Talk:Curvature collineation

Definition of a cc
This is how the Riemann tensor is defined ('1 up and 3 down'). By writing $$\mathcal{L}_X R = 0$$, the definition is ambiguous unless it is specified where the indices are - the positions of the indices matters in the definition of a cc (unless $$\mathcal{L}_X g = 0$$, i.e., unless $$X$$ is Killing, which in general isn't true).


 * You're quite right. Unfortunately the index placement is not enough to be unambiguous since it differs according to which definition of R you use.--MarSch 14:57, 20 Jun 2005 (UTC)

True, but the Riemann tensor is defined to be a 1 up 3 down tensor (remember the idea of taking a vector and parallel transporting it around the manifold in two different ways). Strictly speaking, any raising or lowering of indices changes the Riemann tensor (this is a technicality that many people overlook). To take a simple example, consider a metric tensor (defined as a rank 2 covariant tensor) for a specific spacetime (say exterior Schwarzschild) and write it's components as $$g_{ab}$$; now contract this with $$g^{bc}$$ to get $$\delta_a{}^c$$ - nobody would claim that this ($$\delta_a{}^c$$) still represents the Schwarzschild metric ! Similarly, taking a Riemann tensor (defined to be 1 up 3 down) with components $$R^a{}_{bcd}$$, any tensor contraction of this changes the Riemann tensor to something else - the notation we use is still '$$R$$' but the fact that the indices change position after contraction - which in general means multiplying the Riemann tensor components with non-constant functions of the coordinates - means it is not the same tensor (just like $$\delta_a{}^c$$ is not the same as the Schwarzschild metric) ! In desperation, one may define a new type of symmetry vector field satisfying $$\mathcal{L}_X R_{abcd} =0$$, but this would be different from the cc definition (different tensor). Hope this clears the ambiguity. --Mpatel 6 July 2005 16:56 (UTC)