📂Geometry

Ricci Curvature of Differentiable Manifolds

Definition¹

¹

Given a differentiable manifold $M$ and the tangent space $T_{p}M$ at point $p \in M$, let’s have the function $f$ as follows. For a given $X, Y \in T_{p}M$,

$$\begin{align*} f : T_{p}M &\to T_{p}M \\ Z &\mapsto R(X,Z)Y \end{align*}$$

where $R$ is the Riemann curvature. Then, the Ricci curvature $\Ric : T_{p}M \times T_{p}M \to \mathbb{R}$ at point $p$ is defined as

$$\Ric (X,Y) = \Ric_{p} (X,Y) := \tr f = \tr \left( Z \mapsto R(X,Z)Y \right)$$

where $\tr$ is the trace of a linear transformation.

Description

Since $\Ric$ is bilinear by definition, it’s sufficient to know its value on the basis. Let’s say $\left\{ X_{i} = \dfrac{\partial }{\partial x_{i}}\right\}$ is the basis of $T_{p}M$. Then, by the trace inner product representation,

$$\Ric (X_{i}, X_{j}) = \tr (Z \mapsto R(X_{i}, Z)X_{j}) = g\left( R(X_{i}, X_{k})X_{j}, X_{l}\right)g^{kl}$$

The previous inner product is denoted as $R_{ikjl}$. Therefore,

$$\Ric (X_{i}, X_{j}) = R_{ikjl}g^{kl} = R_{ikj}^{s}g_{sl}g^{kl} = R_{ikj}^{s}\delta_{s}^{k} = R_{ikj}^{k}$$

Since $R_{ijkl} = R_{ijk}^{s}g_{sl}$, the Ricci curvature $\Ric (X_{i}, X_{j}) = R_{ikj}^{k}$ implies taking an average of the second and fourth components of the Riemann curvature $R_{ijkl}$.

$$R_{ijkl}g^{lj} = R_{ijk}^{s}g_{sl}g^{lj} = R_{ijk}^{s}\delta_{s}^{j} = R_{ijk}^{j}$$

If $\left\{ Z_{i} \right\}$ are considered as the orthonormal basis of $T_{p}M$, then since $g^{kl} = \delta_{kl}$,

$$\Ric (X, Y) = g\left( R(X, Z_{k})Y, Z_{l}\right)\delta_{kl} = g\left( R(X, Z_{k})Y, Z_{k}\right) = R(X, Z_{k}, Y, Z_{k}),\quad X, Y \in T_{p}M$$

Moreover, $\Ric (X) := \Ric (X, X)$ is called the Ricci curvature in the direction of $X$ at $p$.