SURFACES OF FINITE CURVATURE

1. Introduction

1.1. Gauss-Bonnet, classical. One of the deepest and most impor-tant theorems in classical differential geometry of surfaces is the the-orem of Gauss–Bonnet. It has a local and a global form and can beformulated for general curves and more geometrically for polygons ortriangles.

We recall the local form for triangles. Let U be a smooth surfacewith a Riemannian metric homeomorphic to the 2-dimensional plane.Classically U is an immersed disc in R3 and the Riemannian metric isinduced by the so called first fundamental form.

There are well-defined notions of lengths of curves and of shortestcurves between points. These shortest curves are smooth solutions ofthe geodesic equation, the so-called geodesics. Given a simple triangle∆ in U , thus simple closed curve built by a concatenation of 3 geodesics,we have well-defined vertices and angles α1, α2, α3 at these vertices.

The theorem of Gauss-Bonnet now tells us that the sum of the anglesin the triangle ∆ can be computed from the so-called Gaussian curva-ture, known also as sectional curvature in Riemannian geometry. Thecurvature (in our case of surfaces) is a real number κ(x) assigned toeach point x ∈ U , which is a measure of how non-Euclidean the metricaround this point behaves infinitesimally (in a precise and non-trivialsense).

Denote by δ(∆) the defect of the triangle ∆ defined as

δ(∆) :=∑

αi − π .

The theorem of Gauss–Bonnet states that

δ(∆) =

∫∆

κ(x) dvol(x) ,

where vol denotes the natural area measure on U .Turning to the global version of the theorem of Gauss–Bonnet assume

now that M is a compact two-dimensional surface with a Riemannianmetric. Again the notions of shortest curves, geodesics, angles betweenthem and curvature are well-defined and M is equipped with its natural

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area measure vol. One can triangulate M into simple geodesic trian-gles, thus one can dissect M into simple triangles, each pair of themintersecting in the empty set, a common side or a common vertex.

Summing the right-hand sides in the local theorem of Gauss-Bonnetone obtains the integral of the curvature∫

M

κ(x) dvol(x)

Summing the left-hand sides we get

2π · V − π · F ,where V and F are the numbers of vertices and triangles in the trian-gulation. In the triangulation, the number F of triangles satisfies

3F = 2E ,

where E is the number of edges. Thus, we arrive at the global formulaof Gauss–Bonnet

2πχ(M) =

∫M

κ(x) dvol(x) ,

with

χ(M) = V − E + F

the Euler characteristic of M .This equality has many different consequences. The first (usually

proven on the way to this formula) is the statement that the Eulercharacteristic V − E + F does not depend on the triangulation. It isa topological invariant of M . In fact it can also be computed by thesame formula using decomposition into polygons different from trian-gles. This invariant equals to 2 on the sphere, 0 for the torus and −2kfor the torus with k handles attached.

Secondly, it proves the highly non-trivial statement, that the inte-gral of the curvature over M does not depend on the choice of theRiemannian metric.

1.2. Gauss-Bonnet, simplicial. Now we slightly change the subjectand consider a simplical surface. For simplicity we consider the bound-ary M of a (possibly non-convex) polyhedron in R3. Subdividing all ofits faces we get a triangulation of M . Now, all triangles involved areEuclidean and the sum of all its defects is 0.

However, we can express this 0 in another way, again first summingup the angles around the vertices. We get

π · F =∑

(αv) ,2

where αv denotes the sum of all angles at the vertex v and the sum-mation is taken over all vertices v in the triangulation.

Write κ(v) = 2π − αv and call it the curvature of the simplicialsurface M at the vertex v, we see∑

v

κ(v) = 2π · V − π · F = 2πχ(M) .

Thus, the formula of Gauss–Bonnet holds true, if we replace thedifferential-geometric measure κ(x) · vol(x) by a discrete measure con-centrated in the vetrices of M and measuring the difference of the totalangle at the vertices from 2π.

1.3. Aim and plan. The central aim of this lecture will be to under-stand the connection between the theorems of Gauss–Bonnet in thediscrete and in the classical setting and extend it to a general theory ofsurfaces for which a version of the theorem holds true for some curva-ture measure, generalizing the classical curvature times area measureand the discrete measure concentrated in the vertices in the simplicialcase.

On the way, we will learn some metric geometry and repeat (orshortly sketch) results in classical differential geometry of surfaces.

We will start with learning some basics about shortest curves andangles.

Since we will work on surfaces almost exclusively, we will need somebasics in classical two-dimensional topology, related to Jordan’s curvetheorem. We will recall that as well.

We will learn and discuss some important metric constructions likegluing and coning, describe simplicial metric spaces and understandthe validity also of the local version of the theorem of Gauss-Bonnet insuch spaces.

After these preparations we will turn to surfaces on which a theoremof Gauss–Bonnet holds true and will try to understand some aspects oftheir theory. This theory has been developed by Alexandrov, Reshet-nyak and Zallgaller and developped further by students of Alexandrov.

The plan is to follow the survey of Y. Reshetnyak, [Res93]. At the be-ginning we will recall some basic material in metric geometry, [BBI01]can be used as a good source.

2. Length spaces

2.1. Curves. Let (X, d) be a metric space. A curve in X is a contin-uous map γ : I → X of an interval I ⊂ R.

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A curve γ : I → X is a reparametrization of another curve γ : J → Xif there exist a non-decreasing surjection s : J → I such that γ = γ s.

A curve is called simple if it injective. It is called closed if I = [a, b]and γ(a) = γ(b). It is called simple closed if I = [a, b], γ is injective on[a, b) and γ(a) = γ(b).

We will often (in particular, if γ is simple or simple closed) identifya curve and its image in X by abuse of notations.

2.2. Length of curves. For a curve γ : I → X the length of γ is

`(γ) = `X(γ) = `d(γ) = sup∑i

d(γ(ti), γ(ti+1)) ,

where the supremum is taken over all sequences t0 ≤ t1 ≤ ... ≤ tn ∈ I.The following properties hold:1) If X = Rn with its Euclidean metric and γ is C1 then

`(γ) =

∫I

|γ′(t)| dt .

The same formula holds for C1 curves in Riemannian manifolds M .2) `(γ) ≥ d(γ(a), γ(b)), for all a, b ∈ I. The curve γ is constant if

and only if `(γ) = 0.3) If I is subdivided in two intervals I = I− ∪ I+ intersecting at

one point, thus γ is represented as a concatenation of γ1 = γ|I− andγ2 = γ|I+ , then

`(γ) = `(γ1) + `(γ2) .

4) Length does not change under reparametrizations and change oforientation.

5) If γ is L-Lipschitz, then `(γ) ≤ L · H1(I).Here and later we denote by Hn (for us only n = 1, 2 are relevant)

the n-dimensional Hausdorff measure. On Rn the measureHn coincideswith the Lebesgue measure.

6) We say that γ is parametrized by arclength if, for any subintervalJ ⊂ I we have `(γ|J) = H1(J).

Any curve parametrized by arclength is 1-Lipschitz.A C1-curve γ : I → Rn (or in a Riemannian manifold M) is parama-

trized by arclength if and only if |γ′(t)| = 1 for all t ∈ I.Every curve of finite length (called a rectifiable curve) admits a

parametrization by arclength. The same is true for locally rectifiablecurves, thus curves, whose restrictions to any compact subinterval of Ihave finite length.

7) If curves γn : I → X converge pointwise to γ : I → X then

`(γ) ≤ lim inf `(γi) .4

The theorem of Arzela-Ascoli implies that any sequence of Lipschitzcurves with the same Lipschitz constant whose images are containedin a compact set always has a convergent subsequence. Together withthe above semicontinuity it allows us to find curves of shortest lengthbetween a pair of points contained in a sufficiently large compact sub-set. Here and later we denote by Br(x) the open metric ball of radiusr around x:

Proposition 2.1. Let the closure Br(x) be compact. Let y ∈ Br(x) bea point connected to x by a curve of length ≤ r. Then there exists acurve of shortest length among all curves connecting x and y in X.

2.3. Length spaces. The metric space (X, d) is called a length spaceif, for all x, y ∈ X

(2.1) d(x, y) = inf`(γ) | γ : [a, b]→ X, γ(a) = x, γ(b) = y

Note that inequality ” ≤ ” is true for all curves γ connecting x and y.We also call d a length metric or intrinsic metric.A few (hopefully well-known) classical examples:1) The Euclidean space, and its convex subsets are length spaces.2) R2 \ 0 is a length space but not R2 without a segment. Here we

consider the subset with the metric restricted from the ambient spaces.3) Any Riemannian manifold (always connected, by default) (M, g),

in particular, a surface M2 in R3 with the first fundamental form g, isa length space with respect to the Riemannian metric dg induced by g.

4) Any normed vector space with the metric induced by the normand any convex subset of a normed space is a length space.

2.4. First properties. A shortest curve γ : [a, b] → X between γ(a)and γ(b) in a length space X satisfies d(γ(t), γ(s)) = `(γ|[s,t]), for alls < t ∈ [a, b]. If, in addition, γ is parametrized by arclength, thend(γ(s), γ(t)) = |t− s|, for all s < t ∈ [a, b].

Shortest curves play in metric geometry the role of segments in Eu-clidean geometry. They will be parametrized by arclength, if not oth-erwise stated. Often, shortest curves in metric geometry are calledgeodesics by an abuse of notation.

Recall that shortest curves in Riemannian geometry solve the geo-desic equation, but solutions of this equation are shortest curves onlyon sufficiently short subintervals.

If X is a locally compact length space then the theorem of Arzela-Ascoli stated above provides the following statement. For every pointx ∈ X there exists some r > 0 such that for every pair of pointsy, z ∈ Br(x) there exists a shortest curve in X connecting y and z.

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Note, that such a shortest curve is contained in B3r(x). In general,such a curve is not contained in Br(x), thus the ball Br(x) is in nosense convex.

In general, there is no kind of uniqueness for shortest curves withprescribed fixed points, even locally.

In length spaces the closure Br(x) of the open ball is the closed ball,thus the set of points y ∈ X such that d(x, y) ≤ r.

A pointwise limit of shortest curves in a length space is a shortestcurve and its limit is the limit of lengths of the approximating curves.

An important consequence of the theorem of Arzela-Ascoli is thefollowing theorem of Hopf–Rinow and Cohn-Vossen, which might beknown from a course in differential geometry.

Theorem 2.2. Let X be a locally compact length space. Then X isa complete metric space if and only if all of its bounded closed subsetsare compact. In this case any pair of points in X is connected by ashortest curve.

3. Constructions of length spaces

3.1. Induced length metric. Let (X, d) be a metric space, let A ⊂ Xbe a subset. We say that A is rectifiably connected if for any x, y ∈ Athere exists a curve of finite length γ : I → A connecting x and y.

The induced intrinsic metric dA on a rectifiably connected subsetA ⊂ X is defined as

dA(x, y) = inf`(γ) | γ : [a, b]→ A, γ(a) = x, γ(b) = y .On A we have the inequality dA ≥ d.

For A = X the equality dA = d is equivalent to the statement thatd is a length metric.

The following Lemma is elementary but non-trivial.

Lemma 3.1. For any rectifiably connected subset A ⊂ X, the metricdA is a length metric on A. Moreover, for any curve γ : I → A, thelength of γ with respect to d and dA coincide.

The main classical example of this construction is the induced in-trinsic metric on a connected submanifold of Rn. This is nothing elseas the Riemannian metric induced by the first fundamental form. Theinvestigation of such metric spaces is the central subject of Riemanniangeometry.

3.2. Cartesian product. Given metric spaces (X, dX) and (Y, dY ) wedefine the product metric on the Cartesian product

Z = X × Y = (x, y) | x ∈ X, y ∈ Y 6

by the formula of Pythagoras

d((x1, y1), (x2, y2)) =√d2X(x1, x2) + d2

Y (y1, y2) .

We say that a curve γ : I → X in a metric space X has constantvelocity v ≥ 0 if, for all s, t ∈ I,

d(γ(s), γ(t)) = v · |t− s| .Any curve of finite length can be reparametrized to have constant ve-locity v.

We observe that if γ1 : I → X is a curve of constant velocity v1 andγ2 : I → Y is a curve of constant velocity v2 then γ = (γ1, γ2) : I →X × Y is a curve of constant velocity

√v2

1 + v22.

Now we can easily see:

Lemma 3.2. If X and Y are length metric spaces then X × Y is alength metric space as well. An arclength parametrized curve in X×Yis a curve of shortest length if and only if its projections to X and Yare curves of shortest length parametrized proportionally to arclength.

Proof. Given two points z1 = (x1, y1) and z2 = (x2, y2), we find curvesγ1 in X and γ2 in Y connecting x1 to x2, respectively, y1 to y2, suchthat `(γ1) is arbitrary close to d(x1, x2) and `(γ2) is arbitrary close tod(y1, y2).

We now reparametrize γi to be defined on [0, 1] and to have constantvelocity `(γ1). Then γ = (γ1, γ2) connects z1 and z2 and has constant

velocity√`2(γ1) + `2(γ2).

Choosing γi to (almost) realize the distance between x1 and x2, rep-sectively, between y1 and y2, we see that `(γ) gets arbitrary close tod(z1, z2). This shows the first statement.

The remaining statements are similar and left to the reader.

3.3. Euclidean cone. Let X be a metric space. The Euclidean coneCon(X) over X is defined as follows. As set Con(X) is

(X × (0,∞)) ∪ o = (X × [0,∞))/X × 0 .Elements of Con(X) are written as (x, t) =: t ·x, with t ≥ 0 and x ∈ X.All points 0 · x are identified to one point o, the origin of the cone.

We define the metric on Con(X) in two steps. In the first step wetruncate the metric on X by declaring

d(x, y) = mind(x, y), π .Thus d and d coincide locally around any point.

We think of X as sitting in the unit sphere, of the distance in X asthe ”angle” and the first step just cuts all the angles by π.

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Now we define the cone metric dc on Con(X) as follows. For anypoints (x, t) and (y, s) in Con(X) we find a pair of points x and y in

S1 ⊂ R2 such that d(x, y) = dS1(x, y).

Then we consider the points t · x and s · y in R2 and set

dc((x, t), (y, s)) := |t · x− s · y| .In particular, dc(o, (x, t)) = t for all (x, t) ∈ Con(X).We have a canonical embedding of X via x → (x, 1) onto the unit

distance sphere in Con(X) around the origin. Under this identification

dc(x, y) corresponds to d(x, y) as arclength to chord-length on the circle.More precisely,

dc(x, y) = 2 · sin(d(x, y)

2) .

By constructionCon(Sn−1) = Rn ,

if Sn−1 is considered with its natural (= of constant curvature one =induced intrinsic) metric.

Any map f : X → Y extends to a natural map Con(f) : Con(X)→Con(Y ) by

Con(f)(tx) := t · f(x)

If f preserves the distances then so does Con(f) and if f is 1-Lipschitzthen so is Con(f). The last statement is an elementary (but not abso-lutely trivial) statement in Euclidean geometry.

Embedding an interval isometrically into S1 we see

Lemma 3.3. If X is isometric to an interval of length α ≤ π thenCon(X) is isometric to the convex subset in R2 bounded by two raysenclosing angle α.

Moreover, by definition, or looking on S0, we see that for any pointsx, y ∈ X with d(x, y) ≥ π the cone Con(x, y) = R is isometricallyembedded into Con(X).

Now we can show:

Proposition 3.4. Con(X) is a metric space, which is a length spaceif so is X.

Proof. Clearly dc is symmetric, non-negative and vanishes only betweenidentical points.

In order to prove the triangle inequality, consider zi = (ti · xi) ∈Con(X) for i = 1, 2, 3. We then find a map f : x1, x2, x3 → S1 sothat for the images xi = f(xi) we have

d(x1, x3) = dS1

(x1, x3) ; d(x1, x2) ≥ dS1

(x1, x2) ; d(x2, x3) = dS1

(x2, x3) .8

Looking at Con(f) and the triangle inequality in R2 we see

dc(z1, z3) = dc(z1, z3) ≤ dc(z1, z2) + dc(z2, z3) ≤ dc(z1, z2) + dc(z2, z3) .

This shows that Con(X) is a metric space.Assume now that X is a length space. Let zi = (ti, xi) ∈ Con(X) be

as above. If t1 · t2 = 0 or d(x1, x2) ≥ π then z1 and z2 are connected bya radial shortest curve with length d(z1, z2) = t1 + t2.

If d(x1, x2) < π, consider a shortest curve γ : I → X if it existsconnecting x1 and x2. If it does not exist apply the subsequent argu-ment to almost shortest curves between x1 and x2. We may assumethat γ is parametrized by arclength, thus γ : I → X is an isometricembedding and so is Con(γ). But Con(I) is a convex subset of R2,which is a geodesic spaces. Connect preimages of zi in Con(I) by a(unique) shortest curve in Con(I) and sent it by Con(γ) to Con(X) toobtain a shortest curve between z1 and z2 of the right length.

3.4. Quotient spaces. Let X be a metric space and let ∼ be an equiv-alence relation on X.

Define the function d∼ : X ×X → [0,∞) as follows.

d∼(x, y) = inf∑

d(qi, pi+1) ,

where the infimum is considered over finite sequences p1, q1, p2, q2, ..., pm, qmsuch that x = p1, y = qm and pi ∼ qi for all i.

Then d∼ ≤ d, d is symmetric and satisfies the triangle inequality.Moreover, d∼(x, y) = 0 if x ∼ y.

(It is easy to see, that d∼ is the largest function with these threeproperties).

We set x ≡ y if d∼(x, y) = 0. It is again an equivalence relation onX, coarser than the equivalence relation ∼.

Define the space X/ ∼ as the set of equivalence classes [x] of the

relation ≡ with the metric d∼([x], [y]) := d∼(x, y). We have:

Lemma 3.5. The space X/ ∼ is a metric space. The canonical mapx→ [x] from X to X/ ∼ is 1-Lipschitz. If X is a length space then sois X/ ∼.

Proof. All statements but the last one follow directly by construction.To find a curve of short d∼-length between [x] and [y], we consider asequence pi, qi as in the definition of d∼(x, y) and connect qi with pi+1 bya curve γi of short length in X. Now observe, that the concatenationof the projections of the curves γi to X/ ∼ is a continuous curve inX/ ∼, whose length can be chosen arbitrary close to d∼(x, y).

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The definition seems complicated but the construction should beintuitively clear, at least in most important examples. Here are somecomments and examples.

1) Let A ⊂ X be a subset. Consider ∼:=∼A by letting two pointsbe equivalent iff they are equal or both contained in A. Then

d∼(x, y) = mind(x, y), d(A, x) + d(A, y) .A very simple (but already very strange) example arises if X = R2 andA = B1(0). Then X/ ∼A is a space homeomorphic to R2 and locallyisometric to R2 outside one ”thick point”, the image of A. Note thatany curve which runs around this point (remember from topology orcomplex analysis what it means) has length at least 2π!

2) It can happen that d∼(x, y) = 0 also if x and y are not equivalentwith respect to ∼. In other words, ∼6=≡. For example, it happens ifA in the previous example is not closed in X.

3) Let X = [0, 2π] and let A = 0, 2π in the first example. ThenX/ ∼A= S1.

4) For X = [0, 1] × [0, 1] and the equivalence relation ∼ identifyingopposite sides of X, it can be shown (Exercise!) that X/ ∼ is the directproduct S1 × S1

5) Let X1, X2 be metric spaces, let A be a subset of X1 and letφ : A → X2 be a map. Consider the metric space X given as thedisjoint union of X1 and X2, where we define the distance betweenpoints in the same Xi as in original space Xi and between x1 ∈ X1 andx2 ∈ X2 to be infinity (by an abuse of definition). Consider now on Xthe equivalence relation ∼φ, where the only non-singleton equivalenceclasses are subsets f−1(x2) ∪ x2 for x2 ∈ φ(A1).

Let us explore the space X/ ∼φ in some special cases:If X2 is a point, then X/ ∼φ is the space X/ ∼A from example (1).If A = X1 and φ is 1-Lipschitz then X/ ∼φ is isometric to X2.Let now φ : A→ φ(A) ⊂ X2 be isometric. Often, one would call the

arising space X/φ the gluing of X1 and X2 along the isometry φ. Inthis case, the natural projection from X to X/ ∼φ sends X1 and X2

isometrically onto their images.For x1 ∈ X1 and x2 ∈ X2 the distance satisfies

d∼(x1, x2) = infa∈A

(d(x1, a) + d(x2, φ(a)) .

4. Examples of gluings. Simplicial complexes

4.1. Observation about quotient spaces. While it is often intu-itively clear, what a quotient space, repsectively a gluing of spacesshould look like, the formal verification is often tedious. Here are a few

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helpful observations. We restrict ourselves mostly to length spaces,even to the most important case of such.

We call a length space X a geodesic space if any pair of points ofX are connected by a shortest curve. Recall that by the theorem ofCohn-Vossen and Hopf–Rinow, every complete, locally compact lengthspace is geodesic.

Lemma 4.1. Let X, Y be geodesic metric spaces, let f : X → Y bea surjective map. Denote by ∼=∼f the equivalence relation on X,whose equivalence classes are exactly the fibers of f . Then f inducesan isometry f : X/ ∼f→ Y if the following conditions hold true.

(1) f is 1-Lipschitz(2) For every shortest curve γ : [a, b]→ Y there exist a = t1 ≤ t2 ≤

... ≤ tm = b and lifts of restrictions γ|[ti,ti+1] to shortest curvesηi : [ti, ti+1] → X of the same length. Thus, `(ηi) = `(γ|[ti,ti+1])and f ηi = γ|[ti,ti+1].

Proof. Since f is 1-Lipschitz, the definition of the distance d∼ in X andthe triangle inequality in Y imply that

d∼(x, x′) ≥ dY (f(x), f(x′)) ,

for all x, x′ ∈ X. Therefore, f(x) = f(x′) if and only if d∼(x, x′) = 0.Thus, f induces a bijective 1-Lipschitz map f : X/ ∼f→ Y .

Now consider arbitrary y, y′ ∈ Y connect them by a geodesic γ.Applying the second condition, subdivide γ and take pi = ηi(ti) andqi = ηi(ti+1). Then, by definition, we see that x = f−1(y) and x′ =f−1(y′) satisfies

d∼(x, x′) ≤ `(γ) = d(y, y′) .

Thus, f is an isometry.

For example, this lemma allows us to verify directly the intuitivelyclear Example (3) in the previous section. It also shows the following.If X is a convex sector in R2 of angle α and ∼ identifies all pairs ofpoints on the boundary rays of X at the same distance from the origin,then the arising space is the Euclidean cone over a circle of length α.

Another useful observation says that under some often encounteredassumptions, one obtains a small neighborhood in the quotient spaceby the quotient of corresponding small neighborhoods. Here is a moreprecise statement, the proof is much less complicated than the state-ment and is left to the reader.

Lemma 4.2. Let X be a metric space and ∼ be an equivalence relation.Let x ∈ X and ε > 0 be given. Assume that the ε-neighborhood Bε in

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X of the equivalence class [x] of x is a union of equivalence classes.Assume that for all z, z′ ∈ Bε with z′ ∼ z we have d([x], z) = d([x], z′).

Then the ball B ε2([x]) in X/ ∼ is isometric to the quotient B ε

2/ ∼ of

the ε2-neighborhood B ε

2of [x] in X.

4.2. Metric graphs. Let Ii be a disjoint union of compact intervals(for simplicity, finite or denumerable). Let ∼ be some equivalencerelation on the union of boundary points of the intervals. We (usually)require, that the endpoints of the same interval are not in the sameequivalence class.

Define now a space by gluing the intervals along the equivalencerelation. Thus, we consider first the disjoint union of the intervals withthe distance between points in different intervals set to be infinity.Then use the equivalence relation and obtain a quotient space. Thearising metric space is called a metric graph. We will mostly restrictourselves to one connected component of it.

In a metric graph, we have edges and vertices. Any point in theinterior of an edge has a neighborhood isometric to an open interval.We have a canonical embedding of any Ij into the arising metric graphZ. Note that the embedding is length preserving but it may not be anisometry (example?!).

Assume now that all intervals Ii which are adjacent to a vertex xhave lengths bounded from below. Then a neighborhood of this pointx in Z is isometric to a small ball in a gluing Y of a number of raysalong the common endpoint. In other words, Y is the cone Con(X)over a discrete metric space X at which all points have distance π fromeach over.

On the other hand, it is not very difficult to prove:

Proposition 4.3. Let X be a locally compact length metric space inwhich every point has a neighborhood isometric to a ball in the coneover a finite space. Then X is a metric graph.

4.3. Simplical complexes in higher dimensions. A general metricsimplicial complex is defined similarly to metric graphs.

We consider a disjoint union of convex Euclidean simplices ∆i (pos-sibly of different dimensions). For any two different simplices i, j werequire the existence of some (possibly empty) face Ai of ∆i and Aj of∆j and an isometry φij : Ai → Aj. Note, that a face here mean a faceof any codimension. It can be the empty set, a vertex, an edge, .... ,or the whole simplex.

Now we construct a space X by gluing the simplices ∆i along theisometries φij.

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We can get a good feeling of the construction by gluing the bound-ary of tetrahedron from 4 triangles or the boundary of cube from 12triangles.

Very often (always for the rest of this course) we will require thatat any point only finitely many triangles are glued together, in otherwords, any equivalence class arising in the course of the constructionis finite.

For any point in any simplex, any sufficiently small ball around anypoint is isometric to a ball in a Euclidean cone. This (and some notquite trivial thoughts) show that in any metric simplical complex asmall ball around any point is isometric to a ball in some Euclideancone. In fact there is a (highly non-trivial) converse:

Theorem 4.4. Let X be a locally compact, complete length space. As-sume that any point in X has a neighborhood isometric to a ball insome Euclidean cone. Then X is isometric to a simplicial complex.

In the lecture, I will explain the proof of the theorem only in thefollowing special situation. A simpicial surface is a locally compactlength space, in which every point has a neighborhood isometric to aball in a cone over some circle or in a cone over an interval.

With a little knowledge in algebraic topology, it is possible to prove,that a metric simplicial complex is a simplicial surface if and only ifit is homeomorphic to a two-dimensional manifold (with boundary),thus if every point has a neighborhood homeomorphic to plane or to ahalfplane.

4.4. Total angles. Let X be a simplicial surface. For any point x ∈ Xa small ball Bε(x) is isometric to the ε-ball in Con(Σx), where Σx is acircle or an interval. The isometry type of Σx is uniquely determined,for instance by the topology and lengths of small distance circles aroundx in X. We will call Σx the space of directions in x. Its length will becalled the total angle at x.

The set of points where Σx is an interval is a closed subset, theboundary ∂X of X. The complement is the interior of the surface

For all but discretely many points x ∈ ∂X, the total angle at x is πand for all but discretely many points in X \ ∂X the total angle at xis 2π.

The corresponding points will be called the (interior, respectivelyboundary) vertices of the surface. At each interior vertex the valueω(x) = 2π− `(Σx) will be called the curvature at x. At each boundaryvertex the value κ(x) = π−`(Σx) will be called the turn of the boundaryat x.

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For instance, consider the boundary X of the tetrahedron. It is asurface without boundary. At each vertex of the tetrahedron the totalangle is π, hence the curvature is π as well.

Consider now the surface Y of the tetrahedron with one face deleted.Then there remains one interior vertex with total angle π and 3 bound-ary vertices with turn π

3at each of them.

5. Gauß–Bonnet in simplicial surfaces (05.05)

5.1. Signed measures. Here and later a signed measure ω on a locallycompact topological space X is a difference of two Borel measures ω1−ω2 each of them being finite on compact subsets. A Borel functionf : X → R is integrable with respect to ω if it is integrable withrespect to ω1 and ω2. And we set∫

f dω :=

∫f dω1 −

∫f dω2 .

In particular, for any Borel subset A contained in a compact subset Kof X the value

ω(A) = ω+(A)− ω−(A) =

∫X

χA dω

is well defined.All continuous functions with compact support are integrable. It is

a deep theorem of Riesz, maybe discussed in a lecture on functionalanalysis, that the set of signed measure is exactly the dual space (withrespect to the right topology) of the topological vector space of contin-uous functions with compact support.

Another non-trivial related statement is that any signed measure ωcan be uniquely represented as a difference of two measures ω = ω+−ω−with the following property. There exist a Borel decomposition X =A+ ∪ A− such that ω+(A−) = ω−(A+) = 0.

These measures ω± are called the positive and the negative part ofthe signed measure ω. The measure

|ω| := ω+ + ω−

is called the variation of the signed measure ω.The principal example one should have in mind is the following. Let

µ be some Borel measure on X, which is finite on compact subsets. Leth : X → R be a Borel function, which is integrable over any compactsubset. Then we have the signed measure h · µ on X such that for any

14

Borel subset A contained in a compact subset of X

h · µ(A) =

∫X

χA d(h · µ) =

∫A

(h) dµ .

Setting A+ = h−1([0,∞)) and A− := h−1((−∞, 0)) we have a disjointdecomposition of X as X = A+ ∪ A−. The positive and the negativeparts of h · µ are exactly the measures

(h · µ)±(B) :=

∫B∩A±

h .

The variation of h · µ is the measure |h| · µ.Indeed, every signed measure arises (non-uniquely) in the way de-

scribed above. Indeed, any signed measure ω can be written as h · |ω|,where h = χA+ − χA− .

5.2. Curvature and turn measures of a simplicial surface. LetM be a simplicial surface with (possibly empty) boundary ∂M . We de-fine two signed measures on M , the curvature measure ω, concentratedin M \ ∂M and the turn κ concentrated on ∂M .

For any subset B contained in a compact subset of X we set

ω(B) :=∑x∈B

ω(x) ,

where ω(x) is the curvature at the point x ∈ X, thus the differencebetween 2π and the total angle in x. We obtain the positive and thenegative part ω± by restricting in the above formula the summation toall points in B with positive, respectively, negative curvature.

Note that the set V of points in X \ ∂X with ω(x) 6= 0 is discrete inX. Thus the following sum of Dirac measure

µ :=∑x∈V

δx

is a locally finite (counting) measure on X. The curvature measure ωis nothing but ω · µ, where we see ω as a locally integrable functionwith respect to µ.

We call ω(B) the curvature of the set B and |ω|(B) the absolutecurvature of B.

Similarly, we define the turn as a measure concentrated on all pointsv ∈ ∂X with non-zero turn κ(v). Again, for any subset B contained ina compact subset of X we define the turn of the boundary of B as

κ(B) :=∑

v∈B∩∂X

κ(v) .

By construction |κ|(X \ ∂X) = |ω|(∂X) = 0.15

5.3. Gauß–Bonnet in simplicial surfaces. We can now formulatethe global theorem of Gauß– Bonnet for simplical surfaces.

Theorem 5.1. Let M be a compact simplical surface with possiblyempty boundary ∂M . Let ω denote the curvature measure and let thesigned measure κ denote the turn of the boundary. Then

ω(M) + κ(∂M) = 2πχ(M) .

The proof was explained in the first section for the case of emptyboundary. The same proof applies here. We write M as any simplicialsurface, as a gluing of Euclidean triangles. The sum s of all angles of alltriangles is exactly π ·F , where F is the number of triangles. Summingthe angles ”‘vertice-wise”, we get s = sint + sbound, with

sint =∑

v∈M\∂M

(2π − ω(v)) and sbound =∑v∈∂M

(π − κ(v)) .

We arrive at

π · F = 2π · Vint − ω(M) + π · Vbound − κ(∂M) ,

where Vint and Vbound denote the number of interior, respectively bound-ary vertices of M . This shows

ω(M) + κ(∂M) = 2π · V − π · Vbound − π · F .

It is now a combinatorial exercise to see that π · Vbound + π · F equals2πE − 2πF , with E the number of edges. This finishes the proof.

5.4. Local version of Gauß–Bonnet. Let again M be a simplicialsurface, let x ∈ M be a point and let Γ be the concatenation of twoshortest curves starting in x. A small ball B around x is subdividedby Γ into two parts Ul and Ur (the ”left” and ”right” connected com-ponents). The closure of this two parts in B is exactly Ul,r ∪ (Γ ∩ B).These are two simplicial surfaces with boundary Γ ∩ B. There are nointerior vertices in Ul,r and at most one boundary vertex x. We havethe turns κl(x) of Γ at x with respect to the surrounding manifold Ul,r.

By construction the sum of the left and the right turns κl(x) +κr(x)is exactly the curvature ω(x) in the original space.

Let now U ⊂ M be an open subset homeomorphic to R2 and letΓ ⊂ U be a simple closed curve. Assume that Γ is a polygon, thus aconcatenation of shortest curves.

By the Jordan curve theorem (see next section), there exists a uniqueconnected component O of U \ Γ homeomorphic to a an open disc.Moreover, O = O ∪ Γ is homeomorphic to a closed disc.

16

The set O in its intrinsic metric is a simplical surface with boundaryΓ. At any point x ∈ Γ we have the turn κ(x) of Γ with respect to Oand the theorem of Gauß–Bonnet applied to O gives us

ω(O) + κl(Γ) = 2π ,

where κl(Γ) is the turn of the polygon Γ inside O.An important point of caution: Unlike in Riemannian geometry,

even for a triangle Γ, the contribution of the sides of the triangles tothe turn of Γ may be non-trivial. The reason is that a shortest curvein M can pass through vertices v of M . However, this can only happenif ω(v) ≤ 0. Also the contribution of such a point to the turn cannotbe positive.

6. Topology of surfaces (07.05)

6.1. Jordan, Schoenflies. In this and the next subsection we discusssome classical results in the theory of surfaces. The proofs are too longand too complictaed to be included here.

The following seemingly trivial theorem, a strengthening of the the-orem of Jordan, due to Schoenflies, is surprisingly difficult to prove.

Theorem 6.1. Let Γ be a simple closed curve in R2. Then there existsa homeomorphism F : R2 → R2 which sends Γ onto S1.

In particular, R2 \ Γ has two components, one bounded and one un-bounded. Both components have Γ as their boundary. The closure ofthe bounded component is homeomorphic to a closed disc.

Also in contrast to higher dimensions simple non-closed curves can-not be wild. This can be deduced by a trick from the above theorem:

Corollary 6.2. For any compact simple curve Γ in R2 there exists asimple closed curve Γ containing Γ. There exists a homeomorphismfrom R2 to R2 which sends Γ to the interval [0, 1] ∈ R.

Thus, whatever we can describe in topological terms for the circleor a segment in R2 we can transform to any arc and an simple closedcurve, called a Jordan curve.

For instance, any arc locally ”‘separates”’ a neighborhood in twoparts and we can talk about convergence of points or curves from agiven side to the arc. In other words, we can locally distinguish leftand right sides of the arc (without knowing which side is left and whichis right). In order to fix right and left we would need an orientation ofR2 and of the arc.

17

6.2. Topological surfaces. A surface with boundary is a metric (orjust metrizable) topological space in which any point has a neighbor-hood homeomorphic to a plane or to a halfplane. Based on the theoremof Schoneflies the following result has been proved by Rado:

Theorem 6.3. Any surface admits a triangulation. In other words,any surface is homeomorphic to some simplicial surface.

The triangulation is definitely not unique, but for compact surfaces,the Euler characteristic of any simplicial surface homeomorphic to itis a topological invariant. Thus χ(M) is well-defined. It essentiallydetermines all of the topology of M as follows from the classificationof surfaces:

Theorem 6.4. Two compact surfaces M1,M2 are homeomorphic if andonly if they have the same number of boundary components, the sameEuler characteristic and are both orientable or non-orientable.

From the theorem about the existence of the triangulation or fromthe classification result above the following consequence can be derived:

Theorem 6.5. Any topological surface admits a smooth atlas. It ad-mits a smooth Riemannian metric and a geodesic triangulation, suchthat the boundary is a disjoint union of geodesic polygons.

7. Comments on convex simplicial surfaces (07.05).

The following result might be intuitively clear. We explain the ideaof the proof, skipping some details

Proposition 7.1. Let X 6= R3 be a convex cone in R3, thus a closedconvex subset invariant under multiplication with non-negative num-bers. Let Y be its boundary equipped with the induced intrinsic metric.Then Y is a Euclidean cone over a circle of length ≤ 2π.

Proof. Y with the restricted metric is a Euclidean cone Con(Z), whereZ is the intersection of Y with the unit sphere. Hence, Y is a Euclideancone with respect to the induced intrinsic metric and we only need toshow that Z has length ≤ 2π (Exercise).Z is the boundary in S2 of the convex subset X ⊂ S2. Approxomat-

ing Z by a polygon, it suffices to prove that a polygonal boundary ofa convex subset of S2 has length at most 2π. This is shown by fixingan upper bound on the number of sides, maximizing the length andproving that this maximal object must have length 2π.

A direct consequence of this statement is:18

Corollary 7.2. Let M be the boundary of a convex polyhedron in R3.Then M is a simplical surface homeomorphic to S2 and the curvaturemeasure ω on M is non-negative.

This is the discrete analogue of the result well-known from a lecturein differential geometry saying that a smooth convex surface in R3 hasnon-negative Gauß curvature everywhere.

The converse is a deep theorem of Alexandrov. Namely, every non-negatively curved compact simplical surface homeomorphic to S2 isisometric to the boundary of a convex polyhedron in R3 uniquely de-termined up to rigid motions.

This result together with the observations that any convex body inR3 can be approximated by convex polyhedra establishes a very preciseconnection between smooth surfaces of non-negative curvature, convexbodies in R3 and simplicial surfaces of non-negative curvature. Thisconnection lies in the origin of the theory of surfaces with boundedintegral curvature, which we want to approach soon.

8. Angles in metric spaces (12.05)

8.1. Definition of angles. For any 3 points A,B,C in a metric spaceX there exists a triangle A, B, C ∈ R2 with side lengths equal tod(A,B), d(B,C), d(C,A), which is unique up to motions of the Eu-clidean plane. It is called the comparison triangle of the triangle (herejust 3 points) A,B,C.

The comparison angle ∠A(B,C) ∈ [0, π] is the angle at A of thiscomparison triangle.

Let now X be a length space and let γi : [0, εi] → X for i = 1, 2 beshortest curves starting in the same point x = γi(0). For any s, t > 0,

consider the comparison angle α(s, t) := ∠x(γ1(s), γ2(t)) and define theupper angle ∠(γ1, γ2) between γ1 and γ2 as

∠(γ1, γ2) := lim sups,t→0

α(s, t) .

We say that the angle between γ1 and γ2 is well-defined (and equals∠(γ1, γ2)) if the lim sup in the definition is really a limit.

First, a simple example. In the Euclidean cone Con(Σ) over any(length) space Σ the angle between any shortest curves γ1, γ2 startingat the origin o of Con(Σ) are well-defined. The curves γi have theform γi(t) = t · xi for some xi ∈ Σ and the angle between γ1 and γ2

equals minπ, dΣ(x1, x2), by the construction of the Euclidena cone.In particular, this applies to all simplical metric spaces at all points ofthe spaces.

19

The following example is much less trivial and requires some knowl-edge in Riemannian geometry. If M is a Riemannian manifold andγ1 and γ2 are arbitrary shortest curves starting in a point x, then γihave well-defined directions vi ∈ TxM and the angle between γ1 andγ2 is exactly the angle between the vectors v1 and v2 in the Eucldieanspace TxM . (Exercise: try to prove it!). Note that the local theoremof Gauß–Bonnet discussed in the first lecture was using exactly thisnotion of angles.

The last statement becomes much easier (Exercise: try to provethis!), if in the definition of angle one does not consider arbitrary limitss, t→ 0 but only such limits for which s/t remains bounded away from0 and∞. This would provide an alternative definition of (upper) anglesand everything what follows would work with that definition as well.

The next example is even less trivial and requires some knowledge inconvex geometry. Let X be a normed vector space. The angle betweenany pairs of shortest curves starting in the same point exist if and onlyif the norm comes from a scalar product (Exercise: try to prove this,at least check it for some examples).

8.2. Space of directions. Let X be a length space, x ∈ X a point.Consider the set Γx of all shortest curves starting in x. Then the upperangle ∠ : Γx × Γx → [0, π] is symmetric, vanishes on the diagonal andsatisfies the triangle inequality (Exercise: Prove that!).

Identify all shortest curves which include an upper angle of 0 anddenote by Σx = Γx/ ∼ the space of all such equivalence classes. Notethat ∠ defines now a metric on Σx. Either the space Σx or its metriccompletion Σx is called the space of directions of the space X at thepoint x

From the above examples, we see that for a Euclidean cone Con(Σ)the space of directions Σo at the origin o of the cone is identified withthe space Σ on which the metric is truncated by π.

For a Riemannian manifold M and any point x ∈ M the space ofdirections Σx = Σx is canonically identified (what does this mean?)with the unit sphere Sn−1 in the Euclidean space TxM .

An easy example of a compact space for which the space of directionsΣx at some point x is not complete is the closed unit disc D ⊂ R2

(Question: why?). Answering this question, you might also recognize inthis example, why it might be more natural to consider the completionΣx instead of Σx.

8.3. First formula of variation. Recall that for a point x ∈ R2,and a shortest curve γ(t) = y + t · v starting at a point y 6= x the

20

function f(t) = d(x, γ(t)) satisfies f ′(0) = − cos(α), where α is theangle between γ and the segment [yx] at the point y (Exercise: Provethat). Remember, that a similar statement holds in any Riemannianmanifold (Question: what statement?).

It turns out that half of this formula holds true in any length metricspace:

Proposition 8.1. Let x, y be points in a length space X. Let γ be anyshortest curve connecting y and x. Let η : [0, ε] → X be any shortestcurve starting in y. Set f(t) = d(x, η(t)). Then the 1-Lipschitz functionf : [0, ε]→ R satisfies

f+(0) := lim supr→0

f(r)− f(0)

r≤ − cos(∠(γ, η)) .

Here is a sketch of the proof (Exercise: try to fill in the details!): Forall small s we have

f(r)−f(0) ≤ (d(x, γ(s))+d(γ(s), η(r))−(d(x, γ(s))+s) = d(γ(s), η(r))−s .

Now, for all s, r, sufficiently small, the comparison angle ∠x(γ(s), η(r)))is almost bounded from above by ∠(γ, η). Now, if r is much smaller thans we can estimate s− d(γ(s), η(r)) by the first formula of variation inthe Euclidean comparison triangle.

9. Main definition (14.05)

9.1. Excess of a triangle. Let X be a length space. A triangle in Xconsists of 3 points A,B,C and three shortest curves [AB], [BC], [CA]connecting them. Note that the sides [AB], [BC], [AC] are not uniquelydetermined by the vertices A,B,C. Nevertheless, we will denote thetriangle consisting of these three sides by ABC.

Let X be locally compact, and U be an open subset of X, such thatU is compact. For all points A,B,C ∈ U whose pairwise distancesare smaller than the distance d(A,B,C, ∂U), the points A,B,C de-termine (at least) one triangle ABC. Moreover, any such triangle iscontained in U (Exercise: Prove the last statements).

Denote by α, β, γ the upper angles of the triangle ABC at A,B,C.Thus, α is the upper angle between the sides [AB] and [AC] at A andso on. The upper excess of the triangle ABC is the quantity

δ(ABC) := α + β + γ − π .

Let α0, β0, γ0 be the comparison angles: α0 = ∠A(B,C) and so on.Then (why?)

δ(ABC) = (α− α0) + (β − β0) + (γ − γ0) .21

9.2. Some notions of convexity. A subset G of a length space Xwill be called convex, if any two points in G are connected in G by ashortest curve. The subset G is called completely convex if, in addition,all shortest curves in X between any pair of points in G is contained inG. (Exercise: what are examples of convex but not completely convexsubsets? Can the whole space X be non-convex?)

(Exercise: An intersection of completely convex subsets is completelyconvex. An intersection of convex subsets may fail to be convex).

Let X be a length space homeomorphic to a surface, let U ⊂ X bean open subset homeomorphic to a disc. Let Γ be a simple closed curveof finite length in U . Let finally C be the closed disc in U bounded byΓ, the Jordan domain of C. We say that C is convex relative to theboundary if the following conditions hold true.

(1) The distance from C to ∂U is at least 4 times the length of Γ,also denoted as the perimeter of C.

(2) For any pair of points A,B on Γ, any simple arc Γ′ outside ofC connecting A and B in U divides Γ in two arcs Γ+ and Γ−. One ofthese parts, say Γ+ is closer to Γ′ (what does it mean precisely?). Wenow require, that for any such arc Γ′, the length of Γ′ is not smallerthan the length of arc Γ+.

Any subset convex relative to its boundary is convex (Exercise!).But it may not be completely convex (Example?). It is slightly morecomplicated to find examples showing that a completely convex subsetC bounded by a rectifiable curve may violate the second conditionabove.

We end this subsection with two examples. Assume first that U isa completely convex open disc in X. Assume moreover, that shortestcurves between any pair of points in U are uniquely determined by theendpoints. (Recall that such an U exists if X has a Riemannian metric.Prove, using Gauß–Bonnet that the uniqueness statement is true if Xis a polyhedral surface and the curvature measure is non-positive).

Then a subset C of U is convex if and only if it is completely convex.Moreover, if a Jordan curve Γ in U bounds a convex subset C andd(C, ∂U) ≥ 4·`(Γ) then C is convex relative to the boundary (Exercise,several steps are needed for the solution).

In the second example we consider the cone X = Con(Σ) over acircle of length 2α < 2π. Take a triangle ABC, with A the vertex ofthe cone d(A,B) = d(A,C) and the (comparison=upper) angle at Aequal to α. Then the triangle is convex but not completely convex norconvex with respect to the boundary.

22

9.3. Simple triangles in surfaces. Let a length space X be home-omorphic to a surface. We call a triangle T = ABC a simple triangleif the sides of T form a simple closed curve which is contained in anopen subset U homeomorphic to a disc. Moreover, we require T to beconvex relative to its boundary in U .

For example, any sufficiently small non-degenerated in any Riemann-ian surface is simple, if it is not contained in a single geodesic.

In a cone over a circle of length ≥ 2π any triangle is simple if the sidesintersect only in vertices (Exercise: why is the additional conditionneeded?). In a cone over a circle of length < 2π there are many non-simple triangles.

9.4. Main definition. We call two simple triangles in a length spaceX homeomorphic to a surface non-overlapping if their interiors (theirJordan domains) do not intersect.

We say that a length space X homeomorphic to a surface is a surfaceof bounded curvature if any point x ∈ X admits an open neighborhoodU homeomorphic to a disc and a constant Ω(U) > 0 with the followingproperty. For any finite sequence of non-overlapping simple trianglesT1, ..., Tm ⊂ U , we have

δ(T1) + ....+ δ(Tm) ≤ Ω(U) .

Note, that we only assume a one-sided bound on the sum of the defects.Rather surprisingly the opposite bound turns out (much much later)to follow as a consequence.

From the theorem of Gauß–Bonnet we deduce that any Riemanniansurface and any simplicial surface have bounded curvature in the senseabove (Exercise).

10. Statement of the main theorems. First steps to theproof.

10.1. Formulation. We now formulate central results of the theory ofAlexandrov–Zallgaler whose proofs will be provided in several subse-quent lectures.

Theorem 10.1. Let X be a surface without boundary with a lengthmetric d of bounded curvature. Then, for any point x ∈ X there existsa neighborhood U of x homeomorphic to a closed disc and a sequenceof metrics dn on U with the following properties.

(1) The metrics dn : U × U → R converge uniformly to the metricd : U × U → R.

23

(2) The metric spaces (U, dn) are simplical surfaces and there existsa constant C > 0 such that

|ωdn|(U) + |κdn|(∂U) ≤ C .

In the above theorem ωdn and κdn denote the curvature measure andthe turn-of-the boundary measure of the metric dn. By |ω| we denotethe variation |ω| := ω+ + ω− of the signed measure ω = ω+ − ω−.

It is possible to deduce from Theorem 10.1 that the metric on Xcan be locally approximated in the same sense by smooth Riemannianmetrics with bounded intregral of the Gaussian curvature and boundedintegral of geodesic curvature of the boundary.

With some more work it is possible to globalize the local statementand to prove the same result for ”large” open subsets U , in particular,if X is compact to the case X = U .

The second theorem is a converse of the first:

Theorem 10.2. Let X be a surface without boundary with a lengthmetric. Assume that every point x ∈ X admits a neighborhood U and asequence of simplicial metrics dn on U converging to d on U uniformly.Assume in addition, that the total positive curvatures (ωdn)+(U) areuniformly bounded. Then X is a surface with bounded curvature.

Note the difference, that in Theorem 10.2 we only require that thepositive parts of the curvature measures are uniformly bounded. Onthe other hand in Theorem 10.1 an existence of converging sequenceswith bounded positive and negative parts of the measure is claimed.

10.2. Very basic ideas of proofs and some problems. The basicidea of the proof of the first theorem is the following one. We will tryto find a triangulation of U in simple triangles of small diameter. Thenwe will construct a simplical surface, by replacing any triangle in U bya Euclidean triangle with the same sidelength. We can see the arisingsimplical surfaces as metrics on the original subset U .

Then we plan to show that these metrics dn converge to the originalmetric uniformly and that they inherit a uniform control on the upperbound of the curvature measures (ωdn)+. In addition, the constructionwill give us an upper bound for (κdn)+. Note that a bound on (ωdn)−

and (κdn)− will automatically follow from Gauß–Bonnet.The implementation of all parts of this program is elementary but

highly non-trivial. As an exercise, try to understand the followingdifficulty in a very innocently looking step of the above program.

Consider R2 with the sup-norm. Take the unit square U = [0, 1] ×[0, 1]. Consider a natural squaring, thus a decomposition of U into

24

axis-parallel squares of sidelength 1n. Now replace any square by the

Euclidean square with the same sidelengths. What is the arising metricdn on U and what is its limit for n→∞?

The naive idea of the proof of the second part would be to show thattriangles converge to triangles and defects converge to defects, underthe convergence of dn → d. Indeed, dn-shortest curves converge to d-shortest curves under this convergence (Easy exercise!). However, theangles may not converge to angles (Exercise: Find in the Euclideancone over a circle of length 6= 2π shortest curves γ±n connecting xnto y±, such that the angles between γ±n do not converge to the anglebetween the limit curves γ± at x = lim(xn).)

The real idea consists in finding some new geometric properties ofsimplical metrics with bounded curvature which are stable enough tosurvive the limiting procedure and which can be used to control thecurvature.

10.3. Weak convergence of measures. Our first aim will be theproof of the second theorem. As a preparation, we discuss convergenceof measures (one should have in mind the positive part of the curvaturemeasures as in Theorem 10.2)

As before all measures considered here are Borel measures which arefinite on compact subsets.

Let U be a locally compact metric space. Let ωn be a sequence ofmeasures on U . We say that the sequence ωn converges weakly to themeasure ω, if for any continuous function f : U → R with compactsupport we have

lim

∫U

f dωn =

∫U

f dω .

If such a weak limit exists it is uniquely defined (Exercise: why?).Some compactness results in functional analysis (Theorem of Banach-

Alaouglu, possibly seen in a lecture on functional analysis or PDE)imply the following

Theorem 10.3. Let the measures ωn be such that ωn(U) is uniformlybounded by a constant. Then the sequence ωn has a weakly convergingsubsequence.

Here are some typical examples of convergence:

Example 10.1. On U = [0, 1] the sum of the Dirac measures

1

n(∑i

δ in)

converges to the Lebesgue measure.25

Example 10.2. For a sequence of non-negative continuous functionsfε : R→ R with support in [−ε, ε] and integral 1, the measures fε · L1

converge to the Dirac measure δ0.

Example 10.3. If continuous functions fj : [0, 1] → [0,∞)] convergeuniformly (or in L1) to a function f : [0, 1] → [0,∞) then fj · L1

converge to f · L weakly.

In general, let the measures ωn weakly converge to ω. Then, for anyclosed set A ⊂ U we have

lim supn→∞

ωn(A) ≤ ω(A) .

and for any open subset O ⊂ U we have

lim infn→∞

ωn(O) ≥ ω(O) .

Exercise: Prove the semicontinuity statements; prove the convergencestatements in the Examples above. Verify that equality does not needto hold in the last two inequalities.

11. Proof of the second main theorem (26.05)

11.1. Beginning of the proof. Thus let x ∈ U ⊂ X be a point ina neighborhood in a surface X as in the formulation of Theorem 10.2.Consider any open subset x ∈ U0 homeomorphic to a disc, such thatU0 is compact.

After choosing a subsequence we may assume that the curvaturemeasures ω+

n converge weakly on U0 to a measure ω.In order to verify that X has bounded curvature, we need to prove

that for a finite number of non-overlapping simple triangles Ti in U0

the sum of the defects∑δ(Ti) ≤ ω(U0) ≤ ω(U0) <∞ .

Since ω is a measure and the set of inner points T 0i are pairwise disjoint,

we have∑ω(T 0

i ) ≤ ω(U0). Therefore, we have reduced the result tothe following version of ”‘limiting Gauß– Bonnet inequality”’:

Proposition 11.1. In the above notation, for any simple triangle Tcontained in U0 we have

ω(T 0) ≥ δ(T ) ,

where T 0 is the set of inner points of T .

In the proof of Proposition 11.1 the assumption that the triangle issimple will not play a role. In fact the result will be deduce for any

26

Jordan triangle T , thus a triangle which is the topological boundary ofa closed disc (which we also denote by T ).

11.2. Majorization. In order to prove Proposition 11.1, we considerthe following concept of majorization.

Let Γ and Γ′ be simple closed curves in metric spaces Y and Y ′

respectively. We call a map F : Y ′ → Y a majorization of Γ if the mapF is 1-Lipschitz and if it sends Γ′ bijectively onto Γ in an arclengthpreserving way.

We will also say in this case that the space Y ′ majorizes the curveΓ, without mentioning Γ′.

For example, the identity map F : Y → Y is a (not very useful)majorization

Let F : Y ′ → Y be a majorization as above. If a part η of Γ is ashortest curve (parametritzed by arclength) between two points A,Bon Γ then its preimage η′ in Γ′ is a shortest curve between its endpointsin Γ′ (Exercise: why is it the case?). Therefore, if Γ is a triangle thenΓ′ is a triangle as well.

We have the following

Lemma 11.2. Let F : Y ′ → Y with F (Γ′) = Γ be a majorization ofa triangle Γ in Y . Then any of the upper angles of the triangle Γ′ isnot smaller than the corresponding upper angle of the triangle Γ. Inparticular, δ(Γ′) ≥ δ(Γ).

Proof. Consider the sides [AB] and [AC] of Γ and corresponding sides[A′B′] and [A′C ′] of Γ′. Consider points X ∈ [AB], Z ∈ [AC] andcorresponding points X ′, Z ′.

Then d(A,X) = d(A′, X ′), d(A,Z) = d(A′, Z ′) and d(X ′, Z ′) ≥d(X,Z). Thus

∠A′(X′, Z ′) ≥ ∠A(X,Z) .

The claim follows by letting X, Y converge to A.

The following elementary but highly non-trivial example of majoriza-tion is important in metric geometry and will be used later:

Lemma 11.3. Let ABCD be a non-convex quadrangle in R2 with anon-convex angle at D. Let A′B′C ′ be the Euclidean triangle with

d(A′, B′) = d(A,B); d(B′, C ′) = d(B,C); d(A′, C ′) = d(A,D)+d(D,C) .

Then there exists a majorization F : R2 → R2 of Γ = ABCD by Γ′ =ABC, which maps R2 onto the closed Jordan domain of the quadrangleABCD.

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Proof. Put the plane into R3, consider a perpenicular ray to the planethrough the point D. Let O = O(t) be a point on this ray at distance tfrom D. Let T (t) be the union of the 3 upper sides of the pyramid withvertex O and base ABCD. (Thus the boundary without the base).

Equipped with the induced intrinsic metric it is a simplical surfacehomeomorphic to the closed disc. It has at most one inner vertex Oand at most 4 boundary vertices A,B,C,D. Note, that D is not avertex (Look at the total angle!).

Thus if the total angle at O equals 2π then T (t) is isometric to theEuclidean triangle A′B′C ′. (Non-trivial exercise!) Note further thatthe orthogonal projection from R3 to R2 provides a majorization ofABCD by T (t).

The total angle at O(t) varies continuously with t. For t → ∞ thisangle converges to 0. On the other hand, for t→ 0 the angle convergesto the sum of the total angle at D of ABCD and π which is larger than2π by concavity at D. (Exercise: prove the last statement!)

By continuity, we find some t0 ∈ (0,∞) for which the total angleat O(t0) equals 2π. The horizontal projection of T (t0) to ABCD is arequired majorization.

11.3. Reduction to a majorization theorem. Based on Lemma11.3 we will prove in the next lectures the following majorization the-orem:

Theorem 11.4. Let T be a simple triangle in the open subset U0 of asurface X as in Proposition 11.1. If ω(T 0) < 2π then in the Euclideancone Y ′ over the circle of length 2π − ω(T 0), there exists a convextriangle T ′ and a majorization F : T ′ → T .

The advantage of this approach to triangles in the ”‘limit space”’ isthat it is fomrulated only in terms of Lipschtiz maps and is thereforestable under limiting procedure. It will allow us to prove the theoremfirst in the simplicial setting and then going to the limit.

Given Theorem 11.4 as granted we can now easily finish the proof ofProposition 11.1 and, therefore, of Theorem 10.2.

Indeed, if ω(T 0) ≥ 2π then ω(T 0) ≥ δ(T ), since the defect is boundedfrom above by 2π (Exercise: check that!). Thus, we may assumeω(T 0) < 2π. Consider the majorizing triangle T ′ in the cone Y ′ pro-vided by Theorem 11.4.

Then δ(T ′) ≥ δ(T ) by Lemma 11.2. On the other hand, by theGauß–Bonnet theorem in Y ′ the defect δ(T ′) is bounded by the totalpositive curvature of the interior (T ′)0 of T ′. The whole curvature in

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Y ′ is concentrated in one point, the tip of the cone and equals ω(T 0).This finishes the proof.

12. Majorization theorem, simplical (28.05).

12.1. Formulation and plan of the proof. Subsequently, we willconsider simplicial discs. We will call a map F : X → Y between twosuch discs a majorization if F is 1-Lipschitz and F : ∂X → ∂Y is anarclength preserving homeomorphism.

The aim of this subsection is to prove the following simplicial versionof Theorem 11.4:

Theorem 12.1. Let X be a simplicial metric space homeomorphic to aclosed disc. Assume that the total positive curvature ω+(X) is less than2π. Then there exists a convex simplicial subset K in the cone Con(Σ)over a circle Σ of length 2π − ω+(X) and a majorization F : K → X.

The proof proceeds by a consecutive surgery on X. The surgerykeeps the boundary segments unchanged and decreases step by stepthe number of singularities, while keeping ω+(X) constant. The spacearising in each step admits a majorization of the predecessor.

The idea is simple but the precise proof is rather long. It relies on 3devices. The first one is the decomposition device explained in Subsec-tion 12.2 below, which allows us to simplify the structure of the space.The second device is Lemma 11.3, which allows us to get rid (with somecare) of non-convex quadrangles within X, and, in particular, to getrid of boundary vertices with negative turn. The last device is a gluingdevice, which allows us to replace 2 vertices of positive curvature by asingle one.

Exercise: Understand the statements of the devices.

12.2. A decomposition device. Let Z be a simplicial surface home-omorphic to a disc. Let η be a simple polygonal curve intersecting∂Z exactly in its endpoints. Then η separates Z into two simplicialsurfaces Z1, Z2 homeomorphic to discs. Let Xi be simplicial discs andlet Fi : Xi → Zi be majorizations (preserving the lengths of boundarycurves.) Let X be the space arising from X1 and X2 by gluing alongthe preimages of η in ∂Xi. Then X is a simplicial disc, and we have acanonical majorization F : X → Z (the gluing of F1 and F2).

Outside of the ”special” curve η1 = η2 = η in X the curvature of Xcoincides with the curvature of the corresponding points of Xi.

At all inner points x of the special curve η ⊂ X the change of thecurvature can be controlled as well, due to Lemma 11.2. We will onlyneed it for the case, when η is a shortest curve in Z or if it consists of

29

two shortest curves meeting at x. If η is a shortest curve in Z, thenalso its preimage is a shortest curve, in particular it does not containpoints with positive curvature. Moreover, if Xi have both non-negativeturn-of-the-boundary measure then x the curvature of the space X atthe point x is zero. If the curvature at x in Z is non-negative and if ηdivides the angle at x in Z in two equal parts, then, applying Lemma11.2 we see that the curvature at x in X is not larger than the curvatureof x in Z.

Similarly, if we find a simple closed polygonal curve η in Z andmajorize the Jordan domain Z0 of η by some space F : X0 → Z0, thencutting out Z0 from Z and gluing X0 instead we obtain a space Xmajorizing Z (Exercise: understand and formalize the statement).

12.3. A gluing device. The following Lemma allows us to assumethat in Theorem 12.1, we have at most one inner vertex of positivecurvature.

Lemma 12.2. Let X be a simplicial disc as in Theorem 12.1. If Xhas k > 1 vertices with positive curvature then we find a majorizationF : Y → X, such that Y has k − 1 vertices of positive curvature andsuch that the ω+(Y ) = ω+(X).

Proof. Let A,B be two vertices in X of positive curvature 2α and 2βrespectively. Connect A and B by a shortest curve [AB]. Note that[AB] does not contain points of positive curvature but A and B.

Take a Euclidean triangle A′B′C ′ such that d(A′, B′) = d(A,B) andthe angles at A and B are α and β. (Exercise: why is this possible?).

Then glue two copies of this triangle along their respective sidesA′C ′ and B′C ′. We obtain a simplicial disc Z with one inner vertexof angle 2π − 2(α + β) and the boundary consisting of two copies ofthe segment [A′B′]. There are two boundary vertices A′ and B′ withtotal angle 2α and 2β, respectively. Finally, we have a 1-Lipschitz mapG : Z → [AB] which sends both copies [A′B′] isometrically onto [AB](Exercise: construct the map G).

Now, we cut our original space X along the segment [AB] to obtain a

new simplicial metric space X, which contains two distinguished copiesof [AB]. If [AB] is contained in the set of inner points (the maincase one should understand first), the arising space is homeomorphicto a closed annulus with boundary consisting of ∂X and the simpleclosed curve built by the two copies of [AB]. (Exercise: understandthe meaning of cutting out and the topological statement above).

30

In the last step of our construction, we ”fill in the hole” by gluing Zto X, by identifying the boundary of Z with the simple closed curveconsisting of the two copies of [AB].

The arising space Y is a simplicial disc. Sending Z by the map Gto [AB] and sending the complement Y \ Z by the identity map weobtain a 1-Lipschitz map F : Y → X which preserves the length of theboundary, hence is a majorization (Exercise: why is it true?).

Finally, the inner vertices of Y in Y \Z coincide with the correspond-ing inner vertices of X. The points A′, B′ have curvature 0 in Y , thusare no vertices (Why?). In Z \A′, B′ ⊂ Y there is exactly one vertexof positive curvature, the vertex C ′ with total curvature 2α + 2β.

Thus, we have majorized X by a simplicial disc with one less vertexof positive curvature and such that the total positive curvature of Xand Y coincide.

Applying this Lemma and proceeding by induction on k, we mayassume from now on, that our space X has at most one inner vertex ofpositive curvature.

12.4. Preparations. We begin with a few general auxiliary state-ments about the geometry of simplicial surfaces.

Lemma 12.3. Let Y be a simplical surface homeomorphic to a disc.Let y ∈ Y be a point and assume that in Y \y the curvature measureis non-positive. Then every point x ∈ Y is connected with y by aunique shortest curve. This shortest curve γx depends continuously onthe endpoint x.

Proof. If two different shortest curves γx, γx exist, find a simple closedcurve build by parts of γx and γx. Apply now the theorem of Gauß-Bonnet to the Jordan domain of this simple closed curve to obtain acontradiction (Exercise: fill the details!).

Since shortest curves converge to shortest curves, uniqueness of γximplies continuity (Exercise: fill the details).

Lemma 12.4. Let Y = Con(Σ) be the Euclidean cone over a circle oflength 2α ≤ 2π. Let K ⊂ Y be a convex simplical subset homeomorphicto a closed disc. Then the turn-of-the-boundary measure κ of the spaceK is non-negative.

On the other hand, let Z be a simplicial surface homeomorphic to thedisc with non-negative turn-of-boundary measure κ ≥ 0. Assume thatZ contains at most one inner vertex and the curvature at this vertex is2π − 2α ≥ 0. Then Z is isometric to a convex subset K in Y .

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Proof. If at a boundary point p ∈ ∂K the turn κK(p) < 0 then, for anypoints on the different boundary segments starting in p any shortestcurve connecting them lies in Y \ K (Exercise: prove the statement,relying on the fact that the curvature of p in Y is non-negative).

Let, on the other hand, Z be as in the formulation of the lemma.Since κ ≥ 0, every shortest curve γ is either completely contained inthe boundary or no point of γ, but possibly its endpoints lie on ∂Z(Exercise: confirm this statement).

Let p be the unique vertex point or any inner point if no inner verticesexist. For any point z ∈ Z there exist a unique shortest curve γz fromp to z by the previous Lemma. Due to the last argument and due tothe fact that there is no branching of shortest curves in Z (exercise:what does it mean and why is it true??), for any points z1, z2 on theboundary ∂Z, the curves γz1 , γz2 and any of the arcs of ∂Z between z1

and z2 constitute a simple closed curve.In particular, if z1, z2 are two consecutive boundary vertices, we ob-

tain a Jordan triangle pz1z2. By construction, the curvature of such atriangle is 0 and the turn-of-the-boundary-measure is concentrated onthe vertices of the triangle.

We claim that the triangle is isometric to a triangle in the Euclideanplane. Once the claim is proven, we get a triangulation of Z by suchtriangles. Now we go around the boundary, and find trinagle b trianglethe same (thus the same simplical surface) in Con(Σ) (Exercise: makea picture!). The arising subset K of Con(Σ) is isometric in its intrinsicmetric to Z and it is convex in Con(Σ) by the first part of the Lemma.

It remains to prove the claim, that a simplicial surface T home-omorphic to a disc, with vanishing curvature and boundary being atriangle, whose turn-of-the-boundary is concentrated in the vertices isisometric to a Euclidean triangle. To prove the claim we just considera triangulation of T in Euclidean triangles and then find the same tri-angulation in a subset of R2 using the assumptions on the curvatureand boundary-of-turn (Exercise: understand the hidden but non-trivialarguments).

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09.06

12.5. Start of the proof. We now begin with the proof of theorem12.1 and start with the case of vanishing curvature, thus we assumethat ω = 0. Then we need to prove that X is majorized by a flatconvex polygon.

By the theorem of Gauß– Bonnet there are at least 3 boundary ver-tices with positive turn. We proceed by induction on the number b ofboundary vertices. If the turn at all boundary vertices is non-negative,the X is itself a convex polygon in R2 and we can use K = X. Inparticular, this covers the case b = 3. We assume the contrary and findsuch X with minimal possible number b > 3.

We find a shortest curve η connecting two boundary-vertices A,B ofX and intersecting the boundary only in A and B (Exercise: how do wefind it?). Then we apply the decomposition device and our inductive

assumption (thus the minimality of b) and find a space X which isglued from two convex polygons in R2 along a common side, such thatX majorizes X.

Note that the space X still has no curvature, the turn of the bound-ary is negative in at most two points (the points arising from A and

B) and X also has at most b boundary vertices.Assume that the turn of the boundary is negative at A. Then con-

sider the quadrangle AA+BA− in X, where A+ and A− are the neigh-boring boundary vertices of A.

This quadrangle is isometric to a non-convex quadrangle in R2 (aris-ing by gluing the Euclidean triangles AA+B and AA−B). To thisquadrangle we find a majorization by a Euclidean triangle A+BA−,due to Lemma 11.3. Replacing AA+BA− by this triangle we obtain aspace X ′ majorizing X (and thus X) which has one boundary vertexless (Exercise: why?). By the minimality of b we can majorize this new

space X ′ and thus X.

12.6. No positive curvature. Assume now that just the positive partω+ of the curvature measure vanishes in X. We then need to prove thatX is majorized by a convex polygon in R2 and proceed by inductionon the number v of inner vertices. The case v = 0 has been verifiedbefore. Assume the contrary and choose a non-majorizable X withminimal possible v.

Shortest curves in X are uniquely determined by endpoints. Fixingp ∈ ∂X, the union of all shortest curves γpq from p to points q ∈ ∂X isall of X. Indeed, otherwise, using continuous dependence of γpq on q wewould find a continuous deformation of ∂X to the point p outside some

33

subset of X. But this is impossible by topological reasons (Exercise:understand the details of the last argument).

Thus we find a vertex O and a shortest curve η connecting p withsome other boundary point q and passing through O. Making η shorter,we may assume that η intersects ∂X only in its boundary points. Thenη decomposes X into two simplical discs X1, X2 both of which have nopositive curvature and with the number of vertices in X1 and in X2

less then v. Due to the minimality of v we find majorizations Yi of Xi

by convex polygons in R2 and thus a majorization of X by the gluingY of Y1 and Y2 along the part of the boundaries corresponding to η.This part is indeed an edge of ∂Y1 and ∂Y2. Thus, the arising space Yhas no inner vertices.

Applying the previous point to the space Y , we obtain a majorizationof Y by a convex polygon C in R2 and, therefore, also of X.

12.7. One positive vertex. It remains to deal with the case of exactlyone positive vertex. First an auxiliary statement:

Lemma 12.5. Let Σ,Σ′ be two circles. Assume that the length of Σis not less than the length of Σ′. Let K be convex simplicial disc inCon(Σ) which contains the vertex A of the cone in its interior. ThenK is majorized by a convex subset K ′ in Σ′.

Proof. If `(Σ) = `(Σ′) we are done. Otherwise, we majorize our convexsubset step by step, each time finding a majorization in a larger cone.

We apply step by step a version of the gluing device from above. Ineach step we connect the vertex A with a boundary vertex B. Thenwe cut X along [AB]. Then we choose

2β := minκ(B), `(Σ)− `(Σ′) .We find a triangle with a side of length d(A,B) and angles α = π −12`(Σ) and β. We glue two such triangles as before and then glue the

arising simplicial disc to X to fill the hole we have produced.This replaces A by a new vertex A and the curvature at this new

vertex has been increased by 2β.The arising space C ′ has non-positive turn-of-the-boundary (Exer-

cise: why?). If β = `(Σ)−`(Σ′) we are done after this step. Otherwise,the space K ′ has one boundary vertex less than K and we finish theproof by induction. (Exercise: what happens at the base of the induc-tion? Gauß–Bonnet should be used!)

Assume now that the space X contains exactly one inner vertex pwith positive curvature. The point p is connected with any point inX by a unique shortest curve. We deduce that the union of shortest

34

curves γpx from p to boundary points x ∈ ∂X covers all of X (Exercise:recall this argument from above). In particular, we find such shorestcurves γpx and γpy such that they start in p in directions dividing thetotal angle at p in two equal halfes.

By running on these two geodesics until the first intersection pointswith the boundary, we obtain a simple polygonal curve η dividing Xinto two halfes X1, X2. Note that both halves do not contain verticesof positive curvature. Thus, we find majorizations of X1 and X2 byconvex polygons Y1 and Y2.

As before, gluing Y1 and Y2 we obtain a simplicial disc Y majorizingX. By construction, the space Y has at most one inner vertex p andat this inner vertex the total angle consists of two sectors both of themhaving at least a total angle of α (Exercise: why?). Thus, the curvatureat p is not larger than the curvature at p.

Thus, we have reduced the statement to the following

Lemma 12.6. Let X be a simplical disc with at most one inner vertexp. Assume that p has non-negative curvature ω0 ≤ ω = 2π − 2α.Assume, moreover, that the shortest curve from p to any point x ∈ ∂Xintersects ∂X only in x.

Then there exists a convex simplicial subset K in the cone Con(Σ),where Σ is a circle of length 2α and a majorization F : K → X.

Proof. Assume that there are boundary vertices with negative turn.Taking such a vertex D and connecting p with its boundary vertices Band C we find a non-convex Euclidean quadrangle triangle pDBC.

Majorizing this quadrangle by a triangle p′, B′, C ′ we majorize Xby a simplicial disc X ′ which has only one inner vertex p′ and suchthat the number of boundary vertices has decreased by one. Moreover,the curvature of p′ in X ′ is not larger than the curvature of p in X(Exercise: why?).

In this way, we reduce the question to the case, when the turn-of-the-boundary is non-negative. Now we can apply Lemma 12.5 and Lemma12.4 to finish the proof.

13. Majorization theorem, general case, 16.06

We return to the proof of Theorem 11.4. Thus, let on a closed discU0 a sequence dn of simplicial metrics be given, which converge in auniform way to a metric d, defining the same topology of a closed disc.

If a sequence of curves Γn : [a, b]→ U0 converge pointwise to a curveΓ then we have the semicontinuity (Exercise: verify the claim!):

lim infn→∞

`dn(Γn) ≥ `d(Γ) .

35

A more difficult exercise: find a sequence dn of metrics converginguniformly to the Euclidean metric on the unit disc, such that for somelinear segment Γ the equality lim `dn(Γ) =∞ holds true.

Lemma 13.1. Let Γ be a Jordan curve in U0. Then, there exist Jordancurves Γn converging to Γ, such that the following holds true. Thecurve Γn is a polygonal curve in the metric dn and the lengths `dn(Γn)converge to `(Γ).

Proof. We assume that Γ is rectifiable, the non-rectifiable case is evensimpler. Parametrize Γ by arclength Γ : [a, b] → U0 and find a suffi-ciently dense set of points a = t1 ≤ t2 ≤ ... ≤ tk = b. Fixing ε andchoosing the set dense enough, we may assume d(Γ(ti),Γi+1) < ε forall i and

`(Γ) ≥k−1∑i=1

d(Γ(ti),Γ(ti+1)) ≥ `(Γ)− ε .

We choose n so that

`(Γ) + ε >∑

dn(Γ(ti),Γ(ti+1)) ≥ `(Γ)− 2ε .

We now construct a graph Γn by connecting all Γ(ti) with Γ(ti+1) bya dn-shortest curve. The arising closed polygonal curve may be nonsimply closed. (Exercise: how could this happen?) However, (theimage of) Γn is a topological graph. (Exercise: why is this the case?What could happen if dn were general and not a simplicial metric?).

Note the the dn-length of this closed curve Γn is at most `(Γ) + ε.Moreover, the curves Γn converge to Γ (Exercise: What does it meanexactly and why is it true?)

Using topological arguments we find a simple closed subgraph Γnin Γn, such that Γn (after a reparametrization) still converges to Γpointwise. (Exercise: make a picture, understand the statement andwhy it sounds very plausible. Difficult exercise: try to understand themissing details).

This curve Γn is the one we have looked for (why?).

Now we turn to the

Proof of Theorem 11.4. Let T be a Jordan triangle in (U0, d). Findsimple closed Jordan curves Tn in U0, such that Tn converges uniformlyto T and `dn(Tn)→ `d(T ).

Let Jn denote the Jordan domain of Tn and J = T 0 the Jordandomain of T . For the (positive part of the ) curvatures ω+

n convergingto a Radon measure ω, we have

36

ω(T0) ≤ lim inf ω+(Jn) ≤ lim supω+(Jn) ≤ ω(T ) .

Exercise: understand this inequality.The difference between, ω(T ) and ω(T0) requires some additional

technical steps. We ignore this difference, and assume that

ω(T 0) = limω+(Jn) .

We apply the simplicial majorization theorem and find, for any n, amajorization Fn : Kn → Jn of the simplical disc Jn, where Kn isa convex polygon in the Euclidean cone over a circle of length 2π −ω+(Jn).

The result now follows from the following compactness statementwhich will be proved in the next section.

Theorem 13.2. Let tn be a sequence of positive real numbers converingto t > 0. Let Yn be the Euclidean cone over the circle of length tn and Ybe the Euclidean cone over the circle of length t. Let Kn be a compactconvex subset of Yn, homeomorphic to a closed disc. Assume that thereexists some ε > 0 such that all Kn contain a ball of radius ε and arecontained in a ball of radius 1

ε.

Then there exists a compact convex subset K of Y , homeomorphicto a closed disc and, for a subsequence ni, surjective maps Gi : K →Kni that are (1 + εi)-Lipschitz continuous, with εi → 0. Moreover,Gi(∂K) = ∂Kni and `(∂Kni) converges to `(∂K).

Exercise: Understand the statement of the theorem!Assuming this theorem, the proof is finished as follows. Replacing

Kn by a subsequence, we may assume that Kn converges to a convexsubset K in the cone over a circle of length 2π − ω(T 0) in the senseof the above theorem. (Exercise: why are the conditions on the ballssatisfied by the sets Kn?)

The compositions FnGn : K → Jn are (1+εn)-Lipschitz continuousmaps from K to (U0, dn). By the theorem of Arzela-Ascoli, the mapsFn Gn converge uniformly to a map H : K → U0. Moreover, this mapH is 1-Lipschitz with respect to the metric d. Furthermore, H(K) = T .

Lastly, H(∂K) = ∂T and

`(∂K) = lim `(∂Kn) = lim `dn(∂Jn) = `(∂T ) .

Since H is 1-Lipschitz, it maps ∂K in an arclength preserving way onto∂T .

Exercise: Understand the rather dense sequence of statements in thetwo paragraphs above and fill the details!

Therefore, H is the required majorization of the triangle T . 37

13.1. Some simplifications in Theorem 13.2. We may assume thateither all Ki contain the origin oi of the cone Yi or none of the sets Ki

contains oi.If none of the sets Ki contain oi, then Ki is isometric to a convex

polygon in R2 (Exercise: why?). In this case it is sufficient to findK ∈ R2 satisfying the conclusion of the theorem (Exercise: why?).Thus, in this case we are reduced to the case Yi = Y = R2. Moreover,we may assume that all Bε(0) ⊂ Ki ⊂ B 1

ε(0) in R2. (Exerise: What

is the statement in comparison to the assumptions and why does ithold?).

If all Ki contain oi, we may assume that Ki ⊂ B 1ε(oi). However,

in general, we cannot assume Bε(oi) ⊂ Ki (Exercise: why?). The casewhen the last condition is violated, requires some additional argumentsand we will not consider it.

Thus, we will only deal with the case

Bε(oi) ⊂ Ki ⊂ B 1ε(oi) .

14. Hausdorff metric and convergence, 18.06

14.1. Hausdorff metric. Definition. The aim of this section is toprove Theorem 13.2. The essential step is a compactness theorem ”‘variant of the so-called Blaschke selection theorem”’, which allows usto find a limit convex set of a sequence of arbitrary convex subsets.

Let now X be an abstract metric space (at the beginning, no topo-logical and no geometric assumption!). For a subset A ⊂ X let dAdenote the distance function to the set A:

dA(x) := infd(a, x) ; a ∈ A .

For two compact subset A,B ⊂ X we define the Hausdorff distancebetween the subsets to be

dH(A,B) := supx∈X|dA(x)− dB(x)| .

Using triangle inequality one verifies (Exercise: verify!) that

dH(A,B) := supx∈A∪B

|dA(x)− dB(x)| = maxsupx∈B

dA(x), supx∈A

dB(x) .

Thus, for R > 0, we have dH(A,B) ≤ R if and only if A is containedin the R-neighborhood of B and B is in the R-neighborhood of A.

Lemma 14.1. The set H(X) of all compact subsets of X with theHausdorff distance dH defined above is a metric space.

38

Below is a list of exercises which should help in developing an un-derstanding of the properties of the Hausdorff metric and convergence.

Exercise: Prove the Lemma above. What goes wrong if we extenddH by the same formulas to non-compact subsets?

Exercise: Prove that X has a natural isometric embedding intoH(X).

Exercise: Prove that the diameter diam : H(X)→ R assigning to asubset its diameter is a 2-Lipschitz map.

Exercise: Prove that the subset E(X) ⊂ H(X) of all finite subsets ofX is dense inH(X). More precisely, for any compact A ⊂ X there exista sequence of finite subsets Ei ⊂ A converging to A in the Hausdorffmetric.

Exercise: Let En the the subset of H(X) which consists of all subsetsofX with at most n elements. Prove that En is a closed subset ofH(X).Does the statement remain true, if ”at most n elements” is replaced by”exactly n elements?

Exercise: Prove that the Lebesgue measure on X = [0, 1] does notdefine a continuous function from H([0, 1]) to [0, 1].

Exercise: Let µ be a finite Borel measure on X. Prove that µ :H(X)→ R is semicontinuous: For Ai → A in H(X) we have

lim supi→∞

µ(Ai) ≤ µ(A) .

Exercise: If Ai → A in H(R2), then the boundaries ∂Ai may notconverge to ∂A.

Exercise: Let X be a compact length space. For A,B ∈ H(X) letM be the set of all midpoints of shortest curves connecting a ∈ A withb ∈ B such that d(a, b) ≤ dH(A,B). Prove that M is compact anddH(A,M) = dH(M,B) = 1

2dH(A,B). What does it imply about H(X)

in combination with the next lemma?

14.2. A compactness theorem.

Lemma 14.2. Let X be complete. Then H(X) is complete.

Proof. Let Ai ∈ H(X) be a Cauchy sequence. We may assume thatdH(Ai, Ai+1) ≤ 2−i.

Let A be the set of all limit points of sequences xj ∈ Aj, for somesequence j →∞. A ”diagonal-limit” argument shows that A is closed,hence complete.

Let xj ∈ Aj be arbitrary. Take xj+1 ∈ Aj+1, closest to xj thenxj+2 ∈ Aj+2 closest to xj+1 and so on. The arising sequence is Cauchy,its limit point x ∈ A satisfies

d(x, xj) ≤ 2−j+1 .39

A similar argument shows that for any x ∈ A and any j we findxj ∈ Aj with d(xj, x) ≤ 2−j+2.

This already shows that dH(A,Aj) ≤ 2−j+2 and thus the convergenceof Aj to A, modulo the statement that A itself is a compact subset.

However, A is complete and it remains to prove that A is totallybounded, meaning that for all ε > 0 we find a covering of A by finitelymany balls of radius ε. (Exercise: recall the compactness criterion usedhere).

We cover Aj for 2−j << ε by m balls with centers pl of radius ε4.

Choosing, for any pl, a point pl ∈ A with distance << ε to p we usethe triangle inequality and see that balls Bε(pl) cover A.

Now we can show:

Theorem 14.3. If X is compact then so is H(X).

Proof. We already know that H(X) is complete. It remains to provethat it is totally bounded.

Given ε > 0 consider a finite subset E ⊂ X which is ε-dense in X,that is the closed ε-neigborhood of E is all of X.

For any subset A of X, consider the subset EA of all points x ∈ Ewith d(x,A) ≤ ε. Then, dH(A,EA) ≤ ε (Exercise: prove that).

The set 2E of all subsets of E is finite. Thus, the finitely many closedballs of radius ε around points in 2E ⊂ H(X) cover H(X). Thus, H(X)is totally bounded.

Exercise: Find a connection between the last theorem and the theo-rem of Arzela–Ascoli on convergence of Lipschitz maps.

The convergence in H(X) is easiest described by the followingExercise: A sequence Ki in H(X) converges to K ∈ H(X) if and

only if ∪iKi ∪K is compact and the set K is exactly the set of limitpoints of sequences (xi) with xi ∈ Ki.

Now it is easy to see, that if Ai ⊂ Bi and Ai converges to A and Bi

converges to B then A ⊂ B.Exercise: Conclude that if Ai converge to A and all Ai are contained

in (respectively, contain) a closed ball of radius r then the same is truefor A.

14.3. Convergence of maps. The following is a version of the the-orem of Arzela–Ascoli, with essentially the same proof. The proof issomewhat technical, it is more important to understand the statementand idea than all steps in the proof.

Lemma 14.4. Let X, Y be metric spaces. Let Ai ∈ H(X) convergeto A and Bi ∈ H(Y ) converge to B. Let Fi : Ai → Bi be a sequence

40

of uniformly continuous maps, thus, for every ε > 0 there exists someδ > 0, such that d(x, z) < δ in Ai implies d(Fi(x), Fi(z)) ≤ ε in Bi.

Then, there is a subsequence Fj converging to a continuous map F :A → B in the sense that for any xj ∈ Aj converging to x ∈ A thesequence Fj(xj) converges to F (x).

Proof. For any n we find some number N(n) and a finite set Anj in

Aj with at most N(n) elements which is 12n

dense in Aj. (Exercise:Why?).

We can furthermore assume that Anj ⊂ An+1j , for all j, n.

Taking a subsequence we may assume that A1j converge to a finite set

A1 ⊂ A. Taking another subsequence we can assume that A2j converges

to A2 and so on. Taking a diagonal sequence we therefore may assumethat Anj converges to a finite subset An of A for all n. Note that An is2−n dense in A (Exercise: confirm that).

We now find at each step n a subsequence of the sequence Fj suchthat Fj : Anj → Bn

j converges to a map F n : An → Bn.Again, by a diagonal argument, we thus find a map F defined on a

dense subset A′ of A and for any x ∈ A′ a sequence xj ∈ Aj convergingto x such that Fj(xj) converges to F (x).

Then the map F : A′ → B has the same modulus of continuity,thus satisfies the same ε− δ-condition as all Fj. Then F sends Cauchysequences to Cauchy sequences and therefore uniquely extends to amap F : A = A′ → B, with the same modulus of continuity.

Finally, the condition that Fj converges to F in the required senseis deduced from the traingle inequality and the convergence of Fj ondense subsets (Technical exercise: verify this last statement).

Exercise: Let γi : [a, b] → X be a sequence of L-Lipschitz curvesand let γ : [a, b] → X be another L-Lipschitz curve. Show that if γiconverges to γ pointwise then the images γi([a, b]) converge to γ([a, b])in the Hausdorff topology. Prove that the converse does not need tohold.

15. Convergence of convex subsets, 23.06

First an Exercise: Let X be a metric space and Ki → K a convergentsequence in H(X). If all Ki are connected then K is connected as well.On the other hand, K can be connected while all Ki are disconnected.Finally, it can happen that all Ki are path-connected but K is not.

Lemma 15.1. Let X be a length space. If Ki ⊂ X are convex com-pact subsets which converge to a compact subset K in H(X) then K isconvex.

41

Proof. For x, y ∈ K we find xi, yi ∈ Ki converging to x and y respec-tively. Consider shortest curves Γi connecting xi and yi. After choosinga subsequence we may assume that Γi converge in H(X) to a subset Γof K (Exercise: why?). This subset Γ is a shortest curve connecting xand y. (Exercise: why?).

On the other hand, it can happen that Ki are completely convexcompact subsets converging to a compact set K which is not completelyconvex (Exercise: recall the definition from Subsection 9.2 and find anExample in the sphere).

Convergence of convex subsets in the Euclidean space and Euclideancones is much more ”well-behaved” than the convergence of generalcompact subsets. The meaning is that much more geometric/topologicalquantities behave ”continuously” under this kind of convergence, as weare going to see.

We prove our results in spaces isometric to cones over a circle X andnot only in Euclidean planes, as we need it for Theorem 13.2. Firstsome general observations about convex subsets in such spaces.

Proposition 15.2. Let K be a compact convex subset of a Euclideancone X over a circle. Assume that K contains the ball Bε(o) aroundthe vertex o of the cone, for some ε > 0.

Then K is homeomorphic to the disc D and the boundary ∂K of Kin X is homeomorphic to the circle S1. The boundary ∂K intersectseach radial ray γ(t) := t · v (starting at o) in exactly one point t0 · vand this point depends continuously on the direction v.

More precisely, the modulus of continuity of the map v → t0(v) de-pends only on ε and an upper bound on the diameter of K.

Exercise: Understand the statement. More difficult Exercise: provethe Proposition.

Proposition 15.3. Let K ⊂ X be a convex compact subset as above.Then the closest point projection Π : X → K is uniquely defined and1-Lipschitz

Proof. By compactness, for every x ∈ X there exists a point p ∈ Kwith d(p, x) = d(x,K).

Assume that there two such points x and y. Fix some shortest curvespx and py. Consider the unique shortest curves ox and oy, which byconvexity are contained in K.

By the first formula of variation ∠(xp, xo) ≥ π2

and ∠(yp, yo) ≥ π2

(Exercise: understand and verify this statement).Consider now the quadrangle pxoy and observe that this quadrangle

is isometric to a quadrangle in the plane (Exercise: verify this). Now42

we use convexity and compactness of K to see that the shortest curveconnecting x and y inside the quadrangle poxy (which might in generalnot be a shortest curve in X!) is completely contained in K. (Non-easyExercise: verify this statement.)

Now, the distance from p to the midpoint m of this shortest curvexy is smaller than the distance from p to x (Exercise: why?), in con-tradiction to the choice of x.

This shows that the closest-point projection Π is well-defined. More-over, it also implies that the map Π is continuous (Exercise: why?). Asimilar argument proves that for any x ∈ X there exists exactly oneshortest curve connecting x and p = Π(x). Exercise: fill the details.

In order to prove that the map Π is 1-Lipschitz it is sufficient (Ex-ercise: why?) to fix x ∈ X and to find some δ > 0 such that for ally ∈ Bδ(x) we have

d(Π(x),Π(y)) ≤ d(x, y) .

Thus we fix such x. We may assume that x is not contained in theinterior of K (Exercise: why?). Since Π is well-defined and continuousand the shortest curve between x and Π(x) is unique, the shortestcurves xΠ(x) and yΠ(y) are sufficiently close to each other, if δ issmall enough (Exercise: why?)

In such a case, the quadrangle xyΠ(y)Π(x) is a Euclidean quadranglein which the angles at Π(x) and Π(y) are at least π

2(Exercise: why?

Use the first variation formula.)Use Euclidean geometry to see that this implies

d(Π(x),Π(y)) ≤ d(x, y) ,

finishing the proof.

Theorem 15.4. Let X be a Euclidean cone over a circle (of somelength). Let Ki be sequence of compact convex subsets of X convergingto a compact convex subset K which contains Bε(0).

Then, for any δ > 0 and all sufficiently large i, the dilated set (1 +δ) ·K contains Ki and (1− δ) ·K is contained in Ki.

Moreover, there exist surjective (1 + δ)-Lipschitz maps F+i : Ki →

K and F−i : K → Ki which send the boundary surjectively onto thecorresponding boundary.

In particular, the areas H2(Ki) converge to H2(K) and the lengthsof the boundaries `(∂Ki) converges to `(∂K).

Exercise: Understand the statements! Recall, what can go wrong inthe non-convex situation.

43

Difficult exercise: Find an example showing that one cannot hope tofind a (1+ε)-biLipschitz map F : Ki → K (hint: smooth-out a triangleand look closely onto the corner.)

Proof. The dilated subset (= scaled by the factor (1 + δ) in radialdirections) (1 + δ) · K contains a neigborhood of K (Exercise: why?You should use Proposition 15.2). Thus, for all i large enough, (1+δ)·Kcontains Ki.

The closest point projection Πi onto Ki is 1-Lipschitz, thus the com-position F−i : K → Ki of the dilation by (1 + δ) and Πi is surjectiveand (1+δ)-Lipschitz. The boundary ∂K is mapped to ∂Ki by this map(Exercise: why?). By topological reasons, this map from one boundaryonto another is surjective (Difficult exercise: why is it the case?).

To prove that (1− δ) ·K is contained in Ki we argue in the followingway. Find a finite, sufficiently dense set of directions vj ∈ ΣoX. Forany of this directions find a point xij in Ki which is sufficiently closeto the boundary point t0(vj) · vj ∈ ∂K lying in the direction of K.Now we use the last statement in Proposition 15.2, applied first to Ki

and then to K, to see that for any direction v ∈ ΣoX, the set Ki

contained a point t · v such that t is ε0-close to the boundary pointt0(v) · v ∈ ∂K. Here, ε0 is any positive number fixed at the beginningof the last argument (Difficult exercise). This implies that (1− δ) ·Kis contained in Ki.

The projection F+i is constructed now as above, as a composition

of a dilation and a projection. Exercise: Verify this and prove theremaining statements.

The subsequent observations are not relevant for our main topic.The above proof works without changes in Euclidean spaces of all di-mensions. This provides an important tool for convex geometry andall of its applications. The space of compact convex subsets C(Rn) ofany Euclidean space is complete with respect to the Hausdorff metricand its bounded balls are compact.

The volume and boundary measure are continuous with respect tothis topology. Moreover, there are many dense subsets which are muchmore well behaved than general convex subsets. For instance, it iseasy to see that convex polyhedra are dense in C(Rn) (Exercise: verifythis statement). This provides a way to prove results for general con-vex subsets including some quantities continuous with respect to theHausdorff topology, first showing the validity on polyhedra and thenextending them by continuity to general convex subsets. This works formost inequalities in convex geometry. Another way of applications, alsouseful for proving such inequalities, for instance the most famous one,

44

the isoperimetric inequality, is to use compactness to find an ”optimalconvex body for the ineqaulity” and then to use variational argumentsto analyze its shape.

16. End of proof of Theorem 13.2, Repetition, 25.05

16.1. End of proof. An additional difficulty in Theorem 13.2 is thatthe convex subsets Ki live in different ambient spaces Yi. In order tocircumvent this difficulty and use Hausdorff coinvergence, we find anatural comparison map between the cones Yi and the ”‘limit”’ coneY , later we will mostly work in a single space Y .

Thus, let Σn = S1tn be a circle of length tn and Σ = S1

t be a circleof length t with tn → t. Then we find a map fi : Σi → Σ which is(1 + δi)-biLipschitz with δi = max t

ti, tit (Exercise: find the map).

Extending the map fi radially we obtain a map fi : Yi = Con(Σi)→Y = Con(Σ) given as fi(r · φ) := r · fi(φ). This map is again (1 + δi)-biLipschitz (Exercise: why? ).

We will replace the convex subsets Ki ⊂ Yi by Ki := fi(Ki) ⊂ Y .

The new compact subsets live in the same space Y . Note that Ki maynot be convex! However, it satisfies the following ”‘almost convexitystatement”: for any x, y ∈ Ki there exists some m ∈ Ki such thatd(x,m), d(y,m) ≤ (1

2+ δ′i) · d(x, y), where δ′i → 0 as i→∞ (Exercise:

determine δ′i in terms of δi).Now we apply the compactness theorem and see that, after choosing

a subsequence, the subsets Ki converge to a subset K ⊂ Y .Looking at the proof of Theorem 15.4, we see that K contains (1−δ)·

Ki and Ki contains (1−δ) ·f−1i (K) for any δ > 0 and all i large enough

(Exercise: confirm this statement!). Now, arguing as in the proof ofTheorem 15.4, we use 1-Lipschitz projections onto convex subsets tofind surjective (1 + δ)-Lipschitz maps Gi : Ki → K which send theboundary surjective to the boundary. Moreover, `(∂Ki) converge to`(∂K), again as in the proof of Theorem 15.4. (Exercise: confirmthat).

This finishes the proof of Theorem 13.2.

16.2. Repetition. Large Exercise: Recall now the long way that hasled us to the proof of Theorem 10.2. Understand the main steps andformulate questions to the general strategy and details.

45

17. Beginn of proof of Theorem 10.1, 30.06

17.1. Plan. Recall that we want to prove that the metric of a surface ofbounded curvature can locally around any point be approximated uni-formly by simplicial metrics of bounded total curvature and boundedtotal variation of turn.

First a simple observation. We may assume that our surface is home-omorphic to an open disc. Indeed, choose an open neighborhood Ohomeomorphic to a disc of any fixed point x ∈ X and replace X bythe subset O with its induced intrinsic metric dO. Locally around x,the metrics dO and d coincide. Thus, if we approximate dO on a smallneighborhood of x by simplical metrics with the required bounds thenalso of d.

The plan of the proof is rather simple. We fix a point x. We find asufficiently small closed neighborhood of x homeomorphic to a closeddisc. We then fix ε > 0 and find a triangluation of this neighborhoodin sufficiently small convex triangles. Then we replace each triangle bya Euclidean triangle with the same sidelength. Finally we will provethat the arising simplical metrics provide the required approximations.

The implementations of all steps in the proof is more or less compli-cated.

17.2. Transit points. In the rest of this section the assumption ofbounded curvature will not play a role. Everything will work for anylength metric on a surface (disc).

We call a point x ∈ X a transit point if it lies as an inner pointon a shortest curve. Recall that in a simplicial surface transit pointsare exactly the points on the boundary with non-positive turn of theboundary and points in the interior with non-negative curvature. Thus,one can expect that this general condition will also play some role ingeneral surfaces of bounded curvature.

The proof of the following Lemma is left as a (non-trivial) Exercise:

Lemma 17.1. Let Γ be a topological arc in X. Then the set of transitpoints contained in Γ is dense in Γ. Thus, the set of transit points inX is dense in X.

17.3. Choice of special shortest curves. The next observation doesnot use any topology of the ambient space and is valid in all metricspaces.

Let p±i be a finite set of pairs of points. Let Γi be a shortest curveconnecting p−i and p+

i . Then the union Γ of the curves Γi contains afamily of shortest curves Γ′i connecting the same endpoints as Γi suchthat the union Γ′ ⊂ Γ of all Γ′i is a topological graph. Thus, any Γ′i

46

intersects any Γ′j in a finite set of connected subsets. The proof of thisstatement proceeds by induction on i.

Exercise: Understand the case of two curves. Use it in the inductionstep to prove the general statement.

The next observation is very two-dimensional. Let p, q be two pointsin a topological disc X with a length metric. Assume that the closedball of radius d(p, q) around p is compact. We fix an orientation on Xwhich allows us to say, for a Jordan curve through p and q what is the”‘right”’ and what is the ”‘left”’ part of the curve. Now, among allshortest curves connecting p and q we can find the ”‘most left”’ andthe ”‘most right”’ one.

Exercise: Understand the statement informally, try to make it formaland to prove it.

17.4. Existence of polygonal domains. Let x ∈ X be a point.Then there exist polygonal Jordan curves Γ, such that every pointof Γ is a transit point of X, the Jordan domain J of Γ contains x andthe diameter of J is arbitrary small.

Indeed, we consider a closed topological disc of small diamater sur-rounding x. Its boundary Γ is a topological circle. We can now finda sufficiently dense sequence of consecutive transit points pi on Γ andconnect any pair of consecutive points pi by a shortest curve. By theprevious arguments, we can choose the curves so that their union is agraph. In this graph every point is a transit point! Using topologicalarguments (Exercise: understand this!) we find in this graph a simpleclosed curve whose Jordan domain contains x.

A word of caution: Note that for a general length metric on a discone cannot expect that the length of Γ is as small as we want. Ex-ercise: Remember a corresponding example! For surfaces of boundedcurvature it is indeed the case, but it is a highly non-trivial statement(which maybe will be shown later).

17.5. Existence of absolutely convex neighborhoods. We call adomain G such that G is homeomorphic to a closed disc absolutelyconvex if it is boundary convex and completely convex.

Let again x ∈ X be a point. We are going to prove the following:

Proposition 17.2. Assume that for any ε > 0 we find a Jordan curveΓε surrounding x, such that Γε has length at most ε and is containedin the ε-ball around x. Then x has arbitrary small absolutely convexneighborhoods G with arbitrary small perimeter `(∂G).

Proof. Fix a closed disc neighborhood Y of x with a polygonal bound-ary. Using compactness we show that there exists some ε > δ > 0

47

with the following property. Any compact subset K of Y of diameter≤ δ which separates x from the boundary ∂Y must be contained in theε-ball around x (Exercise: Verify this statement!).

Fix Γδ ⊂ Y as above. Arguing as above, we can replace Γδ by apolygonal curve all of whose points are transit points.

Consider the annulus A bounded by Γδ and ∂Y . We can now find ashortest closed curve Γ in this annulus which is homotop in A to bothboundary curves (Non-trivial Exercise: verify this statement). By theprevious observation, this curve is contained in the ε-ball around x.(Exercise: why?).

By minimality, this curve Γ is a polygonal curve which is moreoversimple closed (Exercise: why?). Note that the only ”‘corners”’ of Γ hasto lie in ∂A, thus on Γδ. Hence every point of Γ is a transit point.

The Jordan domain G of Γ is boundary convex by the minimality of`(Γ) (Exercise: prove this!).

The curve Γ might be non-uniquely defined and the closed Jordandomain G might be not completely convex (Exercise: what could hap-pen?). However, among all such minimizers we can consider at outer-most one (Exercise: Understand the meaning and the proof!).

The Jordan domain of such an outermost curve is completely convex,thus absolutely convex in Y . Since such Γ has a short length, it is stillfar from ∂Y (he distance to ∂Y can be assumed to be larger then thelength of ∂Y ). Then G is absolutely convex in X as well.

17.6. Subdivision of polygons. We are going to prove:

Lemma 17.3. Let G ⊂ X be a polygonal domain convex relative to itsboundary. Then G is a union of non-overlappng triangles Ti, convexrelative to their boundaries.

Proof. A central observation is the following one. Assume that K is asubset X convex relative to its boundary. Let Γ be a shortest curve inK intersecting ∂K exactly in its boundary points. Then Γ divides Kinto two subsets, both of which are convex relative to their boundaries.

Exercise: Understand the statement and prove it.Now we can proceed inductively by the (minimal) number of vertices

of ∂G. Assuming that there are at least 4 vertices, we fix one vertexand connect it by a shortest curve in K with a non-adjacent vertex.We choose this shortest curve to have a maximal possible intersectionwith ∂G (Exercise: what does this mean? What s it good for?).

This curve subdivides G is polygonal domains. Each of the domainsis boundary convex by the first observation in the proof.

48

Exercise: Analyze different combinatorial situations and prove thatin any case the arising polygons have less vertices than G.

18. Triangulation, 02.07

If you do not have enough time just understand the formulation ofTheorems 18.2, 18.3, 18.4 and go to chapter 19.

Proposition 18.1. Let Y be a topologically closed disc with a lengthmetric. Assume that we have a finite set of absolutely convex closeddiscs Gi ⊂ Y such that the following holds true. Every boundary ∂Gi

is polygonal and the diameter and perimeter of Gi are at most ε.Then there is a finite set of pairwise non-overlapping boundary con-

vex triangles Tj such that ∪Gi = ∪Tj and such that the following holdstrue. For any i, the diameter and the perimeter of Ti are at most ε andevery point in ∂Ti \ ∪∂Gi is a transit point.

Sketch of Proof: We proceed by induction on the number of sets Gi.The case of a single set G1 is covered by Lemma 17.3.

By induction, we can write the union of all but one Gi as a unionof pairwise non-overlapping simple triangles Tj. It remains to show,that the union of the last G = Gn and all Tj can be written as a unionof non-overlapping triangles. We may discard those triangles Tj whichare contained in G or such that G does not intersect the interior of Tj.

For any of the remaining triangles Ti we look at the intersection of∂Ti and ∂G. We find only finitely many arcs ∂T ki of ∂Ti lying outsideof G. It is a formally highly non-trivial exercise to make the followingeasy idea work: for any of these arcs ∂T ki connect the endpoints of thearc (lying in ∂Ti ∩ ∂G) by a shortest curve Γki in Ti. This shortestcurve is contained in G, since G is absolutely convex. Now the curvesΓki for different i do not intersect, since the Ti are non-overlapping.Using this and modifying Γki as in Section 17.3, if needed, we see,that all the shortest curves Γki cut G∪Ti into pairwise non-overlappingpolygons convex relative to their boundaries. Subdividing them furtherwe obtain the required triangles.

By construction, all points in all ∂Tj are either contained in ∪∂Gi orare intrinsic points of some shortest curves added in the course of theconstruction.

Some technical parts of the following theorem will not be proven (noteven sketched):

Theorem 18.2. Let X be a surface of bounded curvature. Then, forany point x ∈ X and any δ > 0 there exists a closed neighborhood G

49

of x with polygonal boundary ∂G consisting only of transit points, suchthat G is absolutely convex, has diameter and perimeter at most δ.

Fixing such G and any ε > 0 we can write G as a union of pairwisenon-overlapping simple triangles Ti of diameter at most ε.

Moreover, we can assume all triangles to be non-degenerate in thesense that the sum of the lengths of any two sides is larger than thethird one.

Proof. We will assume but not prove that any point x ∈ X admits aneighborhood with arbitrary small perimeter (this is rather technical).

Given this fact, Proposition 17.2 directly implies the first statement.Fix such G. Use compactness, to cover G by finitely many absolutely

convex discs of diameter smaller than ε. Apply now Proposition 18.1to write ∪(G ∩Gi) as a union of simple triangles.

It remains to show that the triangles can be chosen to be non-degenerate. This is another non-trivial somewhat technical statementthat will be explained in the lecture.

The aim of the remaining lectures will be the proof of the followingstatements.

Theorem 18.3. Let G be as in Theorem 18.2, consider the decompo-sition of G in pairwise non-overlapping simple triangles Ti of diameterat most ε, as in Theorem 18.2.

Let G be the simplicial disc constructed from this trinagulation viaa replacement of any triangle Ti by its Euclidean comparison triangleTi. Then G has curvature measure and turn of the boundary, witha uniform bound on their total mass (of positive and negative parts),independent of the triangulation.

Theorem 18.4. Let G, G be as in Theorem 18.3. Then, for any”natural” identifications of G with G, the metric d induced on G fromG differs from the original metric d by some amount ρ which goes to 0as ε goes to 0.

Together, these two results would finish the proof of Theorem 10.1.

19. An application of first formula of variation, 02.07

We will show that a control of all defects of ”subtriangles” of anytriangle controls the difference between the angle and the comparisonangle at any vertex. This will imply that if the defects of all subtrianglesof a triangle T are small then T is ”almost Euclidean” in a precise sense.

The topology of the space does not play a role and the statementapplies to all metric spaces.

50

Proposition 19.1. Let T = ABC be a triangle in a length space X.Assume that all pairs of points on T are connected by shortest curves.Then

∠(AB,AC)− ∠BAC ≤ νA(T ) := supM∈AB N∈AC

( inf[MN ]

δ(AMN)) .

The infimum in the formulation above is taken over all shortestcurves connecting M and N . Exercise: make a picture and understandthe statement of the Proposition.

Before going to the proof (which will be explained in the lecture) weset for a triangle T = ABC in X

ν(T ) = maxνA(T ), νB(T ), νC(T ) .Up to the possible choice of different shortest curves (which makes thevalue possibly smaller) the number ν(T ) is supremum of all defectsδ(T ′), where the triangle T ′ has a vertex coinciding with A, B or Cand two of the sides of T ′ adjacent to this vertex are contained in thecorresponding sides of T . Exercise: Make a picture and understandthe definition.

As a direct consequence of Proposition 19.1 we deduce (Exercise:deduce!):

Corollary 19.2. Let T be a triangle in a length space X as in Propo-sition 19.1. Then at any of its vertices the difference between the upperangle α and the comparison angle α is estimated as

α− α ≤ ν(T )

As a special case we obtain an interesting observation (which will notplay any role below), characterizing the so-called spaces of non-positivecurvature:

Corollary 19.3. Let X be a length space in which any pair of points isconnected by a shortest curve. If the upper defects of all triangles arenon-positive then, for any triangle T , the upper angle at any vertex isnot larger than the corresponding comparison angle.

On surfaces of bounded curvature not only the sum of defects butalso the sum of these new invariants ν(T ) is bounded as we see now.The proof of the following Lemma is an Exercise:

Lemma 19.4. Let X be a surface with a length metric and let T ⊂ Xbe a simple triangle. Then

ν(T ) ≤ supδ(T ′) ,51

where T ′ is a triangle as in the definition of ν(T ), which is in additionsimple and is contained in T .

As a consequence we deduce (Exercise: Understand, why the Corol-lary is a consequence of the Lemma):

Corollary 19.5. Let Ti be the triangle in our subdivision of G, as inTheorem 18.2. Then ∑

ν(Ti) ≤ C0 ,

for a constant C0 independent of the subdivision.

07.07We now turn to the proof of Proposition 19.1 and start with a

Lemma. In the Lemma below we denote by β and α the compari-son angles of a triangle ABC at B and A respectively. By β and αwe denote the upper angles at the same vertices. The main step in theproof of Proposition 19.1 is the following

Lemma 19.6. For any ε > 0 there exists some R > 0 with the followingproperty. Let T = ABC be a triangle in a length space X. Assumethat β ≥ β + ε and that α ≥ ε.

Then there exists some point B′ ∈ [AB] such that the comparisonangle α′ at A of the triangle AB′C satisfies

α′ − α > R · (ln(d(A,B))− ln(d(A,B′)) .

The lemma says that under the assumption that the comparisonangle at B is larger than the upper angle, one can increase (by a certainamount) the comparison angle at A by moving B towards A.

Before proving the Lemma we provide

Proof of Proposition 19.1. Assume the contrary and find some ε > 0such that

α− α ≥ sup(δ(T ′)) + 3ε ,

where T ′ is as in the formulation of the Proposition.For T ′ = T this shows that either β − β > ε or the corresponding

inequality holds at the vertex C. Applying Lemma 19.6, we obtainB1 = B′ as in the Lemma. We set C = C1 and consider a shortestcurve C1B1, such that the defect of T1 = A1B1C1 is (sufficiently) closeto infimum between all such curves B1C1.

In the triangle T1 = AB1C1 the upper angle α1 at A has not changedand equals α. The comparison angle α1 has increased by Lemma 19.6:

(19.1) α1 − α > R · ln d(A,B) · d(A,C)

d(A,B1) · d(A,C1)52

Among all pairs B1, C1 satisfying the non-strict version of the aboveinequality we find one with minimal d(A,B1) · d(A,C1). Exercise: whydoes such a pair exist?

Applying Lemma 19.6 again, we deduce that β1 − β1 ≤ ε and thecorresponding inequality holds at C1. Exercise: Understand this andsee why it is good that ln appears in the formula above.

Summing up the angles in the triangle T1 we obtain a contradiction.Exercise: understand this final argument.

Now we proceed to

Proof of Lemma 19.6. Denote by a, b, c the side lengths of ABC.For small h > 0 consider Bh on AB at distance h from B. Note that

ln d(A,B)− ln d(A,Bh) ≤ 2 · hc

for all h small enough (Exercise: why?). Denote by αh the comparison

angle ∠CABh we only need to prove that, for some R = R(ε) > 0 andall sufficiently small h,

αh − α ≥ R · hc.

Consider the comparison triangle ABC of ABC in R2 and let Bh bethe corresponding point on AB.

From the first variation formula in the Euclidean plane and the firstvariation inequality, Proposition 8.1 and the difference of (upper) angles

at B and B we derive the following statement (Exercise: Derive it, bytaking δ = 1

2(cos(β)− cos(β)):

There exists some δ = δ(ε) > 0 such that for all sufficiently small h

d(C,Bh) ≤ d(C, Bh)− δ · h .

The other two sides in the triangles ABhC and ABhC have pairwiseequal length b and c− h respectively.

We apply now the cosine formula to the Euclidean triangle ABhCand the comparison triangle of ABhC to obtain (Exercise: apply andobtain):

cos(αh) ≤ cos(α)− δ · d(B,C) · hd(A,C) · d(A,B)

.

From the assumption β > ε we get a uniform positive lower bound ond(B,C)d(A,C)

(Exercise: check this!).

This finishes the proof (Exercise: why?). 53

References

[BBI01] D. Burago, Y. Burago, and S. Ivanov. A course in metric geometry, vol-ume 33 of Graduate Studies in Mathematics. American Mathematical So-ciety, Providence, RI, 2001.

[Res93] Yu. G. Reshetnyak. Two-dimensional manifolds of bounded curvature. InGeometry, IV, volume 70 of Encyclopaedia Math. Sci., pages 3–163, 245–250. Springer, Berlin, 1993.

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