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.

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