Does an explicit list of 20 independent compenents of Riemman exist?

[ozone]

Hello all,

I have been given a problem where I am asked to calculate "all" the components of the Riemann tensor in a gross non-diagonalized metric. I know there exists at most 20 independent components of Riemann, but I want to actually compose a list of these combinations.

It is easy enough to generate a good quantity of combinations which don't vanish under the antisymmetric exchange, but it is a bit harder to do this and account for the first Bianchi Identity. I made a list of 21 components which I should have to calculate and their could exist more but I realized I have no idea how to figure out how many more I need since there may be redundancies in my current list.

To reiterate; I was wondering if there was a compiled list somewhere for 20 independent components.

Thanks
--Ozone.

PS: If anyone could help me to figure out what the difference in the # of components with the Bianchi Identity applied and with just the block identity//antisymmetry of the Riemann that would be much appreciated.

 

[Russell E]

I know there exists at most 20 independent components of Riemann, but I want to actually compose a list of these combinations... To reiterate; I was wondering if there was a compiled list somewhere for 20 independent components.

You can find a discussion of the 20 independent components of the Riemann tensor in section 8 of this web page:

http://www.mathpages.com/rr/appendix/appendix.htm

 

[WannabeNewton]

Hello all,

I have been given a problem where I am asked to calculate "all" the components of the Riemann tensor in a gross non-diagonalized metric.

My god, what a torturous and arduous exercise to give a poor soul there's nothing worse than coordinate computations!

 

[Bill_K]

To reiterate; I was wondering if there was a compiled list somewhere for 20 independent components.

This is pretty easy. The first pair of indices is antisymmetric (can you say bivector?) therefore can have only 6 independent combinations: 01, 02, 03, 12, 13, 23. Likewise for the last pair.

The Riemann tensor is symmetric on exchanging the first pair with the last pair. So it's a symmetric tensor in the space of bivectors, and therefore has (6 x 7)/2 = 21 possible components. All that's left is the cyclic identity. e.g. R₀₁₂₃ + R₀₃₁₂ + R₀₂₃₁ = 0, leaving you with 20.

 

[ozone]

haha thanks for the help guys, I guess then I know which pair is redundant out of the 21 I came up with. Yes I am not looking forward to calculating this =/ I guess my Prof. must really want to punish his students.

 

[pervect]

If your prof didn't forbid the use of automatic computational software, I'd probably use it. Maxima, for instance. MTW's gravitation has some remarks on efficient ways to do it by hand, though.

 

[robphy]

It's one of those calculations that one should do once (maybe twice) in one's life.