The conformal Laplacian

Let \((X^n,g)\) be a closed Riemannian manifold, \(n\geq3\). We use the conformal Laplacian

\[ L_g:=-c_n\Delta^g+R_g, \qquad c_n:=\frac{4(n-1)}{n-2}. \]

Here \(R_g\) is the scalar curvature of \(g\). With our sign convention,

\[ \int_X uL_gu\,d\mu_g = \int_X c_n|\nabla u|^2+R_gu^2\,d\mu_g. \]

\[ \widehat g=u^{\frac{4}{n-2}}g, \qquad u\gt{}0, \]

then this is the same as writing \(\widehat g=e^{2f}g\) with

\[ e^{2f}=u^{\frac{4}{n-2}}, \qquad f=\frac{2}{n-2}\log u. \]

For a general conformal change \(\widehat g=e^{2f}g\), the scalar curvature formula is

\begin{equation} \tag{6.1.1} \label{eq:psc-scalar-exp-transform} R_{\widehat g} = e^{-2f}\bigl( R_g-2(n-1)\Delta^gf-(n-1)(n-2)|\nabla f|^2 \bigr). \end{equation}

We now plug in \(f=\frac{2}{n-2}\log u\). First,

\[ \nabla f=\frac{2}{n-2}\frac{\nabla u}{u}, \qquad |\nabla f|^2 = \frac{4}{(n-2)^2}\frac{|\nabla u|^2}{u^2}, \]

and

\[ \Delta^g f = \frac{2}{n-2}\Delta^g\log u = \frac{2}{n-2} \left( \frac{\Delta^gu}{u} - \frac{|\nabla u|^2}{u^2} \right). \]

Therefore

\[ \begin{aligned} R_{\widehat g} &= u^{-\frac{4}{n-2}} \biggl[ R_g - \frac{4(n-1)}{n-2} \left( \frac{\Delta^gu}{u} - \frac{|\nabla u|^2}{u^2} \right) - \frac{4(n-1)}{n-2} \frac{|\nabla u|^2}{u^2} \biggr] \\ &= u^{-\frac{4}{n-2}} \left( R_g-\frac{4(n-1)}{n-2}\frac{\Delta^gu}{u} \right). \end{aligned} \]

The two \(|\nabla u|^2/u^2\) terms cancel exactly: the first comes from \(-2(n-1)\Delta^g f\), and the second comes from \(-(n-1)(n-2)|\nabla f|^2\). Since \(c_n=\frac{4(n-1)}{n-2}\), we get

\begin{equation} \tag{6.1.2} \label{eq:psc-conformal-laplacian-transform} R_{\widehat g} = u^{-\frac{n+2}{n-2}} \bigl(-c_n\Delta^gu+R_gu\bigr) = u^{-\frac{n+2}{n-2}}L_gu . \end{equation}

Lemma 6.1.1

If the first eigenvalue of \(L_g\) is positive, then \(X\) admits a metric of positive scalar curvature in the conformal class of \(g\).

Proof. Let \(u\gt{}0\) be the first eigenfunction:

\[ L_gu=\lambda_1u, \qquad \lambda_1\gt{}0. \]

For \(\widehat g=u^{4/(n-2)}g\), formula (6.1.2) gives

\[ R_{\widehat g} = \lambda_1u^{-\frac{4}{n-2}}\gt{}0. \]

◻

Lemma 6.1.2

Let \(X^n\) be closed, \(n\geq3\). If \(R_g\geq0\) and \(R_g\gt{}0\) somewhere, then \(X\) admits a metric of positive scalar curvature.

Proof. We show that \(\lambda_1(L_g)\gt{}0\). For every nonzero \(\phi\),

\[ \int_X \phi L_g\phi\,d\mu_g = \int_X c_n|\nabla\phi|^2+R_g\phi^2\,d\mu_g\geq0. \]

Thus \(\lambda_1\geq0\). If \(\lambda_1=0\), let \(u\gt{}0\) be a first eigenfunction. Then

\[ 0=\int_X c_n|\nabla u|^2+R_gu^2\,d\mu_g . \]

Both terms are nonnegative, so \(\nabla u\equiv0\) and \(R_gu^2\equiv0\). Since \(u\gt{}0\), this forces \(R_g\equiv0\), contradicting the assumption that \(R_g\gt{}0\) somewhere. Hence \(\lambda_1\gt{}0\), and Lemma 6.1.1 applies. ◻

The conformal Laplacian

Gaoming Wang

Assistant Professor