Ricci curvature

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(Redirected from Ricci tensor)

In differential geometry, the Ricci curvature tensor is (0,2)-valent tensor, obtained as a trace of the full curvature tensor. It can be thought of as a Laplacian of the Riemannian metric tensor in the case of Riemannian manifolds.

The Ricci tensor acting on vectors u and v, usually denoted by Ric(u,v), is defined as the trace of the endomorphism

<math>w \mapsto R(w,v)u<math>

where R is the Riemann curvature tensor. In local coordinates, one can write (using the Einstein summation convention)

<math>Ric = R_{ij}dx^i \otimes dx^j<math>


<math>R_{ij} = {R^k}_{ikj}<math>.

On a Riemannian manifold, the Ricci tensor is the only independent trace of the curvature tensor; all others vanish or are equal to the Ricci tensor up to sign. Furthermore, the Ricci tensor on a Riemannian manifold is symmetric in its arguments

<math>Ric(u,v) = Ric(v,u)<math>

This follows from the fact that the Levi-Civita connection is torsion-free.

Ricci curvature can be also explained in terms of the sectional curvature in the following way: for a unit vector v, Ric(v,v) is sum of the sectional curvatures of all the planes spanned by the vector v and a vector from an orthonormal frame containing v (there are n−1 such planes).

One can think of Ricci curvature on a Riemannian manifold, as being an operator on the tangent bundle. That is, one can contract the Ricci tensor with the metric to obtain a (1,1)-valent tensor. In local coordinates,

<math>{R^{i}}_j = g^{ik}R_{kj}<math>

For any vectors u and v the vector Ric(u) satisfies

<math>Ric(u,v) = g(Ric(u), v)<math>

If this operator is just multiplication by a constant, then we have an Einstein manifold. The Ricci curvature is proportional to the metric tensor in this case.

In dimensions 2 and 3 Ricci curvature describes completely the curvature tensor, but in higher dimensions Ricci curvature contains less information. For instance, Einstein manifolds do not have to have constant curvature in dimensions 4 and up.

An explicit expression for the Ricci tensor in terms of the Levi-Civita connection is given in the article on Christoffel symbols.

Applications of the Ricci curvature tensor

The Ricci curvature can be used to define Chern classes of a manifold, which are topological invariants (so independent of the choice of metric).

Ricci curvature is also used in Ricci flow, where a metric is deformed in the direction of the Ricci curvature. On surfaces, the flow produces a metric of constant Gauss curvature and the uniformization theorem for surfaces follows. In dimension 3, it was recently used to give a complete classification of compact 3-manifolds.

Ricci curvature plays an important role in general relativity, where it is the key term in the Einstein field equations.

Global geometry/topology and Ricci curvature

Here is a short list of most basic results on manifolds with positive Ricci curvature.

  1. Myers theorem states that if Ricci curvature is bounded from below on a complete Riemannian manifold by <math>\left(n-1\right)k > 0 \,\!<math>, then its diameter <math>\le \pi/\sqrt{k}<math>, and manifold has to have a finite fundamental group. If the diameter is equal to <math>\pi/\sqrt{k}<math>, then the manifold is isometric to a sphere of a constant curvature k.
  2. Bishop-Gromov inequality states that if Ricci curvature of a complete m-dimensional Riemannian manifold is ≥0 then the volume of a ball is smaller or equal to the volume of a ball of the same radius in Eulidean m-space. More over if <math>v_p(R)<math> denotes the volume of the ball with center p and radius <math>R<math> in the manifold and <math>V(R)=c_m R^m<math> denotes the volume of the ball of radius R in Euclidean m-space then function <math>v_p(R)/V(R)<math> is nonincreasing. (The last inequality can be generalized to arbitrary curvature bound and is the key point in the proof of Gromov's compactness theorem.)
  3. Splitting theorem states that if a complete Riemannian manifold with Ricc ≥ 0 has a straight line (i.e. a geodesic γ such that <math>d(\gamma(u),\gamma(v))=|u-v|<math> for all <math>v,u\in\mathbb{R}<math>) then it is isometric to a product space <math>\mathbb{R}\times L,<math> where <math>L<math> is a Riemannian manifold.

All results above show that positive Ricci curvature does have some geometric meaning, on the contrary negative curvature is not that restrictive, in particular as it was shown by Joachim Lohkamp, any manifold admits a metric of negative curvature.

See also


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