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Does $Graff(n,mathbbR^infty)$ generate all $n$-dimensional closed Riemannian manifolds $M$?
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How does one generate all possible $n$-dimensional simply connected closed Riemannian manifolds $M$ from the affine Grassmannian $Graff(n,V)$? Would $Graff(n,mathbbR^infty)$ suffice? (It seems like we need more for it to be closed and simply connected.)
Thanks in advance!
riemannian-geometry grassmannian
asked Jul 30 at 2:20
Multivariablecalculus
490313
add a comment |Â
up vote
1
down vote
favorite
How does one generate all possible $n$-dimensional simply connected closed Riemannian manifolds $M$ from the affine Grassmannian $Graff(n,V)$? Would $Graff(n,mathbbR^infty)$ suffice? (It seems like we need more for it to be closed and simply connected.)
Thanks in advance!
riemannian-geometry grassmannian
asked Jul 30 at 2:20
Multivariablecalculus
490313
1
What do you mean by generate in this context?
– Lorenzo
Jul 30 at 4:54
As in the affine Grassmannian parametrically exhausts (i.e. it is identically equal to) all possible $n$-dimensional simply connected closed Riemannian manifolds. I think it should be smaller than $Graff(n,mathbbR^n+1)$ if we just need the conditions of closure and simple-connectedness to be met, however.
– Multivariablecalculus
Jul 30 at 5:05
Are you asking about whether every simply connected closed manifold must be homeomorphic to a $Graff(n, mathbbR^n+1)$ for some $n$? (I'm not sure why you are giving them Riemannian metrics...)
– Lorenzo
Jul 30 at 5:09
Yes, this is my question (can they be identified). I suppose the choice of a Riemannian metric is arbitrary (but preferable) @Lorenzo
– Multivariablecalculus
Jul 30 at 5:13
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How does one generate all possible $n$-dimensional simply connected closed Riemannian manifolds $M$ from the affine Grassmannian $Graff(n,V)$? Would $Graff(n,mathbbR^infty)$ suffice? (It seems like we need more for it to be closed and simply connected.)
Thanks in advance!
riemannian-geometry grassmannian
asked Jul 30 at 2:20
Multivariablecalculus
490313
How does one generate all possible $n$-dimensional simply connected closed Riemannian manifolds $M$ from the affine Grassmannian $Graff(n,V)$? Would $Graff(n,mathbbR^infty)$ suffice? (It seems like we need more for it to be closed and simply connected.)
Thanks in advance!
riemannian-geometry grassmannian
asked Jul 30 at 2:20
Multivariablecalculus
490313
edited Jul 31 at 18:48
asked Jul 30 at 2:20
Multivariablecalculus
490313
asked Jul 30 at 2:20
Multivariablecalculus
490313
asked Jul 30 at 2:20
Multivariablecalculus
490313
490313
1
What do you mean by generate in this context?
– Lorenzo
Jul 30 at 4:54
As in the affine Grassmannian parametrically exhausts (i.e. it is identically equal to) all possible $n$-dimensional simply connected closed Riemannian manifolds. I think it should be smaller than $Graff(n,mathbbR^n+1)$ if we just need the conditions of closure and simple-connectedness to be met, however.
– Multivariablecalculus
Jul 30 at 5:05
Are you asking about whether every simply connected closed manifold must be homeomorphic to a $Graff(n, mathbbR^n+1)$ for some $n$? (I'm not sure why you are giving them Riemannian metrics...)
– Lorenzo
Jul 30 at 5:09
Yes, this is my question (can they be identified). I suppose the choice of a Riemannian metric is arbitrary (but preferable) @Lorenzo
– Multivariablecalculus
Jul 30 at 5:13
add a comment |Â
1
What do you mean by generate in this context?
– Lorenzo
Jul 30 at 4:54
As in the affine Grassmannian parametrically exhausts (i.e. it is identically equal to) all possible $n$-dimensional simply connected closed Riemannian manifolds. I think it should be smaller than $Graff(n,mathbbR^n+1)$ if we just need the conditions of closure and simple-connectedness to be met, however.
– Multivariablecalculus
Jul 30 at 5:05
Are you asking about whether every simply connected closed manifold must be homeomorphic to a $Graff(n, mathbbR^n+1)$ for some $n$? (I'm not sure why you are giving them Riemannian metrics...)
– Lorenzo
Jul 30 at 5:09
Yes, this is my question (can they be identified). I suppose the choice of a Riemannian metric is arbitrary (but preferable) @Lorenzo
– Multivariablecalculus
Jul 30 at 5:13
1
1
What do you mean by generate in this context?
– Lorenzo
Jul 30 at 4:54
What do you mean by generate in this context?
– Lorenzo
Jul 30 at 4:54
As in the affine Grassmannian parametrically exhausts (i.e. it is identically equal to) all possible $n$-dimensional simply connected closed Riemannian manifolds. I think it should be smaller than $Graff(n,mathbbR^n+1)$ if we just need the conditions of closure and simple-connectedness to be met, however.
– Multivariablecalculus
Jul 30 at 5:05
As in the affine Grassmannian parametrically exhausts (i.e. it is identically equal to) all possible $n$-dimensional simply connected closed Riemannian manifolds. I think it should be smaller than $Graff(n,mathbbR^n+1)$ if we just need the conditions of closure and simple-connectedness to be met, however.
– Multivariablecalculus
Jul 30 at 5:05
Are you asking about whether every simply connected closed manifold must be homeomorphic to a $Graff(n, mathbbR^n+1)$ for some $n$? (I'm not sure why you are giving them Riemannian metrics...)
– Lorenzo
Jul 30 at 5:09
Are you asking about whether every simply connected closed manifold must be homeomorphic to a $Graff(n, mathbbR^n+1)$ for some $n$? (I'm not sure why you are giving them Riemannian metrics...)
– Lorenzo
Jul 30 at 5:09
Yes, this is my question (can they be identified). I suppose the choice of a Riemannian metric is arbitrary (but preferable) @Lorenzo
– Multivariablecalculus
Jul 30 at 5:13
Yes, this is my question (can they be identified). I suppose the choice of a Riemannian metric is arbitrary (but preferable) @Lorenzo
– Multivariablecalculus
Jul 30 at 5:13
add a comment |Â
1 Answer
1
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up vote
2
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accepted
There are lots of simply connected, closed manifolds which are not Grassmannians of $k$ planes in some $mathbbR^n$.
One way to see this is as follows: the dimension of $G(k,n)$ is $k(n - k)$.
Solving for $k(n -k) = 8$, we get $(k,n) in (1,9), (2,6), (4,6), (8,9) $
However, here are five non-homeomorphic $8$-manifolds that are closed and simply connected:
1) $S^2 times S^2 times S^2 times S^2$
2) $S^8$.
3) $S^3 times S^3 times S^2$
4) $S^4 times S^2 times S^2$
5) $S^6 times S^2$.
Thus, at least one of these is not homeomorphic to a Grassmannian.
To check that these five products of spheres are not homeomorphic to each other, you can use the Poincare polynomial: https://topospaces.subwiki.org/wiki/Poincare_polynomial
answered Jul 30 at 5:35
Lorenzo
11.5k31537
+1 @Lorenzo Thanks! Is there a known expression involving affine Grassmannians that exhausts all simply connected closed manifolds? Going back to my original question, how does one generate all possible $n$-dimensional simply connected closed Riemannian manifolds $M$ from the affine Grassmannian $Graff(n,V)$?
– Multivariablecalculus
Jul 30 at 5:59
1
@Multivariablecalculus I doubt that there's a simple formula... you may want to look at this: en.wikipedia.org/wiki/Classification_of_manifolds
– Lorenzo
Jul 30 at 6:00
1
I should add that spheres (n > 1) are the universal covers of projective spaces , so the manifolds I gave you are products of covers of Grassmannians... (For that matter, I should point out that Grassmannians aren't simply connected in general...)
– Lorenzo
Jul 30 at 6:12
Thank you. Hmm, I wonder then if there is any topological object that parameterizes such manifolds that may be explicitly written down?
– Multivariablecalculus
Jul 30 at 6:14
1
I don't think so ... Classifying manifolds is a really hard problem.
– Lorenzo
Jul 30 at 15:38
 |Â
show 2 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 are lots of simply connected, closed manifolds which are not Grassmannians of $k$ planes in some $mathbbR^n$.
One way to see this is as follows: the dimension of $G(k,n)$ is $k(n - k)$.
Solving for $k(n -k) = 8$, we get $(k,n) in (1,9), (2,6), (4,6), (8,9) $
However, here are five non-homeomorphic $8$-manifolds that are closed and simply connected:
1) $S^2 times S^2 times S^2 times S^2$
2) $S^8$.
3) $S^3 times S^3 times S^2$
4) $S^4 times S^2 times S^2$
5) $S^6 times S^2$.
Thus, at least one of these is not homeomorphic to a Grassmannian.
To check that these five products of spheres are not homeomorphic to each other, you can use the Poincare polynomial: https://topospaces.subwiki.org/wiki/Poincare_polynomial
answered Jul 30 at 5:35
Lorenzo
11.5k31537
+1 @Lorenzo Thanks! Is there a known expression involving affine Grassmannians that exhausts all simply connected closed manifolds? Going back to my original question, how does one generate all possible $n$-dimensional simply connected closed Riemannian manifolds $M$ from the affine Grassmannian $Graff(n,V)$?
– Multivariablecalculus
Jul 30 at 5:59
1
@Multivariablecalculus I doubt that there's a simple formula... you may want to look at this: en.wikipedia.org/wiki/Classification_of_manifolds
– Lorenzo
Jul 30 at 6:00
1
I should add that spheres (n > 1) are the universal covers of projective spaces , so the manifolds I gave you are products of covers of Grassmannians... (For that matter, I should point out that Grassmannians aren't simply connected in general...)
– Lorenzo
Jul 30 at 6:12
Thank you. Hmm, I wonder then if there is any topological object that parameterizes such manifolds that may be explicitly written down?
– Multivariablecalculus
Jul 30 at 6:14
1
I don't think so ... Classifying manifolds is a really hard problem.
– Lorenzo
Jul 30 at 15:38
 |Â
show 2 more comments
up vote
2
down vote
accepted
There are lots of simply connected, closed manifolds which are not Grassmannians of $k$ planes in some $mathbbR^n$.
One way to see this is as follows: the dimension of $G(k,n)$ is $k(n - k)$.
Solving for $k(n -k) = 8$, we get $(k,n) in (1,9), (2,6), (4,6), (8,9) $
However, here are five non-homeomorphic $8$-manifolds that are closed and simply connected:
1) $S^2 times S^2 times S^2 times S^2$
2) $S^8$.
3) $S^3 times S^3 times S^2$
4) $S^4 times S^2 times S^2$
5) $S^6 times S^2$.
Thus, at least one of these is not homeomorphic to a Grassmannian.
To check that these five products of spheres are not homeomorphic to each other, you can use the Poincare polynomial: https://topospaces.subwiki.org/wiki/Poincare_polynomial
answered Jul 30 at 5:35
Lorenzo
11.5k31537
+1 @Lorenzo Thanks! Is there a known expression involving affine Grassmannians that exhausts all simply connected closed manifolds? Going back to my original question, how does one generate all possible $n$-dimensional simply connected closed Riemannian manifolds $M$ from the affine Grassmannian $Graff(n,V)$?
– Multivariablecalculus
Jul 30 at 5:59
1
@Multivariablecalculus I doubt that there's a simple formula... you may want to look at this: en.wikipedia.org/wiki/Classification_of_manifolds
– Lorenzo
Jul 30 at 6:00
1
I should add that spheres (n > 1) are the universal covers of projective spaces , so the manifolds I gave you are products of covers of Grassmannians... (For that matter, I should point out that Grassmannians aren't simply connected in general...)
– Lorenzo
Jul 30 at 6:12
Thank you. Hmm, I wonder then if there is any topological object that parameterizes such manifolds that may be explicitly written down?
– Multivariablecalculus
Jul 30 at 6:14
1
I don't think so ... Classifying manifolds is a really hard problem.
– Lorenzo
Jul 30 at 15:38
 |Â
show 2 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
There are lots of simply connected, closed manifolds which are not Grassmannians of $k$ planes in some $mathbbR^n$.
One way to see this is as follows: the dimension of $G(k,n)$ is $k(n - k)$.
Solving for $k(n -k) = 8$, we get $(k,n) in (1,9), (2,6), (4,6), (8,9) $
However, here are five non-homeomorphic $8$-manifolds that are closed and simply connected:
1) $S^2 times S^2 times S^2 times S^2$
2) $S^8$.
3) $S^3 times S^3 times S^2$
4) $S^4 times S^2 times S^2$
5) $S^6 times S^2$.
Thus, at least one of these is not homeomorphic to a Grassmannian.
To check that these five products of spheres are not homeomorphic to each other, you can use the Poincare polynomial: https://topospaces.subwiki.org/wiki/Poincare_polynomial
answered Jul 30 at 5:35
Lorenzo
11.5k31537
There are lots of simply connected, closed manifolds which are not Grassmannians of $k$ planes in some $mathbbR^n$.
One way to see this is as follows: the dimension of $G(k,n)$ is $k(n - k)$.
Solving for $k(n -k) = 8$, we get $(k,n) in (1,9), (2,6), (4,6), (8,9) $
However, here are five non-homeomorphic $8$-manifolds that are closed and simply connected:
1) $S^2 times S^2 times S^2 times S^2$
2) $S^8$.
3) $S^3 times S^3 times S^2$
4) $S^4 times S^2 times S^2$
5) $S^6 times S^2$.
Thus, at least one of these is not homeomorphic to a Grassmannian.
To check that these five products of spheres are not homeomorphic to each other, you can use the Poincare polynomial: https://topospaces.subwiki.org/wiki/Poincare_polynomial
answered Jul 30 at 5:35
Lorenzo
11.5k31537
answered Jul 30 at 5:35
Lorenzo
11.5k31537
answered Jul 30 at 5:35
Lorenzo
11.5k31537
answered Jul 30 at 5:35
Lorenzo
11.5k31537
11.5k31537
+1 @Lorenzo Thanks! Is there a known expression involving affine Grassmannians that exhausts all simply connected closed manifolds? Going back to my original question, how does one generate all possible $n$-dimensional simply connected closed Riemannian manifolds $M$ from the affine Grassmannian $Graff(n,V)$?
– Multivariablecalculus
Jul 30 at 5:59
1
@Multivariablecalculus I doubt that there's a simple formula... you may want to look at this: en.wikipedia.org/wiki/Classification_of_manifolds
– Lorenzo
Jul 30 at 6:00
1
I should add that spheres (n > 1) are the universal covers of projective spaces , so the manifolds I gave you are products of covers of Grassmannians... (For that matter, I should point out that Grassmannians aren't simply connected in general...)
– Lorenzo
Jul 30 at 6:12
Thank you. Hmm, I wonder then if there is any topological object that parameterizes such manifolds that may be explicitly written down?
– Multivariablecalculus
Jul 30 at 6:14
1
I don't think so ... Classifying manifolds is a really hard problem.
– Lorenzo
Jul 30 at 15:38
 |Â
show 2 more comments
+1 @Lorenzo Thanks! Is there a known expression involving affine Grassmannians that exhausts all simply connected closed manifolds? Going back to my original question, how does one generate all possible $n$-dimensional simply connected closed Riemannian manifolds $M$ from the affine Grassmannian $Graff(n,V)$?
– Multivariablecalculus
Jul 30 at 5:59
1
@Multivariablecalculus I doubt that there's a simple formula... you may want to look at this: en.wikipedia.org/wiki/Classification_of_manifolds
– Lorenzo
Jul 30 at 6:00
1
I should add that spheres (n > 1) are the universal covers of projective spaces , so the manifolds I gave you are products of covers of Grassmannians... (For that matter, I should point out that Grassmannians aren't simply connected in general...)
– Lorenzo
Jul 30 at 6:12
Thank you. Hmm, I wonder then if there is any topological object that parameterizes such manifolds that may be explicitly written down?
– Multivariablecalculus
Jul 30 at 6:14
1
I don't think so ... Classifying manifolds is a really hard problem.
– Lorenzo
Jul 30 at 15:38
+1 @Lorenzo Thanks! Is there a known expression involving affine Grassmannians that exhausts all simply connected closed manifolds? Going back to my original question, how does one generate all possible $n$-dimensional simply connected closed Riemannian manifolds $M$ from the affine Grassmannian $Graff(n,V)$?
– Multivariablecalculus
Jul 30 at 5:59
+1 @Lorenzo Thanks! Is there a known expression involving affine Grassmannians that exhausts all simply connected closed manifolds? Going back to my original question, how does one generate all possible $n$-dimensional simply connected closed Riemannian manifolds $M$ from the affine Grassmannian $Graff(n,V)$?
– Multivariablecalculus
Jul 30 at 5:59
1
1
@Multivariablecalculus I doubt that there's a simple formula... you may want to look at this: en.wikipedia.org/wiki/Classification_of_manifolds
– Lorenzo
Jul 30 at 6:00
@Multivariablecalculus I doubt that there's a simple formula... you may want to look at this: en.wikipedia.org/wiki/Classification_of_manifolds
– Lorenzo
Jul 30 at 6:00
1
1
I should add that spheres (n > 1) are the universal covers of projective spaces , so the manifolds I gave you are products of covers of Grassmannians... (For that matter, I should point out that Grassmannians aren't simply connected in general...)
– Lorenzo
Jul 30 at 6:12
I should add that spheres (n > 1) are the universal covers of projective spaces , so the manifolds I gave you are products of covers of Grassmannians... (For that matter, I should point out that Grassmannians aren't simply connected in general...)
– Lorenzo
Jul 30 at 6:12
Thank you. Hmm, I wonder then if there is any topological object that parameterizes such manifolds that may be explicitly written down?
– Multivariablecalculus
Jul 30 at 6:14
Thank you. Hmm, I wonder then if there is any topological object that parameterizes such manifolds that may be explicitly written down?
– Multivariablecalculus
Jul 30 at 6:14
1
1
I don't think so ... Classifying manifolds is a really hard problem.
– Lorenzo
Jul 30 at 15:38
I don't think so ... Classifying manifolds is a really hard problem.
– Lorenzo
Jul 30 at 15:38
 |Â
show 2 more comments
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1
What do you mean by generate in this context?
– Lorenzo
Jul 30 at 4:54
As in the affine Grassmannian parametrically exhausts (i.e. it is identically equal to) all possible $n$-dimensional simply connected closed Riemannian manifolds. I think it should be smaller than $Graff(n,mathbbR^n+1)$ if we just need the conditions of closure and simple-connectedness to be met, however.
– Multivariablecalculus
Jul 30 at 5:05
Are you asking about whether every simply connected closed manifold must be homeomorphic to a $Graff(n, mathbbR^n+1)$ for some $n$? (I'm not sure why you are giving them Riemannian metrics...)
– Lorenzo
Jul 30 at 5:09
Yes, this is my question (can they be identified). I suppose the choice of a Riemannian metric is arbitrary (but preferable) @Lorenzo
– Multivariablecalculus
Jul 30 at 5:13