\neq 0$ :

where (it's easier to represent it separately) :

"Not much geometry in all that" ? On the contrary, I think.