# 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,

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

Notice that

$\displaystyle |\mathop {{\text{Ric}}}\limits^ \circ {|^2} = {g^{im}}{g^{jn}}{{\mathop {{\text{Ric}}}\limits^ \circ }_{ij}}{{\mathop {{\text{Ric}}}\limits^ \circ} _{mn}}$

and

$\displaystyle\begin{gathered} {g^{im}}{g^{jn}}\left( {\frac{1}{{{n^2}}}{{(R - \overline R )}^2}{g_{ij}}{g_{mn}}} \right) = \frac{1}{{{n^2}}}{(R - \overline R )^2}{g^{im}}{g^{jn}}{g_{ij}}{g_{mn}} \hfill \\\qquad\qquad = \frac{1}{{{n^2}}}{(R - \overline R )^2}\delta _j^m\delta _m^j = \frac{1}{n}{(R - \overline R )^2}. \hfill \\ \end{gathered}$

Lastly

$\displaystyle\begin{gathered} \frac{1}{n}(R - \overline R ){g^{im}}{g^{jn}}({g_{ij}}{{\mathop {{\text{Ric}}}\limits^ \circ }_{mn}} + {g_{mn}}{{\mathop {{\text{Ric}}}\limits^ \circ} _{ij}}) \hfill \\ \qquad= \frac{1}{n}(R - \overline R )\left[ {{g^{im}}{g^{jn}}{g_{ij}}{{\mathop {{\text{Ric}}}\limits^ \circ }_{mn}} + {g^{im}}{g^{jn}}{g_{mn}}{{\mathop {{\text{Ric}}}\limits^ \circ }_{ij}}} \right] \hfill \\ \qquad= \frac{1}{n}(R - \overline R )\left[ {{g^{im}}\delta _i^n{{\mathop {{\text{Ric}}}\limits^ \circ }_{mn}} + \delta _n^i{g^{jn}}{{\mathop {{\text{Ric}}}\limits^ \circ }_{ij}}} \right] \hfill \\ \qquad= \frac{1}{n}(R - \overline R )\left[ {{g^{mn}}{{\mathop {{\text{Ric}}}\limits^ \circ }_{mn}} + {g^{ij}}{{\mathop {{\text{Ric}}}\limits^ \circ }_{ij}}} \right]. \hfill \\ \end{gathered}$

By mean of the traceless, we know that

$\displaystyle {g^{mn}}{{\mathop {{\text{Ric}}}\limits^ \circ} _{mn}} = {g^{ij}}{{\mathop {{\text{Ric}}}\limits^ \circ }_{ij}} = \text{trace}(\mathop {{\text{Ric}}}\limits^ \circ ) = 0.$

Thus we have the desired result.

1. I think a slightly tidier way to prove it, is to see that

$\displaystyle\mbox{Ric} -f g=\stackrel{\circ}{\mbox{Ric}}+(R/n-f)g$

and

$\displaystyle g\perp \stackrel{\circ}{\mbox{Ric}},$

so the identity follows from Pythagoras theorem.

Comment by K — June 1, 2011 @ 19:16

• Ah yes, that’s interesting. I was thinking in a traditional way, by calculus only. Thank you for your beautiful approach.

Comment by Ngô Quốc Anh — June 1, 2011 @ 19:19