I try to get my head around the different kinds of homogi-enities (global similarities of a space ) and these are two of the global similarities I stumble upon and I don't really know the difference.

Feel free to add other global similarities to your answer if you like I would like to know about all


How to say that the "geometry is the same" at every point of a metric space?


Speaking of a connected Riemannian manifold $(M, g)$, the structure is

  • Homogeneous if, for every pair of points $p$ and $q$, there exists an isometry $f:M \to M$ such that $f(p) = q$. (The isometry group acts transitively on points of $M$.)

    Examples include flat Euclidean space, arbitrary complete $1$-dimensional Riemannian manifolds; a sphere or real projective space with the round metric, or a complex projective space with the Fubini-Study metric; real or complex hyperbolic space.

    A Riemannian product of homogeneous spaces is homogeneous. Particularly, an arbitrary product of (complete) real curves, namely a flat torus of arbitrary dimension or a product of a flat torus and a Euclidean space, is homogeneous.

    A Riemannian manifold covered by a homogeneous space is generally not homogeneous, e.g., a compact Riemann surface of genus at least $2$ with a constant-curvature metric.

  • Symmetric if, for every pair of points $p$ and $q$, there exists an isometry $f:M \to M$ such that $f(p) = q$ and $f(q) = p$. (Every pair of points is exchanged by some isometry. Wikipedia's definition differs formally, but is equivalent: Either definition implies geodesic completeness. If $p$ and $q$ are arbitrary, let $\gamma$ be a minimizing geodesic from $p$ to $q$, and let $r$ denote the midpoint; an isometry fixing $r$ and reversing the direction of $\gamma$ exchanges $p$ and $q$.)

  • Isotropic at a point $p$ if for arbitrary unit vectors $v$ and $w$ at $p$, there exists an isometry $f:M \to M$ such that $f_{*}v = w$. (The isotropy group of $(M, g)$ at $p$ acts transitively on the space of directions at $p$.)

    Isotropic if $(M, g)$ is isotropic at $p$ for every $p$ in $M$. (See also Isotropic Manifolds.)

    An arbitrary surface of rotation is isotropic at its pole(s), the point(s) on the axis of symmetry (if any), but not generally isotropic at any other point.

    Isotropic spaces include Euclidean space; a round sphere, or complex projective space with a Fubini-Study metric; real or complex hyperbolic space.

    A product of isotropic manifolds is not generally isotropic, e.g., a flat cylinder, flat torus, product of spheres.

    (Edit: Originally, I claimed that if a product of (connected, positive-dimensional) isotropic Riemannian manifolds is isotropic, then each factor is a Euclidean space. My argument for this claim was specious. I still suspect the assertion is true: A real $2$-plane spanned by unit vectors tangent to the respective factors exponentiates to a flat immersed surface, and isotropy "should" be enough to show that every real $2$-plane exponentiates to a flat immersed surface. I have not, however, found a convincing proof.)

As defined above, homogeneity and symmetry obviously make sense for arbitrary metric spaces. Isotropy requires some kind of local model for which "transitivity of the stabilizer" makes sense.

  • $\begingroup$ I think this is all a bit above my head are there simple models that are homogeneous but not symmetric, symmetric but not isotropic , one of these could be a (normal) torus but which one $\endgroup$ – Willemien Jul 27 '17 at 5:48
  • $\begingroup$ A flat torus is homogeneous (under translations) but not symmetric: If $p$ and $q$ are arbitrary points coming from points $\tilde{p}$ and $\tilde{q}$ in the Euclidean plane, the reflection exchanging $\tilde{p}$ and $\tilde{q}$ does not generally preserve the torus's lattice, so the reflection does not descend to the quotient. // Every symmetric space is isotropic: Briefly, if $v$ and $w$ are arbitrary unit vectors at a point $p$, an isometry exchanging $\exp_{p}(v)$ and $\exp_{p}(w)$ sends $v$ to $w$. $\endgroup$ – Andrew D. Hwang Jul 27 '17 at 11:00
  • $\begingroup$ sorry I was thinking about the surface of a normal (doughnut shaped ) torus because the curvature is not constant it cannot be isotropic I think but you are the expert $\endgroup$ – Willemien Jul 27 '17 at 12:12
  • $\begingroup$ An ordinary doughnut-shaped torus isn't even homogeneous. :) The only isometries are axial rotations. $\endgroup$ – Andrew D. Hwang Jul 27 '17 at 12:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.