# (p.186) Appendix E: More on Congruence and Superposability

# (p.186) Appendix E: More on Congruence and Superposability

In Chapter IV we were interested in those metric spaces with the special feature that specifying the distance relations between the points of a region suffices to fix all of the qualitative geometric facts about that region. To put things a bit more precisely: we were interested in those metric spaces *X* with the feature that whenever subspaces *X* _{1}, *X* _{2} ⊂ *X* are congruent (i.e., there is an isometry *ϕ*: *X* _{1} → *X* _{2}) they are also superposable (i.e., there is a isometry Φ: *X* → *X* with *X* _{2} = Φ (*X* _{1})). This property of metric spaces doesn't have a standard name (and isn't often discussed). I will call it *lability*.

Lability is a very strong condition. This appendix aims to give a feeling for what the condition amounts to, what goes wrong when it fails, and, in general, what modal relationalists are up against. Its primary focus is the Riemannian case (but some remarks will be made about more general cases).

Lability is a close relative to a condition which has received quite a bit of attention. We call a metric space *X fully homogeneous* if any isometry between subspaces can be extended to an isometry of *X* (i.e., whenever subspaces *X* _{1}, *X* _{2} ⊂ *X* are related by an isometry *ϕ*: *X* _{1} → *X* _{2} then there is an isometry Φ: *X* → *X* that extends *ϕ*, in the sense that Φ(*x*) = *ϕ*(*x*) for all *x* ∈ *X* _{1}). A related notion: for *k* = 1, 2,…, we say that *X* is *k‐point homogeneous* if any isometry between *k*‐point subspaces of *X* can be extended to an isometry of *X*.

Clearly, any fully homogeneous space is also labile and *k*‐point homogeneous for any *k*. And for *k* 〉 *m*, any *k*‐point homogeneous space is also *m*‐point homogeneous (of course, one‐point homogeneity is just ordinary homogeneity—the property of there being, for any two points of the space, an isometry that maps the first to the second).

(p.187)
It is known that the only fully homogeneous Riemannian manifolds are the elementary geometries (the Euclidean, hyperbolic, and spherical geometries).^{1} So these spaces are also labile. We will see that they are in fact the only labile Riemannian manifolds.

Our strategy will be to identify several necessary conditions for a Riemannian manifold to be labile, then to observe that they are jointly sufficient, since the elementary geometries are the only Riemannian spaces that satisfy them.

# The Riemannian Case

NECESSARY
CONDITIONS: HOMOGENEITY AND
COMPLETENESS. It is immediate that any labile space is homogeneous (any two one‐point subspaces are congruent, and so superposable). And in the Riemannian setting homogeneity implies completeness. To see this, note that any Riemannian manifold *X* is locally compact.^{2} So for any *x* ∈ *X*, we can find *r* 〉 0 so that *B̄* _{r}(*x*) is compact. If *X* is labile and therefore homogeneous, we can choose *r* to be independent of *x*. But since any Cauchy sequence in *X* is eventually restricted to a ball of radius *r* around some point, any such sequence must converge (since it is eventually confined to a set that is compact and hence complete).

As with each of the conditions we will encounter, it is worthwhile thinking about cases in which these fail—and about what the problem of specifying the superposability type of a one‐point set looks like in such cases.

Beginning with any complete Riemannian manifold, one can create a whole host of incomplete spaces by excising points. So in some sense incompleteness is the norm.

*Example* E.1 (The Punctured Plane). Consider the metric space that results if we remove a single point from the Euclidean plane. For every point *x* in this space, there is one special direction, for which there is an upper bound on how far one can proceed along a straight line in that direction. Two points in this space are superposable if and only if they agree about the magnitude of
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this upper bound. Being told that there is a one‐particle world with the spatial geometry of the punctured plane leaves open infinitely many qualitatively distinct possibilities—parameterized by how far the occupied point is from the “missing point of space.” (Of course, if we had begun instead by removing an asymmetric subset from the plane, then every remaining point would have played a distinct geometric role.) □

Even among complete Riemannian manifolds, inhomogeneity is the norm.

*Example* E.2 (A Crumpled Sphere). Generic Riemannian manifolds are not homogeneous.^{3} A helpful example to picture is any manifold that results from putting gentle bulges or dents in the surface of a sphere in Euclidean space: the resulting geometry fails to be homogeneous. In particular, we can give qualitative characterizations of landmarks on the surface, and in this way distinguish the geometric role played by any point by mentioning its distance from these landmarks. □

*Example* E.3 (The Twisted Cylinder). Take the infinite vertical strip of the Euclidean plane bounded by the lines *x* =−1 and *x* = 1 and identify points on the boundary via (*y*, − 1) ↔ (−*y*, 1) (i.e., give the strip a twist about the *x*‐axis and glue the edges together). The resulting space is a sort of Möbius band. It contains many closed curves that count as straight lines, built out of horizontal line segments in the original strip. All such curves are of length four, except for one special one of length two (corresponding to the segment of the *x*‐axis in the original strip). Two points in this space are superposable if and only if they are the same distance from this distinguished curve. So if we are told that there is a one‐particle world with this structure, we must also be told how far the occupied point lies from the distinguished curve in order to fix all of the qualitative geometric facts. □

NECESSARY
CONDITION: TWO‐POINT
HOMOGENEITY. Any labile Riemannian manifold is also two‐point homogeneous.^{4} To see this, we let (*X, d*) be a labile Riemannian manifold and suppose that we have four
(p.189)
points *x* _{1}, *x* _{2}, *y* _{1}, *y* _{2} ∈ *X* such that *d* (*x* _{1}, *x* _{2}) = *d* (*y* _{1}, *y* _{2}) and then show that there must be an isometry of *X* that maps *x* _{1} to *y* _{1} and *x* _{2} to *y* _{2}. As we have just seen, it follows from the lability of *X* that *X* is complete. Now, in any complete Riemannian manifold, any two points are joined by a geodesic segment that, considered as a metric subspace of the manifold, is isometric to a segment of the real line of length equal to the distance between the points.^{5} Let *γ* _{1} be such a segment that connects *x* _{1} to *x* _{2} and *γ* _{2} be such a segment that connects *y* _{1} to *y* _{2}, with each *γ* _{i} including its initial but not its final point. Then *γ* _{1} and *γ* _{2} determine congruent subsets of *X*: so by the lability of *X*, there must be an isometry Φ : *X* → *X* that maps one on to the other—and given that each segment includes its initial but not its final point, Φ must map *x* _{1} to *y* _{1} and *x* _{2} to *y* _{2}.

The class of two‐point homogeneous Riemannian manifolds is an exclusive club comprising just the classical geometries (elliptic, Euclidean, hyperbolic, spherical) in each dimension and various analogues of elliptic and hyperbolic geometry based on the complex numbers (in even dimensions greater than or equal to four), the quaternions (in dimensions divisible by four and greater than or equal to eight), and the octonions (in dimension sixteen only).^{6}

*Example* E.4 (The Complex Projective Plane). Recall that one way to construct the ordinary elliptic plane (*alias* the real projective plane equipped with a nice metric) has one begin with the real vector space *V* = ℝ^{3} equipped with its Euclidean structure, then identify points of the elliptic plane with one‐dimensional (real) vector subspaces of *V*, equipped with the metric structure induced by their intersection with the unit sphere (itself equipped with the metric structure induced by the ambient Euclidean geometry). The construction of the complex projective plane follows the same template, with the complex numbers substituted for real numbers.^{7}
(p.190)
Introduce an equivalence relation on ℂ^{3} by taking points (*w* _{1}, *w* _{2}, *w* _{3}) and (*z* _{1}, *z* _{2}, *z* _{3}) to be equivalent if there is a non‐zero complex number *λ* such that *w* _{i} = *λ* ∙ *z* _{i} and call the resulting space of equivalence classes ℂ *P* ^{2}. If we take ℂ^{3} to be equipped with its usual inner product, then ℂ *P* ^{2} inherits a metric *d* given by:

(where *x, y* ∈ ℂ^{3}, [*x*] and [*y*] are the corresponding points in ℂ *P* ^{2}, and we use the inner product and associated norm on ℂ^{3}). Equipped with this metric, ℂ *P* ^{2}, is a two‐point homogeneous Riemannian manifold of four (real) dimensions.^{8} □

It is not hard to find spaces that are homogeneous but not two‐point homogeneous.

*Example* E.5 (The Cylinder). Take the vertical strip of the Euclidean plane bounded by the lines *x* = −1 and *x* = 1 and glue the edges together by identifying points on the boundary via (*y*, −1) ↔ (*y*, 1). The result is a cylinder, which we will picture as being embedded in Euclidean three‐space and as having a vertical axis of symmetry. So the horizontal sections are circles. This space is homogeneous: rotations about the axis and translations along the axis both count as symmetries, and by composing these we can map any point to any other point. But it is anisotropic and hence not two‐point homogeneous: let *x* be any point and *y* and *z* both be separated from *x* by a unit of distance with *x* and *y* lying on a horizontal line and *x* and *z* lying on a vertical line; then no isometry can fix *x* while mapping *y* to *z*, for any such isometry would have to induce an isometric mapping from a circle (the geodesic joining *x* and *y*) to an infinite line (the geodesic joining *x* to *z*). If we have a two‐particle world with cylindrical spatial geometry, then in order to fix the totality of qualitative geometric facts about matter we have to be told not only how far apart the particles are, but also what direction they lie in from one another. □

NECESSARY CONDITION: CONSTANT CURVATURE. Any labile Riemannian manifold is a space of constant curvature. Before giving the argument, it will be helpful to recall a few basic facts about Riemannian geometry.

(p.191)
Recall first the notion of sectional curvature. Let (*M, g*) be a Riemannian manifold (i.e., *M* is a manifold and *g* a Riemannian metric tensor). Let *x* be a point in *M* and Π be a tangent plane at *x* (i.e., Π is a two‐dimensional linear subspace of the space of tangent vectors at *x*). For any tangent vector *v* at *x*, we can consider the geodesic in *M* that departs from *x* with velocity given by *v*. If we restrict attention to *v* ∈ Π, then we find that in some sufficiently small neighbourhood of *x*, the points lying on such geodesics form a two‐dimensional surface *S* _{Π} through *x*. We can calculate the Gaussian curvature of *S* _{Π} at *x*: this the real number *k* = *k*(*x*, Π) such that for small *r*, the circumference of any circle of radius *r* about *x* in *S* _{Π} is given by:

(where the ellipsis indicates higher‐order terms in *r*).^{9} We call *k*(*x*, Π) the *sectional curvature* of Π. We can define spaces of constant curvature as those in which *k*(*x*, Π) doesn't depend on either *x* or Π.^{10} It turns out that any Riemannian manifold *M* in which for each *x* ∈ *M, k*(*x*, Π) is independent of Π is in fact a space of constant curvature.^{11}

Recall next that if *f* : *M* → *M* is a smooth map from a manifold to itself that fixes a given *x* ∈ *M*, then *f* induces a linear map *f* _{∗} on the tangent space at *x* (think of the way that a rotation that fixes a given point in Euclidean space nonetheless acts non‐trivially on the space of directions at that point). Since the sectional curvature associated with a tangent plane is definable in terms of the metric structure of the manifold, if *f* : *M* → *M* is an isometry that fixes *x* ∈ *M*, then *f* _{∗} must map any tangent plane at *x* to a tangent plane with the same sectional curvature. Putting this together with the last point of the preceding paragraph, we see that in order to show that a Riemannian manifold (*M, g*) is a space of constant curvature, it suffices to show that for each point *x* and each pair of tangent planes Π_{1} and Π_{2} at *x*, there is an isometry *f* of (*M, g*) such that: (a) *f* fixes *x*; and (b) *f* _{∗} maps Π_{1} to Π_{2}.

Let us now turn to the case of interest. Let (*X, d*) be a labile Riemannian manifold and let *x* be any point in *X* and Π_{1} any tangent plane at *x*. We can choose *ϵ* 〉 0 so that any point *x′* within a ball of radius 6*ϵ* of *x* is connected to *x* by a unique geodesic segment ℓ(*x, x′*) that remains within the ball.^{12}

(p.192) Claim: we can find

y_{1},z_{1}∈Xsuch thatd(x, y_{1}) = 2ϵ, d(x, z_{1}) = 3ϵ, andd(y_{1},z_{1}) = 4ϵ; and such that the tangent vectors to ℓ(x, y_{1}) and ℓ(x, z_{1}) atxlie in Π_{1}.

To see this, let *v* be any unit vector in Π_{1} and let *y* _{1} be the point that lies 2*ϵ* units of distance along the geodesic through *x* with tangent vector *v*. For any unit vector *w* ^{*} in Π_{1}, let *z* ^{*} be the point that lies 3*ϵ* units of distance along the geodesic through *x* with tangent vector *w* ^{*}, and let *g*(*w* ^{*}) = *d* (*y* _{1}, *z* ^{*}). Note that *g* (*w* ^{*}) = 1 if *w* ^{*} = *v* and that *g* (*w* ^{*}) = 5 if *w* ^{*} = −*v*. So there must be some *w* ∈ Π linearly independent of *v* such that *g*(*w*) = 4. So we can take *z* _{1} to be the point that lies 3*ϵ* units of distance along the geodesic through *x* with tangent vector *w*.

Now let Π_{2} be any other tangent plane at *x*. The same argument shows that we can find *y* _{2}, *z* _{2} ∈ *X* such that *d* (*x, y* _{2}) = 2 *ϵ, d* (*x, z* _{2}) = 3 *ϵ*, and *d* (*y* _{2}, *z* _{2}) = 4 *ϵ*; and such that the tangent vectors to ℓ (*x, y* _{2}) and ℓ (*x, z* _{2}) at *x* lie in Π_{2}. So (*x, y* _{1}, *z* _{1}) and (*x, y* _{2}, *z* _{2}) are congruent triples of points—so since *X* is labile, there must be an isometry Φ : *X* → *X* that maps {*x, y* _{1}, *z* _{1}} to {*x, y* _{2}, *z* _{2}}. Given the distances involved, it follows that Φ fixes *x* and maps *y* _{1} to *y* _{2} and *z* _{1} to *z* _{2}. Further, since the tangent vectors to ℓ (*x, y* _{i}) and ℓ (*x, z* _{i}) at *x* span Π_{i} (*i* = 1,2) the map Φ_{∗} that Φ induces on the tangent space at *x* maps Π_{1} to Π_{2}. Since *x*, Π_{1}, and Π_{2} were arbitrary, this suffices to show that *X* is a space of constant curvature.

We have of course already seen examples of Riemannian manifolds that fail to be spaces of constant curvature. For present purposes, it is interesting to note that being two‐point homogeneous and being of constant curvature are logically independent conditions: there exist Riemannian manifolds that are both two‐point homogeneous and of constant curvature (the classical geometries); some that are neither (in generic Riemannian geometries, the sectional curvature varies from tangent plane to tangent plane at a point in a way that varies from point to point); some that are spaces of constant curvature but not two‐point homogeneous (such as the cylinder); and some that are two‐point homogeneous but not spaces of constant curvature (such as the complex, quaternionic, and octonionic analogues of projective and hyperbolic spaces).^{13}

*Example* E.6 (The Complex Projective Plane Once More). The complex projective plane is homogeneous and two‐point homogeneous. So for
(p.193)
one‐ and two‐point regions, congruence implies superposability. But the argument above by which we showed that lability implies constant curvature suffices to show that any space that is not a space of constant curvature is not three‐point homogeneous (since it shows us how to find congruent three‐points sets {*x, y* _{1}, *z* _{1}} and {*x, y* _{2}, *z* _{2}} that cannot be superposed if the sectional curvature varies from tangent plane to tangent plane at *x*).

What sort of additional data, in addition to distances, is required to fix the superposability type of a three‐point set in the complex projective plane? Generically a single additional number will do (for instance, one of the angles that would be enclosed by two sides if the three points were joined to form a geodesic triangle).^{14} But this is only the beginning of the story: the superposability type of a generic region of the complex projective plane is fixed by fixing the superposability type of each four‐point region—but not by fixing the superposability type of each three‐point region.^{15} □

NECESSARY CONDITION: NON‐ELLIPTICITY. Our necessary conditions for lability among Riemannian manifolds are not quite jointly sufficient: the only Riemannian spaces that are complete, homogeneous, two‐point homogeneous, and of constant curvature are the classical geometries (elliptic, Euclidean, hyperbolic, and spherical) in each dimension. But one of these is not like the others.

*Example* E.7 (The Elliptic Plane). Recall once again that the points of the elliptic plane can be represented by lines through the origin in ℝ^{3}, with the distance between points being given by the smaller of the angles that the corresponding lines make at the origin; a line in the elliptic plane corresponds to a plane through the origin in ℝ^{3}.

The elliptic plane is both two‐point homogeneous and a space of constant curvature. But it is not a labile, for there exist three‐point sets in the elliptic plane that are congruent but not superposable.^{16} To see this, consider, first, the points (0, 1, 1), (1, 0, 1), (1, 1, 0) ∈ ℝ^{3}. These three
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points are the vertices of an equilateral triangle on the sphere of radius two centred at the origin in ℝ^{3}; they therefore determine three non‐coplanar lines through the origin; the corresponding points in the elliptic plane are non‐collinear and separated from one another by $\frac{\pi}{3}$. Consider, next, the points (0, 1, −1), (−1, 0, 1), (1, −1, 0) ∈ ℝ^{3}. These determine three coplanar lines through the origin in ℝ^{3}. So the three corresponding points in the elliptic plane are collinear—and they too have mutual distance $\frac{\pi}{3}$ from one another.^{17} But now we have two congruent three‐point sets that cannot be superposed: isometries map straight lines to straight lines, and so cannot map a set of non‐collinear points to a collinear set of points (nor vice versa, since the inverse of any isometry is an isometry).^{18}

Since the elliptic plane is homogeneous and two‐point homogeneous, congruence implies superposability for one‐ and two‐point regions. We have seen that this fails for three‐point regions. What data are required to fix the superposability type of a region consisting of three points, *x, y*, and *z* in the elliptic plane?

Here is something that will suffice: specify the distance *a* = *d*(*x, y*) and *b* = *d*(*x, z*) and an angle *θ* that can be formed at *x* by geodesic segments of length *a* and *b* connecting *x* to *y* and to *z*.^{19} In other words: a form of the side‐angle‐side rule for congruence of triangles holds in the elliptic plane.^{20} In order to see this, it suffices to note the following. Suppose
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that (*x* _{1}, *y* _{1}, *z* _{1}) and (*x* _{2}, *y* _{2}, *z* _{2}) are two triples of points that satisfy the given constraints. Note that for each triple of points, (*x* _{i}, *y* _{i}, *z* _{i}), there are tangent vectors *v* _{i} and *w* _{i} at *x* _{i} with lengths *a* and *b* (respectively) that make an angle *θ*, and such that one can reach *y* _{i} or *z* _{i} by travelling for a unit of time along the geodesic at *x* _{i} with tangent vector *v* _{i} or *w* _{i}. In order to show that (*x* _{1}, *y* _{1}, *z* _{1}) and (*x* _{2}, *y* _{2}, *z* _{2}) are superposable, it suffices to show that there is an isometry of the elliptic plane that maps *x* _{1} to *x* _{2} and whose tangent map sends *v* _{1} to *v* _{2} and *w* _{1} to *w* _{2}. Since the elliptic plane is homogeneous, we can without loss of generality take *x* _{1} = *x* _{2}. So our question is whether given two pairs of tangent vectors at *x* _{1}, such that the corresponding members of each pair have equal length and such that the angles formed by the members of each pair are equal, there exists an isometry of the elliptic plane that fixes *x* _{1} and whose tangent map sends each element of the first pair onto the corresponding member of the second pair. At this point, it is helpful to switch to a different way of thinking of the elliptic plane: as the upper half of the unit sphere, with antipodal points on the equator identified. Since the elliptic plane is homogeneous, we can without loss of generality take *x* _{1} to be the North Pole. Isometries of the elliptic plane that fix *x* _{1} are given by rotations of the hemisphere about the North–South axis and by reflections in vertical planes that pass through the North Pole.^{21} By employing a map of the first sort, then one of the second, we can map *v* _{1} to *v* _{2} and *w* _{1} to *w* _{2}.

Interestingly, it is *almost true* that the superposability type of a region in the elliptic plane is specified by specifying the congruence type of each of its three‐point subregions.^{22} □

In fact, no elliptic space is labile.^{23} Specifying the distances between three points in an elliptic space suffices to determine whether they are collinear if and only if one of the distances is $\frac{\pi}{2}$ or the sum of the distances is less than *π*.^{24} Further, in any elliptic space there are congruent but non‐superposable regions of cardinality *κ* for any cardinal number *κ*, 3 〈 *κ* ≤ *c*.

(p.196) THE FOREGOING CONDITIONS ARE JOINTLY SUFFICIENT. So far we have seen that that any labile Riemannian manifold is a non‐elliptic two‐point homogeneous space of constant curvature (recall that homogeneity and completeness are implied by two‐point homogeneity). Conversely, since the only Riemannian manifolds that satisfy these conditions are the elementary geometries (Euclidean, hyperbolic, and spherical) and these are known to be fully homogeneous (and hence labile), we find that any non‐elliptic two‐point homogeneous space of constant curvature is a labile Riemannian manifold.

# Beyond the Riemannian Case

The picture that emerges from the above discussion holds more generally in the much larger class of locally compact path metric spaces.^{25}

Claim: Let (

X, d) be a locally compact path metric space. The following are equivalent.

(i)

Xis three‐point homogeneous.(ii)

Xis labile.(iii)

Xis fully homogeneous.(iv)

Xis one of the elementary geometries.

To see this, note that independently of the assumption that *X* is a locally compact path metric space, we know that (iv) ⇒ (iii) (the result of Birkhoff cited above), that (iii) ⇒ (ii) (immediate from the definitions), and that (iii) ⇒ (i) (likewise immediate). So we need only satisfy ourselves that (ii) ⇒ (iv) and that (i) ⇒ (iv). The key here is a result of Jacques Tits and Hsien‐Chung Wang, according to which any complete locally compact two‐point homogeneous path metric space is a Riemannian manifold.^{26} Now, the arguments given above showing that for Riemannian manifolds homogeneity implies
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completeness and that lability implies two‐point homogeneity carry over without modification to the setting of locally compact path metric spaces. So any labile locally compact path metric space is both two‐point homogeneous and complete—and thus, by the Tits–Wang result is also a Riemannian manifold. But of course we have already seen that the only labile Riemannian manifolds are the elementary geometries—so (ii) ⇒ (iv). Similarly, if (*X, d*) is a three‐point homogeneous locally compact path metric space, then *X* is both complete and two‐point homogeneous and hence Riemannian. But then it follows more or less immediately that *X* is elementary: we already know that *X* is two‐point homogeneous and the arguments given above to show that a labile Riemannian manifold must be of constant curvature and cannot be elliptic in fact also show that a three‐point homogeneous Riemannian manifold must likewise be of constant curvature and non‐elliptic. So (i) ⇒ (iv).

But (at least some of) conditions (i)–(iv) can come apart when we move beyond locally compact path metric spaces.

There are fully homogeneous spaces other than the elementary geometries: a set *X* equipped with the discrete metric is fully homogeneous if and only if *X* is finite; a subset of the real numbers is fully homogeneous if and only if it has two or fewer members or is isometric to an additive subgroup of the real numbers (e.g., the integers, the rational numbers, the algebraic numbers, the real numbers).^{27}

And there are spaces that are three‐point homogeneous without being labile or fully homogeneous. An infinite‐dimensional Hilbert space *ℋ* is *k*‐point homogeneous for every finite *k* but includes infinite sets that are congruent but not superposable.^{28} For instance, let *S* _{1} be an orthonormal basis for *ℋ* and let *S* _{2} be the set that results from omitting a single element from *S* _{1}. Then *S* _{1} and *S* _{2} are congruent sets that are not superposable (no isometry of *ℋ* could map *S* _{1} onto *S* _{2} because then its image would not be all of *ℋ*).

I am unsure whether there are examples of spaces that are labile but not fully homogeneous.^{29}
(p.198)

## Notes:

(1) See Birkhoff, “Metric Foundations of Geometry. I.”

(2)
See Remark B.3 of Appendix B above. Recall that a metric space *X* is *locally compact* if sufficiently small open metric balls have compact closures. For path metric spaces, local compactness is equivalent to the compactness of all closed bounded subsets; see Plaut, “Metric Spaces of Curvature ≥ *k*,” theorem 8.

(3) Indeed, generic Riemannian manifolds have no non‐trivial symmetries. For references and discussion, see Blair, “Spaces of Metrics and Curvature Functionals,” §§1.1 f.

(4)
For Riemannian manifolds, two‐point homogeneity is equivalent to isotropy (the condition that for any point and any two unit tangent vectors at that point, there be an isometry that fixes the point and whose tangent map sends the first of the given tangent vectors to the second). See Wolf, *Spaces of Constant Curvature,* lemma 8.12.1.

(5)
This is a standard corollary to the Hopf–Rinow theorem; see e.g. §5.8 of Petersen, *Riemannian Geometry*. The result continues to hold in the setting of locally compact path metric spaces; see Burago *et al*., *A Course in Metric Geometry,* theorem 2.5.23.

(6)
See Wolf, *Spaces of Constant Curvature,* §8.12.

(7)
The same holds, *mutatis mutandis,* for higher‐dimensional complex analogues of the elliptic spaces, and for the corresponding quaternionic and octonionic spaces. The same basic strategy leads to various analogues of the hyperbolic spaces, modelled on the following construction of the ordinary hyperbolic plane. Begin with the real vector space *W* = ℝ^{3} equipped with the structure of Minkowski space, and identify points of the hyperbolic plane with the one‐dimensional (real) vector subspaces of *W* that point in timelike directions, equipped with the metric structure induced by their intersection with the unit spacelike hyperboloid (itself equipped with the metric structure induced by the ambient Minkowski geometry). For details, see Busemann, *The Geometry of Geodesics,* §53.

(8)
See e.g. Busemann, *Geometry of Geodesics,* p. 380. Note that ℂ*P* ^{2} arises as the space of states of the spin degrees of freedom of a spin‐one quantum particle.

(9)
See e.g. Bishop and Goldberg, *Tensor Analysis on Manifolds,* §5.14.

(10)
This condition is equivalent to being a connected Riemannian locally isometric to one of the elementary geometries; see e.g. Wolf, *Spaces of Constant Curvature,* §2.4.

(11)
See e.g. Wolf, *Spaces of Constant Curvature,* corollary 2.2.7.

(12)
It suffices that 6*ϵ* be less than the injectivity radius of *X* at *x*; see e.g. §§5.5 and 5.9 of Petersen, *Riemannian Geometry*.

(13)
For the behaviour of the sectional curvatures in the complex projective plane and its ilk, see e.g. Petersen, *Riemannian Geometry,* §§3.5.3 and 8.1.1.

(14) See Brehm, “The Shape Invariant of Triangles and Trigonometry in Two‐Point Homogeneous Spaces.” The same holds true for the other two‐point homogenous spaces that are not spaces of constant curvature.

(15) See Brehm and Et‐Taoui, “Congruence Criteria for Finite Subsets of Complex Projective and Complex Hyperbolic Spaces,” propositions 1 and 2. For results concerning the quaternionic case, see Brehm and Et‐Taoui, “Congruence Criteria for Finite Subsets of Quaternionic Elliptic and Quaternionic Hyperbolic Spaces.”

(16) For this example, see Seidel, “Discrete Non‐Euclidean Geometry,” p. 877.

(17)
To picture what is going on here, picture choosing two points *x* and *y* on the equator of the unit sphere separated by an arc of length $\frac{\pi}{3}$. Now choose a point *z* on the sphere that is $\frac{\pi}{3}$ from each of *x* and *y* and a point *z* ^{*} that is $\frac{2\pi}{3}$ from *x* and $\frac{\pi}{3}$ from *y*. Then *z* ^{*} lies on the equator with *x* and *y* but *z* does not. When we pass to the elliptic plane by identifying antipodal points on the sphere, any distance less than or equal to $\frac{\pi}{2}$ is unaltered while any greater distance *b* is replaced by *π* − *b*; and the resulting points are collinear if and only if the points we started with on the sphere lie on a great circle. So the points in the elliptic plane corresponding to *x, y*, and *z* have distance $\frac{\pi}{3}$ from one another and are non‐collinear; the points in the elliptic plane corresponding to *x, y*, and *z* ^{*} satisfy the same pattern of distance relations but are collinear. For this example, see e.g. Busemann and Kelly, *Projective Geometry and Projective Metrics,* pp. 221 f.

(18)
Don't make too much of the role that collinearity plays here: it is also possible to concoct examples of congruent but non‐superpoasble three‐point subsets of the elliptic plane, neither of which is collinear. For examples, see e.g. Gans, *An Introduction to Non‐Euclidean Geometry,* pp. 240 f.; or Busemann and Kelly, *Projective Geometry,* p. 222.

(19)
Note that in the elliptic plane of unit radius, two points *x* and *y* are connected by a unique shortest geodesic segment if $d\left(x,y\right)\u3008\frac{\pi}{2}$, but are connected by two such segments when $d\left(x,y\right)=\frac{\pi}{2}$.

(20)
Another form of the side‐angle‐side rule fails: the data we have specified do not in general suffice to determine the angles formed if we connect *x, y*, and *z* by geodesic segments. See Gans, *Introduction,* pp. 239 f.

(22)
See Blumenthal, *Theory and Applications of Distance Geometry,* §§81, 92, 93, and 97.

(23)
For the results mentioned in this paragraph, see Blumenthal, “Congruence and Superposability in Elliptic Space” or *Theory and Applications,* ch. X.

(24) So there is a sense in which the problem is a global one: locally, of course, the elliptic space looks like a spherical space (which is fully homogeneous); for certain ‘large’ configurations, one can exploit the global structure of an elliptic space to find congruent but non‐superposable configurations.

(25)
How can a path metric space fail to be locally compact? e.g., by having a lot of holes inconveniently arranged or by being infinite‐dimensional. If one deletes from the ℝ^{2} a sequence of points that converges to (but does not include) the origin, then the resulting space is still a path metric space but fails to be locally compact—because no closed metric ball around the origin is complete. Any Hilbert space *ℋ* is a path metric space; *ℋ* is locally compact if and only if finite‐dimensional (Abraham *et al*., *Manifolds, Tensor Analysis, and Applications,* proposition 2.1.11).

(26)
For discussion and references, see Busemann, *Recent Synthetic Differential Geometry,* §19; and Freudenthal, “Lie Groups in the Foundations of Geometry,” §§2.19 ff.

(27)
Ovchinnikov, “Homogeneity Properties of Some ℓ_{1}‐Spaces,” theorem 2.2.

(28) Birkhoff gives this example, and mentions that the Urysohn space has the same feature; “Metric Foundations,” §6.

(29) Birkhoff asserts that lability is weaker than full homogeneity, but gives no examples; “Metric Foundations,” §6. I have been unable to find any in the literature, so I leave this as a challenge to the reader.