Let $M$ be a smooth, closed manifold of dimension $n>2$. Let $L_g$ be the conformal Laplacian of the metric $g$. That is, $L_g=-\Delta_g + \frac{n-2}{4(n-1)}R_g$, where $R_g$ is the scalar curvature of the metric $g$. Is the set of $C^2$ Riemannian metrics on $M$ such that conformal Laplacian has a trivial kernel dense with respect to the $C^2$ norm?
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$\begingroup$ Is there a geometric interpretation of a nonzero element in the kernel? $\endgroup$– Deane YangCommented Jul 18, 2011 at 3:58
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$\begingroup$ If there is a non-vanishing solution $u$ of $L_{g}u = 0$ then $u^{4/(n-2)}g$ has scalar curvature $0$. $\endgroup$– Dan FoxCommented Jul 18, 2011 at 11:21
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$\begingroup$ Dan, thanks. But what if the sign of $u$ changes somewhere? $\endgroup$– Deane YangCommented Jul 18, 2011 at 12:15
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