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$(M,g)$ is homogeneous if $\forall a,b\in M$ there is an isometry $f:M\to M$ such that $f(a)=b$

$(M, g)$ is called isotropic if $\forall p\in M$ $x,y\in T_pM$ of unit length, there is an isometry $f:M\to M$ s.t. $f_\ast(x)=y$

I know that Every connected isotropic manifold is homogeneous (given 2 points, take the middle of the geodesics connecting them)

Now, are there homogenous manifolds that are not isotropic? Please, explain in detail if so.

Jack
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1 Answers1

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A cylinder $S^1\times \mathbb{R}$ will do the job ! In general, isometries of a product Riemannian manifolds are not the "product of isometries of the factors", but in this simple case I hope we can agree it's true: an isometry of the cylinder is of the type $$(x,y)\mapsto (r(x),\pm y+a),$$ where $r\in O(2)$ and $a\in \mathbb{R}$. If you want a rigorous proof of this fact, check this answer. Clearly the isometries of the cylinder map tangent vectors like $(v,0)$ in tangent vectors of the same kind (the last coordinate stays equal to $0$), so the cylinder is not isotropic, but it's homogeneous.

I want to add an interesting curiosity. Isotropic connected riemannian manifolds are completely classified up to isometry and are:

  • Euclidean spaces $\mathbb{E}^n$,
  • Spheres $\mathbb{S}^n$,
  • Hyperbolic (real) spaces $\mathbb{H}^n$,
  • Hyperbolic complex spaces $\mathbb{H}^n\mathbb{C}$,
  • Hyperbolic quaternionic spaces $\mathbb{H}^n\mathbb{H}$,
  • Hyperbolic octonionic plane $\mathbb{H}^2\mathbb{O}$,
  • Projective real spaces $\mathbb{P}^n\mathbb{R}$,
  • Projective complex spaces $\mathbb{P}^n\mathbb{C}$,
  • Projective quaternionic spaces $\mathbb{P}^n\mathbb{H}$,
  • Projective octonionic plane $\mathbb{P}^2\mathbb{O}$.

While homogeneous Riemannian manifolds are much more! They are precisely the manifolds $G/H$ where $G$ is a Lie group and $H\leq G$ is a compact subgroup (basically a complete classification is near impossible).

Kandinskij
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