Extremal Configurations of Polygonal Linkages

Transcription

1 Extremal Configurations of Polygonal Linkages Dirk Siersma Department of Mathematics University of Utrecht Singularity Conference Bruce 60, Wall 75 Liverpool, June 18-22, 2012

2 Outline Introduction Planar case Properties of moduli spaces Signed Area Morse theory Three Dimensional case Signed volume Topological Morse Function Joint work with G. Khimshiashvili, G. Panina, A. Zhukova

3 Introduction We consider in the plane R 2 and in space R 3 polygonal linkages and (robot) arms. constructed from edges in R d of fixed length L = (l 1,, l n ) which are connected as an open chain or as a polygon. At the vertices we allow movements. We will study several functions on the (moduli) spaces of out objects, e.g: Length of diagonals Area or Volume of a region generated by our objects. Especially we look for geometric criteria for critical points and their Morse types.

4 Moduli spaces We use a set of vectors p 1,, p n, where each vector describes the oriented edge between one vertex and its successor. We get a set of points in S d 1 S d 1. In the case of a linkage we add the relation p p n = 0 We work moduli diagonal SO d -action. The resulting moduli spaces we denote by: A(L) in the arm case M(L) in the linkage case

5 Smoothness; planar case Fix the starting point and direction of the first edge. Proposition A(L) = (S 1 ) n 1, smooth! M(L) is generically a smooth compact n 3-dimensional manifold.

6 For the proof of the second item we consider the function: h : A(L) R defined by the squared length of the connecting vector: h = (p p n ) 2. Statement: The critical points of h are the zero set of h all direction vectors p i are dependent. The Morse index of a linkage L at the critical point is one less than the number of Forward edges. Conclusion: For any non-zero value of h: Fix the connecting edge with the corresponding length and we get the moduli space of a closed linkage. This space is smooth exactly when the edges are not collinear.

7 Linkage examples 4-gons Pentagons Singular examples

8 Signed Area definition For a polygon with the ordered vertices (x 1, y 1 ),, (x n, y n ) we define its signed area by: A = x 1 y 1 x 2 y 2 + x 2 x 3 y 2 y x n 1 x n y n 1 y n + x n x 1 y n y 1 We consider A as function on the moduli spaces: Proposition A : M(L) R is critical in P P is a cyclic polygon,. A : A(L) R is critical in R R is a diacyclic arm..

9 Permutation of edges Lemma If two linkages differ a permutation of the length vector then their critical point theories coincide. (locally a reflection in a miror)

10 Notations for cyclic configurations 2α i = P i OP i+1. ε i = +1if O is left of P i P i+1 ; ε i = 1if O is right of P i P i+1. e(p) is the number of ε i = 1 δ(p) = ε i tan α i d(p) = sign δ(p) ω P winding number of P Hess P (A) Hessian matrix of A at P. H P = ±1 sign of determinant of Hess P (A). µ P = µ P (A) Morse index of A O center of circle.

11 Morse indices Theorem For a generic linkage or arm A is a Morse function; Linkage: m(p) = e(p) 1 2ω(P) if d(p) > 0 m(p) = e(p) 2 2ω(P) otherwise Arms: m(r) = e(r) + 1 2ω(R) if d(r) > 0 m(r) = e(r) 2ω(R) otherwise. NB. Morse index not only combinatoric. As start of a proof one can show: Lemma If e(p) = 0 then m(p) = 2ω(P) 2 If e(p) = n then m(p) = n 1 2ω(P)

12 Morse indices Theorem For a generic linkage or arm A is a Morse function; Linkage: m(p) = e(p) 1 2ω(P) if d(p) > 0 m(p) = e(p) 2 2ω(P) otherwise Arms: m(r) = e(r) + 1 2ω(R) if d(r) > 0 m(r) = e(r) 2ω(R) otherwise. NB. Morse index not only combinatoric. As start of a proof one can show: Lemma If e(p) = 0 then m(p) = 2ω(P) 2 If e(p) = n then m(p) = n 1 2ω(P)

13 Morse indices Theorem For a generic linkage or arm A is a Morse function; Linkage: m(p) = e(p) 1 2ω(P) if d(p) > 0 m(p) = e(p) 2 2ω(P) otherwise Arms: m(r) = e(r) + 1 2ω(R) if d(r) > 0 m(r) = e(r) 2ω(R) otherwise. NB. Morse index not only combinatoric. As start of a proof one can show: Lemma If e(p) = 0 then m(p) = 2ω(P) 2 If e(p) = n then m(p) = n 1 2ω(P)

14 about δ Linkage in circular situation: formula relating length vector and radius: F k (r) = ε i arcsin l i 2r πk = 0 Its derivative is: 1 2r εi tan α i = 1 2r δ(p) Genericity of F k (r) = 0 is equivalent to δ(p) 0.. Cerf s principle: Index constant if no collision of Morse points in a family. Generic colision: Index changes with ±1. Collission means two cyclic configurations come together, so F k (r) = 0 must have a dubble root in r.

15 about δ Linkage in circular situation: formula relating length vector and radius: F k (r) = ε i arcsin l i 2r πk = 0 Its derivative is: 1 2r εi tan α i = 1 2r δ(p) Genericity of F k (r) = 0 is equivalent to δ(p) 0.. Cerf s principle: Index constant if no collision of Morse points in a family. Generic colision: Index changes with ±1. Collission means two cyclic configurations come together, so F k (r) = 0 must have a dubble root in r.

16 About the proof Hessian matrix can be brought into tri-diagonal form. Not clear how to show index formula from that. Stategy: Fix radius and move (l 1,, l n ) to a convex situation. Lemma H(P) = d(p)( 1) e(p)+1 Proof. Take a well choosen transversal path to the generic case. The formula is true in de convex case. Show next that H(P) 1 d(p)( 1) e(p)+1 does not change sign. There are 3 cases to consider: passage through δ = 0 : Cerf principle H(P) changes sign passage through circle center: e and d change sign flip H and e change sign (explicit computation) So we know the Morse index mod 2.

17 About the proof Hessian matrix can be brought into tri-diagonal form. Not clear how to show index formula from that. Stategy: Fix radius and move (l 1,, l n ) to a convex situation. Lemma H(P) = d(p)( 1) e(p)+1 Proof. Take a well choosen transversal path to the generic case. The formula is true in de convex case. Show next that H(P) 1 d(p)( 1) e(p)+1 does not change sign. There are 3 cases to consider: passage through δ = 0 : Cerf principle H(P) changes sign passage through circle center: e and d change sign flip H and e change sign (explicit computation) So we know the Morse index mod 2.

18 About the proof Hessian matrix can be brought into tri-diagonal form. Not clear how to show index formula from that. Stategy: Fix radius and move (l 1,, l n ) to a convex situation. Lemma H(P) = d(p)( 1) e(p)+1 Proof. Take a well choosen transversal path to the generic case. The formula is true in de convex case. Show next that H(P) 1 d(p)( 1) e(p)+1 does not change sign. There are 3 cases to consider: passage through δ = 0 : Cerf principle H(P) changes sign passage through circle center: e and d change sign flip H and e change sign (explicit computation) So we know the Morse index mod 2.

19 About the proof Hessian matrix can be brought into tri-diagonal form. Not clear how to show index formula from that. Stategy: Fix radius and move (l 1,, l n ) to a convex situation. Lemma H(P) = d(p)( 1) e(p)+1 Proof. Take a well choosen transversal path to the generic case. The formula is true in de convex case. Show next that H(P) 1 d(p)( 1) e(p)+1 does not change sign. There are 3 cases to consider: passage through δ = 0 : Cerf principle H(P) changes sign passage through circle center: e and d change sign flip H and e change sign (explicit computation) So we know the Morse index mod 2.

20 Arms in R 3 How to relate a volume to a n-arm in R 3? The following picture where all simplices contain a = b 1 illustrates the below definition.

21 A signed volume Definition Let an n-arm be given by the vectors b 1,, b n. The vertices are O, B 1,, B n. We fix b 1 = a (as before). We denote c k = k i=1 b i (the endpoint of this vector is B k ). The signed volume function is defined as which can be rewritten as: n 1 V = [b 1, c k, c k+1 ], k=1 V = [b 1, b 2, b 3 ]+[b 1, b 2 +b 3, b 4 ]+[b 1, b 2 +b 3 +b 4, b 5 ]+ [b 1, b 2 + b n N.B. Note that this signed volume is essentially the signed area of the projection onto the plane orthogonal to b 1. N.B. V is defined on the parameter space (S 2 ) n 1

22 3 arm in R 3 Proposition The signed area V : S 2 S 2 R has the following critical points: Tri-orthogonal arms (maximum, resp minimum). These are Bott-Morse critical points with transversal index 3 and critical value ± a b c. Isolated points, corresponding to the aligned configurations. Here V has Morse index 2 and the critical value 0. NB. S 2 S 2 is not the moduli space of 3 arms.

23 3 arm in R 3 Proposition The signed area V : S 2 S 2 R has the following critical points: Tri-orthogonal arms (maximum, resp minimum). These are Bott-Morse critical points with transversal index 3 and critical value ± a b c. Isolated points, corresponding to the aligned configurations. Here V has Morse index 2 and the critical value 0. NB. S 2 S 2 is not the moduli space of 3 arms.

24 permutation of arms Lemma Let two arms differ on a permutation of the arms 2,..., n. Then there exits a bijection (by mirror-symmetry ) between their "moduli spaces" which preserves the signed volume function. Consequently this bijection preserves critical points and their local (Morse) types.

25 Theorem The critical points of V up to "mirror-symmetry" are as follows There exits a division of the n-arm into a subarm b 1, a subarm b 2,..., b k and a subarm b k+1,..., b n such that: b 1 is orthogonal to each of b 2,..., b k (which lie in a plane Rb1 ). The vertices of the arm b 2,..., b k lie on a circle, satisfying the closing condition if n k = odd, the diameter condition if n k = even. The arm b k+1,..., b n is a zigzag, which projects orthogonally to the diameter of the circle. special cases

26 Theorem The critical points of V up to "mirror-symmetry" are as follows There exits a division of the n-arm into a subarm b 1, a subarm b 2,..., b k and a subarm b k+1,..., b n such that: b 1 is orthogonal to each of b 2,..., b k (which lie in a plane Rb1 ). The vertices of the arm b 2,..., b k lie on a circle, satisfying the closing condition if n k = odd, the diameter condition if n k = even. The arm b k+1,..., b n is a zigzag, which projects orthogonally to the diameter of the circle. special cases

27 Projection on planes Take a generic plane and consider the signed area PA of the projected arm on this plane. Theorem The critical points of PA up to "mirror-symmetry" are as follows: There exits a division into two subarms b 1,..., b k and b k+1,..., b n, such that: The vertices of the arm b 1, b 2,..., b k lie on a circle in the projection plane, satisfying the closing condition if n k = odd, the diameter condition if n k = even. The arm b k+1,..., b n is a zigzag, which projects orthogonally to the diameter of the circle. Remark: Similar theorem exists for closed linkages.

28 Projection on planes Take a generic plane and consider the signed area PA of the projected arm on this plane. Theorem The critical points of PA up to "mirror-symmetry" are as follows: There exits a division into two subarms b 1,..., b k and b k+1,..., b n, such that: The vertices of the arm b 1, b 2,..., b k lie on a circle in the projection plane, satisfying the closing condition if n k = odd, the diameter condition if n k = even. The arm b k+1,..., b n is a zigzag, which projects orthogonally to the diameter of the circle. Remark: Similar theorem exists for closed linkages.

29 Projection on planes Take a generic plane and consider the signed area PA of the projected arm on this plane. Theorem The critical points of PA up to "mirror-symmetry" are as follows: There exits a division into two subarms b 1,..., b k and b k+1,..., b n, such that: The vertices of the arm b 1, b 2,..., b k lie on a circle in the projection plane, satisfying the closing condition if n k = odd, the diameter condition if n k = even. The arm b k+1,..., b n is a zigzag, which projects orthogonally to the diameter of the circle. Remark: Similar theorem exists for closed linkages.

30 Zigzags. Planar 5 gons. Spatial 5 gons.

31 Exact Morse function for a 3-arm Study the moduli space of n-arms in R n b using the Gram matrix: there is a directy relation with the volume. Given a set of vectors, the Gram matrix G is the matrix of all possible inner products. Let B be the matrix whose columns are the arm vectors b 1,..., b n. Gram matrix:. G = B t B det G = (V ) 2. The Gram matrix is always a positive semi definite symmetric matrix. If G is positive definite it determines the vectors up to isometry (no orientation involved). But not every positive semi definite matrix S (with det S= 0) is a Gram matrix.

32 We consider the 3 dimensional case: a 2 z y det G = z b 2 x y x c 2 = 2xyz a2 x 2 b 2 y 2 c 2 z 2 + a 2 b 2 c 2 = 0 The compact component corresponds to Gram matrices. Figure: Zero locus of the determinant of G. (SINGULAR software.)

33 Proposition The closure of the compact component of G 1 (0, a 2 b 2 c 2 ) is a topological 3-ball. Its boundary is a topological 2-sphere (differentiable outside 4 critical points). Proof. Since we have det G = V 2, the both functions have the same level curves on the domain of common definition. There is one maximum and no other critical points in the interior. Near critical points on G = 0 we have local topological triviality. The above proposition tells that the (unoriented) moduli space of 3-arm is a topological ball. By gluing two copies of the ball along the common boundary we get: Theorem The oriented moduli space of 3-arms in R 3 is a 3-sphere. V is an exact topological Morse function on this space with precisely two Morse critical points.

34 Proposition The closure of the compact component of G 1 (0, a 2 b 2 c 2 ) is a topological 3-ball. Its boundary is a topological 2-sphere (differentiable outside 4 critical points). Proof. Since we have det G = V 2, the both functions have the same level curves on the domain of common definition. There is one maximum and no other critical points in the interior. Near critical points on G = 0 we have local topological triviality. The above proposition tells that the (unoriented) moduli space of 3-arm is a topological ball. By gluing two copies of the ball along the common boundary we get: Theorem The oriented moduli space of 3-arms in R 3 is a 3-sphere. V is an exact topological Morse function on this space with precisely two Morse critical points.

35 References Gibson C., Newstead P., On the geometry of the planar 4-bar mechanism. Acta Appl. Math., 1986, 7, 23, Kapovich M., Millson J., On the moduli space of polygons in the Euclidean plane. J. Differential Geom. Volume 42, Number 1 (1995), Khimshiashvili G., Panina G., Cyclic polygons are critical points of area. Zap. Nauchn. Sem. S.-Peterburg. Otdel. Mat. Inst. Steklov. (POMI), 2008, 360, 8, Khimshiashvili G., Panina G., Siersma D., Zhukova A., Extremal Configurations of Polygonal Linkages, Oberwolfach Preprint, Khimshiashvili G., Panina G., Siersma D., Zhukova A., Critical Configurations of Planar Robot Arms,To appear in Cent. Eur. J. Math., (2012). Panina G., Zhukova A.,Morse index of a cyclic polygon,cent. Eur. J. Math., 9(2) (2011),

36 Proof for diagonal

37 Proof for linkages

38 Proof for arms

39 4gon Moduli space is a circle; in other case (one big arm) two seperate circles.

40 singular example more precise

41 singular example

42 M-example m(p) = e(p) 1 2ω(P) if d(p) > 0 m(p) = e(p) 2 2ω(P) otherwise

43 flip m(p) = e(p) 1 2ω(P) if d(p) > 0 m(p) = e(p) 2 2ω(P) otherwise

44 graph of r

45 Pentagons

46 Critical Arms Solutions in the general case

47 Critical Arms special pictures

48 Critical Arms the first edge is the projection direction

49 Zigzags

50 critical Planar 5-gons

51 critical Spatial 5-gons