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 vacuum 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 vacuum 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.