Independent components of the curvature tenso

[PAllen]

There is nothing unreasonable about the curvature tensor in the presence of torsion. It is still perfectly well-defined. The only difference is that the algebraic Bianchi identity becomes

R^a{}_b \wedge e^b = dT^a + \omega^a{}_b \wedge T^b

instead of

R^a{}_b \wedge e^b = 0

The differential Bianchi identity remains

dR^a{}_b + \omega^a{}_c \wedge R^c{}_b - R^a{}_c \wedge \omega^c{}_b = 0

These are easy to see from Cartan's structure equations.

Ok, then I guess it was just a bias of a particular author not to introduce it for spaces with asymmetric connection. They covered other things, but not curvature for such spaces.

 

[Ben Niehoff]

It does become rather unwieldy to introduce torsion if you are using index notation. Both of the identities in my above post end up getting complicated-looking torsion terms. This may be why the authors avoided it.

The Cartan formalism is much more elegant in this regard, as it is possible to deal with torsion almost as simply as without.

Note also that even with torsion, the curvature tensor still has 20 independent components in four dimensions. While the algebraic Bianchi identity is different, it still imposes the same number of algebraic constraints.

 

[Ben Niehoff]

A possible answer to the discrepancy between the metric degrees of freedom and the number of independent components of the curvature tensor could be this:

It is easiest to think of the curvature 2-form as a map R : T_pM \wedge T_pM \rightarrow \mathrm{End}(T_pM); that is, a map that takes two vectors at the point p and spits back a linear transformation acting on the tangent space at p. This linear transformation represents the change in a vector at p going around a small parallelogram bounded by the two vectors given earlier.

In general, we have R(X,Y) \in GL(n, \mathbb{R}), as the holonomy going around a tiny parallelogram can be any invertible linear transformation. However, when we have a metric, we can measure the lengths of vectors, and if our connection is metric-compatible, then the lengths of vectors are not changed by parallel transport (note, this fact is entirely independent from the presence or absence of torsion). Therefore, in any open patch, we can choose a frame in which R(X,Y) \in SO(n). Such a frame is called orthonormal.

It turns out that the number of generators of SO(n) is n(n-1)/2, which is the same as the number of degrees of freedom in the metric tensor, after taking into account diffeomorphism invariance. Hence the degrees of freedom match.

Since it is always possible to find a local orthonormal frame, we can define a "reduced" curvature tensor via

R(X,Y) = \mathcal{M}^{-1} \tilde{R}(X,Y) \mathcal{M}

where \mathcal{M} is a GL(n,R)-valued function on the manifold that relates the coordinate frame to a local orthonormal frame (such a continuous function always exists on an open patch). Then \tilde{R}(X,Y) \in SO(n) always, and we see that even in a coordinate frame, our original R(X,Y) is actually in a subgroup of GL(n,R) that is isomorphic to SO(n).

I have not, however, been able to find a way to count the second-derivatives of the metric tensor to give a consistent answer (but I only tried for a few minutes).

 

[atyy]

In the Berger reference (post #2), he does say that if you go to the normal coordinates, at the origin all second partials are completely specified by the curvature, and he gives an explicit formula. However, he says that this is because normal coordinates contain information relating the metric and the curvature that isn't found in the curvature alone.

 

[PAllen]

Thanks Ben, that helps a lot.

One little discrepancy is that my old differential geometry book (by Synge and Schild) has a proof that the existence of tiny parallelograms implies a symmetric (but not necessarily metric compatible) connection. Is this proof wrong?

More generally, this clarifies that the assumption of metric compatibility of connection, leading to the special coordinates you describe, allows agreement of degrees of freedom. This supports two hypotheses I threw out earlier:

1) Most arbitrarily constructed Riemann tensors would correspond to a non-metric compatible connection.

2) For a connection to possibly metric compatible is a huge restriction because definition of Christoffel symbols amounts to a large number of differential identities. Just as an arbitrarily chosen array of functions is almost never the partials of some scalar function, an arbitrarily chosen connection can rarely satisfy the conditions for metric compatibility.

 

[Ben Niehoff]

One little discrepancy is that my old differential geometry book (by Synge and Schild) has a proof that the existence of tiny parallelograms implies a symmetric (but not necessarily metric compatible) connection. Is this proof wrong?

I would be curious to know what their definitions are. If you ask me, it depends on how tiny you make the parallelograms. Remember that the definition of a manifold requires that it locally look like R^n, so to zeroth order, tiny parallelograms always close.

Torsion measures the first-order failure of tiny parallelograms to close. Curvature measures the second-order failure, after taking into account torsion.

More generally, this clarifies that the assumption of metric compatibility of connection, leading to the special coordinates you describe, allows agreement of degrees of freedom.

Just to clarify, the orthonormal frame does not have to be a coordinate frame; that is, the basis vectors at each point do not have to be pure partials in any coordinate system. They just need to be continuous (and sufficiently differentiable) from point to point.

This supports two hypotheses I threw out earlier:

1) Most arbitrarily constructed Riemann tensors would correspond to a non-metric compatible connection.

Careful, this depends on how many symmetries you impose. Some of the symmetries of the Riemann tensor are equivalent to metric-compatibility. Some of the symmetries are equivalent to torsion-freeness. Only one symmetry, that R^a{}_{bcd} = -R^a{}_{bdc}, is completely generic.

2) For a connection to possibly metric compatible is a huge restriction because definition of Christoffel symbols amounts to a large number of differential identities. Just as an arbitrarily chosen array of functions is almost never the partials of some scalar function, an arbitrarily chosen connection can rarely satisfy the conditions for metric compatibility.

Yes. In general, if you specify the torsion, and you demand that the connection be metric-compatible, then there is exactly one connection satisfying both statements.

 

[PAllen]

1) Most arbitrarily constructed Riemann tensors would correspond to a non-metric compatible connection.

Careful, this depends on how many symmetries you impose. Some of the symmetries of the Riemann tensor are equivalent to metric-compatibility. Some of the symmetries are equivalent to torsion-freeness. Only one symmetry, that R^a{}_{bcd} = -R^a{}_{bdc}, is completely generic.

I would like to understand this better. When I review the symmetries used to derive n^2(n^2-1)/12 for a curvature tensor based on symmetric connection (for simplicity), none of them seem related metric compatibility. They are true for arbitrary symmetric connection (in the derivation I happen to reading, which doesn't bother defining curvature for non-symmetric connection).

Metric compatibility is introduced as further restriction, and the implication is that there may be huge number of symmetric connections that are not metric compatible.

If I understood your argument matching degrees of freedom for the metric compatible case, a key point was length of vectors was not changed by parallel transport. This underpinned the rest of your argument. I interpreted this as a restrictive requirement on possible more general connections. This requirement seems unrelated to any connection properties used to derive algebraic symmetries of the Riemann tensor.

 

[PAllen]

I would be curious to know what their definitions are. If you ask me, it depends on how tiny you make the parallelograms. Remember that the definition of a manifold requires that it locally look like R^n, so to zeroth order, tiny parallelograms always close.

Torsion measures the first-order failure of tiny parallelograms to close. Curvature measures the second-order failure, after taking into account torsion.

This is consistent with their proof. They require first order closure as part of their definition of 'existence of tiny parallelograms'.

 

[Ben Niehoff]

I would like to understand this better. When I review the symmetries used to derive n^2(n^2-1)/12 for a curvature tensor based on symmetric connection (for simplicity), none of them seem related metric compatibility. They are true for arbitrary symmetric connection (in the derivation I happen to reading, which doesn't bother defining curvature for non-symmetric connection).

Looking at each symmetry of the Riemann tensor:

R^a{}_{bcd} = -R^a{}_{bdc}
is always true.

R_{abcd} = -R_{bacd}
comes from metric compatibility. (Note that this symmetry implies that R(X,Y) belongs to a group isomorphic to SO(n), because the generators of SO(n) are all the antisymmetric n x n matrices). And

R^a{}_{bcd} + R^a{}_{cdb} + R^a{}_{dbc} = 0
comes from torsion-free-ness.

The exchange symmetry

R_{abcd} = R_{cdab}
comes from combining the above symmetries; it offers nothing new.

If I understood your argument matching degrees of freedom for the metric compatible case, a key point was length of vectors was not changed by parallel transport. This underpinned the rest of your argument. I interpreted this as a restrictive requirement on possible more general connections. This requirement seems unrelated to any connection properties used to derive algebraic symmetries of the Riemann tensor.

The length of vectors being unchanged by parallel transport is precisely the metric compatibility condition. It relates to the antisymmetry of the first two indices of the Riemann tensor. This is easy to see mathematically. Start with the metric compatibility condition

\nabla_\lambda g_{\mu\nu} = 0
and work out what this means for the Christoffel symbols. Then put this into the Riemann tensor; it will give you that the first two (lowered) indices are antisymmetric.

 

[PAllen]

Looking at each symmetry of the Riemann tensor:

R^a{}_{bcd} = -R^a{}_{bdc}
is always true.

R_{abcd} = -R_{bacd}
comes from metric compatibility. (Note that this symmetry implies that R(X,Y) belongs to a group isomorphic to SO(n), because the generators of SO(n) are all the antisymmetric n x n matrices). And

R^a{}_{bcd} + R^a{}_{cdb} + R^a{}_{dbc} = 0
comes from torsion-free-ness.

Ah, my mistake (not the author's). For general symmetric connections, the covariant form of curvature tensor doesn't exist. I just missed that the derivations of N^2(N^-1)/12 was using the covariant curvature symmetries which don't exist for general symmetric connections.

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.

 

[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...