- #1
center o bass
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These questions applies to both spatially homogenous cosmological models, and multidimensional Kaluza-Klein theories:
Suppose we have a manifold M, of dimension m, for which there is a transitive group of isometries acting on some n-dimensional homogeneous subspace N of M. Thus there exists a basis of n killing vectors ##\{\xi_1, \ldots, \xi_n\}## on N with structure constants ##C_{ij}^k##.
In this paper (http://arxiv.org/abs/gr-qc/9804043), after Equation (10.2), it is claimed that in order for Hilbert's variational principle
$$\delta \int R \text{vol}^{m}=0$$
to be compatible with the homogeneity of ##N\subset M##, then the volume form ##\text{vol}^n = \omega^1 \wedge \cdots \wedge \omega^n##, where ##\omega^i## are the dual basis form to ##\xi_i## (##\text{vol}^m = \text{vol}^{m-n} \wedge \text{vol}^{n}##), must be invariant with respect to the group of isometries. A requirement that is translated into ##C^k_{kj} = 0##.
Additionally the author claims that ##C^k_{kj} = 0## ensures that ##\mathcal{L}_{\xi_i} \delta g= 0##, where g is the metric tensor on M.
I thus wonder:
1. Why does ##\text{vol}^n = \omega^1 \wedge \cdots \wedge \omega^n## have to be invariant (with respect to the group of isometries) to make the variational Hilbert principle compatible with the homogeneity of N?
2. Why is this only true iff ##C^{k}_{kj} =0##? (Is this related to the Haar measure of the group of isometries being bi-invariant?)
3. How does ##C^k_{kj} = 0## ensure ##\mathcal{L}_{\xi_i} \delta g= 0## ?
Suppose we have a manifold M, of dimension m, for which there is a transitive group of isometries acting on some n-dimensional homogeneous subspace N of M. Thus there exists a basis of n killing vectors ##\{\xi_1, \ldots, \xi_n\}## on N with structure constants ##C_{ij}^k##.
In this paper (http://arxiv.org/abs/gr-qc/9804043), after Equation (10.2), it is claimed that in order for Hilbert's variational principle
$$\delta \int R \text{vol}^{m}=0$$
to be compatible with the homogeneity of ##N\subset M##, then the volume form ##\text{vol}^n = \omega^1 \wedge \cdots \wedge \omega^n##, where ##\omega^i## are the dual basis form to ##\xi_i## (##\text{vol}^m = \text{vol}^{m-n} \wedge \text{vol}^{n}##), must be invariant with respect to the group of isometries. A requirement that is translated into ##C^k_{kj} = 0##.
Additionally the author claims that ##C^k_{kj} = 0## ensures that ##\mathcal{L}_{\xi_i} \delta g= 0##, where g is the metric tensor on M.
I thus wonder:
1. Why does ##\text{vol}^n = \omega^1 \wedge \cdots \wedge \omega^n## have to be invariant (with respect to the group of isometries) to make the variational Hilbert principle compatible with the homogeneity of N?
2. Why is this only true iff ##C^{k}_{kj} =0##? (Is this related to the Haar measure of the group of isometries being bi-invariant?)
3. How does ##C^k_{kj} = 0## ensure ##\mathcal{L}_{\xi_i} \delta g= 0## ?
