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In physics, homogenous is used to mean that some quantity does not change with position.

This is less general than the use of homogeneous in mathematics, I think.

Question: So how does one define a metric space which is "homogeneous" in the sense of physics, i.e. its (metric) geometry is the same at every point?

(E.g. hyperbolic, Euclidean, and spherical geometries.)

There are definitions of a homogeneous metric space (see here or here), but both seem to be more general than what I am trying to describe above.

Attempt: Given a metric space $(X,d)$, denote the isometry group by $Iso(X)$, and for each $x \in X$, denote the subgroup of $Iso(X)$ consisting of elements fixing $x$ ($f(x) = x, f \in Iso(X)$), by $Iso(X,x)$.

Then a metric space "has the same geometry at each point" if and only if, for any two points $x_1, x_2 \in X$, one has that $Iso(X,x_1) \cong Iso(X, x_2)$.

Euclidean space satisfies this condition, since $Iso(X,x) \cong O(n)$ for any point $x$ in $n$-dimensional Euclidean space (I think). I don't know if hyperbolic and spherical geometries satisfy this too.

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The definition of a homogeneous metric space is a little stronger than what you say: a metric space $(X,d)$ is homogeneous if and only if for every $x_1,x_2 \in X$ there exists $f \in Iso(X)$ such that $f(x_1)=x_2$. This implies the identity that you asked for, namely that $Iso(X,x_1) \simeq Iso(X,x_2)$, because one obtains an isomorphism $$A : Iso(X,x_1) \to Iso(X,x_2) $$ using the "conjugation" formula, also called the "adjoint" formula $$A(g) = f \circ g \circ f^{-1} \in Iso(X,x_2) \quad\text{for each}\quad g \in Iso(X,x_1) $$

All the geometries that you mention --- spherical, Euclidean, and hyperbolic --- are homogenous in this stronger sense: one simply works in Euclidean geometry, in spherical geometry, or in some particular model of hyperbolic geometry, to directly construct a desired isometry taking any given point $x_1$ to any other given point $x_2$. Those geometries therefore all satisfy the identity that you ask for, namely $Iso(X,x) \simeq O(n)$ for each $x \in X$, because you can easily use the symmetry of the metric to verify this identity for one particular value of $x$: use the origin in Euclidean space; the north pole in spherical space with the metric expressed in spherical coordinates; or the center of the Poincare disc model of hyperbolic space.

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  • $\begingroup$ Does the extra condition also guarantee (for an arbitrary metric space) that the local similitude groups at every point coincide? (similitude - $f: X \to X$, surjective and such that $d(f(x_1), f(x_2)) = \lambda d(x_1, x_2),\ \lambda > 0$, isometry is the case that $\lambda = 1$, then the local similitude group $Sim(X,x)$ is the same as $Iso(X,x)$ replacing isometry with similitude in the definition.) Since angles are invariants of the local similitude groups, not just of the local isometry groups, if the geometry is to be the same at each point the local sim. grps. need to be preserved too. $\endgroup$ – Chill2Macht Jul 13 '17 at 16:26
  • $\begingroup$ Anyway my point being that the definition proposed in the question doesn't necessarily preserve local similitude groups, but the definition given in the answer is stronger, so it might satisfy this stronger condition, which seemingly is also necessary. $\endgroup$ – Chill2Macht Jul 13 '17 at 16:29
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    $\begingroup$ Yes, the stronger definition preserves angles, and similitudes. Also many other important features (for instance curvature tensors, in the case of Riemannian manifolds). $\endgroup$ – Lee Mosher Jul 13 '17 at 16:53
  • $\begingroup$ Is the following also correct? "The three spaces mentioned are the only types of Riemannian manifolds satisfying this definition, since all Riemannian manifolds are isotropic at each point (unlike general Finslerian manifolds, for which the metric tensor also depends on direction)." $\endgroup$ – Chill2Macht Jul 13 '17 at 18:00
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    $\begingroup$ That statement is a little vague so I am unsure exactly what it is asserting, but there are similar true statements. $\endgroup$ – Lee Mosher Jul 13 '17 at 18:42

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