Ngô Quốc Anh

May 30, 2011

Norm of traceless Ricci tensor

Filed under: Riemannian geometry — Ngô Quốc Anh @ 15:06

In this note, we shall prove the following

\displaystyle {\left| {{\text{Ric}} - \dfrac{{\overline R }}{n}g} \right|^2} = |\mathop {{\text{Ric}}}\limits^ \circ {|^2} + \dfrac{1}{n}{(R - \overline R )^2}

that I have leart from this paper [here]. Here, R is scalar curvature, \mathop {{\text{Ric}}}\limits^ \circ is the traceless Ricci tensor and \overline R denotes the average of R. The proof is simple. First, we have

\displaystyle {\text{Ric}} - \frac{{\overline R }}{n}g = {\text{Ric}} - \frac{R}{n}g + \frac{{R - \overline R }}{n}g = \mathop {{\text{Ric}}}\limits^ \circ + \frac{{R - \overline R }}{n}g.

Therefore,

\displaystyle\begin{gathered} {\left| {{\text{Ric}} - \frac{{\overline R }}{n}g} \right|^2} = {g^{im}}{g^{jn}}{\left( {\mathop {{\text{Ric}}}\limits^ \circ + \frac{{R - \overline R }}{n}g} \right)_{ij}}{\left( {\mathop {{\text{Ric}}}\limits^ \circ + \frac{{R - \overline R }}{n}g} \right)_{mn}} \hfill \\ \qquad= {g^{im}}{g^{jn}}\left( {{{\mathop {{\text{Ric}}}\limits^ \circ }_{ij}}{{\mathop {{\text{Ric}}}\limits^ \circ }_{mn}} + \frac{1}{n}(R - \overline R )({g_{ij}}{{\mathop {{\text{Ric}}}\limits^ \circ }_{mn}} + {g_{mn}}{{\mathop {{\text{Ric}}}\limits^ \circ }_{ij}}) + \frac{1}{{{n^2}}}{{(R - \overline R )}^2}{g_{ij}}{g_{mn}}} \right). \hfill \\ \end{gathered}

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