Properties of curvature tensor in 3 dimensions?

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Discussion Overview

The discussion revolves around the properties of curvature tensors in three dimensions, specifically exploring the relationships between the Ricci tensor, Ricci scalar, and Riemann tensor. The context includes theoretical aspects of general relativity and differential geometry.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant inquires about properties of curvature tensors in three dimensions, suggesting a possible proportionality between the Ricci tensor and Ricci scalar.
  • Another participant notes that in three dimensions, the Ricci tensor has the same number of components as the Riemann tensor and that the Riemann tensor can be expressed in terms of the Ricci tensor.
  • This participant also mentions that the relationship between the Ricci tensor and Ricci scalar can be derived from this expression.
  • A third participant seeks clarification on whether the original question pertains to 2+1 dimensions or 3-dimensional Euclidean space.
  • It is stated that in three dimensions, the Weyl tensor vanishes.

Areas of Agreement / Disagreement

Participants have raised various points regarding the properties of curvature tensors, but there is no consensus on the specific relationships or implications discussed. The conversation includes differing interpretations of dimensionality.

Contextual Notes

The discussion touches on advanced concepts in general relativity and may depend on specific definitions and assumptions regarding dimensionality and curvature.

im_hammer
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Is there any properties with the curvature tensors in 3 dimensions?
(Maybe between the Ricci tensor and the Ricci scalar, they are proportional to each other? )

I heard about it in a lecture, but I can not remember the details. The 3 dimensional case is not discussed in many reference books.

Thank you
 
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In three dimensions, the number of components of the Ricci tensor equals that of the Riemann tensor. The Riemann tensor can be completely expressed in terms of the Ricci tensor (as an exercise, you can try to write it down based on the symmetries). From that you can derive the relation between the Ricci tensor and the Ricci scalar.

These things are often mentioned in books about (2+1)-GR (like the one of Carlip). The number of DOF's of the gravitational field go like (D-3), so in 2+1 dimensions the gravitational field becomes "trivial" if you don't add any extra terms to the Hilbert action.
 
Im_hammer, are you asking about 2+1 dimensions, or 3 Euclidean dimensions?
 
In three dimensions the Weyl tensor vanishes.
 

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