# 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 $b_n =4/(n-2)$.