Ngô Quốc Anh

December 12, 2009

R-G: Traceless Ricci tensor, Einstein tensor, and Schouten tensor

Filed under: Riemannian geometry — Ngô Quốc Anh @ 13:47

Traceless Ricci tensor

Traceless Ricci tensor is defined to be $\displaystyle {B_{ij}} = {R_{ij}} - \frac{1}{n}S{g_{ij}}$.

Clearly, the trace of traceless Ricci tensor $B$ is nothing but $\displaystyle g^{ij}B_{ij}$

which can be estimated as follows $\displaystyle\begin{gathered}{g^{ij}}{B_{ij}} = {g^{ij}}\left( {{R_{ij}} - \frac{1}{n}S{g_{ij}}} \right) \hfill \\\qquad= {g^{ij}}{R_{ij}} - \frac{1}{n}{g^{ij}}S{g_{ij}} \hfill \\\qquad= {g^{ij}}{R_{ij}} - \frac{1}{n}\delta_{ij}S \hfill \\\qquad = 0. \hfill\end{gathered}$

This is why we call $B$ the traceless Ricci tensor.

Einstein tensor

The Einstein tensor $E$ is a rank- $2$ tensor defined over Riemannian manifolds. In index-free notation it is defined as $\displaystyle E={\rm Ric}-\frac{1}{2}gR$.

In component form, the previous equation reads as $\displaystyle E_{ij} = R_{ij} - {1\over2} g_{ij}R$.

The Einstein tensor is symmetric $\displaystyle E_{ij} = E_{ji}$

and, like the stress-energy tensor, divergenceless $\displaystyle E_{ij,j} = 0$.

The divergenceless property will be proved elsewhere later.

Schouten tensor

The Schouten tensor $S$ is defined as $\displaystyle S={\rm Ric}-\frac{1}{2(n-1)}gR$.

In component form, the previous equation reads as $\displaystyle E_{ij} = R_{ij} - {1 \over{2(n-1)}} g_{ij}R$.