Deforming metrics from non-negative to positive Ricci curvature
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Given a closed Riemannian manifold $(M,g)$ with non-negative Ricci curvature and $dimgeq 3$, when can we deform the metric to a positive Ricci curved one?
I know it's impossible in general due to the flat factor in the universal covering. But what about we add some topological restrictions on $M$ like simply connectedness? Are there any positive or negative results on this problem?
( Besides, are there now any examples of simply connected closed manifold with positive scalar curved metric by do not admit a positive Ricci curved metric? )
------------------------------------------------------------update 1------------------------------------------------------------
Thanks to the answer by Robert, I may simplify the problem in the following sense,
Given a simply connected flat Einstein manifold $(M,g)$ with $hatA$-genus non-vanishing, and set $(mathbbS^2,h)$ be the standard unit sphere, can $(Mtimes mathbbS^2, g+h)$ be perturbed to a Ricci-positive manifold? $ $In general, what about changing $(mathbbS^2,h)$ to an arbitrary closed Ricci-positive manifold?
dg.differential-geometry riemannian-geometry ricci-curvature
asked Nov 26 at 4:06
ZHans Wang
484
add a comment |
up vote
9
down vote
favorite
Given a closed Riemannian manifold $(M,g)$ with non-negative Ricci curvature and $dimgeq 3$, when can we deform the metric to a positive Ricci curved one?
I know it's impossible in general due to the flat factor in the universal covering. But what about we add some topological restrictions on $M$ like simply connectedness? Are there any positive or negative results on this problem?
( Besides, are there now any examples of simply connected closed manifold with positive scalar curved metric by do not admit a positive Ricci curved metric? )
------------------------------------------------------------update 1------------------------------------------------------------
Thanks to the answer by Robert, I may simplify the problem in the following sense,
Given a simply connected flat Einstein manifold $(M,g)$ with $hatA$-genus non-vanishing, and set $(mathbbS^2,h)$ be the standard unit sphere, can $(Mtimes mathbbS^2, g+h)$ be perturbed to a Ricci-positive manifold? $ $In general, what about changing $(mathbbS^2,h)$ to an arbitrary closed Ricci-positive manifold?
dg.differential-geometry riemannian-geometry ricci-curvature
asked Nov 26 at 4:06
ZHans Wang
484
It is better to ask the follow up question separately.
– Igor Belegradek
Nov 29 at 22:36
On the bottom of p.3 of arxiv.org/pdf/1607.00657.pdf D.Wraith discusses a related question. He considers the product of a K3 surface with a homotopy $(4n-1)$-sphere $Sigma$ that bounds a parallelizable manifold. (Wraith proved elsewhere that any such sphere admits a metric of $Ric>0$ so $K3timesSigma$ has a metric of $Ricge 0$). Wraith then remarks: "There are no known obstructions to positive Ricci curvature for these manifolds... Nevertheless, the author is tempted to conjecture that no Ricci positive metrics exist".
– Igor Belegradek
Nov 30 at 1:35
add a comment |
up vote
9
down vote
favorite
up vote
9
down vote
favorite
Given a closed Riemannian manifold $(M,g)$ with non-negative Ricci curvature and $dimgeq 3$, when can we deform the metric to a positive Ricci curved one?
I know it's impossible in general due to the flat factor in the universal covering. But what about we add some topological restrictions on $M$ like simply connectedness? Are there any positive or negative results on this problem?
( Besides, are there now any examples of simply connected closed manifold with positive scalar curved metric by do not admit a positive Ricci curved metric? )
------------------------------------------------------------update 1------------------------------------------------------------
Thanks to the answer by Robert, I may simplify the problem in the following sense,
Given a simply connected flat Einstein manifold $(M,g)$ with $hatA$-genus non-vanishing, and set $(mathbbS^2,h)$ be the standard unit sphere, can $(Mtimes mathbbS^2, g+h)$ be perturbed to a Ricci-positive manifold? $ $In general, what about changing $(mathbbS^2,h)$ to an arbitrary closed Ricci-positive manifold?
dg.differential-geometry riemannian-geometry ricci-curvature
asked Nov 26 at 4:06
ZHans Wang
484
Given a closed Riemannian manifold $(M,g)$ with non-negative Ricci curvature and $dimgeq 3$, when can we deform the metric to a positive Ricci curved one?
I know it's impossible in general due to the flat factor in the universal covering. But what about we add some topological restrictions on $M$ like simply connectedness? Are there any positive or negative results on this problem?
( Besides, are there now any examples of simply connected closed manifold with positive scalar curved metric by do not admit a positive Ricci curved metric? )
------------------------------------------------------------update 1------------------------------------------------------------
Thanks to the answer by Robert, I may simplify the problem in the following sense,
Given a simply connected flat Einstein manifold $(M,g)$ with $hatA$-genus non-vanishing, and set $(mathbbS^2,h)$ be the standard unit sphere, can $(Mtimes mathbbS^2, g+h)$ be perturbed to a Ricci-positive manifold? $ $In general, what about changing $(mathbbS^2,h)$ to an arbitrary closed Ricci-positive manifold?
dg.differential-geometry riemannian-geometry ricci-curvature
dg.differential-geometry riemannian-geometry ricci-curvature
asked Nov 26 at 4:06
ZHans Wang
484
asked Nov 26 at 4:06
ZHans Wang
484
edited Nov 29 at 21:19
asked Nov 26 at 4:06
ZHans Wang
484
asked Nov 26 at 4:06
ZHans Wang
484
asked Nov 26 at 4:06
ZHans Wang
484
484
It is better to ask the follow up question separately.
– Igor Belegradek
Nov 29 at 22:36
On the bottom of p.3 of arxiv.org/pdf/1607.00657.pdf D.Wraith discusses a related question. He considers the product of a K3 surface with a homotopy $(4n-1)$-sphere $Sigma$ that bounds a parallelizable manifold. (Wraith proved elsewhere that any such sphere admits a metric of $Ric>0$ so $K3timesSigma$ has a metric of $Ricge 0$). Wraith then remarks: "There are no known obstructions to positive Ricci curvature for these manifolds... Nevertheless, the author is tempted to conjecture that no Ricci positive metrics exist".
– Igor Belegradek
Nov 30 at 1:35
add a comment |
It is better to ask the follow up question separately.
– Igor Belegradek
Nov 29 at 22:36
On the bottom of p.3 of arxiv.org/pdf/1607.00657.pdf D.Wraith discusses a related question. He considers the product of a K3 surface with a homotopy $(4n-1)$-sphere $Sigma$ that bounds a parallelizable manifold. (Wraith proved elsewhere that any such sphere admits a metric of $Ric>0$ so $K3timesSigma$ has a metric of $Ricge 0$). Wraith then remarks: "There are no known obstructions to positive Ricci curvature for these manifolds... Nevertheless, the author is tempted to conjecture that no Ricci positive metrics exist".
– Igor Belegradek
Nov 30 at 1:35
It is better to ask the follow up question separately.
– Igor Belegradek
Nov 29 at 22:36
It is better to ask the follow up question separately.
– Igor Belegradek
Nov 29 at 22:36
On the bottom of p.3 of arxiv.org/pdf/1607.00657.pdf D.Wraith discusses a related question. He considers the product of a K3 surface with a homotopy $(4n-1)$-sphere $Sigma$ that bounds a parallelizable manifold. (Wraith proved elsewhere that any such sphere admits a metric of $Ric>0$ so $K3timesSigma$ has a metric of $Ricge 0$). Wraith then remarks: "There are no known obstructions to positive Ricci curvature for these manifolds... Nevertheless, the author is tempted to conjecture that no Ricci positive metrics exist".
– Igor Belegradek
Nov 30 at 1:35
On the bottom of p.3 of arxiv.org/pdf/1607.00657.pdf D.Wraith discusses a related question. He considers the product of a K3 surface with a homotopy $(4n-1)$-sphere $Sigma$ that bounds a parallelizable manifold. (Wraith proved elsewhere that any such sphere admits a metric of $Ric>0$ so $K3timesSigma$ has a metric of $Ricge 0$). Wraith then remarks: "There are no known obstructions to positive Ricci curvature for these manifolds... Nevertheless, the author is tempted to conjecture that no Ricci positive metrics exist".
– Igor Belegradek
Nov 30 at 1:35
add a comment |
2 Answers
2
active
oldest
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up vote
9
down vote
accepted
There are obstructions. Perhaps the most famous comes from the theorem that, if a compact spin manifold has a metric of positive scalar curvature, then its $hat A$-genus must vanish.
If you take a compact Riemannian spin manifold $(M,g)$ with special holonomy $mathrmG_2$ (in dimension $7$), $mathrmSpin(7)$ (in dimension $8$), or holonomy in $mathrmSU(n)$ (in dimension $2n$) whose $hat A$-genus is nonzero (and there are lots of these, even simply-connected ones), then $g$ will be Ricci-flat and hence will have non-negative Ricci curvature. However, by the above theorem, it cannot carry any metric with positive scalar curvature, let alone a metric with positive Ricci curvature.
answered Nov 26 at 12:33
Robert Bryant
72.4k5213313
This is why I asked the second question. So what about supposing Ricci non-negative and scalar positive on the metric, can it be deformed to a Ricci positive one? In general, what Riemannian manifold could be on the boundary of the space of non-collapsing positive Ricci curved manifolds?
– ZHans Wang
Nov 27 at 5:03
add a comment |
up vote
7
down vote
This is not a complete answer but would be helpful. Here are a few facts:
Theorem (T. Aubin 1979). If the Ricci curvature of a compact Riemannian manifold is
non-negative and positive somewhere, then the manifold carries a metric with
positive Ricci curvature.
Also vanishing the first Betti number is a necessary condition in compact case for admitting strictly
positive Ricci curvature (see on Google books: A Course in Differential Geometry, By Thierry Aubin).
Relation with scalar curvature:
There are still no known examples of simply connected manifolds that admit
positive scalar curvature but not positive Ricci curvatureIf a manifold $M$ cannot have a metric with positive (or zero) Scalar curvature, then it certainly does not admit a metric with positive (zero res.) Ricci curvature.
This paper of G. Perelman is also useful: "Construction of manifolds of positive Ricci curvature with big volume and large Betti numbers". Comparison Geometry. 30: 157–163 Click here for pdf
answered Nov 26 at 6:33
C.F.G
912430
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
There are obstructions. Perhaps the most famous comes from the theorem that, if a compact spin manifold has a metric of positive scalar curvature, then its $hat A$-genus must vanish.
If you take a compact Riemannian spin manifold $(M,g)$ with special holonomy $mathrmG_2$ (in dimension $7$), $mathrmSpin(7)$ (in dimension $8$), or holonomy in $mathrmSU(n)$ (in dimension $2n$) whose $hat A$-genus is nonzero (and there are lots of these, even simply-connected ones), then $g$ will be Ricci-flat and hence will have non-negative Ricci curvature. However, by the above theorem, it cannot carry any metric with positive scalar curvature, let alone a metric with positive Ricci curvature.
answered Nov 26 at 12:33
Robert Bryant
72.4k5213313
This is why I asked the second question. So what about supposing Ricci non-negative and scalar positive on the metric, can it be deformed to a Ricci positive one? In general, what Riemannian manifold could be on the boundary of the space of non-collapsing positive Ricci curved manifolds?
– ZHans Wang
Nov 27 at 5:03
add a comment |
up vote
9
down vote
accepted
There are obstructions. Perhaps the most famous comes from the theorem that, if a compact spin manifold has a metric of positive scalar curvature, then its $hat A$-genus must vanish.
If you take a compact Riemannian spin manifold $(M,g)$ with special holonomy $mathrmG_2$ (in dimension $7$), $mathrmSpin(7)$ (in dimension $8$), or holonomy in $mathrmSU(n)$ (in dimension $2n$) whose $hat A$-genus is nonzero (and there are lots of these, even simply-connected ones), then $g$ will be Ricci-flat and hence will have non-negative Ricci curvature. However, by the above theorem, it cannot carry any metric with positive scalar curvature, let alone a metric with positive Ricci curvature.
answered Nov 26 at 12:33
Robert Bryant
72.4k5213313
This is why I asked the second question. So what about supposing Ricci non-negative and scalar positive on the metric, can it be deformed to a Ricci positive one? In general, what Riemannian manifold could be on the boundary of the space of non-collapsing positive Ricci curved manifolds?
– ZHans Wang
Nov 27 at 5:03
add a comment |
up vote
9
down vote
accepted
up vote
9
down vote
accepted
There are obstructions. Perhaps the most famous comes from the theorem that, if a compact spin manifold has a metric of positive scalar curvature, then its $hat A$-genus must vanish.
If you take a compact Riemannian spin manifold $(M,g)$ with special holonomy $mathrmG_2$ (in dimension $7$), $mathrmSpin(7)$ (in dimension $8$), or holonomy in $mathrmSU(n)$ (in dimension $2n$) whose $hat A$-genus is nonzero (and there are lots of these, even simply-connected ones), then $g$ will be Ricci-flat and hence will have non-negative Ricci curvature. However, by the above theorem, it cannot carry any metric with positive scalar curvature, let alone a metric with positive Ricci curvature.
answered Nov 26 at 12:33
Robert Bryant
72.4k5213313
There are obstructions. Perhaps the most famous comes from the theorem that, if a compact spin manifold has a metric of positive scalar curvature, then its $hat A$-genus must vanish.
If you take a compact Riemannian spin manifold $(M,g)$ with special holonomy $mathrmG_2$ (in dimension $7$), $mathrmSpin(7)$ (in dimension $8$), or holonomy in $mathrmSU(n)$ (in dimension $2n$) whose $hat A$-genus is nonzero (and there are lots of these, even simply-connected ones), then $g$ will be Ricci-flat and hence will have non-negative Ricci curvature. However, by the above theorem, it cannot carry any metric with positive scalar curvature, let alone a metric with positive Ricci curvature.
answered Nov 26 at 12:33
Robert Bryant
72.4k5213313
answered Nov 26 at 12:33
Robert Bryant
72.4k5213313
answered Nov 26 at 12:33
Robert Bryant
72.4k5213313
answered Nov 26 at 12:33
Robert Bryant
72.4k5213313
72.4k5213313
This is why I asked the second question. So what about supposing Ricci non-negative and scalar positive on the metric, can it be deformed to a Ricci positive one? In general, what Riemannian manifold could be on the boundary of the space of non-collapsing positive Ricci curved manifolds?
– ZHans Wang
Nov 27 at 5:03
add a comment |
This is why I asked the second question. So what about supposing Ricci non-negative and scalar positive on the metric, can it be deformed to a Ricci positive one? In general, what Riemannian manifold could be on the boundary of the space of non-collapsing positive Ricci curved manifolds?
– ZHans Wang
Nov 27 at 5:03
This is why I asked the second question. So what about supposing Ricci non-negative and scalar positive on the metric, can it be deformed to a Ricci positive one? In general, what Riemannian manifold could be on the boundary of the space of non-collapsing positive Ricci curved manifolds?
– ZHans Wang
Nov 27 at 5:03
This is why I asked the second question. So what about supposing Ricci non-negative and scalar positive on the metric, can it be deformed to a Ricci positive one? In general, what Riemannian manifold could be on the boundary of the space of non-collapsing positive Ricci curved manifolds?
– ZHans Wang
Nov 27 at 5:03
add a comment |
up vote
7
down vote
This is not a complete answer but would be helpful. Here are a few facts:
Theorem (T. Aubin 1979). If the Ricci curvature of a compact Riemannian manifold is
non-negative and positive somewhere, then the manifold carries a metric with
positive Ricci curvature.
Also vanishing the first Betti number is a necessary condition in compact case for admitting strictly
positive Ricci curvature (see on Google books: A Course in Differential Geometry, By Thierry Aubin).
Relation with scalar curvature:
There are still no known examples of simply connected manifolds that admit
positive scalar curvature but not positive Ricci curvatureIf a manifold $M$ cannot have a metric with positive (or zero) Scalar curvature, then it certainly does not admit a metric with positive (zero res.) Ricci curvature.
This paper of G. Perelman is also useful: "Construction of manifolds of positive Ricci curvature with big volume and large Betti numbers". Comparison Geometry. 30: 157–163 Click here for pdf
answered Nov 26 at 6:33
C.F.G
912430
add a comment |
up vote
7
down vote
This is not a complete answer but would be helpful. Here are a few facts:
Theorem (T. Aubin 1979). If the Ricci curvature of a compact Riemannian manifold is
non-negative and positive somewhere, then the manifold carries a metric with
positive Ricci curvature.
Also vanishing the first Betti number is a necessary condition in compact case for admitting strictly
positive Ricci curvature (see on Google books: A Course in Differential Geometry, By Thierry Aubin).
Relation with scalar curvature:
There are still no known examples of simply connected manifolds that admit
positive scalar curvature but not positive Ricci curvatureIf a manifold $M$ cannot have a metric with positive (or zero) Scalar curvature, then it certainly does not admit a metric with positive (zero res.) Ricci curvature.
This paper of G. Perelman is also useful: "Construction of manifolds of positive Ricci curvature with big volume and large Betti numbers". Comparison Geometry. 30: 157–163 Click here for pdf
answered Nov 26 at 6:33
C.F.G
912430
add a comment |
up vote
7
down vote
up vote
7
down vote
This is not a complete answer but would be helpful. Here are a few facts:
Theorem (T. Aubin 1979). If the Ricci curvature of a compact Riemannian manifold is
non-negative and positive somewhere, then the manifold carries a metric with
positive Ricci curvature.
Also vanishing the first Betti number is a necessary condition in compact case for admitting strictly
positive Ricci curvature (see on Google books: A Course in Differential Geometry, By Thierry Aubin).
Relation with scalar curvature:
There are still no known examples of simply connected manifolds that admit
positive scalar curvature but not positive Ricci curvatureIf a manifold $M$ cannot have a metric with positive (or zero) Scalar curvature, then it certainly does not admit a metric with positive (zero res.) Ricci curvature.
This paper of G. Perelman is also useful: "Construction of manifolds of positive Ricci curvature with big volume and large Betti numbers". Comparison Geometry. 30: 157–163 Click here for pdf
answered Nov 26 at 6:33
C.F.G
912430
This is not a complete answer but would be helpful. Here are a few facts:
Theorem (T. Aubin 1979). If the Ricci curvature of a compact Riemannian manifold is
non-negative and positive somewhere, then the manifold carries a metric with
positive Ricci curvature.
Also vanishing the first Betti number is a necessary condition in compact case for admitting strictly
positive Ricci curvature (see on Google books: A Course in Differential Geometry, By Thierry Aubin).
Relation with scalar curvature:
There are still no known examples of simply connected manifolds that admit
positive scalar curvature but not positive Ricci curvatureIf a manifold $M$ cannot have a metric with positive (or zero) Scalar curvature, then it certainly does not admit a metric with positive (zero res.) Ricci curvature.
This paper of G. Perelman is also useful: "Construction of manifolds of positive Ricci curvature with big volume and large Betti numbers". Comparison Geometry. 30: 157–163 Click here for pdf
answered Nov 26 at 6:33
C.F.G
912430
answered Nov 26 at 6:33
C.F.G
912430
answered Nov 26 at 6:33
C.F.G
912430
answered Nov 26 at 6:33
C.F.G
912430
912430
add a comment |
add a comment |
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It is better to ask the follow up question separately.
– Igor Belegradek
Nov 29 at 22:36
On the bottom of p.3 of arxiv.org/pdf/1607.00657.pdf D.Wraith discusses a related question. He considers the product of a K3 surface with a homotopy $(4n-1)$-sphere $Sigma$ that bounds a parallelizable manifold. (Wraith proved elsewhere that any such sphere admits a metric of $Ric>0$ so $K3timesSigma$ has a metric of $Ricge 0$). Wraith then remarks: "There are no known obstructions to positive Ricci curvature for these manifolds... Nevertheless, the author is tempted to conjecture that no Ricci positive metrics exist".
– Igor Belegradek
Nov 30 at 1:35