# 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

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

## December 12, 2013

### An upper bound for the total integral of the Q-curvature in the non-negative Yamabe invariants

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 6:24

As we have already discussed once that a natural conformally invariant in dimension four is the following

$\displaystyle Q_g=-\frac{1}{12}(\Delta\text{Scal}_g -\text{Scal}_g^2 +3|{\rm Ric}_g|^2)$

which is commonly refered to the Q-curvature of metric $g$, see this topic. Note that, under a conformal change of the metric $\widetilde g =e^{2u}g$, the quantity $Q$ transforms according to

$\displaystyle 2Q_{\widetilde g}=e^{-4u}(P_gu+2Q_g)$

where $P=P_g$ denotes the Paneitz operator with respect to $g$. Keep in mind that the Paneitz operator is conformally invariant in the sense that

$\displaystyle P_{\widetilde g}=e^{-4u}P_g$

for any conformal metric $\widetilde g =e^{2u}g$. For any $g$, the operator $P_g$ acts on a smooth function u on M via the following rule

$\displaystyle {P_g}(u) = \Delta _g^2u + {\rm div}\left( {\frac{2}{3}\text{Scal}_g - 2{\rm Ric}_g} \right)du$

which plays a similar role as the Laplace operator in dimension two. Observe that $dv_{\widetilde g} = e^{4u}dv_g$, therefore, a simple calculation shows

$\displaystyle \int_M Q_{\widetilde g}dv_{\widetilde g}=\int_M Q_{\widetilde g}e^{4u}dv_g=\int_M Q_g dv_g.$

Hence the total integral $\int_M Q_g dv_g$ is conformally invariant.