# Ngô Quốc Anh

## February 22, 2015

### The conditions (NN), (P), (NN+) and (P+) associated to the Paneitz operator for 3-manifolds

Filed under: Uncategorized — Tags: — Ngô Quốc Anh @ 18:54

Of recent interest is the prescribed Q-curvature on closed Riemannian manifolds since it involves high-order differential operators.

In a previous post, I have talked about prescribed Q-curvature on 4-manifolds. Recall that for 4-manifolds, this question is equivalent to finding a conformal metric $\widetilde g =e^{2u}g$ for which the Q-curvature of $\widetilde g$ equals the prescribed function $\widetilde Q$? That is to solving

$\displaystyle P_gu+2Q_g=2\widetilde Q e^{4u},$

where for any $g$, the so-called Paneitz operator $P_g$ acts on a smooth function $u$ on $M$ via

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

which plays a similar role as the Laplace operator in dimension two and the Q-curvature of $\widetilde g$ is given as follows

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

Sometimes, if we denote by $\delta$ the negative divergence, i.e. $\delta = - {\rm div}$, we obtain the following formula

$\displaystyle {P_g}(u) = \Delta _g^2u + \delta \left( {\frac{2}{3}{R_g} - 2{\rm Ric}_g} \right)du.$

Generically, for $n$-manifolds, we obtain

$\displaystyle Q_g=-\frac{1}{2(n-1)} \Big(\Delta\text{Scal}_g - \frac{n^3-4n^2+16n-16}{4(n-1)(n-2)^2} \text{Scal}_g^2+\frac{4(n-1)}{(n-2)^2} |{\rm Ric}_g|^2 \Big)$

and

$\displaystyle {P_g}(u) = \Delta _g^2u + {\rm div}\left( { a_n {R_g} + b_n {\rm Ric}_g} \right)du + \frac{n-4}{2} Q_g u,$

where $a_n = -((n-2)^2+4)/2(n-1)(n-2)$ and $b_n =4/(n-2)$.

## August 29, 2014

### Prescribed Q-curvature and scalar curvature problems in the null case

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 20:24

On a 2-dimensional compact Riemannian manifold $(M, g)$ without boundary, the prescribed scalar curvature problem in the flat case is equivalent to solving the following PDE

$\displaystyle -\Delta_g u = fe^{2u}$

with $f$ is a given non-constant smooth function on $M$ and $\Delta_g$ is the Laplace-Beltrami operator associated with the metric $g$.

Simply by integrating both sides of the PDE, it is immediate to see that if $u$ solves the PDE, it would satisfy $\int_M f e^{2u} dv =0$; hence the candidate function $f$ must change sign in $M$. In their elegant paper published in 1974, Kazdan and Warner showed that in addition to the sign-changing property of $f$, it must also satisfy the following inequality

$\displaystyle \int_M f dv <0.$

This is just a simple observation from integration by parts if we multiply both sides of the PDE by $e^{-2u}$. Interestingly, Kazdan and Warner were able to show that the above two properties are also sufficient in the sense that it is enough to prove that the PDE is solvable.

In higher dimensions, perhaps, the most natural generalization of the operator $\Delta_g$ is the well-known Paneitz operator and its corresponding Q-curvature, see this link.

Assume that $(M,g)$ is a compact Riemannian 4-manifold without boundary. We denote by $P_g^4$ the so-called Paneitz operator acting on any smooth function $u$ via the following rule

$\displaystyle P_g^4(u) = \Delta _g^2u + {\rm div}\left( {\frac{2}{3}{R_g} - 2{\rm Ric}_g} \right)du ,$

where by ${\rm Ric}$ and $R$ we mean the Ricci tensor and the scalar curvature of $g$, respectively.

## 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.