How can I Bel-decompose the Riemann tensor over SO(3,3)?

[deSitter]

Hello,

I need to Bel-decompose the Riemann tensor built over SO(3,3). Does anyone have a decent reference? Funny how this topic is elusive. Is this in Wald?

Thanks in advance.

-drl

 

[Bill_K]

Not sure what you want, but I'll take a stab at it. You have R_abcd, and you want to split it into parts according to whether the indices lie in the first or second 3-spaces. Bel calls the parts 'electric' and 'magnetic', and SO(3,3) is like Minkowski space except that time is three dimensional. Whatever :-)

Consider first one antisymmetric pair of indices by itself, ab. They can lie in either 11, 12 or 22. That's three cases. Likewise for the second pair, cd. So you have a 3 x 3 matrix of cases, except that R_abcd is symmetric on the two pairs, so the 3 x 3 matrix is symmetrical and there are really only 6 cases: 1111, 1112, 1122, 1212. 1222 and 2222. The decomposition will therefore have six parts.

Is this anywhere near what you want?

 

[deSitter]

Not sure what you want, but I'll take a stab at it. You have R_abcd, and you want to split it into parts according to whether the indices lie in the first or second 3-spaces. Bel calls the parts 'electric' and 'magnetic', and SO(3,3) is like Minkowski space except that time is three dimensional. Whatever :-)

Consider first one antisymmetric pair of indices by itself, ab. They can lie in either 11, 12 or 22. That's three cases. Likewise for the second pair, cd. So you have a 3 x 3 matrix of cases, except that R_abcd is symmetric on the two pairs, so the 3 x 3 matrix is symmetrical and there are really only 6 cases: 1111, 1112, 1122, 1212. 1222 and 2222. The decomposition will therefore have six parts.

Is this anywhere near what you want?

Not really - there is a definite procedure here for carving up the Riemann tensor, sort of like the Ricci decomposition, but I'll be damned if I can remember the details. You need a time-like congruence, which should still exist for SO(3,3) because there is still a light cone. Well I can't remember where I saw this, and need the details. There are actually 4 pieces in general, only 3 survive in 4-d, where they represent tidal distortion and tendency to rotate in the congruence, and the sectional curvatures of the space-like surface orthogonal to the time-like congruence.

edit: Ok I found a paper by Deser talking about it - great! The Bel-Robinson tensor in 4-d is basically sort of like the Maxwell energy tensor in reference to Fmn, that is, you can write Maxwell as

Tmn = Fma Fan + F*ma F*an

and the Bel-Robinson tensor is

Bmnab = Rkmjn Rkajb + R*kmjn R*kajb

where R*mnab = 1/2 epsilon_mnrs Rrsab

I'm speculating that if you think of Rmnab as a thing have two 2-form indices ("surface tensor of 2nd rank" as Pauli called it), then the analogy is exact.

paper by Deser is here

http://arxiv.org/abs/gr-qc/9901007

-drl

 

[Bill_K]

The timelike congruence X_a used in 4 dimensions is really a projection operator. You need to use two projectors, one projecting on the space dimensions, the other on the time dimension(s). In 4-d, since time is one-dimensional, the projectors can be written in terms of a timelike congruence X_a, namely P_ab = X_aX_b and Q_ab = n_ab - X_aX_b. The Riemann tensor is decomposed by applying one of these projectors to each of the four indices. Taking into account the symmetry of the indices, you wind up with a decomposition into three independent parts.

What I'm saying is that you do likewise in SO(3,3). You still have P_ab and Q_ab, but there's no X_a any longer because 'time' is 3-dimensional. (I suppose if you really insisted on it you could use three congruences and define P_ab = X_aX_b + Y_aY_b + Z_aZ_b.) Project each index onto either the timelike subspace or the spacelike subspace. You'll wind up with six independent parts. For example the first one which I wrote as 1111 earlier would be R_abcdP_aiP_bjP_ckP_dl. The next one would be R_abcdP_aiP_bjP_ckQ_dl, and so on.

 

[deSitter]

The timelike congruence X_a used in 4 dimensions is really a projection operator. You need to use two projectors, one projecting on the space dimensions, the other on the time dimension(s). In 4-d, since time is one-dimensional, the projectors can be written in terms of a timelike congruence X_a, namely P_ab = X_aX_b and Q_ab = n_ab - X_aX_b. The Riemann tensor is decomposed by applying one of these projectors to each of the four indices. Taking into account the symmetry of the indices, you wind up with a decomposition into three independent parts.

What I'm saying is that you do likewise in SO(3,3). You still have P_ab and Q_ab, but there's no X_a any longer because 'time' is 3-dimensional. (I suppose if you really insisted on it you could use three congruences and define P_ab = X_aX_b + Y_aY_b + Z_aZ_b.) Project each index onto either the timelike subspace or the spacelike subspace. You'll wind up with six independent parts. For example the first one which I wrote as 1111 earlier would be R_abcdP_aiP_bjP_ckP_dl. The next one would be R_abcdP_aiP_bjP_ckQ_dl, and so on.

Good points, thanks for the grist :)

-drl