Contraction between Levi-Civita symbol and Riemann tensor

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SUMMARY

The contraction of the Levi-Civita symbol with the Riemann tensor, specifically εμνρσ Rμνρσ, equals zero due to the symmetries inherent in the Riemann tensor. The discussion emphasizes utilizing the first Bianchi identity, Rμ[νρσ]=0, and rearranging indices of the Levi-Civita symbol to achieve the proof. Participants confirmed that understanding these symmetries is crucial for solving the problem effectively.

PREREQUISITES
  • Understanding of the Riemann tensor and its symmetries
  • Familiarity with the Levi-Civita symbol and its properties
  • Knowledge of the first Bianchi identity in differential geometry
  • Basic concepts of tensor calculus
NEXT STEPS
  • Study the properties of the Riemann tensor in detail
  • Learn about the Levi-Civita symbol and its applications in tensor analysis
  • Explore the implications of the first Bianchi identity in general relativity
  • Investigate advanced topics in differential geometry related to tensor contractions
USEFUL FOR

This discussion is beneficial for physicists, mathematicians, and students engaged in general relativity and differential geometry, particularly those focusing on tensor analysis and the properties of curvature in spacetime.

mhob
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How to proof that
εμνρσ Rμνρσ =0 ?

Thanks.
 
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Last edited by a moderator:
ShayanJ said:
Use the symmetries of the Riemann tensor!
I'm not sure Rμνρσ = Rμ(νρ)σ or not? If so, the problem solved for me.

Thanks.
 
I thought those symmetries would help, but it seems they don't. Start with the first Bianchi identity and contract all the indices with a Levi-Civita symbol. Rearranging the indices of the Levi-Civita symbol in two of the terms will give you what you want.
 
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Likes   Reactions: mhob
OK, I know use the Bianchi identity
Rμ[νρσ]=0
Thanks!
ShayanJ!
 

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