1. Characterizing Generic Global Rigidity
Steven J. Gortler
(with A. Healy and D. Thurston)

2. Global Rigidity Given a graph: “G”, Given a “framework”: p, in Rd R2
For some fixed d R2

3. Global Rigidity
p is GR in Rd : there is no second framework in Rd with the same edge lengths R2
globally rigid

4. Global Rigidity
p is GR in Rd : there is no second framework in Rd with the same edge lengths R2

5. Global Rigidity
p is GR in Rd : there is no second framework in Rd with the same edge lengths R2

6. Global Rigidity
p is GR in Rd : there is no second framework in Rd with the same edge lengths
Else: p is GF in Rd R2

7. Global Rigidity (notes)
Edge crossing is allowed
Euclidean: rot, trans, reflect is not considered different R2

8. Global Rigidity (notes)
Locally flexible => globally flexible
Easier to characterize R2

9. Motivation: Distance Geometry
Input: some pairwise distances
Output: geometric framework/Eucl
Chemistry, sensor networks

10. Motivation: Distance Geometry
Input: some pairwise distances
Output: geometric framework/Eucl
GR: Well posed-ness of problem
GR: Divide and conquer

11. Motivation: Distance Geometry
Molecule problem
MDS with partial information
Rank constrained distance matrix completion

12. Generic
Given a graph G, and framework p, the GR problem is NP-Hard [Saxe ‘79]

13. Generic
Given a graph G, and framework p, the GR problem is NP-Hard [Saxe ‘79]
The reductions all involve special “coincidences” in the framework R1

14. Generic
Given a graph G, and framework p, the GR problem is NP-Hard [Saxe ‘79]
The reductions all involve special “coincidences” in the framework R1

15. Generic
Given a graph G, and framework p, the GR problem is NP-Hard [Saxe ‘79]
Problem seems simpler if we assume no coincidences R1
globally rigid

16. Generic
Given a graph G, and framework p, the GR problem is NP-Hard [Saxe ‘79]
Problem seems simpler if we assume no coincidences R1
globally rigid
globally flexible

17. Generic
Perhaps the problem is easier if we assume that the input framework is “generic”
Think: randomly perturbed R1
globally rigid
globally flexible

18. Generic
Perhaps the problem is easier if we assume that the input framework is “generic”
In 1D, a generic framework is GR iff the graph is 2-connected R1
globally rigid
globally flexible

19. Generic
Perhaps the problem is easier if we assume that the input framework is “generic”
In 1D, a generic framework is GR iff the graph is 2-connected
So GGR in R1 is a property of the graph alone R1
globally rigid
globally flexible

20. History of GGR
CC: (Connelly condition) Sufficient for all d [’89, H‘95, ‘05]
HC: (Hendrickson condition) Necessary for all d [’88, ’92]

21. History of GGR
CC: (Connelly condition) Sufficient for all d [’89, H‘95, ‘05]
HC: (Hendrickson condition) Necessary for all d [’88, ’92]
HC = CC (nec & suff) for d=2 [JJ’05]

22. History of GGR
CC: (Connelly condition) Sufficient for all d [’89, H‘95, ‘05]
HC: (Hendrickson condition) Necessary for all d [’88, ’92]
HC not sufficient for d >= [C ’91]
HC = CC (nec & suff) for d=2 [JJ’05]
K5,5

23. History of GGR
CC: (Connelly condition) Sufficient for all d [’89, H‘95, ‘05]
HC: (Hendrickson condition) Necessary for all d [’88, ’92]
HC not sufficient for d >= [C ’91]
HC = CC (nec & suff) for d=2 [JJ’05]
K5,5

24. History of GGR
CC: (Connelly condition) Sufficient for all d [’89, H‘95, ‘05]
HC: (Hendrickson condition) Necessary for all d [’88, ’92]
HC not sufficient for d >= [C ’91]
HC = CC (nec & suff) for d=2 [JJ’05]
CC is necessary for all d [this work]

25. Main result
[Connelly] If CC is satisfied by a generic framework in Rd, it is globally rigid in Rd .
[Thm] If CC is not satisfied by a generic framework in Rd then it is globally flexible in Rd

26. Main result
[Connelly] If CC is satisfied by a generic framework in Rd, it is globally rigid in Rd .
[Thm] If CC is not satisfied by a generic framework in Rd then it is globally flexible in Rd
Note: CC can be tested with an efficient randomized algorithm.

27. Main result
[Connelly] If CC is satisfied by a generic framework in Rd, it is globally rigid in Rd .
[Thm] If CC is not satisfied by a generic framework in Rd then it is globally flexible in Rd
Note: CC test gives same answer for all generic frameworks of G in Rd

28. Main result
[Connelly] If CC is satisfied by a generic framework, it is globally rigid.
[Thm] If CC is not satisfied by a generic framework in Rd then it is globally flexible in Rd
[Cor] A graph is either GR for all generic fmwks in Rd, or is not GR for any generic fmwk in Rd.
So GGR in Rd is a property of the graph alone

29. On to the condition….

30. Stress Vector satisfied by a framework
A real number wuv on each edge euv
Σv wuv [p(v) – p(u)] = 0 a b g c h d f R2 e
p(u)

31. Stress Vector satisfied by a framework
A real number wuv on each edge euv
Σv wuv [p(v) – p(u)] = 0 g c f R2 e
p(u)

32. Stress Vector satisfied by a framework
A real number wuv on each edge euv
Σv wuv [p(v) – p(u)] = 0 b
p(u) h f R2

33. Stress Vector satisfied by a framework
Equivalent to (symmetrically) writing each vertex as an affine comb of its nbrs
[1/ Σv wuv] Σv wuv p(v) = p(u) g c f R2 e
p(u)

34. Stress Vector satisfied by a framework
Equivalent to equilibrium point of quadratic spring/strut energy (no pins)
E(p)= Σu Σv wuv |p(v) – p(u)|2 R2

35. Stress vectors: easy facts
Stress vectors satisfied by a fixed p form a linear space W(p) p

36. Stress vectors: easy facts
Stress vectors satisfied by a fixed p form a linear space W(p)
Any affine transform ‘T(p)’ will satisfy all stresses in W(p)
T(p) p

37. Stress vectors: easy facts
For some graphs, there are even more fmwks than the affine transforms of p,
that still satisfy all stresses in W(p) p
not an affine tform of p

38. Connelly’s Condition
CC := The only fmwks that satisfy all of W(p) are the affine transforms of p
This will somehow describe GGR!

39. Stresses and lengths What is the relationship between stress vectors
Affine invariant
….. and lengths
Euclidean invariant

40. The mapping “L” “L” is the mapping from d-dim fmwks to Re
Describing each edge’s squared length L Re R2

41. The mapping “L” “L” is the mapping from d-dim fmwks to Re
Describing each edge’s squared length L Re R2

42. The set “M”
“M” is the image of “L”
All possible measurements L Re R2

43. The set “M” “M” is the image of “L” Re R2 All possible measurements
A semi-algebraic set
A smooth manifold a.e. L Re R2

44. Lengths and stresses: the connection
At a generic fmwk p,
span L* = the tangent space of M at L(p) L R2 Re

45. Lengths and stresses: the connection
At a generic fmwk p,
span L* = the tangent space of M at L(p)
W(p) spans the normal space of M at L(p)
[Maxwell] L R2 Re

46. Lengths and stresses: the connection
So a generic fmwk that satisfies all the same stresses as p must have the same normal space L R2 Re

47. Lengths and stresses: the connection
So a generic fmwk that satisfies all the same stresses as p must have the same normal space L R2 Re

48. Lengths and stresses: the connection
So a generic fmwk that satisfies all the same stresses as p must have the same tangent space L R2 Re

49. Lengths and stresses: the connection
So a generic fmwk that satisfies all the same stresses as p must have the same tangent space
M is a cone: λM = M
Same tangent space, not just parallel L R2 Re

50. Lengths and stresses: the connection
So a generic fmwk that satisfies all the same stresses as p must have the same tangent space
This is the key connection
We will return to this in the proof L R2 Re

51. On to the proof…

52. Degree mod two thm

53. Degree mod two thm 2

54. Degree mod two thm 2 4

55. Degree mod two thm Typical version:
Given smooth map from a compact manifold to a connected manifold of same dimension
Every generic point in the range has the same number of pre-images mod 2
The creases are singular
This number {0,1} is called the degree

56. Degree mod two thm General version:
Given smooth map from a compact manifold to a connected manifold of same dimension
Every generic point in the range has the same number of pre-images mod 2
proper

57. Our plan 2 4

58. Our plan 2 4

59. Our plan Assume (!CC) Start with the map L Define a domain
Equate Euclidean transforms in the domain
Define range
Connected smooth manifold
Will need to remove some singularities while maintaining connectivity of range and properness of map
Show the map has degree 0
Each point in image has multiple pre-images
Framework is globally flexible

60. Domain Start with stress satisfiers: A(p)
Frameworks that satisfy all of the stresses that are satisfied by p
Affine tforms of p plus maybe more

61. Domain Start with stress satisfiers: A(p)
Mod out Euclideans: A(p)/Eucl

62. Domain Start with stress satisfiers: A(p)
Mod out Euclideans: A(p)/Eucl
Result is smooth manifold + singularities

63. Domain Start with stress satisfiers: A(p)
Mod out Euclideans: A(p)/Eucl
Result is smooth manifold + singularities
Singularities: fmwks stabilized by a n.t. euclidean.

64. Domain Start with stress satisfiers: A(p)
Mod out Euclideans: A(p)/Eucl
Result is smooth manifold + singularities
Singularities: deficient affine span.

65. Lemma 1
Lemma: If (!CC)
A(p) is “big”
then the singularities of A(p)/Eucl are of co-dimension >= 2.
Proof: counting

66. Codimension and cutting
The singular co-dim will carry over to the range
Removal a co-dimension 2 set does not disconnect
Needed for degree thm
co-dim 2

67. Codimension and cutting
The singular co-dim will carry over to the range
Removal a co-dimension 2 set does not disconnect
Needed for degree thm
co-dim co-dim 1

68. The range Let B(p) := L(A(p)) Re
Achievable measurements of stress satisfiers
Some subset of M L Re

69. The range For the degree to be well defined Re
Need to include B(p) as a full dimensional subset of a connected range manifold L Re

70. The range For the degree to be well defined To show the degree is 0 Re
Need to include B(p) as a full dimensional subset of a connected range manifold
To show the degree is 0
Sufficient for range to include pts not in B(p) L Re

71. The range For the degree to be well defined To show the degree is 0 Re
Need to include B(p) as a full dimensional subset of a connected range manifold
To show the degree is 0
Sufficient for range to include pts not in B(p) L Re

72. The range For the degree to be well defined To show the degree is 0 Re
Need to include B(p) as a full dimensional subset of a connected range manifold
To show the degree is 0
Sufficient for range to include pts not in B(p) L Re

73. The range
So we need to understand the shape of B(p) L Re

74. Gauss fiber
Digression
Gauss fiber: points with same (not just parallel) tangent as chosen point … … Re

75. Gauss fiber
Digression
Gauss fiber thm: The Gauss fiber at a generic point of an irreducible algebraic variety is an affine space … … Re
1d fiber

76. Gauss fiber
Digression
Gauss fiber thm: The Gauss fiber at a generic point of an irreducible algebraic variety is a affine space
1d non-affine fiber Re

77. Gauss fiber
Digression
Gauss fiber thm: The Gauss fiber at a generic point of an irreducible algebraic variety is a affine space
1d non-affine fiber, exceptional Re

78. Gauss fiber
Digression
Gauss fiber thm: The Gauss fiber at a generic point of an irreducible algebraic variety is a affine space
0-d fiber Re

79. The range Recall “the connection” B(p) is a gauss fiber in M Re
Same stresses = same tangent in M
B(p) is a gauss fiber in M L Re

80. The range Recall “the connection” B(p) is a gauss fiber in M
Same stresses = same tangent in M
B(p) is a gauss fiber in M
M is not an irreducible algebraic variety
But it is a full dimensional semi-algebraic subset of one L Re

81. Lemma 2 Lemma: B(p) is a flat space Re
Full dimensional subset of an affine space L Re

82. Lemma 2 Lemma: B(p) is a flat space
Full dimensional subset of an affine space
So define the range to be this whole affine space L Re

83. Lemma 2 Lemma: B(p) is a flat space
Full dimensional subset of an affine space
So define the range to be this whole affine space L Re

84. Lemma 2 Lemma: B(p) is a flat space
Full dimensional subset of an affine space
M is contained in first octant, the affine space is not
The degree will be zero L Re

85. Lemma 2 Lemma: B(p) is a flat space Re
Full dimensional subset of an affine space
Note: domain and range have same dimension L Re

86. Last step
Remove the images of the singularities of A(p)/Eucl from the range and their pre-images
Range remains connected if (!CC)
Domain and range are smooth manifolds L Re

87. Last step
Remove the images of the singularities of A(p)/Eucl from the range and their pre-images
Range remains connected if (!CC)
Domain and range are smooth manifolds L Re

88. Last step
Remove the images of the singularities of A(p)/Eucl from the range and their pre-images
Can now apply degree thm 4 L 2 Re

89. Main Theorem 4 L 2 Re

90. Main Theorem 4 L 2 Re

91. Main Theorem
[Thm] If CC is not satisfied by a generic framework in Rd then it is globally flexible in Rd 4 L 2 Re

92. Review Defined a domain
A(p)/Eucl
Created some singularities
Defined same dimensional connected smooth range
Affine space containing B(p) (due to flatness)
Removed singularities
To get smooth domain manifold
Maintained range connectedness (due to high co-dim if !CC)
Now we there is a well defined degree
Need to know degree is 0 (due to flatness)

93. Deep Breath…..

94. Algorithm
Input: Graph, d

95. Algorithm Input: Graph, d Pick random framework p in Rd
Compute stress vector space W(p)
Linear algebra
Pick a random stress vector w from W(p)
Compute m: dimensionality of fmwks that satisfy w
If m = d(d+1) output “GR in Rd”
If m > d(d+1) output “GGR in Rd”

96. Algorithm With high probability, p will behave like a generic fmwk
With high probability, fmwks that satisfy w will satisfy all of W(p)
Can be done with integer linear algebra
The exceptions satisfy a “low” degree polynomial
No false positives
GGR in RP

97. One more breath…

98. Bonus result If a graph is generically globally flexible
One can continuously flex in one higher dimension back down to second framework

99. Future work More algebraic: More combinatorial:
Not just the L function on graphs
More general field
More general metric signature
More combinatorial:
Deterministic efficient algorithm

100. Future work
Given lengths, it is NP-hard to figure out the framework in Rd
Semi-definite programming will typically find solution in Rv
But sometimes it happens to give an answer in Rd
These are frameworks for which higher dimensions don’t help
Can these cases be nicely characterized?