Ricci tensor: symmetric or not?

A diagonal metric is not sufficient.So far the Einstein's tensor is defined from the Ricci tensor (the generalization) in the same way as in the Riemannian geometry. It is in the expression of the energy-momentum tensor that torsion intervenes. So I think I have the answer!In summary, the question is asking whether the Ricci tensor is symmetric or anti-symmetric in a torsion-free affine connection. The confusion arises from two different references giving different answers, one stating that it is anti-symmetric and the other stating that it is symmetric. However, the correct answer is that the Ricci tensor is symmetric, as demonstrated in a two-dimensional space that is locally Euclidean and in general relativity.
  • #1
member 11137
I am really confused and the question can appear to be trivial or stupid:

Is the Ricci tensor symmetric or anti-symmetric in a torsion-free affine connection?

I am full of troubles since two different references gives two different answers (sorry no one is in english language but one of them is Lichnerowicz and the latter indicates an antisymmetric Ricci tensor whilst the other -based on the tensor calculus and on the expression of the curvature tensor Rabcd- tells me that Ricci is symmetric) !

Thanks for some feed-back and sorry if it is not the good place to post.
 
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  • #2
It's symmetric. Here are a couple of simple examples.

In a two-dimensional space that's locally Euclidean, the only intrinsic measure of curvature is the Gaussian curvature K, and the Ricci tensor is diag(K,K).

In general relativity, the Ricci tensor is closely related to the Einstein tensor, and the Einstein tensor is proportional to the stress-energy tensor, which is symmetric.
 
  • #3
My textbooks all agree that the Ricci is symmetric.
 
  • #4
pervect said:
My textbooks all agree that the Ricci is symmetric.

You have chance (!) or I missed something in Lichnerowicz [01]. Anyway, thanks to you and to bcrowell.

Just to develop a little bit more the reason of my confusion. The chapter I have read in [1] concerns a change of perspective. In a first attempt equations of the field (Einstein) are written for any affine connection. The author then tries to rewrite these equations for a torsion-free connection. This yield a relation between the old Ricci tensor and the new one. In that relation a covariant vector (anyone) appears in such a manner that the tensor becomes antisymmetric... It is also explained that that tensor is closely related to the Einstein tensor... The title of § 80 reads: "The antisymmetric contracted curvature tensor and the Einstein's tensor"...

So where is the mistake?
 
  • #5
Where in Lichnerowicz do you find antisymmetry? In "Theorie globale des connexions...? Which section? Which formula? Or somewhere else?
 
  • #6
"The title of § 80 reads: "The antisymmetric contracted curvature tensor and the Einstein's tensor"... "

What is the rest of the quote?
 
  • #7
Ricci rotation coefficients (connection)? Just guessing.
 
  • #8
George Jones said:
Ricci rotation coefficients (connection)? Just guessing.

I think you must be very very closed to the correct explanation. But it does not make it clearer for me... may I get some ligth about this from you?

The book is: "Théories relativistes de la gravitation et de l'électromagnétisme"; Collège de France, introduction from Darmois; Masson and Co editors, Paris. A very old book (1955) but with very interesting demonstrations. One feels that many ways were open for physics at this time. The Einstein-Schrödinger, the Einstein and the Klein Kaluza theory are explained... a.s.a... Progresses have been made in between, of course.
 
  • #9
I have the book. Please, tell me which page...
 
  • #10
arkajad said:
I have the book. Please, tell me which page...

Oh, really; nice! Ok: § 79 page 266 and around (+ or -). Thx if you can help me a little bit. I have in between "googled". Ricci coefficients and Christoffel's symbols of second kind: related? I thought that these symbols were symmetric! Oh, I think I must make a stop.
 
  • #11
But you have a particular connection with torsion there (only "vecteur covariant de torsion est nul").

Ricci tensor is symmetric for the Levi-Civita connection (thus no torsion).
 
  • #12
arkajad said:
But you have a particular connection with torsion there (only "vecteur covariant de torsion est nul").

Ricci tensor is symmetric for the Levi-Civita connection (thus no torsion).

I'm sure you meant Christoffel connection--I also confuse the two.

And yes, that must be it. In defining the covariant derivative, there are about half a dozen constraints such as linearity, the Leibniz product rule and metric compatibility. There are those that single-out one unique connection, one of which is the Christoffel connection such that it is symmetric in its lower indices (torsion free).
 
  • #13
It is called http://en.wikipedia.org/wiki/Levi-Civita_connection" [Broken].
 
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  • #14
arkajad said:
It is called http://en.wikipedia.org/wiki/Levi-Civita_connection" [Broken].

OK. I'm always getting the terms mixed up. Did I say that already? Anyway, it seems "Levi-Civita connection", "Christoffel connection", and sometimes the "Riemannian connection" are variously used among authors to mean the same thing.
 
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  • #15
arkajad said:
It is called http://en.wikipedia.org/wiki/Levi-Civita_connection" [Broken].

If you read the wikipedia lik attentively you will certainly state, as me, that Christoffel's symbols have a symmetry on the subscripts: {abd} = {bad}. So your explanation does not hold.

But if you read arXiv: gr-qc/9912087v1 21 Dec 1999, equation (2.1), you get the answer I think.
 
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  • #16
Blackforest said:
If you read the wikipedia lik attentively you will certainly state, as me, that Christoffel's symbols have a symmetry on the subscripts: {abd} = {bad}.

Of course they do - for the torsion-free Levi-Civita connection in a coordinate basis.
 
  • #17
arkajad said:
But you have a particular connection with torsion there (only "vecteur covariant de torsion est nul").

Ricci tensor is symmetric for the Levi-Civita connection (thus no torsion).

Yes if you mean the sentence page 266 after (78-4). It is, in § 78, the starting point of the problematic. The author begins with a affine connection "L" without torsion. But the purpose of § 79 is to find a new connection (the gamma one) and to connect it with the L connection. The result is (I go over the steps) an antisymmetric Ricci tensor (79-9) and (79-10).
 
  • #18
Blackforest said:
The author begins with a affine connection "L" without torsion.

Where do you find that the connection L is assumed to be without torsion?
 
  • #19
arkajad said:
Where do you find that the connection L is assumed to be without torsion?

page 264 after (78-4)!
 
  • #20
You must distinguish between torsion tensor [tex]S^{\alpha}_{\beta\gamma}[/tex] and covariant torsion vector (64-1)

[tex]S_\alpha=S^\rho_{\alpha\rho}[/tex].

Connection can have zero torsion vector, but non-zero torsion tensor. An that is what Lichnerowicz is examining.
 
  • #21
arkajad said:
You must distinguish between torsion tensor [tex]S^{\alpha}_{\beta\gamma}[/tex] and covariant torsion vector (64-1)

[tex]S_\alpha=S^\rho_{\alpha\rho}[/tex].

Connection can have zero torsion vector, but non-zero torsion tensor. An that is what Lichnerowicz is examining.

Thanks +++. So far I understand § 64, a confusion in the language is possible since a Ricci tensor can be the generalization of the Ricci tensor of the Riemannian geometry (§ 64 b). The latter is symmetric but its generalization is not necessarily symmetric (the sentence under 64-4). This is the answer... so far.

Super help.

That means one must really state with precision the geometry one is working with...
 
  • #22
Alright. If you will ever need some other help - don't hesitate. And Lichnerowicz is not very easy to read...
 
  • #23
arkajad said:
Alright. If you will ever need some other help - don't hesitate. And Lichnerowicz is not very easy to read...

Alright... I think that I need some more help concerning that point. Is is correct to say that in Levi-Civita connection non symmetry of the Ricci tensor is equivalent to discontinouus metric? Does the concept of discontinouus metric have any relevant signification for physics? I have tried to find some explanations in the work of H. Kleinert (the new equivalence principle) who insist on the unicity of the metric but on the multivaluated tetrads. I don't have a clear overview on that problematic of the discontinuities...
 
  • #24
Where did you read, I mean in mathematically coherent text, about "discontinuous metric"?
 
  • #25
arkajad said:
Where did you read, I mean in mathematically coherent text, about "discontinuous metric"?

It is an article (arXiv... sorry I don't remember the n°; researched in Google) with title: The non holonomic mapping principle (or something similar). I shall try to give the exact reference in a few days. Any way I effectively got a problem with the coherence of some equations in that reference. In fact I was looking for works where that item (discontinuities of the metric) has still been explored with the hope to make progresses in my own approach. In between I could discover that such discontinuities are motivating the Regge calculus... So I come back later for a discussion.

Thanks
 
  • #26
There is nothing about discontinuous metric in http://users.physik.fu-berlin.de/~kleinert/258/258j.pdf" [Broken]. There is just a standard talk about torsion. So, where did you read about discontinuous metric.
 
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  • #27
arkajad said:
There is nothing about discontinuous metric in http://users.physik.fu-berlin.de/~kleinert/258/258j.pdf" [Broken]. There is just a standard talk about torsion. So, where did you read about discontinuous metric.

This is true. Equation (18) is the proof for that. (Apart from this, the problem of coherence I mentionned is the r.h.t of (24)). The reason why this article is interesting is that it is mixing a discontinuity: (7) and a continuity (18), thus a pedagogic illustration of the "boarder" I tried to explore. This was motivating my two questions: see post 23. Is there real contexts where (18) would not be true? I got a part of the answer in some arXiv articles but as said before I don't really have a picture (a concrete example where I could apply calculations and test hypotheses) for such situations. I cannot and I will not develop more extensively because it would hurt the rules of the forum.
 
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  • #28
Blackforest said:
Is there real contexts where (18) would not be true?

As far as I know this subject is open. I myself would like to know the answer. But, suppose, one can develop such a theory. What would be the gain?
 
  • #29
arkajad said:
As far as I know this subject is open. I myself would like to know the answer. But, suppose, one can develop such a theory. What would be the gain?

The gain? The initial question about the symmetry of the Ricci tensor appears to be a strategic and crucial one for the understanding of the generalized theory of relativity - alias GTR- (which is the main purpose of that subforum here). As far I understand this point (symmetry or not of the Ricci tensor), it is exactly situated (in the construction of theories) at the limit between the well accepted GTR and extensions of it. The most popular of the viable extensions is the Einstein Cartan Theory (ECT)... and the latter has some very useful applications (the gain)... but here is also exactly the point where we leave the territory of the allowed discussions.

Thanks for the help.
 

1. Is the Ricci tensor symmetric?

Yes, the Ricci tensor is symmetric. This means that the values on either side of the diagonal are equal, making it a square matrix.

2. What does it mean for the Ricci tensor to be symmetric?

A symmetric Ricci tensor means that the curvature of space-time is symmetrical and does not favor any particular direction. This is a key component of Einstein's theory of general relativity.

3. How is the Ricci tensor related to the Riemann curvature tensor?

The Ricci tensor is a contraction of the Riemann curvature tensor, which is a mathematical object used to describe the curvature of space-time. The Ricci tensor is found by summing the components of the Riemann tensor along one of the indices.

4. What does the Ricci tensor tell us about the geometry of space-time?

The Ricci tensor gives us information about the curvature of space-time. It helps us understand how objects with mass and energy interact with the space-time around them, and how this curvature affects the motion of particles.

5. Why is the Ricci tensor important in Einstein's theory of general relativity?

The Ricci tensor is a key component of Einstein's theory of general relativity, which describes the relationship between gravity and the curvature of space-time. It helps us understand how mass and energy affect the geometry of space-time, and how this affects the motion of objects in the universe.

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