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Let $(M^n,g)$ be a closed (compact, without boundary) Riemannian manifold and let $p\in M$. Let $$ \square: =c\Delta+S $$ be the conformal Laplacian of $(M,g)$. Here $c=4\frac{n-1}{n-2}$ is a constant, $\Delta$ is the Laplacian and $S$ is the scalar curvature defined by $g$. Lastly, let $G_p:M\setminus \{p\}\to \mathbb{R}$ be the Green function of $\square$, so $\square\, G_p=0$ everywhere on $M\setminus \{p\}$. See here for a discussion of the conformal Laplacian $\square$ and its Green function $G_p$.

If $(M,g)$ has positive scalar curvature, then the generalized stereographic projection of $(M,g)$ is defined to be the Riemannian manifold $(\hat{M},\hat{g})$, where $$ \hat{M}=M\setminus \{p\} \qquad \text{and} \qquad \hat{g}=G_p^{k-2}g, $$ and $k=2n/(n-2)$ is a constant. This is well defined when $(M,g)$ has positive scalar curvature, since in this case the Green function $G_p$ is strictly positive.

Question 1: If $n\geq 4$, is the Riemannian metric $\hat{g}$ on $\hat{M}$ complete?

I suspect this is the case as Lee and Parker prove that metric $\hat{g}$ is asymptotically flat of order at least 2.

My second question is related to the curvature properties of $(\hat{M},\hat{g})$. Note that $(\hat{M},\hat{g})$ has zero scalar curvature, as follows from the fact that $\square\, G_p=0$.

Question 2: How is the Ricci curvature of $\hat{g}$ related to the Ricci curvature of $g$? More specifically, if $g$ has Ricci curvature bounded above or below, does this give bounds on the Ricci curvature of $\hat{g}$?

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  • $\begingroup$ A naive question : is classical stereographic projection from the sphere minus one of its pole to a plane really a particular case of what you explain ? $\endgroup$ – Jean Marie Jan 12 at 22:40
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    $\begingroup$ You can establish an asymptotic estimate for $G_p$ near $p,$ which in particular lets you bound it from below by $r^k$ for some constant $k(n).$ (You should definitely be able to find this in Lee-Parker.) You can then verify completeness of the metric by using this bound to show that the length of any curve approaching $p$ is infinite. $\endgroup$ – Anthony Carapetis Jan 13 at 7:08
  • $\begingroup$ @JeanMarie: this is explained very nicely in Section 6 of Lee and Parker's paper (see hyperlink in my question). $\endgroup$ – rpf Jan 16 at 15:30

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