1. How to Get Where you Want to Go
ABSTRACT
How to Get Where you Want to Go
(A Primer on Topological Robotics)
Petar Pavešić, University of Ljubljana
Following the seminal work of Micheal Farber the last years have witnessed several interesting applications of methods from homotopy theory to problems in robotics. In this introductory talk we will discuss some basic concepts of topological robotics such as configuration spaces, navigation plans and topological complexity.
Provo 1

2. HOW TO GET WHERE YOU WANT TO GO
CONFIGURATION SPACES
1. CONFIGURATION SPACES
Topological robotics is one of the several new areas of application of algebraic topology methods to practical problems. Other examples include persistent homology, statistical topology, directed topology, concurrency, ...
Algebraic topology enters robotics through the notion of configuration space.
The configuration space of a mechanical device is the space of all its possible states.
These states are usually described by finitely many real numbers, so the configuration space may be viewed as a subspace of some n. 2

3. EXAMPLES mechanical device configuration space
HOW TO GET WHERE YOU WANT TO GO
CONFIGURATION SPACES
EXAMPLES
mechanical device
configuration space 3

4. How do we get configuration spaces other than tori?
HOW TO GET WHERE YOU WANT TO GO
CONFIGURATION SPACES
S1×S1×S1×S1×S1×S1
How do we get configuration spaces other than tori?
By restricting the movement of the components.
OBSTACLES 2π 4

5. The configuration space of the 3-bar linkage is S1.
HOW TO GET WHERE YOU WANT TO GO
CONFIGURATION SPACES x
LINKAGES
For each admissible angle of the left rod there are two possible admissible angles of the right rod...
...except at the two extreme admissible values, where there is only one admissible angle on the right.
The configuration space of the 3-bar linkage is S1. 5

6. The configuration space of the 4-bar linkage is S2.
HOW TO GET WHERE YOU WANT TO GO
CONFIGURATION SPACES x
The configuration space of the 4-bar linkage is S2. x x
The configuration space of the 3-arm ‘spider’ is S2.
The configuration space of the 4-arm ‘spider’ is the torus. 6

7. Theorem (M. Kapovich, J.L.Millson, 2002)
HOW TO GET WHERE YOU WANT TO GO
CONFIGURATION SPACES
By constructing more complicated linkages we get even more complicated configurations spaces. How general can they be?
Theorem (M. Kapovich, J.L.Millson, 2002)
For every smooth compact manifold M there is a linkage whose configuration space is homeomorphic to a disjoint union of a finite number of copies of M. 7

8. 2. TOPOLOGICAL COMPLEXITY
HOW TO GET WHERE YOU WANT TO GO
TOPOLOGICAL COMPLEXITY
2. TOPOLOGICAL COMPLEXITY
Let X be the configuration space of some device.
Continuous motion of the device is represented by a path α: IX.
We will always assume that X is path-connected. x y
A navigation plan for X is a rule that takes as input a pair of points x,y in X, and returns as output a path α in X starting at x and ending at y. XI
X×X p
Formally speaking, a navigation plan is a section of the evaluation fibration p: XIX×X, p(α)=(α(0),α(1)). 8

9. We are interested in continuous navigation plans s: X×X  XI .
HOW TO GET WHERE YOU WANT TO GO
TOPOLOGICAL COMPLEXITY
We are interested in continuous navigation plans s: X×X  XI .
However, a continuous navigation plan for X exists if and only if X is contractible.
In fact, given a continuous s: X×X  XI, the map x s(x, x0) determines a contraction of X to x0. x0
Conversely, given a contraction of X to x a canonical path from x to y is obtained by joining the given path from x to x0 to the inverse of the path from y to x0. x0 x y 9

10. HOW TO GET WHERE YOU WANT TO GO
TOPOLOGICAL COMPLEXITY
On a non-contractible space it is not possible to find a navigation plan that depends continuously on the input data. XI
X×X p F sF
For non-contractible spaces we are lead to consider partial navigation plans, i.e. continuous sections
M. Farber (2003)
The topological complexity of X is the minimal number of continuous partial navigation plans needed to describe all possible navigation plans for X.
More precisely:
TC(X) is the minimal n such that there is a decomposition of X×X as a disjoint union X×X = F1 ... Fn of ENRs that admit continous sections si: Fi  XI of the fibration p. 10

11. Navigation plans on the circle S1:
HOW TO GET WHERE YOU WANT TO GO
TOPOLOGICAL COMPLEXITY x
Navigation plans on the circle S1:
Divide S1× S1 into:
U={(x,y) | x+ y  0}: non-antipodal points
sU (x,y):= shortest path on S1 from x to y
V={(x,y) | x+ y =0}: antipodal points
sV (x,y):= path on S1 from x to y in the positive sense
Hence TC(S1)=2 11

12. Definition of complexity reminds the Schwarz genus of the fibration p.
HOW TO GET WHERE YOU WANT TO GO
TOPOLOGICAL COMPLEXITY
OBSERVATIONS:
Definition of complexity reminds the Schwarz genus of the fibration p.
Schwarz genus or sectional category of a fibration p: EB is the minimal n such that there exists an open cover U1,...,Un, for B admitting partial continuous sections si:Ui  E for the projection p.
Theorem (M. Farber, 2003)
For nice spaces (manifolds, polyhedra) the topological complexity of X coincides with the Schwarz genus of the fibration p: XIX×X.
TC(X) is a homotopy invariant of X.
TC(X)=1 if and only if X is contractible. 12

13. RELATION WITH THE LUSTERNIK-SCHNIRELMANN CATEGORY
HOW TO GET WHERE YOU WANT TO GO
TOPOLOGICAL COMPLEXITY
RELATION WITH THE LUSTERNIK-SCHNIRELMANN CATEGORY
Recall: cat(X), the Lusternik-Schnirelmann category of X is the minimal n such that there exists an open cover U1,...,Un of X with sets that are contractible in X (i.e. Ui X are null-homotopic).
e.g. cat(sphere)=2, cat(torus)=3, cat(n-dim projective space)=n+1
If U X×X is null-homotopic, then p: XIX×X admits a section over U.
cat(X)  TC(X)  cat(X×X),
cat(X)  TC(X)  2 cat(X)1 13

14. U={(x,y) | x+ y  0} (non-antipodals),
HOW TO GET WHERE YOU WANT TO GO
TOPOLOGICAL COMPLEXITY x
cat(S2)=2  2  TC(S2)  3.
2 or 3?
Divide S2× S2 into:
U={(x,y) | x+ y  0} (non-antipodals),
sU (x,y):= shortest path on S2 from x to y N
V={(x,y) | x+ y =0} (antipodals),
sV (x,y):= path on the great circle on S2
from x through north pole N to y
Doesn’t work for the pair (N,-N), so we need
W={(N,-N)}, sW (N,-N):= any path on S2 connecting N to –N
Can we do better? 14

15. we see that Ker p*= Ker *.
HOW TO GET WHERE YOU WANT TO GO
TOPOLOGICAL COMPLEXITY
Enters cohomology: XI
X×X p U s i
H*( XI )
H*( X×X ) p*
H*(U) s* i*
H*( X×X,U ) j*
Ker p* Ker i*
= Im j*
If X×X has a cover U1,...,Un admitting local sections, then every product of n elements of Ker p* is in the image of an element in H*( X×X,U1...  Un )=0, therefore Ker p* is a nilpotent ideal of H*( X×X ) of order n. XI
X×X p X  
From the diagram
we see that Ker p*= Ker *.
nil(Ker *)  TC(X) 15

16. For x  H2(S2) we have x×11×x  Ker *  H*(S2× S2).
HOW TO GET WHERE YOU WANT TO GO
TOPOLOGICAL COMPLEXITY x
For x  H2(S2) we have x×11×x  Ker *  H*(S2× S2).
(x×11×x ) 2= 2 x×x  0 implies nil(Ker *)3, therefore TC(S2)=3. x
TC(S1)=2, TC(S2)=3, TC(S3)=?
U: non-antipodal pairs of points on S3
sU (x,y):= shortest path on S3 from x to y
V: antipodal pairs of points on S3
sV (x,y):= path on a great circle in S3 from x to y, direction is
determined by some non-vanishing vector field
TC(S3)= 2, and similarly TC(S2n-1) =2 x
For TC(S2n) category estimate gives TC(S2n)  3, while cohomology estimate gives TC(S2n)  3.
TC(S2n) =3 16

17. Further results (using the same methods):
HOW TO GET WHERE YOU WANT TO GO
TOPOLOGICAL COMPLEXITY
Further results (using the same methods):
TC(Mg)=
3, g =0,1
5, g > 1
Mg closed orientable surface of genus g
TC(Γ)=
2, Γ has one cycle
3, otherwise
1, Γ a tree
Γ finite connected graph
TC((S2n)k) =
k+1, n odd
2k+1, n even
TC(SO(3))= TC(P3)=3 17

18. Is it possible to correct this anomalous behaviour?
HOW TO GET WHERE YOU WANT TO GO
TOPOLOGICAL COMPLEXITY
An open problem:
In general a navigation plan sends the robot around even when the starting and ending points coincide. x
e.g. the obvious navigation plan for contractible spaces always sends the robot around! x0
Is it possible to correct this anomalous behaviour? x0
Idea: modify the given navigation plan
by truncating the paths at a level that depends on the distance between the starting and ending point. 18

19. Is TC(X) = TC(X) for every simplicial complex X?
HOW TO GET WHERE YOU WANT TO GO
TOPOLOGICAL COMPLEXITY
For general spaces:
A pointed navigation plan on U  X×X is a map s: X×X XI such that
a) s(x,y) is a path from x to y for all (x,y)U
b) s(x,x) is the constant path for all xX.
The pointed topological complexity TC( X ) is the minimal number of pointed navigation plans needed to connect all pairs of points in X .
Iwase – Sakai (2008) proved that on polyhedra TC( X ) = TC( X ).
In February 2012 they retracted the claim, proved that TC(X)  TC(X)  TC(X)+1,
and that the equality holds under some additional assumptions.
In August 2012 Dranishnikov proved TC(X) = TC(X) if TC(X)>dim(X)/(conn(X)+1).
Is TC(X) = TC(X) for every simplicial complex X? 19

20. THANK YOU! 20