What is Ricci tensor: Definition and 58 Discussions

In differential geometry, the Ricci curvature tensor, named after Gregorio Ricci-Curbastro, is a geometric object which is determined by a choice of Riemannian or pseudo-Riemannian metric on a manifold. It can be considered, broadly, as a measure of the degree to which the geometry of a given metric tensor differs locally from that of ordinary Euclidean space or pseudo-Euclidean space.
The Ricci tensor can be characterized by measurement of how a shape is deformed as one moves along geodesics in the space. In general relativity, which involves the pseudo-Riemannian setting, this is reflected by the presence of the Ricci tensor in the Raychaudhuri equation. Partly for this reason, the Einstein field equations propose that spacetime can be described by a pseudo-Riemannian metric, with a strikingly simple relationship between the Ricci tensor and the matter content of the universe.
Like the metric tensor, the Ricci tensor assigns to each tangent space of the manifold a symmetric bilinear form (Besse 1987, p. 43). Broadly, one could analogize the role of the Ricci curvature in Riemannian geometry to that of the Laplacian in the analysis of functions; in this analogy, the Riemann curvature tensor, of which the Ricci curvature is a natural by-product, would correspond to the full matrix of second derivatives of a function. However, there are other ways to draw the same analogy.
In three-dimensional topology, the Ricci tensor contains all of the information which in higher dimensions is encoded by the more complicated Riemann curvature tensor. In part, this simplicity allows for the application of many geometric and analytic tools, which led to the solution of the Poincaré conjecture through the work of Richard S. Hamilton and Grigory Perelman.
In differential geometry, lower bounds on the Ricci tensor on a Riemannian manifold allow one to extract global geometric and topological information by comparison (cf. comparison theorem) with the geometry of a constant curvature space form. This is since lower bounds on the Ricci tensor can be successfully used in studying the length functional in Riemannian geometry, as first shown in 1941 via Myers's theorem.
One common source of the Ricci tensor is that it arises whenever one commutes the covariant derivative with the tensor Laplacian. This, for instance, explains its presence in the Bochner formula, which is used ubiquitously in Riemannian geometry. For example, this formula explains why the gradient estimates due to Shing-Tung Yau (and their developments such as the Cheng-Yau and Li-Yau inequalities) nearly always depend on a lower bound for the Ricci curvature.
In 2007, John Lott, Karl-Theodor Sturm, and Cedric Villani demonstrated decisively that lower bounds on Ricci curvature can be understood entirely in terms of the metric space structure of a Riemannian manifold, together with its volume form. This established a deep link between Ricci curvature and Wasserstein geometry and optimal transport, which is presently the subject of much research.

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  1. L

    Ricci tensor along a Killing vector

    In Carrol's text, he shows that the covariant derivative of the Ricci scalar is zero along a Killing vector. He then goes on to say something about how this intuitively justifies our notion of geometry not changing along a Killing vector. This same informal reasoning would seem to imply that...
  2. Philosophaie

    Ricci Tensor from Schwarzschild Metric

    Looking for the Schwarzschild Solution for this equation: ds^2 = -A(r) / c^2 * dr^2 - r^2 / c^2 *(d\\theta^2 +(sin(\\theta))^2 *d\\phi^2) + B(r) * dt^2 where A(r) = 1 / (1-2*m/r) And B(r) = (1-2*m/r) From this can be calculated the co- and contra-varient metric tensors and...
  3. J

    Understanding the Ricci and Riemann Curvature Tensors in Tensor Calculus

    The Ricci Tensor comes from the Riemann Curvature Tensor: R^{\beta}_{\nu\rho\sigma} = \Gamma^{\beta}_{\nu\sigma,\rho} - \Gamma^{\beta}_{\nu\rho,\sigma} + \Gamma^{\alpha}_{\nu\sigma}\Gamma^{\beta}_{\alpha\ rho} - \Gamma^{\alpha}_{\nu\rho}\Gamma^{\beta}_{\alpha\sigma} The Ricci Tensor just...
  4. P

    Ricci tensor equals a constant times the metric

    Hello; What does it mean physically if I have? R_{a b} = Ag_{a b} I think it means that my manifold is an n-sphere (i.e. if A is positive), or it is AdSn (i.e. if A is negative). Is this correct?
  5. Mentz114

    Solving Ricci Tensor Problem: Schwarzschild Metric

    Starting with this definition of the Reimann tensor R^a_{mbn}=\Gamma ^{a}_{mn,b}-\Gamma ^{a}_{mb,n}+\Gamma ^{a}_{rb}\Gamma ^{r}_{mn}-\Gamma ^{a}_{rn}\Gamma ^{r}_{mb} Can I contract on indices a,b and r to get R_{mn} ? It bothers me that the expression on the right is not symmetric in...
  6. W

    Ricci Tensor & Trace: Exploring the Relationship Between Tensors & Traces

    Just wondering if Traces can be applied to tensors. If the Ricci tensor is Rii then is sums over diagonal elements. So technically, can you say the trace of the Riemann tensor is the Ricci tensor?
  7. S

    Proving the Relation Between Weyl Tensor, Ricci Tensor & Scalar

    Hello, I wish to show that on 3-dimensional manifolds, the weyl tensor vanishes. In other words, I want to show that the curvature tensor, the ricci tensor and curvature scalar hold the relation Please, if anyone knows how I can prove this relation or refer to a place which proves the...
  8. Oxymoron

    Why Must We Contract Two Indices to Form the Ricci Tensor?

    Producing the Ricci tensor On a pseudo-Riemannian manifold we can contract the Riemann curvature tensor to form the Ricci tensor. In this process of contraction we sum over two indices to make a (3-1)-tensor into a (2-0)-tensor. My question is, why must we contract two indices? Why can't we...
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