2012-10-15 02:29:21 8 Comments

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?

### Related Questions

#### Sponsored Content

#### 2 Answered Questions

### [SOLVED] Darboux-like theorems

**2016-11-15 06:07:07****user44191****615**View**15**Score**2**Answer- Tags: dg.differential-geometry riemannian-geometry sg.symplectic-geometry kahler-manifolds

#### 1 Answered Questions

### [SOLVED] Who first proved that a vanishing Riemann tensor is sufficient for the existence of Euclidean coordinates?

**2017-03-17 15:39:56****Igor Khavkine****582**View**10**Score**1**Answer- Tags: reference-request riemannian-geometry ho.history-overview curvature

#### 1 Answered Questions

### [SOLVED] Does every smooth manifold admit a metric with bounded geometry and uniform growth?

**2016-07-16 18:44:48****user44172****691**View**5**Score**1**Answer- Tags: dg.differential-geometry gt.geometric-topology mg.metric-geometry riemannian-geometry differential-topology

#### 3 Answered Questions

### [SOLVED] Classification of natural invariants of Riemannian structures

**2015-04-04 09:14:35****makt****564**View**9**Score**3**Answer- Tags: dg.differential-geometry riemannian-geometry differential-topology

#### 3 Answered Questions

### [SOLVED] Changing coordinates so that one Riemannian metric matches another, up to second derivatives

**2010-01-20 23:08:37****Tom LaGatta****1690**View**7**Score**3**Answer- Tags: riemannian-geometry dg.differential-geometry

## 1 comments

## @Peter Michor 2012-10-15 08:01:26

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).