| |
| | The Bianchi identities |
 | | I first show that cycling among the first three terms and summing the results delivers zero: from this, we have that the triply-symmetric and triply-antisymmetric parts of the Riemann tensor are both zero. |
 | | When we contract Riemann(Q) with a [R,R]-rank tensor, g, which Q (of rank R) annihilates, we get a [R,R,R,R]-rank tensor Riemann(Q)·g which is [1,0,2,3]-antisymmetric, [0,1,3,2]-antisymmetric and annihilated by (τ[0,1,2] +τ[1,2,0] +τ[2,0,1]). |
 | | A length, however, is arrived at by integrating the square root of the result of applying a metric to the (tensor) square of displacement along the line. |
| www.chaos.org.uk /~eddy/math/smooth/bianchi.html (1145 words) |
|