Independent components of the curvature tenso

[TrickyDicky]

Good to know I'm not the only one not enlightened by the cryptic posts of George Jones.


PAllen, if I may go back to the begining, do you have some reference where they talk about those "six functional degrees of freedom" of the Riemann tensor. In all the books I look all I get is the equation for the 20 independent components.

 

[PAllen]

Good to know I'm not the only one not enlightened by the cryptic posts of George Jones.


PAllen, if I may go back to the begining, do you have some reference where they talk about those "six functional degrees of freedom" of the Riemann tensor. In all the books I look all I get is the equation for the 20 independent components.

The issue is exactly all I see are derivations of the 20 for the Riemann tensor, and 10/6 for the symmetric metric. I just asked myself (a few days ago) how to reconcile these since a choice of metric completely determines a Riemann (given a metric, you compute a single Riemann). I looked in all my books and searched the internet and could not find anyone else even raising this issue. I've given it some thought myself, but so far have not made any headway. Both derivations seem valid (and must be, given that they are fully established math), but I can't see how to make them fit together.

I am sure there is an explanation, I just have not found it yet (including considering all the answers here so far).

 

[atyy]

1. The metric determines the curvature. if you specify 10 metric components, you immediately specify all 20 curvature components because 10 metric components specify 100 second partials.

2. The curvature does not determine the metric. This is not obvious to me, and is my reading of Berger's text, but can be motivated by noting that 20 curvature components is fewer than 100 second partials.

If you have one function f(x,y), and you differentiate to get one function g(x,y)=∂f/∂x, g(x,y) isn't sufficient to determine f(x,y), since you can integrate to get G(x,y)+z(y), where the "constant of integration" z(y) is not determined by g(x,y). Having *two* first derivatives g(x,y)=∂f/∂x and h(x,y)=∂f/∂y would help to recover *one* f(x,y).

 

[PAllen]

1. The metric determines the curvature. if you specify 10 metric components, you immediately specify all 20 curvature components because 10 metric components specify 100 second partials.

2. The curvature does not determine the metric. This is not obvious to me, and is my reading of Berger's text, but can be motivated by noting that 20 curvature components is fewer than 100 second partials.

If you have one function f(x,y), and you differentiate to get one function g(x,y)=∂f/∂x, g(x,y) isn't sufficient to determine f(x,y), since you can integrate to get G(x,y)+z(y), where the "constant of integration" z(y) is not determined by g(x,y). Having *two* first derivatives g(x,y)=∂f/∂x and h(x,y)=∂f/∂y would help to recover *one* f(x,y).

But this analysis supports the hypothesis of mine you disagreed with earlier. Just with partials of a scalar function: one function on R^n, n partials, but almost all random choices of n functions cannot be partials of one scalar. The equivalent reasoning would seem imply that 'almost all' possible curvature tensors meeting Bianchi and subscript interchange symmetries would fail to correspond to a symmetric metric space.

And my problem remains that no one knowledgeable has:

1) either agreed with this hypothesis and helped explain that is fine and well understood.
or
2) given any explanation of where this reasoning breaks down. The explanations offered so far are either irrelevant or support the hypothesis which seems surprising to me (because I've never seen it hinted at; and then I would expect someone to be able to point to what are the extra differential identities that exclude most possible Riemann tensors).

 

[atyy]

But this analysis supports the hypothesis of mine you disagreed with earlier. Just with partials of a scalar function: one function on R^n, n partials, but almost all random choices of n functions cannot be partials of one scalar. The equivalent reasoning would seem imply that 'almost all' possible curvature tensors meeting Bianchi and subscript interchange symmetries would fail to correspond to a symmetric metric space.

There are 100 second partials in the case of the metric, not just 2 first partials as for f(x,y).

So in the case of f(x,y), one must specify "more than 1 derivative but fewer than 2". Similarly, for the metric, one would expect to have to specify "more than 10 but fewer than 100".

 

[PAllen]

There are 100 second partials in the case of the metric, not just 2 first partials as for f(x,y).

That should make the problem much worse, not better. There are effectively 100 identities to be satisfied rather than 2. Take George's trivial case of f(x,y)= xy . [f,x ; f,y] = (y,x) works. But (x,y) does not. Similarly, a random choice of possible partials for a metric should have probability of 0 of being valid.

 

[atyy]

That should make the problem much worse, not better. There are effectively 100 identities to be satisfied rather than 2. Take George's trivial case of f(x,y)= xy . [f,x ; f,y] = (y,x) works. But (x,y) does not. Similarly, a random choice of possible partials for a metric should have probability of 0 of being valid.

20 < 100, you are more in the situation of having under specified than over specified.

 

[PAllen]

20 < 100, you are more in the situation of having under specified than over specified.

But the 100 is fictitious. If the 100 partials are derived from a metric with 6 functional degrees of freedom, all together, they still only have 6 functional degrees of freedom.

You haven't addressed in any my argument from mapping curvature tensors to metric tensors.

I have another thought, leading me more strongly to the point of view that 'almost all' well formed Riemann tensors will not correspond to any symmetric metric space. Consider the definition of a metric compatible connection: the Christoffel symbol definitions. This definition amounts to a large number of differential identities on the connection. Most arbitrary choices functions arranged to formally look like a symmetric connection cannot be a valid metric compatible connection. Maybe these are the implicit differential identities that support the view that most formally well constructed Riemann tensors don't correspond to symmetric metric spaces.

 

[atyy]

But the 100 is fictitious. If the 100 partials are derived from a metric with 6 functional degrees of freedom, all together, they still only have 6 functional degrees of freedom.

The point of the f(x,y) example is if you specify derivatives, if you freely specify 1 you are underspecified. Of course, if you specify 2, you are over specified, so the second one cannot be "free". So the answer is something like "1.5". Similarly in the case of the metric, we'd exect more than 10 and fewer than 100.

 

[atyy]

That should make the problem much worse, not better. There are effectively 100 identities to be satisfied rather than 2. Take George's trivial case of f(x,y)= xy . [f,x ; f,y] = (y,x) works. But (x,y) does not. Similarly, a random choice of possible partials for a metric should have probability of 0 of being valid.

I agree with the spirit of your point, but for this particular example, why not f(x,y)=(x²+y²)/2

 

[PAllen]

I agree with the spirit of your point, but for this particular example, why not f(x,y)=(x²+y²)/2

Oops, tried to over simplify. I was going to write (x,xy) but 'simpified' too much at the last second.

 

[PAllen]

Maybe the following the issue here (can someone confirm this?):

Call the set of continuous real valued functions of one real variable F. Call the set of 100X100 arrays of continuous functions of 100 real variables G. There exist one-one mappings between these sets (same strength of infinity). However, if one imposes dimensional structures on these sets, and asks for a continuous mapping given structure added to the sets, then it is not possible. Is that what is going on here? That for the structure implied by saying Riemann has 20 independent components, and suitable structure on possible symmetric metrics, there does not exist a continuous mapping; but there can be an arbitrary mapping.

 

[atyy]

If you specify 10 metric components, then you have 0 (not 20) freedom to specify the curvature.

If you specify 0 metric components, then you have the freedom to specify 20 curvature components. In the Riemannian (as opposed to semi-Riemannian, for which I don't know the result) case, specifying the curvature is insufficient to specify the metric (even up to isometry or "diffeomorphism").

 

[Samshorn]

A formula I know for the number of functionally independent components of the curvature tensor is: (n^2)(n^2 -1)/12. It gives 1 for n=2, 6 for n=2, 20 for n=4. However, for a metric space (with symmetric metric), the curvature tensor is completely specified by the metric tensor...

and its derivatives...

For n=4, there are only 10 different components of the metric...

and each of those has first and second partial derivatives with respect to each of the four coordinates...

...one can argue there are only 6 functions of the manifold needed to specify the geometry...

Right, 6 functions, which have 6 numerical values at any given point, but we also have the values of the first and second partial derivatives of the functions at that point, and of course the derivatives at a point are independent of the absolute values of the function at that point.

Think of it this way: To represent all the necessary information for the curvature tensor at a given point, we need the coefficients of the series expansions of those six metric functions about the given point up to second order in each of the coordinates - not just the constant coefficients at that point. Note that these expansion coefficients themselves are different, depending on which point we expand around. Imagine expanding each metric function about each point of the manifold, up to second order in each of the coordinates. So you have a continuous family of expansions. Each coefficient is then a scalar field, which can be represented by a function of the coordinates. So we really have to consider a huge number of potentially independent functions. The question then becomes, in view of this huge number, why are there ONLY 20 independent components of the curvature tensor.

How does one square this with 20 functionally independent components of the curvature tensor?

Once you realize how much information from those six functions is needed to define the curvature tensor, I think your puzzlement will be reversed, and you'll be wondering how on Earth the number of independent components is as low as 20. This is explained in lots of places on the web, for example, see the last section of this web page:

http://www.mathpages.com/rr/appendix/appendix.htm

 

[PAllen]

Thanks, this is helpful, but still a few questions.

Right, 6 functions, which have 6 numerical values at any given point, but we also have the values of the first and second partial derivatives of the functions at that point, and of course the derivatives at a point are independent of the absolute values of the function at that point.

But the derivatives are completely determined by the function. I cannot say choose function f, and independently choose g as its first partial, for example.

If at a given point, I choose a functional value and set of derivatives, fine, but this greatly constrains choices at nearby points.

Think of it this way: To represent all the necessary information for the curvature tensor at a given point, we need the coefficients of the series expansions of those six metric functions about the given point up to second order in each of the coordinates - not just the constant coefficients at that point. Note that these expansion coefficients themselves are different, depending on which point we expand around. Imagine expanding each metric function about each point of the manifold, up to second order in each of the coordinates. So you have a continuous family of expansions. Each coefficient is then a scalar field, which can be represented by a function of the coordinates. So we really have to consider a huge number of potentially independent functions. The question then becomes, in view of this huge number, why are there ONLY 20 independent components of the curvature tensor.
Once you realize how much information from those six functions is needed to define the curvature tensor, I think your puzzlement will be reversed, and you'll be wondering how on Earth the number of independent components is as low as 20. This is explained in lots of places on the web, for example, see the last section of this web page:

http://www.mathpages.com/rr/appendix/appendix.htm

I am well aware of the derivation of the 20. My confusion is relating this to the ability associate a Riemann tensor (considered as a tensor function the whole manifold) to symmetric metric. If one considers F to be the space of all scalar functions of a manifold (a function space), then I see this map from F^6 to F^20 that is expected to cover the latter. I am now thinking this is the core confusion - that there is a contradiction only in context of a choice of topology for each, then asking for the mapping to be continuous against the topology chosen for each. Otherwise, nothing prevents a mapping from the set of f:R->R to the set of all Riemann tensors on a 1000 dimensional manifold. They both have the same cardinality (aleph-2).

 

[Samshorn]

But the derivatives are completely determined by the function. I cannot say choose function f, and independently choose g as its first partial, for example.

Given a function v(x), suppose a system at any given x is characterized by the values v, 2v, and 3v. This system has three components, but only one independent component, because once you tell me the value of (say) v at a given point, the values of 2v and 3v are fully determined. We cannot arbitrarily choose three numbers a0,a1,a2 and claim that these characterize the state of a possible system at a given point, because in general we can’t find a value of v such that v = a0, 2v = a1, and 3v = a2.

But now suppose the system at any given x is characterized by the values v, dv/dx, and d^2v/dx^2. This system too has three components, but how many independent components? Still just one? No, in general it has three independent components. We can arbitrarily choose three numbers a0,a1,a2 and find a function v(x) such that v = a0, dv/dx = a1, and d^2/dx^2 = a2 at the given point. For example, at the origin we can expand v(x) = a0 + a1 x + a2/2 x^2 + …, and we can freely choose a0, a1, and a2. Likewise we can expand the function about any point, and choose the coefficients of the expansion independently. So the system is characterized by three independent components at any given point.

If at a given point, I choose a functional value and set of derivatives, fine, but this greatly constrains choices at nearby points.

True, but the question of how the values and derivatives of a function (or the components of a tensor) at one point constrain the values at neighboring points is completely different from the question of how many independent components there are at a given point. By specifying a value of a continuous function at x0, the value at x0+dx for sufficiently small dx is constrained to approach the value at x0, but we don't thereby say the function has zero degrees of freedom. This is just the definition of continuity.

 

[PAllen]

Given a function v(x), suppose a system at any given x is characterized by the values v, 2v, and 3v. This system has three components, but only one independent component, because once you tell me the value of (say) v at a given point, the values of 2v and 3v are fully determined. We cannot arbitrarily choose three numbers a0,a1,a2 and claim that these characterize the state of a possible system at a given point, because in general we can’t find a value of v such that v = a0, 2v = a1, and 3v = a2.

But now suppose the system at any given x is characterized by the values v, dv/dx, and d^2v/dx^2. This system too has three components, but how many independent components? Still just one? No, in general it has three independent components. We can arbitrarily choose three numbers a0,a1,a2 and find a function v(x) such that v = a0, dv/dx = a1, and d^2/dx^2 = a2 at the given point. For example, at the origin we can expand v(x) = a0 + a1 x + a2/2 x^2 + …, and we can freely choose a0, a1, and a2. Likewise we can expand the function about any point, and choose the coefficients of the expansion independently. So the system is characterized by three independent components at any given point.



True, but the question of how the values and derivatives of a function (or the components of a tensor) at one point constrain the values at neighboring points is completely different from the question of how many independent components there are at a given point. By specifying a value of a continuous function at x0, the value at x0+dx for sufficiently small dx is constrained to approach the value at x0, but we don't thereby say the function has zero degrees of freedom. This is just the definition of continuity.

But in all my posts I was talking about functional degrees of freedom (I was explicit in quite a few). I was wrestling with the seeming discrepancy between a choice of 6 functions to characterize a symmetric metric space versus 20 to not quite characterize it. Plus the mapping argument. The analogy I saw with partials is that if you pick n functions 'at random' to be the partials of some unknown scalar function of n variables, the probability that there is any solution is zero (in the sense that any greater value is wrong).

So far, the only way I see out of my conundrum is the one I proposed on strength of infinity. Going with this, I would say it follows that there isn't a meaningful sense in which 'nearby' metrics must have 'nearby Riemann tensors'; and without this, there is no obstacle to the mapping except for cardinality; and cardinality is clearly aleph-2 for both conceivable symmetric metrics, and conceivable Riemann tensors.

 

[PAllen]

Let me try to restate, as clearly as possible, the open question in my mind (open, because no answers so far address it):

1) Suppose you take 20 random continuous scalar functions of a differentiable manifold (coordinate patch(es), but no geometric structure assigned yet). You build a 'formal' curvature tensor out of them - simply a 4 subscript array meeting all of the purely algebraic symmetries of the Riemann tensor (skew symmetry, interchange symmetry, first Bianchi identity; these alone are sufficient to derive the n^2(n^2-1)/12 algebraically independent components). This is possible for absolutely any choice of 20 functions.

2) What is the likelihood you can find symmetric tensor on the same manifold such that formal curvature tensor would correspond to this symmetric tensor treated as a metric?

My main intuition is that the probability is zero (not that there are no solutions, but that any nonzero probability will be too high). This statement obviously implies some ability apply a measure to the space of candidate Riemann tensors. The basis for this intuition is both the discrepancy in functional degrees of freedom (20 versus 6), and the fact that for the n partials of a function of n variables, picking n random functions as possible partials has zero probability of being the n partials of any possible scalar function.

However I see that this intuition could be wrong if, for some reason, the mapping between a Riemann tensor and its corresponding symmetric metric is of the 'exotic' flavor that only cares about strength of infinity. This would be a surprising fact to me.

I have reviewed books and searched, and cannot find anything referencing this particular question. I have also thought about it myself, to no firm conclusion.

 

[Samshorn]

Let me try to restate, as clearly as possible, the open question in my mind (open, because no answers so far address it):

1) Suppose you take 20 random continuous scalar functions of a differentiable manifold (coordinate patch(es), but no geometric structure assigned yet). You build a 'formal' curvature tensor out of them - simply a 4 subscript array meeting all of the purely algebraic symmetries of the Riemann tensor (skew symmetry, interchange symmetry, first Bianchi identity; these alone are sufficient to derive the n^2(n^-1)/12 algebraically independent components). This is possible for absolutely any choice of 20 functions.

2) What is the likelihood you can find symmetric tensor on the same manifold such that formal curvature tensor would correspond to this symmetric tensor treated as a metric?

My main intuition is that the probability is zero (not that there are no solutions, but that any nonzero probability will be too high). This statement obviously implies some ability apply a measure to the space of candidate Riemann tensors. The basis for this intuition is both the discrepancy in functional degrees of freedom (20 versus 6), and the fact that for the n partials of a function of n variables, picking n random functions as possible partials has zero probability of being the n partials of any possible scalar function.

However I see that this intuition could be wrong if, for some reason, the mapping between a Riemann tensor and its corresponding symmetric metric is of the 'exotic' flavor that only cares about strength of infinity. This would be a surprising fact to me.

I see what you're getting at, although I don't think it exactly represents a connundrum about the numbers of algebraically independent components of the metric and curvature tensors. I think there are a few issues involves in what you're saying: (1) constrained degrees of freedom are still degrees of freedom, so the fact that the sets of parameters satisfying all the constraints may be a small portion of an infinite parameter space doesn't really argue that they are not degrees of freedom, (2) assigning a uniform probability measure to the real numbers is impossible, so the concept of "randomly selecting" functions is very dubious anyway, and perhaps most importantly (3) the concept of "functional degrees of freedom" needs to be defined. Obviously if we didn't limit ourselves to continuous functions, then the concept of "functional degrees of freedom" would be totally meaningless, because we can trivially encode any number of functions into a single function.

Even for continuous functions we would have to define precisely what we mean. For example, one could simply define "functional degrees of freedom" to mean the smallest number of continuous functions from which the target system of functions can be inferred. (This is analogous to definitions of complexity of binary strings as the size of the smallest Turing machine that could generate the string.) With this definition, obviously the number of functional degrees of freedom of the curvature tensor is no more than 6 (for all I know, it could even be less), but this isn't necessarily inconsistent with the fact that the curvature tensor has (up to) 20 algebraically independent components.

Since the Riemann tensor is defined in terms of derivatives of the metric tensor, it isn't surprising that it has more algebraically independent components. For example, the function F=(f,f') with f(x)=ln(x) can be something like ln(x) + 1/x, which can't be expressed algebraically in terms of f alone. But the topic of algebraic independence is quite a bit different from the (somewhat dubious) concept of "functional degrees of freedom". I think the answer to questions about this requires a more precise definition of the concept. Also, I don't think the fact that the parameters may be constrained to a finite region of an infinite parameter space is very relevant, so I wouldn't read much into any attempts to evaluate the "probability" that a "randomly selected" set of functions would satisfy the constraints. In any case, that would be another separate issue, even if it could be given a well-defined meaning.

 

[PAllen]

I see what you're getting at, although I don't think it exactly represents a connundrum about the numbers of algebraically independent components of the metric and curvature tensors. ...

Thanks for the thoughtful response. For me, that is enough to be done with this thread. Getting away from the word random, my basic intuition is almost certainly correct (though not proven):

The statement of 20 algebraically independent components of Riemann is obviously correct and largely unrelated to the question of how likely a choice of 20 arbitrary functions is to lead to a valid Riemann tensor for some unknown metric. Such an approach will 'almost always fail', in practice. Further, this statement is in no way inconsistent with the known fact that given a valid Riemann tensor, it is possible for two (or more) metrics not connected by diffeomorphism to lead to the same Riemann tensor.

 

[Troponin]

PAllen, if you carefully thought about it I think you'd remember the connection used in GR is not only metric compatible but also symmetric. That 's all you need to go from 20 to 10 components IMO.

That's the first thing I thought when he specified GR.

There was a discussion here a few months ago on vanishing Ricci tensors and what it says about "flat" spacetime in n-dimensions in comparison to a vanishing/non-vanishing Curvature tensor that (I think) led to some discussion on independent components.
If it wasn't in that thread, it was in one close enough to the same time period for my brain to lump them together. I'll see if I can find it.

 

[PAllen]

That's the first thing I thought when he specified GR.

There was a discussion here a few months ago on vanishing Ricci tensors and what it says about "flat" spacetime in n-dimensions in comparison to a vanishing/non-vanishing Curvature tensor that (I think) led to some discussion on independent components.
If it wasn't in that thread, it was in one close enough to the same time period for my brain to lump them together. I'll see if I can find it.

Well, the derivation of 20 depends (as Trickydicky noted later) on symmetric connection. The twenty algebraically independent components is true whether or not the connection is metric compatibly; all that is required is that the connection be symmetric. And if the connection is not symmetric, you can't derive a curvature tensor with 'reasonable properties' (or at least so says one of my old differential geometry books).

 

[WannabeNewton]

Well, the derivation of 20 depends (as Trickydicky noted later) on symmetric connection. The twenty algebraically independent components is true whether or not the connection is metric compatibly; all that is required is that the connection be symmetric. And if the connection is not symmetric, you can't derive a curvature tensor with 'reasonable properties' (or at least so says one of my old differential geometry books).

I remember that for one of my exercises from a text I had to show that, without imposing the torsion - free condition on the affine connection, the usual definition of the Riemann tensor (in terms of commutator of parallel transport) picked up a term of the form $\bigtriangledown _{[X, Y]}Z$ for vector fields X, Y, Z. I don't see what unreasonable properties this implies though.

 

[PAllen]

I remember that for one of my exercises from a text I had to show that, without imposing the torsion - free condition on the affine connection, the usual definition of the Riemann tensor (in terms of commutator of parallel transport) picked up a term of the form $\bigtriangledown _{[X, Y]}Z$ for vector fields X, Y, Z. I don't see what unreasonable properties this implies though.

Well, a lot of the 'meaning' of curvature tensor revolves around its relation to the properties of infinitesimal parallelograms. Without symmetric connection, you can't define infinitesimal parallelograms.

 

[Ben Niehoff]

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.

 

[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.