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Let $M$ be a Riemannian manifold and $G$ a closed connected subgroup of isometries of $M$. Call the pair $(M,G)$ an isotropic pair if $G$ acts transitively on the sphere bundle $SM$. As an example, the pair $(S^6,SO(7))$ is isotropic, but also $(S^6,G_2)$.

I am looking for a reference for the classification of all isotropic pairs. In several places in the literature, the possible manifolds $M$ are given (in the non-flat case these are symmetric spaces of rank one), but not the corresponding groups $G$.The best reference seems to be J. Tits paper Espaces homogènes et isotropes, et espaces doublement homogènes from 1954, but it is not so easy to understand his terminology.

Questions:

  1. Is the following list complete?
  2. There seems to be no negatively curved versions of the pairs $(S^6,G_2)$ and $(S^7,Spin(7))$. What is the reason for this?

Flat case:

  • $\mathbb{R}^n$ with the group generated by translations and $SO(n), U(n/2), SU(n/2), Sp(n/4), Sp(n/4) U(1), Sp(n/4) Sp(1)$
  • $\mathbb{R}^7$ with the group generated by translations and $G_2$
  • $\mathbb{R}^8$ with the group generated by translations and $Spin(7)$
  • $\mathbb{R}^{16}$ with the group generated by translations and $Spin(9)$

Symmetric space case (each with $G$ equal to the connected component of the trivial element of the isometry group):

  • $S^n$ sphere
  • $\mathbb{RP}^n$ real projective space
  • $\mathbb{H}^n$ real hyperbolic space
  • $\mathbb{CP}^n$ complex projective space
  • $\mathbb{CH}^n$ complex hyperbolic space
  • $\mathbb{HP}^n$ quaternionic projective space
  • $\mathbb{HH}^n$ quaternionic hyperbolic space
  • $\mathbb{OP}^2$ octonionic projective plane
  • $\mathbb{OH}^2$ octonionic hyperbolic plane

Other cases:

  • $(\mathbb{RP}^6,G_2)$ and $(S^6,G_2)$
  • $(\mathbb{RP}^7,Spin(7))$ and $(S^7,Spin(7))$
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  • $\begingroup$ I possibly misunderstood your definition: why $R^n$ with $SO(n)$ is isotropic? How does $SO(n)$ act on $R^n$? $\endgroup$ Sep 19, 2014 at 9:24
  • $\begingroup$ @VladimirSMatveev: The OP is using a different notation in his listing for the flat cases, listing only the isotropy subgroup, not the whole group $G$. He should explain this. $\endgroup$ Sep 19, 2014 at 11:16
  • $\begingroup$ @VladimirSMatveev: Thank you Vladimir, I edited the question and put the correct group in this case. $\endgroup$ Sep 19, 2014 at 12:33

1 Answer 1

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The reason there are no 'negatively curved' analogs of $(S^6,\mathrm{G}_2)$ or $\bigl(S^7,\mathrm{Spin}(7)\bigr)$ is that, in each of these cases of homogeneous Riemannian manifolds $G/H$, the corresponding $H$-structure ($H=\mathrm{SU}(3)$ in the first case, $H=\mathrm{G}_2$ in the second) is not torsion-free.

In fact, there is, up to scale, only one possible torsion tensor in each of these cases that is $H$-invariant, and the Cartan structure equations imply that the $H$-invariant part of the Riemann curvature tensor must be a quadratic expression in the torsion tensor. Since $x^2 = (-x)^2 >0$, changing the sign of the torsion tensor won't change the sign of the $H$-invariant part of the curvature tensor. That's why you can't just reverse it and get compatible structure equations (which is what works and gives rise to duality in the case of symmetric spaces).

Your list does look complete to me. I think you could prove it relatively easily if you are willing to accept the known list of Lie groups acting transitively on spheres.

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