Ricci curvature tensor

This article defines a tensor (viz a section on a tensor bundle over the manifold) of type (0,2)

Description

Given data

A manifold $M$ with a linear connection $\nabla$ on it. For instance, we may take a Riemannian manifold and consider the Levi-Civita connection on it.

Definition part

The Ricci curvature tensor is a $(0,2)$-tensor that takes as input two vector fields and outputs a scalar function, as follows.

Let $X$ and $Y$ be two vector fields. Then consider the map that sends a vector field $Z$ to $R(X,Z)Y$ (here $R$ denotes the Riemann curvature tensor).

This is a linear map. The Ricci curvature function $Ric(X,Y)$ is defined as the trace of this map.

The Ricci curvature function is a tensor because the Riemann curvature function is also a tensor.

Properties

Symmetry

The Ricci curvature function is symmetric, viz $Ric(X,Y)=Ric(Y,X)$.

Related notions

Ricci curvature

The Ricci curvature is closely related to the Ricci curvature function and is defined as follows. It associates to each one-dimensional subspace, the value $Ric(X,X)$ for any unit vector in that one-dimensional subspace.

The following turns out to be a way of computing the Ricci curvature of a one-dimensional space. Take a unit vector in that subspace, and complete it to an orthonormal basis of the tangent space at the point. Now, add up the values of the sectional curvature for the subspaces spanned by this unit vector with each other unit vector in that orthonormal basis.

Note that the Ricci curvature at one-dimensional spaces determines the Ricci curvature tensor by means of polarization.

Related properties of metrics

• Einstein metric is one where the Ricci curvature tensor is a constant multiple of the metric tensor.
• Ricci-flat metric is one where the Ricci curvature tensor vanishes identically (or equivalently, the Ricci curvature is zero for all one-dimensional subspaces)

Flow

Given any flow of a metric, we get a corresponding flow of the Ricci curvature tensor. Note that the Ricci curvature tensor is of the same type (a $(0,2)$-tensor) as the metric tensor, hence we can actually compare the metric tensor with the Ricci curvature tensor.