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Yamabe problem ensures that for any Riemannian metric $g$, in its conformal class $[g]$ there always exists a metric $\bar g$ whose scalar curvature $\bar R$ is constant.

I was wondering whether this is a possible way to find metrics with positive scalar curvature? i.e., for a Riemannian metric $g$, in its conformal class $[g]$ does there exist a metric $\bar g$ whose scalar curvature $\bar R$ is positive? If $R$ is positive, this trivial. If $R$ is negative, this is not possible by the conformal change of scalar curvature. If $R$ is positive somewhere and negative somewhere, is it possible to find a solution?

In other words, suppose $\phi>0$, is there a sufficient condition for $f$ such that $$-\Delta u+fu=\phi$$ has a positive solution?

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    $\begingroup$ I don't understand how your displayed equation has anything to do with the paragraphs above. The conformal change equation for scalar curvature is well-known (see Wikipedia or this article), and is definitely non-linear. It differs from your linear equation. Did you mean to have a term $u^q$ for some $q$ on the right hand side? $\endgroup$ – Willie Wong Jun 22 '15 at 9:09
  • $\begingroup$ I use $e^{2h}$ as the conformal factor, and after some change of variable we get an equation of the form that I wrote. $\endgroup$ – littlelittlelittle Jun 24 '15 at 16:23
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First observe that by the positive solution to the Yamabe problem we can assume that the background metric $g$ has constant scalar curvature $S$. The conformal change of metric formula gives that the new scalar curvature $\tilde{S}$ is given by $$ \frac{4(n-1)}{n-2} \triangle_g u + S u = \tilde{S} u^{p-1} $$ where $p = 2n/(n-2)$. When $M$ is compact, we can integrate over $M$ and obtain that $$ \int S u = \int \tilde{S} u^{p-1} $$ Observing that the conformal factor needs to be positive, this means that

Theorem: if $g$ has zero or negative Yamabe invariant, there does not exist a conformal metric with positive scalar curvature.


For the positive Yamabe class case, we have the following (slightly old) reference (so I don't know if there are more developments). On page 211 of Aubin's Some nonlinear problems in Riemannian Geometry you find the assertions that

Theorem (p289 of Aubin's Yamabe paper) If the compact manifold $(M,g)$ is not conformal to the sphere with the standard metric, there exists a constant $k > 1$ (which depends on the manifold) such that any smooth $f$ satisfying $$ 0 < \sup f \leq k \inf f$$ is the scalar curvature of some metric conformal to $g$.

Theorem (Escobar-Schoen) When $n = 3$ and $(M,g)$ not conformal to the sphere with the standard metric, if $\sup f > 0$ then $f$ is the scalar curvature of some metric conformal to $g$.

There are also some other sufficient conditions mentioned in the book which I won't take the time to list here.

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