Talk:Ricci curvature

message: The article on the trace-free Ricci curvature is silly as it stands. It assumes the manifold has dimension 4, but never says so!!
In dimension n, the correct formula is Ric - \frac{S}{n}g. 
70.18.30.252 11:11, 7 August 2006 (UTC)
replies: text: It might be best to just start with a general relativity section in this article. If and when that section grows too large we can move it to its own article. For now I agree that trace-free Ricci tensor should be merged in here. -- Fropuff 16:46, 31 January 2006 (UTC)
replies: text: What fropuff said. --MarSch 13:19, 23 April 2006 (UTC)
thread_title: Proposed merger

message: It seems to me that the statement "Because the Levi-Civita connection is torsion-free, the Ricci tensor of a Riemannian manifold is symmetric, in the sense that
Ric(ξ,η) = Ric(η,ξ)" is not correct.
In fact, a symmetric linear connection (torsion free) does not have necessarily a symmetric Ricci curvature as the following example shows:
In R^n, let's define a symmetric linear connection (covariant derivative) by the formula:
(\nabla_{A} B) = B'.A + Γ(A,B)
where Γ(A,B)(x) = (A·x)(B·x)u, where · stands for the standard scalar product on R^{n} and u is a fixed vector.
One can easily compute the Ricci curvature of that symmetric linear connection and find:
Ric(A,B) = (u·x)(A·B) - (A·x)(B·u)
which is not symmetric because of the term (A·x)(B·u).
Hence I would suggest to replace that statement by:
"The Ricci tensor of a connection derived from a metric is always symmetric, in the sense that
Ric(ξ,η) = Ric(η,ξ)".
Boris Kolev (France)
The correct general statement, now incorporated into the article,  is that any torsion-free connection with SL(n,R) holonomy has symmetric Ricci tensor. This holonomy condition is equivalent to saying that the connection is compatible with a volume form. It implies that 
{R^k}_{kij}= 0
since the curvature tensor is always  a 2-form with values in the Lie algebra of the holonomy group. However, 
{R^k}_{kij}+{R^k}_{ijk}+{R^k}_{jki}=0
for any torsion-free connection, as follows by contracting the first Bianchi identity. Hence 
{R}_{ij}-{R}_{ji}={R^k}_{kij}=0
as claimed. Voila!
Riemannian Geometer, 3/29/09  —Preceding unsigned comment added by 71.245.247.96 (talk) 00:39, 30 March 2009 (UTC)
thread_title:  A small erroneous statement 

message: I suspect that the reference of Ricci's paper was a bad OCR job from the EOM. I think venelo should be veneto, as in "Atti R. Inst. Venelo" maybe stands for "Atti del Reale Istituto Veneto di Scienze, Lettere ed Arti", though there are a couple of similarly named journals. —The preceding unsigned comment was added by Lionelbrits (talk • contribs) 16:36, 14 April 2007 (UTC)
thread_title:  Reference 

message: The introduction of the article seems to say that the Ricci tensor is symmetric for all pseudo-Riemannian manifolds. In a book I'm reading at the moment, it says that an affine connection ∇ with zero torsion has symmetric Ricci tensor if and only if ∇ is locally equi-affine. Where we call an affine connection locally equi-affine if around each point x of M there is a parallel volume form, i.e. a non-vanishing n-form ω such that . Which one is correct, the article or the book? It seems to me that there are some missing hypotheses in the article's statement. — Fly by Night (talk) 19:23, 25 June 2011 (UTC)
replies: text: In differential geometry, "curvature of a Riemannian manifold" means "curvature of the Levi-Civita connection".  There is no missing hypothesis.  The article discusses the non-symmetric case further down, but it would be very inappropriate to emphasize this in the lead, given the relative WP:WEIGHT.  Sławomir Biały (talk) 21:15, 25 June 2011 (UTC)
thread_title:  Symmetric? 

message: I know this is generally acceptable, but saying that the Riemann curvature tensor is a (1,3)-tensor is a bit confusing here, since it is actually used as a (0,2)-tensor with values in Hom(T_pM,T_pM) (this is also stated in the "Riemann curvature tensor" wiki page, by the way). At least it should be said that R can be viewed as (1,3) tensor using the identification between linear maps and (1,1)-tensors. Otherwise, Ric could be defined as tr^{1,2}R=-tr^{1,3}R. — Preceding unsigned comment added by 151.15.165.83 (talk • contribs) 2014-04-11T18:54:28
:I would support clarifying the article on this point.  Sławomir Biały (talk) 20:38, 11 April 2014 (UTC)
thread_title:  Definition of Riemann curvature tensor 

message: In this article, it is said that the Ricci tensor
"represents the amount by which the volume of a geodesic ball in a curved Riemannian manifold deviates from that of the standard ball in Euclidean space".
But there is the exact same description in the article about the Ricci scalar...
Since there are spaces where the Ricci tensor isn't zero and the Ricci scalar is (the product of a constant positive curvature space with a constant negative curvature space at same scale), this explanation can't hold for both... 
My guess is the scalar mesures the change of form while the tensor mesures the change of volume. 
Except that volume difference can be expressed both in terms of the scalar (https://en.wikipedia.org/wiki/Scalar_curvature#Direct_geometric_interpretation) and in terms of the tensor in this article (https://en.wikipedia.org/wiki/Ricci_curvature#Direct_geometric_meaning).
How come ? Is there an error somewhere or am I wrong ? JohannCR (talk) 14:47, 19 December 2016 (UTC)
replies: text: The direct meaning is clarified in the section Direct geometric meaning.  I've added this to the lead to help clarify the difference with the scalar curvature. Sławomir Biały (talk) 15:53, 19 December 2016 (UTC)
replies: text: Oh, so that means the Ricci scalar really measures differences in volume "globaly" (meaning the whole ball), while the Ricci tensor measures it only partially or depending on direction. So in a non Ricci-flat space with a null Ricci scalar, a ball could be deformed into an ellipsoid of same volume (hence null Ricci scalar) but in the direction of the longer semi-axis there would be more volume in the wedge of the deformed ball (compared to euclidian space), while in the direction of the shorter semi-axis there would be a reduction of volume of the wedge. Am I correct ? Thanks for the clarification ! JohannCR (talk) 21:09, 19 December 2016 (UTC)
thread_title:  Confusion over geometric interpretation 

message: Does anyone else think the "conical volume" part of the first sentence of the article should be removed? It deserves to be in the article (and more precisely), but it seems misleading as a context-setter since that characterization is (so far as I know) not so useful and just a little eldritch
Also I added a bit to the definition section, including a new part where I try to explain Ricci curvature (non-geometrically) to someone only familiar with linear algebra and multivariable calculus. Hope it's successful
Gumshoe2 (talk) 08:34, 5 May 2020 (UTC)
thread_title:  the first sentence