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Recall that a Riemannian manifold $(M,g)$ is isotropic if for any $p\in M$ and any unit vectors $v,w\in T_pM$ there is an isometry $f:M\to M$ such that $f_\ast(v)=w.$ Recall also that $(M,g)$ is homogeneous if for any $p,q\in m$ there is an isometry $f:M\to M$ such that $f(p)=q.$

I would appreciate answers to any of the following questions:

  1. I know that the simply connected constant curvature spaces are isotropic. What are some other examples of isotropic manifolds?

  2. I suspect that the flat torus is not isotropic and that the product metric on $M\times N$ is not isotropic if $M$ and $N$ are not isometric. Are these true?

  3. Is there some way to relate transitivity of the holonomy group on the unit sphere in $T_pM$ to isotropy?

  4. Are isotropic manifolds and symmetric spaces related?

  5. What are some results connecting homogeneity and isotropy? Homogeneous isotropic manifolds are important in relativity; are these classified, at least in dimension 3? When can we conclude that a homogeneous manifold is isotropic? When can we conclude that an isotropic manifold is homogeneous?

  6. What are some references for reading about isotropic manifolds? I've tried using google and a few Riemannian geometry and general relativity books (Petersen, Wald, Choquet-Bruhat) but I haven't been able to find very much.

youler
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4 Answers4

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Here's a proof that all isotropic manifolds are homogeneous. Given any $p$ and $q$ in $M$, let $\gamma:[0,2]\rightarrow M$ be a minimizing geodesic from $p$ to $q$. Set $r = \gamma(1)$. So, following $\gamma'(1)$ along for one unit of time lands you at $q$ while following it backwards for one unit o ftime lands you at $p$.

By assumption, there is an isometry $f$ for which $d_r f$ maps $\gamma'(1) \in T_r M$ to $-\gamma'(1)$. Then, by uniqueness of geodesics, we have \begin{align*} f(q) &= f(\exp_r ( \gamma'(1)))\\ &= \exp_r(d_r f \, \gamma'(1)) \\ &= \exp_r(-\gamma'(1)) \\ &= p.\end{align*}

As other have pointed out, among, say, compact simply connected homogeneous spaces, isotropic spaces are very rare. In fact, given such a homogeneous space $G/H$ (where we assume wlog $H$ and $G$ share no common normal subgroups of positive dimension), this space is isotropic iff the induced action of $H$ on $T_{eH} G/H$ is transitive on the unit sphere. Under these assumptions, one can prove, for example, that the universal cover of $H$ has at most two factors. (In fact, those $H$ which act effectively and transitively on a sphere have been completely classified.)

I don't know of a classification of when $G/H$ has $H$ acting transitively on the unit sphere, but beyond $\mathbb{C}P^n$, it also happens for $\mathbb{H}P^n$ and $\mathbb{O}P^2$ (the compact, rank one, symmetric spaces). I don't know any other examples of homogeneous spaces for which $H$ acts transitively on the sphere.

Jason DeVito - on hiatus
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  • Thanks for your reply. But sorry, I'm somewhat weak on algebra. What does it mean for a Lie group to have two factors, and how significant of a restriction is that? Also do you know where I can find a proof of the rank one symmetric spaces being isotropic? – youler Jul 22 '13 at 02:58
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    The universal cover of a compact Lie group is isomorphic to a Lie group of the form $G_1 \times G_2\times G_3 \times \ldots \times G_n$ where each $G_n$ is one of ${\mathbb{R}, SO(n), Sp(n), SU(n), G_2, F_4, E_6, E_7, E_8}$. The point is that if $G$ and $H$ don't share any nontrivial normal subgroups (which can always be assumed: If $N$ is normal in both, then $G/H \cong (G/N)/(H/N)$), then the assumption that $H$ acts transitively on a sphere implies that the univeral cover is built out of at most 2 of those things. Then, for example, $H = SO(7)\times SU(6) \times Sp(3)$ cannot appear. – Jason DeVito - on hiatus Jul 22 '13 at 14:24
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    In fact, while I don't know of a classification of $G/H$ where $H$ acts transitively on a sphere, I don't think such a classification would be too hard to carry out in practice. The list of $H$s which act transitively on a sphere is known. Further, if $H$ acts transitively on $S^n$, then $\dim G - \dim H = n+1$, so this puts severe constraints on what $G$s one must consider for a given $H$. – Jason DeVito - on hiatus Jul 22 '13 at 14:27
  • how do you know that given any two points there are a minimizing geodesic join both? – L.F. Cavenaghi Sep 18 '16 at 23:42
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    @frusciante14: I guess I was assuming the Riemmanina metric was complete, then using the Hopf Rinow theorem. But here's a sketch that isotropic implies complete. Assume some geodesic $\exp(tv)$ based at $p$ can't be continued for all time. Now, at the point $q = \exp(0.5 v)$, the geodesic in the direction of $w = \exp'(0.5v)$ can only run for an even shorter time. Thus, the geodesic starting at $q$ in the direction of $-w$ runs for only the shorter time, by isotropicness. But after the shorter time, it's at $p$, so can continue a bit further. – Jason DeVito - on hiatus Sep 19 '16 at 02:18
  • trank you @JasonDeVito. – L.F. Cavenaghi Sep 19 '16 at 14:58
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There is a class of spaces variously known as (1) rank one symmetric spaces and/or (2) two-point homogeneous spaces. These are isotropic (and of course homogeneous). A key example to think about are complex projective spaces and complex hyperbolic spaces. Using some of these terms as keywords you should be able to make more progress. Certainly Helgason's book is a must.

Mikhail Katz
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  • Thanks for your quick reply. Are there many examples of isotropic spaces which are not homogeneous? All of the examples of isotropic manifolds I know of are also homogeneous. – youler Jul 20 '13 at 21:18
  • Merely assuming that at every point there is an isometry integrating the antipodal map already gives you a symmetric space by definition, so I am tempted to answer "no", but I have not thought of these issues recently and hope one of my colleagues corrects me if I am ovelooking some aspect of this. – Mikhail Katz Jul 20 '13 at 21:22
  • Complete symmetric spaces are always homogeneous. Proof: Let $x,y\in M$, a complete symmetric space. Let $\gamma:[0,1]\rightarrow M$ be a minimizing geodesic connecting $x$ and $y$ (which exists by Hopf-Rinow, since $M$ is assumed to be complete). Then at the point $p = \gamma(1/2)$, the antipodal isometry at $p$ maps $x$ to $y$. – Jason DeVito - on hiatus Nov 17 '16 at 02:50
  • @JasonDeVito, that's precisely what I wrote: "No" means that there are no examples of isotropic spaces that are not homogeneous. – Mikhail Katz Dec 01 '16 at 12:09
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    @Mikhail: I think we are one the same page. I was trying to reassure that you were not "overlooking some aspect of this." – Jason DeVito - on hiatus Dec 01 '16 at 14:16
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I'll leave a more detailed, systematic answer to an expert, but "morally speaking" it's harder for a Riemannian manifold to be isotropic than to be homogeneous; a few examples may indicate why:

  • The Fubini-Study metric on $\mathbf{CP}^n$ is isotropic, but does not have constant (real) sectional curvature.

  • A point in a flat torus has finite stabilizer in the isometry group, so is far from isotropic. (An isometry of a $2$-torus lifts to a lattice-preserving isometry of $\mathbf{R}^2$.)

  • Flat $\mathbf{R}^n$ is isotropic, but has trivial holonomy. A constant-curvature oriented surface of genus $g \geq 2$ has discrete isometry group but holonomy $SO(2)$.

  • I don't have these at hand to verify their aptness, but you might try Einstein Manifolds by Besse, The Shape of Space by Weeks (a delightful read in any event), Foundations of Differential Geometry by Kobayashi and Nomizu, or Spaces of Constant Curvature by Wolf.

Andrew D. Hwang
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A perfect reference for isotropic manifolds is the book by Joe Wolf, Spaces of Constant Curvature. In Chapter 8, the Riemannian case is treated, showing in particular that an isotropic Riemannian manifold is either euclidean or a Riemannian symmetric space of rank 1. The whole chapter 12 is devoted to locally isotropic spaces.

I.Agricola
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