Ricci tensor equals zero implies flat splace?

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SUMMARY

The Ricci tensor equaling zero does not imply a flat Riemannian manifold. There exist both compact and non-compact Ricci flat manifolds that cannot be endowed with flat Riemannian metrics. For compact manifolds, a flat Riemannian metric necessitates a zero Euler characteristic, as the Euler class is a polynomial function of the curvature 2 form. Notably, Calabi-Yau manifolds can exhibit non-zero Euler characteristics, demonstrating the complexity of this relationship.

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  • Understanding of Ricci tensor and its implications in differential geometry
  • Familiarity with Riemannian and semi-Riemannian manifolds
  • Knowledge of Euler characteristic and its relation to curvature
  • Basic concepts of Calabi-Yau manifolds and their properties
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Mathematicians, physicists, and students of differential geometry seeking to deepen their understanding of the implications of the Ricci tensor in manifold theory.

Abrahamsk8
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Hi, my question is the title, if Ricci tensor equals zero implies flat space? Thanks for your help
 
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If by flat space you mean a flat Riemannian manifold, the answer is no. There are examples of both compact and non-compact Ricci flat manifolds that cannot be given flat Riemannian metrics.

One thing to note is that for compact manifolds,a flat Riemannian metric implies that the Euler characteristic is zero - because the Euler class can be expressed as a polynomial in the curvature 2 form. If you look around you will find Calabi -Yau manifolds with non-zero Euler characteristic.
 
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As a physics application, in the semi-Riemannian case, a vacuum has a Ricci tensor equal to zero. A vacuum doesn't have to be flat. A vacuum be curved by gravity. An example would be the space above the Earth's surface where you're sitting right now.
 

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