Generalized stereographic projection
$begingroup$
Let $(M^n,g)$ be a closed (compact, without boundary) Riemannian manifold and let $pin M$. Let
$$
square: =cDelta+S
$$
be the conformal Laplacian of $(M,g)$. Here $c=4frac{n-1}{n-2}$ is a constant, $Delta$ is the Laplacian and $S$ is the scalar curvature defined by $g$. Lastly, let $G_p:Msetminus {p}to mathbb{R}$ be the Green function of $square$, so $square, G_p=0$ everywhere on $Msetminus {p}$. 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
$$
hat{M}=Msetminus {p} qquad text{and} qquad hat{g}=G_p^{k-2}g,
$$
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 $ngeq 4$, 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 $hat{g}$?
geometry differential-geometry riemannian-geometry smooth-manifolds
asked Jan 12 at 22:14
rpfrpf
1,090512
$endgroup$
add a comment |
$begingroup$
Let $(M^n,g)$ be a closed (compact, without boundary) Riemannian manifold and let $pin M$. Let
$$
square: =cDelta+S
$$
be the conformal Laplacian of $(M,g)$. Here $c=4frac{n-1}{n-2}$ is a constant, $Delta$ is the Laplacian and $S$ is the scalar curvature defined by $g$. Lastly, let $G_p:Msetminus {p}to mathbb{R}$ be the Green function of $square$, so $square, G_p=0$ everywhere on $Msetminus {p}$. 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
$$
hat{M}=Msetminus {p} qquad text{and} qquad hat{g}=G_p^{k-2}g,
$$
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 $ngeq 4$, 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 $hat{g}$?
geometry differential-geometry riemannian-geometry smooth-manifolds
asked Jan 12 at 22:14
rpfrpf
1,090512
$endgroup$
$begingroup$
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
1
$begingroup$
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
add a comment |
$begingroup$
Let $(M^n,g)$ be a closed (compact, without boundary) Riemannian manifold and let $pin M$. Let
$$
square: =cDelta+S
$$
be the conformal Laplacian of $(M,g)$. Here $c=4frac{n-1}{n-2}$ is a constant, $Delta$ is the Laplacian and $S$ is the scalar curvature defined by $g$. Lastly, let $G_p:Msetminus {p}to mathbb{R}$ be the Green function of $square$, so $square, G_p=0$ everywhere on $Msetminus {p}$. 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
$$
hat{M}=Msetminus {p} qquad text{and} qquad hat{g}=G_p^{k-2}g,
$$
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 $ngeq 4$, 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 $hat{g}$?
geometry differential-geometry riemannian-geometry smooth-manifolds
asked Jan 12 at 22:14
rpfrpf
1,090512
$endgroup$
Let $(M^n,g)$ be a closed (compact, without boundary) Riemannian manifold and let $pin M$. Let
$$
square: =cDelta+S
$$
be the conformal Laplacian of $(M,g)$. Here $c=4frac{n-1}{n-2}$ is a constant, $Delta$ is the Laplacian and $S$ is the scalar curvature defined by $g$. Lastly, let $G_p:Msetminus {p}to mathbb{R}$ be the Green function of $square$, so $square, G_p=0$ everywhere on $Msetminus {p}$. 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
$$
hat{M}=Msetminus {p} qquad text{and} qquad hat{g}=G_p^{k-2}g,
$$
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 $ngeq 4$, 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 $hat{g}$?
geometry differential-geometry riemannian-geometry smooth-manifolds
geometry differential-geometry riemannian-geometry smooth-manifolds
asked Jan 12 at 22:14
rpfrpf
1,090512
asked Jan 12 at 22:14
rpfrpf
1,090512
edited Jan 16 at 15:28
rpf
asked Jan 12 at 22:14
rpfrpf
1,090512
asked Jan 12 at 22:14
rpfrpf
1,090512
asked Jan 12 at 22:14
rpfrpf
1,090512
1,090512
$begingroup$
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
1
$begingroup$
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
add a comment |
$begingroup$
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
1
$begingroup$
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
$begingroup$
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
$begingroup$
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
1
1
$begingroup$
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
$begingroup$
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
$begingroup$
@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
add a comment |
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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
1
$begingroup$
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