# Applicable Geometry MATH 455 Lec 01 Tuesdays, web455/ Geometry MATH 455 Lec 01 Tuesdays, Thursdays 2:55 { 4:10P Malott Hall 238 Victor Alexandrov alexandrov@math. alex@math.nsc.ru Prerequisite: good introduction to linear alge-bra (e.g., MATH 221, 223 ...

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Applicable GeometryMATH 455 Lec 01Tuesdays, Thursdays 2:55 4:10PMalott Hall 238Victor Alexandrovalexandrov@math.cornell.edualex@math.nsc.ruPrerequisite: 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/web455Grading 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].1Important: Return your homework on Tuesday.My office hours at MT 438:Monday, Friday 6:30 7:30 PMSyllabusTheory of convex bodies:general properties of convex bodies (week 1;pages 47),Hellys theorem (week 2; pages 812),Minkowski addition (weeks 34; pages 1326).Combinatorics of convex polytopes:introd. to graph theory (week 5; pages 2729),Eulers theorem (week 5; pages 2931),Steinitzs theorem (week 7; pages 3543).Regular polytopes (week 6; pages 3234).Metric theory of polytopes:Cauchy rigidity theorem (week 8; pages 4447),H. Minkowskis and A.D. Alexandrovs unique-ness theorems (week 9; pages 4850),H. Minkowskis and A.D. Alexandrovs existencetheorems (weeks 1011; pages 5358).Rigidity theory:flexible polyhedra and rigidity theory (weeks 1213; pages 6270).2Recommended 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, 1993.3Week 1. General properties of convex bodiesDef.: 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 i 0 for i =1, 2, . . . , k andki=1i = 1, thenki=1iai 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 andonly if its diagonals intersect inside the quadri-lateral.4Def.: If u Rd and u 6= 0, thenH = {x Rd|(x, u) = }is called a hyperplane with normal vector u, andeither of the sets{x Rd|(x, u) } or {x Rd|(x, u) }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. Choose5x A and x A such that|x x| = minyA,yA|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 byH(A, x) = sup{(y, x)|y A}.Properties:(1) The supporting function is positively ho-mogeneous, i.e.,H(A, x) = H(A, x) for all 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), thenH(T (A), x) = H(A, x) + (x, y).6(4) If T : Rd Rd is a homothetic transfor-mation (i.e., T (x) = x for all x Rd and someconstant 0), thenH(T (A), x) = H(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 Rdwhich 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.7Week 2. Hellys theorem and its applicationsAdv.: Bradley Forrest btf4@cornell.edu is theteaching assistant for MATH 455. His office hoursare Monday 122 & Thursday 12 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.8Theorem 2.1 (Hellys 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, . . . , An1, 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, . . . , An1, An (that is toA1, . . . , An1, 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 npoints can be enclosed in a circle of radius 1.Theorem 2.2 (Jungs 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 1/3.9Proof is based on the homework: it suffice toshow that Theorem is true for three points.Note: In the Jungs theorem, the radius 1/3can be replaced by no smaller number.Note: Hellys theorem in the plane is obviouslyfalse for infinite families of sets.Theorem 2.3 (Hellys 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, yi) and (xi, yi ) be the end points of the i-thsegment, where yi < yi and i = 1, 2, . . . , n. Notethat a line y = kx + b intersects the i-th segmentiffyi kxi + b and yi kxi + b.10But these formulas determine a strip on (k, b)-plane and, according to the conditions of the the-orem, every three strips intersect. Applying Hellystheorem 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 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 (Hellys 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 5j=1Aj = . Let11A = 4j=1Aj. Note A 6= and A A5 = . LetH be a plane that separates A and A5. Notethat the four sets Aj H satisfy the conditionsof Helleys 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, . . . , An1, 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, . . . , An1, An (that isto A1, . . . , An1, An, An+1). Q.E.D.Homework 2 (5 points): Formulate and prove atheorem for R3, which is analogous to the JungsTheorem 2.2.12Week 3. Minkowski additionDef.: For two (not necessarily convex) sets A,Bin Rd the setA + 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, thenT (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.13Remark (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 + . . .+trAris 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) = supyA+B(y, x) == supuA,vB(u + v, x) = supuA(u, x) + supvB(v, x) == HA(x) + HB(x) for all x Rd. Q.E.D.14Def.: If H is a supporting hyperplane of theclosed convex set A, we call F = AH 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 Aand (y, x) = (u, x) for some y Aand (z, x) (v, x) for all z Band (z, x) = (v, x) for some z B}= {w Rd |w = u + vfor some u A HA(x), v B HB(x)}= FA + FB. Q.E.D.15Theorem 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, . . . , Anbe 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 din t1, . . . , tn,V (t1A1 + + tnAn)=np1,...,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 findt1A1+ +tnAn = [t1x1+ +tnxn, t1y1+ +tnyn]16V (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. SetA = t1A1 + + tnAn ( = (t1, . . . , n))F(x) = A HA (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 hyperplaneHA (x) and 0 A . We decompose A into pyra-mids with appex 0 over the faces and obtainVd(A) =1dmj=1HA (xj)Vd1(F(xj))17=1dmj=1(t1HA1(xj)+ + tnHAn(xj))Vd1(F(xj))By inductive assumption, Vd1(F(xj)) is eitherzero or a homogeneous polynomial of degree d1in t1, . . . , tn. Q.E.D.Def.: Arranging the coefficients on the right sideof the formulaV (t1A1 + + tnAn)=np1,...,pd=1V (Ap1, . . . , Apd)tp1 tpd,such thatV (A(p1), . . . , A(pd)) = V (Ap1, . . . , Apd)for any permutation 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).18Homework 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+(a+b+c)t1t22+43t32.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) =3 (a + b + c),V (A2, A2, A2) =43 .Lemma 4.1: The mixed volume is linear in thefollowing sense:V (1A1 + 1A1, A2, . . . , Ad)= 1V (A1, A2, . . . , Ad) + 1V (A1, A2, . . . , Ad)for any 1 0 and 1 0.19Proof: Evaluate the coefficients of the termst1t2 . . . td in the identical polynomialsVd(t1(1A1 + 1A1) + t2A2 + + tdAd)andVd((t11)A1 + (t11)A1 + t2A2 + + tdAd)Q.E.D.Lemma 4.2: If A1 A1, then A1+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.20Homework 1 (5 points): Verify that, ifA{(x, y)||x| 1, |y| 1}andB{(x, y)|x2 + y2 1}then the function f : [0, 1] R defined asf(t) =V2((1 t)A + tB) t [0, 1],is concave.Theorem 4.3 (The BrunnMinkowski 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) =di=1ai, Vd(B) =di=1bi,Vd(A + B) =di=1(ai + bi),21By the arithmetic-geometric mean inequality( di=1aiai + bi)1/d+( di=1biai + bi)1/d 1ddi=1aiai + bi+1ddi=1biai + bi= 1.Step 2 (is called a HadwigerOhmann 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 thatVd(A)Vd(A)=Vd(B)Vd(B).Note that A++B+ {xd > 0}, A+B {xd 2 we shall apply a reduction of type 0or 0 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 0) or a trivalentvertex (use 0).43Week 8. Theorem of CauchyTheorem (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 Cauchys 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 R3with 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,44there 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 8.1N 4v. (1)Let fk denote the number of k-gonal 2-faces of G.ThenN 2f3 + 4f4 + 4f5 + 6f6 + . . . . (2)Counting the number of incidences of edges andk-gonal faces we obtain2e =k3kfk.The total number of faces isf =k3fk.Combining these expressions with Eulers inequal-ity v e + f 2 yields4v8 k32(n2)fn = 2f3+4f4+6f5+. . . (3)45Comparing (2) and (3), we get N 4v8 whichcontradicts (1).Lemma 8.1 (Cauchy): Let a convex polygon(planar or spherical) v1v2 . . . vn be transformedinto another convex polygon v1v2 . . . vn in such away that the lengths of the sides do not change.If the angles at the vertices v2 . . . vn1 under thistransformation either all increase, or if part ofthe angles increase while all the remaining onesstay unchanged, then the length of the side vnv1increases.Proof: The lemma is clear for triangles.Suppose lemma holds true for all polygons withfewer number of sides and only the angle6 vi1vivi+1 at the vertex vi changes. In 4v1vivnthe 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.46In 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.47Week 9. A.D. Alexandrovs andH. Minkowskis uniqueness theoremsTheorem (A.D. Alexandrovs 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 =12(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 P1/2by 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.48Proof: 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 Q1and 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 Q1which is shorter than the parallel side of Q2.Theorem (H. Minkowskis uniqueness theorem):If each (d1)-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 (d1)-dimensionalvolumes and parallel outer outer normals, andconversely, then the two polytopes are equal andparallel.Remarks: (1) For d = 2 Minkowskis theoremis trivial. (2) For d = 3 it is an immediate con-siquence of A.D. Alexandrovs theorem. (3) Itsays nothing about the combinatorial structureof the polytopes.49Homework 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 of50the 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 Hellys 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 +Qstands 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 can51be 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.52Week 10. H. Minkowskis existence theoremTheorem (H. Minkowskis existence theorem):If n1, . . . , nm are unit vectors in Rd, d 2, notpointed in some closed half-space, and F1, . . . , Fmare positive numbers such thatmi=1Fini = 0, ()then there exists a closed convex polytope in Rdwhose (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 be amapping from A into B s.t.(1) each connected component of B containsimages of points in A;(2) is one-to-one;(3) is continuous;(4) if points Bm (m = 1, 2, . . .) of the manifoldB 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, (A) = B, i.e., allpoints of the manifold B are images of some pointsof the manifold A.53Proof: The set (A) is an open and closed sub-set of B intersecting each connected componentof B. Q.E.D.Outline of the proof of Minkowskis 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. associates with every class of equivalent poly-topes the set of areas of faces: : A B.Lemma 10.1 implies (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.54Proof: 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). Taking into accountf < Fi, we get hi < F3/2/(2f). Q.E.D.Minkowskis proof of his existence theorem:Fix a set F 01 , . . . , F0m satisfying (). Among allpolytopes with prescribed normals satisfying thecondition ihiF0i = 1find a polyhedron of greatest volume.By Lemma 10.3, the volume of the polytopeis a differentiable function of h1, . . . , hm. By theLagrange multiple rule, at a maximum point wemust havehi(V (h1, . . . , hm) + iF 0i hi) = 0,i = 1, . . . , m. Taking into account that V/hi =F 1i , we find that there exists a polytope P1 s.t.F 1i = F0i , where = 1/. The polytopeP 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.55Def:. Let P and P be convex polytipes in Rdwith non-empty interiors. Let N = {n1, . . . , nm}be the set of outward unit vectors to (d 1)-dimensional faces of P and N = {n1, . . . , nm}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 eachn 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 (d1)-dimensionalfaces and F#(n) is the (d1)-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).56P# is called the Blaschke sum of the polytopes Pand 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 Afor which the necessary extremum conditions co-incide with B0. Then the problem is reduced tochecking that f(A) actually attains a minimum(or maximum).57Week 11. A.D. Alexandrovs existence theoremDef. (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 2.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.58Homework 1 (5 points): Give an example ofshortest arcs on a nonconvex polyhedron in R3which 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. Alexandrovs 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) 2. 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 R3f06. Hence A is a manifoldand dimA = dimB.59Let : A B maps a polyhedron to its devel-opment. is one-to-one according to Cauchys unique-ness theorem. 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 that(A) = B. Q.E.D.Remark: There is no theorem for many-dimensionalpolytopes which is similar to A.D. Alexandrovsexistence 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-drovs existence theorem on page 59. Calculatethe dimensions of A and B. Should we expect a4D-version of A.D. Alexandrovs existence the-orem?60There other important theorems that can be provedin the same way as Cauchy uniqueness theoremand A.D. Alexandrov existence theorem:Theorem (Andreevs 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 (Andreevs 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 3 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 ;(2) if three faces are pairwise adjacent, but notconcurrent, then the sum of the dihedral anglesbetween them is less than ;(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 /2.61Week 12. Flexible polyhedraDef. 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 (d2)-dimensional face Fof P , let VF be the (d2)-dimensional volume ofF and F be the angle between inward unit nor-mal vectors nF and nF to two (d1)-dimensionalfaces of P which are adjacent to F . The quantityFVF ( F )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.62Theorem (R. Alexander; 1985): Total mean cur-vature of every flexible polyhedron in Rd (d 3)is constant during a flex.Proof: First, prove the formuladFdt= (dnFdt,mF ) + (dnFdt,mF ),where mF and mF are unit vectors in the hy-perplanes containing the two (d1)-dimensionalfaces adjacent to F that are perpendicular to F ;(, ) stands for the scalar product in Rd.Note that the unit vectors nF , nF , mF and mF lyein the 2-plane which is the orthogonal comple-ment to F . Introduce a coordinate system in .If and are the angles between mF (and mFrespectively) and the first axe of that coordinatesystem, thenF = ,mF = (cos , sin ),mF = (cos , sin ),nF = ( sin , cos ), dnFdt= (cos , sin )ddt,nF = (sin , cos ),dnFdt= (cos , sin )ddt,(dnFdt,mF ) + (dnFdt,mF ) = ddt+ddt= dFdt.63Now rearrange terms in the expressionddtFVF ( F ) = FVFdFdt=FVF (dnFdt,mF ) +FVF (dnFdt,mF )=12GFGVF (dnFdt,mF )+12GFGVF (dnFdt,mF )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 (d1)-dimensional faces G. Then proceed as follows=G(dnFdt,FGVFmF ) +G(dnFdt,FGVFmF )and use the fact thatFGVFmF = 0. ()Hence, the derivative of the total curvature equalszero. Q.E.D.Note: () shows that Alexanders 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 the64apex. 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 istaken over all (d 1)-dimensional faces of P .Note: The oriented volume equals the usualvolume 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.65Remark: The above polynomial S, called Sabitovpolynomial, for tetrahedra is given by the follow-ing formula known as the CayleyMenger deter-minant:V 2 =12d(d!)2det0 1 1 1 11 0 l201 l202 l20d1 l210 0 l212 l21d1 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 geometryV =1d!detp(1)0 p(2)0 p(d)0 1p(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-66pose of the determinant obtained from it by in-terchanging the last two columns we haveV 2 =1(d!)2det(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) =12[(pj, pj) + (pk, pk) l2jk]in the last determinant, subtract from the jthrow the product of the last row by (pj1, pj1)/2and get () after reductions.Example: ConnellyWalz 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 smallpolyhedra can be rearranged in such a way thatprovide a partition of P .67Theorem: 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 i(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() = 0 consider theDehn invariant defined by the formulaDf(P (t)) =i`if(i(t)).Hilberts 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.68Note: 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 stationaryunder the action of v;(3) v is not generated by a movement of P asa 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 Cauchys arguments. Q.E.D.69Theorem (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 algebraicvarietyrankRM = 6which 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 i at a vertexvi of a polytope in R3 is the difference between2 and the sum of face angles adjacent to vi. Forevery convex polytope in R3, prove the DescartesTheorem: i = 4where the sum is taken over all vertices.702. (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 Pand 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 hi 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 ) =12i`ihi =12i`ihiandV (P ) =12i`ihi.4. (5 points) Use the results of Problem 3 toprove that the mixed area of convex polygons Pand P with pairwise parallel and similarly di-rected sides where P is circumscribed about a2I.e., the mixed volume for planar convex sets.71circle 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 BrunnMinkowski 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 ConnellyWalz 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

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