Applicable Geometry

MATH 455 Lec 01

Tuesdays, Thursdays 2:55 – 4:10P

Malott Hall 238

Victor Alexandrovalexandrov@math.cornell.edu

alex@math.nsc.ru

Prerequisite: good introduction to linear alge-bra (e.g., MATH 221, 223, 231, or 294) and somemodest knowledge of calculus of several variables(e.g., I suppose you know a definition and basicproperties of continuous functions).

Lecture notes1 are available athttp://www.math.cornell.edu/∼web455

Grading policy: The course grade will be basedon your score out of 550 points. One (take home)prelim is 120 points, the final (take home) examis 240 points, the homework is 120 points, classparticipation is 70 points.

1 ‘Putting things in writing is safer than simply speaking to people. . . ’[Planning in Advance].

1

Important: Return your homework on Tuesday.

My office hours at MT 438:Monday, Friday 6:30 – 7:30 PM

Syllabus

Theory of convex bodies:•general properties of convex bodies (week 1;pages 4–7),•Helly’s theorem (week 2; pages 8–12),•Minkowski addition (weeks 3–4; pages 13–26).

Combinatorics of convex polytopes:•introd. to graph theory (week 5; pages 27–29),•Euler’s theorem (week 5; pages 29–31),•Steinitz’s theorem (week 7; pages 35–43).

Regular polytopes (week 6; pages 32–34).

Metric theory of polytopes:•Cauchy rigidity theorem (week 8; pages 44–47),•H. Minkowski’s and A.D. Alexandrov’s unique-ness theorems (week 9; pages 48–50),•H. Minkowski’s and A.D. Alexandrov’s existencetheorems (weeks 10–11; pages 53–58).

Rigidity theory:•flexible polyhedra and rigidity theory (weeks 12–13; pages 62–70).

2

Recommended literature:

1. Yaglom, I.M.; Boltyanskii, V.G. Convexfigures. New York: Holt, Rinehart and Winston,1961.

2. Lyusternik, L.A. Convex figures and poly-hedra. New York: Dover Publications, 1963.

3. Coxeter, H.S.M. Regular polytopes. 2nded. New York: The Macmillan Company, 1963.

4. Grunbaum, B. Convex polytopes. 2nd ed.New York: Springer, 2003.

5. Alexandrov, A.D. Convex polyhedra. Berlin:Springer, 2005.

6. Graver, J.; Servatius, B.; Servatius, H.Combinatorial rigidity. Providence: AmericanMathematical Society, 1993.

3

Week 1. General properties of convex bodies

Def.: A set A ⊂ Rd is convex if for each pair ofdistinct points a, b ∈ A the closed segment withendpoints a and b is contained in A.

Examples:(1) the empty set ∅;(2) any single point;(3) any linear subspace (including Rd);(4) every (closed or open) triangle in Rd.

Properties:(1) If {An} is any family of convex sets in Rd,

then their intersection ∩An is also convex.(2) If A is convex, ai ∈ A and λi ≥ 0 for i =

1, 2, . . . , k and

k∑

i=1λi = 1, then

k∑

i=1λiai ∈ A.

(3) If A is convex, both its closure cl A and itsinterior intA are convex.

(4) If T : Rd → Rd is an affine transformation,and A ⊂ Rd is convex, than T (A) is convex.

Homework 1 (5 points):A quadrilateral on the plane is convex if and

only if its diagonals intersect inside the quadri-lateral.

4

Def.: If u ∈ Rd and u 6= 0, then

H = {x ∈ Rd|(x, u) = α}is called a hyperplane with normal vector u, andeither of the sets

{x ∈ Rd|(x, u) ≥ α} or {x ∈ Rd|(x, u) ≤ α}is called a closed half-space determined (or bounded)by H. If the inequalities are strict, the half-spaces are called open.

Def.: We say that a hyperplane H supports aconvex set A provided clA ∩ H 6= ∅ and A iscontained in one of the half-spaces bounded byH.

Def.: Two subsets A and A′ of Rd are said tobe separated (strictly separated) by a hyperplaneH provided A is contained in one of the closed(open) half-spaces determined by H while A′ iscontained in the other.

Theorem 1.1: If A and A′ are convex subsets ofRd such that A′ is bounded and cl A ∩ clA′ = ∅,then A and A′ may be strictly separated by ahyperplane.

Proof: It is sufficient to prove Theorem 1.1 forthe case when A and A′ are closed sets. Choose

5

x ∈ A and x′ ∈ A′ such that

|x− x′| = miny∈A,y′∈A′

|y − y′|.

Note that hyperplanes H and H ′, orthogonal to[x, x′] and passing through x (respectively, x′),determine an open slab which contains no pointsof A ∪ A′.

Def.: Let A ⊂ Rd be a nonempty set. The sup-porting function H(A, x) of A is defined for allx ∈ Rd by

H(A, x) = sup{(y, x)|y ∈ A}.

Properties:

(1) The supporting function is positively ho-mogeneous, i.e.,

H(A, λx) = λH(A, x) for all λ ≥ 0, x ∈ Rd.

(2) The supporting function is convex, i.e.,

H(A, x+y) ≤ H(A, x)+H(A, y) for all x, y ∈ Rd.

(3) If T : Rd → Rd is a translation (i.e., T (x) =x + y for all x ∈ Rd and some constant vectory ∈ Rd), then

H(T (A), x) = H(A, x) + (x, y).

6

(4) If T : Rd → Rd is a homothetic transfor-mation (i.e., T (x) = λx for all x ∈ Rd and someconstant λ ≥ 0), then

H(T (A), x) = λH(A, x).

(5)|H(A, x)|/√

(x, x) is the distance from theorigin 0 to the supporting plane H of the convexset A with the outward normal vector x.

(6) If A,A′ are nonempty, closed convex setsin Rd such that H(A, x) = H(A′, x) for everyx ∈ Rd, then A = A′.

(7) Each closed, convex subset of Rd is theintersection of all the closed (or of all the open)half-spaces of Rd which contain the set.

(8) Each open, convex set in Rd is the inter-section of all the open half-spaces containing it.

Def.: The convex hull convA of a subset A ofRd is the intersection of all the convex sets in Rd

which contain A.

Def.: The convex hull of a finite set is called apolytope.

Homework 2 (5 points): Prove that every poly-tope is the intersection of a finite number of closedhalf-spaces.

7

Week 2. Helly’s theorem and its applications

Adv.: Bradley Forrest 〈btf4@cornell.edu〉 is theteaching assistant for MATH 455. His office hoursare Monday 12–2 & Thursday 1–2 in Malott 114.

Lemma: Let four convex sets Ai, i = 1, . . . , 4,be given in the plane, each three of which have acommon point. Then all four sets have at leastone common point.

Note: Lemma is obviously false for non-convexsets.

Proof: Let xi be a common point of the set

{∪4j=1Aj} \ Ai.

Since x1, x2, x3 ∈ A4, the entire triangle x1x2x3 iscontained in A4. Similarly, x1x2x4 ⊂ A3, x1x3x4 ⊂A2, x1x3x4 ⊂ A1.Consider two cases:

(1) One of the points x1, x2, x3, x4 belongs tothe triangle formed by the other three. [This verypoint belongs to the intersection.]

(2) None of the points x1, x2, x3, x4 belongs tothe triangle formed by the other three. [In thiscase the the points x1, x2, x3, x4 are vertices of aconvex quadrilateral and the intersection of itsdiagonals belongs to the intersection.] Q.E.D.

8

Theorem 2.1 (Helly’s theorem in the plane):Let n convex sets Ai, i = 1, . . . , n, be given inthe plane, and suppose each three of them havea common point. Then all n sets have a commonpoint.

Proof is by induction: According to Lemma,Theorem is true for n = 4. Suppose Theoremis true for some n and prove it for n + 1.Put An = An ∩An+1. Then every three of n con-vex sets A1, . . . , An−1, An have a common point[this is obvious for any three sets which are dif-ferent from An and this follows from Lemma forAi, Aj, An (that is Ai ∩Aj ∩An ∩An+1 6= ∅)]. Bythe inductive hypothesis, there is a point belong-ing to all the figures A1, . . . , An−1, An (that is toA1, . . . , An−1, An, An+1). Q.E.D.

Homework 1 (5 points): Let n points be givenin the plane such that each three of them can beenclosed in a circle of radius 1. Prove that all n

points can be enclosed in a circle of radius 1.

Theorem 2.2 (Jung’s theorem). Let n pointsbe given in the plane such that each pair of themare at the distance of at most 1 from each other.Then all these points can be enclosed in a circleof radius 1/

√3.

9

Proof is based on the homework: it suffice toshow that Theorem is true for three points.

Note: In the Jung’s theorem, the radius 1/√

3can be replaced by no smaller number.

Note: Helly’s theorem in the plane is obviouslyfalse for infinite families of sets.

Theorem 2.3 (Helly’s planar theorem for boundedsets): If a finite or countable family {A1, . . . , An, . . .}of closed, bounded, plane, convex sets is givensuch that each three have a common point, thenthere exists a point which belongs to all the sets.

Proof: Let xn ∈ A1∩ . . .∩An. Choose a conver-gent subsequence xn1

, . . . , xnk, . . . and put x∗ =

limk→∞ xnk. Then x∗ ∈ A1∩. . .∩An∩. . .. Q.E.D.

Theorem 2.4: In the plane, n parallel line seg-ments are given such that for each three of themthere exists a line that intersects the three linesegments. Then there also exists a line that in-tersects all the line segments.

Proof: Introduce a system of coordinates. Let(xi, y

′i) and (xi, y

′′i ) be the end points of the i-th

segment, where y′i < y′′i and i = 1, 2, . . . , n. Notethat a line y = kx + b intersects the i-th segmentiff

y′i ≤ kxi + b and y′′i ≤ kxi + b.

10

But these formulas determine a strip on (k, b)-plane and, according to the conditions of the the-orem, every three strips intersect. Applying Helly’stheorem in the plane, we conclude the all thestrips intersect. Q.E.D.

Application (to the theory of best approxima-tions): If for each three points of the intervala ≤ x ≤ b there exists a line which approximatesthe function y = f(x) with exactness up to ε atthese three points, then there exists a line ap-proximating the function to the given degree odaccuracy on the entire interval a ≤ x ≤ b.

Remark: A Helly-type theorem is a statementof the following type: A family of elements hasa certain property whenever each of its subfami-lies, containing not more than a fixed number ofelements, has this property.

Theorem 2.5 (Helly’s theorem in space): If inRd a finite number of bounded convex sets aregiven, each d + 1 of which have a common point,then all these sets have a common point.

Proof (for d = 3 only): The crucial step is toprove Theorem for n = 5. Let xi be a commonpoint of the intersection of all the sets A1, A2,A3, A4, A5 besides Ai. Suppose ∩5

j=1Aj = ∅. Let

11

A = ∩4j=1Aj. Note A 6= ∅ and A ∩ A5 = ∅. Let

H be a plane that separates A and A5. Notethat the four sets Aj ∩ H satisfy the conditionsof Helley’s theorem in the plane (Theorem 2.1).Thus A ∩H 6= ∅. A contradiction.The rest part of the proof is like for Theorem 2.1:Put An = An ∩An+1. Then every three of n con-vex sets A1, . . . , An−1, An have a common point[this is obvious for any three sets which are dif-ferent from An and this follows from Lemma forAi, Aj, An (that is Ai ∩ Aj ∩ An ∩ An+1 6= ∅)].By the inductive hypothesis, there is a point be-longing to all the sets A1, . . . , An−1, An (that isto A1, . . . , An−1, An, An+1). Q.E.D.

Homework 2 (5 points): Formulate and prove atheorem for R3, which is analogous to the Jung’sTheorem 2.2.

12

Week 3. Minkowski addition

Def.: For two (not necessarily convex) sets A,B

in Rd the set

A + B = {x + y | x ∈ A, y ∈ B}is called the Minkowski sum or, briefly, the sumof A and B.

Examples: The sum of two triangles in a planeis either a triangle, a quadrangle, a pentagon, ora hexagon.

Lemma 3.1: If A and B are both convex, thenA + B is convex.

Proof: Let x0, x1 ∈ A and y0, y1 ∈ B.Then, for 0 ≤ t ≤ 1,

t(x0 + y0) + (1− t)(x1 + y1) =

=(tx0 + (1− t)x1

)+

(ty0 + (1− t)y1

)∈ A + B.

Q.E.D.

Lemma 3.2: If T denotes a translation, then

T (A) + B = T (A + B) = A + T (B).

Proof: Let T (u) = u + v. Then T (A) + B == {x+y|x ∈ T (A), y ∈ B} = {u+v+y|u ∈ A, y ∈B} = T

({u + y|u ∈ A, y ∈ B}

)= T (A + B).

Q.E.D.

13

Remark (visualization of Minkowski sum): HoldA in one of its points, x say, and move A bytranslations such that x attains all points of B.Then, the translates of A cover A + B.

Def.: If t is a real number and A ∈ Rd is a set,then, we call tA = {tx | x ∈ A} a multiple of A.

Remark: t may be negative. However, the set(−1)A (denoted, by definition, as −A) is not thenegative of A with respect to Minkowski addition:A + (−A) 6= {0}.Def.: If t1, . . . , tr ∈ R and A1, . . . , Ar ∈ Rd,then, t1A1 + . . .+ trAr is called a linear combina-tion of of A1, . . . , Ar.

Lemma 3.3: If A1, . . . , Ar ∈ Rd are convex andt1, . . . , tr are real numbers, then, t1A1 + . . .+trAr

is convex.

Theorem 3.4: If HA, HB are the support func-tions of the convex sets A, B, then, HA + HB isthe support function of A + B:

HA+B = HA + HB.

Proof: HA+B(x) = supy∈A+B

(y, x) =

= supu∈A,v∈B

(u + v, x) = supu∈A

(u, x) + supv∈B

(v, x) =

= HA(x) + HB(x) for all x ∈ Rd. Q.E.D.

14

Def.: If H is a supporting hyperplane of theclosed convex set A, we call F = A∩H a face ofA. A face, consisting of one point only, is calleda vertex.

Theorem 3.5: If F is a face of A + B, then,there exist faces FA and FB of A and B suchthat F = FA + FB. In particular, each vertex ofA + B is the sum of vertices of A and B.

Proof: For x ∈ Rd we denote by HA(x), HB(x),and HA+B(x) the supporting hyperplanes withthe normal vector x of A, B, and A + B respec-tively. Let x be s.t. F = (A+B)∩HA+B(x). Weset FA = A ∩HA(x) and FB = B ∩HB(x).Then we obtain F = (A + B) ∩HA+B(x)

= {w ∈ Rd |w = u + v for some u ∈ A, v ∈ Band (y + z, x) ≤ (w, x) for all y ∈ A, z ∈ B and(y∗ + z∗, x) = (w, x) for some y∗ ∈ A, z∗ ∈ B}

= {w ∈ Rd |w = u + v for some u ∈ A, v ∈ Band (y, x) ≤ (u, x) for all y ∈ A

and (y∗, x) = (u, x) for some y∗ ∈ A

and (z, x) ≤ (v, x) for all z ∈ B

and (z∗, x) = (v, x) for some z∗ ∈ B}= {w ∈ Rd |w = u + v

for some u ∈ A ∩HA(x), v ∈ B ∩HB(x)}= FA + FB. Q.E.D.

15

Theorem 3.6: Minkowski sum of any two poly-tops is a polytope.

Proof follows from Theorem 3.5 since the sumof two vertices is a vertex and each vertex of thesum is obtained in this way. Q.E.D.

Homework 1 (5 points): Find two 3-simplicesin R3 whose Minkowski sum has 16 vertices.

Theorem 3.7 (H. Minkowski): Let A1, . . . , An

be convex polytopes in Rd and ti ≥ 0, i = 0, . . . , n.Then, the volume V (t1A1 + · · · + tnAn) of thelinear combination t1A1 + · · · + tnAn is eitherzero or a homogeneous polynomial of degree d

in t1, . . . , tn,

V (t1A1 + · · ·+ tnAn)

=n∑

p1,...,pn=1V (Ap1

, . . . , Apn)tp1

· · · tpn,

where the summation being carried out indepen-dently over the pi.

Proof is by induction in dimension d.

For d = 1, Ai are the intervals [xi, yi] (or pointsif xi = yi). We find

t1A1+· · ·+tnAn = [t1x1+· · ·+tnxn, t1y1+· · ·+tnyn]

16

V (t1A1 + · · ·+ tnAn)

= (t1y1 + · · ·+ tnyn)− (t1x1 + · · ·+ tnxn)

= t1(y1 − x1) + · · ·+ tn(yn − xn)

= t1V (A1) + · · ·+ tnV (An)

which is either zero or a homogeneous polynomialof degree 1 in t1, . . . , tn.

Suppose the theorem is true for dimension d− 1instead of d.

Let x ∈ Rd, HAj(x) be the supporting hyper-planes, and Fj = Fj(x) = Aj ∩ HAj(x) be thefaces, j = 1, . . . , n. Set

Aτ = t1A1 + · · ·+ tnAn (τ = (t1, . . . , n))

Fτ(x) = Aτ ∩HAτ (x)

Then, according to Theorem 3.5,

FA(x) = t1F1(x) + · · ·+ tnAn(x).

Volumes do not change under translations. Sowe can assume that all Aj lie in the hyperplaneHAτ (x) and 0 ∈ Aτ . We decompose Aτ into pyra-mids with appex 0 over the faces and obtain

Vd(Aτ) =1

d

m∑

j=1HAτ

(xj)Vd−1(Fτ(xj))

17

=1

d

m∑

j=1(t1HA1

(xj)+ · · ·+ tnHAn(xj))Vd−1(Fτ(xj))

By inductive assumption, Vd−1(Fτ(xj)) is eitherzero or a homogeneous polynomial of degree d−1in t1, . . . , tn. Q.E.D.

Def.: Arranging the coefficients on the right sideof the formula

V (t1A1 + · · ·+ tnAn)

=n∑

p1,...,pd=1V (Ap1

, . . . , Apd)tp1

· · · tpd,

such that

V (Aπ(p1), . . . , Aπ(pd)) = V (Ap1, . . . , Apd

)

for any permutation π of p1, . . . , pd, we callV (Ap1

, . . . , Apd) the (d-dimensional) mixed vol-

ume of Ap1, . . . , Apd

.

Lemma 3.8: The d-dimensional mixed volumeof d copies of a convex body A, V (A, . . . , A),equals the volume of A, V (A).

Proof: tdV (A) = V (tA) = tdV (A, . . . , A).Q.E.D.

Example: For planar convex sets we haveV (A1 + A2) = V (A1) + 2V (A1, A2) + V (A2).

18

Homework 2 (5 points): LetA1 = conv {0, e1, e2, } be a triangle in R2;and A2 = [0, e1] be a line segment.Calculate V (A1, A2).

Week 4. Minkowski addition (a continuation)

Example: ForA1 = {(x, y, z)|0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c}and A2 = {(x, y, z)|x2 + y2 + z2 ≤ 1}

V (t1A1 + t2A2)

= (abc)t31+2(ab+bc+ca)t21t2+π(a+b+c)t1t22+

4π

3t32.

Hence, V (A1, A1, A1) = abc,V (A1, A1, A2) = V (A1, A2, A1)

= V (A2, A1, A1) = 23(ab + bc + ca),

V (A1, A2, A2) = V (A2, A1, A2)= V (A2, A2, A1) = π

3 (a + b + c),V (A2, A2, A2) = 4π

3 .

Lemma 4.1: The mixed volume is ‘linear’ in thefollowing sense:

V (α1A1 + α′1A′1, A2, . . . , Ad)

= α1V (A1, A2, . . . , Ad) + α′1V (A′1, A2, . . . , Ad)

for any α1 ≥ 0 and α′1 ≥ 0.

19

Proof: Evaluate the coefficients of the termst1t2 . . . td in the identical polynomials

Vd(t1(α1A1 + α′1A′1) + t2A2 + · · ·+ tdAd)

and

Vd((t1α1)A1 + (t1α′1)A

′1 + t2A2 + · · ·+ tdAd)

Q.E.D.

Lemma 4.2: If A′1 ⊂ A1, then A′

1+A2 ⊂ A1+A2.

Proof follows from the visualization of Minkowskisum. Q.E.D.

Def.: The Hausdorff distance of A1 and A2 isdefined by d(A1, A2)

= inf{t ≥ 0|A1 ⊂ A2 + tB and A2 ⊂ A1 + tB},where B is a unit ball in Rd.

Remark: The Minkowski addition and, thus,mixed volume, is continuous w.r.t. the Haus-dorff distance. Hence Minkowski Theorem 3.8and Lemmas 3.8 and 4.1 hold true for arbitraryconvex bodies.

20

Homework 1 (5 points): Verify that, if

A{(x, y)||x| ≤ 1, |y| ≤ 1}and

B{(x, y)|x2 + y2 ≤ 1}then the function f : [0, 1] → R defined as

f(t) =√

V2((1− t)A + tB) t ∈ [0, 1],

is concave.

Theorem 4.3 (The Brunn–Minkowski inequal-ity): Let A and B be non-empty sets in Rd whichhave volume. Then

(Vd(A + B))1/d ≥ (Vd(A))1/d + (Vd(B))1/d.

Remark: For convex bodies equality holds if andonly if A and B are homothetic.

Proof is in 3 steps.

Step 1: prove the inequality for A and B be-ing boxes, i.e., rectangular parallelepipeds whosesides are parallel to the coordinate hyperplanes(side length is ai and bi):

Vd(A) =d∏

i=1ai, Vd(B) =

d∏

i=1bi,

Vd(A + B) =d∏

i=1(ai + bi),

21

By the arithmetic-geometric mean inequality

( d∏

i=1

ai

ai + bi

)1/d+

( d∏

i=1

bi

ai + bi

)1/d

≤ 1

d

d∑

i=1

ai

ai + bi+

1

d

d∑

i=1

bi

ai + bi= 1.

Step 2 (is called a Hadwiger–Ohmann cut): provethe inequality for A and B being finite unions ofboxes.

Translate A in such a way that the hyperplanexd = 0 separates two boxes in A. Let A+ =A ∩ {x ∈ Rd|xd > 0}. (A− is defined similarly.)Translate B so that

Vd(A±)

Vd(A)=

Vd(B±)

Vd(B).

Note that A++B+ ⊂ {xd > 0}, A−+B− ⊂ {xd <0}, and that the numbers of boxes in A+ ∪ B+

and A− ∪ B− are both smaller than the numberof boxes in A ∪ B. By induction on the latternumber and Step 1, we have

Vd(A + B) ≥ Vd(A+ + B+) + Vd(A− + B−)

≥(Vd(A+)1/d + Vd(B+)1/d

)d

+(Vd(A−)1/d + Vd(B−)1/d

)d

22

= Vd(A+)(1+

Vd(B)1/d

Vd(A)1/d

)d+Vd(A−)

(1+

Vd(B)1/d

Vd(A)1/d

)d

= Vd(A)(1+

Vd(B)1/d

Vd(A)1/d

)d=

(Vd(A)1/d+Vd(B)1/d

)d.

Step 3: approximate bounded measurable sets byfinite unions of boxes. Is standard for the theoryof integral and is omitted here. Q.E.D.

Theorem 4.4: Let A, B be non-empty boundenconvex sets in Rd. Then the functionf : [0, 1] → R defined by the equation

f(t) = V1/dd

((1− t)A + tB

)

is concave.

Proof: Let x, y ∈ [0, 1]. Let λ, µ ≥ 0 withλ + µ = 1. Apply Theorem 4.3 to the setsλ((1− x)A + xB) and µ((1− y)A + yB):

f(λx+µy) = V1/dd

((1−(λx+µy))A+(λx+µy)B

)

= V1/dd

(λ((1− x)A + xB) + µ((1− y)A + yB)

)

≥ λV1/dd ((1− x)A + xB) + µV

1/dd ((1− y)A + yB)

= λf(x) + µf(y).

Q.E.D.

23

Theorem 4.5 (Minkowski’s inequality for mixedvolumes): Let A and B be convex bodies in Rd.Then

V (A, . . . , A, B) ≥(Vd(A)

)(d−1)/d(Vd(B)

)1/d,

with equality holding if and only if A and B arehomothetic.

Proof: Define a function f : [0, 1] → R by theequation

f(t) = V1/dd ((1− t)A + tB) for0 ≤ t ≤ 1.

Then

fd(t) = Vd(A)(1−t)d+d·V (A, . . . , A, B)(1−t)d−1t

· · ·+ Vd(B)td,

whence(fd(t)

)′(0) = d · f (d−1)(0) · f ′(0)

and

f ′(0) =V (A, . . . , A,B)− Vd(A)

V(d−1)/dd (A)

.

Theorem 4.4 implies that f ′(0) ≥ f(1)− f(0).

We omit the proof for the case of equality.

Q.E.D.

24

Theorem 4.6 (Alexandrov – Fenchel’s quadraticinequality): Let A, B be non-empty boundenconvex sets in Rd. Then

V (A, . . . , A,B)2 ≥ V (A, . . . , A, A,A)

×V (A, . . . , A, B, B).

Proof: According to Theorem 4.4, the functionf : [0, 1] → R defined by the equation

f(t) = V1/dd

((1− t)A + tB

)

is concave. Hence, f ′′(0) ≤ 0. Q.E.D.

Homework 2 (5 points): Let A be a compactconvex set in Rd and B be a unit ball in Rd.Prove that

S(A) = d · V (A, . . . , A, B),

where S(A) stands for the (d − 1)-dimensionalsurface area of the boundary of A.

25

Theorem 4.7 (isoperimetric inequality):Every convex body A in Rd has a surface areagreater or equal than that of a ball rB with thesame volume.

Proof: Let Vd(A) = Vd(rB). ThenVd(rB) = rdVd(B), S(A) = d · V (A, . . . , A, B),S(rB) = d · V (rB, . . . , rB, B) and

S(A) = d·V (A, . . . , A,B) ≥ d·V (d−1)/dd (A)V

1/dd (B)

= d·rd(d−1)/d·V 1−1/dd (B)V

1/dd (B) = d·r(d−1)·Vd(B)

= d · r(d−1) · V (B, . . . , B) = d · V (rB, . . . , rB, B)

= S(rB). Q.E.D.

Remark: Taking into consideration that in theMinkowski’s inequality for mixed volumes equal-ity holds if and only if the bodies are homothetic,we may conclude that every convex body A in Rd

has a surface area greater than that of a ball rB

with the same volume, unless it is itself a ball.

26

Combinatorial properties ofconvex polytopes

Week 5. Introduction to graph theory

Def.: Formally, a graph is a binary relation onthe set of vertices. In formally, a graph is a math-ematical object composed of points, known asvertices or nodes and lines connecting some (pos-sibly empty) subset of them, known as edges.

Def.: Let P be a convex polytope. The verticesand the edges of P form an (abstract, finite, undi-rected, simple) graph, called the graph of P anddenoted by G(P ).

Def.: A graph G is planar if there exists a draw-ing of G in the plane in which no two edges in-tersect in a point other than a vertex of G, whereeach edge is a simple arc. Such a drawing of a pla-nar graph G is called a plane representation of G.In this case, we also say that G has been embed-ded in the plane. A plane graph is a planar graphthat has been already embedded in the plane.

Examples: K4 and the graph of any polytopeare planar.

Lemma 5.1: K5 is nonplanar.

27

Proof: Assume the contrary, namely, K5 is pla-nar and v1, . . . , v5 be the vertices of its embed-ding. The cycle C = v1v2v3v4v1 (as a closedJordan curve) divides the plane into two parts.Suppose v5 belong to the interior of C (the othercase can be studied similarly). Draw the edgesv5v1, v5v2, v5v3, and v5v4. None of the remain-ing two edges v1v3 and v2v4 can be drawn in theinterior of C. Thus v1v3 lies in the exterior ofC. Then one of v2 and v4 belongs to the interiorof the cycle C1 = v1v5v3v1 and the other to itsexterior. Hence, v2v4 cannot be drawn withoutviolating planarity. Q.E.D.

Lemma 5.2: K3,3 is nonplanar.

Proof: Assume the contrary, namely, K3,3 isplanar and u1, u2, u3 and v1, v2, v3 be its bipar-tition in a plane representation. The cycle C =u1v1u2v2u3v3u1 (as a closed Jordan curve) dividesthe plane into two parts. Suppose the edge u1v2

lies in the interior of C (the other case can bestudied similarly). None of the remaining twoedges u2v3 and u3v1 can be drawn in the interiorof C without crossing the edge u1v2. Draw u2v3

in the exterior of C. Then one of v1 and u3 be-longs to the interior of the cycle C1 = u1v2u2v3u1

and the other to its exterior. Hence, v1u3 cannotbe drawn without violating planarity. Q.E.D.

28

Theorem 5.3 (Kuratowski): A graph is planarif, and only if, it has no subgraph homeomorphicto K5 or K3,3.

Theorem 5.4 (Euler formula): For a connectedplane graph G, f0 − f1 + f2 = 2, where f0, f1,and f2 denote the number of vertices, edges, andfaces respectively.

Proof by induction on f2.If f2 = 1, then G is a tree and f1 = f0−1. Hence,f0 − f1 + f2 = 2.Assume f0−f1+f2 = 2 is true for all plane graphswith f2 ≤ f − 1 faces, f ≥ 2. Suppose G has f

faces. Since f ≥ 2, G contains a cycle C. Let e

be an edge of C. Then e belongs to exactly twofaces, say F1 and F2, of G and the deletion of efrom G results in the formation of a single facefrom F1 and F2. Also, since e is not a cut edge ofG, G\e is connected and has f2−1 faces. By theinduction hypothesis, f0− (f1−1)+(f2−1) = 2.Q.E.D.

Lemma 5.5: If G is a simple planar graph withat least three vertices, then f1 ≤ 3f0 − 6.

Homework (5 points): Show that the comple-ment of a simple planar graph with 11 vertices isnonplanar.

29

Def.: A subset V ′ of the vertex set V (G) of aconnected graph G is a vertex cut of G, if G \ V ′

is disconnected; it is a k-vertex cut if |V ′| = k.

Def.: For nontrivial connected graph G havinga pair of nonadjacent vertices, the minimum kfor which there exists a k-vertex cut is called thevertex connectivity or simply the connectivity ofG.

Def.: Dual of a plane graph.

Theorem 5.6 (Euler – Poincare relation): LetP be a non-empty polytope in Rd. Then

f−1(P )− f0(P ) + · · ·+ (−1)d+1fd(P ) = 0,

where fk(P ) denotes the number of k-faces of P

and f−1(P ) = 1.

Proof: . . . Q.E.D.

Remark: Is the Euler – Poincare relation theonly linear equation satisfied by the numbers offaces of various dimensions of every polytope witha given dimension?

30

Theorem 5.7: Suppose that α0, . . . , αd are realnumbers s.t. the numbers fk(P ) of the k-faces ofany polytope P satisfy the equation

α−1f−1(P ) + α0f0(P ) + · · ·+ αdfd(P ) = 0.

Then (α−1, α0, . . . , αd) is proportional to(1,−1, . . . , (−1)d+1).

Proof: Use induction. Calculate the numbers offaces (d+1)-dimensional pyramid and bipyramidwith an arbitrary base. Q.E.D.

Theorem 5.8 (Dehn – Sommerville equations):Let P be a simplicial polytope in Rd. Then

d−1∑

j=k

(−1)jCk+1j+1 fj(P ) = (−1)d−1fk(P )

for k = −1, 0, . . . , d − 2. Here Ckn is a binomial

coefficient defined by (a + b)n =∑n

k=0 Cknakbn−k.

Homework 2 (5 points): Write all Dehn – Som-merville equations for 3- and 4-dimensional sim-plicial polytopes.

31

Week 6. Regular polytopes

Notation: {p} stands for a regular polygon.

Def.: A convex polyhedron in R3 is said to beregular if its faces are regular and equal, while itsvertices are surrounded alike.

Def.: {p, q} is the Schlafly symbol for a regu-lar polytope whose faces are {p}’s, q surroundingeach vertex.

Lemma 6.1: The Schlafly symbol {p, q} cannothave any other values than {3, 3}, {3, 4}, {4, 3},{3, 5}, {5, 3}.Proof: Solid angle at a vertex is convex and hasq face-angles, each (1− 2/p)π. Hence,q(1− 2/p)π < 2π or (p− 2)(q − 2) < 4. Q.E.D.

Lemma 6.2: For each of the Schlafly symbols{3, 3}, {3, 4}, {4, 3}, {3, 5}, {5, 3} there exists aregular polytope

Proof by constructing a regular tetrahedron, oc-tahedron, cube, icosahedron, and dodecahedron.Q.E.D.

32

Homework 1 (5 points): Take a cube with anedge 2 and construct segments with the endpointsat the edges such that they are parallel for par-allel faces and perpendicular otherwise. On eachsegment, mark 2 points at the distance (3−√5)/2from those edges of the cube to which this seg-ment is orthogonal. Prove that, cutting parts offthe cube by planes that pass through triples ofpoints according to the combinatorial structureof an icosahedron, we get a regular icosahedron.

Remark: In the definition of the regular poly-tope, all three conditions are necessary. (Con-sider a triangular bipiramid, a prism, and a diphe-noid.)

Def.: The vertex figure at the vertex O of a poly-tope in R3 is the polygon whose vertices are themidpoints of all edges through O.

Def.: A polyhedron is regular if its faces andvertex figures are all regular.

Lemma 6.3: The two definitions of regular poly-topes are equivalent.

Proof: Check that all sides, faces, and dihedralangles are equal. Q.E.D.

Def.: A flag (A,AB,ABC . . .) of a convex poly-tope in R3 is the figure consisting of a vertex A,

33

an edge AB containing this vertex, and a faceABC . . . containing this edge.

Def.: The polyhedron is said to be regular if itssymmetry group is transitive on its flags.

Remark: Finite groups, generated by reflections;Coxeter group is a group with the representation(RmRn)

qmn = I where qmn = 1 so that generatorsare involutory; Dynkin diagrams.

Remark: Regular polytopes in four dimensions

5-cell, α4: {3, 3, 3} 5 10 10 516-cell, β4: {3, 3, 4} 8 24 32 16Tessaract, γ4: {4, 3, 3} 16 32 24 824-cell: {3, 4, 3} 24 96 96 24600-cell: {3, 3, 5} 120 720 1200 600125-cell: {5, 3, 3} 600 1200 720 120

Remark: Regular polytopes in d dimensions(d ≥ 5)

regular simplex, αd: {3d−1} d + 1; Cj+1d+1; d + 1

cross-polytope, βd: {3d−2, 4} 2d; 2j+1Cj+1d ; 2d

measure polytope, γd: {4, 3d−2} 2d; 2d−jCjd; 2d

Homework 2 (5 points): Which d-dimensionalregular polytopes are simplicial for d ≥ 5? WriteDehn – Sommerville equations for them.

34

Week 7. Theorem of Steinitz

Lemma: Graph of any polytope with non-emptyinterior in R3 is planar and 3-connected.

Proof: Planarity was already explained. Let’sprove that the graph G of any polytope P withnon-empty interior in Rd is d-connected.

Let v1, . . . , vd−1 be some d−1 vertices of P and letG∗ be the subgraph of G obtained by omission ofv1, . . . , vd−1 and all the edges incident with them.

Denote A = aff {v1, . . . , vd−1} and consider twocases: (i) A ∩ intP = ∅ and (ii) A ∩ intP 6= ∅.(i): Let F = A∩P be the face of P determined byv1, . . . , vd−1, dim F ≤ d−2; let H be a supportinghyperplane of P s.t. H ∩ P = F ; let H+ be theother supporting hyperplane of P parallel to H.

∀ vertex v ∈ P either v ∈ H+ or ∃ a vertexv′ ∈ P , joined by an edge to v, s.t. v′ is nearerto H+ than v.

Hence, each vertex of P which is not in v1, . . . , vd−1,is connected by a path in G∗ to some vertex ofH+∩P . Since H+∩P is a polytope, its graph is aconnected subgraph of G∗; hence G∗ is connectedin case (i).

35

(ii): Let H be any hyperplane containing A andat least one vertex vd which is not in A.

Denote by H+ and H− the two supporting hy-perplanes of P parallel to H. Apply the samearguments as in case (i) separately to the partof P contained in the slab determined by H andH+ and to the part of P contained in the slabdetermined by H and H−.

Therefore each of the corresponding graphs isconnected, and since they have a common vertexvd the graph G∗ is connected in case (ii). Q.E.D.

Theorem (Steinitz’s theorem): For every planar3-connected graph G, with 6 or more edges, thereexists a polytope in R3 whose graph is isomorphicto G.

Proof is by induction on the number of edges e

of G.

Base of induction: the only 3-connected pla-nar graph with 6 edges is K4 which is realizableas a simplex.

Outline of the step of induction:

(i) Show that each graph G considered has 3-valent elements (i.e., 3-valent vertices or triangu-lar faces).

36

(ii) For any given G and any 3-valent elementof G construct a new graph G∗, also planar, 3-connected, and with 6 or more edges, s.t. fromany realization of G∗ by a 3-polytope P ∗ a 3-polytope P realizing G may be constructed. Thisprocedure is called elementary transformations orreductions.

(iii) If G contains a 3-valent vertex incident toa triangular face then an elementary transforma-tion of G yields a G∗ which has less edges than G

and induction takes over. If G does not containsuch an incidence, show that there exists a finitesequence of elementary transformations s.t. thetransformed graph contains a 3-vertex incident toa triangular face and the former arguments work.

(i) (Double counting of incidences): Let v,e, and f stand for the numbers of vertices, edges,and 2-faces of P . Let vk denote the number of k-valent vertices, while fk is the number of k-gonal2-faces of P . Thus

v =∑

k≥3vk and f =

∑

k≥3fk.

Counting the number of incidences of edges andk-gonal faces we obtain

2e =∑

k≥3kfk.

37

Similarly, counting incidences of edges and k-valent vertices we get

2e =∑

k≥3kvk.

Combining these expressions with Euler’s equa-tion v − e + f = 2 yields

∑

k≥3kvk +

∑

k≥3kfk = 4e

= 4v + 4f − 8 = 4∑

k≥3vk + 4

∑

k≥3fk − 8

or

v3 + f3 = 8 +∑

k≥5(k − 4)(vk + fk) ≥ 8.

Therefore every 3-polytope P has at least eight3-valent elements. Case (i) is done.

Homework 1 (5 points): Prove that if a convexpolytope with non-empty interior in R3 has morefaces than vertices, it has at least six triangularfaces.

(ii) (Reductions): In either of the following twocases we say that a graph G∗ is obtained from agraph G by an elementary transformation (or areduction).

38

(1): A trivalent vertex of G and the edges inci-dent to it are deleted, and the three vertices con-nected to it in G are pairwise connected by ‘new’edges (unless some of them are already connectedin G). The four possible cases are represented ina figure and denoted by ωi, i = 0, 1, 2, 3.

(2): The 3 edges of a trigonal face are deleted andthe 3 vertices joined to a ‘new’ node. If any 2-valent node would result it is omitted and the twoedges incident to it are replaced by a single edge.The four possible elementary transformations ηi,i = 0, 1, 2, 3, of this type are indicated in a figure.

Properties of elementary transformations:

• every graph G∗ obtained from a planar graphG by an elementary transformation is planar;

• if G is 3-connected, G∗ is 3-connected also.

If P ∗ is any 3-polytope with non-empty interiorwith graph G∗, describe the construction of apolytope P with graph G:

• for ηi: cut off the ‘new’ vertex of P ∗;

39

• for ω1, ω2, ω3: P is the convex hull of theunion of P ∗ with an appropriate point v, which isbeyond the facet of P ∗, and beneath all the otherfaces of P ∗ in case of an ω3 reduction, while inthe case of ω2 or ω1 reductions v is beneath allbut one (or two) of them;

• for ω0: P is the convex hull of the union of P ∗

with the point v, determined as the intersectionof the planes of the three 2-faces of P ∗ adjacentto the ‘new’ triangle, provided v is beyond thistriangular face of P ∗. However . . .

Therefore, in all cases if we can realize G∗ as apolytope, we can realize G also. Since G∗ con-tains i edges less than G if it is obtained from G

by ωi or ηi, it follows that the inductive proof ofthe theorem is completed for all G to which ωi

or ηi is applicable, i = 1, 2, 3. Thus we have toprove that

(iii): given a graph G to which none of the reduc-tions ωi or ηi, i = 1, 2, 3, is applicable, it is pos-sible to obtain a graph to which some of them areapplicable by performing on G a finite sequenceof operations ω0 and η0.

40

Homework 2 (5 points): Prove Steinitz’s the-orem for 3-valent graphs G by showing that asingle ‘elementary operation’ θ is needed, whichconsists of deleting an adge AB and amalgamat-ing the two pairs of edges incident to A and B totwo edges. Namely, show that:

(i) every 3-valent, 3-connected planar graph G

with more than 6 edges containes an edge AB towhich θ is applicable; the resulting graph G∗ is3-valent and 3-connected;

(ii) from any realization of G∗ by a 3-polytopeP ∗ a realization of G by a 3-polytope P may beconstructed.

Description of step (iii): Let G be a plane3-connected graph; define a new graph I(G) asfollows:

The vertices of I(G) are (interior) points of G,one on each edge. The vertices are joined by anedge iff the two edges of G corresponding to themhave a common vertex and are incident to thesame face in G.

Properties: I(G) is planar; 3-connected; everyvertex has valence 4; the faces of I(G) are in one-to-one correspondence with the union of the setof faces of G and the set of vertices of G, i.e.,f(I(G)) = f(G) + v(G); a vertex and a face of

41

G are incident iff the corresponding faces of I(G)have a common edge; a k-gonal face of I(G) cor-responds to a k-gonal face of G or to a k-valentvertex of G.

Def.: An edge AB of a 4-valent, 3-connected,planar graph C has a direct extension BC pro-vided the edges AB and BC separate the othertwo edges incident to B.

Def.: A path A0A1 . . . An in C where the vertexAk−1 is joined to Ak by an edge, k = 1, . . . , n, iscalled a geodesic arc provided Ak−1Ak has AkAk+1

as direct extension, for 1 ≤ k < n; for a closedgeodesic, A0 = An, and An−1An has A0A1 as di-rect extension.

Def.: A subgraph L of C is called a lens if

(α): L consists of a simple closed path Z:A0A1 . . . AnB0B1 . . . BmA0 (called the boundaryof L) and all the vertices and edges of C con-tained in one of the connected components of thecomplement of Z (called inner vertices and edgesof Z).

(β)): Z is formed by two geodesic arcsA0A1 . . . AnB0 and B0B1 . . . BmA0, such that noinner edge of Z is incident to the poles A0 and B0

of Z. (See fugures. . . )

42

Def.: A lens L in C is called indecomposable pro-vided no lens of C is properly contained in L.

Property of indecomposable lens: Every inde-composable lens L contains a triangular face in-cident to the boundary Z of L.

Basic def.: Define g(G) as the minimal numberof faces in an indecomposable lens L in I(G).

In case (iii) induction is in g(G).

If g(G) = 2, there is only one possibility for theindecomposable lens and G contains a triangularface incident to a trivalent vertex; thus one of theelementary transformations ωi or ηi, i = 1, 2, 3,may be applied to G.

If g(G) > 2 we shall apply a reduction of type ω0

or η0 to the graph G in order to obtain a graphG∗ with g(G∗) < g(G).

Consider an indecomposable lens L in I(G) withg(G) faces. Let T be a triangle in L, incident tothe boundary of L. Consider two cases: T corre-sponds to a triangular face (use η0) or a trivalentvertex (use ω0).

43

Week 8. Theorem of Cauchy

Theorem (Cauchy; 1813): Two convex polyhe-dra in R3 with corresponding equal (congruent)and similarly situated faces have equal dihedralangles between their respective faces (and, thus,are either congruent or symmetrical.

Remark: A similar theorem holds true in 3-dimensional spherical and hyperbolic spaces.

Homework 1 (5 points): Using the remark aboveprove that Cauchy’s theorem holds true for poly-topes in R4.

Outline of the proof of the theorem for R3:

Let P and P ′ be two convex polyhedra in R3

with corresponding equal (congruent) and simi-larly situated faces. Mark an edge of P with aplus sign if the dihedral angle at this edge of thepolyhedron P is greater than the correspondingdihedral angle of P ′, and, in the contrary case,with a minus sing. Leave the remaining edgesunmarked.

If there is no edge marked, the theorem is proved.So, suppose there is a vertex of P out of whichpasses a marked edge. By Lemma 8.1 below,

44

there must pass out of this vertex at least 4 markededges; moreover on a consecutive circuit of theseedges we must find no fewer than 4 changes fromplus to minus and from minus to plus.

The set of all marked edges form a planar graphG. Let v, e, and f stand for the numbers ofvertices, edges, and 2-faces of G. Let N denotesthe total number of sign changes. By Lemma 8.1

N ≥ 4v. (1)

Let fk denote the number of k-gonal 2-faces of G.Then

N ≤ 2f3 + 4f4 + 4f5 + 6f6 + . . . . (2)

Counting the number of incidences of edges andk-gonal faces we obtain

2e =∑

k≥3kfk.

The total number of faces is

f =∑

k≥3fk.

Combining these expressions with Euler’s inequal-ity v − e + f ≥ 2 yields

4v−8 ≥ ∑

k≥32(n−2)fn = 2f3+4f4+6f5+. . . (3)

45

Comparing (2) and (3), we get N ≤ 4v−8 whichcontradicts (1).

Lemma 8.1 (Cauchy): Let a convex polygon(planar or spherical) v1v2 . . . vn be transformedinto another convex polygon v′1v

′2 . . . v′n in such a

way that the lengths of the sides do not change.If the angles at the vertices v2 . . . vn−1 under thistransformation either all increase, or if part ofthe angles increase while all the remaining onesstay unchanged, then the length of the side vnv1

increases.

Proof: The lemma is clear for triangles.

Suppose lemma holds true for all polygons withfewer number of sides and only the angle6 vi−1vivi+1 at the vertex vi changes. In 4v1vivn

the lengths of the sides v1vi and v1vn do notchange but the angle at the vertex vi increases;thus the length of the side v1vn also increases.

46

In the general case, we increase one angle afteranother, leaving the remaining angles unchanged.Q.E.D.

Remark: In this proof there is a defect whichwas recognized and corrected by E. Steinitz: byincreasing one angle in a convex n-gone we mayarrive at a non-convex polygon.

47

Week 9. A.D. Alexandrov’s andH. Minkowski’s uniqueness theorems

Theorem (A.D. Alexandrov’s uniqueness theo-rem): Let there be given two convex polyhedrain R3 such that each 2-face of one correspondsto a 2-face of the other with parallel outer nor-mal, and conversely. If each pair of correspond-ing faces has the property that neither face can bestrictly imbedded in the other by a parallel trans-lation, then the polyhedra are equal and parallel.

Homework 1 (5 points): By choosing appropri-ate parallelepipeds, prove that the above theoremis not true in R4.

Proof of the theorem: consider the Minkowskisum P1/2 = 1

2(P1 + P2). Each edge of P1/2 is gen-erated by a pair of edges of P1 and P2 (probably ofzero length, i.e., of a vertex). Mark sides of P1/2

by pluses or minuses and use arguments from theproof of the Cauchy theorem and the followinglemma 9.1. Now edges are marked. Q.E.D.

Lemma 9.1: If two convex polygons cannot beimbedded one in the other by a parallel transla-tion, then the differences of the lengths of theirsides with parallel outer normals change sign noless then 4 times on a circuit of these polygons.

48

Proof: Obtain a contradiction in the both cases(a) no changes of sign and (b) two changes ofsign. Use Lemmas 9.2 and 9.3. Q.E.D.

Lemma 9.2: If in a polygon P1 all the sides,with the possible exception of one side l, are lessthan those of a polygon P2, then P1 can by aparallel translation be strictly imbedded in P2.

Lemma 9.3: Let two convex polygonal lines Q1

and Q2 lie in an angle with vertex O, having theirend points on the sides of this angle and turningtheir convex sides toward O. Then if any rayfrom O meets Q1 before Q2, there is a side of Q1

which is shorter than the parallel side of Q2.

Theorem (H. Minkowski’s uniqueness theorem):If each (d−1)-dimensional face of a convex poly-tope in Rd, d ≥ 2, corresponds to a (d − 1)-dimensional face of another convex polytope suchthat the two faces have equal (d−1)-dimensionalvolumes and parallel outer outer normals, andconversely, then the two polytopes are equal andparallel.

Remarks: (1) For d = 2 Minkowski’s theoremis trivial. (2) For d = 3 it is an immediate con-siquence of A.D. Alexandrov’s theorem. (3) Itsays nothing about the combinatorial structureof the polytopes.

49

Homework 2 (5 points): Among convex poly-topes with no more that 6 vertices in R3, findtwo polytopes with parallel 2-dimensional facesand non-isomorphic graphs.

Take home prelim (due date 04/10/2007)

1. (10 points) Determine all convex subsets A ofR2, for which R2 \ A is also convex.

2. (10 points) Prove that, for a bounded set A ⊂R2 with non-empty interior, the following twoconditions are equivalent:

(a) A is convex;(b) every line passing through an arbitrary in-

terior point of A intersects the boundary of A intwo points.

3. (10 points) Prove that the greatest distancebetween two points of a convex set A ⊂ R2 isidentical with the greatest distance between par-allel supporting lines.

4. (10 points) Prove that the support function ofa non-empty convex set is affine (that is, linearplus a constant) if and only if the set consists ofa single point.

5. (10 points) Determine all 3-faces of the 4-polytope P ⊂ R4, defined as the convex hull of

50

the ten points (0, 0, 0, 0), (0, 1, 0, 1), (0, 1, 1, 0),(0, 0, 1, 1), (±1, 1, 0, 0), (±1, 0, 1, 0), (±1, 0, 0, 1).

6. (50 points) Prove that inside every closedbounded convex set A ⊂ R2 with non-empty in-terior there exists a point O such that every chordab of A which passes through O is divided intotwo segments aO and Ob, each of whose lengthsis not less then 1/3 the length of the segment ab.(Hint: (a) for every boundary point a considerthe set Aa of all points c ∈ A such that, if c lies ona chord ab for some boundary point b ∈ A, then|ac| ≤ (2/3)|ab|; (b) prove that Aa is a convex setand reduce problem 6 to the following statement:∩Aa 6= ∅; (c) now use Helly’s theorem to provethe latter statement.)

7. (10 points) Prove that, in problem 6 above,the constant 1/3 cannot be improved.

8. (10 points) For any pair P,Q of polytopes inRd prove f0(P +Q) ≤ f0(P ) ·f0(Q), where f0(P )stands for the number of vertices of P and P +Q

stands for the Minkowski sum of P and Q.

9. (50 points) Prove that every convex polygoncan be represented as the Minkowski sum of tri-angles and segments.(Hint: prove that every convex polygon A can

51

be represented as the sum A1 + B, where A1 isa convex polygon with fewer sides than A and Bis a segment or a triangle. Distinguish two cases:(a) A has parallel sides and (b) A has no parallelsides.)

10. (10 points) Prove that the representationconstructed in problem 9 in not unique, i.e., thatsome polytopes can be represented in several waysas sums of triangles.(Hint: Find an example among pentagons.)

11. (10 points) Prove that if in two convex poly-topes in R3 the faces in pairs have parallel outernormals and equal perimeters, then the polytopesare equal and parallel.

52

Week 10. H. Minkowski’s existence theorem

Theorem (H. Minkowski’s existence theorem):If n1, . . . , nm are unit vectors in Rd, d ≥ 2, notpointed in some closed half-space, and F1, . . . , Fm

are positive numbers such that

m∑

i=1Fini = 0, (∗)

then there exists a closed convex polytope in Rd

whose (d − 1)-dimensional faces have outwardnormals ni and (d− 1)-dimensional volumes Fi.

Lemma 10.1 (Mapping Lemma): Let A and Bbe two manifolds of dimension d. Let ϕ be amapping from A into B s.t.

(1) each connected component of B containsimages of points in A;

(2) ϕ is one-to-one;(3) ϕ is continuous;(4) if points Bm (m = 1, 2, . . .) of the manifold

B are images of points Am of the manifold A andBm → B, then there is a point A in A whose im-age is B and for which there exists a subsequenceAmi

of Am converging to A.Under these conditions, ϕ(A) = B, i.e., all

points of the manifold B are images of some pointsof the manifold A.

53

Proof: The set ϕ(A) is an open and closed sub-set of B intersecting each connected componentof B. Q.E.D.

Outline of the proof of Minkowski’s existencetheorem:

Let B be the set of all collections of positivenumbers F1, F2, . . . , Fm (for some fixed vectorsn1, n2, . . . , nm) satisfying (∗). B is a manifoldof dimension m− d.

Let A be the set of classes of convex polytopesin Rd (polytopes are said to be equivalent if oneis obtained from the other by a parallel displace-ment). A polytope may be treated as a set of val-ues of its supporting function h1, h2, . . . , hm andA is a manifold of dimension m− d.

ϕ associates with every class of equivalent poly-topes the set of areas of faces: ϕ : A → B.

Lemma 10.1 implies ϕ(A) = B, i.e., for ev-ery set of positive numbers F1, F2, . . . , Fm sat-isfying (∗), there exists a polytope. Conditions(1)–(3) are obviously satisfied. Condition (4) fol-lows from Lemma 10.2 below. Q.E.D.

Lemma 10.2: Each convex polytope in R3 ofarea ≤ F and with all faces of area ≥ f has sup-port numbers bounded by a quantity dependingonly on F and f , provided that the origin liesinside the polytope.

54

Proof: Assume the origin lies inside the givenpolytope P ; Fi is the area and hi is the supportfunction of a face of P . Then P contains a pyra-mid of altitude hi and base Fi. Thus V ≥ Fihi/3.On the other hand, the isoperimetric inequalityyields V ≤ F 3/2/(6

√π). Taking into account

f < Fi, we get hi < F 3/2/(2√

πf). Q.E.D.

Minkowski’s proof of his existence theorem:Fix a set F 0

1 , . . . , F 0m satisfying (∗). Among all

polytopes with prescribed normals satisfying thecondition ∑

i

hiF0i = 1

find a polyhedron of greatest volume.By Lemma 10.3, the volume of the polytope

is a differentiable function of h1, . . . , hm. By theLagrange multiple rule, at a maximum point wemust have

∂

∂hi(V (h1, . . . , hm) + λ

∑

i

F 0i hi) = 0,

i = 1, . . . , m. Taking into account that ∂V/∂hi =F 1

i , we find that there exists a polytope P 1 s.t.µF 1

i = F 0i , where µ = −1/λ. The polytope√

µP 1 is a desired one. Q.E.D.

Lemma 10.3: The volume of the polytope is adifferentiable function of h1, . . . , hm. Q.E.D.

55

Def:. Let P and P ′ be convex polytipes in Rd

with non-empty interiors. Let N = {n1, . . . , nm}be the set of outward unit vectors to (d − 1)-dimensional faces of P and N ′ = {n′1, . . . , n′m′}be the set of outward unit vectors to (d − 1)-dimensional faces of P ′. For every vector n ∈ N ,let F (n) be the (d − 1)-dimensional volume ofthe face of P with the outward normal vector n.Similarly, for every vector n ∈ N ′, let F ′(n) bethe (d− 1)-dimensional volume of the face of P ′

with the outward normal vector n.By definition put N# = N ∪ N ′ and, for each

n ∈ N#, define the number F#(n) as follows:

F#(n) = F (n) + F ′(n), if n ∈ N ∩N ′;

F#(n) = F (n), if n ∈ N and n /∈ N ′;

F#(n) = F ′(n), if n /∈ N and n ∈ N ′.

Note that the sets of unit vectors N# and ofpositive numbers F#(n) satisfy the conditions ofthe Minkowski existence theorem. Hence, thereexists a convex polytope P# s.t. N# is the setof its outward unit vectors to (d−1)-dimensionalfaces and F#(n) is the (d−1)-dimensional volumeof the face of P# with the outward normal vectorn. According to Minkowski uniqueness theorem,such a polytope is unique (up to a translation).

56

P# is called the Blaschke sum of the polytopes P

and P ′, denoted by P# = P#P ′.

Home work (10 points): Prove that the Blaschkesum coincides with the Minkowski sum for convexpolygons and differs from it for convex polytopesin R3.

Remark: The above idea is used in many proofsof existence of an object A satisfying given con-ditions B0. We choose such a function f of A

for which the necessary extremum conditions co-incide with B0. Then the problem is reduced tochecking that f(A) actually attains a minimum(or maximum).

57

Week 11. A.D. Alexandrov’s existence theorem

Def. (naive definition of the development of apolyhedron): union of faces endowed with thegluing (or pairing) rule.

Problem: Describe developments of convex 3Dpolyhedra.

Necessary conditions for a development:

(1) ‘sphere homeomorphic’: f0 − f1 + f2 = 2;

(2) ‘convex’: sum of face angles around eachvertex ≤ 2π.

Remark: Faces of the development may be di-vided into smaller parts, when realized as a poly-hedron.

Def. (formal definition of the development of apolyhedron): a class of isometric metric spaceseach of which is (1) homeomorphic to a sphere;(2) locally isomorphic to the plane except a fi-nite set of points; (3) convex (or of positive cur-vature).

Remark: Doubly-covered convex polygon ob-tained by gluing together two superposed equalpolygons are considered developments.

58

Homework 1 (5 points): Give an example ofshortest arcs on a nonconvex polyhedron in R3

which issue from a single point, overlap over somesegment, and further diverge forming a ‘fork.’

Def.: A development is said to be realizable ifthere is a convex polytope with an isometric de-velopment.

Theorem (A.D. Alexandrov’s existence theorem):Every sphere homeomorphic development of pos-itive curvature is realizable.

Proof is based of the Mapping Lemma 10.1.

Let B be the set of all developments with pre-scribed number of vertices f0. WLOG devel-opments are supposed to be triangulated. Sidelengths li of the triangles are used to determinea development. li are non-negative and satisfyinequalities li + lj ≤ lk and

∑arccos(l2i + l2j −

l2k)/(2lilj) ≤ 2π. Open subsets of Rf1 may beused to parameterize B and, thus, it may be con-sidered as a manifold of dimension f1 = 3f0 − 6.

Let A be the set of all classes of isometric 3D-polytopes with prescribed number of vertices f0.Every point in B has a neighborhood homeomor-phic to a ball in R3f0−6. Hence A is a manifoldand dimA = dimB.

59

Let ϕ : A → B maps a polyhedron to its devel-opment.

• ϕ is one-to-one according to Cauchy’s unique-ness theorem.

• ϕ is continuous.

• if points Bm (m = 1, 2, . . .) of the manifold Bare images of points Am of the manifold A andBm → B, then there is a point A in A whose im-age is B and for which there exists a subsequenceAmi

of Am converging to A.

• each connected component of B contains im-ages of points in A.

Now Mapping Lemma 10.1 implies thatϕ(A) = B. Q.E.D.

Remark: There is no theorem for many-dimensionalpolytopes which is similar to A.D. Alexandrov’sexistence theorem.

Homework 2 (5 points): For four-dimensionalpolytopes, define A and B similarly to the defini-tions in the outline of the proof of A.D. Alexan-drov’s existence theorem on page 59. Calculatethe dimensions of A and B. Should we expect a‘4D-version’ of A.D. Alexandrov’s existence the-orem?

60

There other important theorems that can be provedin the same way as Cauchy uniqueness theoremand A.D. Alexandrov existence theorem:

Theorem (Andreev’s uniqueness theorem):A bounded acute-angled polyhedron in the Loba-chevskij space Hd for d ≥ 3 is defined by its com-binatorial structure and dihedral angles uniquelyup to a motion.

Theorem (Andreev’s existence theorem):A bounded acute-angled polyhedron of a givensimple combinatorial type, different from that oftetrahedron or a triangular prism, and havinggiven dihedral angles exists in the Lobachevskijspace Λ3 if and only if the following conditionsare satisfied:

(1) if three faces meet at a vertex, then the sumof the dihedral angles between them is greaterthan π;

(2) if three faces are pairwise adjacent, but notconcurrent, then the sum of the dihedral anglesbetween them is less than π;

(3) if four faces are ‘cyclically’ adjacent (likethe lateral faces of a quadrilateral prism), thenat least one of the dihedral angles between themis different from π/2.

61

Week 12. Flexible polyhedra

Def. A (non-convex) polyhedron in R3 is flexibleif its spatial form can be changed continuouslywithout changing shapes or sizes of faces (onlydihedral angles are subject to change).

Ex.: Bricard octahedra of the first and secondtype; the Connelly flexible sphere; the Steffenflexible polyhedron with 9 vertices.

Homework 1 (10 points): Prove that there is aflexible (intersection-free) polyhedron which tilesR3.

Def. Let P ⊂ Rd be a closed orientable polyhe-dral surface. For every (d−2)-dimensional face F

of P , let VF be the (d−2)-dimensional volume ofF and θF be the angle between inward unit nor-mal vectors nF and n′F to two (d−1)-dimensionalfaces of P which are adjacent to F . The quantity

∑

F

VF (π − θF )

is called the total mean curvature of P .

Note: This definition is independent of subdivi-sions of P and agree with the standard definitionfor smooth surfaces.

62

Theorem (R. Alexander; 1985): Total mean cur-vature of every flexible polyhedron in Rd (d ≥ 3)is constant during a flex.

Proof: First, prove the formula

−dθF

dt= (

dnF

dt,mF ) + (

dn′Fdt

,m′F ),

where mF and m′F are unit vectors in the hy-

perplanes containing the two (d−1)-dimensionalfaces adjacent to F that are perpendicular to F ;(·, ·) stands for the scalar product in Rd.

Note that the unit vectors nF , n′F , mF and m′F lye

in the 2-plane Π which is the orthogonal comple-ment to F . Introduce a coordinate system in Π.If ϕ and ψ are the angles between mF (and m′

F

respectively) and the first axe of that coordinatesystem, then

θF = ϕ− ψ,

mF = (cos ϕ, sin ϕ),

m′F = (cos ψ, sin ψ),

nF = (− sin ϕ, cos ϕ),dnF

dt= −(cos ϕ, sin ϕ)

dϕ

dt,

n′F = (sin ψ,− cos ψ),dn′Fdt

= (cos ψ, sin ψ)dψ

dt,

(dnF

dt,mF ) + (

dn′Fdt

,m′F ) = −dϕ

dt+

dψ

dt= −dθF

dt.

63

Now rearrange terms in the expression

d

dt

∑

F

VF (π − θF ) = −∑

F

VFdθF

dt

=∑

F

VF (dnF

dt,mF ) +

∑

F

VF (dn′Fdt

,m′F )

=1

2

∑

G

∑

F⊂G

VF (dnF

dt,mF )+

1

2

∑

G

∑

F⊂G

VF (dn′Fdt

,m′F )

in such a way that summation is taken, first, overall (d − 2)-dimensional faces F of every (d − 1)-dimensional face G of P and then over all (d−1)-dimensional faces G. Then proceed as follows

=∑

G

(dnF

dt,

∑

F⊂G

VFmF ) +∑

G

(dn′Fdt

,∑

F⊂G

VFm′F )

and use the fact that∑

F⊂G

VFmF = 0. (∗)

Hence, the derivative of the total curvature equalszero. Q.E.D.

Note: (∗) shows that Alexander’s theorem is aconsequence of Stokes Theorem.

Def. Let P be an oriented closed polyhedron inRd. Let F be a (d− 1)-dimensional face of P ; itsorientation induces an orientation of the pyramidwith F being the base and the origin O being the

64

apex. Denote by vF the volume of that pyramid,if this orientation agrees with the orientation ofRd; in the opposite case denote vF negative thatvolume. Now, the oriented volume of P is definedby the formula V (P ) =

∑vF where the sum is

taken over all (d− 1)-dimensional faces of P .

Note: The oriented volume equals the ‘usual’volume for polyhedra without self-intersections(under a suitable choice of orientation) and is in-dependent of the choice of the origin O.

Homework 2 (5 points): Compute the orientedvolume of Bricard octahedron of the second type.

Theorem (I. Sabitov; 1996): The oriented vol-ume of every flexible polyhedron in R3 is constantduring a flex.

Proof is based on the following lemma: For ev-ery class of combinatorially equivalent polyhedrawith triangular faces in R3 there exists a non-zero polynomial S s.t. the oriented volume of apolyhedron P of that class is a root of S and thecoefficients of S are polynomials in edge lengthsof P . Q.E.D.

65

Remark: The above polynomial S, called Sabitovpolynomial, for tetrahedra is given by the follow-ing formula known as the Cayley–Menger deter-minant:

V 2 =1

2d(d!)2 det

0 1 1 1 · 11 0 l201 l202 · l20d

1 l210 0 l212 · l21d

1 l220 l221 0 · l22d

· · · · · ·1 l2d0 l2d1 l2d2 · 0

. (∗)

Here V stands for the volume of the tetra-hedron in Rd with vertices pj = (p

(1)j , . . . , p

(d)j ),

j = 0, . . . , d, and l2jk = |pj − pk|2 stand for thesquared edge lengths.

To prove (∗) start with a well-know formula ofelementary analytic geometry

V =1

d!det

p(1)0 p

(2)0 · · p

(d)0 1

p(1)1 p

(2)1 · · p

(d)1 1

· · · · · ·p

(1)d p

(2)d · · p

(d)d 1

.

The determinant is unaltered in value by board-ing it with (d + 2)th row and column, with ‘in-tersecting’ element 1, and remaining elements 0.Multiplying this boarded determinant by the trans-

66

pose of the determinant obtained from it by in-terchanging the last two columns we have

V 2 =1

(d!)2 det

(p0, p0) (p0, p1) · · (p0, pd) 1(p1, p0) (p1, p1) · · (p1, pd) 1· · · · · ·

(pd, p0) (pd, p1) · · (pd, pd) 11 1 · · 1 0

,

where (pj, pk) stands for the scalar product in Rd.Now substitute

(pj, pk) =1

2[(pj, pj) + (pk, pk)− l2jk]

in the last determinant, subtract from the jthrow the product of the last row by (pj−1, pj−1)/2and get (∗) after reductions.

Example: Connelly–Walz flexible polyhedral sur-face in R4.

Open problem: Do there exist flexible polyhe-dra in Rd for d ≥ 5?

Def.: Two polyhedra, P and P ′, are called tobe scissor congruent if P can be cut into a fi-nite number of ‘small’ polyhedra and these ‘small’polyhedra can be rearranged in such a way thatprovide a partition of P ′.

67

Theorem: Two polyhedra in R3 are scissor con-gruent if and only if they have the same volumeand the same Dehn invariants.

Def.: Let P (t) be a flexible polyhedron in R3; let`i stands for the length of its ith edge and ϕi(t)stands for its dihedral angle associated with itsith edge. For every function f : R → R s.t.f(x+y) = f(x)+f(y) and f(π) = 0 consider theDehn invariant defined by the formula

Df(P (t)) =∑

i

`if(ϕi(t)).

Hilbert’s third problem: Do there exist poly-hedra in R3 of the same volume which are NOTscissor congruent?

Answer: Yes. Regular tetrahedron and cube ofthe same volume are not scissor congruent.

Open problem: If a polyhedron P in R3 is ob-tained from another polyhedron P ′ by means ofa continuous flex, then P and P ′ are scissor con-gruent. In other words: is every Dehn invariantconstant during a flex?

Partial answer: Yes for all Bricard octahedra.

68

Note: Every deformation of a polyhedron (whichis not necessarily a flex) defines a vector fieldof velocity vectors at the initial moment. If thedeformation is a flex, that vector field preserveslength of every curve ‘up to order 1’.

Def.: A polyhedron P is said to be infinitesi-mally flexible if there exists a vector field v s.t.

(1) v is linear on each face;(2) the length of every curve on P is stationary

under the action of v;(3) v is not generated by a movement of P as

a rigid body.

Def.: Rigidity matrix RM .

Theorem: A polyhedron in Rd is infinitesimallyrigid if and only if the dimension of the null-spaceof its rigidity matrix is ≥ d(d + 1)/2.

Theorem: Every strictly convex polyhedron inRd (d ≥ 3) is infinitesimally rigid.

Proof: Use Cauchy’s arguments. Q.E.D.

69

Theorem (H. Gluck; 1972): Given a combinato-rial structure of a sphere-homeomorphic polyhe-dra in R3, almost all polyhedra of that structureare not flexible.

Proof: In fact almost all are not infinitesimallyflexible because all not infinitesimally flexible poly-hedra constitute the complement to an algebraicvariety

rankRM = 6

which differs from the whole space because thereis a convex polytope of the same combinatorialstructure and the latter is not flexible. Q.E.D.

Open problem: Given a combinatorial struc-ture of a polyhedron of arbitrary genus in R3, isit true that almost all polyhedra of that structureare not flexible?

Take home final exam (due date 05/11/2007)

1. (10 points) The angular defect δi at a vertexvi of a polytope in R3 is the difference between2π and the sum of face angles adjacent to vi. Forevery convex polytope in R3, prove the DescartesTheorem: ∑

δi = 4π

where the sum is taken over all vertices.

70

2. (50 points) Prove that every (bounded, con-vex) polytope P in R3 admits only finitely manydevelopments whose vertices correspond to thevertices of P and whose edge lengths are boundedabove by a fixed number.

3. (50 points) Let P and P ′ be convex polygonsin R2 with parallel and similarly directed sides.Suppose the origin is an interior point of both P

and P ′ . Let `i and hi stand for the length andsupporting number (i.e., the value of the supportfunction) of the ith side of P and, similarly, `′iand h′i stand for the length and supporting num-ber parallel side of P ′. Prove that the mixedarea2 V (P, P ′) and area V (P ) are given by theformulas:

V (P, P ′) =1

2

∑

i

`′ihi =1

2

∑

i

`ih′i

and

V (P ) =1

2

∑

i

`ihi.

4. (5 points) Use the results of Problem 3 toprove that the mixed area of convex polygons P

and P ′ with pairwise parallel and similarly di-rected sides where P ′ is circumscribed about a

2I.e., the mixed volume for planar convex sets.

71

circle of radius 1, is numerically equal to half ofthe length of the perimeter of P .

5. (20 points) Use the results of Problem 4 andthe Brunn–Minkowski inequality to prove the fol-lowing Theorem of Lhuilier: Among all convexpolygons the sides of which have given directionsand the perimeters of which are of length 1, thatpolygon circumscribed about a circle has greatestarea. Formulate (but do not prove) the similarstatement for convex polytopes in R3.

6. (15 points) Prove that the Connelly–Walz flex-ible polyhedral surface in R4 is combinatoriallyequivalent to the regular cross-polytope in R4,i.e., to the boundary of the convex hull of the 8points (±1, 0, 0, 0), (0,±1, 0, 0), (0, 0,±1, 0), and(0, 0, 0,±1).

7. (10 points) Give an example of a sphere-homeomorphic polyhedral surface in R3 which isinfinitesimally flexible but not flexible.

72