A formula I know for the number of functionally independent components of the curvature tensor is: (n^2)(n^2 -1)/12. It gives 1 for n=2, 6 for n=2, 20 for n=4.(adsbygoogle = window.adsbygoogle || []).push({});

However, for a metric space (with symmetric metric), the curvature tensor is completely specified by the metric tensor. For n=4, there are only 10 different components of the metric, and one can argue there are only 6 functions of the manifold needed to specify the geometry, as all metric representations connected by diffeomorphism represent the same geometry.

How does one square this with 20 functionally independent components of the curvature tensor?

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Independent components of the curvature tenso

Loading...

Similar Threads - Independent components curvature | Date |
---|---|

I Number of independent components of the Riemann tensor | Mar 27, 2017 |

I What are the independent components of the Riemann tensor | Jul 16, 2016 |

No. of Independent Components of Riemann Tensor in Schwartzchild Metric | Jan 14, 2012 |

Counting independent components | Sep 16, 2010 |

Independent Components in Riemann-Christoffel Tensor | Sep 11, 2009 |

**Physics Forums - The Fusion of Science and Community**