A spacetime can be foliated by spacelike hypersurfaces which are related by an isometry $\partial_t$.

  1. Why is $\partial_{t}$ called an isometry of the spacetime?

  2. How are the spacelike hypersurfaces related by the isometry $\partial_{t}$?

The answers here and here talk about the foliation of spacetime into spacelike hypersurfaces indexed by a time coordinate $t$, but make no reference to the associated isometry.

  • $\begingroup$ $\partial_t$ being an isometry means your theory is invariant wrt time translations: any spacelike slice looks the same as any other spacelike slice $\endgroup$
    – Kosm
    May 6, 2017 at 3:48

1 Answer 1


An isometry vector is a Killing vector field. It does not exist in every spacetime, when it does the spacetime is said to have a timelike symmetry, and as in Noether's theorem energy is conserved. Otherwise it is not. An example where there is no timelike symmetry is the cosmological Big Bang model, the FLRW solution.

There is no timelike symmetry because the universe is expanding and the curvature changes over time. You can't find a timelike coordinate where that is not so.

But even then, $\partial_{t}$ is a vector which is perpendicular to the spacelike hypersurfaces we use at constant cosmic times, i.e. , constant t. That's how the hypersurfaces are selected, those with every vector spacelike and perpendicular to $\partial_{t}$.

You can find spacelike hypersurfaces in any spacetime. The initial value problem is well defined in general relativity, even in spacetimes with no isometry. The ADM formalism worked it out, and the metric and its derivatives evolve over time. See it in Wikipedia at https://en.m.wikipedia.org/wiki/ADM_formalism

You don't need an isometry to have foliations of spacetime with spacelike hypersurfaces, at least if you exclude some anomalies like some topological anomalies or singularities, maybe also horizons.


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