# Talk:Ricci curvature

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## Ricci tensor as a Laplacian of the metric

How exactly can the Ricci tensor be thought of as a Laplacian of the metric? Certainly, not as the covariant Laplacian ∇*∇ since ∇g = 0 for the Levi-Civita connection, and not as the Hodge-Laplacian (dd* + d*d) which is defined only on differential forms. I can't find a statement like this in any of my Riemannian geometry books. -- Fropuff 18:17, 2004 Oct 8 (UTC)

it is bit informal, in the exponential coordiantes at one point Ricc_{ij} will be usual laplasian of g_{ij} at this point. —Preceding unsigned comment added by Tosha (talkcontribs) 01:16, 10 Oct 2004 (UTC)
The question is a bit ironic, given that I entered this page to point out that the close relation of the Ricci tensor (belonging to the Levi-Civita connection) to the Hodge de Rham operator needs to be highlighted. It's not actually necessary to make explicit reference to a trace operator or tensor contraction, since one has the general identity $\Delta dx^{\mu} = -R^{\mu}_{\nu} dx^{\nu}$. For more general one-forms, the residual of the Hodge de Rham on one-forms $(\Delta a)_{\mu} - \Delta (a_{\mu}) = -R^{\nu}_{\mu} a_{\nu}$ gives you an expression that also characterizes the Levi-Civita connection's Ricci tensor.
For the Ricci tensor belonging to more general connections, another definition is used, which actually goes well beyond the connections associated with tangent spaces and even beyond connections associated with principal_bundles; applying to general connections defined in the arena of jet bundles. This requires the use of the Froelicher-Nijenhuis bracket and the Hodge dual.
It's also worth pointing out that when you're dealing with a metrical connection, you can still write down an expression in terms of this connection alone that gives you the Ricci tensor associated with the Levi-Civita connection. In Einstein-Cartan gravity, if you use this "corrected" Ricci tensor and relate it through Einstein's equation to the corresponding "corrected" stress tensor, the effect will almost be equivalent to the spin correction employed on the canonical stress tensor in classical field theory. This is used to derive the "effective" stress tensor associated with an Einstein-Cartan source. So, the two varieties of Ricci tensor (the Levi-Civita one and the non-symmetric one associated with the native connection) should really be regarded as independent objects, rather than as two incarnations of the same concept.
—The preceding unsigned comment was added by 129.89.32.117 (talk) 01:36, 20 June 2007

The correct statement is that in harmonic coordinates, the components of the Ricci tensor satisfy

$Ric_{jk} = -\frac{1}{2}\Delta g_{jk} + Q(g, \partial g)$,

where $Q(g, \partial g)$ is quadratic in the components of g, and also involves first derivatives of these components, and where $\Delta$ is the Laplace-Beltrami operator of $g$. Here the phrase harmonic cordinates means we are using a coordinates chart in which each coordinate $x^j$ satisfies $\Delta x^j=0$. This formula was used by DeTurck to show that any (Riemannian) Einstein metric is real-analytic in harmonic coordinates, and is also used extensively e.g. in proving Anderson's pre-compactness theorems for the space of Einstein metrics.

70.18.30.252 11:11, 7 August 2006 (UTC) Riemannian geometer

BTW, I do not like your edit very much, you made it more formal and therefore less accessible to beginers. In addition you use nonstandard convention, usually Ricci tensor is an opreator on tangent space and not a quadratic form. Tosha 01:16, 10 Oct 2004 (UTC)

I disagree, I think the Ricci tensor is more generally regarded as a bilinear form than an operator on the tangent space. Besides, one needs a metric in order to define it as a operator. The definition of the bilinear form is independent of the metric (i.e. it works for any connection, not just the Levi-Civita connection) and thus more fundamental. -- Fropuff 02:32, 2004 Oct 10 (UTC)

## Proposed merger

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) Riemannian geometer

The article Ricci tensor quite properly focuses on Riemannian geometry. If the merger took place, for stylistic reasons it may not be good to insist on clearly distinguishing between properties relevant to gtr and those irrelevant to gtr. So I'd suggest instead that whoever wanted to merge write an article on Ricci tensor in general relativity and merge the traceless Ricci tensor into that article. If you can't guess what I have in mind for Ricci tensor in general relativity, maybe best leave it to me (but be very patient, it might take 12 months for me to get to this, since there are so many more urgent tasks awaiting me just in the gtr pages.)---CH 08:36, 31 January 2006 (UTC)

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)
What fropuff said. --MarSch 13:19, 23 April 2006 (UTC)

## Is this English?

This is one of the most difficult articles on Wikipedia! As Ricci tensors are used to formulate the Einstein field equations, they are of fundamental importance even to laymen. Can someone make this article a little more accessible? At least tell us poor people without graduations in physics/math what a Ricci tensor is? Loom91 18:11, 15 March 2006 (UTC)

One simple way of explaining the Ricci tensor is to say that in geodesic normal coordinates the volume form is

[ 1 - (1/6) Ric_ij x^i x^j + O (|x|^3)] dx^1 ... dx^n

In other words, the Ricci tensor measures whether the volume of a geodesic cone is larger or smaller than what it would be in Euclidean space. I have appended this to the article, though it might be better to move it to the beginning of the article. (However, I will leave it to someone with administrative privileges to decide whether or not this is a good idea.)

I don't know why someone thought they should discuss conformal changes in this article, but in any case the gven formula for the change of the Ricci tensor under conformal change is absolutely incorrect! The leading factor of e^{2f} simply should not be there. The correct formula can, for example, be found at the top of page 59 of Besse's "Einstein Manifolds." --- A Riemannian geometer. 8/5/06

I agree, but since I am on the verge of quitting WP I am not about to try to fix this. The basic problem is that the Ricci tensor is fundamental in both Riemannian geometry and Lorentzian geometry (particularly for the gtr application), but this background tends to pull the article in somewhat directions from the point of view of notation and what to emphasize. (Buzzwords from Riemannian geometry include Ricci flow and comparison theorems such as the Bishop-Gromov comparison theorem, while buzzwords from general relativity include Raychaudhuri equation. A key point here is the profound implications of positive definite signature, in the case of Riemannian geometry.)
To (not) answer the question asked, the Ricci tensor is a kind of "trace" of the Riemann tensor (which has various purely geometric interpretations). Therefore, the components of the Ricci tensor can be thought of as a kind of "average" of Riemann components. In the case of Lorentzian manifolds interpreted as models in general relativity, the Riemann and Ricci tensor also acquire some physical interpretations. See for example Gravitational curvature by Theodore Fraenkel, an undergraduate level gtr textbook which offers a particularly extensive discussion of some of the best known interpretations.---CH 19:58, 15 March 2006 (UTC)
Yes, this is a nasty article. Oh well. Chris -- don't leave. Just try to invest less time and emotional energy into WP, so you don't have to get angry and feel like quiting. linas 00:44, 16 March 2006 (UTC)

## Joachim Lohkamp's Result

Couldn't get a readable version of his result, but it can't be right as quoted. The Cartan-Hadamard theorem states that any manifold that can be given a metric of nonpositive scalar curvature must have universal cover R^n. S^n (n>2) is a counterexample. But (I believe) the result might be right for AVERAGE curvature. —The preceding unsigned comment was added by Oz john (talkcontribs) 10:22, 4 April 2006.

No, no, no! Lohkamp's theorem is correct as stated. Cartan-Hadamard is about negative SECTIONAL curvature. Scalar curvature is at the other end of the spectrum. The existence of metrics on negative SCALAR curvature on any manifold of dimension > 2 was proved independently by Aubin and by Berard-Bergery in in the mid 1970's. Ricci is harder, but already the existence negative Ricci metrics on any 3-manifold (and in particular, on the 3-sphere) was already proved by Gao and Yau in the early 1980's. This is a good case to try to wrap your mind around, because in dimension 3, the Ricci tensor algebraically determines the sectional curvatures ---- but negative Ricci does not imply negative sectional! ---Riemannian geometer, 8/5/06

## A small erroneous statement

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)

## Reference

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 (talkcontribs) 16:36, 14 April 2007 (UTC).

## Geometrical Meaning

As for the taylor-expansion of the metric

$g_{ij} = \delta_{ij} - \frac{1}{3}R_{ikj\ell}x^kx^\ell + O(|x|^3)$

I d like to have a hint how to derive this. Somehow it appears from nowhere. —Preceding unsigned comment added by Dbfrosch (talkcontribs) 16:17, 27 October 2009 (UTC)

Hint: Take a geodesic γ at P and a Jacobi field J along γ with initial conditions $J(0)=0, J'(0)=\xi.\,$ Then compute the fourth order Taylor series of $\langle J(t),J(t)\rangle$ about t=0 to get
$\langle J(t), J(t)\rangle = t^2\langle\xi,\xi\rangle - \frac{t^4}{3}\langle R(\xi,\dot{\gamma})\dot{\gamma},\xi\rangle + O(t^5).$
Finally apply the polarization identity to obtain the required result. 71.182.247.220 (talk) 21:04, 27 October 2009 (UTC)
This can be incorporated into the article. I'm not certain precisely how to do it, though. 71.182.247.220 (talk) 16:16, 28 October 2009 (UTC)
Sorry for the late reply here. I haven't been monitoring my talk page. Take a geodesic γ(t) with initial conditions $\gamma'(0) = v^i\partial/\partial x^i$. In normal coordinates, $J(t) = t \xi^i\partial/\partial x^i$. So the left-hand side of the above equation is
$t^2\xi^i\xi^j\langle \partial/\partial x^i, \partial/\partial x^j\rangle_{\gamma(t)}=t^2 g_{ij}(\gamma(t))\xi^i\xi^j.$
So by polarizing on the ξi, we get
$g_{ij}(\gamma(t)) = \delta_{ij} - \frac{t^2}{3}R_{ikj\ell}v^kv^\ell + O(t^3).$
Now the normal coordinates of the point γ(t) are precisely xj(γ(t)) = tvj, which gives the result. 71.182.244.158 (talk) 01:23, 9 November 2009 (UTC)

Thanks, I was wondering the same, and sure enough there was info on the talk page! Gotta love wikipedia. I also found a reference discussing higher order corrections as well if anyone is interested [1]. If people think it is worthwhile, I'll link it in the article. DefnerBly (talk) 08:26, 23 December 2013 (UTC)

## Symmetric?

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 ∇ω = 0. Which one is correct, the article or the book? It seems to me that there are some missing hypotheses in the article's statement. 19:23, 25 June 2011 (UTC)

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)

## Definition of Riemann curvature tensor

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 (talkcontribs) 2014-04-11T18:54:28

I would support clarifying the article on this point. Sławomir Biały (talk) 20:38, 11 April 2014 (UTC)