Physics Asked on August 14, 2021
In a paper about the splitting of the space-time [for reference "THE STRUCTURE OF SPACE–TIME: RELATIVITY GROUPS" (International Journal of Geometric Methods in Modern Physics Vol. 3, No. 3 (2006) 591–603)], it identifies the reference frame with a $(1,1)$-tensor field $Rinmathcal{T}^1_1(M)$, $M$ is the space-time, and satisfying the condition $R^2=R$. For convenience, it decompose $R$ into the tensor product of a vector field $Gamma$ and a one-form $alpha$
$$R=alpha otimes Gamma$$
and the requirement $R^2=R$ is equivalent to the condition $alpha[Gamma]=1$.
My question is: how do I prove this last equivalence? I have no clue about how to compute $R^2$.
A $(1,1)$-tensor of the form $alphaotimes v$ eats a vector $x$ and spits out the vector $alpha(x) v$, where $alpha(x)$ is just a number. If you feed the result to $alphaotimes v$ again, you'll get the answer you're looking for.
Correct answer by J. Murray on August 14, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP