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)