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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 ofmetric spaces

Jeremy Wong

April 30, 2013

Bounded extrinsic curvature of subsets of metric spaces

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+1 or dimX = 2,then it has no geodesic branching.

If the subset has constant extrinsic curvature, then the subset has no branching, and has itselfa 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 ingeneral.

2010 Mathematics Subject Classification. Primary 53C70 Secondary 53C20 53C24 51K10 53B2553A04Keywords: 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)-convexif 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, a0-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 orderterms. 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, 2013work partially supported by NSF stipend, NSC grant 101-2115-M-002 -001.

c©2013

1

2 JEREMY WONG

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 pointsx, 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 analternative definition for the special case of hypersurfaces might be possible). Just as the definitionof Alexandrov curvature, it is defined locally, rather than infinitesimally, as the shape operator ofa smooth submanifold would be.

This extrinsic notion of curvature bound for a subset is related to other notions in variouscontexts: a pointwise bound on the norm |IIZ↪→X | of second fundamental form, positive reach, andλ-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 (λ2

24 , 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 unionsof sets which are closed manifolds or manifolds with C1 boundary of various dimensions which arejoined along codimension ≥ 1 subsets of their boundaries. A (C, ρ)-convex subset need not havepositive 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 ofsecond fundamental form which are preserved in taking limits. For general metric spaces, one has

Proposition 1. Suppose XiGH−→ 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, ZiGH−→ 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. Example 1 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 aλ-concave function f with λ < 0, in terms of given gradient bounds on the f . The lower boundequality case was also characterized in [5]. Similarly, at least for boundaries ∂X of CBB spacesX, a related notion of base-angle extrinsic curvature was shown to lead to comparison results withballs in standard model spaces.

It is sensible to consider an opposite inequality of the form dX + Cd3X + o(d3X) ≤ dZ (whatone might call ”strictly positively convex”), which, in the special case of submanifolds, is basicallyequivalent to the uniformly positive condition that IIZ↪→X ≥ const ≥ 0. The latter notationmeans that all principal curvatures are uniformly bounded from below. Informally, one can see

BOUNDED EXTRINSIC CURVATURE OF SUBSETS OF METRIC SPACES 3

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 continuityan asymptotic direction giving rise (at least in manifolds) to an asymptotic curve. Such a curvewould be a geodesic (or close to a geodesic, at small scale) in the ambient space, and contradictthe ”strictly positively convex” hypothesis.

For C-convexity, one can see the variation [0, Cd3X + o(d3X)] in the difference dZ − dX byconsidering Z-geodesics in a non-convex domain. Either a geodesic segment lies entirely in theinterior of the domain (in which case effectively dZ = dX), or its interior intersects the boundaryof Z along a non-trivial portion, where the boundary has some concavity. By continuity, thereis a whole family of Z-geodesics whose endpoints interpolate between these two extreme types ofgeodesics, 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 possiblebifurcations along the set of geodesic terminals (or boundary). Also, using a conformal stereographicprojection, 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 stillcan produce branching. Thus, being strictly positively C-convex (or having a two-sided pinchingC1d

3X ≤ dZ − dX ≤ C2d

3X 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 Theorem 2 and Propositions 4 and 5. 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 abovedefinition 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 metricdZ , 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 theinduced 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.

4 JEREMY WONG

Note that curvZ ≥ k holds for a general constantly C-convex subset when C = 0 (and thehigher 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 hada two-sided bound |IIZ↪→X | ≤ λ (with λ2 = 24C) for the second fundamental form, then a lowerbound to the sectional curvature of Z would be k − 24C, by the Gauss equations, but in highercodimensions 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 boundedfrom below by k, then Z would have positive semi-definite second fundamental form, and a lowersectional curvature bound ≥ k + 24C.

The lower bound of k − 48C in (iii), which does not require the ambient space X to be amanifold, is sharp for general codimensions.

Recall that one of the charaterizations (or definitions) of a space of curvature bounded frombelow 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 ≥ ∠Xk pqrand(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]Xand 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, ∠

Xk pqr 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 anontrivial first order condition of the space. This assumption implies that geodesics in X havewell-defined directions. On the other hand, angles, in the above sense (sometimes called angles inthe strict sense), do not exist in normed vector spaces wherein the norm does not come from aninner product.

Condition (ii), that the sum of adjacent angles equals π, by itself rules out the more extremetypes of branching, but not all branching. For example, manifolds-with-boundary satisfy (ii) butin 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. Forgeneral 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 thedefinition 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 certainelements which could potentially be adapted to more general spaces in order to show that the subset

BOUNDED EXTRINSIC CURVATURE OF SUBSETS OF METRIC SPACES 5

has no branching. In the context of ambient spaces with a lower Alexandrov curvature bound, itprovides 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 doesthe 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 ε distance from y, one does

not simultaneously have |zx| = ε+ |xy| and |vx| = ε+ |xy|(This rules out branching in X)2) angles exist3) Let x, y ∈ X. For any other three distinct points z, v, w in X each at ε distance from y,

which form angles ∠xyz ≈ π > π/2, ∠xyv ≈ π > π/2, and ∠xyw ≈ π > π/2 at y, the intersectionof the three associated pairwise X-equidistants E(v, z), E(w, v), E(z, w) does not contain x.

and4) first variation formula holds in X

(1) is clearly necessary (see Figure 1), 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 beslightly different than saying that X itself has no branching, since here the point y is not requiredto lie on a geodesic connecting x with one of the other points (see Fig. 4). Condition (3) rulesout tri-branching or trifurcation, but is not equivalent to it. Condition (3) could be considered asa rough analogue of having sectional curvatures bounded from below, along [xy]. Compare withLemma 1 below.

X

x yZ

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 assumedangles in X varied continuously, then (3) would imply (1). To see this, take v = w, resp. v → w.

6 JEREMY WONG

If one did not require angles be close to π in (3) (or at least strictly larger than π/2), then itwould be possible for (1) to hold, while (3) did not. See Figure 4. It is not known whether (1) and(2) imply (3). CBB spaces satisfy (1)-(4).

Theorem 2(i) is proven in the next section, and parts (ii) and (iii) are shown in the second-to-last section. Before discussing Theorem 2 in more detail, it seems worthwhile to examine a specialcase, 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 geometriccircles, but because of torsion can also be helices, or have variable torsion, such as Salkowski curves[23, 17]. A classification of curves in R3 of constant geodesic curvature (up to rigid motion) dependson 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 ofsubmanifold 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-dimensionalmanifolds 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, theyare equivalent in the smooth manifold setting if ρ is sufficiently small (relative to X and C, say forinstance, ρ < min{inj(X), π√

k+24C}, 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 hasconstant 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 =∫ t0

[a2 + b2

]1/2 ≥ 2π(a2 + b2

)1/2could 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 thehigher-order terms are fixed in advance, hence bounded).

However, for large ρ in this example, an exception occurs when b = 0, which corresponds to astandard circle. Then Z is totally geodesic in X. In particular, when b = 0 and t = 2π one hasdZ(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 theinduced intrinsic metric, since otherwise one could take an immersion having self-intersections, inwhich 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 satisfydZ(x, y) = dX(x, y) + Cd3X(x, y) exactly for all close points x, y. If 0 ≤ C < 1

24 , then curvessatisfying this would lie off a circle of radius 1 (except at a point of tangency).

BOUNDED EXTRINSIC CURVATURE OF SUBSETS OF METRIC SPACES 7

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 theremarks 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 allx, y ∈ Z with dZ(x, y) < ρ, where f is the function which gives arclength of a constantly C-curvedcircle 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 circleIn 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)-convexlike 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 wasnot 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, isessential for the conclusion.

On the other hand, usually one is interested in only requiring C-convexity locally, so that takesρ to be small.

Proposition 2. Suppose Z ⊂ Rn+1 (n ≥ 2) is a connected C2 hypersurface (possibly withboundary, 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 = 1r2

= |II|2 = 24C.

Proof. The equality analogue of property (i) is that |IIZ↪→X | = λ if and only if Z is constantly

(C = λ2

24 , 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 assumen ≥ 2.

Since |IIZ↪→X(u, u)| is independent of the vector u (and the basepoint in Z), Z is a constantλ-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 theradius 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, sonone is left to contribute to geodesic curvature (w.r.t. Z). (Recall that the curvature as a curvein Rn+1 can be decomposed into normal and tangent parts to Z). Hence ∂Z cannot be anywhereconcave (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:

8 JEREMY WONG

Proof. Suppose Z is a smooth embedded hypersurface (without boundary) of Rn+1 and 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 inthe ambient space. These arcs have a common tangent plane, under the assumption that Z is asmooth 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 fundamentalform, whereas asymptotic directions have zero second fundamental form. Recall that for smoothhypersurfaces, 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 planeat x, in the sense of having points on both sides of the plane, arbitrarily close to x, then one cannothave 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 tangentto minimal geodesics, are not necessarily themselves geodesics. However, they are approximatelyclose, using points close to x.

Therefore if C > 0, Z must be strictly locally convex in X = Rn+1. In particular, Z then hasstrictly positive curvature, since it is a smooth hypersurface. �

Remark: In the above proof, smoothness was invoked. However, it is known that C-convex subsetsof 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, toobtain 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 satisfydZ = dX + o(d3X) in a neighborhood of a point, and thus are relatively flat to high order in theneighborhood, but were not totally geodesic everywhere.

Of course one knows in the smooth case that hypersurfaces which have zero second fundamentalform at every point and direction are totally geodesic (equivalently, locally convex in the metricspace sense), and therefore have zero sectional curvature. Thus, if one looks in a neighborhood ofall points and uses Property (i), then when C = 0, one deduces that Z ⊆ Rn+1 is totally geodesicand curvZ = 0.

However, without invoking smoothness, it is harder to show that if dZ = dX + o(d3X) holds fora non-smooth subset Z ⊂ X near every point (for X being a more general ambient space), thendZ = dX near every point.

Despite the utility of having positive-definite fundamental form (or C > 0), or being locallyconvex (in the sense of lying on one side of a supporting hyperplane in X = Rn+1), it is not alwaysneeded in order to have a lower curvature bound. There is an example due to Sacksteder of thesmooth surface z = x3(1 + y2) (for |y| < 1/2) in R3 with non-negative sectional curvature on aneighborhood of the origin, and non-positive-definite second fundamental form, which is not locallyconvex, 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 submanifoldwhich is totally umbilic and has parallel mean curvature [19].

BOUNDED EXTRINSIC CURVATURE OF SUBSETS OF METRIC SPACES 9

Circles of submanifolds of space forms (i.e. geodesics in the submanifold which are also roundgeometric circles in the ambient manifold) have been studied in [2], [1], etc. However, these worksassumed 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 Figure 2. For reference, an X-geodesicis 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 constantC-convexity equality property (for its endpoints). See the last figure in Figure 3. Z includes atleast the shaded region. In the case of X = R2, one can take the geodesic ending on a point in themiddle of Z-geodesic side indicated. That the geodesic does not satisfy the constant C-convexityequality is clear.

(For X = Rn, the only way a collection of Z-geodesics from the point x to points on the thirdside could all satisfy the constant extrinsic curvature condition is if they would comprise part of asphere. 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) = 0for two Z-geodesics α and β. But the same argument shows that no two Z-geodesics can meet andmake 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 commonpoint.

10 JEREMY WONG

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 whichdoes not satisfy the constant C-convexity equality, as in the above outline for X = R2, is moredifficult.

For another difficulty, two different C-convex Z-segments in X need not be interchangeable viaan isometry of X. For example, consider a spherical lune bounded by two geodesic arcs emanatingfrom a point x, along which are isometrically attached two intrinsically flat rectangles via thoseedges. This can be realized in R3 as X = ∂N , where N := {x ∈ R3 : d(x, [0, 1] × [0, 1]) < ε} is aneighborhood of a planar square. Then curvX ≥ 0. If now one considers Z to have two geodesicsemanating 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 whichare constantly curved. In this example, however, one can find a Z-geodesic starting from x lyingin 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 curvaturebound 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 allowedto depend not only on the distance, but also on the particular points involved. In other words, thecoefficients 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. Supposeotherwise.

Step 1: Given constant curvature curves [xyz]Z and [xyw]Z coinciding from x to y, andbranching 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 theconstant 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.

BOUNDED EXTRINSIC CURVATURE OF SUBSETS OF METRIC SPACES 11

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 whichspans the hypothetical branching geodesics. For sufficiently small radius ε > 0, there are infinitelymany points in this filling which are at distance ε from y.

In particular, there are at least n + 1 distinct points of Z all ε-equidistant from y, wheren = 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 rdepending on the point.

However, according to Lemma 1 below, the set of points equidistant (for any r) to all the n+ 1points must be finite or discrete. Therefore [xy]Z cannot lie entirely in the equidistant. Hence thereis no branching in Z.

�

Lemma 1 (Triangulation-type lemma). Suppose curvX ≥ k and dimX = n. Let {xi}Ni=1 beN distinct points in X (in general position, generic with respect to the a.e. defined Riemannianstructure [21]).

If N = n+ 1, then the set of all points equidistant from all the x′is (if nonempty) is a finite ordiscrete 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 minimalgeodesic, for i = 1, . . . , N . The set of vectors [pxi]

′ in the tangent cone TxX then consists of Ndistinct 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 thesubset Z. Instead of taking n + 1 points ε-equidistant from y initially, one can produce a finitesequence of at most n + 1 triples (z, v, w) of points getting sufficiently close to y, and look at theequidistant E := E(w, v) ∩ E(w, z) ∩ E(z, v) = {x ∈ X : |xw| = |xz| = |xv|} where z and w lie inthe 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 v′

together with points z′ and w′ on the original branches, such that z′, v′, w′ are distinct, equidistantto y, and such that either together with y and x they satisfy the five-point condition, or theassociated E′ = E(w′, v′) ∩ E(w′, z′) ∩ E(z′v′) is distinct from E as a set. One can repeat this atmost n more times if necessary.

One can informally see why three points are necessary as follows. Suppose X = Rn. Supposeequidistant E were defined by only two distinct points. The retraction as above is analogous to

12 JEREMY WONG

x y

z v w

z′ v′ w′

E

E′

Figure 4. The intersection of all equidistants E, E′, . . ., is the supposed branchpoint y

providing a one-parameter rotation of the E. If dimE = n − 1, then a rotation may send E ontoitself, so that E ∩ E′ is not lower dimensional. On the other hand, having three distinct points,dimE = n − 2. A rotation of such dimensional E will produce E′ such that E ∩ E′ is (n − 3)-dimensional, E ∩ E′ ∩ E′′ is (n − 4)-dimensional, etc. Eventually one reaches a zero-dimensionalintersection, 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 provinga 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} ≡isom

Sn(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 exist 9, so one could ask whether there is another way to prove a lowercurvature bound.

In the case of round spheres in Rn+1, or constantly C-convex subsets in Alexandrov spaces ofcurvature 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 forZ (whose sidelengths are equal to those in the Z triangle), the vertices can be made to overlap. SeeFigure 5. Here, the sides corresponding to Z-geodesics are circular arcs (whose radius is completelydetermined by C and k).

Straightening the sides of this non-geodesic triangle, one may have ∠Xk less than or greater than

∠Zk . But by Lemma 3, one can find a new model space (typically with very negative curvature k′) to

re-accommodate the straightened triangle, so that ∠Zk′ ≤ ∠

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.

BOUNDED EXTRINSIC CURVATURE OF SUBSETS OF METRIC SPACES 13

Figure 5.

2.3. Existence. Uniqueness of Z geodesics starting in a given direction is addressed below onbranching.

Remarks on existence:The question of global existence of subspaces satisfying the hypotheses of Theorem 2 seems to havea negative answer in general, if one also demands Z be geodesically extendible and Z be compactand 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 (possibilynoncompact) can locally be found which passes through x and has constant geodesic curvature. Intwo-dimensional surfaces, geodesically extendible Z will typically not be compact or embedded (inthe sense of having the induced intrinsic metric) unless C is large.

However, there are many positive examples of existence. Constant extrinsic curvature naturallyarises 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 symmetricspace setting, G-equivariant isometric immersions of rank one symmetric spaces G/K into arbitrary

Riemannian homogeneous spaces G/K 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 ε > 0 and two unit-speedZ-geodesics γ1 and γ2 : (−ε, ε) −→ Z such that γ1(t) = γ2(t) for all t ∈ (−ε, 0], γ1(0) = γ2(0) = z,and γ1|(0,ε) and γ2|(0,ε) are (non-trivial) disjoint subsegments.

14 JEREMY WONG

For manifolds with C2 smooth metric, geodesics are unique (do not branch), by uniqueness ofsolutions 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 ofvertices.

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 coneover 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 Lemma 2(i).Part (ii) follows from [25, Prop.3].

�

Part (ii) entails that if γ is a Z-geodesic, one can define γ′, as an element of TxX.As a corollary of Proposition 3, angles in Z satisfy the triangle inequality, since the triangle

inequality holds in TX.

Definition. Let ε > 0. A unit-speed geodesic segment γ : [a, b] −→ Z is ε-extendible if thereexists a unit-speed geodesic γ : [a, b+ ε] −→ Z such that γ(t) = γ(t) for all t ∈ [a, b].

Z is uniformly extendible if all its geodesic segments are ε-extendible for some ε > 0 independentof 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].

BOUNDED EXTRINSIC CURVATURE OF SUBSETS OF METRIC SPACES 15

On the other hand, if curvZ ≤ k and Z is a homology manifold (meaning Hm(Z,Z−{x};Z) ∼= Zfor all x ∈ Z for some m), then Z is geodesically extendible [10, Prop. 5.12] (respectively, uniformlyextendible, 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 geodesicallyconvex subset in the sense that if x, y ∈ X \S, then (all, minimal) geodesics from x to y lie entirelyin X \ S (see [21])

Example 2. If X = cone× [0, 1] is a metric product of a singular cone having a definite vertexangle with an interval, then Z could lie in the singular subset {v}× [0, 1] of X, where v is the vertexof 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 isreplaced by any manifold M with bilaterally bounded curvature K− ≤ KM ≤ K+, a geodesicallyextendible (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 Proposition 4’s conclusion, althoughthe proof used structure afforded by upper curvature bounds.

Alternatively a somewhat different method can also be used, by replacing the CAT (k) boundassumption there with uniform extendible assumption. Some ideas in the proof below are similarto Theorem 2, and parts of [15, Proposition 1.4], but there is less (albeit still some) dependency onupper curvature bounds. The proof is similar to an open-closed argument, in which C-convexityyields 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 isometricallyC-convexly embedded.

[xz]Z is not necessarily unique, but one can choose one. Because of the C-convexity, if t ischosen sufficiently small, then [α(t)β(t)]Z intersects both sides α and β in an angle ∠X = π

2 ± τ(t),where τ(t) denotes a quantity tending to 0 as t tends to 0. Using the fact that the ambient spaceis Rn, one can see this by noting that the osculating circles to the curves α and β at the branchpoint approximate them arbitrarily closely, when t is sufficiently small. Their radii is definable interms 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 coplanarand lying on opposite sides of a line in that plane, but in this case one obtains π

2 + τ(t). Since Z isuniformly extendible, there exists an r > 0 such that [α(t)β(t)]Z can be extended r units distanceon both sides.

16 JEREMY WONG

Figure 6. Given an initial pair of Z-geodesics forming a branch, the C-convexitycondition 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 cannotbe 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 itsinterior. This is due in part because each geodesic comprising it is C1,1, and in part becausegeodesics 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 subsegmentof either α or β could be shortened, which contradicts the assumption that α and β are minimizing.

�

BOUNDED EXTRINSIC CURVATURE OF SUBSETS OF METRIC SPACES 17

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 representativeshaded part. As t→ 0, the angle between the transversal segment and α or β tendsto π

2 .

Remark: Proposition 4 is purely local, and holds when Rn is replaced by a general Riemannianmanifold 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 isextendible, it is locally uniformly extendible. Combining this with Proposition 4, one obtains thatZ 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 isknown 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 Corollary 1 in 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 Berestovskiı,according to which curvZ ≤ k and Z being a Busemann geodesic space implies that Z has thestructure of a C1 submanifold.

The result of Proposition 4 is valid when the ambient space is a two-dimensional Alexandrovspace of curvature bounded below:

18 JEREMY WONG

Proposition 5. If Z ⊆ X is (C, ρ)-convex, uniformly extendible, curvX ≥ k, dimX = 2, thenZ has no branching.

At first glance, one apparent possible difficulty, which was not present in Proposition 4, is thata supposed branch point could be the limit of a sequence of other branch points. See Figure 8below. In other words, the angle estimate π/2 + τ(t) might not hold. Actually, in the exampleshown in Figure 8, that limit is not a branch point according to the definition, since the nontrivialgeodesics σ1 and σ2 do not have disjoint interiors. On the other hand, one can see that in thisexample that Z is not (C, ρ)-convex.

A second apparent possible difficulty is that (C, ρ)-convex Z-geodesics in X are perhaps notreadily 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 circlestransversely, no matter how small their length

One proof of Proposition 5 uses Proposition 4 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 ispossible since X is two-dimensional. Again by approximation, one can replace these pairs by C∞

pairs (X ′i, Z′i), for which curvZi is uniformly bounded from below. Then curvZ ≥ constant (see

Proposition 1).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 Proposition 4,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 ofβ as a set when it comes around, it is necessary in this case at each step to choose t at most halfwaybetween 0 and T , where T is a point at which α becomes β, since otherwise [α(T/2)α(T )β(T/2)]Zcould provide a minimal Z-segment joining α(T/2) with β(T/2). This ensures that a Z-geodesicdistinct 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).

BOUNDED EXTRINSIC CURVATURE OF SUBSETS OF METRIC SPACES 19

Therefore one can repeat the construction of a net of geodesics in Z spanning the originalbranching pair of geodesics, and the result will be sufficiently dense so that the result, by closednessof 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 areused 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 hyperbolicspace, 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 k′ ≤ k and ∆xyz ⊂M2(k) is arbitrary.Then

0 ≤ ∠kxyz − ∠k′xyz ≤ c(k, k′, R)

where R depends on the area of a Euclidean triangle with the same sidelengths as ∆xyz, and onthe 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) isgreater than that of M2(k′).

The term on the right-hand side can depend on the aspect ratio of the triangle, so is notuniformly depending on the largest sidelength. By Sublemma 1, each of the model space angles∠kxyz and ∠k′xyz can first be compared to Euclidean angles, and then cos θ − cos θ′ = (k − k′) ·(a4 + b4 + c4 − 2a2b2 − 2a2c2 − 2b2c2

)24ab

+ (k2 − k′2) · O(6) gives the bound c(k, k′, R) for θ − θ′ =

∠kxyz − ∠k′xyz. �

Let a = |xy|X , b = |xz|X , c = |yz|X .If k > 0 and k′ = 0, a special case of Lemma 3 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)︸ ︷︷ ︸

≤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 benon-positive, by the triangle inequality.

From this, if k ≥ 0, then cos θ ≤ cos θ′, or θ ≥ θ′, where θ′ is the comparison angle of thetriangle 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.

20 JEREMY WONG

On the other hand, if k is fixed and all sides change by the same multiplicative factor, then ofcourse 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 functionssuch that the angle should increase if each sidelength is increased by the same function. For givensidelengths one can sometimes find such functions, though rarely do they have the property for allsufficiently 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 −mCfor the curvature and f(x) = x + Cx3 + o(x3) for the sidelengths provides monotonicity of modelangles, 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 k′ = k (respectively k′ = k − 12C, k′ = k − 48C), then

∠Zk′xyz ≤ ∠

Xk xyz

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, k′ > 0. When one allowsk < 0 or k′ < 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 trigonometricfunctions).

Suppose f(x) = x + Cx3 + o(x3) and k′ = k −mC, where m is a constant to be determinedsoon.

Let a = |yx|X , b = |yz|X , c = |xz|X . Corresponding lengths in the subset Z are f(a), f(b), andf(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(∠Zk′xyz

)=

cos(√

k′ (c+ Cc3 + o(c3)))− cos

(√k′ (a+ Ca3 + o(a3))

)cos(√

k′ (b+ Cb3 + o(b3)))

sin(√

k′ (a+ Ca3 + o(a3)))

sin(√

k′ (b+ Cb3 + o(b3)))

Analytic expansion in a, b, c, treating k, k′, and C as parameters, yields

BOUNDED EXTRINSIC CURVATURE OF SUBSETS OF METRIC SPACES 21

cos(∠Zk′xyz

)− cos

(∠Xk xyz

)=

1

2ab·[(

1

12k′ − 1

12k + C

)a4 +

(1

6k − 1

6k′ − 2C

)a2b2 +

(1

6k − 1

6k′ + C

)a2c2

+

(1

12k′ − 1

12k + C

)b4 +

(1

6k − 1

6k′ + C

)b2c2 +

(− 1

12k +

1

12k′ − 2C

)c4

+O(6)

]Taking k′ = k − 12C,

cos(∠Zk′xyz

)− cos

(∠Xk xyz

)=

C

2ab·[3(a2 + b2 − c2)c2 +O(6)

]If k′ = k −mC, then

cos(∠Zk′xyz

)− cos

(∠Xk xyz

)=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 possiblyalso powers of C, m, and/or k.

Upon rewriting and keeping the lowest order terms, this is

∠Xk xyz − ∠

Zk′xyz =

1

ab· C ·

[(1

2− m

24

)(a2 − b2

)2+

(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 −m24) ≥ 0

(i.e, m ≤ 12) and (1

2+m

12

)(a2 + b2)−

(1 +

m

24

)c2 > 0(4)

Hence

22 JEREMY WONG

When m ≈ −∞, (4), then (3), is positive when c2 & 2(a2 + b2). But c ≤ a + b ≤ 2

√a2 + b2

2implies c2 ≤ 2(a2 + b2), so this case is not possible.

When m = −48, (3) is positive when c2 > 72(a2 + b2), which is not possible.

When m = −25, it is positive when c2 > 38(a2 + b2), which is not possible.When m = −24, it cannot always be positiveWhen m = −12, it cannot always be positiveWhen m = −6, it cannot always be positiveWhen m = −5, it is positive when c2 < 2

19(a2 + b2).

When m = −3, it is positive when c2 < 27(a2 + b2).

When m = 0, it is positive when c2 < 12(a2 + b2).

When m = 12, it is positive when c2 < a2 + b2.Thus, according to the above, (3) is positive if −6 < m ≤ 12, and c2 is sufficiently small in

comparison with a2 + b2 (and a, b, c are all sufficiently small relative to 1, k,m,C).In a worst case, if m ≤ 12 , taking a = b in (3) shows that (4) is necessary for positivity of the

lowest order term in (3).In fact (3) is still strictly positive for all m > 12 when weak inequality holds, namely, when

c2 ≤ a2 + b2. One can see this as follows. View

h(a, b, c,m) :=

(1

2− m

24

)(a2 − b2

)2+

(1

2+m

12

)(a2 + b2)c2 −

(1 +

m

24

)c4

as a function quadratic in c2. Then h > 0 if and only if its discriminant is non-negative (so that hhas at least one real root) and either

(1 +m/24) ≥ 0 (i.e., m ≥ −24) and c < c0or (1 +m/24) ≤ 0 (i.e., m ≤ −24) and c > c0where c20 = c20(a, b) is the root of h given by

c20 :=1

2(1 +m/24)

(1

2+m

12

)(a2 + b2) +

[((1

2+m

12)(a2 + b2)

)2

− 4(

1 +m

24

)(−(

1

2− m

24

))(a2 − b2)2)

]1/2=

m+ 6

m+ 24(a2 + b2) +

2

m+ 24

√(9a2 + 9b2)2 + ((m+ 6)2 − 182)a2b2

=m+ 6

m+ 24(a2 + b2) +

2

m+ 24

√(9a2 − 9b2)2 + (m+ 6)2a2b2

(Since the discriminant is clearly non-negative, the roots are always real. Moreover the lesserroot value is non-positive if (12 + m

12) ≥ 0 and 4(1 + m

24

) (12 −

m24

)≥ 0 i.e., if −6 ≤ m ≤ 12.

For m > 12, the lesser root value can be positive, but fortunately in this case the triangleinequalities a − b ≤ c and b − a ≤ c, when combined into a single inequality (a− b)2 ≤ c2, implythat c2 is greater than such a root.)

Now

when m = 0, c20 =1

4(a2+b2)+ 1

12

√(9a2 − 9b2)2 + (6)2a2b2 ≥ 1

4(a2+b2)+

1

4(a2+b2) =

1

2(a2+b2)

when m = 12, c20 =m+ 6

m+ 24(a2 + b2) +

2

m+ 249(a2 + b2) = a2 + b2

BOUNDED EXTRINSIC CURVATURE OF SUBSETS OF METRIC SPACES 23

whenm = 48, c20 =3

4(a2+b2)+

1

36

√(9a2 − 9b2)2 + (54)2a2b2 ≥ 3

4(a2+b2)+

6

4ab =

(√3

2(a+ b)

)2

As m increases, the range for c2 guaranteeing positivity of h increases even more, since∂

∂mc20 >

0.In the limit as m −→∞,

c20 −→ a2 + b2 + 2ab = (a+ b)2(5)

�

As a corollary, the expansion (3) in the proof of Lemma 4 gives the following expected result,which says that the higher order terms do not influence curvature.

Corollary 2. If curvX ≥ k and Z ⊆ X is constantly 0-convex, then curvZ ≥ k.

Proof. Since C = 0, expansion (3) gives ∠Zk′xyz ≤ ∠

Xk xyz for k′ = k and for all x, y, z ∈ Z

contained in a small neighborhood of an arbitrary point (with no other restrictions on their mutualdistances). Hence using the characterization of lower curvature bound involving quadruples ofpoints (y;x, z, w),

∠Zk xyz + ∠

Zk zyw + ∠

Zk wyx ≤ ∠

Xk xyz + ∠

Xk zyw + ∠

Xk wyx

≤ 2π

which means curvZ ≥ k.�

Proof of Theorem 2(ii). Claim: For any fixed quadruple (y;x, z, w) in Z, there exists afinite (though possibly extremely large, negative) k′ such that

∠Zk′xyz + ∠

Zk′zyw + ∠

Zk′wyx ≤ 2π

To see this, first note that model angles are continuous with respect to the points x, y, z, w (aswell as model space curvature k′).

Now either ∠Xk xyz = π or ∠

Xk xyz < π.

If ∠Xk xyz = π then ∠

Zk′xyz ≤ ∠

Xk xyz would automatically hold for any k′, since model angles

are by convention/definition never bigger than π.

If ∠Xk xyz < π this means model X-triangles satisfy strict inequality c < a + b in the notation

of Lemma 4. (recall that ∠Xk xyz = π if and only if ∠Xxyz = π)

In this case, by (5) of Lemma 4 one can choose a (potentially extremely large, though finite)positive value of m, hence a potentially extremely large (though finite) value of k′ := k − mC,(depending on the points (y;x, z, w)) such that

∠Zk′xyz ≤ ∠

Xk xyz

24 JEREMY WONG

In order to guarantee that model triangles corresponding to Z-triangles always exist, it is withoutloss of generality to assume that k′ ≤ 0. Similarly, a k′ can be found such that at the same timeone has

∠Zk′zyw ≤ ∠

Xk zyw

and ∠Zk′wyx ≤ ∠

Xk wyx

Adding these three inequalities gives

∠Zk′xyz + ∠

Zk′zyw + ∠

Zk′wyx ≤ ∠

Xk xyz + ∠

Xk zyw + ∠

Xk wyx ≤ 2π

which holds since curvX ≥ k. By decreasing k′ slightly if necessary, using Lemma 3, one thenobtains strict inequality

∠Zk′xyz + ∠

Zk′zyw + ∠

Zk′wyx < 2π

Since this is strict and Z is locally compact, one can choose (an even larger, negative, thoughfinite) k′ uniformly so that

∠Zk′xyz + ∠

Zk′zyw + ∠

Zk′wyx ≤ 2π

holds for all points in a small neighborhood of a given basepoint. This means Z has a locallower curvature bound of k′ (depending on the neighborhood). Thus if Z is moreover compact,curvZ > −∞ globally.

�

One sees that the inequalities in Lemma 4 involving sidelengths allow certain ranges of angles.For example, by the law of cosines, when k = 0 (Euclidean space) and a = b,

c2 ≤ 12(a2 + b2) corresponds to angles less than π/3,

c2 ≤ a2 + b2 corresponds to acute angles,c2 ≤ 3

4(a+ b)2 corresponds to angles less than 2π/3, and so on.The geometric significance of this is that it allows Lemma 4 to be applied to tripods.

Corollary 3 (Theorem 2(iii)). If curvX ≥ k, Z ⊆ X is constantly C-convex for someC ≥ 0, and Z itself happens to be intrinsically a C2 submanifold (possibly with boundary), thencurvZ ≥ k − 48C.

Proof. By Theorem 2(i), Z has no branching, so that if it has a boundary, the boundary mustbe locally convex (with respect to the interior of Z).

If an interior sectional curvature of Z were at some point (the point lying either in the interioror on the boundary of Z) strictly less than k − 48C, then by continuity the sectional curvatureswould also be strictly less than k − 48C in some open Z-neighborhood of that point, say less thank′ := k − 48C − ε for some ε > 0. It is clear that one can choose a quadruple of points (y;x, z, w)in this neighborhood with

|xz|2X ≤3

4(|yx|X + |yz|X)2

|zw|2X ≤3

4(|yz|X + |yw|X)2

|wx|2X ≤3

4(|yw|X + |yx|X)2

BOUNDED EXTRINSIC CURVATURE OF SUBSETS OF METRIC SPACES 25

(e.g., a nearly equilateral standard tripod w.r.t. X metric) such that the total subtended modelspace angle is

∠Zk′xyz + ∠

Zk′zyw + ∠

Zk′wyx > 2π

Specifically, one can take y ∈ Z to belong to the interior of Z, and if the sectional curvature ofthe manifold Z corresponding to a tangent plane is strictly less than k − 48C, then one can takethe points x, z, w ∈ Z to lie in the image of the exponentiated plane.

But by Lemma 4, and the quadruple condition for having a lower curvature bound,

∠Zk′xyz + ∠

Zk′zyw + ∠

Zk′wyx ≤ ∠

Xk xyz + ∠

Xk zyw + ∠

Xk wyx

≤ 2π

Contradiction.Thus curvZ ≥ k − 48C − ε. But if ε > 0 is arbitrary, this implies curvZ ≥ k − 48C.

�

k−48C is the optimal lower bound and cannot be taken larger, for a general constantly C-convexsubset Z ⊆ X (making no assumptions about Z being a codimension one subset or hypersurface).The reason can be seen by examining the smooth submanifold case. According to [18, Theorem

5.8], for any constant c > 0, there exists a (λ =√c

2√2)-isotropic embedding of a flat torus T 2 in the

four-real-dimensional complex projective space CP 2(c) of constant holomorphic sectional curvaturec (sectional curvatures between c/4 and c).

Thus it is constantly C-convex, where 24C = λ2. Then 48C = 2λ2, and the Gauss relation forthe sectional curvatures

0 = KT 2 ≥ KCP 2(c) − 2λ2 ≥ c4 − 2

( √c

2√2

)2= 0

becomes an equality.

Remark: It seems that Theorem 2(iii) is likely to be true assuming only constant C-convexity for thesubset, without assuming it has any a priori regularity. This warrants further study of Toponogov’stheorem.

3.2. Arc-chord angles. Leaving aside constant curvature, but considering C-convexity still,here are some angles estimates, which refer to arc/chord angles.

It is possible to establish an upper base angle estimate for any arc-chord pair consisting of [xy]Zand [xy]X , namely

∠X([xy]X , [xy]Z) ≤ τ(|xy|),(6)

where x ∈ Z is fixed, x is a strained point, and τ denotes a non-negative function which tends tozero as its argument tends to zero.

It holds only when x is strained, or has a neighborhood with an upper curvature bound insome sense. Without such a type of upper curvature bound, X may have arbitrarily small geodesicbigons (or arbitrarily small nearly geodesic bigons, taking C to be 0 or close to 0, respectively).

26 JEREMY WONG

Sublemma 2. Suppose curvX ≥ k, x ∈ X and a−1, a1 ∈ X with ∠Xa−1xa1 ≥ π − δ, b ∈ Xsuch that |xb|X < ε

4 min{|xa−1|X , |xa1|X}. Then

0 ≤ ∠Xa1xb− ∠Xa1xb < 2 max{ε, δ}.

�The following was shown in [25], but a slightly different proof is given here.

Proposition 6. Suppose Z ⊆ X, curvX ≥ k, curvZ ≥ k, X geodesically extendible, dZ ≤dX + Cd3X locally. Then

∠X([xy]X , [xy]Z) ≤ τ(|xy|X)

for all x, y ∈ Z.

Proof. Let v := x ∈ Xvi := y ∈ X, i fixed for the moment.Consider a segment [vvi]X .Since X geodesically extendible, there exists v′i ∈ X such that [v′ivvi]X is a segment, with

|vv′i|X = |vvi|X > 0, ∠Xv′ivvi = π (initially restricting y to be at a closer distance to x if necessary)Taking δ = 0 in Sublemma 2,

∠Xvivvji ≤ ∠

Xk vivv

ji + 2ε(7)

for all vji ∈ [vvi]Z with |vvji |X < ε4 |vvi|X . Then

∠X([vvi]X , [vvi]Z) = limvji→v

vji∈[vvi]Z

∠X([vvi]X , [vvji ]Z) since [vvji ]Z ⊆ [vvi]Z

≤ limj→∞∠Xvivv

ji by definition

≤ limj→∞

(∠Xk vivv

ji + 2ε

)if |vvji |X ≤

ε

4|vvi|X , by (7)

≤ limj→∞

(∠Zk vivv

ji + τQ

(|vvi|X + |vvji |X

)+ 2ε

)if

1

Q≤|vvi|3/2X

|vvji |X≤ Q, by Lemma 5

≤ limj→∞

(∠Zvivv

ji + τQ

(|vvi|X + |vvji |X

)+ 2ε

)since curvZ ≥ k

= limj→∞

(τQ

(|vvi|X + |vvji |X

)+ 2ε

)since ∠Zvivv

ji = 0

≤ τ(|vvi|X)

For the last step here, if |vvi|X = 1i , choose vji ∈ [vvi]Z such that |vvji |X = 1

i3/2.

Then|vvji |X|vvi|X

=1

i1/2−→ 0 whence can take ε −→ 0, yet

1

Q≤|vvji |X|vvi|3/2X

≡ 1 ≤ Q.

�

Lemma 5. Suppose Z ⊆ X both have curv ≥ k or both have curv ≤ k, and dZ ≤ dX + Cd3Xlocally, x, y, z ∈ Z, and either

BOUNDED EXTRINSIC CURVATURE OF SUBSETS OF METRIC SPACES 27

(i) 1Q ≤

|xy|X|yz|X ≤ Q, or

(ii) 1Q ≤

|xy|3/2X|yz|X ≤ Q for some Q > 0. Then

cos(∠Zxyz

)≤ cos

(∠Xxyz

)+ τQ(|xy|X + |yz|X)

In particular, ∠Zxyz ≥ ∠Xxyz − τQ(|xy|X + |yz|X).

Proof. If k > 0 one must also assume so that comparison triangle ∆Zxyz exists.

For below, assume for simplicity k = 0.Denote the sidelengths of ∆Xxyz and ∆Zxyz by

aX = |yx|X bX = |yz|X cX = |xz|Xresp. aZ = |yx|Z bZ = |yz|Z cZ = |xz|Z

Then

aX ≤ aZ ≤ aX + Ca3X

bX ≤ bZ ≤ bX + Cb3XcX ≤ cZ

so

cos(∠Zxyz

)=a2Z + b2Z − c2Z

2aZbZ≤

(aX + Ca3X)2

+ (bX + Cb3X)2 − c2X

2aXbX

=a2X + b2X − c2X

2aXbX+

2Ca4X + C2a6X + 2Cb4X + C2b6X2aXbX

=a2X + b2X − c2X

2aXbX+ C ′

a4X + b4XaXbX

where C ′ = (2C + C2), provided a ≤ 1 and b ≤ 1

= cos(∠Xxyz

)+ C ′

(a2X

(aXbX

)+ b2X

(bXaX

))≤ cos

(∠Xxyz

)+ C ′ (aXQ+ bXQ) if

1

Q≤ aXbX≤ Q or

1

Q≤a3/2X

bX≤ Q

= cos(∠Xxyz

)+ τQ(aX + bX)

�

Remark: If dZ ≤ dX + Cd3X then a bi-Lipschitz relation 1Q ≤

|xy|X|yz|X ≤ Q between lengths in X can

be reformulated equivalently as a bi-Lipschitz relation 1Q1≤ |xy|Z|yz|Z ≤ Q1 between lengths in Z, when

distances are small in comparison with C, and where Q and Q1 are constant multiples (by a factorof 2) of each other. However, one needs the full strength of C-convexity to be able to relate anglesto each other as in Lemma 5.

28 JEREMY WONG

Acknowledgements: The author thanks the organizers of the seminar proceedings for suggestingto write up the proceedings.

Notes

1 One always has dX ≤ dZ . Although the strong version does not imply the weak version, theyare so called because the conditions in the weak version actually allows more types of subsets tooccur. [15] used the terminology (C, 2, ρ)-embedded instead of strong (C, ρ)-convexity.

2 There, the curves were posited to have bounded extrinsic curvature, and certain metricalproperties on width, base angles, arc/chord length ratio were derived. The equality case there (asin Schur and Schmidt-type theorems) involved equality of length, rather than equality attainmentof curvature bound.

3 One could also consider a more general relation dZ(x, y) = f(dX(x, y)) where f : [0, ε) → Ris a smooth, increasing and convex function, defined on a neighborhood [0, ε) of the origin, andsatisfies f ′(0) = 1, f(0) = 0, f(x) > 0 for x > 0.

However, the most typical and important example is the function which gives the arclengthcorresponding to a given chordlength dX , along a circle of curvature λ =

√24C in a model space

of constant curvature k: When k = 0 (the Euclidean plane R2), this function is given by f(x) =2

λsin−1

(λ

2x

). For general k, such an f is f(x) = x+

λ2

24x3 +

9λ4 + 8λ2 · k1920

x5 + o(x5).

4 So-called helical immersions (usually in the setting of complex geometry) have been studiedby authors of [2], which involved the study of circular paths in the domain manifold being mappedonto helicies in the target manifold.

5 Lemma 25 from [24] says: If Nm is a manifold of constant curvature, Mn ⊂ Nm is an immersedtotally umbilic submanifold, then either Mn is totally geodesic or M lies in some (n+1)-dimensionaltotally geodesic submanifold of N .

Theorem 26 ([24]): In particular, if n ≥ 2 and Mn ⊂ Rm, is connected, immersed, totallyumbilic, then either Mn ⊆ Rn or Mn ⊆ Sn ⊂ Rn+1 ⊂ Rm.

The proofs are local, applying to the interior of M , and so M is allowed to have boundary.

6 In [13] it was shown that isometric immersions Mn ↪→ Rn+k which map geodesics into 2-dimensional planes in Rn+k must have weakly 1/4-pinched sectional curvature. In fact complex

NOTES 29

projective spaces satisfy this. B.-Y. Chen [12, Theorem 6.1] has noted that there exists an isometric

embedding of CPn with its standard metric into Rn(n+2), using eigenfunctions associated to thefirst non-zero eigenvalue of the Laplacian on functions. One considers the composite map f :

CPn −→minimal

Sn(n+2)−1(√

n2(n+1)

)−→

totally umbilicRn(n+2). Since the image of geodesics are curves

with constant geodesic curvature, the immersion is in particular constantly C-convex. But inthe present situation (Proposition 2), those projective spaces cannot occur because the additionalhypersurface assumption gives all sectional curvatures equal to a constant.

By the way, in high codimensions, (without going to codimension 1, for which k′ = k + 24Cwould be optimal) this special type of immersion is directly related to why one sees k′ = k + 6C

appearing in Lemma 4 (stated in the next-to-last section) as the borderline between having ∠Zk′ ≤

∠Xk hold for some c2 ≤ q(a2 + b2) (q > 0) and not.

Recall that, by the Gauss inequality in upper curvature bound setting [4], if curvX ≤ Kand Z ⊆ X is C-convex (but not necessarily constantly C-convex nor a smooth manifold), thencurvZ ≤ K + 24C.

Hence if the curvature of the immersion of Z = CPn in Euclidean space were strictly greaterthan k + 6C, then it would be strictly quarter-pinched (at at least some point(s) and directions),which since it is also simply-connected would be impossible (for n ≥ 2) by the Sphere Theorem.

7 In [6], it was shown that Z must be the unit hypersphere Sn(1) if it is assumed that Z is aclosed C2-smooth hypersurface of Rn+1 such that

(i) the curvature (w.r.t X = Rn+1) of any geodesic curve in Z is ≥ 1,

and (ii) Z contains a curve with constant curvature 1 (w.r.t. X) and length π.

8 However, assumption (*) probably implies that the coefficients in the higher terms dependonly on the distance, and not on any of the points involved. To see this, suppose y, z, v, w ∈ Z arefour points with |yz|X = |vw|X = r, say.

To show that (*) probably implies o(|yz|3X) = o(|vw|3X), imagine a sequence of segments[xixi+1]X of X-length r (with xi ∈ Z) laid end to end, joining y to v, with x0 = y and xm+1 = v.Then |yz|X = |vw|X = |xixi+1| = r for all i, so o(|yz|3X) = o(|x0x1|3X) = · · · = o(|xmv|3X) =o(|vw|3X).

If r is sufficiently small, then it seems likely that such a sequence exists, since Z is at leastpath-connected. However, if v ∈ Z does not have a nice neighborhood in Z, then it is not clearhow to close the chain, i.e. find xm.

9 Here is a simple example: Let X be the double of the first quadrant of R2 along its boundary,the non-negative x- and y-axes. Let Z be a circular arc contained in the first quadrant not passingthrough the origin. Thus, X is metrically a cone and Z is constantly C-convex for some C > 0.One can place many positive curvature vertices arbitrarily close to Z, resulting in a new space (call

30 NOTES

it X again) for which curvX is bounded from below, but there is no open neighborhood of Z in Xon which orthogonal (nearest point) projection is defined.

10 One can see lack of monotonicity perhaps most simply by considering Z to be a round circlein the Euclidean plane X = R2, and taking an inscribed X-triangle. The corresponding Z-triangleis degenerate (thence so is its Euclidean model triangle) and has an angle π, which corresponds toan angle in X strictly less than π.

Let us verify more generally lack of monotonicity of angles under the specified change in side-lengths. When k ∈ R is fixed and all sidelengths are changed from x to x+Cx3 +o(x3), the changein cosine of angles is

cos(∠Zk xyz

)− cos

(∠Xk xyz

)=

C

2ab

[a4 + b4 − 2c4 − 2a2b2 + a2c2 + b2c2 +O(6)

]=

C

2ab

(a2 − b2)2 + c2(a2 + b2 − c2)− c4︸ ︷︷ ︸+O(6)

The lowest order expression underlined would be minimal when a = b = c/2 (corresponding toa degenerate isosceles triangle with angle equal to π), and has there the value of −3

4c4, which is

strictly negative.

Note that the lowest order expression does not explicitly involve k or C (aside from the coef-ficient in front). So for any chosen C > 0, if a, b, c are all sufficiently small but comprise the side-

lengths of such a degenerate (or nearly degenerate) X-triangle, then cos(∠Zk xyz

)−cos

(∠Xk xyz

)≥

0 does not necessarily hold, which means that the model angle is not necessarily monotone decreas-ing.

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NOTES 31

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Department of Mathematics, National Taiwan UniversityE-mail address: jwong@math.ntu.edu.tw