Lower and Upper Bound for Ricci Curvature

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As mentioned in first chapter of John M. Lee: Riemannian Geometry, one of our goal in differential geometry is connecting geometry and topology. For this reason it is natural to compare curvature quantities with its correspondence in model spaces; e.g. $Ric geq k g, Ricgeq0$. But I never seen any result about $Ricleq kg$ or $Ric geq -k g$ for some positive constance $k$.
Does this conditions deduced from earlier one? or topology of this manifolds are so complicated to handle?
differential-geometry differential-topology riemannian-geometry curvature
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As mentioned in first chapter of John M. Lee: Riemannian Geometry, one of our goal in differential geometry is connecting geometry and topology. For this reason it is natural to compare curvature quantities with its correspondence in model spaces; e.g. $Ric geq k g, Ricgeq0$. But I never seen any result about $Ricleq kg$ or $Ric geq -k g$ for some positive constance $k$.
Does this conditions deduced from earlier one? or topology of this manifolds are so complicated to handle?
differential-geometry differential-topology riemannian-geometry curvature
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
As mentioned in first chapter of John M. Lee: Riemannian Geometry, one of our goal in differential geometry is connecting geometry and topology. For this reason it is natural to compare curvature quantities with its correspondence in model spaces; e.g. $Ric geq k g, Ricgeq0$. But I never seen any result about $Ricleq kg$ or $Ric geq -k g$ for some positive constance $k$.
Does this conditions deduced from earlier one? or topology of this manifolds are so complicated to handle?
differential-geometry differential-topology riemannian-geometry curvature
As mentioned in first chapter of John M. Lee: Riemannian Geometry, one of our goal in differential geometry is connecting geometry and topology. For this reason it is natural to compare curvature quantities with its correspondence in model spaces; e.g. $Ric geq k g, Ricgeq0$. But I never seen any result about $Ricleq kg$ or $Ric geq -k g$ for some positive constance $k$.
Does this conditions deduced from earlier one? or topology of this manifolds are so complicated to handle?
differential-geometry differential-topology riemannian-geometry curvature
differential-geometry differential-topology riemannian-geometry curvature
asked Sep 2 at 5:37
C.F.G
1,3151521
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There can be no topological consequences of estimates of the form $Ric le k g$ or $Ric ge -k g$ (at least in dimensions greater that $2$), because this paper by Joachim Lohkamp shows that every smooth manifold of dimension at least $3$ admits a complete metric whose Ricci curvature is bounded between two negative constants.
Addition:
Your comments suggest that you may have some misunderstanding about what these inequalities mean. Here are some remarks that might help to clarify what's going on.
- First, an equality like $Ricle k$ or $Ric ge k$ is really shorthand for $Ricle kg$ or $Ric ge kg$. Equivalent statements are that $Ric(v,v)le k$ (resp. $ge k$) for every unit vector $v$; or $Ric(v,v) le k|v|^2$ (resp. $ge k|v|^2$) for every tangent vector $v$. For example, if an $n$-dimensional Riemannian manifold has sectional curvatures bounded above by a constant $c$, then its Ricci curvature is bounded above by $(n-1)c$.
- Second, note that the metrics whose existence is guaranteed by Lohkamp satisfy $Ric le -k_2g$ for some positive constant $k_2$. By transitivity, such a metric therefore also satisfies $Ric le k g$ for every positive constant $k$, so there is no topological consequence that can be deduced from any upper bound on Ricci curvature.
- Next, it certainly does not follow from Lohkamp's theorem (and it's not true) that no manifold admits a metric whose Ricci curvature is positive somewhere and negative somewhere else, as you seemed to suggest in a comment. In fact, every manifold admits such a metric -- you can just use a bump function to "paste in" a positively curved metric in one open set, and paste in a negatively curved metric in a different open set.
- Finally you asked what topological consequences can be deduced from estimates of the form $Kle k$ or $Kge -k$ (where $K$ represents sectional curvature). The answer is none -- this paper by R. E. Greene showed that every manifold admits a complete metric satisfying both of these inequalities.
Hope this helps.
Do you mean that all manifolds admit a metric with $Ricleq k$ or $Ric geq -k$? what we can say about topological properties of manifolds with $Kleq k$ or $Kgeq -k$?
â C.F.G
Sep 3 at 5:40
@C.F.G: Lohkamp's theorem says that every manifold admits a metric with $-k_1gle Ric le -k_2g < 0$ for positive constants $k_1,k_2$. Any such metric satisfies both of the hypotheses you asked about: $Ric le -k_2g$ and $Ric ge -k_1g$.
â Jack Lee
Sep 3 at 15:38
@C.F.G.: What do you mean by $K$?
â Jack Lee
Sep 3 at 15:38
Dear Prof. Jack Lee, note that one of my hypotheses is $Ric le k_2g$ and not $Ric le -k_2g$. $K$ denotes sectional curvature.
â C.F.G
Sep 4 at 5:21
Is the hypotheses $Ric geq -k_2$ equivalent to negative Ricci curvature as you mentioned above? i.e. No manifolds admit some positive somewhere and negative Ricci curvature somewhere.
â C.F.G
Sep 4 at 5:27
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
There can be no topological consequences of estimates of the form $Ric le k g$ or $Ric ge -k g$ (at least in dimensions greater that $2$), because this paper by Joachim Lohkamp shows that every smooth manifold of dimension at least $3$ admits a complete metric whose Ricci curvature is bounded between two negative constants.
Addition:
Your comments suggest that you may have some misunderstanding about what these inequalities mean. Here are some remarks that might help to clarify what's going on.
- First, an equality like $Ricle k$ or $Ric ge k$ is really shorthand for $Ricle kg$ or $Ric ge kg$. Equivalent statements are that $Ric(v,v)le k$ (resp. $ge k$) for every unit vector $v$; or $Ric(v,v) le k|v|^2$ (resp. $ge k|v|^2$) for every tangent vector $v$. For example, if an $n$-dimensional Riemannian manifold has sectional curvatures bounded above by a constant $c$, then its Ricci curvature is bounded above by $(n-1)c$.
- Second, note that the metrics whose existence is guaranteed by Lohkamp satisfy $Ric le -k_2g$ for some positive constant $k_2$. By transitivity, such a metric therefore also satisfies $Ric le k g$ for every positive constant $k$, so there is no topological consequence that can be deduced from any upper bound on Ricci curvature.
- Next, it certainly does not follow from Lohkamp's theorem (and it's not true) that no manifold admits a metric whose Ricci curvature is positive somewhere and negative somewhere else, as you seemed to suggest in a comment. In fact, every manifold admits such a metric -- you can just use a bump function to "paste in" a positively curved metric in one open set, and paste in a negatively curved metric in a different open set.
- Finally you asked what topological consequences can be deduced from estimates of the form $Kle k$ or $Kge -k$ (where $K$ represents sectional curvature). The answer is none -- this paper by R. E. Greene showed that every manifold admits a complete metric satisfying both of these inequalities.
Hope this helps.
Do you mean that all manifolds admit a metric with $Ricleq k$ or $Ric geq -k$? what we can say about topological properties of manifolds with $Kleq k$ or $Kgeq -k$?
â C.F.G
Sep 3 at 5:40
@C.F.G: Lohkamp's theorem says that every manifold admits a metric with $-k_1gle Ric le -k_2g < 0$ for positive constants $k_1,k_2$. Any such metric satisfies both of the hypotheses you asked about: $Ric le -k_2g$ and $Ric ge -k_1g$.
â Jack Lee
Sep 3 at 15:38
@C.F.G.: What do you mean by $K$?
â Jack Lee
Sep 3 at 15:38
Dear Prof. Jack Lee, note that one of my hypotheses is $Ric le k_2g$ and not $Ric le -k_2g$. $K$ denotes sectional curvature.
â C.F.G
Sep 4 at 5:21
Is the hypotheses $Ric geq -k_2$ equivalent to negative Ricci curvature as you mentioned above? i.e. No manifolds admit some positive somewhere and negative Ricci curvature somewhere.
â C.F.G
Sep 4 at 5:27
 |Â
show 3 more comments
up vote
2
down vote
accepted
There can be no topological consequences of estimates of the form $Ric le k g$ or $Ric ge -k g$ (at least in dimensions greater that $2$), because this paper by Joachim Lohkamp shows that every smooth manifold of dimension at least $3$ admits a complete metric whose Ricci curvature is bounded between two negative constants.
Addition:
Your comments suggest that you may have some misunderstanding about what these inequalities mean. Here are some remarks that might help to clarify what's going on.
- First, an equality like $Ricle k$ or $Ric ge k$ is really shorthand for $Ricle kg$ or $Ric ge kg$. Equivalent statements are that $Ric(v,v)le k$ (resp. $ge k$) for every unit vector $v$; or $Ric(v,v) le k|v|^2$ (resp. $ge k|v|^2$) for every tangent vector $v$. For example, if an $n$-dimensional Riemannian manifold has sectional curvatures bounded above by a constant $c$, then its Ricci curvature is bounded above by $(n-1)c$.
- Second, note that the metrics whose existence is guaranteed by Lohkamp satisfy $Ric le -k_2g$ for some positive constant $k_2$. By transitivity, such a metric therefore also satisfies $Ric le k g$ for every positive constant $k$, so there is no topological consequence that can be deduced from any upper bound on Ricci curvature.
- Next, it certainly does not follow from Lohkamp's theorem (and it's not true) that no manifold admits a metric whose Ricci curvature is positive somewhere and negative somewhere else, as you seemed to suggest in a comment. In fact, every manifold admits such a metric -- you can just use a bump function to "paste in" a positively curved metric in one open set, and paste in a negatively curved metric in a different open set.
- Finally you asked what topological consequences can be deduced from estimates of the form $Kle k$ or $Kge -k$ (where $K$ represents sectional curvature). The answer is none -- this paper by R. E. Greene showed that every manifold admits a complete metric satisfying both of these inequalities.
Hope this helps.
Do you mean that all manifolds admit a metric with $Ricleq k$ or $Ric geq -k$? what we can say about topological properties of manifolds with $Kleq k$ or $Kgeq -k$?
â C.F.G
Sep 3 at 5:40
@C.F.G: Lohkamp's theorem says that every manifold admits a metric with $-k_1gle Ric le -k_2g < 0$ for positive constants $k_1,k_2$. Any such metric satisfies both of the hypotheses you asked about: $Ric le -k_2g$ and $Ric ge -k_1g$.
â Jack Lee
Sep 3 at 15:38
@C.F.G.: What do you mean by $K$?
â Jack Lee
Sep 3 at 15:38
Dear Prof. Jack Lee, note that one of my hypotheses is $Ric le k_2g$ and not $Ric le -k_2g$. $K$ denotes sectional curvature.
â C.F.G
Sep 4 at 5:21
Is the hypotheses $Ric geq -k_2$ equivalent to negative Ricci curvature as you mentioned above? i.e. No manifolds admit some positive somewhere and negative Ricci curvature somewhere.
â C.F.G
Sep 4 at 5:27
 |Â
show 3 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
There can be no topological consequences of estimates of the form $Ric le k g$ or $Ric ge -k g$ (at least in dimensions greater that $2$), because this paper by Joachim Lohkamp shows that every smooth manifold of dimension at least $3$ admits a complete metric whose Ricci curvature is bounded between two negative constants.
Addition:
Your comments suggest that you may have some misunderstanding about what these inequalities mean. Here are some remarks that might help to clarify what's going on.
- First, an equality like $Ricle k$ or $Ric ge k$ is really shorthand for $Ricle kg$ or $Ric ge kg$. Equivalent statements are that $Ric(v,v)le k$ (resp. $ge k$) for every unit vector $v$; or $Ric(v,v) le k|v|^2$ (resp. $ge k|v|^2$) for every tangent vector $v$. For example, if an $n$-dimensional Riemannian manifold has sectional curvatures bounded above by a constant $c$, then its Ricci curvature is bounded above by $(n-1)c$.
- Second, note that the metrics whose existence is guaranteed by Lohkamp satisfy $Ric le -k_2g$ for some positive constant $k_2$. By transitivity, such a metric therefore also satisfies $Ric le k g$ for every positive constant $k$, so there is no topological consequence that can be deduced from any upper bound on Ricci curvature.
- Next, it certainly does not follow from Lohkamp's theorem (and it's not true) that no manifold admits a metric whose Ricci curvature is positive somewhere and negative somewhere else, as you seemed to suggest in a comment. In fact, every manifold admits such a metric -- you can just use a bump function to "paste in" a positively curved metric in one open set, and paste in a negatively curved metric in a different open set.
- Finally you asked what topological consequences can be deduced from estimates of the form $Kle k$ or $Kge -k$ (where $K$ represents sectional curvature). The answer is none -- this paper by R. E. Greene showed that every manifold admits a complete metric satisfying both of these inequalities.
Hope this helps.
There can be no topological consequences of estimates of the form $Ric le k g$ or $Ric ge -k g$ (at least in dimensions greater that $2$), because this paper by Joachim Lohkamp shows that every smooth manifold of dimension at least $3$ admits a complete metric whose Ricci curvature is bounded between two negative constants.
Addition:
Your comments suggest that you may have some misunderstanding about what these inequalities mean. Here are some remarks that might help to clarify what's going on.
- First, an equality like $Ricle k$ or $Ric ge k$ is really shorthand for $Ricle kg$ or $Ric ge kg$. Equivalent statements are that $Ric(v,v)le k$ (resp. $ge k$) for every unit vector $v$; or $Ric(v,v) le k|v|^2$ (resp. $ge k|v|^2$) for every tangent vector $v$. For example, if an $n$-dimensional Riemannian manifold has sectional curvatures bounded above by a constant $c$, then its Ricci curvature is bounded above by $(n-1)c$.
- Second, note that the metrics whose existence is guaranteed by Lohkamp satisfy $Ric le -k_2g$ for some positive constant $k_2$. By transitivity, such a metric therefore also satisfies $Ric le k g$ for every positive constant $k$, so there is no topological consequence that can be deduced from any upper bound on Ricci curvature.
- Next, it certainly does not follow from Lohkamp's theorem (and it's not true) that no manifold admits a metric whose Ricci curvature is positive somewhere and negative somewhere else, as you seemed to suggest in a comment. In fact, every manifold admits such a metric -- you can just use a bump function to "paste in" a positively curved metric in one open set, and paste in a negatively curved metric in a different open set.
- Finally you asked what topological consequences can be deduced from estimates of the form $Kle k$ or $Kge -k$ (where $K$ represents sectional curvature). The answer is none -- this paper by R. E. Greene showed that every manifold admits a complete metric satisfying both of these inequalities.
Hope this helps.
edited Sep 8 at 18:57
answered Sep 2 at 17:49
Jack Lee
25.6k44362
25.6k44362
Do you mean that all manifolds admit a metric with $Ricleq k$ or $Ric geq -k$? what we can say about topological properties of manifolds with $Kleq k$ or $Kgeq -k$?
â C.F.G
Sep 3 at 5:40
@C.F.G: Lohkamp's theorem says that every manifold admits a metric with $-k_1gle Ric le -k_2g < 0$ for positive constants $k_1,k_2$. Any such metric satisfies both of the hypotheses you asked about: $Ric le -k_2g$ and $Ric ge -k_1g$.
â Jack Lee
Sep 3 at 15:38
@C.F.G.: What do you mean by $K$?
â Jack Lee
Sep 3 at 15:38
Dear Prof. Jack Lee, note that one of my hypotheses is $Ric le k_2g$ and not $Ric le -k_2g$. $K$ denotes sectional curvature.
â C.F.G
Sep 4 at 5:21
Is the hypotheses $Ric geq -k_2$ equivalent to negative Ricci curvature as you mentioned above? i.e. No manifolds admit some positive somewhere and negative Ricci curvature somewhere.
â C.F.G
Sep 4 at 5:27
 |Â
show 3 more comments
Do you mean that all manifolds admit a metric with $Ricleq k$ or $Ric geq -k$? what we can say about topological properties of manifolds with $Kleq k$ or $Kgeq -k$?
â C.F.G
Sep 3 at 5:40
@C.F.G: Lohkamp's theorem says that every manifold admits a metric with $-k_1gle Ric le -k_2g < 0$ for positive constants $k_1,k_2$. Any such metric satisfies both of the hypotheses you asked about: $Ric le -k_2g$ and $Ric ge -k_1g$.
â Jack Lee
Sep 3 at 15:38
@C.F.G.: What do you mean by $K$?
â Jack Lee
Sep 3 at 15:38
Dear Prof. Jack Lee, note that one of my hypotheses is $Ric le k_2g$ and not $Ric le -k_2g$. $K$ denotes sectional curvature.
â C.F.G
Sep 4 at 5:21
Is the hypotheses $Ric geq -k_2$ equivalent to negative Ricci curvature as you mentioned above? i.e. No manifolds admit some positive somewhere and negative Ricci curvature somewhere.
â C.F.G
Sep 4 at 5:27
Do you mean that all manifolds admit a metric with $Ricleq k$ or $Ric geq -k$? what we can say about topological properties of manifolds with $Kleq k$ or $Kgeq -k$?
â C.F.G
Sep 3 at 5:40
Do you mean that all manifolds admit a metric with $Ricleq k$ or $Ric geq -k$? what we can say about topological properties of manifolds with $Kleq k$ or $Kgeq -k$?
â C.F.G
Sep 3 at 5:40
@C.F.G: Lohkamp's theorem says that every manifold admits a metric with $-k_1gle Ric le -k_2g < 0$ for positive constants $k_1,k_2$. Any such metric satisfies both of the hypotheses you asked about: $Ric le -k_2g$ and $Ric ge -k_1g$.
â Jack Lee
Sep 3 at 15:38
@C.F.G: Lohkamp's theorem says that every manifold admits a metric with $-k_1gle Ric le -k_2g < 0$ for positive constants $k_1,k_2$. Any such metric satisfies both of the hypotheses you asked about: $Ric le -k_2g$ and $Ric ge -k_1g$.
â Jack Lee
Sep 3 at 15:38
@C.F.G.: What do you mean by $K$?
â Jack Lee
Sep 3 at 15:38
@C.F.G.: What do you mean by $K$?
â Jack Lee
Sep 3 at 15:38
Dear Prof. Jack Lee, note that one of my hypotheses is $Ric le k_2g$ and not $Ric le -k_2g$. $K$ denotes sectional curvature.
â C.F.G
Sep 4 at 5:21
Dear Prof. Jack Lee, note that one of my hypotheses is $Ric le k_2g$ and not $Ric le -k_2g$. $K$ denotes sectional curvature.
â C.F.G
Sep 4 at 5:21
Is the hypotheses $Ric geq -k_2$ equivalent to negative Ricci curvature as you mentioned above? i.e. No manifolds admit some positive somewhere and negative Ricci curvature somewhere.
â C.F.G
Sep 4 at 5:27
Is the hypotheses $Ric geq -k_2$ equivalent to negative Ricci curvature as you mentioned above? i.e. No manifolds admit some positive somewhere and negative Ricci curvature somewhere.
â C.F.G
Sep 4 at 5:27
 |Â
show 3 more comments
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