# Ngô Quốc Anh

## January 8, 2014

### Conformal metric having strictly negative scalar curvature in a given region

Filed under: PDEs, Riemannian geometry — Ngô Quốc Anh @ 0:39

Let $(M,g)$ be a closed (compact without boundary) Riemannian manifold of dimension $3$ with a $C^2$ metric $g$ of arbitrary Yamabe class. Suppose $\Omega \subset M$ be an open subset of $M$ with regular boundary $\partial\Omega$ and with $M\backslash (\Omega \cup \partial \Omega )$ non-empty and open in $M$.

In this note, we mention a very interesting result basically due to O’Murchadha-York from here and Isenberg from here. The result says that there exists a conformal metric $\widehat g \in [g]$ such that

$\displaystyle \text{Scal}_{\widehat g} < -\xi<0$

in $\Omega$ for some constant $\xi>0$ to be specify later. The novelty of this result is that although the metric $g$ may be of positive Yamabe class which tells us that it is impossible to construct a conformal metric which is everywhere negative, it is possible to make it negative in a proper subset of $M$. A proof for this result goes as follows:

(1) First, since $g \in C^2$ and $\text{Scal}_g$ involves up to second order derivatives, it is clear that $\text{Scal}_g$ is continuous in $M$. This and the compactness of $M$ tell us that $\text{Scal}_g$ is bounded from above, say by some constant $K>0$, i.e.

$\displaystyle \text{Scal}_g < K$

everywhere in $M$.

(2) We now conclude that the constant $K$ in the preceding estimate can be any constant sitting in $(0,1)$ for suitable choice of conformal metric. Indeed, recall that for any number (i.e. vanishing Laplacian) $\alpha$, the metric $\alpha^4g$ obeys the following

$\displaystyle \text{Scal}_{\alpha^4 g}=\alpha^{-4}\text{Scal}_g.$

Hence, choosing $\alpha = (2K)^{1/4}$, one can easily check that

$\displaystyle\text{Scal}_{\alpha^4 g}<\frac 12.$

(3) We now consider the following Dirichlet problem on $\Omega$

$\displaystyle\begin{array}{rcl} \Delta_{\alpha^4g} u &=& \dfrac{u}{8} \quad \text{ in }\Omega , \hfill \\ u &=& 1, \quad \text{ on }\partial \Omega . \hfill \\ \end{array}$

The existence, uniqueness, and positivity of $u$ is clear. In addition, the solution $u$ is at least continuous in $\Omega$. Hence, there is some constant $k>0$ such that $u in $\Omega$.

(4) We now choose a conformal factor $\theta$ in such a way that $\theta \equiv u$ in $\Omega \cup \partial \Omega$. Since $M\backslash (\Omega \cup \partial \Omega )$ is open, such a $\theta$ exists. Finally, we simply set $\widehat g = \theta^4 (\alpha^4g)$ and do some calculation as shown below:

$\displaystyle \text{Scal}_{\widehat g}=-\theta^{-5} \Big( 8\Delta_{\alpha^4g}\theta -\theta\text{Scal}_{\alpha^4 g} \Big).$

Obviously,

$\displaystyle \text{Scal}_{\widehat g}\big|_{\Omega}=\Big(\theta^{-4}\text{Scal}_{\alpha^4 g}-8\theta^{-5}\Delta_{\alpha^4g}\theta \Big)\big|_{\Omega}=u^{-4}\Big( \text{Scal}_{\alpha^4 g}-1\Big).$

Thanks to $\text{Scal}_{\alpha^4 g}<\frac 12$ and $u in $\Omega$, there holds

$\displaystyle \text{Scal}_{\widehat g}

Thus, $\xi=\frac 12 k^{-4}$.