Ricci curvature

Definition

In terms of Ricci curvature tensor

Let $M$ be a differential manifold and $g$ a Riemannian metric on $M$. The Ricci curvature on $g$ is a function from $\mathbb {P} (TM)$ to <math\R</math> (tangent directions at points, to real numbers) that associates to a particular tangent direction the value $Ric(X,X)$ where $X$ is a unit tangent vector in that direction.

In terms of sectional curvature

Another way of defining the Ricci curvature is in terms of the sectional curvature. Let $(M,g)$ be a Riemannian manifold, $p\in M$ and $X$ a unit tangent vector at $p$. Let $e_{1},e_{2},\ldots e_{n}$ be an orthonormal basis at $p$ such that $e_{1}=X$. Then the Ricci curvature of $X$ is defined as:

$\sum _{i=2}^{n}K(e_{1},e_{i})$

By $K(e_{1},e_{i})$ is meant the sectional curvature of the plane spanned by $e_{1}$ and $e_{i}$.

In terms of Riemann curvature tensor

We now define the Ricci curvature directly in terms of the Riemann curvature tensor, and this definition explains both the above definitions. The Ricci curvature at a point, for a tangent direction with unit tangent vector $X$, is defined as:

$Tr(Z\mapsto R(X,Z)X)$

or equivalently, if we choose an orthonormal basis with $X=e_{1}$ as:

$\sum _{i=2}^{n}R(X,e_{i},X,e_{i})$

This gives the above two definitions.

Facts

The Ricci curvature tensor is determined by the Ricci curvature

By the polarization trick, we can compute the Ricci curvature tensor from the Ricci curvature. This is based on the following facts:

• The Ricci curvature tensor is symmetric
• A symmetric bilinear form is completely determined by the values it takes on pairs $(X,X)$ because of the identity:

$b(X,Y)=1/2(b(X+y,X+Y)-b(X,X)-b(Y,Y))$

• In particular, it is determined by the values taken at all pairs $(X,X)$ for a unit vector $X$ because every vector is a scalar multiple of a unit vector