Independent components of the curvature tenso

[PAllen]

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. For n=4, there are only 10 different components of the metric, and one can argue there are only 6 functions of the manifold needed to specify the geometry, as all metric representations connected by diffeomorphism represent the same geometry.

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

[atyy]

http://books.google.com/books/about/A_panoramic_view_of_Riemannian_geometry.html?id=d_ SsagQckaQC, p215:

"one cannot recover from the Rijkh, which form a total of d2(d2 − 1)/12 numbers, all of the second derivatives ∂i∂jgkh, which form a total of (d(d + 1)/2)2 numbers. ... is there enough room between (d(d+1)/2)2 and d2(d2 − 1)/12 to find a nonisometric map between two metrics which still preserves the whole curvature tensor? The subject was initiated quite recently: we know now many examples of nonisometric Riemannian manifolds admitting diffeomorphisms preserving their respective curvature tensors."

[PAllen]

http://books.google.com/books/about/A_panoramic_view_of_Riemannian_geometry.html?id=d_ SsagQckaQC, p215:

"one cannot recover from the Rijkh, which form a total of d2(d2 − 1)/12 numbers, all of the second derivatives ∂i∂jgkh, which form a total of (d(d + 1)/2)2 numbers. ... is there enough room between (d(d+1)/2)2 and d2(d2 − 1)/12 to find a nonisometric map between two metrics which still preserves the whole curvature tensor? The subject was initiated quite recently: we know now many examples of nonisometric Riemannian manifolds admitting diffeomorphisms preserving their respective curvature tensors."

Interesting, but sort of the opposite direction I'm coming from. Roughly, they seem to talk about the possibility of different manifolds having the same curvature tensor. I'm thinking there are too many curvature tensors for the number of distinct manifolds (when the latter are limited to symmetric metric, with 'metric compatible' connection). I'm sure there is a resolution, but I couldn't find it either in my books or my own searches of the internet. I'm thinking that the generic (n^2)(n^2-1)/12 formula does not account for hidden dependencies in connection values due to the fact that a metric compatible connection is completely determined by metric components modulo diffeomorphism.

Thanks for anything you can find.

[TrickyDicky]

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. For n=4, there are only 10 different components of the metric, and one can argue there are only 6 functions of the manifold needed to specify the geometry, as all metric representations connected by diffeomorphism represent the same geometry.

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

You need only ten because you are thinking of the vacuum solution with the ten components of Ricci tensor curvature vanishing , therefore only has the ten nonzero Weyl components.
There are no known solutions in general relativity with particles stress-energy.

[PAllen]

You need only ten because you are thinking of the vacuum solution with the ten components of Ricci tensor curvature vanishing , therefore only has the ten nonzero Weyl components.
There are no known solutions in general relativity with particles stress-energy.

I don't see vaccuum as having anything to do with it - symmetry says the metric has only 10 different components. Then coordinate conditions (which amount to modulo diffeomorphism) say the number of distinct geometries would be specified by 6 functions. Whether exact solutions are known is also completely unrelated - this is an implicit function argument.

[WannabeNewton]

What conditions are you talking about? \triangledown _{\nu }G^{\mu \nu } = 0?

[PAllen]

What conditions are you talking about? \triangledown _{\nu }G^{\mu \nu } = 0?

No, on top of the field equations, you can specify 4 additional, rather arbitrary conditions without loss of generality. These are called coordinate conditions. They arise because any given metric 'expression' is connected to family of geometrically equivalent metrics by coordinate transformation, defined by 4 continuous functions on the manifold. This implies 6 independent functions are sufficient to specify all distinct geometries defined by symmetric metric tensor (in 4-d).

[TrickyDicky]

I don't see vaccuum as having anything to do with it - symmetry says the metric has only 10 different components. Then coordinate conditions (which amount to modulo diffeomorphism) say the number of distinct geometries would be specified by 6 functions. Whether exact solutions are known is also completely unrelated - this is an implicit function argument.

Sorry I might not have read your question correctly, I assumed you were talking about GR.
The metric tensor is not the curvature tensor, you need additional info, the connection, to obtain the curvature tensor.

[PAllen]

Sorry I might not have read your question correctly, I assumed you were talking about GR.
The metric tensor is not the curvature tensor, you need additional info, the connection, to obtain the curvature tensor.

I am talking about GR. In GR, the choice is made to use a metric compatible connection. This is fancy talk for the connection is completely specified by the metric with the standard Christoffel formula. Thus there cannot be more geometrically distinct connections than metric tensors. Then, the curvature tensor is completely determined by these connections. Thus 6 continuous functions should be enough to generate all geometrically distinct curvature tensors. I'm looking for how to coordinate this fact with the curvature tensor is normally said to have 20 functionally independent components (not 4^4 = 256) in dim=4.

[WannabeNewton]

No, on top of the field equations, you can specify 4 additional, rather arbitrary conditions without loss of generality. These are called coordinate conditions. They arise because any given metric 'expression' is connected to family of geometrically equivalent metrics by coordinate transformation, defined by 4 continuous functions on the manifold. This implies 6 independent functions are sufficient to specify all distinct geometries defined by symmetric metric tensor (in 4-d).

Right if you specify initial conditions for the metric you can still make arbitrary coordinate transformations in the future so you end up with 4 arbitrary functional degrees of freedom. Isn't this made explicit by \triangledown _{\nu }G^{\mu \nu } = 0?

[PAllen]

Right if you specify initial conditions for the metric you can still make arbitrary coordinate transformations in the future so you end up with 4 arbitrary functional degrees of freedom. Isn't this made explicit by \triangledown _{\nu }G^{\mu \nu } = 0?

I'm not sure. This is four equations, but my understanding is that this divergence equation is true by construction of G(mu,nu) from the full curvature tensor. Maybe it is equivalent to the ability specify 4 coordinate conditions. I'll have to think on that. Post any further thoughts of yours on this, as well.

[TrickyDicky]

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.

[atyy]

Hmmm, Berger says that 20 curvature conditions are too few to determine the metric, whereas you say those are too many (ie. the 20 cannot all be independent)?

Naively, I'd think it'd be too few, maybe like there are more component Maxwell equations than E and B field components, and yet the fields are underspecified, since one has to add in boundary conditions.

Berger does comment that if we go to Riemann normal coordinates, then at the origin the curvature components do determine all second partials of the metric. However, this is because going to normal coordinates brings in additional information beyond the curvature (I'm not sure what exactly this additional information is).

[PAllen]

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.

Right, I thought about that later, but didn't mention it. So then there is only the discrepance from 10 to 6.

[Edit: wait, I'm not sure it decreases it by that much. You may be right. Can you post an argument or reference that symmetric metric reduces independent curvature components that much?]

[PAllen]

Hmmm, Berger says that 20 curvature conditions are too few to determine the metric, whereas you say those are too many (ie. the 20 cannot all be independent)?

Naively, I'd think it'd be too few, maybe like there are more component Maxwell equations than E and B field components, and yet the fields are underspecified, since one has to add in boundary conditions.

Berger does comment that if we go to Riemann normal coordinates, then at the origin the curvature components do determine all second partials of the metric. However, this is because going to normal coordinates brings in additional information beyond the curvature (I'm not sure what exactly this additional information is).

The whole point of my question is how to square arguments like this with the fact that (in GR at least), the metric completely determines the curvature tensor and has only 6 functional degrees of freedom.

[atyy]

Hmm, let's see: the metric defines the curvature, but the curvature does not define the metric.

I thought you were asking about the second part of that statement, but you are asking about the first?

[TrickyDicky]

Right, I thought about that later, but didn't mention it. So then there is only the discrepance from 10 to 6.

[Edit: wait, I'm not sure it decreases it by that much. You may be right. Can you post an argument or reference that symmetric metric reduces independent curvature components that much?]
I don't have a reference, but you certainly save all the torsion tensor components of the curvature, it is kind of obvious.

[TrickyDicky]

From ten to six usually the argument is used that four of the ten would be fixed by the test particle coordinates, but this would only work for vacuum solutions, as I explained above no other kind of solutions are known for test particle stress-energy so no one cares much about that discrepancy (two body problem is still not solved in GR).

[PAllen]

I don't have a reference, but you certainly save all the torsion tensor components of the curvature, it is kind of obvious.

I'ts obvious to me you save some. It's not obvious to me that you get from 20 to 10 just by assuming symmetric connection.

[PAllen]

From ten to six usually the argument is used that four of the ten would be fixed by the test particle coordinates, but this would only work for vacuum solutions, as I explained above no other kind of solutions are known for test particle stress-energy so no one cares much about that discrepancy (two body problem is still not solved in GR).

I don't see the 10 to 6 argument as having anything to do with vaccuum/non-vaccuum. It is derived assuming nothing other than that the metric is symmetric, and you have diffeomorphism invariance.

[PAllen]

Hmm, let's see: the metric defines the curvature, but the curvature does not define the metric.

I thought you were asking about the second part of that statement, but you are asking about the first?

I am asking you one can really say there are 20 functionally independent components of curvature when it is completely derived from tensor with 6 functional degrees of freedom. So, in least in the context of assuming a symmetric metric tensor (which leads to the 6 figure), this suggests to me there must be a great many implicit dependencies between the curvature components, beyond the 20 figure derived from Bianchi identities and interchange symmetries.

[TrickyDicky]

I don't see the 10 to 6 argument as having anything to do with vaccuum/non-vaccuum. It is derived assuming nothing other than that the metric is symmetric, and you have diffeomorphism invariance.

I provided you with the argument usually given.
Do you have any reference with the 10 to 6 components derivation? If so, can you share it? But then why would you ask for it if you already have that derivation.

[TrickyDicky]

I'ts obvious to me you save some. It's not obvious to me that you get from 20 to 10 just by assuming symmetric connection.

Well, it turns out not even that you save some should be obvious because the Riemann tensor formula that gives the 20 components is constructed accounting already for the fact that the Levi civita connection is used.
Sorry about the misleading lead.
I'm really curious about this too, hopefully somebody will enlighten us.

[George Jones]

If f\left(x,y\right) = xy, are \partial_x f and \partial_y f independent?

[TrickyDicky]

Surely not, and? You mean I was right, then?

[PAllen]

I provided you with the argument usually given.
Do you have any reference with the 10 to 6 components derivation? If so, can you share it? But then why would you ask for it if you already have that derivation.

I am not asking about the 10 to 6, I understand that fully (I think). I am asking about how to understand this fact along with the derivations that there are 20 independent components to the curvature tensor (which I also understand, read in isolation). Maybe something as simple as " Oh, this is well known to the experts, it is covered in such and such".

As for the 10 to 6, the books I have don't formally derive, the simply justify with the following type of argument (in several variants):

Supposing you have the set of all possible symmetric metric tensor representations. Pick one; then all other representations connected to this one by diffeomorphism are describing the same geometry. Diffeomorpthisms are described by a 4 functions in 4-dim. Thus 4 identities can be added without loss of generality (effectively picking one representation from those equivalent by diffeomorphism). Four identities applied to 10 functions leaves you 6 functional degrees of freedom.

[PAllen]

If f\left(x,y\right) = xy, are \partial_x f and \partial_y f independent?

Well, x and y (the partials) are independent, but you cannot arbitrarily choose functions for the partials and have them satisfy the condition that they represent partials. You have only one functional degree of freedom due the requirement that they be partials of one function.

So is this mostly a matter of terminology? Different senses of the word independent? What is a clear terminology to use here?

I mean, by one set of arguments, we have there there 20 independent components to the curvature tensor. By another set of arguments, we conclude that there are only 6 functional degrees of freedom in the choice distinguishable metric spaces with symmetric metric and metric based connection. So am I right that most possible arbitrary functional choices for the curvature tensor [even after satisfying Bianchi identities and interchange symmetries] would fail to be the curvature tensor for any symmetric metric space? In the same way the two arbitrary functions are unlikely to be partial derivatives of any possible single function?

[George Jones]

Here, I think "independent" means "linearly independent" in some appropriate space.

[pervect]

Choose some coordinate basis.

Then you can write the components \Gamma_{\alpha\,\beta\,\gamma} in terms of the metric coefficients and their derivatives.

Proceeding on, you can write \Gamma^{alpha){}_{\beta\,\gamma} in terms of the previous plus the metric, then you can write R^{\alpha}{}_{\beta\,\gamma\,\delta) in terms of \Gamma_{\alpha\,\beta\,\gamma} and it's derivatives \partial_{\mu} \Gamma_{\alpha\,\beta\,\gamma}

So the first fundamental question you need to answer is George's - simplifying it to one variable, are f(x) and \delta_{x} f(x)) independent?

I'm not sure I know what context you're thinking about the question in but in the context of the evolution of f(x) evolving as an initial value problem satisfying a second-order differential equation, they're generally independent.

This is IMO a reasonable context, because if you think of some globally hyperbolic space-time evolving in terms of an initial value formulation, the evolving metric satisfies some second order hyperbolic quasi-linear differential equation

But that doesn't mean it's the context you were thinking of the problem in. But if you consider the remarks and answer George's question, you should be able to answer your own.

[TrickyDicky]

Ok so the symmetric connection and symmetric metric reduces the 20 components to 6, not to ten, then?

[DrGreg]

Here, I think "independent" means "linearly independent" in some appropriate space.

If f\left(x,y\right) = xy, are \partial_x f and \partial_y f independent?

I think these are pertinent questions to try to get to the bottom of what is meant by "independent" in this thread.

So I could put it another way to PAllen: if f is a scalar field on the Riemannian manifold \mathbb{R}^2, how many independent components does f have? And how many does \nabla_{\alpha}f have?

[PAllen]

I think these are pertinent questions to try to get to the bottom of what is meant by "independent" in this thread.

So I could put it another way to PAllen: if f is a scalar field on the Riemannian manifold \mathbb{R}^2, how many independent components does f have? And how many does \nabla_{\alpha}f have?

I am asking about correct terminology because I don't know, and it seems some books are sloppy in what they mean in different cases.

However, my own personal concept would be:

f has one one indpendent component, del f formally has 2, but only one functional degree of freedom because of the identities it must follow. In fact, if you pick components of del f randomly, the result will 'almost never' be valid.

Be that as it may, I am asking one very specific, non-terminilogy question that seems to be true but is surprising to me:

Imagine the set of all 4-subscript tensors (using archaic language) in dim 4 that satisfy all of identities and subscript interchange symmetries used in deriving the figure of 20. Pick one at random. Is it true, that with probability of 1 it will fail to be the curvature tensor of any symmetric metric space?

[atyy]

Imagine the set of all 4-subscript tensors (using archaic language) in dim 4 that satisfy all of identities and subscript interchange symmetries used in deriving the figure of 20. Pick one at random. Is it true, that with probability of 1 it will fail to be the curvature tensor of any symmetric metric space?

It's the other way: it will correspond to at least one metric, but maybe more than one.

Roughly, there are 100 (or something like that) second partials of the metric to determine, and 20 isn't enough.

[TrickyDicky]

This is the funniest thread I've participated in.

[PAllen]

It's the other way: it will correspond to at least one metric, but maybe more than one.

Roughly, there are 100 (or something like that) second partials of the metric to determine, and 20 isn't enough.

Well, the we have a contradiction. Six functional degrees of freedom (10 components plus 4 identities) determine all possible symmetric metric tensors. From each metric, there is exactly one curvature tensor (this is not inconsistent with saying some curvature tensors may result from more than one non-isometric metric). If I map these metrics to their corresponding curvature tensor, how do they cover a space of 20 independent functions?

Something doesn't make sense, and so far none of the answers help. The example of the gradient doesn't help because there it is true that most random choices fail; and for the gradient there is differential identity that makes the counting consistent. The analogy for the curvature tensor is that there should be at least 10 identities on top of the well known ones to make the counting match.

[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 continous 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 continous 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.

[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. :rofl:

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 Relativity: an introduction to special and general relativity (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? :biggrin:

[Ben Niehoff]

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