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.

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