PREPRINT
國立臺灣大學 數學系 預印本 Department of Mathematics, National Taiwan University
www.math.ntu.edu.tw/ ~ mathlib/preprint/2013- 08.pdf
Bounded extrinsic curvature of subsets of metric spaces
Jeremy Wong
April 30, 2013
Jeremy Wong
Abstract. Subsets of Alexandrov spaces of curvature bounded below with bounded extrinsic cur- vature are studied. If the subset is a geodesically extendible length space in X = Rn+1or dimX = 2, then it has no geodesic branching.
If the subset has constant extrinsic curvature, then the subset has no branching, and has itself a lower curvature bound. If the subset has constant extrinsic curvature and is a smooth manifold (possibly with boundary), then it has an explicit intrinsic lower curvature bound which is sharp in general.
2010 Mathematics Subject Classification. Primary 53C70 Secondary 53C20 53C24 51K10 53B25 53A04
Keywords: branching, extrinsic curvature, geodesically extendible, Alexandrov space, lower curva- ture bound, Gauss equations
1. Introduction
Definition. [15] A subspace Z of a metric space X is called strongly (resp. weakly) (C, ρ)-convex if the metrics satisfy
dZ(x, y) ≤ dX(x, y) + Cd3X(x, y) for all x, y ∈ Z with dX(x, y) < ρ
(resp. for all x, y ∈ Z with dZ(x, y) < ρ).
In the definition, ρ > 0 is a function on X which may vary from point to point, and 0 ≤ C < ∞ is a constant. If ρ is not emphasized, one can call the condition merely C-convex. For instance, a 0-convex subset is simply a locally convex subset. ρ plays the role of a type of injectivity radius.
Another version (used in [4]) is
dZ ≤ dX + Cd3X + o(d3X) (1)
which is more suitable for obtaining a smaller constant C at the expense of adding higher order terms. Throughout the paper, unless otherwise noted, this latter version, dZ ≤ dX+ Cd3X + o(d3X) on all pairs of points (x, y) sufficiently Z-close, is used. 1
Last typeset: July 2, 2013
work partially supported by NSF stipend, NSC grant 101-2115-M-002 -001.
2013c
1
In this paper X and Z will usually be assumed to have at least the structure of a length space.
Here dZ denotes an intrinsic metric on Z, for which the distance dZ(x, y) between two points x, y ∈ Z is the infimum of the lengths of paths in Z joining x and y.
Note that C-convexity for subsets does not represent a signed curvature quantity (although an alternative definition for the special case of hypersurfaces might be possible). Just as the definition of Alexandrov curvature, it is defined locally, rather than infinitesimally, as the shape operator of a smooth submanifold would be.
This extrinsic notion of curvature bound for a subset is related to other notions in various contexts: a pointwise bound on the norm |IIZ,→X| of second fundamental form, positive reach, and λ-convexity of distance-like functions.
Properties:
(i) If X Riemannian manifold and Z ⊂ X a closed embedded submanifold then |IIZ,→X| ≤ λ if and only if Z is (λ242, ρ)-convex for some sufficiently small ρ > 0
(ii) If X Riemannian manifold, the condition is equivalent to positive reach [15, Theorem 1.3]
Subsets of Rn+1 with uniformly positive reach can include, for example, finite disjoint unions of sets which are closed manifolds or manifolds with C1 boundary of various dimensions which are joined along codimension ≥ 1 subsets of their boundaries. A (C, ρ)-convex subset need not have positive reach nor admit a supporting ball at each point, for a general ambient space. For example, take Z to be a generator of a cone over a circle of length less than 2π.
A motivation for considering (C, ρ)-convexity is that one would like to consider analogues of second fundamental form which are preserved in taking limits. For general metric spaces, one has
Proposition 1. Suppose Xi GH
−→ X and Zi −→ Z as subsets.
(1) [15, p.208] Zi strongly (C, ρ)-convex =⇒ Z strongly (C, ρ)-convex (2) [25] Zi weakly (C, ρ)-convex, Zi GH
−→ Z =⇒ Z weakly (C, ρ)-convex
Note that it could happen that the limit set Z was (C, ρ)-convex but each Zi was not (C, ρ)- convex. Example1 below shows how the property can suddenly appear in the limit.
For further discussion of (C, ρ)-convexity in relation to curves (especially in CBA spaces), see [3]. 2
Concerning spaces for which one can define λ-concave functions (such as Alexandrov spaces), one has
Theorem 1. [5]. Sublevel sets of λ-concave functions are (C, ρ)-convex for some C = C(λ).
In [5] sharp lower bounds on extrinsic curvature were found for a level or sublevel set of a λ-concave function f with λ < 0, in terms of given gradient bounds on the f . The lower bound equality case was also characterized in [5]. Similarly, at least for boundaries ∂X of CBB spaces X, a related notion of base-angle extrinsic curvature was shown to lead to comparison results with balls in standard model spaces.
It is sensible to consider an opposite inequality of the form dX + Cd3X + o(d3X) ≤ dZ (what one might call ”strictly positively convex”), which, in the special case of submanifolds, is basically equivalent to the uniformly positive condition that IIZ,→X ≥ const ≥ 0. The latter notation means that all principal curvatures are uniformly bounded from below. Informally, one can see
why this would be so instead of the condtion |IIZ,→X| ≥ const, because, in the contrary case, e.g. saddle-surfaces, if two principal curvatures had opposite signs, there would exist by continuity an asymptotic direction giving rise (at least in manifolds) to an asymptotic curve. Such a curve would be a geodesic (or close to a geodesic, at small scale) in the ambient space, and contradict the ”strictly positively convex” hypothesis.
For C-convexity, one can see the variation [0, Cd3X + o(d3X)] in the difference dZ − dX by considering Z-geodesics in a non-convex domain. Either a geodesic segment lies entirely in the interior of the domain (in which case effectively dZ = dX), or its interior intersects the boundary of Z along a non-trivial portion, where the boundary has some concavity. By continuity, there is a whole family of Z-geodesics whose endpoints interpolate between these two extreme types of geodesics, which explains why there is a range, i.e. why one has an inequality.
Question: Given a space with curvX ≥ k, and Z ⊂ X a subset with dZ ≤ dX + Cd3X + o(d3X) locally, under what conditions does curvZ ≥ c(k, C)?
The above example illustrates that dZ ≤ dX + Cd3X + o(d3X) is insufficient, due to possible bifurcations along the set of geodesic terminals (or boundary). Also, using a conformal stereographic projection, a two-dimensional region in R2 between two disjoint spheres is mapped onto a two- dimensional region in S2(1) between two disjoint spheres in S2(1), but the geodesic terminals still can produce branching. Thus, being strictly positively C-convex (or having a two-sided pinching C1d3X ≤ dZ− dX ≤ C2d3X for some constants C1 < C2) by itself also does not produce an intrinsic lower curvature bound curvZ ≥ const.
In this note, two main additional assumptions are separately considered: constancy of extrinsic curvature, and geodesic extendibility.
Main results are Theorem2and Propositions4and5. Common to each of these is the existence of a filling between two geodesics in the subspace.
2. Constant extrinsic curvature
When Z has constant relative curvature in a metric space sense (i.e. equality in the above definition of C-convexity), then one can say more about its geometry.
Definition. If X is a metric space and Z ⊆ X a subspace having an induced intrinsic metric dZ, say Z is constantly C-convex (or more precisely, has constant extrinsic curvature equal to
√
24C) if
dZ(x, y) = dX(x, y) + Cd3X(x, y) + o(d3X(x, y))
for all x, y ∈ Z with dZ(x, y) < ρ, where ρ > 0 is sufficiently small (depending on Z, or equivalently, on X and C). 3
Theorem 2. Suppose curvX ≥ k, X is locally compact, Z ⊆ X is a closed subset with the induced length space structure, and Z is constantly C-convex.
Then
(i) Z has no branching.
(ii) If furthermore Z is compact, then curvZ > −∞.
(iii) If Z is assumed to be a C2 smooth manifold (not necessarily geodesically extendible), then curvZ ≥ k − 48C.
Note that curvZ ≥ k holds for a general constantly C-convex subset when C = 0 (and the higher order terms are 0), since then Z is totally geodesic.
In the smooth setting, if Z were a smooth hypersurface of a smooth manifold X, and one had a two-sided bound |IIZ,→X| ≤ λ (with λ2 = 24C) for the second fundamental form, then a lower bound to the sectional curvature of Z would be k − 24C, by the Gauss equations, but in higher codimensions the optimal lower bound would be k − 48C.
On the other hand, greater (positive) pinching leads to greater rigidity. If one had equality
|IIZ,→X| = λ for a smooth hypersurface of a manifold which had sectional curvatures bounded from below by k, then Z would have positive semi-definite second fundamental form, and a lower sectional curvature bound ≥ k + 24C.
The lower bound of k − 48C in (iii), which does not require the ambient space X to be a manifold, is sharp for general codimensions.
Recall that one of the charaterizations (or definitions) of a space of curvature bounded from below is as follows.
Definition. A length space X has curvX ≥ k if
(i) for any point of X there is some neighborhood such that for all triangles ∆pqr contained inside, ∠Xpqr exists and ∠Xpqr ≥ ∠Xkpqr
and
(ii) for all geodesics [pqr] and [qs], with q 6= p, r, s, there holds ∠Xpqs + ∠Xsqr = π.
The angle ∠Xpqr = ∠X([qp]X, [qr]X) is defined using parametrized X-geodesics α = [qp]X and β = [qr]X via lim
t,u→0∠α(t)qβ(u). Here ∠α(t)qβ(u) denotes the usual angle in the Euclidean plane between line segments [qα(t)] and [qβ(u)], where |qα(t)| = |qα(t)| and |qβ(u)| = |qβ(u)|
and |α(t)β(u)| = |α(t)β(u)|. Similarly, ∠Xkpqr denotes the angle of the model triangle in the model space of constant curvature k, having the same corresponding sidelengths as the triangle ∆pqr ⊂ X.
Beyond Alexandrov spaces, angles also exist in a broader class of spaces, including polyhedral manifolds (topological manifolds with polyhedral metric). The condition of angles existing is a nontrivial first order condition of the space. This assumption implies that geodesics in X have well-defined directions. On the other hand, angles, in the above sense (sometimes called angles in the strict sense), do not exist in normed vector spaces wherein the norm does not come from an inner product.
Condition (ii), that the sum of adjacent angles equals π, by itself rules out the more extreme types of branching, but not all branching. For example, manifolds-with-boundary satisfy (ii) but in general can have branching. Conditions (i) and (ii) together rule out all branching.
Conversely, if a length space X has no branching and angles are defined, then (ii) holds. For general spaces for which angles are defined, one only has the triangle inequality ∠Xpqs+∠Xsqr ≥ π.
Theorem 2 has two main components. One is quantitative, dealing with condition (i) of the definition of lower curvature bound. The other is more qualitative, and involves condition (ii).
While the qualitative component does not by itself give an effective curvature bound, it has certain elements which could potentially be adapted to more general spaces in order to show that the subset
has no branching. In the context of ambient spaces with a lower Alexandrov curvature bound, it provides a partial answer to the following general open-ended
Question: Given a metric space X and a C-convex subset Z ⊂ X, under what conditions does the subset have no branching?
Necessary conditions on the ambient space X (at least for the proof given) are:
1) Let x, y ∈ X. For any other two distinct points z, v in X each at distance from y, one does not simultaneously have |zx| = + |xy| and |vx| = + |xy|
(This rules out branching in X) 2) angles exist
3) Let x, y ∈ X. For any other three distinct points z, v, w in X each at distance from y, which form angles ∠xyz ≈ π > π/2, ∠xyv ≈ π > π/2, and ∠xyw ≈ π > π/2 at y, the intersection of the three associated pairwise X-equidistants E(v, z), E(w, v), E(z, w) does not contain x.
and
4) first variation formula holds in X
(1) is clearly necessary (see Figure1), and (2) and (4) are natural assumptions.
Perhaps one could call (3) a five-point condition. Unlike the five-point condition for constancy of intrinsic curvature, it is suited for constancy of extrinsic curvature of subsets. It seems to be slightly different than saying that X itself has no branching, since here the point y is not required to lie on a geodesic connecting x with one of the other points (see Fig. 4). Condition (3) rules out tri-branching or trifurcation, but is not equivalent to it. Condition (3) could be considered as a rough analogue of having sectional curvatures bounded from below, along [xy]. Compare with Lemma1 below.
X
x Z y
Figure 1. Z is totally geodesic, but both X and Z have branching.
If one did not require distinctness of v from z and w in (3), or if one additionally assumed angles in X varied continuously, then (3) would imply (1). To see this, take v = w, resp. v → w.
If one did not require angles be close to π in (3) (or at least strictly larger than π/2), then it would be possible for (1) to hold, while (3) did not. See Figure4. It is not known whether (1) and (2) imply (3). CBB spaces satisfy (1)-(4).
Theorem2(i) is proven in the next section, and parts (ii) and (iii) are shown in the second-to- last section. Before discussing Theorem2 in more detail, it seems worthwhile to examine a special case, when the ambient space X is a finite-dimensional Euclidean space.
2.1. Euclidean space. Curves in R3 with constant geodesic curvature need not be geometric circles, but because of torsion can also be helices, or have variable torsion, such as Salkowski curves [23,17]. A classification of curves in R3of constant geodesic curvature (up to rigid motion) depends on torsion. If geodesic curvature and torsion (and also all higher order torsions, for curves in Rn+1) are specified, then the curve is determined up to rigid motion, by the fundamental theorem of submanifold theory.
4
If Z itself is a one-dimensional curve, the conclusion of a lower curvature bound in the Alexan- drov sense for Z is a matter of convention because triangles therein are degenerate. One-dimensional manifolds will in this paper be considered as spaces with curv ≥ 0.
Remark: On the size of ρ.
For smooth submanifolds of a manifold, if ρ is large, having equality always holding in (1) is in general stronger than having constant operator norm of second fundamental form. However, they are equivalent in the smooth manifold setting if ρ is sufficiently small (relative to X and C, say for instance, ρ < min{inj(X),√k+24Cπ }, if the sectional curvatures of X are ≥ k.)
To see why they are non-equivalent for large ρ, let Z be a standard helix x(t) = (a cos(t), a sin(t), bt) on the two-dimensional circular cylinder X in R3 of radius a, for some constants a and b. Z has constant curvature λ = |a|/(a2+ b2).
If b > 0 were strictly positive, and t = 2π say, then dX(x(0), x(t)) = 2πb, whereas dZ(x(0), x(t)) = arclength =Rt
0a2+ b21/2
≥ 2π a2+ b21/2
could be much larger (if a b) so that one does not have dZ ≤ dX+ Cd3X + o(d3X) (assuming C and the other coefficients on the right-hand side in the higher-order terms are fixed in advance, hence bounded).
However, for large ρ in this example, an exception occurs when b = 0, which corresponds to a standard circle. Then Z is totally geodesic in X. In particular, when b = 0 and t = 2π one has dZ(x(0), x(t)) = 0 = dX(x(0), x(t)).
Note also that for the equivalence in the smooth setting, it is necessary for Z to have the induced intrinsic metric, since otherwise one could take an immersion having self-intersections, in which case, |IIZ,→X| ≤ λ would hold, but dZ ≤ dX + Cd3X + o(d3X) would not.
Remark: Another important point relates to the higher order terms in the C-convexity condition.
For example, no matter what C is chosen, a standard round circle S1(1) ⊂ R2 does not satisfy dZ(x, y) = dX(x, y) + Cd3X(x, y) exactly for all close points x, y. If 0 ≤ C < 241, then curves satisfying this would lie off a circle of radius 1 (except at a point of tangency).
Although the higher-than-cubic order terms influence the possibility of global embeddability, the curvature estimate is local, and does not depend on the higher-than-cubic terms. See the remarks on p.13 concerning existence.
The above discussion about curves is summarized in
Lemma. If Z = γ is a constantly C-convex curve in Rn+1 with dZ(x, y) = f (dX(x, y)) for all x, y ∈ Z with dZ(x, y) < ρ, where f is the function which gives arclength of a constantly C-curved circle in R2 having a chordlength of dX(x, y), and γ has length > √π
24C, and ρ is sufficiently large, then γ is either (i) a line (if C = 0) or (ii) a standard round circle
In particular, γ lies in a 2-dimensional subspace R2⊂ Rn+1.
In other words, long curves in Rn+1 with non-zero torsion cannot be constantly (C, ρ)-convex like round circles, if ρ is sufficiently large relative to C (though they can be (C, ρ)-convex).
If C > 0, then the curve, since turning at a definite rate (i.e. having a definite curvature), torsions notwithstanding, would have to eventually turn around by π. If it went past that, but was not extendible beyond some point, then it is plausible that equality would not hold.
The particular choice of f in the above Lemma, which includes specific higher-order terms, is essential for the conclusion.
On the other hand, usually one is interested in only requiring C-convexity locally, so that takes ρ to be small.
Proposition 2. Suppose Z ⊂ Rn+1 (n ≥ 2) is a connected C2 hypersurface (possibly with boundary, but closed as a subset of Rn+1), C > 0, and Z is constantly C-convex.
Then Z is isometric to a convex closed domain in a round sphere Sn(r).
In particular, curvZ = r12 = |II|2 = 24C.
Proof. The equality analogue of property (i) is that |IIZ,→X| = λ if and only if Z is constantly (C = λ242, ρ)-convex.
When n = 1, the first statement of the Proposition is true, since it is known that curves in R2 of constant nonvanishing geodesic curvature are portions of standard round circles. So assume n ≥ 2.
Since |IIZ,→X(u, u)| is independent of the vector u (and the basepoint in Z), Z is a constant λ-isotropic hypersurface, in the sense of [20, 16], by definition. By Theorem 2 of [16], Z ⊂ Rn+1 must be an extrinsic sphere (meaning totally umbilic and with parallel mean curvature vector).
The interior of such an extrinsic sphere is known to be an open subset of either a hyperplane Rn (if C = 0) or a standard round sphere (if C > 0). [24, Lemma 25, Theorem 26, p.73]5
In the latter case, the extrinsic radius of the sphere must be r = 1/√
24C, since this is also the radius of a great circle arc which is totally geodesic in the sphere.
To check that the boundary of Z (when nonempty) is convex, one must note that all curvature (w.r.t. Rn+1) of any geodesic in ∂Z is already being used up to be on the surface of a sphere, so none is left to contribute to geodesic curvature (w.r.t. Z). (Recall that the curvature as a curve in Rn+1 can be decomposed into normal and tangent parts to Z). Hence ∂Z cannot be anywhere
concave (w.r.t. the interior of Z).
6 7
One can prove a weaker statement that curvZ ≥ 0, using essentially only the constant C- convexity condition, and continuity:
Proof. Suppose Z is a smooth embedded hypersurface (without boundary) of Rn+1and C > 0.
The key observation is that one considers all minimal Z-geodesics passing through a given point, one by one. Since they are minimal geodesics in Z, they are individually C-convexly embedded in the ambient space. These arcs have a common tangent plane, under the assumption that Z is a smooth hypersurface. They vary continuously in the Hausdorff topology on Rn+1 and in the C2 topology in a neighborhood of the basepoint, as their endpoint varies.
If Z were a smooth saddle surface with sectional curvature K ≤ 0, then there would exist asymptotic directions. Principal directions correspond to maximal norm of second fundamental form, whereas asymptotic directions have zero second fundamental form. Recall that for smooth hypersurfaces, the second fundamental form varies continuously in the direction u ∈ TxZ.
The corresponding statement using C-convexity is that, if the surface crossed the tangent plane at x, in the sense of having points on both sides of the plane, arbitrarily close to x, then one cannot have dZ = dX + Cd3X + o(d3X) for some C > 0 and dZ = dX + o(d3X) holding simultaneously.
Now distance dZ must be measured in Z. Also, lines of curvatures, while initially tangent to minimal geodesics, are not necessarily themselves geodesics. However, they are approximately close, using points close to x.
Therefore if C > 0, Z must be strictly locally convex in X = Rn+1. In particular, Z then has strictly positive curvature, since it is a smooth hypersurface. Remark: In the above proof, smoothness was invoked. However, it is known that C-convex subsets of Euclidean space are at least C1,1 [15, p.203].
Remark: Definiteness (C > 0, rather than C ≥ 0) was important for the above proof just given, to obtain the contradiction.
If one assumed C = 0, but only looked in a neighborhood of a single fixed point (as the proof above did), then there are examples such as the graph of z = (x2+ y2)2 in R3 which satisfy dZ = dX + o(d3X) in a neighborhood of a point, and thus are relatively flat to high order in the neighborhood, but were not totally geodesic everywhere.
Of course one knows in the smooth case that hypersurfaces which have zero second fundamental form at every point and direction are totally geodesic (equivalently, locally convex in the metric space sense), and therefore have zero sectional curvature. Thus, if one looks in a neighborhood of all points and uses Property (i), then when C = 0, one deduces that Z ⊆ Rn+1 is totally geodesic and curvZ = 0.
However, without invoking smoothness, it is harder to show that if dZ = dX+ o(d3X) holds for a non-smooth subset Z ⊂ X near every point (for X being a more general ambient space), then dZ= dX near every point.
Despite the utility of having positive-definite fundamental form (or C > 0), or being locally convex (in the sense of lying on one side of a supporting hyperplane in X = Rn+1), it is not always needed in order to have a lower curvature bound. There is an example due to Sacksteder of the smooth surface z = x3(1 + y2) (for |y| < 1/2) in R3 with non-negative sectional curvature on a neighborhood of the origin, and non-positive-definite second fundamental form, which is not locally convex, in the sense that it crosses its tangent plane at a point.
Remark: In some works, a sphere or circle in a Riemannian manifold is defined as a submanifold which is totally umbilic and has parallel mean curvature [19].
Circles of submanifolds of space forms (i.e. geodesics in the submanifold which are also round geometric circles in the ambient manifold) have been studied in [2], [1], etc. However, these works assumed the subset was a smooth (C2) submanifold to begin with.
First proof of Theorem 2, in special case X = R2. First consider X = R2. The C- convexity condition precludes isolated branches such as in Figure2. For reference, an X-geodesic is also shown along with the branch.
Figure 2.
Figure 3. The C-convexity condition forces the existence of other Z-geodesics, given an initial pair of Z-geodesics forming a branch or (extrinsic, i.e. w.r.t. X) definite angle
Suppose such a branch existed.
By a limiting argument using midpoints, the C-convexity equality condition and Cauchy- completeness of Z, this yields the existence of a Z-geodesic which does not satisfy the constant C-convexity equality property (for its endpoints). See the last figure in Figure 3. Z includes at least the shaded region. In the case of X = R2, one can take the geodesic ending on a point in the middle of Z-geodesic side indicated. That the geodesic does not satisfy the constant C-convexity equality is clear.
(For X = Rn, the only way a collection of Z-geodesics from the point x to points on the third side could all satisfy the constant extrinsic curvature condition is if they would comprise part of a sphere. However, no open subset of a two-dimensional sphere isometrically embeds into R2.)
Hence no Z-geodesic can branch. More exactly, one cannot have ∠Z(α, β) = π and ∠X(α, β) = 0 for two Z-geodesics α and β. But the same argument shows that no two Z-geodesics can meet and make definite angles ∠Z(α, β) = π and 0 < ∠X(α, β) < π either. Thus either ∠Z(α, β) = π and
∠X(α, β) = π or both are strictly < π, for any two distinct Z-geodesics emanating from a common point.
In the former case, Z is locally one- or two-dimensional, and in the latter case, locally two- dimensional (possibly with non-empty (and then necessarily convex) frontier).
In any case, curvZ ≥ 0.
For more general, even two-dimensional, spaces X producing the requisite Z-geodesic which does not satisfy the constant C-convexity equality, as in the above outline for X = R2, is more difficult.
For another difficulty, two different C-convex Z-segments in X need not be interchangeable via an isometry of X. For example, consider a spherical lune bounded by two geodesic arcs emanating from a point x, along which are isometrically attached two intrinsically flat rectangles via those edges. This can be realized in R3 as X = ∂N , where N := {x ∈ R3 : d(x, [0, 1] × [0, 1]) < } is a neighborhood of a planar square. Then curvX ≥ 0. If now one considers Z to have two geodesics emanating from x, each of which lies entirely in a respective rectangular piece and making a con- stant turn in X, together with the domain inbetween, then there are many geodesics of Z which are constantly curved. In this example, however, one can find a Z-geodesic starting from x lying in one of the intrinsically flat rectangular pieces, which is not constantly curved.
Proof of Theorem 2(i). Part (i) follows from part (ii), since having a finite lower curvature bound implies no branching, and Z, being closed in locally compact X, is locally compact.
In the definition of constantly C-convex, the coefficients of the higher order terms are allowed to depend not only on the distance, but also on the particular points involved. In other words, the coefficients in the o(d3X) term can vary from point to point, so that one may have for example
o(|yz|3X) 6= o(|vw|3X) even if |yz|X = |vw|X, or
o(|yz|3X) 6= o(|yw|3X) even if |yz|X = |yw|X.
However, an independent proof of part (i) will be given under the additional assumption that for any points y, z, w ∈ Z close to each other, |yz|X = |yw|X ⇐⇒ the respective higher (*)
order terms in the definition of constantly C-convexity satisfy o(|yz|3X) = o(|yw|3X).
Note that a priori this still allows the coefficients of the higher order terms to vary in y.8
A Second proof of Theorem 2(i), under (*). Z has no branching, as follows. Suppose otherwise.
Step 1: Given constant curvature curves [xyz]Z and [xyw]Z coinciding from x to y, and branching at y, it can be assumed, by restriction, that |yz|Z = |yw|Z and z 6= w. Then |xyz|Z =
|xyw|Z ⇐⇒ |yz|Z = |yw|Z since the segment from x to y is shared ⇐⇒ |yz|X = |yw|X by the constant curvature hypothesis and (*).
On the other hand, again by the constant curvature hypothesis and (*), |xyz|Z = |xyw|Z ⇐⇒
|xz|X = |xw|X. Thus x, y, z, w ∈ Z must be such that x and y are equidistant from w and z (w.r.t.
the metric of X). Likewise, every point in the segment [xy]Z is equidistant from w and z (w.r.t.
the metric of X).
Thus Z-equidistants are X-equidistants.
Step 2: By the same argument as in the proof of the case of Theorem 2 above, (c.f. Propo- sition 4, 5) using the constant extrinsic curvature assumption, one can find a filling near y which spans the hypothetical branching geodesics. For sufficiently small radius > 0, there are infinitely many points in this filling which are at distance from y.
In particular, there are at least n + 1 distinct points of Z all -equidistant from y, where n = dimX (the dimension n is finite since X is locally compact and has a lower curvature bound).
By step 1, any point of [xy]Z must be r-equidistant to each of these n + 1 points, for some r depending on the point.
However, according to Lemma1below, the set of points equidistant (for any r) to all the n + 1 points must be finite or discrete. Therefore [xy]Z cannot lie entirely in the equidistant. Hence there is no branching in Z.
Lemma 1 (Triangulation-type lemma). Suppose curvX ≥ k and dimX = n. Let {xi}Ni=1 be N distinct points in X (in general position, generic with respect to the a.e. defined Riemannian structure [21]).
If N = n + 1, then the set of all points equidistant from all the x0is (if nonempty) is a finite or discrete set.
Here equidistant means r-equidistant for some (any) r > 0,
Proof. Suppose |pxi| = r for all i = 1, . . . , N , for some r > 0. Join p to xi with a minimal geodesic, for i = 1, . . . , N . The set of vectors [pxi]0 in the tangent cone TxX then consists of N distinct vectors in general position, since curvX is bounded from below.
Since N = n + 1, it is impossible to have ∠X(γ, [pxi]) = π2 for all i, or ∠X(γ, [pxi]) = π for all i, where γ is a curve in the equidistant set.
The former corresponds to staying in the r-level (equidistant) set, and the latter corresponds to remaining equidistant, but increasing the common distances r.
Therefore the equidistant set to the given set of points {xi} must be finite or discrete. Remark: Here is a perhaps more axiomatic alternative approach to show lack of branching in the subset Z. Instead of taking n + 1 points -equidistant from y initially, one can produce a finite sequence of at most n + 1 triples (z, v, w) of points getting sufficiently close to y, and look at the equidistant E := E(w, v) ∩ E(w, z) ∩ E(z, v) = {x ∈ X : |xw| = |xz| = |xv|} where z and w lie in the branches and v is third point distinct from w and z which lies in the filling.
At first, E may contain all of [xy]Z, as shown in Figure 4.
Restricting lengths if necessary (by rechoosing points closer to y) one can produce a new point v0 together with points z0 and w0 on the original branches, such that z0, v0, w0 are distinct, equidistant to y, and such that either together with y and x they satisfy the five-point condition, or the associated E0 = E(w0, v0) ∩ E(w0, z0) ∩ E(z0v0) is distinct from E as a set. One can repeat this at most n more times if necessary.
One can informally see why three points are necessary as follows. Suppose X = Rn. Suppose equidistant E were defined by only two distinct points. The retraction as above is analogous to
x y
z v w
z0 v0 w0 E E0
Figure 4. The intersection of all equidistants E, E0, . . ., is the supposed branch point y
providing a one-parameter rotation of the E. If dimE = n − 1, then a rotation may send E onto itself, so that E ∩ E0 is not lower dimensional. On the other hand, having three distinct points, dimE = n − 2. A rotation of such dimensional E will produce E0 such that E ∩ E0 is (n − 3)- dimensional, E ∩ E0∩ E00 is (n − 4)-dimensional, etc. Eventually one reaches a zero-dimensional intersection, which cannot contain all of [xy]Z.
2.2. Model space. For round spheres Sn(r) in Rn+1, there are two known methods for proving a lower curvature bound. One is the Gauss equations, which depend on the surface being C2 intrinsically. The second method involves submersions. Considering Rn+1\{0} ≡
isomSn(1)×φ(0, ∞), where φ(r) = r, projection to the fiber Sn(r) ≡ Sn(1) ×φ{r} yields a lower bound on its curvature.
But strictly speaking, nearest-point projections in a local neighborhood of general subsets of a metric space do not a priori exist9, so one could ask whether there is another way to prove a lower curvature bound.
In the case of round spheres in Rn+1, or constantly C-convex subsets in Alexandrov spaces of curvature bounded below (satisfying additional regularity assumptions), there is such as method, which is used in Theorem 2, and can be roughly described as follows.
If one considers the model triangles for X in the model space for X, and a modified triangle for Z (whose sidelengths are equal to those in the Z triangle), the vertices can be made to overlap. See Figure5. Here, the sides corresponding to Z-geodesics are circular arcs (whose radius is completely determined by C and k).
Straightening the sides of this non-geodesic triangle, one may have ∠Xk less than or greater than
∠Zk. But by Lemma3, one can find a new model space (typically with very negative curvature k0) to re-accommodate the straightened triangle, so that ∠Zk0 ≤ ∠Xk . These two operations are combined in Lemma 4.
Finally, one can use the quadruple criterion, involving sum of three model angles being less than 2π, to obtain a lower curvature bound for Z.
Figure 5.
2.3. Existence. Uniqueness of Z geodesics starting in a given direction is addressed below on branching.
Remarks on existence:
The question of global existence of subspaces satisfying the hypotheses of Theorem2seems to have a negative answer in general, if one also demands Z be geodesically extendible and Z be compact and codimension-one.
In the manifold case, if one considers smooth metrics on R2 and takes C = 0 in the above, then one can find a metric with positive sectional curvature, so that there is no closed geodesic (Klingenberg). Likewise, many other manifolds admit no closed, totally geodesic hypersurface.
If X is a manifold, then it is known that for any given point x ∈ X, an immersed curve (possibily noncompact) can locally be found which passes through x and has constant geodesic curvature. In two-dimensional surfaces, geodesically extendible Z will typically not be compact or embedded (in the sense of having the induced intrinsic metric) unless C is large.
However, there are many positive examples of existence. Constant extrinsic curvature naturally arises as an extremal in many variational setups. For one of numerous representives in the literature, one could see for example [22] and its variational technique for existence of embedded small- constant curvature curves on smooth, strictly convex two-dimensional spheres. In the symmetric space setting, G-equivariant isometric immersions of rank one symmetric spaces G/K into arbitrary Riemannian homogeneous spaces eG/ eK provide other examples of constant C-convexity (see [18]
for some of these).
In order to further study extrinsic curvature, it is worthwhile to assume it is merely bounded (i.e. to allow the C-convexity inequality), but assume additional hypotheses. The notion of ex- tendibility is considered in §3.
3. Branching and extendability
Definition. A point z ∈ Z is a branch point of Z if there exist some > 0 and two unit-speed Z-geodesics γ1 and γ2: (−, ) −→ Z such that γ1(t) = γ2(t) for all t ∈ (−, 0], γ1(0) = γ2(0) = z, and γ1|(0,) and γ2|(0,) are (non-trivial) disjoint subsegments.
For manifolds with C2 smooth metric, geodesics are unique (do not branch), by uniqueness of solutions to the corresponding differential equations. This is not necessarily guaranteed if the met- rics are not sufficiently smooth. Alexandrov spaces of curvature bounded below have no branching.
Example 1. Consider a round circle circumscribed in a two-dimensional planar convex polyg- onal domain. Let X be the double of this polygonal region, and Z be the union of the two corre- sponding circles. Z has branch points at places where it is tangent to the polygonal ridge.
One can truncate the corners, so that the polygonal perimeter acquires a countable dense set of vertices.
The assumption that Z is a closed subset of this limit entails that points of Z with zero X- distance are identified. Thus, in such a limit pair, Z is considered to have no branching.
The tangent cone TxX exists and is the pointed Gromov-Hausdorff limit of a sequence of scaled- up X. It is possible to define TxZ for any x ∈ Z. By passing to the limit,
Lemma 2. (i) TxZ is a convex subset of TxX.
(ii) Z has a terminal at x iff TxZ has a terminal at the origin
Hence if TxX is a cone over a C1 smooth manifold M and Z is extendible, then TxZ is a cone over a (closed) submanifold of M .
Whenever defined, one always has ∠X ≤ ∠Z on pairs of Z-geodesics.
Proposition 3. Suppose curvX ≥ k is locally compact and Z ⊆ X is a (C, ρ)-convex subset.
(i) If Z has branching, then it can only have 0-extrinsic branching
(meaning ∠X(γ1, γ2) = 0 for any two Z-geodesics γ1, γ2 which form a branch).
(ii) ∠Z = ∠X for any pair of Z-geodesics. Thus angles in Z are defined.
Proof. Part (i) is due to Lemma2(i).
Part (ii) follows from [25, Prop.3].
Part (ii) entails that if γ is a Z-geodesic, one can define γ0, as an element of TxX.
As a corollary of Proposition 3, angles in Z satisfy the triangle inequality, since the triangle inequality holds in T X.
Definition. Let > 0. A unit-speed geodesic segment γ : [a, b] −→ Z is -extendible if there exists a unit-speed geodesic eγ : [a, b + ] −→ Z such that eγ(t) = γ(t) for all t ∈ [a, b].
Z is uniformly extendible if all its geodesic segments are -extendible for some > 0 independent of the geodesic.
Subsets which are geodesically extendible are closed subsets (Cauchy-complete) by Hopf-Rinow.
Geodesic extendibility gives rise to what essentially amounts to an (nonpointwise, integral) upper curvature bound, since long segments do not minimize distance, in space with curvX ≥ k > 0.
It is known that spaces Z with bilaterally bounded curvature (−k ≤ curvZ ≤ k for some k > 0) admit a C0 Riemannian metric tensor with associated C1-differentiable manifold structure [9].
Also, geodesically extendible Alexandrov spaces of curvature bounded from below are C1 man- ifolds (precisely, have C0 metric tensors and associated C1 differentiable structures) [7].
On the other hand, if curvZ ≤ k and Z is a homology manifold (meaning Hm(Z, Z −{x}; Z) ∼= Z for all x ∈ Z for some m), then Z is geodesically extendible [10, Prop. 5.12] (respectively, uniformly extendible, if the CAT (k) radius of Z has a uniform lower bound).
Recall that there are spaces which are geodesically extendible, but do not have curv ≤ k.
Recall that convex surfaces may have a dense set of vertices. The complement of the singular set S = {x ∈ X : TxX 6≡ Rn}, where TxX is the tangent cone at x of X, is open, and a geodesically convex subset in the sense that if x, y ∈ X \ S, then (all, minimal) geodesics from x to y lie entirely in X \ S (see [21])
Example 2. If X = cone × [0, 1] is a metric product of a singular cone having a definite vertex angle with an interval, then Z could lie in the singular subset {v} × [0, 1] of X, where v is the vertex of the cone. This is a convex subset of X (C = 0).
However, Z-geodesics cannot cross or pass through X-terminals, in the same way as that X- geodesics do not pass through X-geodesic terminals. This can be seen from Proposition 3.
Let us now consider the absence of terminals, i.e., the condition of extendibility.
Proposition 4. If Z ⊆ Rn is (C, ρ)-convex, uniformly extendible then Z has no branching.
This follows from [15, Proposition 1.4]. In an earlier preprint, he proved that when Rn is replaced by any manifold M with bilaterally bounded curvature K− ≤ KM ≤ K+, a geodesically extendible (C, ρ)-convex subset satisfies c−(K−, C) ≤ curvZ ≤ c+(K+, C) for some constants c− and c+. Having a lower curvature bound for Z is stronger than Proposition4’s conclusion, although the proof used structure afforded by upper curvature bounds.
Alternatively a somewhat different method can also be used, by replacing the CAT (k) bound assumption there with uniform extendible assumption. Some ideas in the proof below are similar to Theorem2, and parts of [15, Proposition 1.4], but there is less (albeit still some) dependency on upper curvature bounds. The proof is similar to an open-closed argument, in which C-convexity yields closedness and extendibility yields openness.
Proof of Proposition 4. Assume by way of contradiction that there existed a branch.
Assume as usual that all geodesic parametrized by unit speed.
Given minimal Z geodesics α and β which form a branch at point y ∈ Z, and x ∈ α, z ∈ β, there must exist a Z-geodesic [xz]Z distinct from α ∪ β, since the curve α ∪ β is not isometrically C-convexly embedded.
[xz]Z is not necessarily unique, but one can choose one. Because of the C-convexity, if t is chosen sufficiently small, then [α(t)β(t)]Z intersects both sides α and β in an angle ∠X = π2± τ (t), where τ (t) denotes a quantity tending to 0 as t tends to 0. Using the fact that the ambient space is Rn, one can see this by noting that the osculating circles to the curves α and β at the branch point approximate them arbitrarily closely, when t is sufficiently small. Their radii is definable in terms of the C1,1 norm of Z-geodesics, which is defined and bounded from above [15, Theorem 1.2].
Then, since the geodesics are unit speed and the distances to their common point are the same, an upper bound to the angle of intersection would occur when the osculating circles were coplanar and lying on opposite sides of a line in that plane, but in this case one obtains π2+ τ (t). Since Z is uniformly extendible, there exists an r > 0 such that [α(t)β(t)]Z can be extended r units distance on both sides.
Figure 6. Given an initial pair of Z-geodesics forming a branch, the C-convexity condition forces the existence of other Z-geodesics.
Consider points α(t1) and β(t1). The angles between the Z-geodesic sides α ∪ β ∪ [xz]Z are all
< π. C-convexity forces there to be a geodesic connecting α(t1) and β(t1), but this curve cannot be α|[0,t1]∪ β[0,t1]nor α|[t1,t]∪ [xz]Z∪ β|[t1,t]because the X and Z angles are not equal (cf. Prop.3).
Therefore there exists a distinct geodesic segment [α(t1)β(t1)]. This segment can also be ex- tended uniformly past its endpoints.
In fact one can repeat this for a dense set of t’s, since the angle estimate π2+ τ (t) only becomes closer to π/2 on smaller scales.
Subdividing the quadrilaterals yields triangles, and then one can continue this subdivision indefinitely (concretely, using for example, midpoints, which yields smaller-perimeter triangles).
The Cauchy completion (w.r.t. metric of X) of the limiting result mesh contains a patch (at least two-dimensional) containing α and β. More precisely, a topological submanifold, C1,1 in its interior. This is due in part because each geodesic comprising it is C1,1, and in part because geodesics are extendible by assumption.
Recall that Z was assumed closed subset.
Because the ambient space is Rn, either α or β cannot then be a minimal geodesic in Z, because the interiors of α and β lie in the interior Z, and there is a shorter curve joining endpoints.
This is clear when n = 2. When n > 2 however, Z could be a curved surface of some dimension, such as part of a sphere, for example.
Consider the segment [α(t)β(t)]Z. For small t, this is close to [α(t)β(t)]Rn. Their extensions in respective spaces are also close.
But the extended Rn segments, when projected to osculating plane, have a vector projection which lies on the ”inward” side of the osculating circles, for sufficiently small t. So some subsegment of either α or β could be shortened, which contradicts the assumption that α and β are minimizing.
α(t)
β(t)
Figure 7. Two views of fixed geodesics α and β forming part of a branch in Z ⊂ Rn. A transversal Z-geodesic from α(t) to β(t), when extended, must lie in representative shaded part. As t → 0, the angle between the transversal segment and α or β tends to π2.
Remark: Proposition 4 is purely local, and holds when Rn is replaced by a general Riemannian manifold M , since manifolds are locally Euclidean (i.e. C∞-close on small neighborhoods).
Since Rn is locally compact and Z ⊆ Rn is a closed subset, Z is locally compact. Thus if it is extendible, it is locally uniformly extendible. Combining this with Proposition 4, one obtains that Z has no branching. Hence one has the following
Corollary 1. If Z ⊆ Rn is a (C, ρ)-convex Busemann geodesic space, then Z is a C1 sub- manifold of Rn (without boundary).
By definition (see [11]) a Busemann geodesic space is an abstract locally compact Menger- convex (thereby geodesic) metric space for which geodesics are extendible and do not branch. It is known that such a space must be a topological manifold if its Hausdorff dimension is 1, 2, 3, or 4 (see [8] for an overview and the references therein).
One can obtain the Corollary1in a different way, using a main theorem of [14, 4], that (C, ρ)- convex subsets of Rn inherit an upper curvature bound, together with a result of Berestovski˘ı, according to which curvZ ≤ k and Z being a Busemann geodesic space implies that Z has the structure of a C1 submanifold.
The result of Proposition 4 is valid when the ambient space is a two-dimensional Alexandrov space of curvature bounded below:
Proposition 5. If Z ⊆ X is (C, ρ)-convex, uniformly extendible, curvX ≥ k, dimX = 2, then Z has no branching.
At first glance, one apparent possible difficulty, which was not present in Proposition4, is that a supposed branch point could be the limit of a sequence of other branch points. See Figure 8 below. In other words, the angle estimate π/2 + τ (t) might not hold. Actually, in the example shown in Figure 8, that limit is not a branch point according to the definition, since the nontrivial geodesics σ1 and σ2 do not have disjoint interiors. On the other hand, one can see that in this example that Z is not (C, ρ)-convex.
A second apparent possible difficulty is that (C, ρ)-convex Z-geodesics in X are perhaps not readily seen to be C1 or C1,1, unlike the case of (C, ρ)-convex Z-geodesics in Rn.
...
Figure 8. curvX ≥ 0, yet the connecting Z-segments do not intersect the circles transversely, no matter how small their length
One proof of Proposition5uses Proposition4 and the fact that the pair (X, Z) can be approx- imated by smooth Riemannian manifold with subset pairs (Xi, Zi) for which Xi converges to X, Zi converges to Z, Zi are (C, ρ)-convex, and Zi have the structure of C1 submanifolds. This is possible since X is two-dimensional. Again by approximation, one can replace these pairs by C∞ pairs (Xi0, Zi0), for which curvZi is uniformly bounded from below. Then curvZ ≥ constant (see Proposition1).
However, it is preferable to give a different proof which is more intrinsic, not relying on approx- imation from outside.
Proof of Proposition 5. Starting from a supposed branch, then exactly as in Proposition4, one can still obtain a dense net of Z geodesics in X, which spans the branches of the branch.
Denseness follows from two-dimensionality of X.
Since the geodesic α may turn quickly in X and actually overlap or coincide with all or part of β as a set when it comes around, it is necessary in this case at each step to choose t at most halfway between 0 and T , where T is a point at which α becomes β, since otherwise [α(T /2)α(T )β(T /2)]Z
could provide a minimal Z-segment joining α(T /2) with β(T /2). This ensures that a Z-geodesic distinct from α|[0,T /2]∪ β[0,T /2]and α|[T /2,T ]∪ β|[T /2,T ] (written as sets) exists. In this case, however, branches (∠Z = 0) or angles (∠Z > 0) appear in two new points, namely α(T /2) and β(T /2).
Therefore one can repeat the construction of a net of geodesics in Z spanning the original branching pair of geodesics, and the result will be sufficiently dense so that the result, by closedness of Z in X, is a two-dimensional C1 surface, namely X itself. 3.1. Angles. This section consists of several auxiliary results concerning angles, which are used elsewhere in the paper, but which themselves have some interest.
If the model space is changed from a Euclidean space to a strongly negatively curved hyperbolic space, any triangle in X will have a definite excess in comparison (refering to each individual vertex angle excess ∠X − ∠Xk).
Lemma 3 (change of model space (curvature)). Suppose k0≤ k and ∆xyz ⊂ M2(k) is arbitrary.
Then
0 ≤ ∠kxyz − ∠k0xyz ≤ c(k, k0, R)
where R depends on the area of a Euclidean triangle with the same sidelengths as ∆xyz, and on the reciprocal of the product of adjacent sidelengths, namely |xy||yz|.
Proof. The first inequality always holds, since the curvature of the model space M2(k) is greater than that of M2(k0).
The term on the right-hand side can depend on the aspect ratio of the triangle, so is not uniformly depending on the largest sidelength. By Sublemma 1, each of the model space angles
∠kxyz and ∠k0xyz can first be compared to Euclidean angles, and then cos θ − cos θ0 = (k − k0) · a4+ b4+ c4− 2a2b2− 2a2c2− 2b2c2
24ab + (k2− k02) · O(6) gives the bound c(k, k0, R) for θ − θ0 =
∠kxyz − ∠k0xyz.
Let a = |xy|X, b = |xz|X, c = |yz|X.
If k > 0 and k0= 0, a special case of Lemma3 is Sublemma 1.
cos θ = cos(√
k c) − cos(√
k a) cos(√ k b) sin(√
k a) sin(√ k b)
= a2+ b2− c2
2ab + 1
24 1 ab
k a4+ b4+ c4− 2a2b2− 2a2c2− 2b2c2
| {z }
≤0
+ k2O(6)
Proof. The underlined term equals a2(a2− b2− c2) + b2(−a2+ b2− c2) + c2(−a2− b2+ c2)
= −(a + b + c)(−a + b + c)(a − b + c)(a + b − c) or −16Area(∆0xyz)2 by Heron, so it must be non-positive, by the triangle inequality.
From this, if k ≥ 0, then cos θ ≤ cos θ0, or θ ≥ θ0, where θ0 is the comparison angle of the
triangle in R2.
When the model angle is π (c = a + b), it remains so after changing the curvature.
For a, b, c fixed, θ is monotone increasing in k.
On the other hand, if k is fixed and all sides change by the same multiplicative factor, then of course this is merely a scaling, which could be incorporated instead into k. Under the transformation (a, b, c) 7→ (ta, tb, tc), t ≥ 1, θ increases.
Still fixing the model space and its curvature k, one might ask if there are other functions such that the angle should increase if each sidelength is increased by the same function. For given sidelengths one can sometimes find such functions, though rarely do they have the property for all sufficiently small sidelengths. By analogy with smooth manifolds, it seems that, in general spaces, global scalings are the only such functions. Under the transformation (a, b, c) 7→ (f (a), f (b), f (c)), where f (x) = x + Cx3+ o(x3), the angle θ is not monotone for all feasible initial triples (a, b, c). 10 However, the observation to be made is that the combination of transformations k 7→ k − mC for the curvature and f (x) = x + Cx3+ o(x3) for the sidelengths provides monotonicity of model angles, restricting to certain triples (a, b, c). More precisely,
Lemma 4. Suppose k ∈ R, curvX ≥ k, C ≥ 0 and Z ⊆ X is constantly C-convex
(meaning equality holds in (1), i.e., dZ(x, y) = f (dX(x, y)), for all points x, y ∈ Z sufficiently close, where f (x) = x + Cx3+ o(x3)).
If k0 = k (respectively k0 = k − 12C, k0 = k − 48C), then
∠Zk0xyz ≤ ∠Xkxyz
for all x, y, z ∈ Z with x, z in a sufficiently small neighborhood of y such that |xz|2X ≤ 12 |yx|2X + |yz|2X (respectively such that |xz|2X ≤ |yx|2X + |yz|2X, |xz|2X ≤ 34(|yx|X + |yz|X)2 ).
Proof. In what follows below, assume for notation’s sake that k, k0 > 0. When one allows k < 0 or k0 < 0, sin and cos would need to be replaced by sinh and cosh counterparts, respectively, but the analysis is similar (or one could cite analytic continuation of the generalized trigonometric functions).
Suppose f (x) = x + Cx3+ o(x3) and k0 = k − mC, where m is a constant to be determined soon.
Let a = |yx|X, b = |yz|X, c = |xz|X. Corresponding lengths in the subset Z are f (a), f (b), and f (c). The model angles in respective model spaces are determined by
cos
∠Xk xyz
= cos(√
k c) − cos(√
k a) cos(√ k b) sin(√
k a) sin(√ k b)
cos
∠Zk0xyz
=
cos√
k0 (c + Cc3+ o(c3))
− cos√
k0 (a + Ca3+ o(a3))
cos√
k0 (b + Cb3+ o(b3)) sin√
k0 (a + Ca3+ o(a3)) sin√
k0 (b + Cb3+ o(b3))
Analytic expansion in a, b, c, treating k, k0, and C as parameters, yields
cos
∠Zk0xyz
− cos
∠Xkxyz
= 1
2ab· 1
12k0− 1 12k + C
a4+ 1 6k −1
6k0− 2C
a2b2+ 1 6k − 1
6k0+ C
a2c2 + 1
12k0− 1 12k + C
b4+ 1 6k −1
6k0+ C
b2c2+
−1 12k + 1
12k0− 2C
c4 + O(6)
Taking k0 = k − 12C,
cos
∠Zk0xyz
− cos
∠Xkxyz
= C
2ab·3(a2+ b2− c2)c2+ O(6) If k0 = k − mC, then
cos
∠Zk0xyz
− cos
∠Xkxyz
= C ab· 1
2− m 24
a4+
−1 + m 12
a2b2+ 1 2 + m
12
a2c2 + 1
2 − m 24
b4+ 1 2+ m
12
b2c2+
−1 − m 24
c4 +1
8(k − mC) a6+1
8(−k + mC) a4b2+
−1 2C + 1
12(−k + mC)
a4c2 + 1
8(−k + mC) a2b4+
−1 2C +1
6(k − mC)
a2b2c2 +
C + 5
24(−k + mC)
a2c4+1
8(k − mC) b6+
−1 2C + 1
12(−k + mC)
b4c2 +
C + 5
24(−k + mC)
b2c4+
−1 2C + 1
6(k − mC)
c6 + 1
360
k − mC 2
m −3a6+ 3a4b2+ 7a4c2+ 3a2b4+ 10a2b2c2− 5a2c4− 3b6 +7b4c2− 5b2c4+ c6 +O(8)
(2)
where the ”O(8)” term contains monomials of at least 8th total degree in a, b, c, as well as possibly also powers of C, m, and/or k.
Upon rewriting and keeping the lowest order terms, this is
∠Xkxyz − ∠Zk0xyz = 1
ab· C · 1 2− m
24
a2− b22
+ 1 2 + m
12
(a2+ b2)c2− 1 + m
24
c4+ O(6)
(3)
and the low order part inside brackets is always strictly positive if (12 −24m) ≥ 0 (i.e, m ≤ 12) and
1 2 + m
12
(a2+ b2) − 1 + m
24
c2> 0 (4)
Hence
