aa r X i v : . [ m a t h . M G ] S e p GEODESIC COMPLEXITY FOR NON-GEODESIC SPACES
DONALD M. DAVIS
Abstract.
We define the notion of near geodesic between pointswhere no geodesic exists, and use this to define geodesic complexityfor non-geodesic spaces. We determine explicit near geodesics andgeodesic complexity in a variety of cases, including one in whichthe geodesic complexity exceeds the topological complexity. Introduction
In [6], Recio-Mitter defined the geodesic complexity GC( X ) of a metric space X to be the minimal number k such that X × X can be partitioned into k + 1 ENRs E i , 0 ≤ i ≤ k , such that on each E i there is a continuous map s i : E i → P X , calleda geodesic motion planning rule (GMPR) on E i , such that, for all ( x , x ) ∈ E i , s i ( x , x ) is a (minimal) geodesic from x to x . This was an analogue of Farber’snotion of topological complexity T C ( X ) ([4]), which applied to any topological space X and did not require that s i ( x , x ) be a geodesic. Clearly TC( X ) ≤ GC( X ) forany metric space X . These notions are of particular interest if X is a space ofconfigurations of one or more robots.A geodesic space is one in which for all pairs ( x , x ) of points, there is a geodesicfrom x to x . According to this definition, GC( X ) = ∞ if X is not geodesic. In[2] and [3], some non-geodesic spaces X were replaced by homotopically equivalentgeodesic spaces, whose GC was computed and interpreted as representing GC( X ).This seems reasonable since TC is a homotopy invariant. In some cases, e.g. F ( R n , F ( X,
2) denote the space of ordered pairs of distinct points of X , with theinduced metric from X × X . Here we use the ℓ metric on X × X , although othermetrics, such as ℓ , are possible. In [2], we replaced the non-geodesic space F ( R n , Date : September 28, 2020.
Key words and phrases. geodesic, configuration space, topological robotics.2000
Mathematics Subject Classification : 53C22, 55R80, 55M30, 68T40. by the homotopically equivalent geodesic space F ε ( R n , x , x )satisfying d ( x , x ) ≥ ε . We determined explicit geodesics in F ε ( R n , F ( R n − { x } , a, a ′ ) , ( b, b ′ )) in whichthe linear path from ( a, a ′ ) to ( b, b ′ ) passes through some point ( c, c ) and those wherethe linear path from a to b , or from a ′ to b ′ , passes through x . It was the complexity ofconsidering a geodesic equivalent of this space that led the author to the considerationsof this paper.For a topological space Y , P ( Y ) = Y I denotes the free path space with thecompact-open topology, and P ( Y ; y , y ) the subspace consisting of paths from y to y . Definition 1.1.
Let X be a metric space whose completion X is geodesic. The setof points ( x , x ) of X × X for which there is no geodesic from x to x is called the nogeo set of X . If x and x are in the nogeo set of X , a near geodesic from x to x is a map φ : I → P ( X ; x , x ) satisfying i. φ (0) is a geodesic in X from x to x ; ii. φ ((0 , ⊂ P ( X ; x , x ) ; iii. if s n → , then length( φ ( s n )) → length( φ (0)) . Definition 1.2. If E is contained in the nogeo set of X , a near geodesic motionplanning rule (NGMPR) on E is a continuous map Φ from E to P ( X ) I such that, forall ( x , x ) ∈ E , Φ( x , x ) is a near geodesic from x to x . The geodesic complexityGC( X ) is defined as the smallest k such that X × X can be partitioned into ENRs E , . . . , E k such that each E i has either a GMPR or NGMPR. It is also allowed that E i be the union of topologically disjoint sets, of which one has a GMPR and the othera NGMPR. Recall that two subsets are topologically disjoint if the closure of each is disjointfrom the other. Then continuous maps on each can be combined on the union.Note that TC( X ) ≤ GC( X ) since if Φ is a NGMPR on E , then the map E → P ( X )defined by ( x , x ) Φ( x , x )( ) is a motion planning rule on E . In this paper, weshow that GC( X ) = TC( X ) for the following non-geodesic spaces: R n − Q with n ≥ Q a finite subset, F ( R n ,
2) and F ( R n − { x } ,
2) with n ≥
2, the unordered
EODESIC COMPLEXITY FOR NON-GEODESIC SPACES 3 configuration space C ( R − { x } , F ( Y, Y is a graph with exactlyone essential vertex, of order 3. In Theorem 3.1, we show that GC( X ) > TC( X ) for X = F ( R − Q,
2) if Q is a finite subset with at least two points.We thank David Recio-Mitter for some helpful comments.2. Examples in Euclidean spaces
In this section, we determine explicit NGMPRs in a number of examples, in each ofwhich we have GC( X ) = TC( X ). In each of these examples, we find it convenient tolet g : I → I denote a continuous function such as g ( t ) = sin( πt ) or g ( t ) = 1 − | t − | satisfying g − (0) = { , } . Theorem 2.1.
For n ≥ and x ∈ R n , GC( R n − { x } ) = TC( R n − { x } ) = ( n even n odd.Proof. Here X = R n . By a linear homeomorphism of R n , we may assume x = 0.The nogeo set E is { ( a, b ) : a = λb, λ < } . We use linear paths as a GMPR on thecomplement of E , called the geoset.If n is even, let v be a unit vector field on S n − . A NGMPR on E is given byΦ( a, b )( s )( t ) = (1 − t ) a + tb + s · g ( t ) · v (cid:0) b − a k b − a k (cid:1) . (2.2)If n is odd, let v be a unit vector field on S n − − { e } and let E ′ = { ( a, b ) : b − a = ke , k > } . Then (2.2) is a NGMPR on E − E ′ , while Φ( a, b )( s )( t ) = (1 − t ) a + tb + s · g ( t ) · e isa NGMPR on E ′ . Thus GC( R n − { } ) ≤ n is even (resp. odd). Equalityfollows from the well-known value ([4]) of TC( S n − ) since R n − { } ≃ S n − . Theorem 2.3.
Let X = R n − Q , where Q is a finite set with at least two points.Then GC( X ) = TC( X ) = 2 .Proof. Again X = R n . First let n be even. We split the nogeo set as E ∪ E , where E is the set of ( a, b ) for which exactly one point of Q lies on the segment ab , while DONALD M. DAVIS E is those for which two or more points of Q lie on the segment. A NGMPR on eachis given byΦ( a, b )( s )( t ) = (1 − t ) a + tb + δ · s · g ( t ) · v (cid:0) b − a k b − a k (cid:1) , (2.4)where δ = min(1 , d ( ab, Q − ( Q ∩ ab ))), a continuous function on X × X . Note thatthe function Φ just defined is not continuous on E ∪ E because of the role of δ .[[If ( a n , b n ) ∈ E , all with a n b n passing through a point x ∈ Q , have the propertythat a n → a , b n → b , and d ( a n b n , x ′ ) → x ′ ∈ Q ∩ ab with x ′ = x , then δ ( a n , b n ) → a, b ) ∈ E with δ ( a, b ) = 0.]]If n is odd, let V be a unit vector which is not realizable as ( x ′ − x ) / k x ′ − x k for any x, x ′ ∈ Q , and let v be a unit vector field on S n − − { V } . Let E be the set of ( a, b )for which exactly one point of Q lies on the segment ab , and ( b − a ) / k b − a k 6 = V ,and define Φ on E using (2.4). Let E be the set of ( a, b ) for which two or morepoints of Q lie on ab or exactly one point of Q lies on ab and ( b − a ) / k b − a k = V .These two portions of E are topologically disjoint. Define Φ( a, b ) on E using (2.4),interpreting v ( V ) to be any particular vector orthogonal to V .Using linear geodesics on the complement of the nogeo set, we obtain GC( X ) ≤ X ) = 2 ([4]). Theorem 2.5.
For n ≥ , GC( F ( R n , F ( R n , ( n even n odd.Proof. We have X = R n × R n . Since F ( R n , ≃ S n − , it suffices to prove the upperbound. The nogeo set E is { (( a, a ′ ) , ( b, b ′ )) : b − b ′ = λ ( a − a ′ ) , λ < } , and we uselinear geodesics on its complement.If n is even, a NGMPR on E is given byΦ(( a, a ′ ) , ( b, b ′ ))( s )( t ) = (cid:0) (1 − t ) a + tb, (1 − t ) a ′ + tb ′ + s · g ( t ) · v (cid:0) b ′ − b k b ′ − b k (cid:1)(cid:1) . (2.6)For a point on a path in the homotopy of (2.6) to have two components equal wouldrequire that v (cid:0) b ′ − b k b ′ − b k (cid:1) is a scalar multiple of b ′ − b , which cannot happen.If n is odd, decompose the nogeo set into subsets determined by whether or not( b ′ − b ) / k b ′ − b k = e . We use a vector field on S n − − { e } in (2.6) for one, and canreplace the v ( − ) expression by e for the other. EODESIC COMPLEXITY FOR NON-GEODESIC SPACES 5
Theorem 2.7. If X = F ( R n − { x } , , then GC( X ) = TC( X ) = ( n even n odd.Proof. Again X = R n × R n . We say that paths γ and γ ′ collide if γ ( t ) = γ ′ ( t ) forsome t ∈ I . Segments ab and a ′ b ′ collide iff b ′ − b = λ ( a ′ − a ) for some λ <
0. Weinitially partition X × X into seven sets, on each of which we will define a GMPRor NGMPR when n is even. Sets C , C , and C x consist of those (( a, a ′ ) , ( b, b ′ )) forwhich segments ab and a ′ b ′ collide and ( C ) neither segment contains x , ( C ) oneof the segments contains x and the other segment has positive length, and ( C x )they collide at x . Sets E j , j ∈ { , , } , consist of those (( a, a ′ ) , ( b, b ′ )) which do notcollide and j of the segments contain x . The set L is those (( a, a ′ ) , ( b, b ′ )) such thateither a = b and these lie on a ′ b ′ , or a ′ = b ′ and are on ab .Note that C and E are topologically disjoint, as are C and E , and also C x and L . Indeed, each set has a property, preserved under limits, which is not true of anyelement of the paired set. Once we have noted the GMPR and NGMPRs on each ofthe seven regions when n is even, the domains E , C ∪ E , C ∪ E , and C x ∪ L imply GC( X ) ≤ n is even. Since TC( X ) = 3 when n is even ([5]), we obtainthe result in this case.We use the linear geodesic on E . For E and E , let δ = min(1 , d ( ab, a ′ b ′ )), where d ( ab, a ′ b ′ ) is the minimum distance for corresponding values of t . On E , we useΦ(( a, a ′ ) , ( b, b ′ ))( s )( t ) = (cid:18) (1 − t ) a + tb + δ · s · g ( t ) · v (cid:0) b − a k b − a k (cid:1) , (1 − t ) a ′ + tb ′ + δ · s · g ( t ) · v (cid:0) b ′ − a ′ k b ′ − a ′ k (cid:1)(cid:19) . (2.8)Because both ab and a ′ b ′ pass through x , both b − a and b ′ − a ′ must be nonzero.On E , we modify this formula by removing the δsgv term in the component whichdid not pass through x .On C , we use the NGMPR of (2.6) with the s · g ( t ) multiplied by an additionalfactor min(1 , d ( ab, x )). On C , we useΦ(( a, a ′ ) , ( b, b ′ ))( s )( t ) = (cid:18) (1 − t ) a + tb, (1 − t ) a ′ + tb ′ + s · g ( t ) · b − a k b − a k (cid:19) , (2.9) DONALD M. DAVIS when a ′ b ′ passes through x , and a similar formula when ab passes through x . Thesecurves do not pass through x since x cannot be written as the sum of a point on a ′ b ′ plus a nonzero multiple of b − a . Since b − b ′ is a scalar multiple of a − a ′ , if thecomponents of (2.9) were to collide, b − a would be a nonzero multiple of a − a ′ , whichit is not.On L , we use (2.6) when x lies on a ′ b ′ , as does a = b , with an obvious modificationif ab and a ′ b ′ play opposite roles. On C x we use, similarly to C ,Φ(( a, a ′ ) , ( b, b ′ ))( s )( t ) = (cid:18) (1 − t ) a + tb + s · g ( t ) · b ′ − b k b ′ − b k , (1 − t ) a ′ + tb ′ + s · g ( t ) · a − b k a − b k (cid:19) . (2.10)For these to collide, we would need ab to be parallel to aa ′ , which it isn’t. Also notethat b − a is nonzero since otherwise we would have a = b = x , which cannot happen.When n is odd, we no longer have a vector field on S n − . For the cases that usedsuch a vector field, we use a vector field on S n − − { pt } . By choosing the excludedpoint differently in different cases, we can arrange it so that all the excluded casesare topologically disjoint, and so can be combined into one additional domain, againagreeing with the known result for TC. Indeed, choose vectors V i , 1 ≤ i ≤
4, in S n − such that if i = j , then V i = ± V j . Our fifth domain is( C ∩ { b ′ − b ∈ h V i} ) ∪ ( E ∩ { b − a or b ′ − a ′ ∈ h V i} ) ∪ ( E ∩ { b − a or b ′ − a ′ ∈ h V i} ) ∪ ( L ∩ { b ′ − b ∈ h V i} ) , where h V i denotes the span of a vector V . Theorem 2.11.
For the unordered configuration space X = C ( R − x , , GC( X ) = TC( X ) = 2 . Proof.
Here we use results from [2, Prop 4.3,(4.1)] that in R × R , d (( a, a ′ ) , ( b, b ′ )) = d (( a, a ′ )( b ′ , b )) iff aa ′ ⊥ bb ′ , and if d (( a, a ′ ) , ( b, b ′ )) < d (( a, a ′ )( b ′ , b )), then ab and a ′ b ′ do not collide. Also, X = R × R / ( a, a ′ ) ∼ ( a ′ , a ).We first consider pairs ( { a, a ′ } , { b, b ′ } ) with d (( a, a ′ ) , ( b, b ′ )) = d (( a, a ′ )( b ′ , b )), andlabel them so that d (( a, a ′ ) , ( b, b ′ )) < d (( a, a ′ )( b ′ , b )). Let E denote the set of thosefor which neither ab nor a ′ b ′ passes through x and use the linear GMPR on E . Let EODESIC COMPLEXITY FOR NON-GEODESIC SPACES 7 E denote the set of those for which exactly one of ab and a ′ b ′ passes through x , and,if x ∈ ab , useΦ( { a, a ′ } , { b, b ′ } )( s )( t ) = (cid:0) (1 − t ) a + tb + δ · s · g ( t ) · v (cid:0) a − b k a − b k (cid:1) , (1 − t ) a ′ + tb ′ (cid:1) (2.12)with δ = min(1 , d ( ab, a ′ b ′ )), with obvious reversal if instead x ∈ a ′ b ′ . Let E denotethose for which ab and a ′ b ′ both pass through x (but not for the same t ), and use(2.8) as the NGMPR.Now we consider pairs ( { a, a ′ } , { b, b ′ } ) for which aa ′ ⊥ bb ′ , so d (( a, a ′ ) , ( b, b ′ )) = d (( a, a ′ )( b ′ , b )). Note that ab and a ′ b ′ do not collide, nor do ab ′ and a ′ b . Let Y denote the set of those for which none of the segments ab , a ′ b , ab ′ , and a ′ b ′ passthrough x . A GMPR on Y can be obtained by choosing ab so that bb ′ is a 90-degree counterclockwise rotation from aa ′ , and using the linear path from ( a, a ′ ) to( b, b ′ ). Let Y denote the set of those such that exactly one of the pairs { ab, a ′ b ′ } and { ab ′ , a ′ b } has neither segment passing through x . Use the linear path on that pair asa GMPR. Let Y denote the set of those for which the pairs { ab, a ′ b ′ } and { ab ′ , a ′ b } have one segment each passing through x . Let the shorter segment containing x be ab . Then the ordered points b ′ bx a or a ′ ax b form a segment with aa ′ (resp. bb ′ )perpendicular to it. We can use (2.12) as a NGMPR on Y .We can use E , E ∪ Y , E ∪ Y , and Y as our four domains, since E and Y aretopologically disjoint, as are E and Y .3. Example when
GC( X ) > TC( X )In this section, we show that GC( X ) > TC( X ) when X = F ( R − Q,
2) if Q is afinite subset with at least two points. Theorem 3.1.
Let Q be a finite subset of R n with at least two points, and X = F ( R n − Q, . If n is even, then GC( X ) ≤ , with equality obtained when n = 2 . By [5], TC( F ( R n − Q, DONALD M. DAVIS
Proof.
We first show that GC( X ) ≤ n is even. We partition X into 18 subsets,on each of which there is a GMPR or NGMPR. Then we will group them into sixcollections of topologically disjoint subsets.We use the word ‘collide” as in the proof of Theorem 2.7. Recall that ab and a ′ b ′ collide iff b − b ′ is a negative multiple of a − a ′ . There are sets E , E , E , E , , E , ,and E , , in which the segments ab and a ′ b ′ do not collide, and the subscripts indicatehow many points of Q lie on each segment, with “2” referring to “2 or more.” Forexample, E consists of those (( a, a ′ ) , ( b, b ′ )) for which ab and a ′ b ′ do not collide andone of these segments contains two or more points of Q , while the other has none. Ifthe segments intersect at a point of Q (for differing values of the parameter t ), thenthat point counts for both lines. So, for example, E , consists both of noncollidingelements where the segments do not meet at a point of Q and each contains a pointof Q , and those where the two segments meet at a point of Q , and neither containsother points of Q . If Q has only two or three points, some of these sets can be empty.There are also sets C , C , C , C , , C , , and C , , in which the segments collide,but not at a point of Q , and the subscripts have the same meaning as before. Forthese, there is not the issue of classifying what happens if segments intersect at apoint of Q . For these C -sets, we exclude colliding elements in which a , a ′ , b , and b ′ are collinear.Next we have sets Y j , j = 0 , ,
2, in which the segments collide at a point of Q ,and j of the segments contain one or more additional points of Q . Again we excludethe case in which a , a ′ , b , and b ′ are collinear. Finally we have the linear cases L j , j = 0 , ,
2, in which the four points are collinear and aa ′ and bb ′ have oppositedirections, and j points of Q lie on ab ∪ a ′ b ′ .On E , we use the linear geodesic. On the other E sets, we use formulas like (2.8),using just the linear part when a segment does not contain any points of Q , andmodifying δ to equal min(1 , d , d ), where d is the distance between the parametrizedsegments, and d is the distance from the segment to the nearest point of Q not onit.Formulas for near geodesics on the C and Y sets are similar to those that workedfor the C sets in the proof of Theorem 2.7. We use a factor δ = min(1 , d ( ab, Q − ( Q ∩ ab ))) on the first component, and an analogue on the second. Incorporating EODESIC COMPLEXITY FOR NON-GEODESIC SPACES 9 that, we use an analogue of (2.6) on C , of (2.9) on C and C , and of (2.10) on C , , C , , C , , and each Y j . On each set L j , we can useΦ(( a, a ′ ) , ( b, b ′ ))( s )( t ) = (cid:18) (1 − t ) a + tb + δ · s · g ( t ) · v (cid:0) b ′ − b k b ′ − b k (cid:1) , (1 − t ) a ′ + tb ′ − δ · s · g ( t ) · v (cid:0) b ′ − b k b ′ − b k (cid:1)(cid:19) . We can group these into six collections of topologically disjoint subsets as follows. E , E ∪ C , E ∪ E , ∪ C ,E , ∪ C ∪ C , ∪ Y ∪ L ,E , ∪ C , ∪ Y ∪ L , C , ∪ Y ∪ L . To show two sets are topologically disjoint, we usually show that each has a propertywhich is preserved under limits of sequences and is not shared by any element of theother set. For example, in the fifth of the above sets, we could use “collinear” for L and “collide at a point of Q ” for Y . The limit of a sequence of E , sets could collide,but would be either in C , or Y . Similarly, the limit of a sequence of C , sets couldbe in Y , but not in Y . Since two points determine a line, the limit of a sequence of C , sets cannot be in Y .Now we prove GC( X ) ≥ n = 2. Any element in the nogeo set is the limit ofelements of the geoset E . Thus E is not topologically disjoint from any subset of thenogeo set. We will show that the assumption that the nogeo set can be partitioned S ⊔ S ⊔ S ⊔ S with a NGMPR Φ j on each S j leads to a contradiction.We may assume that the points ( − , −
1) and (1 ,
1) are in Q . Our diagrams willsuggest that there are no other nearby points of Q , but if there are, the argument isnot affected. We start with the element x of X × X defined by x = (( a, a ′ ) , ( b, b ′ )) = (((2 , , ( − , − , (( − , − , (2 , . Let S be the S j which contains x . Define pr : F ( R − Q, → R by pr ( a, a ′ ) = a ,and pr similarly. Note that pr Φ ( x )(1) is a path in R − Q from (2 ,
2) to ( − , − − , −
1) on either the left or right side. We assume without loss of generality that it passes on the right side, and in Figure3.2 we sketch a possible such curve. There are only four homotopy types of possible curves pr Φ ( x )(1), determined by whether they pass on the left or right of ( − , − , Figure 3.2. Diagram to help show x n S . • (1 , • a n a = b ′ a ′ = bb n pr Φ ( x )(1)We consider elements x n = (( a n , a ′ ) , ( b n , b ′ )) with a n approaching a from the right, a n b n passing through (1 , a n (resp. b n ) at y = 2 (resp. − x n ’s is in S , then pr Φ ( x n )(1) converges uniformly to pr Φ ( x )(1), so for n sufficiently large, pr Φ ( x n )(1) passes to the right of ( − , − Φ ( x n )(0), which is the segment a n b n , and this homotopy is notallowed to pass through ( − , − x n ’s beingin S . Since there are only four sets S j , we can find a subsequence h x n i all in thesame set S j , and j = 1, so we may assume all x n are in S .For each x n , consider a sequence of elements x n ,n = (( a n ,n , a ′ ) , ( b n ,n , b ′ )) con-verging to x n with segment a n ,n b n ,n parallel to a n b n and on the on the side ofthe segment a n b n opposite to the side on which pr Φ ( x n )(1) passes (1 , y -level ±
2. See Figure 3.3.
EODESIC COMPLEXITY FOR NON-GEODESIC SPACES 11
Figure 3.3. Diagram to help show x n ,n S . a n b n a n ,n b n ,n •• pr Φ ( x n )(1)Similarly to the argument following Figure 3.2, if all x n ,n are in S , the curvespr Φ ( x n ,n )(1) converge uniformly to pr Φ ( x n )(1) and so must eventually pass(1 ,
1) on the same side as pr Φ ( x n )(1). But they are homotopic rel endpoints topr Φ ( x n ,n )(0), which is just the segment a n ,n b n ,n . This cannot be done by ahomotopy which is not allowed to pass through (1 , S . For each n , there must be infinitely many n with x n ,n in the same S j , andthen there must be infinitely many n ’s with the same j . So we may renumber andassume all x n ,n are in the same S j , j = 2. If j = 1, then a diagonal sequence x n ,n converges to x and has pr Φ ( x n ,n )(1) converging uniformly to pr Φ ( x )(1). We getthe same contradiction as for x n ’s above: the homotopy rel endpoints is not allowedto pass through ( − , −
1) but its two ends pass through opposite sides of ( − , − a n b n these parallellines passed; it depended on the relative position of pr Φ ( x )(1) and the segment a n b n . Therefore j = 1, so we may choose j = 3.Now we begin to incorporate the second component. For each ( n , n ), we considera sequence x n ,n ,m = (( a n ,n , a ′ m ) , ( b n ,n , b ′ m )) converging to x n ,n with all points at y -level ± a ′ m b ′ m passing through (1 ,
1) and passing on the side of( − , −
1) opposite to that of the curve pr Φ ( x n ,n )(1). See Figure 3.4. Figure 3.4. Diagram to help show x n ,n ,m S . b ′ m b ′•• pr Φ ( x n ,n )(1) a ′ m a ′ There must be infinitely many n for which there are infinitely many n for whichthere are infinitely many m with x n ,n ,m in the same S j . We renumber to restrictto these values. By the sort of homotopy argument employed earlier, j = 3. If j = 2, since x n ,n ,n → x n , then pr Φ ( x n ,n ,n )(1) → pr Φ ( x n )(1) uniformly, sothese must also eventually pass on the same side of (1 ,
1) as pr Φ ( x n )(1), and hencecannot be in S by the homotopy argument used before, since pr Φ ( x n ,n ,n )(0) = a n ,n b n ,n passes (1 ,
1) on the opposite side of pr Φ ( x n )(1). See Figure 3.3. So j = 2.A similar argument shows that j = 1 and hence j = 4; i.e., all x n ,n ,m (afterrenumbering) are in S . Here is the argument. The sequence x n ,n ,n converges to x ,so pr Φ ( x n ,n ,n )(1) converges uniformly to pr Φ ( x )(1) and hence eventually passeson the right of ( − , − Φ ( x n ,n ,n )(0) =pr Φ ( x n ,n )(0) passes ( − , − − , − S ∪ S ∪ S ∪ S , which will complete the proof.For each ( n , n , m ), we consider a sequence of elements x n ,n ,m,m ′ = (( a n ,n , a ′ m,m ′ ) , ( b n ,n , b ′ m,m ′ ))approaching x n ,n ,m , with segments a ′ m,m ′ b ′ m,m ′ parallel to a ′ m b ′ m and on the sideof a ′ m b ′ m opposite to the side where pr Φ ( x n ,n ,m )(1) passes (1 , y -components at ±
2. If x n ,n ,m,m ′ were in S , then pr Φ ( x n ,n ,m,m ′ )(1) would pass(1 ,
1) on this opposite side for m ′ sufficiently large, but the homotopy frompr Φ ( x n ,n ,m,m ′ )(0) = a ′ m,m ′ b ′ m,m ′ EODESIC COMPLEXITY FOR NON-GEODESIC SPACES 13 is not allowed to pass through (1 , x n ,n ,m,m ′ are in the same S j , and now we know j = 4.If x n ,n ,m,m ′ ∈ S , we obtain a contradiction by an argument almost identical tothat used above to show x n ,n ,m S , since for sufficiently large m ′ , the segments a ′ m,m ′ b ′ m,m ′ and a ′ m b ′ m lie on the same side of ( − , − x n ,n ,m,m ′ ∈ S , then x n ,n ,n ,n → x n and we obtain a contradiction similar to that used in showing x n ,n ,m S , using pr Φ , and a similar modification to a previous argument shows x n ,n ,m,m ′ S . 4. Configuration spaces of graphs
Configuration spaces F ( G,
2) of graphs G , as studied in [3], are handled somewhatdifferently than the cases considered above. In [3], F ( G,
2) was given the subspacemetric from G × G , where G × G had either the ℓ or ℓ metric, using distance in thegraph G . Thus, using ℓ , d (( a , a ) , ( b , b )) = p d ( a , b ) + d ( a , b ) .When X = F ( G, X equals G × G and is geodesic. However,certain geodesics in X cannot be approximated by paths in X . For example, let G be the Y -graph Y , and suppose a , a ′ , b ′ , and b are on the same edge at distance 1,1 + δ , 1 + 2 δ , and 1 + 3 δ , respectively, from the vertex v .The geodesic in X from ( a, a ′ ) to ( b, b ′ ) is the linear path of length δ √
10. However,since direct motion will involve a collision, there is no path in X whose length is closeto this. A short path in X from ( a, a ′ ) to ( b, b ′ ) is one that moves from ( a, a ′ ) backjust beyond v onto the two empty arms, and from there to ( b, b ′ ), with length slightlygreater than p δ ) + p (1 + δ ) + (1 + 2 δ ) .We use the intrinsic metric d I on X = F ( G, d I ( x, y ) is the infimumof the d -lengths of paths in X from x to y . For the spaces considered here, this metricinduces the same topology as does the original metric since if ε < d ( a, a ′ ) / √
2, the ε -balls around ( a, a ′ ) in the two topologies are equal. This is because linear motionfrom points in the balls avoids collision. This also implies that lengths of curves usingthe d I metric equal the d -length because length is determined from arbitrarily smallsegments. It is easy to see that, at least when G is a tree, the completion of F ( G, F ( G,
2) with the points ( v, v ) when v is at an essential vertex adjoined. Since the intrinsic and ℓ topologies are the same, we again haveTC( F ( G, ≤ GC( F ( G, F ( Y,
2) is a length space when we use the intrinsic metric. The Hopf-Rinow Theorem says that the completion of a length space is geodesic if it is locallycompact. The advantage of a length space is that geodesics in its completion can beapproximated by paths in the space.Since the definition of GC in [6] only applies well to geodesic spaces, to considerGC( F ( Y, F ( Y,
2) by the homotopically equivalent sub-space F ε ( Y,
2) which consisted of points ( a, a ′ ) satisfying d ( a, a ′ ) ≥ ε . For example,in the diagram at the left in Figure 4.1, there is no geodesic in F ( Y,
2) from ( a, a ′ )to ( b, b ′ ) because in the linear motion from ( a, a ′ ) to ( b, b ′ ), the first particle wouldovertake the second, which is not allowed. In [3], we represented paths in the graphby paths in the xy plane, where, in this case, the x -axis corresponds to the two upperarms on the graph, with the vertex at 0, and the y -axis, similarly, to the arms on theleft. This is shown on the right side of Figure 4.1, in which the interior of the shadedregion is excluded, as those points do not satisfy d ( a, a ′ ) ≥ ε . The representation ofthe geodesic in F ε ( Y,
2) is indicated. It corresponds to the path in the graph whichgoes from ( a, a ′ ) to the point (0 , ε ), with the ε on the bottom arm, and from there to( b, b ′ ). Figure 4.1. An element in F ε ( Y, and a representation of its path • • • • aa ′ b ′ b y x ( a, a ′ ) ( b, b ′ ) • • y x From our near-geodesic point of view, the shaded region shrinks to the line y = x <
0, and the geodesic in F ( Y,
2) is represented by the path from ( a, a ′ ) to (0 ,
0) to
EODESIC COMPLEXITY FOR NON-GEODESIC SPACES 15 ( b, b ′ ). On the graph, we go from ( a, a ′ ) to ( v, v ) to ( b, b ′ ). Note that ( v, v ) ∈ F ( Y, s = ε could be chosen to berepresented by the path on the right side of Figure 4.1.The nogeo set of F ( Y,
2) consists of three types of elements. One is the type justdiscussed. A second, rather similar, type has a and b ′ on one arm, with b and a ′ occupying the other two arms with the second particle passing through the vertexfirst. The geodesic in F ( Y,
2) has the two particles hitting the vertex at the sametime, and then moving on to b and b ′ . The third type is elements where all fourpoints are on just one or two arms, and the orientation of aa ′ is opposite to thatof bb ′ . In either case, we move from ( a, a ′ ) to ( v, v ) and then to ( b, b ′ ) in F ( Y, s first moves the firstparticle s units onto the arm clockwise from the occupied arm, and simultaneouslythe second particle s units onto the other arm, and then on to ( b, b ′ ). If two arms areoccupied, the choice of which particle to move onto the free arm is done in the sameway as the choice in [3] of which particle to move ε units onto the free arm was made.One can check that we obtain an NGMPR on the nogeo set, and a GMPR on itscomplement. Thus we have reproved the following result of [3]. Theorem 4.2. If Y denotes the Y graph, then GC( F ( Y, F ( Y, . We see that for these graph configuration spaces, the analysis of GC is very closelyrelated to the analysis of GC( F ε ( G, References [1] A.Bianchi and D.Recio-Mitter,
Topological complexity of unordered configura-tion spaces of surfaces , Alg Geom Topology (2019) 1358–1384.[2] D.M.Davis, Geodesics in the configuration spaces of two points in R n , arXiv :2001.00850.[3] D.M.Davis, M.Harrison, and D.Recio-Mitter, Two robots moving geodesicallyon a tree , arXiv :2006.14772.[4] M.Farber, Topological complexity of motion planning , Discr Comp Geom (2003) 211–221.[5] M.Farber, M.Grant, and S.Yuzvinsky, Topological complexity of collision-freemotion planning algorithms in the presence of multiple moving obstacles ,Comtemp Math Amer Math Soc (2007) 75–83.[6] D.Recio-Mitter,
Geodesic complexity of motion planning , arXiv :2002.07693. Department of Mathematics, Lehigh University, Bethlehem, PA 18015, USA
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