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. Prerequisite: good introduction to linear alge-bra (e.g., MATH 221, 223

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<ul><li><p>Applicable Geometry</p><p>MATH 455 Lec 01</p><p>Tuesdays, Thursdays 2:55 4:10P</p><p>Malott Hall 238</p><p>Victor</p><p></p><p>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).</p><p>Lecture notes1 are available at</p><p>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.</p><p>1 Putting things in writing is safer than simply speaking to people. . . [Planning in Advance].</p><p>1</p></li><li><p>Important: Return your homework on Tuesday.</p><p>My office hours at MT 438:Monday, Friday 6:30 7:30 PM</p><p>Syllabus</p><p>Theory 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).</p><p>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).</p><p>Rigidity theory:flexible polyhedra and rigidity theory (weeks 1213; pages 6270).</p><p>2</p></li><li><p>Recommended literature:</p><p>1. Yaglom, I.M.; Boltyanskii, V.G. Convexfigures. New York: Holt, Rinehart and Winston,1961.</p><p>2. Lyusternik, L.A. Convex figures and poly-hedra. New York: Dover Publications, 1963.</p><p>3. Coxeter, H.S.M. Regular polytopes. 2nded. New York: The Macmillan Company, 1963.</p><p>4. Grunbaum, B. Convex polytopes. 2nd ed.New York: Springer, 2003.</p><p>5. Alexandrov, A.D. Convex polyhedra. Berlin:Springer, 2005.</p><p>6. Graver, J.; Servatius, B.; Servatius, H.Combinatorial rigidity. Providence: AmericanMathematical Society, 1993.</p><p>3</p></li><li><p>Week 1. General properties of convex bodies</p><p>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.</p><p>Examples:(1) the empty set ;(2) any single point;(3) any linear subspace (including Rd);(4) every (closed or open) triangle in Rd.</p><p>Properties:(1) If {An} is any family of convex sets in Rd,</p><p>then their intersection An is also convex.(2) If A is convex, ai A and i 0 for i =</p><p>1, 2, . . . , k and</p><p>k</p><p>i=1i = 1, then</p><p>k</p><p>i=1iai A.</p><p>(3) If A is convex, both its closure cl A and itsinterior intA are convex.</p><p>(4) If T : Rd Rd is an affine transformation,and A Rd is convex, than T (A) is convex.Homework 1 (5 points):</p><p>A quadrilateral on the plane is convex if andonly if its diagonals intersect inside the quadri-lateral.</p><p>4</p></li><li><p>Def.: If u Rd and u 6= 0, thenH = {x Rd|(x, u) = }</p><p>is called a hyperplane with normal vector u, andeither of the sets</p><p>{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.</p><p>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.</p><p>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.</p><p>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.</p><p>Proof: It is sufficient to prove Theorem 1.1 forthe case when A and A are closed sets. Choose</p><p>5</p></li><li><p>x A and x A such that|x x| = min</p><p>yA,yA|y y|.</p><p>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</p><p>H(A, x) = sup{(y, x)|y A}.</p><p>Properties:</p><p>(1) The supporting function is positively ho-mogeneous, i.e.,</p><p>H(A, x) = H(A, x) for all 0, x Rd.</p><p>(2) The supporting function is convex, i.e.,</p><p>H(A, x+y) H(A, x)+H(A, y) for all x, y Rd.</p><p>(3) If T : Rd Rd is a translation (i.e., T (x) =x + y for all x Rd and some constant vectory Rd), then</p><p>H(T (A), x) = H(A, x) + (x, y).</p><p>6</p></li><li><p>(4) If T : Rd Rd is a homothetic transfor-mation (i.e., T (x) = x for all x Rd and someconstant 0), then</p><p>H(T (A), x) = H(A, x).</p><p>(5)|H(A, x)|/</p><p>(x, x) is the distance from theorigin 0 to the supporting plane H of the convexset A with the outward normal vector x.</p><p>(6) If A,A are nonempty, closed convex setsin Rd such that H(A, x) = H(A, x) for everyx Rd, then A = A.</p><p>(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.</p><p>(8) Each open, convex set in Rd is the inter-section of all the open half-spaces containing it.</p><p>Def.: The convex hull convA of a subset A ofRd is the intersection of all the convex sets in Rd</p><p>which contain A.</p><p>Def.: The convex hull of a finite set is called apolytope.</p><p>Homework 2 (5 points): Prove that every poly-tope is the intersection of a finite number of closedhalf-spaces.</p><p>7</p></li><li><p>Week 2. Hellys theorem and its applications</p><p>Adv.: Bradley Forrest is theteaching assistant for MATH 455. His office hoursare Monday 122 &amp; Thursday 12 in Malott 114.</p><p>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.</p><p>Note: Lemma is obviously false for non-convexsets.</p><p>Proof: Let xi be a common point of the set</p><p>{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:</p><p>(1) One of the points x1, x2, x3, x4 belongs tothe triangle formed by the other three. [This verypoint belongs to the intersection.]</p><p>(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.</p><p>8</p></li><li><p>Theorem 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.</p><p>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.</p><p>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.</p><p>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/</p><p>3.</p><p>9</p></li><li><p>Proof is based on the homework: it suffice toshow that Theorem is true for three points.</p><p>Note: In the Jungs theorem, the radius 1/</p><p>3can be replaced by no smaller number.</p><p>Note: Hellys theorem in the plane is obviouslyfalse for infinite families of sets.</p><p>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.</p><p>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.</p><p>Proof: Introduce a system of coordinates. Let(xi, y</p><p>i) and (xi, y</p><p>i ) be the end points of the i-th</p><p>segment, where yi &lt; yi and i = 1, 2, . . . , n. Note</p><p>that a line y = kx + b intersects the i-th segmentiff</p><p>yi kxi + b and yi kxi + b.10</p></li><li><p>But 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.</p><p>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.</p><p>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.</p><p>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 = . Let</p><p>11</p></li><li><p>A = 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.</p><p>Homework 2 (5 points): Formulate and prove atheorem for R3, which is analogous to the JungsTheorem 2.2.</p><p>12</p></li><li><p>Week 3. Minkowski addition</p><p>Def.: For two (not necessarily convex) sets A,Bin Rd the set</p><p>A + B = {x + y | x A, y B}is called the Minkowski sum or, briefly, the sumof A and B.</p><p>Examples: The sum of two triangles in a planeis either a triangle, a quadrangle, a pentagon, ora hexagon.</p><p>Lemma 3.1: If A and B are both convex, thenA + B is convex.</p><p>Proof: Let x0, x1 A and y0, y1 B.Then, for 0 t 1,</p><p>t(x0 + y0) + (1 t)(x1 + y1) ==</p><p>(tx0 + (1 t)x1</p><p>)+</p><p>(ty0 + (1 t)y1</p><p>) A + B.</p><p>Q.E.D.</p><p>Lemma 3.2: If T denotes a translation, then</p><p>T (A) + B = T (A + B) = A + T (B).</p><p>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</p><p>({u + y|u A, y B}</p><p>)= T (A + B).</p><p>Q.E.D.</p><p>13</p></li><li><p>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.</p><p>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.</p><p>Lemma 3.3: If A1, . . . , Ar Rd are convex andt1, . . . , tr are real numbers, then, t1A1 + . . .+trAris convex.</p><p>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:</p><p>HA+B = HA + HB.</p><p>Proof: HA+B(x) = supyA+B</p><p>(y, x) =</p><p>= supuA,vB</p><p>(u + v, x) = supuA</p><p>(u, x) + supvB</p><p>(v, x) =</p><p>= HA(x) + HB(x) for all x Rd. Q.E.D.14</p></li><li><p>Def.: 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.</p><p>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.</p><p>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}</p><p>= {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}</p><p>= {w Rd |w = u + vfor some u A HA(x), v B HB(x)}</p><p>= FA + FB. Q.E.D.</p><p>15</p></li><li><p>Theorem 3.6: Minkowski sum of any two poly-tops is a polytope.</p><p>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.</p><p>Homework 1 (5 points): Find two 3-simplicesin R3 whose Minkowski sum has 16 vertices.</p><p>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,</p><p>V (t1A1 + + tnAn)</p><p>=n</p><p>p1,...,pn=1V (Ap1, . . . , Apn)tp1 tpn,</p><p>where the summation being carried out indepen-dently over the pi.</p><p>Proof is by induction in dimension d.</p><p>For d = 1, Ai are the intervals [xi, yi] (or pointsif xi = yi). We find</p><p>t1A1+ +tnAn = [t1x1+ +tnxn, t1y1+ +tnyn]</p><p>16</p></li><li><p>V (t1A1 + + tnAn)= (t1y1 + + tnyn) (t1x1 + + tnxn)</p><p>= t1(y1 x1) + + tn(yn xn)= t1V (A1) + + tnV (An)</p><p>which is either zero or a homogeneous polynomialof degree 1 in t1, . . . , tn.</p><p>Suppose the theorem is true for dimension d 1instead of d.</p><p>Let x Rd, HAj(x) be the supporting hyper-planes, and Fj = Fj(x) = Aj HAj(x) be thefaces, j = 1, . . . , n. Set</p><p>A = t1A1 + + tnAn ( = (t1, . . . , n))F(x) = A HA (x)</p><p>Then, according to Theorem 3.5,</p><p>FA(x) = t1F1(x) + + tnAn(x).</p><p>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 obtain</p><p>Vd(A) =1</p><p>d</p><p>m</p><p>j=1HA (xj)Vd1(F(xj))</p><p>17</p></li><li><p>=1</p><p>d</p><p>m</p><p>j=1(t1HA1(xj)+ + tnHAn(xj))Vd1(F(xj))</p><p>By inductive assumption, Vd1(F(xj)) is eitherzero or a homogeneous polynomial of degree d1in t1, . . . , tn. Q.E.D.</p><p>Def.: Arranging the coefficients on the right sideof the formula</p><p>V (t1A1 + + tnAn)</p><p>=n</p><p>p1,...,pd=1V (Ap1, . . . , Apd)tp1 tpd,</p><p>such that</p><p>V (A(p1), . . . , A(pd)) = V (Ap1, . . . , Apd)</p><p>for any permutation of p1, . . . , pd, we callV (Ap1, . . . , Apd) the (d-dimensional) mixed vol-ume of Ap1, . . . , Apd.</p><p>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).</p><p>Proof: tdV (A) = V (tA) = tdV (A, . . . , A).Q.E.D.</p><p>Example: For planar con...</p></li></ul>