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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 $Hol(G/SO(n))=SO(n)=\{g\in G:gx=x\}$ for some $x\in G/SO(n)$.

Cross-posted on MSE.

Sorry for all of the questions packed in; thanks!

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    $\begingroup$ I suppose that $Hol$ means the holonomy group of some connection on the tangent bundle, but which connection do you mean? (I didn't down vote, but to me the ambiguity in the choice of connection is the only aspect of the question which is unclear.) $\endgroup$
    – Ben McKay
    Jul 29, 2018 at 10:31
  • $\begingroup$ @BenMcKay Sorry, yes $Hol$ is the holonomy group of the Levi-Civita connection $\nabla^{\text{L.C.}}$ (i.e. Riemannian holonomy). $\endgroup$ Jul 29, 2018 at 14:48
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    $\begingroup$ No, no and no. $\operatorname{Hol}(G/H)=H $ does not hold for all homogeneous spaces, but for symmetric homogeneous spaces, and those come with a canonical metric. $\endgroup$
    – abx
    Jul 29, 2018 at 17:22
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    $\begingroup$ @abx Sorry, it was a genuine mistake. Are there non-symmetric spaces $G/H$ such that $Hol(G/H)=H$ (which are also not necessarily homogeneous)? $\endgroup$ Jul 29, 2018 at 17:32
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    $\begingroup$ No. Please look at Berger's classification of holonomy group. $\endgroup$
    – abx
    Jul 29, 2018 at 18:42

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