Ngô Quốc Anh

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.

Here we have already used the fact that $\int_M P_g(u)dv_g=0$ since, by the divergence theorem, we know that

$\displaystyle \int_M \Delta_g^2 u dv_g=0$

and

$\displaystyle \int_M {\rm div}\left( \Big( {\frac{2}{3}\text{Scal}_g g - 2{\rm Ric}_g}\Big) \nabla_g u\right) dv_g=0.$

We now cover the following beautiful result due to Gursky published in 1999 in CMP. Before doing so, let us denote by $\kappa_g$ the following

$\displaystyle \kappa_g = \int_M Q_g dv_g.$

We also denote by $\mathcal Y(g)$ the so-called Yamabe invariant given by

$\displaystyle \mathcal Y(g)=\inf_{\widetilde g = e^{2w}g}\left( \int_M \text{Scal}_{\widetilde g} dv_{\widetilde g}\right)\left( \int_M dv_{\widetilde g}\right)^{-1/2}.$

We shall prove

Theorem (Gursky). Let $(M^4, g)$ be a smooth compact fout-dimensional Riemannian manifold. If $\mathcal Y(g) \geqslant 0$ then $\kappa_g \leqslant 8\pi^2$.

Gursky’s proof is quite nice since it makes use of the subcritical equations similarly to the Yamabe approach.

Proof. First, we let $u_k$ solve the following subcritical equation

$-6\Delta_g u_p + \text{Scal}_gu_p = \mu_pu_p^{p-1}$

for each $p \in [2,4)$. Since $\mu_p$ is continuous from the left and non-decreasing by Aubin’s result, we may choose a sequence $p_k \nearrow 4$ such that $\mu_k := \mu_{p_k} \nearrow \mathcal Y(g)$. Let $u_k=u_{p_k}$ and $g_k = u_k^2g$. Then the scalar curvature $\text{Scal}_g$ of $g_k$ is given by

$\displaystyle \text{Scal}_k=\mu_ku_k^{p_k-4}.$

If we let $E_k$ denote the trace-free Ricci tensor of $g_k$, then (and this is the key point) as $\kappa_g$ is a conformal invariant, we have

$\displaystyle \kappa_g = \kappa_{g_k}=\int_M \left(-\frac 14|E_k|^2+\frac 1{48}\text{Scal}_k^2 \right)dv_{g_k}.$

The trace-free Ricci tensor of a metric $g$ is defined to be the Ricci tensor $\text{Ric}_g$ subtracts its trace. Mathematically, it is given by

$\displaystyle {\mathop \text{Ric}\limits^ \circ}_g= \text{Ric}_g -\frac{g}{4}\text{Scal}_g .$

Using the formula for $Q_g$, we obtain

$\begin{array}{lcl} {Q_g} &=& \displaystyle - \frac{1}{12}(\Delta\text{Scal}_g - {\text{Scal}_g^2} + 3|{\text{Ric}}|^2) \hfill \\ &=& \displaystyle - \frac{1}{12}\Delta {\text{Scal}}_g + \frac{1}{12}{{\text{Scal}}_g^2} - \frac{1}{4}|{\mathop \text{Ric}\limits^ \circ}_g + \frac{g}{4}{\text{Scal}}_g|^2 \hfill \\&=& \displaystyle - \frac{1}{12}\Delta {\text{Scal}}_g + \frac{1}{12}{{\text{Scal}}_g^2} - \frac{1}{4}|{\mathop \text{Ric}\limits^ \circ}_g|^2-\frac{1}{16}|\text{Scal}_g|^2 \hfill \\& =& \displaystyle - \frac{1}{12}\Delta {\text{Scal}}_g+ \frac{1}{48}{{\text{Scal}}_g^2} - \frac{1}{4}|{\mathop \text{Ric}\limits^ \circ}_g|^2. \end{array}$

which immediately implies that

$\displaystyle \int_M Q_g dv_g = \int_M \left( -\frac 14 |{\mathop \text{Ric}\limits^ \circ}_g|^2+\frac 1{48}\text{Scal}_g^2\right) dv_g.$

Using the formula for $\text{Scal}_k$ shown above, we can estimate

$\begin{array}{lcl} {\kappa _g} &=& \displaystyle - \frac{1}{4}\int_M {|{E_k}{|^2}d{v_{{g_k}}}} + \frac{1}{{48}}\int_M {S_k^2d{v_{{g_k}}}} \hfill \\ &\leqslant & \displaystyle\frac{1}{{48}}\int_M {{S_k}^2d{v_{{g_k}}}} \hfill \\ &=& \displaystyle\frac{1}{{48}}\mu _k^2\int_M {u_k^{2{p_k} - 4}d{v_g}} \hfill \\ &\leqslant &\displaystyle\frac{1}{{48}}\mu _k^2{\left( {\int_M {u_k^{{p_k}}d{v_g}} } \right)^{\frac{{2({p_k} - 2)}}{{{p_k}}}}}{\left( {\int_M {d{v_g}} } \right)^{\frac{{4 - {p_k}}}{{{p_k}}}}}. \end{array}$

Taking the limit as $k\to \infty$, we obtain

$\displaystyle \kappa_g \leqslant \frac 1{48} (\mathcal Y(g))^2.$

By the energy estimate of Aubin, $\mathcal Y(g) \leqslant \mathcal Y(g_{\mathbb S^4})=8\sqrt 6 \pi$. Thus, if $\mathcal Y(g) \geqslant 0$, we get

$\kappa_g \leqslant 8\pi^2$

as claimed.