# Ngô Quốc Anh

## December 13, 2013

### Norm of some 2-tensors involving the Ricci curvature tensor

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 4:09

Following the previous note where we discuss about the norm of the tensor $\text{Ric}-\frac 1n \overline{\text{Scal}}g$ in terms of the trace-free Ricci tensor, i.e. the following identity

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

holds. Today, we derive some further identites involving the Ricci tensor $\text{Ric}$. As we have already seen from the previous notes, the trace-free Ricci tensor is given by

$\displaystyle {\mathop \text{Ric}\limits^ \circ} = \text{Ric} -\frac{g}{n}\text{Scal}.$

First, we calculate $\text{Ric}-\frac gn \text{Scal}$. For simplicity, we denote by $R$ and $S$ the Ricci tensor and the scalar curvature respectively. By definition, we have

$\begin{array}{lcl} \displaystyle {\left| {R - \frac{g}{n}S} \right|^2} &=& \displaystyle {g^{im}}{g^{jn}}{\left( {R - \frac{g}{n}S} \right)_{ij}}{\left( {R - \frac{g}{n}S} \right)_{mn}} \hfill \\ &=& \displaystyle {g^{im}}{g^{jn}}\left( {{R_{ij}}{R_{mn}} - \frac{S}{n}({g_{ij}}{R_{mn}} + {g_{mn}}{R_{ij}}) + \frac{{{S^2}}}{{{n^2}}}{g_{ij}}{g_{mn}}} \right) \hfill \\ &=& \displaystyle {\left| R \right|^2} - \frac{S}{n}{g^{im}}{g^{jn}}({g_{ij}}{R_{mn}} + {g_{mn}}{R_{ij}}) + \frac{{{S^2}}}{{{n^2}}}{g^{im}}\underbrace {{g^{jn}}{g_{ij}}}_{\delta _i^n}{g_{mn}} \hfill \\ &=& \displaystyle {\left| R \right|^2} - \frac{S}{n}{g^{im}}\underbrace {{g^{jn}}{g_{ij}}}_{\delta _i^n}{R_{mn}} - \frac{S}{n}{g^{im}}\underbrace {{g^{jn}}{g_{mn}}}_{\delta _m^j}{R_{ij}} + \frac{{{S^2}}}{{{n^2}}}\underbrace {{g^{im}}{g_{mi}}}_n \hfill \\ &=& \displaystyle {\left| R \right|^2} - \frac{S}{n}\underbrace {{g^{nm}}{R_{mn}}}_S - \frac{S}{n}\underbrace {{g^{ij}}{R_{ij}}}_S + \frac{{{S^2}}}{n} \hfill \\ &=& \displaystyle {\left| R \right|^2} - \frac{{{S^2}}}{n}. \end{array}$

Thus, we have just shown that

$\displaystyle \boxed{\left|\text{Ric}-\frac gn \text{Scal}\right|^2=|\text{Ric}|^2-\frac 1n |\text{Scal}|^2}.$

We now calculate $\text{Ric}-\frac 1n \overline{\text{Scal}}g$ in terms of the Ricci curvature tensor, not its trace-free. Again, we use $\overline S$ to denote $\overline{\text{Scal}}$. We now write

$\begin{array}{lcl} \displaystyle {\left| {R - \frac{g}{n}\overline S } \right|^2} &=& \displaystyle {g^{im}}{g^{jn}}{\left( {R - \frac{g}{n}\overline S } \right)_{ij}}{\left( {R - \frac{g}{n}\overline S } \right)_{mn}} \hfill \\ &=& \displaystyle {g^{im}}{g^{jn}}\left( {{R_{ij}}{R_{mn}} - \frac{{\overline S }}{n}({g_{ij}}{R_{mn}} + {g_{mn}}{R_{ij}}) + \frac{{{{\overline S }^2}}}{{{n^2}}}{g_{ij}}{g_{mn}}} \right) \hfill \\ &=& \displaystyle {\left| R \right|^2} - \frac{{\overline S }}{n}{g^{im}}{g^{jn}}({g_{ij}}{R_{mn}} + {g_{mn}}{R_{ij}}) + \frac{{{{\overline S }^2}}}{{{n^2}}}{g^{im}}\underbrace {{g^{jn}}{g_{ij}}}_{\delta _i^n}{g_{mn}} \hfill \\ &=& \displaystyle {\left| R \right|^2} - \frac{{2\overline S }}{n}S + \frac{{{{\overline S }^2}}}{n}. \hfill \\ \end{array}$

Thus, we have just proved that

$\displaystyle \boxed{{\left| {{\text{Ric}} - \dfrac{g}{n}\overline{\text{Scal}}} \right|^2} = |\text{Ric}|^2 -\frac 2n\overline{\text{Scal}}\text{Scal} + \frac 1n \overline{\text{Scal}}^2}.$

In the same way, we can calculate

$\begin{array}{lcl} \displaystyle {\left| R \right|^2} &=& \displaystyle {\left| {\mathop R\limits^ \circ + \frac{g}{n}S} \right|^2} \hfill \\ &=& \displaystyle {g^{im}}{g^{jn}}{\left( {\mathop R\limits^ \circ + \frac{g}{n}S} \right)_{ij}}{\left( {\mathop R\limits^ \circ + \frac{g}{n}S} \right)_{mn}} \hfill \\ &=& \displaystyle {g^{im}}{g^{jn}}\left( {{{\mathop R\limits^ \circ }_{ij}}{{\mathop R\limits^ \circ }_{mn}} + \frac{S}{n}({g_{ij}}{{\mathop R\limits^ \circ }_{mn}} + {g_{mn}}{{\mathop R\limits^ \circ }_{ij}}) + \frac{{{S^2}}}{{{n^2}}}{g_{ij}}{g_{mn}}} \right) \hfill \\ &=& \displaystyle {\big| {\mathop R\limits^ \circ } \big|^2} + \frac{S}{n}\underbrace {{g^{im}}{g^{jn}}{g_{ij}}}_{{g^{mn}}}{(R - \frac{g}{n}S)_{mn}} + \frac{S}{n}\underbrace {{g^{im}}{g^{jn}}{g_{mn}}}_{{g^{ij}}}{(R - \frac{g}{n}S)_{ij}} + \frac{{{S^2}}}{{{n^2}}}\underbrace {{g^{im}}{g^{jn}}{g_{ij}}{g_{mn}}}_n \hfill \\ &=& \displaystyle {\big| {\mathop R\limits^ \circ } \big|^2} + \frac{{2{S^2}}}{n} - \frac{{{S^2}}}{{{n^2}}}{g^{mn}}{g_{mn}} - \frac{{{S^2}}}{{{n^2}}}{g^{ij}}{g_{ij}} + \frac{{{S^2}}}{n} \hfill \\ &=& \displaystyle {\big| {\mathop R\limits^ \circ } \big|^2} + \frac{{{S^2}}}{n}. \end{array}$

Thus, we obtain

$\displaystyle \boxed{|\text{Ric}|^2 = \big| {\mathop R\limits^ \circ } \big|^2+\frac 1n \text{Scal}^2}.$

For a smooth function $\phi$, instead of $\text{Ric}$, one can define the so-called Bakry-Emery Ricci curvature tensor via

$\text{Ric}_\phi = \text{Ric} - \text{Hess}(\phi)$

where the hessian of $\phi$ is commonly given by

$\text{Hess}(\phi)(X,Y)=\langle \nabla_X \phi,Y\rangle.$

Clearly, $\Delta := \text{trace}(\text{Hess}(\phi))$. We shall discuss some computation involving $\text{Ric}_\phi$ in the future.