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I am facing some problems in understanding what is the importance of a Killing vector field? I will be grateful if anybody provides an answer, or refer me to some review or books.

Lorenzo B.
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anotherone
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2 Answers2

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In terms of classical general relativity: Einstein's equations $$ G_{ab} = 8\pi T_{ab} $$ can be formulated, in local coordinates, as a system of second order partial differential equations for the metric unknown $g_{ab}$. The matter field equations further generate some family of partial differential equations.

Given a continuous symmetry (as guaranteed by a Killing vector field), one has tools and tricks one can use to help solve the PDEs.

  • Noether's theorem tells us that for Einstein's equation (which admits a Lagrangian formulation), associated to each Killing vector field $X^a$ is a conservation law. One can simply see this by considering the current $$ J^{(X)}_a = T_{ab} X^b $$ Its divergence is $$ \nabla^a J^{(X)}_a = \nabla^a T_{ab} X^b + T_{ab} \nabla^a X^b $$ The first term vanishes since the energy momentum tensor is divergence free. Using that the energy-momentum tensor is symmetric, we write $$ \nabla^a J^{(X)}_a = \frac12 T_{ab} \left( \nabla^a X^b + \nabla^b X^a\right) $$ As a consequence of Killing's equations, if $X^a$ is a Killing field, the term inside the parenthesis evaluates to 0. So $J^{(X)}$ is divergence free. Applying Stokes' theorem we then see a conservation law.

  • Symmetry reduction: given a continuous symmetry for a PDE, we can try to perform a symmetry reduction of the equations. This reduces the number of independent variables for the PDE, and often makes it easier to see an exact solution (or to examine features of symmetric solutions). For a survey of how symmetry can help, I recommend checking out Exact Solutions of Einstein's Field Equations by Stephani et al (Cambridge University Press). Chapters 8, 9, 10, and all of Part II of that book addresses the use of symmetry groups to help solve Einstein's equations.

Willie Wong
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The application of the Stokes' theorem in this case is slightly indirect, as the last equation contains covariant derivative as opposed to the partial derivative over a coordinate in the standard conservation law. This is not a problem, however, because

$$ 0 = \nabla_\nu (X^{\mu} T_{\mu}^{\nu}) = \frac{1}{\sqrt{-g}} \partial_\nu( \sqrt{-g} \ X^{\mu} T_{\mu}^{\nu}), $$ and so the conserved current includes the extra factor of

$$ {\sqrt{-g}} $$

user46694
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