Space-like Killing vector of Robertson-Walker metric?

In the book "Kinetic theory in the expanding Universe" (J. Bernstein, 1988, Camb. Univ. Press), it was stated that

"for nonstationary Robertson-Walker matrixes [sic] there is no spacelike Killing vector."

(page 6, footnote.)

But we know this is not true, since one can see that the generators of spatial translations/rotations are manifestly space-like. (For example, see R. Maartens and S.D. Maharaj, Class. Quantum Grav. 3 (1986) 1005-1011, equation (1.7)).

Was the author of the book wrong, or was it a slip of the hand, where "spacelike" should have been "timelike" (but that would amount to tautology anyway, since "nonstationary" is defined in terms of the absence of a time-like Killing vector).

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From the book by Kolb and Turner "The Early Universe", chapter 3.5:

In the strictest mathematical sense it is not possible for the Universe to be in thermal equilibrium, as the FRW cosmological model does not possess a time-like Killinig vector.

(Emphasis is mine). So it confirms the "slip of the hand" hypothesis.

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Thank you for that reference! –  zcma Apr 1 '11 at 13:08

The de Sitter metric $$ds^2~=~dt^2~-~e^{\sqrt{\Lambda/3}t}(dr^2~+~r^2d\Omega^2)$$ has this time dependent factor. This prevents a time-like Killing vector, for any vector formed by $\sqrt{g_{rr}^{-1}}\partial/\partial t$, or something similar, will not constant on a timelike spacetime vector. The same holds for spacelike directions, for space and time are interchangeable in $g_{rr}$. This means global conservation laws are not defined. Energy conservation in general relativity is a funny thing, and is not something which can be established. In the case of cosmologies, these metrics FLRW etc, are type O Petrov-Penrose-Pirani solutions, which have no isometries or Killing vectors.

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You can write the de-Sitter metric as $ds^{2}=-(1-\frac{1}{3}\Lambda)dt^{2} + \frac{dr^{2}}{1-\frac{1}{3}\Lambda}+r^{2}d\Omega^{2}$, which has an explicit timelike killing vector. The coordinate transformation is easy--Let $r->Re^{\sqrt{\frac{\Lambda}{3}}\frac{t}{2}}$, and $t->T+f(R)$, where f(R) is chosen to cancel the cross-term. –  Jerry Schirmer Mar 31 '11 at 13:22
True, but this is a special coordinate chart. Also the metric term is $g_{tt}~=~(1~-~\Lambda x^2/3)$ –  Lawrence B. Crowell Mar 31 '11 at 14:39
@Lawrence: Yes. typo. But the Robertson-Walker-style expression of the metric is also a special coordinate charte--the true De-Sitter space doesn't have a $t=0$ singularity. –  Jerry Schirmer Mar 31 '11 at 15:20
Both of you are writing metrics on a portion of the full de Sitter spacetime. Four-dimensional de Sitter has topology $R^1 \times S^3$ and is conveniently realized as a hyperboloid in $R^5$. Various combinations of the coordinates on the hyperboloid give the metrics above, but will only cover part of the hyperboloid. There may be timelike Killing vectors on the patches associated with those metrics, but there is no global timelike Killing vector. –  Robert McNees Mar 31 '11 at 19:03
Exactly. The solution in other coordinates or slicings involves hyperbolic trigonometric functions. The exponential form is a flat coordinate, and the form Schirmer uses are static coordinates. There is no global timelike killing vector. –  Lawrence B. Crowell Mar 31 '11 at 19:57