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I know that curvature $R=\nabla ^{2}$ is power of connection. So why not consider third power of connection and so on? Is it because "all the information" are already captured by connection and curvature? If it is true. In what sense?

user928408
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community May 19 '23 at 09:37
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    Does this answer your question? What comes after the curvature tensor in "higher derivatives"? – Dietrich Burde May 19 '23 at 09:43
  • I mean connection is an operator from tensor field of type$(0,1)$ to tensor field of type$(1,1)$. and curvature is apply twice of connection and Leibniz rule to get an operator from tensor field of type$(0,1)$ to tensor field of type$(2,1)$, so why not apply three times to get an operator from tensor field of type$(0,1)$ to tensor field of type$(3,1)$ and so on? – user928408 May 19 '23 at 09:59
  • Such things appear when people find them useful. Third and fourth order invariants play a significant role in conformal geometry. – Deane May 19 '23 at 14:41

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