Mathematical Proof: Does Euclidian Geometry Hold in Riemannian?

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SUMMARY

Riemannian geometry with zero curvature is equivalent to Euclidean geometry for two and three dimensions, meaning that proofs established in Euclidean geometry are valid in this specific Riemannian context. However, these proofs do not hold in Riemannian spaces with non-zero curvature. This distinction is crucial for understanding the applicability of geometric proofs across different types of geometrical frameworks.

PREREQUISITES
  • Understanding of Riemannian Geometry
  • Familiarity with Euclidean Geometry
  • Knowledge of curvature tensors
  • Basic concepts of mathematical proofs
NEXT STEPS
  • Research the properties of Riemannian spaces with non-zero curvature
  • Study the implications of curvature tensors in geometry
  • Examine specific examples of Euclidean proofs and their Riemannian counterparts
  • Explore the concept of proofs without words in mathematical literature
USEFUL FOR

Mathematicians, geometry enthusiasts, and educators interested in the relationship between Euclidean and Riemannian geometries.

Patrick Sossoumihen
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Some of proofs without words have been described using Euclidian Geometry, do those proofs still hold alike in Riemmanian Geometry?
 
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Patrick Sossoumihen said:
Some of proofs without words have been described using Euclidian Geometry, do those proofs still hold alike in Riemmanian Geometry?
Hi Patrick:

Riemannian space with zero curvature is the same as Euclidean space (at least for 2 and 3 dimensions). So in principle any Euclidean Geometry proof would also be valid in a Riemannian space with a zero curvature tensor. It would likely not be valid in a Riemannian space with non-zero curvature.

If you can post some examples of particular Euclidean Geometry proofs, I might be able to offer a more specific answer.

Hope this helps.

Regards,
Buzz
 

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