Balancing location of points on a boundary of region Takeshi Tokuyama Tohoku University Joint work with many collaborators

Prologue: balancing in real life

Center of mass and barycenter

A very special case: Antipodal pair For any bounded region Q with boundary C and a target point p in Q, there are two points (antipodal pair) on C such that their midpoint becomes p Balancing location of two points with a same weight. How to prove it??? (easy if shown, but difficult to find by yourself) Famous related theorem: Borsuk-Ulam theorem: Any continuous function F from d-dimensional sphere to d-dimensional real space has a point x such that F(x)= F(-x) On the earth, there are opposite points with the same temperature and the same height. Many applications (Monograph of Matousek)

Another special case: Motion of robot arm

Minkowski sum ⊕ ⊕ ⊕

⊕⊕ ⊕⊕ ⊕ ⊕

Balancing location on a boundary of region

Ｋｅｙ ｌｅｍｍａ If we have a heavier weight on the boundary and the other in the interior of Q, then we can move both to the boundary keeping the center of mass

Proof

Algorithm for non-monopolistic weight set (k=3) Take any line L through the target point p Place the heaviest weight w(1) at the nearest intersection q(1) of L and the boundary The other two points w(2) and w(3) are located at the same interior point q(2,3) such that they are balancing (i.e. center of mass is at p) This is possible since weights are not monopolistic Consider the heaviest point and the second heaviest point, and move both to the boundary (using the lemma) keeping the balance Consider the second heaviest point and the third heaviest point, and move both to the boundary All three points are on the boundary !

Locate three weights w(1), w(2), and w(3) on the boundary such that their weighted barycenter is at p p

Take any line L through the target point p Place the heaviest weight w(1) at the nearest intersection q(1) of L and the boundary The other two points w(2) and w(3) are located at the same interior point q(2,3) such that they are balancing Consider the heaviest point and the second heaviest point, and move both to the boundary keeping the balance Consider the second heaviest point and the third heaviest point, and move both to the boundary All three points are on the boundary ! p L q(1)

Take any line L through the target point p Place the heaviest weight w(1) at the nearest intersection q(1) of L and the boundary The other two points w(2) and w(3) are located at the same interior point q(2,3) such that they are balancing Consider the heaviest point and the second heaviest point, and move both to the boundary keeping the balance Consider the second heaviest point and the third heaviest point, and move both to the boundary All three points are on the boundary ! p q(1) q(2,3)

Take any line L through the target point p Place the heaviest weight w(1) at the nearest intersection q(1) of L and the boundary The other two points w(2) and w(3) are located at the same interior point q(2,3) such that they are balancing Consider the heaviest point and the second heaviest point, and move both to the boundary keeping the balance Consider the second heaviest point and the third heaviest point, and move both to the boundary All three points are on the boundary !

Take any line L through the target point p Place the heaviest weight w(1) at the nearest intersection q(1) of L and the boundary The other two points w(2) and w(3) are located at the same interior point q(2,3) such that they are balancing Consider the heaviest point and the second heaviest point, and move both to the boundary keeping the balance (Use our Key Lemma) Consider the second heaviest point and the third heaviest point, and move both to the boundary All three points are on the boundary !

Take any line L through the target point p Place the heaviest weight w(1) at the nearest intersection q(1) of L and the boundary The other two points w(2) and w(3) are located at the same interior point q(2,3) such that they are balancing Consider the heaviest point and the second heaviest point, and move both to the boundary keeping the balance Consider the second heaviest point and the third heaviest point, and move both to the boundary All three points are on the boundary !

Take any line L through the target point p Place the heaviest weight w(1) at the nearest intersection q(1) of L and the boundary The other two points w(2) and w(3) are located at the same interior point q(2,3) such that they are balancing Consider the heaviest point and the second heaviest point, and move both to the boundary keeping the balance Consider the second heaviest point and the third heaviest point, and move both to the boundary All three points are on the boundary ! Exercise: Generalize it to general k weights case

What happens in three dimensions?

Impossible in 2D in general Exercise: Give a counter example In 3D, possible! Equilateral triangle is “special”. No analogue for other triangle shape than the equilateral triangle Exercise: Give a counter example

famous problem Toeplitz’ Square Peg Problem (1911) Given a closed curve in the plane, is it always possible to find four points on the curve forming vertices of a square? Yes, if the curve is smooth!, Unknown for a general curve Exercise: Find a peg position of the following shape

famous problem Toeplitz’ Square Peg Problem (1911) Given a closed curve in the plane, is it always possible to find four points on the curve forming vertices of a square? Yes, if the curve is smooth!, Unknown for a general curve

Existence of tripodal location Consider the nearest and farthest points n and f from p. Consider a path A= { a(t): 0 < t <1} from n to f,such that n = a(0) and t=a(1). We fix “base hyperplane H(t)” through a(t) and p which is continuous in t. (This is called “vector field” in mathematics) For each t, we define base triangle T(t, 0) = ( a(t), b(t ）, c(t)): Equilateral triangle on H(t) locating a vertex a(t) and center at p

Existence of tripodal location Consider the nearest and farthest points n and f from p. Consider a path A= { a(t): 0 < t <1} from n to f,such that n = a(0) and f=a(1). We fix “base hyperplane H(t)” through a(t) and p which is continuous in t. (This is called “vector field” in mathematics) For each t, we define base triangle T(t, 0) = ( a(t), b(t ）, c(t)): Equilateral triangle on H(t) locating a vertex a(t) and center at p n p f a(t) p n f p n f p

Existence of tripodal location Consider the nearest and farthest points n and f from p. Consider a path A= { a(t): 0 < t <1} from n to f,such that n = a(0) and f=a(1). We fix “base hyperplane H(t)” through a(t) and p which is continuous in t. (This is called “vector field” in mathematics) For each t, we define base triangle T(t, 0) = ( a(t), b(t ）, c(t)): Equilateral triangle on H(t) locating a vertex a(t) and center at p

Existence of tripodal location Consider the nearest and farthest points n and f from p. Consider a path A= { a(t): 0 < t <1} from n to f on the boundary,such that n = a(0) and t=a(1). We fix “base hyperplane H(t)” through a(t), which is continuous in t. For each t, we define base triangle T(t, 0) = ( a(t), b(t ）, c(t)): Equilateral triangle on H(t) locating a vertex a(t) and center at p T(t, θ ）： Triangle obtained by rotating T(t, 0) by θ about the line a(t)p Sign(t, θ ） = (+,+) if both b(t) and c(t) are outside of Q = (-, +) if b(t) is inside Q and c(t)is outside = (0,+ ) if b(t) is on the boundary and c(t) is outside ETC We should show that there exists t and θ such that Sign (t, θ) = (0,0)

Existence of tripodal location Consider the nearest and farthest points n and f from p. Consider a path G= { a(t): 0 < t <1} from n to f,such that n = a(0) and f=a(1). We fix “base hyperplane H(t)” through a(t) and p, which is continuous in t. For each t, we define base triangle T(t, 0) = ( a(t), b(t ）, c(t)): Equilateral triangle on H(t) locating a vertex a(t) and center at p T(t, θ ）： Triangle obtained by rotating T(t, 0) by θ about the line a(t)p Sign(t, θ ） = (+,+) if both b(t) and c(t) are outside of Q = (-, +) if b(t) is inside Q and c(t)is outside = (0,+ ) if b(t) is on the boundary and c(t) is outside ETC Obseration: Sign(0, θ ） = (+,+) (or (+,0) or (0,+) or (0,0) ) Sign (1, θ ） ＝ (-, -) (or (-,0), or (0,-), or (0,0))

Transition of signature (+,+) cannot change to (-,-) directly if t (or θ ） continuously changes (+,+)  (+,0)  (+,-)  (0,-)  (-,-) (+,+)  (00)  (-,-) Form a walk a graph G from (+,+) to (-,-) if we fix θ and move t from 0 to 1 A walk on a circle C if (0,0) does not appear.

Transition of signature Form a walk a graph G from (++) to (--) if we fix θ and move t from 0 to 1 A walk on a circle C if (0,0) does not appear. Key observation: If we change θ, the parity of the number of the red edges on the walk is unchanged.

Transition of signature Key observation: If we change θ, the parity of the number of the red edges on the walk is unchanged. Each walk use red and green odd times in total Consider T(t, θ ） and T(t, -θ ）, then they have opposite signature We get contradiction  (0,0) must exists.

OPEN PROBLEM Is it true that we have balancing location of four points such that they form a equilateral tetrahedron ???? How about in d-dimensions??

Can we find balancing location on edges? In two dimensional space, boundary of a polygon consists of edges, and we can find antipodal pair on edges. In higher dimensional case, boundary and edge-skeleton (union of edges) are different. Conjecture: Given a convex polytope P in d-dimensional space and a target point p in P, can we locate d points on the edge-skeleton such that their center of mass is p? Conjecture is true if d is a power of 2 Recent result by Michael Dobbins: true if d is 2 i 3 j Beautiful proof using characteristic class of vector bundle Need mathematical tools such as homology theory

Balancing location on edges in 4- dim Set p = o (origin), and consider Q = P ∩ (-P) Q has a vertex v in S 2 (P) ∩ S 2 (-P) (S k (P) is the k-skeleton of P) Caution: S 1 (P) ∩ S 3 (-P) may be empty. This means P has antipodal pair each point on 2-dim faces Each point 2-dim face has antipodal pair on edges Center of mass of these four points is o

Future direction Major conjectures remain unsolved Existence of d-dimensional version of the tripodal location Balancing location on 1-skeleton for a nonconvex 3-dim polytope Relation to Borsuk-Ulam theorem Modern interpretation of Borsuk-Ulam theorem Euler class of vector bundle on d-dim sphere with Z 2 action is nontrivial Extension to fiber bundle of topological space with group action We need collaboration with researchers on algebraic topology More general location problems (higher motion etc)

Epilogue Importance of International Collaboration Collaborators: Luis Barba (ULB, Bellguim and Carleton, Canada), Jean Lou de Carufel (Carleton), Otfried Cheong(KAIST, Korea), Michael Dobbins(POSTECH, Korea), Rudolf Fleischer (Fukdan, China and Gutech, Oman), Akitoshi Kawamura(U. Tokyo, Japan), Matias Korman (Katalonia Polytech, Spain), Yoshio Okamoto (UEC, Japan), Janos Pach (EPFL Swiss, and Reny Inst, Hungary), Yan Tang (Fudan), Sander Verdonschot(Carleton), Tianhao Wang (Fudan) 12 collaborators from 10 countries I thank to organizers of international workshops, which were essential to accomplish this work.

Huge database of unfolding: How to handle? u.ac.jp/horiyama/research/unfoldin g/catalog.new/index.ja.html u.ac.jp/horiyama/research/unfoldin g/catalog.new/index.ja.html Concluding remark: We need to be clever to handle geometric computation with help of power of mathematics!