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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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Is there a geometric interpretation of a nonzero element in the kernel? – Deane Yang Jul 18 '11 at 3:58
    
If there is a non-vanishing solution $u$ of $L_{g}u = 0$ then $u^{4/(n-2)}g$ has scalar curvature $0$. – Dan Fox Jul 18 '11 at 11:21
    
Dan, thanks. But what if the sign of $u$ changes somewhere? – Deane Yang Jul 18 '11 at 12:15

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