Is there a class of non-symmetric spaces $G/SO(n)$ such that $Hol(G/SO(n))=SO(n)$?

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Is there a certain class (or unique construction of a set) of simply connected non-symmetric spaces $G/SO(n)$ such that the Riemannian holonomy is $Hol(G/SO(n))=SO(n)$? (That is, is there a way to write $G$ as a unique Lie group or a class of groups to guarantee this, and must there be infinitely many such non-symmetric spaces?)




Can we enumerate all such possible candidates or classify them (or is this even possible)? If this is plausible, it would be interesting to see how it extends to other quotients like $G/SU(n)$, $G/Sp(n)$, etc. I would like to enumerate all such spaces, but there is still the question of whether there are infinitely-many or finitely-many such (unique) spaces. Since $G/SO(n)$ is simply connected by assumption, the holonomy and isotropy group are equal so $Hol(G/SO(n))=SO(n)=gin G:gx=x$ for some $xin G/SO(n)$.



Cross-posted on MO.



Thanks in advance!







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    Is there a certain class (or unique construction of a set) of simply connected non-symmetric spaces $G/SO(n)$ such that the Riemannian holonomy is $Hol(G/SO(n))=SO(n)$? (That is, is there a way to write $G$ as a unique Lie group or a class of groups to guarantee this, and must there be infinitely many such non-symmetric spaces?)




    Can we enumerate all such possible candidates or classify them (or is this even possible)? If this is plausible, it would be interesting to see how it extends to other quotients like $G/SU(n)$, $G/Sp(n)$, etc. I would like to enumerate all such spaces, but there is still the question of whether there are infinitely-many or finitely-many such (unique) spaces. Since $G/SO(n)$ is simply connected by assumption, the holonomy and isotropy group are equal so $Hol(G/SO(n))=SO(n)=gin G:gx=x$ for some $xin G/SO(n)$.



    Cross-posted on MO.



    Thanks in advance!







    share|cite|improve this question























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite












      Is there a certain class (or unique construction of a set) of simply connected non-symmetric spaces $G/SO(n)$ such that the Riemannian holonomy is $Hol(G/SO(n))=SO(n)$? (That is, is there a way to write $G$ as a unique Lie group or a class of groups to guarantee this, and must there be infinitely many such non-symmetric spaces?)




      Can we enumerate all such possible candidates or classify them (or is this even possible)? If this is plausible, it would be interesting to see how it extends to other quotients like $G/SU(n)$, $G/Sp(n)$, etc. I would like to enumerate all such spaces, but there is still the question of whether there are infinitely-many or finitely-many such (unique) spaces. Since $G/SO(n)$ is simply connected by assumption, the holonomy and isotropy group are equal so $Hol(G/SO(n))=SO(n)=gin G:gx=x$ for some $xin G/SO(n)$.



      Cross-posted on MO.



      Thanks in advance!







      share|cite|improve this question














      Is there a certain class (or unique construction of a set) of simply connected non-symmetric spaces $G/SO(n)$ such that the Riemannian holonomy is $Hol(G/SO(n))=SO(n)$? (That is, is there a way to write $G$ as a unique Lie group or a class of groups to guarantee this, and must there be infinitely many such non-symmetric spaces?)




      Can we enumerate all such possible candidates or classify them (or is this even possible)? If this is plausible, it would be interesting to see how it extends to other quotients like $G/SU(n)$, $G/Sp(n)$, etc. I would like to enumerate all such spaces, but there is still the question of whether there are infinitely-many or finitely-many such (unique) spaces. Since $G/SO(n)$ is simply connected by assumption, the holonomy and isotropy group are equal so $Hol(G/SO(n))=SO(n)=gin G:gx=x$ for some $xin G/SO(n)$.



      Cross-posted on MO.



      Thanks in advance!









      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Jul 29 at 18:17
























      asked Jul 28 at 18:12









      Multivariablecalculus

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