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On Gravitation Page 286 Exercise 11.7, it mentioned a very interesting operator called "Jacobi curvature operator" $${\cal J}(u,v) n\equiv \frac{1}{2} [{\cal R}(n,u)v +{\cal R}(n,v)u ]$$ where it "contained the same information continent as Riemann", and in component form (Eq. 11.36) $$J^\mu_{\nu \alpha \beta} =\frac{1}{2} (R^\mu_{\alpha\nu\beta} +R^\mu_{\beta\nu\alpha})$$

It was also shown that Jacob could be directly used for geodesic deviation, since $${\cal J}(u,u) n ={\cal R}(n,u)u$$ a related post could be found here: Are the Jacobi equation and the geodesic deviation equation related?.

However, why it seemed that Riemann was much popular than Jacobi, and that people seemed to use Jacobi much rarely?

Qmechanic
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ShoutOutAndCalculate
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1 Answers1

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Well, I would also see the Riemann tensor as a "primary" object and the Jacobi curvature tensor as the secondary one. This is readily seen in the definition of the Jacobi curvature tensor; i.e., it is the symmetrisation of the second and the fourth indices, on which Riemann tensor possesses no symmetry.

I think if you would see the Jacobi tensor as the primary definition then you would get into "problems" when you want to define it as a measure for the intrinsic curvature by parallel transporting a vector along a path. Therefore, you would have to introduce the Riemann curvature tensor again.

But, why they call it Jacobi curvature operator? I think the reason is that you could rewrite the Jacobi (see e.g. the post you added) $$ \nabla_u^2 n + {\cal R} (n ,u) u = 0 $$ in the symmetric form $$ \nabla_u^2 n +{\cal J}(u ,u) n = 0 \,. $$ Then, one could write it as an operator acting on $n$: $$ [ \nabla_u^2 +{\cal J}(u ,u) ] n = 0 \,. $$ It means the Jacobi equation will become a zero eigensolution problem.

You can see that this definition has other motivations than the Riemann curvature tensor. So I think the Riemann curvature tensor is indeed more fundamental.

Maybe there is a fundamental geometrical explanation for this. But, I'm not aware of that.

Astrolabe
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