General Relativity: Spatial Inversion Symmetry

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SUMMARY

This discussion focuses on the implications of spatial inversion symmetry in General Relativity, specifically transformations of the form x → -x. The user inquires whether cross terms vanish under such transformations when a Killing vector is present, and whether the proof involving the metric tensor g_{xy} is valid. The conversation also explores the relationship between spatial inversion symmetry and rotational symmetries, particularly in the context of local versus global definitions in arbitrary spacetimes.

PREREQUISITES
  • Understanding of General Relativity concepts, particularly Killing vectors.
  • Familiarity with tensor calculus and metric tensors.
  • Knowledge of symmetry operations in physics, specifically spatial inversion.
  • Basic comprehension of local versus global properties in manifold theory.
NEXT STEPS
  • Research the role of Killing vectors in General Relativity.
  • Study the implications of spatial inversion symmetry on metric tensors.
  • Examine the relationship between symmetries and conservation laws in physics.
  • Explore the concept of orientability in differential geometry.
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The discussion is beneficial for theoretical physicists, mathematicians specializing in differential geometry, and students studying advanced concepts in General Relativity.

Slereah
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I am unable to find any source describing at any lengths transformations of the type

[itex]x \rightarrow -x[/itex]

beyond the case of static spacetime (and even in the case of static spacetime, it is rarely the fundamental definition, which tends to be more along the lign of orthogonality with an hyperplane, [itex]\xi_{[a} \nabla_b \xi_{c]}[/itex]).

So my question is, if I have a space reversal symmetry, do the cross terms of that coordinates always vanish (if that coordinate also has a Killing vector at least), and is the most obvious proof ([itex]x \rightarrow -x, \ g_{xy} dx dy = -g_{xy} dx dy \rightarrow g_{xy} = 0[/itex] the correct one. Also, does it mean that all rotational symmetries have no cross terms, since they have [itex]\mathbb{Z}_2[/itex] as a subgroup?

If there is such a symmetry without a corresponding translation symmetry, does it just imply that the cross terms are odd functions?
 
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You certainly have this symmetry locally in GR. Globally, there is not even any way to define the operation for an arbitrary spacetime. For example, there is not guaranteed to be a coordinate chart that covers all of spacetime, nor do you have any guarantee that the spacetime is orientable.
 
Sure, but let's say we have a spacetime that can admit such a symmetry. Let's say [itex]\mathbb{R} \times \mathbb{R}^3[/itex], with a distribution of matter symmetrical by reflection along some plane.
 

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