I've thought of a new way (at least I never read it anywhere) of counting the independent components of the Riemann tensor, but I am not sure whether my arguments are valid, so I would like to ask whether my argument is sound or total bonkers.(adsbygoogle = window.adsbygoogle || []).push({});

The Riemann tensor gives the deviation of a vector A when transported around a loop with vectors u and v:

[tex] \delta A^\alpha = -R^\alpha_{\beta\mu\nu} A^\beta u^\mu v^\nu[/tex]

A parallel transport is only non-trivial if u and v are different (since u=v just means going forward and backward along a line). This also shows that the tensor is antisymmetric in the final two indices because reversing the vectors means travelling along the loop in the opposite direction. There are six possible planes that can be spanned by u and v (01, 02, ..., 23).

If I fix the plane, what the Riemann tensor does is create an infinitesimal rotation of the initial vector A, resulting in the deviation. An infinitesimal Rotation is antisymmetric (so we get antisymmetry in the first two indices), in 4 D it has 6 independent components.

This leaves me with 6x6=36 independent components.

To get rid of the remaining 16, I use the argument presented in this paper by Ollivier

https://pdfs.semanticscholar.org/6398/77480ace10bb6ad30adc5a17c16b3caa6105.pdf

see fig. 5 and 6.

What this shows is that one can understand the first Bianchi identity geometrically by constructing a cube with parallel transport of vectors and noting that the endpoints of different paths will form a triangle.

In normal derivations of the number of independent components, the Bianchi identity can only eliminate one of them because the symmetry [tex]R_{\alpha\beta\mu\nu}=R_{\mu\nu\alpha\beta}[/tex] has already been established. Since I did not do this, I can exploit the Bianchi identity using the cube construction from the Ollivier-paper. There are four distinct cubes in 4D, each of them yields one Bianchi-identity which fixes four components up to a total of 16.

Thus, 16 components of the Riemann tensor have been fixed this way, leaving 20 as it should be.

I realize that the argument has the big disadvantages of (i) being a bit heuristic and (ii) not establishing the symmetry between the first and last two indices. Nevertheless, I find it interesting - if it is correct.

I'd be very grateful if one of the experts could tell me whether this argument actually works out or where I made a mistake.

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# I Number of independent components of the Riemann tensor

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