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Can scalar curvature and diameter control volume?
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Scalar curvature can control the volume of geodesic ball locally, however, it can not bound the diameter. As for as I know, the example for a manifold with a large scalar curvature and volume has large diameter.
Comparing with the n sphere with the standard metric, if a smooth manifold has scalar curvature larger than this sphere, and the diameter of the manifold is smaller, should the volume of the manifold be smaller than the sphere?
riemannian-geometry
asked 3 hours ago
Yiyue Zhang
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up vote
4
down vote
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Scalar curvature can control the volume of geodesic ball locally, however, it can not bound the diameter. As for as I know, the example for a manifold with a large scalar curvature and volume has large diameter.
Comparing with the n sphere with the standard metric, if a smooth manifold has scalar curvature larger than this sphere, and the diameter of the manifold is smaller, should the volume of the manifold be smaller than the sphere?
riemannian-geometry
asked 3 hours ago
Yiyue Zhang
211
New contributor
Yiyue Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
As a follow up, to build a counter-example I ended up needing there to be some negative Ricci curvature. I experimented a little bit and couldn't figure out how to build any counter-examples with non-negative Ricci curvature. I wonder if it might be possible to get some volume control with non-negative Ricci, positive scalar and bounded diameter that's better than what the volume comparison theorem states.
– Gabe K
10 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Scalar curvature can control the volume of geodesic ball locally, however, it can not bound the diameter. As for as I know, the example for a manifold with a large scalar curvature and volume has large diameter.
Comparing with the n sphere with the standard metric, if a smooth manifold has scalar curvature larger than this sphere, and the diameter of the manifold is smaller, should the volume of the manifold be smaller than the sphere?
riemannian-geometry
asked 3 hours ago
Yiyue Zhang
211
New contributor
Yiyue Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Scalar curvature can control the volume of geodesic ball locally, however, it can not bound the diameter. As for as I know, the example for a manifold with a large scalar curvature and volume has large diameter.
Comparing with the n sphere with the standard metric, if a smooth manifold has scalar curvature larger than this sphere, and the diameter of the manifold is smaller, should the volume of the manifold be smaller than the sphere?
riemannian-geometry
riemannian-geometry
asked 3 hours ago
Yiyue Zhang
211
New contributor
Yiyue Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Yiyue Zhang
211
New contributor
Yiyue Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Yiyue Zhang
211
New contributor
Yiyue Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Yiyue Zhang
211
asked 3 hours ago
Yiyue Zhang
211
211
New contributor
Yiyue Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Yiyue Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Yiyue Zhang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
As a follow up, to build a counter-example I ended up needing there to be some negative Ricci curvature. I experimented a little bit and couldn't figure out how to build any counter-examples with non-negative Ricci curvature. I wonder if it might be possible to get some volume control with non-negative Ricci, positive scalar and bounded diameter that's better than what the volume comparison theorem states.
– Gabe K
10 mins ago
add a comment |Â
As a follow up, to build a counter-example I ended up needing there to be some negative Ricci curvature. I experimented a little bit and couldn't figure out how to build any counter-examples with non-negative Ricci curvature. I wonder if it might be possible to get some volume control with non-negative Ricci, positive scalar and bounded diameter that's better than what the volume comparison theorem states.
– Gabe K
10 mins ago
As a follow up, to build a counter-example I ended up needing there to be some negative Ricci curvature. I experimented a little bit and couldn't figure out how to build any counter-examples with non-negative Ricci curvature. I wonder if it might be possible to get some volume control with non-negative Ricci, positive scalar and bounded diameter that's better than what the volume comparison theorem states.
– Gabe K
10 mins ago
As a follow up, to build a counter-example I ended up needing there to be some negative Ricci curvature. I experimented a little bit and couldn't figure out how to build any counter-examples with non-negative Ricci curvature. I wonder if it might be possible to get some volume control with non-negative Ricci, positive scalar and bounded diameter that's better than what the volume comparison theorem states.
– Gabe K
10 mins ago
add a comment |Â
2 Answers
2
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oldest
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up vote
3
down vote
A connected sum of two spheres can be done with arbitrary thin neck while keeping the lower bound on scalar curvature.
Take a sphere of radius $tfrac110$ and attacch to it 1000000 spheres of the same radius by very thin necks.
You get this way a space of small diameter, large scalar curvature and huge volume.
answered 1 hour ago
Anton Petrunin
25.8k578193
add a comment |Â
up vote
2
down vote
In general, the scalar curvature does not give enough control to get these sorts of volume estimates. However, if you use the Ricci curvature instead, you can get these kinds of estimates using the volume comparison theorem or something similar.
For an example of why scalar curvature bounds and diameter bounds are not enough to compare the volume to a sphere, consider the manifold $M_1$ which is the metric product of a ball of radius r in hyperbolic space $mathbbH^2$ (with sectional curvature 1) and the sphere $mathbbS^2(r)$. For the second manifold, consider the $4$-sphere $mathbbS^4(s)$.
The volume of $M_1$ is $sinh(t) times 4 pi r^2$ while its diameter is $sqrtpi^2 r^2+ 4t^2 < pi r +2 t$. The scalar curvature is $2(frac1r^2-1)$. Whenever $r<1$, the scalar curvature of $M_1$ is positive. Meanwhile, the volume of the $4$-sphere is $frac83pi^2 s^4$, its diameter is $pi s$, and its scalar curvature is $frac12s^2$.
To force our first manifold to have much larger volume then the second, we set $s$ very large so that the scalar curvature is less than $1$.
We also set $r=1/2$ and $t=s-1/2$. In this case, the scalar curvature of $M_1$ is 6 and the diameter of $M_1$ is strictly smaller than that of $mathbbS^4(s)$. However, since hyperbolic sine grows exponentially in $t$ whereas the volume of the $4$-sphere grows as $s^4$, for sufficiently large $s$ the volume of $M_1$ is larger than that of the sphere. By setting $s$ arbitrarily large, we can force $M_1$ to have much larger volume than $mathbbS^4$.
answered 57 mins ago
Gabe K
581212
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
A connected sum of two spheres can be done with arbitrary thin neck while keeping the lower bound on scalar curvature.
Take a sphere of radius $tfrac110$ and attacch to it 1000000 spheres of the same radius by very thin necks.
You get this way a space of small diameter, large scalar curvature and huge volume.
answered 1 hour ago
Anton Petrunin
25.8k578193
add a comment |Â
up vote
3
down vote
A connected sum of two spheres can be done with arbitrary thin neck while keeping the lower bound on scalar curvature.
Take a sphere of radius $tfrac110$ and attacch to it 1000000 spheres of the same radius by very thin necks.
You get this way a space of small diameter, large scalar curvature and huge volume.
answered 1 hour ago
Anton Petrunin
25.8k578193
add a comment |Â
up vote
3
down vote
up vote
3
down vote
A connected sum of two spheres can be done with arbitrary thin neck while keeping the lower bound on scalar curvature.
Take a sphere of radius $tfrac110$ and attacch to it 1000000 spheres of the same radius by very thin necks.
You get this way a space of small diameter, large scalar curvature and huge volume.
answered 1 hour ago
Anton Petrunin
25.8k578193
A connected sum of two spheres can be done with arbitrary thin neck while keeping the lower bound on scalar curvature.
Take a sphere of radius $tfrac110$ and attacch to it 1000000 spheres of the same radius by very thin necks.
You get this way a space of small diameter, large scalar curvature and huge volume.
answered 1 hour ago
Anton Petrunin
25.8k578193
answered 1 hour ago
Anton Petrunin
25.8k578193
answered 1 hour ago
Anton Petrunin
25.8k578193
answered 1 hour ago
Anton Petrunin
25.8k578193
25.8k578193
add a comment |Â
add a comment |Â
up vote
2
down vote
In general, the scalar curvature does not give enough control to get these sorts of volume estimates. However, if you use the Ricci curvature instead, you can get these kinds of estimates using the volume comparison theorem or something similar.
For an example of why scalar curvature bounds and diameter bounds are not enough to compare the volume to a sphere, consider the manifold $M_1$ which is the metric product of a ball of radius r in hyperbolic space $mathbbH^2$ (with sectional curvature 1) and the sphere $mathbbS^2(r)$. For the second manifold, consider the $4$-sphere $mathbbS^4(s)$.
The volume of $M_1$ is $sinh(t) times 4 pi r^2$ while its diameter is $sqrtpi^2 r^2+ 4t^2 < pi r +2 t$. The scalar curvature is $2(frac1r^2-1)$. Whenever $r<1$, the scalar curvature of $M_1$ is positive. Meanwhile, the volume of the $4$-sphere is $frac83pi^2 s^4$, its diameter is $pi s$, and its scalar curvature is $frac12s^2$.
To force our first manifold to have much larger volume then the second, we set $s$ very large so that the scalar curvature is less than $1$.
We also set $r=1/2$ and $t=s-1/2$. In this case, the scalar curvature of $M_1$ is 6 and the diameter of $M_1$ is strictly smaller than that of $mathbbS^4(s)$. However, since hyperbolic sine grows exponentially in $t$ whereas the volume of the $4$-sphere grows as $s^4$, for sufficiently large $s$ the volume of $M_1$ is larger than that of the sphere. By setting $s$ arbitrarily large, we can force $M_1$ to have much larger volume than $mathbbS^4$.
answered 57 mins ago
Gabe K
581212
add a comment |Â
up vote
2
down vote
In general, the scalar curvature does not give enough control to get these sorts of volume estimates. However, if you use the Ricci curvature instead, you can get these kinds of estimates using the volume comparison theorem or something similar.
For an example of why scalar curvature bounds and diameter bounds are not enough to compare the volume to a sphere, consider the manifold $M_1$ which is the metric product of a ball of radius r in hyperbolic space $mathbbH^2$ (with sectional curvature 1) and the sphere $mathbbS^2(r)$. For the second manifold, consider the $4$-sphere $mathbbS^4(s)$.
The volume of $M_1$ is $sinh(t) times 4 pi r^2$ while its diameter is $sqrtpi^2 r^2+ 4t^2 < pi r +2 t$. The scalar curvature is $2(frac1r^2-1)$. Whenever $r<1$, the scalar curvature of $M_1$ is positive. Meanwhile, the volume of the $4$-sphere is $frac83pi^2 s^4$, its diameter is $pi s$, and its scalar curvature is $frac12s^2$.
To force our first manifold to have much larger volume then the second, we set $s$ very large so that the scalar curvature is less than $1$.
We also set $r=1/2$ and $t=s-1/2$. In this case, the scalar curvature of $M_1$ is 6 and the diameter of $M_1$ is strictly smaller than that of $mathbbS^4(s)$. However, since hyperbolic sine grows exponentially in $t$ whereas the volume of the $4$-sphere grows as $s^4$, for sufficiently large $s$ the volume of $M_1$ is larger than that of the sphere. By setting $s$ arbitrarily large, we can force $M_1$ to have much larger volume than $mathbbS^4$.
answered 57 mins ago
Gabe K
581212
add a comment |Â
up vote
2
down vote
up vote
2
down vote
In general, the scalar curvature does not give enough control to get these sorts of volume estimates. However, if you use the Ricci curvature instead, you can get these kinds of estimates using the volume comparison theorem or something similar.
For an example of why scalar curvature bounds and diameter bounds are not enough to compare the volume to a sphere, consider the manifold $M_1$ which is the metric product of a ball of radius r in hyperbolic space $mathbbH^2$ (with sectional curvature 1) and the sphere $mathbbS^2(r)$. For the second manifold, consider the $4$-sphere $mathbbS^4(s)$.
The volume of $M_1$ is $sinh(t) times 4 pi r^2$ while its diameter is $sqrtpi^2 r^2+ 4t^2 < pi r +2 t$. The scalar curvature is $2(frac1r^2-1)$. Whenever $r<1$, the scalar curvature of $M_1$ is positive. Meanwhile, the volume of the $4$-sphere is $frac83pi^2 s^4$, its diameter is $pi s$, and its scalar curvature is $frac12s^2$.
To force our first manifold to have much larger volume then the second, we set $s$ very large so that the scalar curvature is less than $1$.
We also set $r=1/2$ and $t=s-1/2$. In this case, the scalar curvature of $M_1$ is 6 and the diameter of $M_1$ is strictly smaller than that of $mathbbS^4(s)$. However, since hyperbolic sine grows exponentially in $t$ whereas the volume of the $4$-sphere grows as $s^4$, for sufficiently large $s$ the volume of $M_1$ is larger than that of the sphere. By setting $s$ arbitrarily large, we can force $M_1$ to have much larger volume than $mathbbS^4$.
answered 57 mins ago
Gabe K
581212
In general, the scalar curvature does not give enough control to get these sorts of volume estimates. However, if you use the Ricci curvature instead, you can get these kinds of estimates using the volume comparison theorem or something similar.
For an example of why scalar curvature bounds and diameter bounds are not enough to compare the volume to a sphere, consider the manifold $M_1$ which is the metric product of a ball of radius r in hyperbolic space $mathbbH^2$ (with sectional curvature 1) and the sphere $mathbbS^2(r)$. For the second manifold, consider the $4$-sphere $mathbbS^4(s)$.
The volume of $M_1$ is $sinh(t) times 4 pi r^2$ while its diameter is $sqrtpi^2 r^2+ 4t^2 < pi r +2 t$. The scalar curvature is $2(frac1r^2-1)$. Whenever $r<1$, the scalar curvature of $M_1$ is positive. Meanwhile, the volume of the $4$-sphere is $frac83pi^2 s^4$, its diameter is $pi s$, and its scalar curvature is $frac12s^2$.
To force our first manifold to have much larger volume then the second, we set $s$ very large so that the scalar curvature is less than $1$.
We also set $r=1/2$ and $t=s-1/2$. In this case, the scalar curvature of $M_1$ is 6 and the diameter of $M_1$ is strictly smaller than that of $mathbbS^4(s)$. However, since hyperbolic sine grows exponentially in $t$ whereas the volume of the $4$-sphere grows as $s^4$, for sufficiently large $s$ the volume of $M_1$ is larger than that of the sphere. By setting $s$ arbitrarily large, we can force $M_1$ to have much larger volume than $mathbbS^4$.
answered 57 mins ago
Gabe K
581212
answered 57 mins ago
Gabe K
581212
answered 57 mins ago
Gabe K
581212
answered 57 mins ago
Gabe K
581212
581212
add a comment |Â
add a comment |Â
Yiyue Zhang is a new contributor. Be nice, and check out our Code of Conduct.
Yiyue Zhang is a new contributor. Be nice, and check out our Code of Conduct.
Yiyue Zhang is a new contributor. Be nice, and check out our Code of Conduct.
Yiyue Zhang is a new contributor. Be nice, and check out our Code of Conduct.
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As a follow up, to build a counter-example I ended up needing there to be some negative Ricci curvature. I experimented a little bit and couldn't figure out how to build any counter-examples with non-negative Ricci curvature. I wonder if it might be possible to get some volume control with non-negative Ricci, positive scalar and bounded diameter that's better than what the volume comparison theorem states.
– Gabe K
10 mins ago