Metric of positive curvature and Homology groupDoes the curvature determine the metric?Torus with positive sectional curvature.Fundamental group and curvatureWhether there is a Riemannian metric on $S^2times S^2$ with positive scalar curvature?Does there exist a known example of Riemannian manifold who its sectional curvature admit both zero and positive values?Does positive Yamabe invariant imply every metric in that conformal class has positive scalar curvature?Positive scalar curvature metric on $S^4$Positive scalar curvature in dimension 4Constant scalar curvature with positive Ricci curvatureHopf Conjecture about Curvature and Topology
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Metric of positive curvature and Homology group
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Metric of positive curvature and Homology group
Does the curvature determine the metric?Torus with positive sectional curvature.Fundamental group and curvatureWhether there is a Riemannian metric on $S^2times S^2$ with positive scalar curvature?Does there exist a known example of Riemannian manifold who its sectional curvature admit both zero and positive values?Does positive Yamabe invariant imply every metric in that conformal class has positive scalar curvature?Positive scalar curvature metric on $S^4$Positive scalar curvature in dimension 4Constant scalar curvature with positive Ricci curvatureHopf Conjecture about Curvature and Topology
$begingroup$
Is it possible to decide that whether the manifold $M$ admit a metric of positive curvature by knowing all Homology group of $M$?
differential-geometry differential-topology riemannian-geometry
asked 5 hours ago
C.F.GC.F.G
1,4971821
$endgroup$
add a comment |
$begingroup$
Is it possible to decide that whether the manifold $M$ admit a metric of positive curvature by knowing all Homology group of $M$?
differential-geometry differential-topology riemannian-geometry
asked 5 hours ago
C.F.GC.F.G
1,4971821
$endgroup$
add a comment |
$begingroup$
Is it possible to decide that whether the manifold $M$ admit a metric of positive curvature by knowing all Homology group of $M$?
differential-geometry differential-topology riemannian-geometry
asked 5 hours ago
C.F.GC.F.G
1,4971821
$endgroup$
Is it possible to decide that whether the manifold $M$ admit a metric of positive curvature by knowing all Homology group of $M$?
differential-geometry differential-topology riemannian-geometry
differential-geometry differential-topology riemannian-geometry
asked 5 hours ago
C.F.GC.F.G
1,4971821
asked 5 hours ago
C.F.GC.F.G
1,4971821
asked 5 hours ago
C.F.GC.F.G
1,4971821
asked 5 hours ago
C.F.GC.F.G
1,4971821
asked 5 hours ago
C.F.GC.F.G
1,4971821
1,4971821
add a comment |
add a comment |
2 Answers
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No. For example, $S^9$ admits a metric of constant positive sectional curvature, but there is an exotic $9$-sphere $Sigma$ (i.e. $Sigma$ is a smooth manifold homeomorphic to $S^9$, but not diffeomorphic to it) which does not even admit a metric of positive scalar curvature. As $S^9$ and $Sigma$ are homeomorphic, they have the same homology groups.
answered 4 hours ago
Michael AlbaneseMichael Albanese
65.4k15101319
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Definition. A manifold $M$ is called an $n$-dimensional homology sphere if its homology groups are isomorphic to that of $S^n$ (i.e. $M$ is connected and its homology is nonzero only in degrees 0 and n).
There are many examples of integer homology spheres in dimension 3 which are apherical (i.e. have contractible universal covering space). The simplest example I know is obtained as follows: Take the trefoil knot $Ksubset S^3$, let $N$ be the complement in $S^3$ to a regular neighborhood of $K$. The boundary of $N$ is $T^2$. Now, glue two copies of $N$ by swapping the meridians and longitudes. MV sequence shows that the resulting manifold $M$ is a 3-dimensional homology sphere. With a bit more work, one verifies that $M$ is aspherical.
There are many other examples like this, for instance, among
hyperbolic 3-manifolds.
According to Thm. 8.1 in
M. Gromov, H.B. Lawson, Positive scalar curvature and the Dirac operator on complete Riemannian manifolds, Inst. Hautes Etudes Sci. Publ. ´Math. No. 58 (1983), 83–196.
such a manifold cannot admit a metric of positive scalar curvature.
answered 4 hours ago
Moishe KohanMoishe Kohan
49.5k346112
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2 Answers
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2 Answers
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$begingroup$
No. For example, $S^9$ admits a metric of constant positive sectional curvature, but there is an exotic $9$-sphere $Sigma$ (i.e. $Sigma$ is a smooth manifold homeomorphic to $S^9$, but not diffeomorphic to it) which does not even admit a metric of positive scalar curvature. As $S^9$ and $Sigma$ are homeomorphic, they have the same homology groups.
answered 4 hours ago
Michael AlbaneseMichael Albanese
65.4k15101319
$endgroup$
add a comment |
$begingroup$
No. For example, $S^9$ admits a metric of constant positive sectional curvature, but there is an exotic $9$-sphere $Sigma$ (i.e. $Sigma$ is a smooth manifold homeomorphic to $S^9$, but not diffeomorphic to it) which does not even admit a metric of positive scalar curvature. As $S^9$ and $Sigma$ are homeomorphic, they have the same homology groups.
answered 4 hours ago
Michael AlbaneseMichael Albanese
65.4k15101319
$endgroup$
add a comment |
$begingroup$
No. For example, $S^9$ admits a metric of constant positive sectional curvature, but there is an exotic $9$-sphere $Sigma$ (i.e. $Sigma$ is a smooth manifold homeomorphic to $S^9$, but not diffeomorphic to it) which does not even admit a metric of positive scalar curvature. As $S^9$ and $Sigma$ are homeomorphic, they have the same homology groups.
answered 4 hours ago
Michael AlbaneseMichael Albanese
65.4k15101319
$endgroup$
No. For example, $S^9$ admits a metric of constant positive sectional curvature, but there is an exotic $9$-sphere $Sigma$ (i.e. $Sigma$ is a smooth manifold homeomorphic to $S^9$, but not diffeomorphic to it) which does not even admit a metric of positive scalar curvature. As $S^9$ and $Sigma$ are homeomorphic, they have the same homology groups.
answered 4 hours ago
Michael AlbaneseMichael Albanese
65.4k15101319
answered 4 hours ago
Michael AlbaneseMichael Albanese
65.4k15101319
answered 4 hours ago
Michael AlbaneseMichael Albanese
65.4k15101319
answered 4 hours ago
Michael AlbaneseMichael Albanese
65.4k15101319
65.4k15101319
add a comment |
add a comment |
$begingroup$
Definition. A manifold $M$ is called an $n$-dimensional homology sphere if its homology groups are isomorphic to that of $S^n$ (i.e. $M$ is connected and its homology is nonzero only in degrees 0 and n).
There are many examples of integer homology spheres in dimension 3 which are apherical (i.e. have contractible universal covering space). The simplest example I know is obtained as follows: Take the trefoil knot $Ksubset S^3$, let $N$ be the complement in $S^3$ to a regular neighborhood of $K$. The boundary of $N$ is $T^2$. Now, glue two copies of $N$ by swapping the meridians and longitudes. MV sequence shows that the resulting manifold $M$ is a 3-dimensional homology sphere. With a bit more work, one verifies that $M$ is aspherical.
There are many other examples like this, for instance, among
hyperbolic 3-manifolds.
According to Thm. 8.1 in
M. Gromov, H.B. Lawson, Positive scalar curvature and the Dirac operator on complete Riemannian manifolds, Inst. Hautes Etudes Sci. Publ. ´Math. No. 58 (1983), 83–196.
such a manifold cannot admit a metric of positive scalar curvature.
answered 4 hours ago
Moishe KohanMoishe Kohan
49.5k346112
$endgroup$
add a comment |
$begingroup$
Definition. A manifold $M$ is called an $n$-dimensional homology sphere if its homology groups are isomorphic to that of $S^n$ (i.e. $M$ is connected and its homology is nonzero only in degrees 0 and n).
There are many examples of integer homology spheres in dimension 3 which are apherical (i.e. have contractible universal covering space). The simplest example I know is obtained as follows: Take the trefoil knot $Ksubset S^3$, let $N$ be the complement in $S^3$ to a regular neighborhood of $K$. The boundary of $N$ is $T^2$. Now, glue two copies of $N$ by swapping the meridians and longitudes. MV sequence shows that the resulting manifold $M$ is a 3-dimensional homology sphere. With a bit more work, one verifies that $M$ is aspherical.
There are many other examples like this, for instance, among
hyperbolic 3-manifolds.
According to Thm. 8.1 in
M. Gromov, H.B. Lawson, Positive scalar curvature and the Dirac operator on complete Riemannian manifolds, Inst. Hautes Etudes Sci. Publ. ´Math. No. 58 (1983), 83–196.
such a manifold cannot admit a metric of positive scalar curvature.
answered 4 hours ago
Moishe KohanMoishe Kohan
49.5k346112
$endgroup$
add a comment |
$begingroup$
Definition. A manifold $M$ is called an $n$-dimensional homology sphere if its homology groups are isomorphic to that of $S^n$ (i.e. $M$ is connected and its homology is nonzero only in degrees 0 and n).
There are many examples of integer homology spheres in dimension 3 which are apherical (i.e. have contractible universal covering space). The simplest example I know is obtained as follows: Take the trefoil knot $Ksubset S^3$, let $N$ be the complement in $S^3$ to a regular neighborhood of $K$. The boundary of $N$ is $T^2$. Now, glue two copies of $N$ by swapping the meridians and longitudes. MV sequence shows that the resulting manifold $M$ is a 3-dimensional homology sphere. With a bit more work, one verifies that $M$ is aspherical.
There are many other examples like this, for instance, among
hyperbolic 3-manifolds.
According to Thm. 8.1 in
M. Gromov, H.B. Lawson, Positive scalar curvature and the Dirac operator on complete Riemannian manifolds, Inst. Hautes Etudes Sci. Publ. ´Math. No. 58 (1983), 83–196.
such a manifold cannot admit a metric of positive scalar curvature.
answered 4 hours ago
Moishe KohanMoishe Kohan
49.5k346112
$endgroup$
Definition. A manifold $M$ is called an $n$-dimensional homology sphere if its homology groups are isomorphic to that of $S^n$ (i.e. $M$ is connected and its homology is nonzero only in degrees 0 and n).
There are many examples of integer homology spheres in dimension 3 which are apherical (i.e. have contractible universal covering space). The simplest example I know is obtained as follows: Take the trefoil knot $Ksubset S^3$, let $N$ be the complement in $S^3$ to a regular neighborhood of $K$. The boundary of $N$ is $T^2$. Now, glue two copies of $N$ by swapping the meridians and longitudes. MV sequence shows that the resulting manifold $M$ is a 3-dimensional homology sphere. With a bit more work, one verifies that $M$ is aspherical.
There are many other examples like this, for instance, among
hyperbolic 3-manifolds.
According to Thm. 8.1 in
M. Gromov, H.B. Lawson, Positive scalar curvature and the Dirac operator on complete Riemannian manifolds, Inst. Hautes Etudes Sci. Publ. ´Math. No. 58 (1983), 83–196.
such a manifold cannot admit a metric of positive scalar curvature.
answered 4 hours ago
Moishe KohanMoishe Kohan
49.5k346112
answered 4 hours ago
Moishe KohanMoishe Kohan
49.5k346112
answered 4 hours ago
Moishe KohanMoishe Kohan
49.5k346112
answered 4 hours ago
Moishe KohanMoishe Kohan
49.5k346112
49.5k346112
add a comment |
add a comment |
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