Independent components of the curvature tenso

[Ben Niehoff]

diffeomorphism to an orthonormal frame

I would be careful with the language here, as these two concepts have little to do with one another. In this context, you mean a passive diffeomorphism, or a coordinate change. But a choice of frame has, in general, nothing to do with coordinates. A frame is a collection of smooth vector fields on some open subset U such that the vector fields are linearly independent at every point in U. As I mentioned earlier, a frame need not be a coordinate frame (in fact, a frame is a coordinate frame if and only if the Lie brackets between all pairs of vector fields vanish).

Let me see if I can understand your argument on degrees of feedom as follows:

Despite the 20 algebraically independent components of a Riemann tensor satisfying the common symmetries (including, blush, those of the covariant form which imply metric compatibility), and despite the fact that diffeomorphism invariance would not normally reduce degrees of freedom so much, under these assumptions there is, in fact, always a diffeomorphism to an orthonormal frame, and in such a frame the Riemann tensor must take a special form which has only (n)(n-1)/2 degrees of freedom.

Replacing the word "diffeomorphism" with "an invertible linear transformation on each tangent space, chosen continuously on some open patch U", I would say yes.

In fact, I think it can be worded a better way: Given a metric-compatible connection, R(X,Y) at any point P is an element of the group that preserves a metric sphere in the tangent space at P (where a sphere is defined as the locus of points of distance 1 from the origin, according to the inner product given by the metric tensor). The group that preserves a unit sphere in R^n is precisely SO(n), which has dimension n(n-1)/2.

 

[PAllen]

removed oops post.

Can you explain 14? I naively see 40 = 4 * 10 equations.

 

[Ben Niehoff]

Whoops. I multiplied 4 x 10 and got 14.

I've deleted the previous message. :P

 

[PAllen]

I guess I am still confused about something here, and different responders here have had different intuitions about this; and as informative as your information is, I don't see the answer to this.

To be very concrete, let's say I have very large library of elementary functions (x, x^, x^3, ... ,sin x,cos x, sinh x, cosh x, exp(x), ln(x), ... ), and rules for combining them, and including various constants, and randomly substituting variables. These combined with a random number source (including random number of elements to combine) produces such things as:

3x^137 sinh(exp (cos y^42))

Pick 20 functions of 4 variables from this random function supplier. Arrange them to meet all the algebraic requirements of a Riemann tensor for dim 4 [this is possible for any choice]. My intuition remains that the probability that the result is a valid, metric compatible Riemann tensor for any possible metric is vanishingly small.

 

[Ben Niehoff]

Right, I understand the conundrum. We know that metric-compatibility (for some metric) implies anti-symmetry in the first two (lowered) indices; then, given fixed torsion, the Riemann tensor has 20 algebraically-independent components.

The question is, given antisymmetry in the first two (lowered) indices, and a second algebraic constraint related to a fixed torsion tensor (thus reducing the Riemann tensor to 20 components), does it always follow that the resulting tensor must be the curvature derived from some metric (and the given torsion)?

I'm not entirely sure. It may well be that a tensor with the same symmetries as Riemann is not necessarily a curvature tensor. I would guess that there are additional constraints that must be satisfied, due to the relations that must exist between second partials of the metric.

One has to be careful counting degrees of freedom, however, because differential constraints are weaker than algebraic ones (due to the fact that differentiation destroys information). For example, one function of 4 variables has 4 first partials, but there are 6 pairwise differential constraints between them!

 

[atyy]

I'm still bothered by the differential constraints, which, contrary to what I had thought, but now agree with PAllen, aren't ruled out by the fact that specifying all curvature components is under-specifying the metric in some cases.

A reference that may be useful but I haven't read:
Determination of the metric tensor from components of the Riemann tensor
C B G McIntosh and W D Halford
J. Phys A: Math. Gen. 14 (1981) 2331-2338.

 

[atyy]

I haven't read this either, but a paper that cites the McIntosh & Halford one is:
S. Brian Edgar
Sufficient conditions for a curvature tensor to be Riemannian and for determining its metric
J. Math. Phys. 32, 1011 (1991); doi:10.1063/1.529376.

 

[atyy]

This article states that PAllen's intuition, that additional constraints are needed for the curvature to be integrated to a metric, is correct.

Hernando Quevedo
Determination of the metric from the curvature
General Relativity and Gravitation, Volume 24, Number 8, 799-819, DOI: 10.1007/BF00759087

Let Tabcd be the components of a tensor defined on a coordinate domain of M. ... It is easy to see that this tensor possesses the same symmetry properties as those of the curvature tensor. ... find a metric tensor gij (or equivalently a tetrad vga) in terms of a and b such that its curvature tensor Rabca coincides with the tensor Tabca. Note that the manifold M does not carry any connection; it is, therefore, not possible to say whether or not Tabcd is a curvature tensor. This is true only if Tabcd satisfies the Bianchi identities which involve its covariant derivatives. Indeed, the Bianchi identities can be interpreted as the integrability condition of the differential equations relating the components of a curvature tensor with the metric components (see, for instance, Ref. 25).

Quevedo's Reference 25 is Stephani's http://books.google.com/books?id=WAW-4nd-OeIC&source=gbs_navlinks_s". On p143 Stephani writes "The determination of the metric from a specified curvature tensor amounts ... to the solution of a system of twenty second-order differential equations for the ten metric compoenents ... In general such a system will posess no solutions ..." He then lists explicitly the additional differential constraints needed.

PAllen, much thanks for bringing this up! I had asked myself this question before, and resolved it wrongly based on my reading of Berger (who says nothing wrong, I just over-interpreted him). Unfortunately, all this now irrelevant, since Lorentzian spacemites don't exist, right?

 

[Ben Niehoff]

Ah, Lorentzian spacemites, foul vermin of the universe...