This is quite well-known: the ONLY metric invariants are curvature, its higher derivatives, and any possible contractions between them.

The meaning of an invariant is, to put it simply, a tensor that is decided by the metric in a "canonical" way, but is independent on local coordinates.

So my question is how such a result is proved?

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    $\begingroup$ Isn't the metric an invariant of the metric? $\endgroup$ Oct 15, 2012 at 2:35
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    $\begingroup$ I think that what you are trying to ask about is the theorem of Weyl that describes the scalar differential polynomial invariants of a metric in terms of contractions of the metric and its inverse with its curvature and its covariant derivatives. (By the way, you have to thrown in the inverse of the determinant of the metric, too, or else you don't get anything.) The proof is by means of Gauss normal coordinates, the Bianchi identities, and the invariant theory of representations of $O(n)$. Probably, the best reference is Weyl's original paper, the title and date of which I do not have handy. $\endgroup$ Oct 15, 2012 at 3:50
  • $\begingroup$ You might find this talk amusing if you are interested in this sort of thing: youtube.com/watch?v=AnjF72ED-m0 $\endgroup$ Oct 15, 2012 at 15:23

1 Answer 1


See section 33 of the book: Ivan Kolár, Jan Slovák, Peter W. Michor: Natural operations in differential geometry. Springer-Verlag, Berlin, Heidelberg, New York, (1993), (pdf).


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