Interpreting the scalar curvature in a semi-Riemannian manifold
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Background:
Let $M$ be a smooth Riemannian manifold of dimension $n$ and scalar curvature $R$ (with respect to the Levi-Civita connection). Let $m in M$ and let $B$ be the geodesic ball of radius $r$ centered at $m$. That is,
$B = { Exp_m(v) | vin T_mM, ||v||< r } $
Then for small $r$, the volume of $B$ is:
begin{equation}
Vol(B) = (constant) r^n left[1 - frac{R}{6(n+2)}r^2 + O(r^4) right]
end{equation}
where the constant depends only on $n$, and $R$ is evaluated at $m$. So $R$ basically tells us the difference between the volume of a small ball in $M$ and the volume of a small ball in Euclidean $mathbb{R}^n$ (to leading order in the radius $r$).
Questions:
1) Is there any generalization of this to the case of a semi-Riemannian manifold?
2) If not, is there an analogous result for the case of a Lorentzian manifold (that is, a semi-Riemannian manifold whose metric signature has $n-1$ pluses and one minus)?
3) If not, is there some other result that gives a nice way to interpret the scalar curvature $R$ in a semi-Riemannian (or Lorenztian) manifold?
Comment:
The problem I see is this: to define a geodesic ball, we want to start with a ball "of radius $r$" in $T_mM$. But the metric is indefinite, so there is no norm. I looked in Barrett O'Niell's book "Semi-Riemannian Geometry" but did not find the answer there.
Thanks in advance for your help!
differential-geometry riemannian-geometry
edited Nov 16 at 10:16
Kei
277
asked Jan 22 '12 at 4:50
marlow
18311
add a comment |
up vote
12
down vote
favorite
Background:
Let $M$ be a smooth Riemannian manifold of dimension $n$ and scalar curvature $R$ (with respect to the Levi-Civita connection). Let $m in M$ and let $B$ be the geodesic ball of radius $r$ centered at $m$. That is,
$B = { Exp_m(v) | vin T_mM, ||v||< r } $
Then for small $r$, the volume of $B$ is:
begin{equation}
Vol(B) = (constant) r^n left[1 - frac{R}{6(n+2)}r^2 + O(r^4) right]
end{equation}
where the constant depends only on $n$, and $R$ is evaluated at $m$. So $R$ basically tells us the difference between the volume of a small ball in $M$ and the volume of a small ball in Euclidean $mathbb{R}^n$ (to leading order in the radius $r$).
Questions:
1) Is there any generalization of this to the case of a semi-Riemannian manifold?
2) If not, is there an analogous result for the case of a Lorentzian manifold (that is, a semi-Riemannian manifold whose metric signature has $n-1$ pluses and one minus)?
3) If not, is there some other result that gives a nice way to interpret the scalar curvature $R$ in a semi-Riemannian (or Lorenztian) manifold?
Comment:
The problem I see is this: to define a geodesic ball, we want to start with a ball "of radius $r$" in $T_mM$. But the metric is indefinite, so there is no norm. I looked in Barrett O'Niell's book "Semi-Riemannian Geometry" but did not find the answer there.
Thanks in advance for your help!
differential-geometry riemannian-geometry
edited Nov 16 at 10:16
Kei
277
asked Jan 22 '12 at 4:50
marlow
18311
In the $(k+1)$ hyperbolic space $H^{k+1}$ we have that $mathrm{vol}(B_r)=pi^{k/2}Gamma(k/2)e^{kr}/k!+O(re^{(k-2)r})$. That helps?
– emiliocba
Jan 22 '12 at 6:02
@emiliocba I'm afraid I don't see it. A hyperbolic space is an example of a Riemannian manifold, isn't it? The scalar curvature may be negative, but the metric is still positive definite.
– marlow
Jan 22 '12 at 16:36
add a comment |
up vote
12
down vote
favorite
up vote
12
down vote
favorite
Background:
Let $M$ be a smooth Riemannian manifold of dimension $n$ and scalar curvature $R$ (with respect to the Levi-Civita connection). Let $m in M$ and let $B$ be the geodesic ball of radius $r$ centered at $m$. That is,
$B = { Exp_m(v) | vin T_mM, ||v||< r } $
Then for small $r$, the volume of $B$ is:
begin{equation}
Vol(B) = (constant) r^n left[1 - frac{R}{6(n+2)}r^2 + O(r^4) right]
end{equation}
where the constant depends only on $n$, and $R$ is evaluated at $m$. So $R$ basically tells us the difference between the volume of a small ball in $M$ and the volume of a small ball in Euclidean $mathbb{R}^n$ (to leading order in the radius $r$).
Questions:
1) Is there any generalization of this to the case of a semi-Riemannian manifold?
2) If not, is there an analogous result for the case of a Lorentzian manifold (that is, a semi-Riemannian manifold whose metric signature has $n-1$ pluses and one minus)?
3) If not, is there some other result that gives a nice way to interpret the scalar curvature $R$ in a semi-Riemannian (or Lorenztian) manifold?
Comment:
The problem I see is this: to define a geodesic ball, we want to start with a ball "of radius $r$" in $T_mM$. But the metric is indefinite, so there is no norm. I looked in Barrett O'Niell's book "Semi-Riemannian Geometry" but did not find the answer there.
Thanks in advance for your help!
differential-geometry riemannian-geometry
edited Nov 16 at 10:16
Kei
277
asked Jan 22 '12 at 4:50
marlow
18311
Background:
Let $M$ be a smooth Riemannian manifold of dimension $n$ and scalar curvature $R$ (with respect to the Levi-Civita connection). Let $m in M$ and let $B$ be the geodesic ball of radius $r$ centered at $m$. That is,
$B = { Exp_m(v) | vin T_mM, ||v||< r } $
Then for small $r$, the volume of $B$ is:
begin{equation}
Vol(B) = (constant) r^n left[1 - frac{R}{6(n+2)}r^2 + O(r^4) right]
end{equation}
where the constant depends only on $n$, and $R$ is evaluated at $m$. So $R$ basically tells us the difference between the volume of a small ball in $M$ and the volume of a small ball in Euclidean $mathbb{R}^n$ (to leading order in the radius $r$).
Questions:
1) Is there any generalization of this to the case of a semi-Riemannian manifold?
2) If not, is there an analogous result for the case of a Lorentzian manifold (that is, a semi-Riemannian manifold whose metric signature has $n-1$ pluses and one minus)?
3) If not, is there some other result that gives a nice way to interpret the scalar curvature $R$ in a semi-Riemannian (or Lorenztian) manifold?
Comment:
The problem I see is this: to define a geodesic ball, we want to start with a ball "of radius $r$" in $T_mM$. But the metric is indefinite, so there is no norm. I looked in Barrett O'Niell's book "Semi-Riemannian Geometry" but did not find the answer there.
Thanks in advance for your help!
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
edited Nov 16 at 10:16
Kei
277
asked Jan 22 '12 at 4:50
marlow
18311
edited Nov 16 at 10:16
Kei
277
asked Jan 22 '12 at 4:50
marlow
18311
edited Nov 16 at 10:16
Kei
277
edited Nov 16 at 10:16
Kei
277
edited Nov 16 at 10:16
Kei
277
277
asked Jan 22 '12 at 4:50
marlow
18311
asked Jan 22 '12 at 4:50
marlow
18311
asked Jan 22 '12 at 4:50
marlow
18311
18311
In the $(k+1)$ hyperbolic space $H^{k+1}$ we have that $mathrm{vol}(B_r)=pi^{k/2}Gamma(k/2)e^{kr}/k!+O(re^{(k-2)r})$. That helps?
– emiliocba
Jan 22 '12 at 6:02
@emiliocba I'm afraid I don't see it. A hyperbolic space is an example of a Riemannian manifold, isn't it? The scalar curvature may be negative, but the metric is still positive definite.
– marlow
Jan 22 '12 at 16:36
add a comment |
In the $(k+1)$ hyperbolic space $H^{k+1}$ we have that $mathrm{vol}(B_r)=pi^{k/2}Gamma(k/2)e^{kr}/k!+O(re^{(k-2)r})$. That helps?
– emiliocba
Jan 22 '12 at 6:02
@emiliocba I'm afraid I don't see it. A hyperbolic space is an example of a Riemannian manifold, isn't it? The scalar curvature may be negative, but the metric is still positive definite.
– marlow
Jan 22 '12 at 16:36
In the $(k+1)$ hyperbolic space $H^{k+1}$ we have that $mathrm{vol}(B_r)=pi^{k/2}Gamma(k/2)e^{kr}/k!+O(re^{(k-2)r})$. That helps?
– emiliocba
Jan 22 '12 at 6:02
In the $(k+1)$ hyperbolic space $H^{k+1}$ we have that $mathrm{vol}(B_r)=pi^{k/2}Gamma(k/2)e^{kr}/k!+O(re^{(k-2)r})$. That helps?
– emiliocba
Jan 22 '12 at 6:02
@emiliocba I'm afraid I don't see it. A hyperbolic space is an example of a Riemannian manifold, isn't it? The scalar curvature may be negative, but the metric is still positive definite.
– marlow
Jan 22 '12 at 16:36
@emiliocba I'm afraid I don't see it. A hyperbolic space is an example of a Riemannian manifold, isn't it? The scalar curvature may be negative, but the metric is still positive definite.
– marlow
Jan 22 '12 at 16:36
add a comment |
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In the $(k+1)$ hyperbolic space $H^{k+1}$ we have that $mathrm{vol}(B_r)=pi^{k/2}Gamma(k/2)e^{kr}/k!+O(re^{(k-2)r})$. That helps?
– emiliocba
Jan 22 '12 at 6:02
@emiliocba I'm afraid I don't see it. A hyperbolic space is an example of a Riemannian manifold, isn't it? The scalar curvature may be negative, but the metric is still positive definite.
– marlow
Jan 22 '12 at 16:36