Mixing
Show that the rational cohomology ring $H^*(M;mathbbQ)$ needs at least two generators
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
Let $M$ be a simply connected closed Riemannian manifold. How does one find a condition that may be imposed on $M$ (perhaps on the curvature of $M$ and on torsion) which guarantees that the rational cohomology ring $H^*(M;mathbbQ)=bigoplus_kinmathbbNH^k(M;mathbbQ)$ needs at least two generators? That is, how does one force $M$ not to have rational cohomology that is the quotient of a polynomial ring?
Cross-posting on MO: https://mathoverflow.net/questions/306490/show-that-the-rational-cohomology-ring-hm-mathbbq-needs-at-least-two-ge
Any help would be much appreciated. Thanks in advance!
group-theory algebraic-topology homology-cohomology
asked Jul 20 at 4:42
Multivariablecalculus
495313
add a comment |Â
up vote
5
down vote
favorite
Let $M$ be a simply connected closed Riemannian manifold. How does one find a condition that may be imposed on $M$ (perhaps on the curvature of $M$ and on torsion) which guarantees that the rational cohomology ring $H^*(M;mathbbQ)=bigoplus_kinmathbbNH^k(M;mathbbQ)$ needs at least two generators? That is, how does one force $M$ not to have rational cohomology that is the quotient of a polynomial ring?
Cross-posting on MO: https://mathoverflow.net/questions/306490/show-that-the-rational-cohomology-ring-hm-mathbbq-needs-at-least-two-ge
Any help would be much appreciated. Thanks in advance!
group-theory algebraic-topology homology-cohomology
asked Jul 20 at 4:42
Multivariablecalculus
495313
2
If $M$ is oriented then there are at least two non-trivial classes in $H^*(M;mathbbQ)$. More generally you can consider its Betti numbers.
– Tyrone
Jul 20 at 9:47
3
@Tyrone I think the question is how to force $M$ to not have polynomial cohomology ring.
– Mike Miller
Jul 20 at 14:01
What type of conditions are you looking for? Curvature or what else?
– Thomas Rot
Jul 20 at 14:47
@Thomas Rot Yes, curvature in particular.
– Multivariablecalculus
Jul 20 at 16:24
@Mike Miller Yes, I would like to force $M$ not to have a truncated polynomial cohomology ring.
– Multivariablecalculus
Jul 20 at 16:25
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let $M$ be a simply connected closed Riemannian manifold. How does one find a condition that may be imposed on $M$ (perhaps on the curvature of $M$ and on torsion) which guarantees that the rational cohomology ring $H^*(M;mathbbQ)=bigoplus_kinmathbbNH^k(M;mathbbQ)$ needs at least two generators? That is, how does one force $M$ not to have rational cohomology that is the quotient of a polynomial ring?
Cross-posting on MO: https://mathoverflow.net/questions/306490/show-that-the-rational-cohomology-ring-hm-mathbbq-needs-at-least-two-ge
Any help would be much appreciated. Thanks in advance!
group-theory algebraic-topology homology-cohomology
asked Jul 20 at 4:42
Multivariablecalculus
495313
Let $M$ be a simply connected closed Riemannian manifold. How does one find a condition that may be imposed on $M$ (perhaps on the curvature of $M$ and on torsion) which guarantees that the rational cohomology ring $H^*(M;mathbbQ)=bigoplus_kinmathbbNH^k(M;mathbbQ)$ needs at least two generators? That is, how does one force $M$ not to have rational cohomology that is the quotient of a polynomial ring?
Cross-posting on MO: https://mathoverflow.net/questions/306490/show-that-the-rational-cohomology-ring-hm-mathbbq-needs-at-least-two-ge
Any help would be much appreciated. Thanks in advance!
group-theory algebraic-topology homology-cohomology
asked Jul 20 at 4:42
Multivariablecalculus
495313
edited Jul 23 at 15:47
asked Jul 20 at 4:42
Multivariablecalculus
495313
asked Jul 20 at 4:42
Multivariablecalculus
495313
asked Jul 20 at 4:42
Multivariablecalculus
495313
495313
2
If $M$ is oriented then there are at least two non-trivial classes in $H^*(M;mathbbQ)$. More generally you can consider its Betti numbers.
– Tyrone
Jul 20 at 9:47
3
@Tyrone I think the question is how to force $M$ to not have polynomial cohomology ring.
– Mike Miller
Jul 20 at 14:01
What type of conditions are you looking for? Curvature or what else?
– Thomas Rot
Jul 20 at 14:47
@Thomas Rot Yes, curvature in particular.
– Multivariablecalculus
Jul 20 at 16:24
@Mike Miller Yes, I would like to force $M$ not to have a truncated polynomial cohomology ring.
– Multivariablecalculus
Jul 20 at 16:25
add a comment |Â
2
If $M$ is oriented then there are at least two non-trivial classes in $H^*(M;mathbbQ)$. More generally you can consider its Betti numbers.
– Tyrone
Jul 20 at 9:47
3
@Tyrone I think the question is how to force $M$ to not have polynomial cohomology ring.
– Mike Miller
Jul 20 at 14:01
What type of conditions are you looking for? Curvature or what else?
– Thomas Rot
Jul 20 at 14:47
@Thomas Rot Yes, curvature in particular.
– Multivariablecalculus
Jul 20 at 16:24
@Mike Miller Yes, I would like to force $M$ not to have a truncated polynomial cohomology ring.
– Multivariablecalculus
Jul 20 at 16:25
2
2
If $M$ is oriented then there are at least two non-trivial classes in $H^*(M;mathbbQ)$. More generally you can consider its Betti numbers.
– Tyrone
Jul 20 at 9:47
If $M$ is oriented then there are at least two non-trivial classes in $H^*(M;mathbbQ)$. More generally you can consider its Betti numbers.
– Tyrone
Jul 20 at 9:47
3
3
@Tyrone I think the question is how to force $M$ to not have polynomial cohomology ring.
– Mike Miller
Jul 20 at 14:01
@Tyrone I think the question is how to force $M$ to not have polynomial cohomology ring.
– Mike Miller
Jul 20 at 14:01
What type of conditions are you looking for? Curvature or what else?
– Thomas Rot
Jul 20 at 14:47
What type of conditions are you looking for? Curvature or what else?
– Thomas Rot
Jul 20 at 14:47
@Thomas Rot Yes, curvature in particular.
– Multivariablecalculus
Jul 20 at 16:24
@Thomas Rot Yes, curvature in particular.
– Multivariablecalculus
Jul 20 at 16:24
@Mike Miller Yes, I would like to force $M$ not to have a truncated polynomial cohomology ring.
– Multivariablecalculus
Jul 20 at 16:25
@Mike Miller Yes, I would like to force $M$ not to have a truncated polynomial cohomology ring.
– Multivariablecalculus
Jul 20 at 16:25
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
I can't imagine any kind of curvature condition which would work. I will use the phrase "standard examples" to refer to a space diffeomorphic to a sphere or projective space over $mathbbC,mathbbH$ or $mathbbO$. This are certainly the most well known examples of manifolds having the kinds of cohomology rings you are trying to avoid.
For example, one could demand negative/zero/positive sectional curvature. But the first two cases can't arise on a closed simply connected manifold, by Cartan-Hadamard Theorem. This leaves the case of positive sectional curvature - but then all of the standard examples admit such a metric.
So, perhaps one could demand negative/zero/positive Ricci curvature. Again, in the positive case, all the standard examples admit such a metric. Further, Lohkamp has shown that every manifold of dimension at least $3$ admits a metric of negative Ricci curvature. And it seems to be open whether or not $S^n$ admits a Ricci flat metric (see this MO question). I do not know about Ricci flat metrics on any other space with singly generated rational cohomology.
Finally, in the realm of scalar curvature, all the standard examples admit metrics on positive scalar curvature. By Kazdan-Warner, they each admit scalar flat and scalar negative metrics.
To further complicate matters, there are known examples of exotic spheres which do not admit metrics of positive scalar curvature (so don't admit metrics of positive sectional or Ricci curvature). So it doesn't seem like any of the usual curvature conditions can limit the rational cohomology ring like you would like.
answered Jul 23 at 19:23
Jason DeVito
29.5k473129
Thank you. My suggestion of a condition on sectional or Ricci curvature was only speculative and tentative. It does not have to necessarily be a condition on curvature.
– Multivariablecalculus
Jul 23 at 19:43
1
Sure, and there are also notions of curvature I didn't cover (Almost non-negative, bi-orthogonal sectional curvature, etc), so perhaps one of those does what you want.
– Jason DeVito
Jul 23 at 21:03
I am by no means an expert, so I am not too familiar with such curvatures. Does reducible holonomy say anything about one of these curvatures? @JasonDeVito
– Multivariablecalculus
Jul 23 at 21:18
1
I know almost nothing about those other kinds of curvatures, sorry!
– Jason DeVito
Jul 24 at 3:21
add a comment |Â
up vote
2
down vote
The following is Ian Agol's answer on my cross-post on MO: If $M$ is simply-connected and has reducible holonomy, then a theorem of de Rham implies that $M$ is a product, and hence does not have homology generated by one element. Moreover, coupling Berger's classification with de Rham's decomposition theorem, one obtains a classification of reducible holonomy groups by requiring that each factor is one of the examples coming from the list.
answered Jul 23 at 20:34
Multivariablecalculus
495313
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
I can't imagine any kind of curvature condition which would work. I will use the phrase "standard examples" to refer to a space diffeomorphic to a sphere or projective space over $mathbbC,mathbbH$ or $mathbbO$. This are certainly the most well known examples of manifolds having the kinds of cohomology rings you are trying to avoid.
For example, one could demand negative/zero/positive sectional curvature. But the first two cases can't arise on a closed simply connected manifold, by Cartan-Hadamard Theorem. This leaves the case of positive sectional curvature - but then all of the standard examples admit such a metric.
So, perhaps one could demand negative/zero/positive Ricci curvature. Again, in the positive case, all the standard examples admit such a metric. Further, Lohkamp has shown that every manifold of dimension at least $3$ admits a metric of negative Ricci curvature. And it seems to be open whether or not $S^n$ admits a Ricci flat metric (see this MO question). I do not know about Ricci flat metrics on any other space with singly generated rational cohomology.
Finally, in the realm of scalar curvature, all the standard examples admit metrics on positive scalar curvature. By Kazdan-Warner, they each admit scalar flat and scalar negative metrics.
To further complicate matters, there are known examples of exotic spheres which do not admit metrics of positive scalar curvature (so don't admit metrics of positive sectional or Ricci curvature). So it doesn't seem like any of the usual curvature conditions can limit the rational cohomology ring like you would like.
answered Jul 23 at 19:23
Jason DeVito
29.5k473129
Thank you. My suggestion of a condition on sectional or Ricci curvature was only speculative and tentative. It does not have to necessarily be a condition on curvature.
– Multivariablecalculus
Jul 23 at 19:43
1
Sure, and there are also notions of curvature I didn't cover (Almost non-negative, bi-orthogonal sectional curvature, etc), so perhaps one of those does what you want.
– Jason DeVito
Jul 23 at 21:03
I am by no means an expert, so I am not too familiar with such curvatures. Does reducible holonomy say anything about one of these curvatures? @JasonDeVito
– Multivariablecalculus
Jul 23 at 21:18
1
I know almost nothing about those other kinds of curvatures, sorry!
– Jason DeVito
Jul 24 at 3:21
add a comment |Â
up vote
4
down vote
accepted
I can't imagine any kind of curvature condition which would work. I will use the phrase "standard examples" to refer to a space diffeomorphic to a sphere or projective space over $mathbbC,mathbbH$ or $mathbbO$. This are certainly the most well known examples of manifolds having the kinds of cohomology rings you are trying to avoid.
For example, one could demand negative/zero/positive sectional curvature. But the first two cases can't arise on a closed simply connected manifold, by Cartan-Hadamard Theorem. This leaves the case of positive sectional curvature - but then all of the standard examples admit such a metric.
So, perhaps one could demand negative/zero/positive Ricci curvature. Again, in the positive case, all the standard examples admit such a metric. Further, Lohkamp has shown that every manifold of dimension at least $3$ admits a metric of negative Ricci curvature. And it seems to be open whether or not $S^n$ admits a Ricci flat metric (see this MO question). I do not know about Ricci flat metrics on any other space with singly generated rational cohomology.
Finally, in the realm of scalar curvature, all the standard examples admit metrics on positive scalar curvature. By Kazdan-Warner, they each admit scalar flat and scalar negative metrics.
To further complicate matters, there are known examples of exotic spheres which do not admit metrics of positive scalar curvature (so don't admit metrics of positive sectional or Ricci curvature). So it doesn't seem like any of the usual curvature conditions can limit the rational cohomology ring like you would like.
answered Jul 23 at 19:23
Jason DeVito
29.5k473129
Thank you. My suggestion of a condition on sectional or Ricci curvature was only speculative and tentative. It does not have to necessarily be a condition on curvature.
– Multivariablecalculus
Jul 23 at 19:43
1
Sure, and there are also notions of curvature I didn't cover (Almost non-negative, bi-orthogonal sectional curvature, etc), so perhaps one of those does what you want.
– Jason DeVito
Jul 23 at 21:03
I am by no means an expert, so I am not too familiar with such curvatures. Does reducible holonomy say anything about one of these curvatures? @JasonDeVito
– Multivariablecalculus
Jul 23 at 21:18
1
I know almost nothing about those other kinds of curvatures, sorry!
– Jason DeVito
Jul 24 at 3:21
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
I can't imagine any kind of curvature condition which would work. I will use the phrase "standard examples" to refer to a space diffeomorphic to a sphere or projective space over $mathbbC,mathbbH$ or $mathbbO$. This are certainly the most well known examples of manifolds having the kinds of cohomology rings you are trying to avoid.
For example, one could demand negative/zero/positive sectional curvature. But the first two cases can't arise on a closed simply connected manifold, by Cartan-Hadamard Theorem. This leaves the case of positive sectional curvature - but then all of the standard examples admit such a metric.
So, perhaps one could demand negative/zero/positive Ricci curvature. Again, in the positive case, all the standard examples admit such a metric. Further, Lohkamp has shown that every manifold of dimension at least $3$ admits a metric of negative Ricci curvature. And it seems to be open whether or not $S^n$ admits a Ricci flat metric (see this MO question). I do not know about Ricci flat metrics on any other space with singly generated rational cohomology.
Finally, in the realm of scalar curvature, all the standard examples admit metrics on positive scalar curvature. By Kazdan-Warner, they each admit scalar flat and scalar negative metrics.
To further complicate matters, there are known examples of exotic spheres which do not admit metrics of positive scalar curvature (so don't admit metrics of positive sectional or Ricci curvature). So it doesn't seem like any of the usual curvature conditions can limit the rational cohomology ring like you would like.
answered Jul 23 at 19:23
Jason DeVito
29.5k473129
I can't imagine any kind of curvature condition which would work. I will use the phrase "standard examples" to refer to a space diffeomorphic to a sphere or projective space over $mathbbC,mathbbH$ or $mathbbO$. This are certainly the most well known examples of manifolds having the kinds of cohomology rings you are trying to avoid.
For example, one could demand negative/zero/positive sectional curvature. But the first two cases can't arise on a closed simply connected manifold, by Cartan-Hadamard Theorem. This leaves the case of positive sectional curvature - but then all of the standard examples admit such a metric.
So, perhaps one could demand negative/zero/positive Ricci curvature. Again, in the positive case, all the standard examples admit such a metric. Further, Lohkamp has shown that every manifold of dimension at least $3$ admits a metric of negative Ricci curvature. And it seems to be open whether or not $S^n$ admits a Ricci flat metric (see this MO question). I do not know about Ricci flat metrics on any other space with singly generated rational cohomology.
Finally, in the realm of scalar curvature, all the standard examples admit metrics on positive scalar curvature. By Kazdan-Warner, they each admit scalar flat and scalar negative metrics.
To further complicate matters, there are known examples of exotic spheres which do not admit metrics of positive scalar curvature (so don't admit metrics of positive sectional or Ricci curvature). So it doesn't seem like any of the usual curvature conditions can limit the rational cohomology ring like you would like.
answered Jul 23 at 19:23
Jason DeVito
29.5k473129
answered Jul 23 at 19:23
Jason DeVito
29.5k473129
answered Jul 23 at 19:23
Jason DeVito
29.5k473129
answered Jul 23 at 19:23
Jason DeVito
29.5k473129
29.5k473129
Thank you. My suggestion of a condition on sectional or Ricci curvature was only speculative and tentative. It does not have to necessarily be a condition on curvature.
– Multivariablecalculus
Jul 23 at 19:43
1
Sure, and there are also notions of curvature I didn't cover (Almost non-negative, bi-orthogonal sectional curvature, etc), so perhaps one of those does what you want.
– Jason DeVito
Jul 23 at 21:03
I am by no means an expert, so I am not too familiar with such curvatures. Does reducible holonomy say anything about one of these curvatures? @JasonDeVito
– Multivariablecalculus
Jul 23 at 21:18
1
I know almost nothing about those other kinds of curvatures, sorry!
– Jason DeVito
Jul 24 at 3:21
add a comment |Â
Thank you. My suggestion of a condition on sectional or Ricci curvature was only speculative and tentative. It does not have to necessarily be a condition on curvature.
– Multivariablecalculus
Jul 23 at 19:43
1
Sure, and there are also notions of curvature I didn't cover (Almost non-negative, bi-orthogonal sectional curvature, etc), so perhaps one of those does what you want.
– Jason DeVito
Jul 23 at 21:03
I am by no means an expert, so I am not too familiar with such curvatures. Does reducible holonomy say anything about one of these curvatures? @JasonDeVito
– Multivariablecalculus
Jul 23 at 21:18
1
I know almost nothing about those other kinds of curvatures, sorry!
– Jason DeVito
Jul 24 at 3:21
Thank you. My suggestion of a condition on sectional or Ricci curvature was only speculative and tentative. It does not have to necessarily be a condition on curvature.
– Multivariablecalculus
Jul 23 at 19:43
Thank you. My suggestion of a condition on sectional or Ricci curvature was only speculative and tentative. It does not have to necessarily be a condition on curvature.
– Multivariablecalculus
Jul 23 at 19:43
1
1
Sure, and there are also notions of curvature I didn't cover (Almost non-negative, bi-orthogonal sectional curvature, etc), so perhaps one of those does what you want.
– Jason DeVito
Jul 23 at 21:03
Sure, and there are also notions of curvature I didn't cover (Almost non-negative, bi-orthogonal sectional curvature, etc), so perhaps one of those does what you want.
– Jason DeVito
Jul 23 at 21:03
I am by no means an expert, so I am not too familiar with such curvatures. Does reducible holonomy say anything about one of these curvatures? @JasonDeVito
– Multivariablecalculus
Jul 23 at 21:18
I am by no means an expert, so I am not too familiar with such curvatures. Does reducible holonomy say anything about one of these curvatures? @JasonDeVito
– Multivariablecalculus
Jul 23 at 21:18
1
1
I know almost nothing about those other kinds of curvatures, sorry!
– Jason DeVito
Jul 24 at 3:21
I know almost nothing about those other kinds of curvatures, sorry!
– Jason DeVito
Jul 24 at 3:21
add a comment |Â
up vote
2
down vote
The following is Ian Agol's answer on my cross-post on MO: If $M$ is simply-connected and has reducible holonomy, then a theorem of de Rham implies that $M$ is a product, and hence does not have homology generated by one element. Moreover, coupling Berger's classification with de Rham's decomposition theorem, one obtains a classification of reducible holonomy groups by requiring that each factor is one of the examples coming from the list.
answered Jul 23 at 20:34
Multivariablecalculus
495313
add a comment |Â
up vote
2
down vote
The following is Ian Agol's answer on my cross-post on MO: If $M$ is simply-connected and has reducible holonomy, then a theorem of de Rham implies that $M$ is a product, and hence does not have homology generated by one element. Moreover, coupling Berger's classification with de Rham's decomposition theorem, one obtains a classification of reducible holonomy groups by requiring that each factor is one of the examples coming from the list.
answered Jul 23 at 20:34
Multivariablecalculus
495313
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The following is Ian Agol's answer on my cross-post on MO: If $M$ is simply-connected and has reducible holonomy, then a theorem of de Rham implies that $M$ is a product, and hence does not have homology generated by one element. Moreover, coupling Berger's classification with de Rham's decomposition theorem, one obtains a classification of reducible holonomy groups by requiring that each factor is one of the examples coming from the list.
answered Jul 23 at 20:34
Multivariablecalculus
495313
The following is Ian Agol's answer on my cross-post on MO: If $M$ is simply-connected and has reducible holonomy, then a theorem of de Rham implies that $M$ is a product, and hence does not have homology generated by one element. Moreover, coupling Berger's classification with de Rham's decomposition theorem, one obtains a classification of reducible holonomy groups by requiring that each factor is one of the examples coming from the list.
answered Jul 23 at 20:34
Multivariablecalculus
495313
answered Jul 23 at 20:34
Multivariablecalculus
495313
answered Jul 23 at 20:34
Multivariablecalculus
495313
answered Jul 23 at 20:34
Multivariablecalculus
495313
495313
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2857279%2fshow-that-the-rational-cohomology-ring-hm-mathbbq-needs-at-least-two-ge%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
If $M$ is oriented then there are at least two non-trivial classes in $H^*(M;mathbbQ)$. More generally you can consider its Betti numbers.
– Tyrone
Jul 20 at 9:47
3
@Tyrone I think the question is how to force $M$ to not have polynomial cohomology ring.
– Mike Miller
Jul 20 at 14:01
What type of conditions are you looking for? Curvature or what else?
– Thomas Rot
Jul 20 at 14:47
@Thomas Rot Yes, curvature in particular.
– Multivariablecalculus
Jul 20 at 16:24
@Mike Miller Yes, I would like to force $M$ not to have a truncated polynomial cohomology ring.
– Multivariablecalculus
Jul 20 at 16:25