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.NegativeScalarCurvatureInOmega

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<k 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<k in \Omega, there holds

\displaystyle \text{Scal}_{\widehat g}<u^{-4}(-\frac 12)<-\frac 12 k^{-4}.

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

Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Blog at WordPress.com.

%d bloggers like this: