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




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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














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




share|cite|improve this question

















  • 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












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




share|cite|improve this question













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





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 31 at 18:48
























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












  • 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










1 Answer
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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






share|cite|improve this answer





















  • +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











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1 Answer
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active

oldest

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active

oldest

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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






share|cite|improve this answer





















  • +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















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






share|cite|improve this answer





















  • +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













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






share|cite|improve this answer













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







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











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

















  • +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













 

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