aa r X i v : . [ m a t h . DG ] N ov On the Holonomic Equivalence of Two Curves
Tamer Tlas
Abstract
Given a principal G -bundle P → M and two C curves in M with co-inciding endpoints, we say that the two curves are holonomically equivalentif the parallel transport along them is identical for any smooth connectionon P . The main result in this paper is that if G is semi-simple, then thetwo curves are holonomically equivalent if and only if there is a thin, i.e.of rank at most one, C homotopy linking them. Additionally, it is alsodemonstrated that this is equivalent to the factorizability through a tree ofthe loop formed from the two curves and to the reducibility of a certaintransfinite word associated to this loop. The curves are not assumed to beregular. Let M be a smooth manifold, G a Lie group and P → M a smoothprincipal G -bundle. Let A be the space of smooth connections on P and G the space of smooth gauge transformations. It is a well-knownfact that A / G is one of the richest and most interesting objects to study,with applications ranging from knot invariants to high energy physics.Unfortunately, although A is an affine space, the quotient is, due to thenonlinearity of the action of G , not even an infinite dimensional manifold.It is natural to try to find a description of A on which the action of G would be more manageable.Fix a point o ∈ M and suppose γ is a loop based at o , i.e. it is a,sufficiently differentiable, map from I = [0 , → M , and which satisfies γ (0) = γ (1) = o . Any element A ∈ A associates to this loop a bijectionfrom the fiber over o to itself, and if one fixes some point in this fiber,this bijection can be identified with an element of G . This element shallbe denoted by U ( γ, A ) and is called the holonomy of A around the loop γ . The key point is that G acts very transparently on the holonomy. Theholonomy is simply conjugated, with the conjugating element being thesame for all the loops based at o . It is thus natural to look for a formulationof gauge theory where the holonomies are taken as elementary objects andthe connection, and maybe even the bundle, are reconstructed from themafterwards if needed, and indeed, there is a long history of attempts inthis direction dating back to [1].Let us denote the set of loops based at o by L . It is easy to see thatnot any function from L to G is a holonomy of some connection. Thereason is that if one defines the product of two loops in the usual way then U ( γ · γ , A ) = U ( γ , A ) U ( γ , A ). This equation means that, had L been a group, an element of A would simply be a homomorphism from γ · γ ( t ) = (cid:26) γ (2 t ) t ∈ [0 , ] γ (2 t − t ∈ [ , his group to G , and an element of A / G would just be the conjugacyclass of this homomorphism. But, the set of loops is not a group, as theproduct of two loops is neither associative nor does it have an identity.The way to solve this problem is of course clear: one should take anappropriate quotient of the set of loops as is done similarly in elementaryhomotopy theory. It is evident, however, that taking the quotient of L by the standard homotopy is inconsistent, as in general, two homotopicloops will have different holonomies. Thus, we are faced with the problemof finding an equivalence relation ∼ on L such that {L , ·} / ∼ is a groupand moreover γ ∼ γ = ⇒ U ( γ , A ) = U ( γ , A ) , ∀ A ∈ A . In fact, it is natural to require the implication to go in the oppositedirection as well, since otherwise it would mean that a connection is nota homomorphism from {L , ·} / ∼ to G but from a further quotient of it.The above motivates the following Definition 1.
Two loops γ , γ ∈ L which satisfy U ( γ , A ) = U ( γ , A ) for all A ∈ A are called holonomically equivalent. Let us stress the point that in this definition we assume P → M , andthus G , to be fixed.In view of the discussion above, we are tasked with finding necessaryand sufficient conditions for holonomic equivalence. The answer to thisproblem depends on the class of paths under consideration. In general,finding a sufficient condition is much easier than obtaining a necessary one.Let us discuss the different classes of paths which have been consideredin the literature so far as well as the different associated sufficiency andnecessity conditions. Chronologically:i- In [1], the loops were assumed to be piecewise differentiable. Theequivalence relation was that γ ∼ γ if γ · γ factors through afinite tree, [1]. γ stands for the loop γ traversed in the oppositedirection. Necessity of this relation for holonomic equivalence wasnot considered, only sufficiency.ii- In [2], the loops were assumed to be piecewise C as well as regular,i.e. that ˙ γ ( t ) = 0 for all t ∈ I . The equivalence relation was thesame as in [i]. It was shown there that two curves are equivalent ifand only if all the Chen’s iterated integrals of the loop γ · γ vanish.If γ is a loop, then its iterated integrals are Z dt ˙ γ µ ( t ) Z dt Z t ds ˙ γ µ ( t ) ˙ γ ν ( s ) Z dt Z t ds Z s dr ˙ γ µ ( t ) ˙ γ ν ( s ) ˙ γ ρ ( r )...Using standard properties of the iterated integrals, it is not verydifficult to show that the above set of expressions vanishes if and This is not the formulation that was used in [1] but is in fact equivalent to it. The sameis true, where applicable, in the other cases. nly if for any collection of smooth 1-forms w , w , w . . . (which caneven be taken to be matrix-valued), the following integrals vanish Z dt w µ ( γ ( t )) ˙ γ µ ( t ) Z dt Z t ds w µ ( γ ( t )) w ν ( γ ( s )) ˙ γ µ ( t ) ˙ γ ν ( s ) Z dt Z t ds Z s dr w µ ( γ ( t )) w ν ( γ ( s )) w ρ ( γ ( r )) ˙ γ µ ( t ) ˙ γ ν ( s ) ˙ γ ρ ( r )...Recalling that U ( γ, A ) has an expression in terms of the path-orderedexponential of A = A µ dx µ I + Z dtA µ ( γ ( t )) ˙ γ µ ( t )+ Z dt Z t dsA µ ( γ ( t )) A ν ( γ ( s )) ˙ γ µ ( t ) ˙ γ µ ( s )+ . . . it is clear that the fact that all the iterated integrals vanish impliesthat the holonomy around γ is trivial (equal to the identity) for everyconnection. If we know that the U ( A, γ ) = I for every A and forevery G , then one can show that this implies that all the iteratedintegrals vanish [3]. However, if one is only given that U ( A, γ ) = I for every A and a specific G then the triviality of the holonomy is ana priori weaker statement, since one can only conclude (by a scalingargument) that certain sums of iterated integrals vanish. In viewof the above, the necessity condition that has been proven in [2] isweaker than what we seek if we are concerned with the holonomicequivalence for curves.iii- In [4], the loops are assumed to be piecewise analytic and the equiv-alence relation is the same as in [i]. Both sufficiency and necessityare demonstrated.iv- In [5], the loops are assumed to be C ∞ and have all their derivativesvanish at the endpoints. Two loops are declared to be equivalent if γ · γ is homotopic to the constant loop via a thin homotopy. Onlysufficiency is shown.v- In [6], the loops are assumed to be of bounded variation and γ ∼ γ if γ · γ factors through a tree. It is shown that two curves areequivalent if and only if all the Chen’s iterated integrals vanish. Thesame remarks as the ones in [ii] on the relation of this to holonomicequivalence apply here.In this manuscript we demonstrate that if G is a semi-simple Lie group,and γ , γ are two C loops having vanishing derivative at the endpointsthen the following are equivalent:a- γ and γ are holonomically equivalent.b- γ · γ factors through a tree. c- There is a thin C homotopy between the constant loop and γ · γ . See definition 3. See definition 4. - A certain transfinite word (to be defined later) associated to γ · γ is reducible.A few remarks are in order: • In view of the fact that a piecewise C curve can be reparametrizedto become C , it follows that the same statements are true (with anappropriate modification of [c]) for piecewise C loops. • We required the vanishing of the derivative of the loops at the end-points to ensure that their product is C , as this is the most usefulclass of loops if we are interested in a loop-based description of gaugetheory [5]. If we do not require this, then γ · γ is piecewise C andthe previous remark applies. • We have only stated the result for loops, but an essentially identicalresult (see the corollary after theorem 3) applies if γ and γ are twocurves such that thir initial points as well as their final ones coincide.In what follows, we assume that all the maps are C unless statedotherwise.Let us begin with a certain decomposition of curves which is of interestin its own right: Theorem 1.
Given a curve γ : I → M , there is a collection of mutuallydisjoint sets { A n } ∞ n =0 , such that:a- A is closed, while A n is open for n > , and S ∞ n =0 A n = I .b- If t ∈ A n then γ − ( γ ( t )) ⊂ A n , moreover if n > then the cardinal-ity of γ − ( γ ( t )) is equal to n .c- ˙ γ ( t ) = 0 if t ∈ A n for n > , while ˙ γ ( t ) = 0 for t in the interior ofany connected component of A .d- The A n ’s being open are a union of disjoint open intervals. γ re-stricted to any such interval is an embedding. Moreover, any twosuch embeddings are either disjoint or identical (possibly after areparametrization and a switch in orientation).Proof. Let C be the set of all critical points of γ , and let C ′ = γ − ( γ ( C )).Note that C ′ is closed and that it contains all the points t ∈ I such that γ − ( γ ( t )) has infinite cardinality. For assume that t is such a point, thenthis means that γ − ( γ ( t )) contains a limit point t . It is obvious that˙ γ ( t ) = 0, i.e. t ∈ C . It follows that t ∈ γ − ( γ ( t )) ⊂ C ′ .Also, ˙ γ ( t ) = 0 for t in the interior of any connected component of C ′ . To see this, assume that there is t ∈ C ′ with ˙ γ ( t ) = 0 and forsome δ >
0, ( t − δ, t + δ ) ⊂ C ′ . It follows that γ is an embedding ina neighborhood of t , ( t − δ ′ , t + δ ′ ). This in particular implies thatthe 1-dimensional Hausdorff measure of γ ( C ′ ) is strictly greater than 0,since γ ( C ′ ) contains γ (( t − δ ′ , t + δ ′ )), a set diffeomorphic to an interval.However, γ ( C ′ ) = γ ( C ), and the 1-dimensional Hausdorff measure of γ ( C )is equal to 0 by the strong version of Sard’s theorem [7]. Therefore thereare no open intervals contained in C ′ where ˙ γ does not vanish.It follows from above that if t / ∈ C ′ , then γ − ( γ ( t )) is a finite set.For any n >
0, let B n = { t / ∈ C ′ : γ − ( γ ( t )) has cardinality n } . Let A n = B ◦ n (the interior) for n > A = I − S ∞ n =1 A n . It is clearthat the A n ’s are mutually disjoint and that (a) holds.Let us prove (c). By construction, we have that ˙ γ ( t ) = 0 if t ∈ A n for n >
0. To prove the rest of (c) assume that there is a point t ∈ A ◦ such hat ˙ γ ( t ) = 0. Since C ′ does not contain intervals on which ˙ γ does notvanish, and since C ′ is closed, it follows that A − C ′ contains an openinterval. We will be done if we show that A − C ′ does not contain openintervals. To see this, note that A − C ′ = S ∞ n =1 ( B n − B ◦ n ) and assumethat t ∈ B m . We need the following claim: If s ∈ B m then there is a δ > such that ( s − δ, s + δ ) ∩ (cid:16) S ∞ n = m +1 B n ∪ C ′ (cid:17) = φ. Assuming this statement for the moment, note that it implies thatif there is an open interval which is contained in A − C ′ , then it is infact contained in S mn =1 ( B n − B ◦ n ). This interval cannot contain pointsfrom B , since in view of the claim above it would contain a subintervalcontained in B , and thus in B ◦ which is a contradiction. Similarly, itcannot contain points from B , B , . . . , B m , for the same reason. Thussuch an interval cannot exist and so A − C ′ contains no open intervals.Now, to prove the claim above, assume the converse, then there isa sequence of points { s k } ∞ k =1 ⊂ S ∞ n = m +1 B n ∪ C ′ which converges to s .Since C ′ is closed, we can assume without loss of generality that thissequence is in fact contained in S ∞ n = m +1 B n . It follows that γ − ( γ ( s k ))always has cardinality at least equal to m + 1. For every k select m + 1points from γ − ( γ ( s k )). In this way, we get m + 1 sequences in I . Passingto subsequences we can assume that they all converge to L , . . . L m +1 respectively. Moreover, these limits must in fact be different from eachother, since otherwise it would mean that one of these limits is in C and thus in C ′ , but this would imply that s ∈ C ′ since γ ( L ) = · · · = γ ( L m +1 ) = γ ( s ). Thus L , . . . , L m +1 are all different from each other.But this means that γ − ( γ ( s )) has cardinality at least equal to m + 1,which contradicts the fact that s ∈ B m . Thus the claim above is true andthe proof of (c) is complete.It would follow from the definitions of B n and A n , that (b) holds ifwe manage to show that if t ∈ B ◦ n , then every element of γ − ( γ ( t )) ∈ B ◦ n . To see that this is true, let { t , t , . . . , t n } = γ − ( γ ( t )). By con-struction we have that ˙ γ ( t ) = 0 if t ∈ B n . It follows that there arepositive δ , δ , . . . , δ n such that γ is an embedding when restricted to( t i − δ i , t i + δ i ) , i = 1 , . . . , n , and ( t − δ , t + δ ) ⊂ B n . We now makethe following claim: There is a positive δ < δ such that γ restricted to ( t − δ , t + δ ) isonto the set γ (cid:0) ( t − δ, t + δ ) (cid:1) . To see that this is the case, assume the converse. Then, there is asequence { s k } ∞ k =1 ⊂ ( t − δ , t + δ ) such that γ − ( γ ( s k )) / ∈ ( t − δ , t + δ )for all k . Since { s k } ∞ k =1 ⊂ B n , it follows that for every k there is atleast one element of γ − ( γ ( s k )) outside the set S ni =1 ( t i − δ i , t i + δ i ). Thissequence of elements has a subsequence which converges to a point t n +1 / ∈{ t , . . . , t n } , and continuity of γ implies that γ ( t n +1 ) = γ ( t ). But thiswould mean that γ − ( γ ( t )) has cardinality at least equal to n + 1 whichcontradicts the fact that t ∈ B n . Thus the claim above is true.Using the fact that γ restricted to ( t − δ , t + δ ) is an embedding, wehave that the map γ − ◦ γ : ( t − δ, t + δ ) → ( t − δ , t + δ ) is C , takes t to t and has a non-vanishing derivative. Thus the image of ( t − δ, t + δ )contains an open interval containing t , which in turn means that thereis an open interval around t which is contained in B n . Thus t ∈ B ◦ n .Needless to say, the same argument applies to t , t , . . . , t n , and the proofof (b) is complete. Note that we have proven in fact that if t , t ∈ A n and γ ( t ) = γ ( t ), then there are δ, δ ′ > γ (( t − δ, t + δ )) = (( t − δ ′ , t + δ ′ ))It remains to prove (d). Since A n is an open subset of I , then itis equal to a countable union of disjoint open intervals. Let ( a, b ) beone such interval. Then γ is in fact injective on this interval. To seethis, suppose that t , t ∈ ( a, b ) with γ ( t ) = γ ( t ). We know that γ isinjective on a neighborhood of t . Thus, let t ′ be the largest of the pointsin [ t , t ] such that γ is injective on [ t , t ′ ). Let t ′ be the point in [ t , t ′ )for which γ ( t ′ ) = γ ( t ′ ). Note, that γ is, by construction, injective on( t ′ , t ′ ). Consider the set D consisting of all the points t ∈ ( a, b ) such that ∃ s ∈ [ t ′ , t ′ ] with γ ( t ) = γ ( s ). This set is clearly closed in the subspacetopology, being the preimage of the closed set γ ([ t ′ , t ′ ]). This set is alsoopen. To see this, note that if γ ( s ) = γ ( t ) and s ∈ ( t ′ , t ′ ), then we havean open interval around t whose image under γ coincides with the imageof an interval around s which may be taken to be a subset of ( t ′ , t ′ ).Therefore, this interval around t is a subset of D . Suppose now that γ ( t ) = γ ( t ′ ) = γ ( t ′ ), then we know that there is an open interval around t whose image under γ coincides with the image of an open interval around t ′ and with the image of an open interval around t ′ . Then the image ofhalf of the interval around t (either the one less or equal or the one biggeror equal to t ) coincides with the image of the half of the interval around t ′ which is contained in [ t ′ , t ′ ] while the other half of the interval around t coincides with the image of the half of the interval around t ′ which iscontained in [ t ′ , t ′ ] (We used injectivity of γ on ( t ′ , t ′ ) in a crucial wayhere). Therefore, the image of the interval around t coincides with theimage of a subset of [ t ′ , t ′ ] and thus this interval around t is contained in D . We thus have that D is both open and closed in ( a, b ) and thus, beingnonempty, is in fact equal to the whole set. Consider now a sequence { t n } ∞ n =1 converging to a . Then there is a sequence { s n } ∞ n =1 ⊂ [ t ′ , t ′ ]whose image under γ coincides with the image of { t n } ∞ n =1 . Passing to asubsequence and taking the limit we get a point in [ t ′ , t ′ ] whose imageunder γ is equal to γ ( a ). We thus have that a ∈ A n , which contradicts(b).Summarizing the above, we have shown that γ restricted to ( a, b ) is aninjective immersion. To see that it is an embedding, note that γ ( a ) , γ ( b ) ∈ A . We now have two cases:i- γ ( a ) = γ ( b ). In this case γ restricted to [ a, b ] is a topological embed-ding (being an injective continuous map from a compact space to aHausdorff one).ii- γ ( a ) = γ ( b ). In this case γ restricted to [ a, b ] is a topological em-bedding of S .It follows that in both cases γ restricted to ( a, b ) is a smooth embed-ding.To finish, assume that t ∈ ( a , b ) and t ∈ ( a , b ), with ( a , b ) , ( a , b ) being two intervals making up A n , with γ ( t ) = γ ( t ). Con-sider the set of points in ( a , b ) whose image under γ is contained inthe image of ( a , b ). Again, this set is closed (in the subspace topol-ogy of ( a , b ), being the preimage of γ ([ a , b ])), as well as open (by thefact above about the existence of intervals with coinciding images). Thus γ (( a , b )) ⊂ γ (( a , b )), and obtaining the opposite inclusion via a simi-lar argument we arrive at the fact that γ (( a , b )) = γ (( a , b )). We thushave two embeddings of intervals whose images coincide. Since these are -dimensional manifolds, the two embeddings are related by a change ofparametrization and possibly a switch in orientation.The above theorem shows that any C curve behaves in a rather simpleway once A is ignored. This removed set is small topologically in the sensethat it is sufficient to know the definition of the curve on its complementto be able to extend it uniquely by continuity to the whole interval. Notehowever, that neither A nor its image can be assumed to be small in themeasure-theoretic sense. In fact, the Hausdorff dimensions of these twosets may be equal to one.Using the above theorem we can associate a transfinite word to any C curve. Fix an orientation for each one of the disjoint arcs formingthe curve. Let T be the set of all the intervals making up S ∞ n =1 A n withlinear order inherited from R . Choose a countable alphabet A , one letterfor each one of the arcs. The word associated to the curve is just themap taking every element of T either to the letter corresponding to thearc or to its inverse depending on whether the arc is traversed by γ alongthe chosen orientation when restricted to this element of T or oppositely.This set of maps can be multiplied and reduced in essentially the sameway as the set of finite words [8].We can now make an important Definition 2.
A ‘whisker’ is a curve whose reduced word is trivial (empty). A curve ‘has whiskers’ if its word is different from its reduced word.
It is easy to see that a transfinite word reduces to a trivial one if andonly if every finite truncation of it (i.e. the word obtained by keeping onlyfinitely many of the letters) is reducible. Moreover, this is equivalent withthe statement that there is a pairing between any letter in the word withits inverse such that if a appears before b in the word then the a − thatis paired with it appears after the corresponding b − .Before we state the next theorem, let us formally define thin homotopyand trees: Definition 3. If γ and γ are two curves such that γ (0) = γ (0) , γ (1) = γ (1) , ˙ γ (0) = ˙ γ (0) = ˙ γ (1) = ˙ γ (1) = 0 , then we say that these twocurves are thinly homotopic if there is a map H : I × I → M such that H ( t,
0) = γ ( t ) H ( t,
1) = γ ( t ) H (0 , s ) = γ (0) = γ (0) H (1 , s ) = γ (1) = γ (1) ∂H ( t, ∂s = ∂H ( t, ∂s = ∂H (0 , s ) ∂t = ∂H (1 , s ) ∂t = 0 rank ( H ) ≤ Intuitively it is a homotopy which ‘sweeps’ zero area and which ‘stops’,‘comes to a halt’ at the edges.
Definition 4.
Let l be the space of absolutely convergent sequences withits natural norm. A tree is a compact, path-connected, simply connectedsubspace of l consisting of a closed, totally disconnected set whose ele-ments are called vertices, and a countable collection of straight line seg-ments called edges. All the points in an edge have all coordinates constantbut one. Moreover, this special coordinate for one edge is different from The name ‘whisker’ is motivated by the geometry of the curve whose word is abb − c . hat for any other edge. The endpoints of any edge are contained in theset of vertices, the interiors of the edges are disjoint from each other andfrom the vertices. Note that according to this definition, a tree is not nec-essarily a CW -complex. We shall assume that we have a preferred vertexwhich will be called the root of the tree. We are now ready to state our next
Theorem 2.
A curve with whiskers is thinly homotopic to a curve with-out.Proof.
We proceed in four steps:
Step 1: Any whisker factors through a tree. More precisely, there is atree T and two maps ˜ γ : I → T and f : T → M , such that γ = f ◦ ˜ γ with ˜ γ continuous and f Lipschitz with constant 1.
Consider the word that corresponds to the curve. We know that everyletter appears a finite number of times together with its inverse. Replaceany letter which is repeated n times with n different letters, making sureto preserve the pairing between the letter and its inverse. Of course,with this every letter appears precisely twice (once itself and once as aninverse). Whenever needed, interchange the letter with its inverse so thatthe letter appears first. Of course, the resulting word is still reducible.Consider now the following geometric arrangement:Embed I in the closed upper half plane in the natural way (on the x -axis). For any open interval in one of the sets { A n } ∞ n =1 , we know thatthere is another open interval (in the same A n in fact) such that if the firstone corresponds to a letter then the second one corresponds to the inverseof the letter. Associate to every such pair the open set contained betweenthe semicircle in the upper half plane based at the outer endpoints of thetwo intervals and the semicircle based at the inner endpoints. We definea partial order on these open semi-annuli by saying that the one before isthe one ‘above’. Figure 1 should make this clear. We will often use theword ‘above’ instead of ‘before’ or ‘preceds’ and the word ‘below’ insteadof ‘later’, ‘after’ or ‘succeeds’.Perform this for every pair of intervals. Note that the sets obtained fordifferent pairs do not intersect. Take now the set of all these semi-annulias well as the set { ( x, y ) ∈ R : y ≥ , ( x − ) + y > } . Extendthe partial order on the semi-annuli to include this set by making it theminimal element. We will also call this minimal element a semi-annulus.Let us analyze how a path connected component of the complement ofthis set would look like.Let V be such a component. Consider the set of all semi-annuli aboveit. If there is a latest one, whose smaller boundary circle has equation( x − x ′ ) + y = ρ with ( x ′ ,
0) being its center and ρ its radius, then V cancontain only points satisfying ( x − x ′ ) + y ≤ ρ . If, on the other hand,there is no latest one, consider the net of inner semi-circles of these semi-annuli (the index set of the net being the semi-annuli under considerationwith their partial order). For each element in this net, consider the threenets which are the net of the left endpoints, right endpoints and radii ofthe semicircles. Since the first two are monotone, they converge and thusthe third one converges too and together the limits of all three define a More precisely, we could say that the one before is the one which if combined with itsreflection across the x -axis will contain the other one. Here we are abusing the terminology slightly, since the partial order was only defined forsemi-annuli, but the meaning of this statement should be clear. a − b b − Figure 1: Three semi-annuli are shown. If we denote the vertically shaded oneby x , the horizontally shaded one by y and the diagonally shaded region by z ,then z preceds x which is succeeded by y . In fact, z is the minimal semi-annulus.Note that all the three regions are open sets in the subspace topology of theclosed upper half-plane. semi-circle. This is because if these three nets are l α , r α , ρ α , then theysatisfy 2 ρ α = r α − l α , and so, their limits will also satisfy this relation. If x ′ = lim α ( r α + l α ) and ρ = lim α ρ α , then it is easy to see that V canonly contain points satisfying ( x − x ′ ) + y ≤ ρ .Now, if there is no semi-annulus which is contained in ( x − x ′ ) + y ≤ ρ , then V is in fact equal to this set (restricted to nonnegative y ’s ofcourse). For reasons which will be apparent later we shall call such V ’s‘tip’ regions.If there is a semi-annulus which is contained in { ( x, y ) : y ≥ , ( x − x ′ ) + y ≤ ρ } , consider all semi-annuli which precede it but whichsucceed all the semi-annuli preceding V . Performing the same analysis asabove, either taking a minimal semi-annulus (with respect to the partialorder) or taking a net with it being ordered oppositely to the semi-annuli,we can see that V can only contain points which satisfy ( x − x ) + y ≥ ρ for some x and ρ .Now, if there are no semi-annuli contained in { ( x, y ) : y ≥ , ( x − x ′ ) + y ≤ ρ } which do not succeed ( x − x ) + y = ρ then of course V is the intersection of { ( x, y ) : y ≥ , ( x − x ′ ) + y ≤ ρ } and { ( x, y ) : y ≥ , ( x − x ) + y ≥ ρ } . Such V ’s will be called ‘corner’ regions.Finally, in the other case (i.e. when there are semi-annuli containedin { ( x, y ) : y ≥ , ( x − x ′ ) + y ≤ ρ } etc.) repeat the procedure whichled to ( x − x ) + y = ρ starting from another semi-annulus and obtainanother semicircle ( x − x ) + y = ρ such that V is constrained to lieabove it. Note that the new semi-circle is disjoint from the first one.Proceeding in this fashion, we arrive at a (possibly countable) collectionof semicircles such that V is the region above them. Such V ’s will be ip regionsCorner regionsBranch regions Figure 2: This is a drawing of the different vertex regions. All the regions areclosed and so the boundaries are assumed to be included. Note that a tip regioncan be just a point on the x -axis, while a corner region may be just a circle.The shown possibilities are not exhaustive.10 alled ‘branch’ regions. Note that the set of branch regions is countable,due to the fact that each such region has a nonempty interior. This is atechnical point which will be relevant later.The reader should consult figure 2 for drawings of the different cases.We shall call the tip, corner and branch regions collectively ‘vertex’ re-gions. We shall extend the partial order defined on the semi-annuli to thevertex regions in the obvious way as we have done above. Note that thereis a minimal (in the set of vertices) vertex region: it is the region whichis bounded from above by the minimal semi-annulus. We shall call thisvertex region, the root region.It is not difficult, though a little tedious, to see that, for each of thecases above, the curve takes all the points of V which lie on the x -axisto the same image. This follows from the fact that the curve restrictedto an interval in A n traverses the same image as when restricted to theinterval which is paired with it, from the fact that the derivative of thecurve vanishes in the interior of A (the complement of S ∞ n =1 A n ) andfrom continuity (this is needed when taking limits of the nets).Before we define our tree, we still need to make some further refine-ments. Consider any point in A . Then this point naturally splits theword associated to the curve into two parts, the one ‘before’ and the one‘after’. Note that both the words before and after are the same for all thepoints in A which are in the same connected component, and thus we cantalk about the words before and after a component of A . Given a word,we can talk about an initial/final segment of it, which is defined to be therestriction of the map defining the word to an initial/final segment of thelinearly ordered set defining this word. Also, note that the intersection ofany corner region with the x -axis always has two connected components.Using the above we can now distinguish two types of corner regions: • ‘Spurious’ : These are the ones which satisfy the following two con-ditions: – Let us say the two components of this region on the x -axisare C and C with C being to the left of C . Then thereis a final segment of the word before C which is paired withan initial segment of the word after C . Similarly, there is aninitial segment of the word after C which is paired with thefinal segment of the word before C . – The curve is C if parametrized by proper length in the neigh-borhood of the images of both C and C . • ‘True’ : All of the rest of the corner regions.Having classified all the corner regions into the above two classes,we redefine the decomposition of the domain of the curve by fusing allthose paired intervals which are interspersed by the (restriction to the x -axis of the) spurious corner regions. More precisely, if w is paired with w − and all the regions between the letters of w (and consequently of w − ) are of the spurious corner type, we replace w by the smallest openinterval containing all the intervals corresponding to all the letters of w and perform the same for w − . We then replace the word w with oneletter and w − with its inverse, keeping them paired together. We proceedrecursively until there are no more spurious corner regions left. Thus since Proper length is defined of course by l ( t ) = R t p ˙ γ ( s ) ds . Note that a sufficient conditionfor the curve to be C as a function of the proper length is the non vanishing of ˙ γ , but thatthis is not necessary. e only have true corner regions left, we shall simply call them cornerregions from now on.Now, let us define the tree. It shall be embedded as a subspace of l . Let the letters appearing in the word of the curve be given by thesequence { a n } ∞ n =1 , and let the proper length of the arcs correspondingto them be { l n } ∞ n =1 (we will often abuse notation and say that ‘ l i is thelength of a i ’ or it is ‘the length of the semi-annulus corresponding to a i ’).To any vertex region, associate the point in l whose i -th coordinate is l i if the semi-annulus corresponding to a i is above this region, and is 0otherwise. The set of all such points will be the vertices of the tree (hencethe name of these regions). The edges of the tree will be in one to onecorrespondence with the letters of the word. The edge corresponding to a i will stretch from the vertex corresponding to the vertex region whichbounds the semi-annulus associated to a i from above to the one associatedto the vertex region bounding it from below along the i -th axis. We shalldenote the tree by T . Note that excluding the endpoints of any edge, therest of the edge is an open subset of T , and thus the set of vertices is closedin T . It is not hard to see that the vertices are a totally disconnected setas, for any two vertex regions, there is a semi-annulus which is above oneof these regions which is not above the other one. The names for thevertex regions discussed above should be self-explanatory now, as the tipregions correspond to the tips of the tree. The corner regions correspondto corners and the branch regions correspond to the points where the treesplits into several branches. We shall call the origin of l the root of thetree, and of course, it corresponds to the root region. Intuitively, T is justthe Poincar´e dual of the collection of vertex regions and semi-annuli.Let us show that T is compact. Let { x n } ∞ n =1 be a sequence of elementsin T . We need to show that it has a convergent subsequence in T . It issufficient to handle the case when this sequence has no subsequence whichis contained inside an edge of T , for then this subsequence will converge toan element in this edge as an edge is compact. Therefore, without loss ofgenerality, assume that no two x n ’s are in the same edge or vertex. Definenow a sequence { t n } ∞ n =1 of elements of I by letting t n be the left endpointof the interval corresponding to the letter corresponding to the edge in thecase x n belongs to the interior of an edge, while in case x n is a vertex, let t n be any of the points in the corresponding vertex region which are in I .The sequence { t n } ∞ n =1 has a convergent subsequence { t n k } ∞ k =1 such that t n k → t ∈ A . Let x be the vertex corresponding to the vertex regionto which t belongs. We then have that x n k → x . The reason is that if s , s ∈ I are in vertex regions, then the l distance between the verticescorresponding to these two regions is less than or equal to | R s s p ˙ γ ( s ) ds | as this integral is less than or equal to the sum of the lengths of all thesemi-annuli which precede one of the regions without preceding the otherone. Thus T is compact. Note that essentially the same proof shows thatany sequence of vertices has a subsequence converging to a vertex andthus the set of vertices is compact as well.To show that T is a tree we need to show that it is path-connectedand simply connected. It is obvious that if for any vertex we have acontinuous path connecting the root to it, then T is connected. Pick avertex, or equivalently, a vertex region. Then there is a subcollection ofsemi-annuli which are above it. Let L be the sum of the l i ’s correspondingto the letters corresponding to these semi-annuli, and define the path σ ( s )from [0 , L ] to T in the following way:Consider the set of all the vertices preceding the vertex in question. ote that this set of points is in fact a closed subset of T . This is becauseit is equal to the intersection of two closed sets: the set of all vertices withthe hyperplane which has all coordinates equal to 0 for all the semi-annulinot preceding the vertex in question. Now, for every vertex region v above the vertex in question, let s v be the sum of the lengths of the semi-annuli above v . We thus obtain a linearly ordered set of real numberswhich is closed as the map ( α , α , . . . ) → α + α + . . . is a continuousmap from l to R . Note that this set of vertices is in fact linearly orderedand that if v precedes v with no other vertices in between then there isan edge whose initial point is v and final point v . These facts are easyconsequences of the geometry of the vertex regions. Let s ∈ [0 , L ]. If thereis a v such that s v = s , let σ ( s v ) = v . Otherwise, s is in the complementof the closed set and thus belongs to an interval ( s v , s v ) such that thereare no s v ’s in it. In this case, let σ take s to the edge linking v and v ,going along it (in the direction of v ) a distance s − s v .From the construction of σ it follows easily that | σ ( s ) − σ ( s ) | = | s − s | , and thus σ is (uniformly) continuous and T is path connected.In step 3, we shall construct an explicit deformation retraction of T toits root, which will imply that T is simply connected.Having defined our tree, let us show how the curve factorizes throughit. Let us define ˜ γ : I → T first. Note that any point t ∈ I belongseither to a vertex region or to a semi-annulus. In the case it belongsto a vertex region let ˜ γ ( t ) be the vertex corresponding to that region.In the other case, if t is in the earlier interval of the intersection of thesemi-annulus with I , denoted by ( t , t ), let ˜ γ ( t ) be the point on the edgecorresponding to the semi-annulus whose distance from its initial vertex(i.e. the vertex corresponding to the vertex region which contains { t } )is equal to R tt p ˙ γ ( s ) ds . If t is in the second interval, which we will alsodenote by ( t , t ), ˜ γ ( t ) is defined via the same formula but the distance isnow measured from the final vertex (the one corresponding to t ). It is nothard to see that ˜ γ satisfies the inequality | ˜ γ ( t ) − ˜ γ ( t ) | ≤ R t t p ˙ γ ( s ) ds from which continuity of ˜ γ follows. Note that we can say that ˜ γ ( t ) is infact C when t maps to an interior of an edge. Let us now define the map f : T → M . For any vertex x , let f ( x )be the image by γ of the intersection of the vertex region associated to v with the x -axis. Let x be on an edge, and suppose the distance from x to the initial vertex of the edge is equal to l . Suppose that the leftinterval forming intersection of the semi-annulus corresponding to thisedge with the x -axis is ( t , t ). Let f ( x ) = γ ( t ( x )) where t ( x ) is such that R t ( x ) t p ˙ γ ( s ) ds = l . It is straightforward to check that indeed, γ = f ◦ ˜ γ ,and that | f ( x ) − f ( y ) | ≤ | x − y | . Step 2: Any whisker is thinly homotopic to a whisker which factorsthrough the same tree and which has a vanishing derivative at every pointwhich is mapped to a vertex.
Let V be the preimage of the set of vertices under ˜ γ . Since ˜ γ is con-tinuous and the set of vertices is closed, we have that V is closed. It isnot hard to see that for any corner region at least one of its components Any other vertex region is going to be either below the vertex in question, and thusseparated from it by some semi-annulus and will have the l coordinate corresponding to itnonzero; or it is going to be ‘next’ to it (as are the bottom two regions in figure 2) in whichcase it is also not difficult to see that there will be a semi-annulus above it which is not abovethe first vertex. When ˜ γ maps into an edge, it is essentially a function into R as only one component ischanging, and thus we can talk about its differentiability. s in C (the set of critical points, where we are using the same notationas in theorem 1. Recall here that we assume that we have gotten rid ofthe spurious corners).Now, define the maps l : I → [0 , L ] and ˆ γ : [0 , L ] → M , via l ( t ) = Z t p ˙ γ ( s ) ds, ˆ γ = γ ◦ l − , where of course L is the total length of the curve, and ˆ γ is just the curveparametrized by proper length. We are abusing the notation somewhat,since l is not invertible, and thus should be read as a preimage, but thisdoes not cause any problems since γ is of course constant on the preimageof any point by l . The reader should note that ˆ γ is a Lipschitz map (withconstant 1) which is in fact C at every point which is not a critical valuefor l (since there l is invertible).Consider an interval ( a, b ) forming the compelment of C . It is clearthat ˆ γ is a C diffeomorphism if restricted to ( a + δ, b − δ ) for any (suf-ficiently small) δ >
0. Since any corner point is a subset of γ − (cid:0) γ ( C ) (cid:1) ,it is clear that the set of the images of the corners which is contained inˆ γ (cid:0) ( a + δ, b − δ ) (cid:1) has 1-dimensional Hausdorff measure zero (Sard’s theoremagain). Which in turn implies that ˆ γ − of this set has measure 0 in [0 , L ].Since the interval and δ > l has measure 0.It is not difficult to see that the points which map to tips are in C , for ina neighborhood of any such point the curve is at least two-to-one. Considernow the branch points. Recall that any branch region intersected with the x -axis has exactly two components. If either of these two components ismore than a point, then that component is of course contained in C .It follows from this, and from the fact that the set of branch regions iscountable, that the set of points which map to the branch points andwhich are not in C is countable.Therefore, we have that l ( V ) is a closed set and is of measure 0. Wenow need the following Lemma. If S ⊂ I is any countable set such that S ∪ { } ∪ { } is closedthen there is a C , monotone increasing, surjective function ψ : I → I ,whose set of critical values is precisely S ∪ { } ∪ { } . Moreover one canassume that A = sup x ∈ I | ψ ′ ( x ) | ≥ is independent of the set S .Proof. It is a minor tweak of the construction presented in [9]. See theAppendix for details.It follows at once, by scaling and shifting, that the same can be saidfor any other interval J , and that the same bound is in fact independentnot only of S , but also of which interval is under consideration. Let usdenote such an adjusted function by ψ J,S .Now, define the function φ : [0 , L ] → [0 , L ] by: φ ( s ) = s If s ∈ l ( C ). ψ [ a,b ] ,l ( V ) ∩ [ a,b ] If t ∈ [ a, b ] where ( a, b ) is one of the open intervalsmaking up [0 , L ] − l ( C ). n other words, the function φ fixes all the points in l ( C ) and has avanishing derivative at any point which corresponds to a branch point. Itis obvious that φ is C at any s / ∈ l ( C ). Moreover, it is easy to see that φ is in fact Lipschitz (with constant A ) for, let 0 ≤ s ≤ s ≤ [0 , L ], then φ ( s ) − φ ( s ) = | φ ([ s , s ]) | = | φ ( l ( C ) ∩ [ s , s ]) | + | φ ([ s , s ] − l ( C )) | = | l ( C ) ∩ [ s , s ] | + | φ ([ s , s ] − l ( C )) |≤ | l ( C ) ∩ [ s , s ] | + A | [ s , s ] − l ( C ) | . ≤ A ( | l ( C ) ∩ [ s , s ] | + | [ s , s ] − l ( C ) | ) = A ( s − s ) . where above | · | stands for the Lebesgue measure on R .Define now the homotopy H : I × I → M via H ( t, r ) = ˆ γ ◦ ρ ◦ l, where ρ ( s, r ) = (1 − r ) s + rφ ( s ) is just the straight line homotopybetween the identity and φ . Let us show that H is C . It is clear that atany point t / ∈ C , the above function has continuous partial derivatives,since the three constituent functions are continuously differentiable at thecorresponding points. Let t ∈ C , then it follows that H ( t + δt, r ) − H ( t , r ) = o ( δt ), as this is an immediate consequence of the fact that l has a vanishing derivative at any such point and from the fact that both ρ and ˆ γ are Lipschitz. Also, at such points we trivially have H ( t , r + δr ) − H ( t , r ) = 0 = o ( δr ). Therefore, H has both partials everywhere.Moreover they both vanish when t ∈ C and are continuous when t / ∈ C .Now fix t ∈ C . Note that ∂H∂t = ˆ γ ′ ∂ρ∂s l ′ (1)at any point t / ∈ C . However, the first two multiplicands are boundedeverywhere (the differentiated functions are Lipschitz) while l ′ vanishesat every point in C and is continuous everywhere. It follows thatlim t → t , t/ ∈ C ∂H∂t = 0 , uniformly in r , and since ∂H∂t = 0 at any t ∈ C , we have that ∂H∂t iscontinuous everywhere. Note that (1) implies that | ∂H∂t | ≤ A sup t ∈ I | l ′ ( t ) | ,where the constant A was defined above.Consider now the other partial. If t / ∈ C , then this partial is given by ∂H∂r = ˆ γ ′ (cid:0) φ ( l ( t )) − l ( t ) (cid:1) . (2)Note that if t ∈ C , then φ ( l ( t )) − l ( t ) = ( t − t )(( φ ◦ l ) ′ ( τ ) − l ′ ( τ )) forsome τ between t and t by mean value theorem. Here we used the factthat φ ◦ l is in fact differentiable, despite φ by itself being only Lipschitzas it is not differentiable precisely at the points where l has a vanishingderivative. Using the fact that φ is Lipschitz and that it is differentiableeverywhere away from C , it follows that φ ( l ( t )) − l ( t ) → t → t .Utilizing the boundedness of ˆ γ ′ again, we get that ∂H∂r is continuous ev-erywhere. Therefore H is C . Also, (2) and mean value theorem againimply that | ∂H∂r | ≤ A sup t ∈ I | l ′ ( t ) | .The fact that H ( t,
0) = γ ( t ) is obvious, and that H ( t,
1) is a whiskerwhich factors through the same tree is immediate from the monotonicity f φ . It is clear that H ( I × I ) ⊂ γ ( I ), which forces H to have rank atmost 1.Finally, to get a thin homotopy we should have the appropriate deriva-tives vanish at the boundary of I × I . It is obvious that the deriva-tive vanishes at the two vertical edges forming ∂ ( I × I ). If we let, e.g.,˜ H ( t, r ) = H ( t, ψ ( r )) we see that ˜ H is the thin homotopy that we want. Step 3: Any whisker is thinly homotopic to the constant map.
The reader is urged at this point to read the Appendix as a closelyrelated argument will be utilized in this step.It was proven in steps 1 and 2 that γ factors through a tree (we haven’tproven yet that it is simply connected), i.e. γ = f ◦ ˜ γ , where f is Lipschitzwith constant 1, ˜ γ is continuous everywhere, C in the interior of everyedge, and without loss of generality is o ( δt ) when it is mapped to a vertexas we can assume that we have thinly homotoped the curve to one whichhas a vanishing derivative at any point mapping to a vertex.Consider now the set of edges of the tree. Arrange them so thattheir lengths { l n } ∞ n =1 are non-increasing. Assume that P ∞ n =1 l n = L ′′ .Associate to every edge an interval l ′ n such that P ∞ n =1 l ′ n = L ′ < ∞ , l n ≤ l ′ n and lim n →∞ l n l ′ n = 0.Recall that in step 1 we have constructed for any vertex x a continuouspath σ : [0 , | x | ] → T , linking the root to this vertex. This path satisfiedthe property that | σ ( s ) − σ ( t ) | = | s − t | , and in particular | σ ( s ) | = s . Sinceevery edge begins with a vertex, it is trivial to define an analogous mapwhen x is in the interior of any edge of T . Thus, for any x ∈ T , we havea path denoted by σ x : [0 , | x | ] → T , such that σ x ( s ) is the point on thispath a distance s from the root. It is obvious that σ x is Lipschitz withconstant 1 and if σ x ( s ) is in the interior of an edge, then σ x is C at sucha point.Note that if x is a vertex, then | x | is the sum of the lengths of all theedges traversed by σ x . Define | x | ′ to be the sum of the l ′ n ’s of the edgesin this path. For any vertex x , denote by ρ x the function which takes[0 , | x | ′ ] → [0 , | x | ], which is C , monotone increasing, and when s = | y | ′ for some vertex y along this path, ρ x ( s ) = | y | and dds ρ x ( s ) = 0. This isdone by mapping s = | y | ′ to s = | y | for any vertex y on this path and byinterpolating by a (scaled and shifted) version of the same monotone, C function with vanishing derivative at the endpoints. The proof that theresulting function is C is a version of the proof in the Appendix.The crucial fact about σ x and ρ x is that if y is a vertex on the path σ x then ρ x = ρ y and σ x = σ y on the common domains as is evident fromthe construction of these functions. This fact guarantees the consistencyof the definition of the map χ which shall be given momentarily.Let χ : T × L ′ → T be defined in the following way: • If x is a vertex, let χ ( x, r ) = (cid:26) x if r ≤ L ′ − | x | ′ , σ x ◦ ρ x ( L ′ − r ) if r ≥ L ′ − | x | ′ . • If x is a point in the interior of an edge e whose initial vertex is v and whose final vertex is v , and | x − v | = α | v − v | , let χ ( x, r ) = x if r ≤ L ′ − | x | ′ , x + α χ ( x , r ) if L ′ − | x | ′ ≤ r ≤ L ′ − | x | ′ .χ ( x , r ) if L ′ − | x | ′ ≥ r. ote that the above map is in fact a continuous deformation retractionof T to its root, and thus we have shown that T is simply connected.Now, define the homotopy H : I × L ′ → M via: H = f ◦ χ ◦ ˜ γ. Roughly speaking, T is contracted within itself using χ , such that thetips ‘stop’ whenever they pass by a vertex. It is clear, intuitively, that H is C as the only troublesome points are the vertices and the holonomy‘comes to a halt’ at all such points, in both variables. At any rate, it is nothard, though quite tedious, to verify this rigorously. First, by consideringthe different cases, one can show ∂H∂t and ∂H∂s exist everywhere. There arefive cases to consider for each partial, corresponding to the five possibilitiesin the definition of χ . It is easy to see that both partials vanish wheneverthe image of a point under χ ◦ ˜ γ is a vertex, and that they satisfy theinequalities | ∂H ( t,s ) ∂t | ≤ | ˙ γ ( t ) | and | ∂H ( t,s ) ∂s | ≤ ρ ′ x ( L ′ − s ) where x is anyvertex succeeding ˜ γ ( t ). Using the fact that ∂H ( t ,s ) ∂t = 0 if t is in avertex region and the first inequality, it is easy to see that this partial iscontinuous. Continuity of the other partial is a little more subtle: oneneeds to use the fact that H ( t , s ) is a C function as a function of s if t and then considering the various possibilities for the structure of theword around t , essentially, whether there are infinitely many letters inany neighbourhood of t or not.It is trivial to check that H ( t,
0) = γ and H ( t, L ′ ) is the constant path,and it is obvious that H has rank at most 1 as its image is contained in γ ( I ).Finally, scaling and composing H with a function which vanishes atthe endpoints we obtain our thin homotopy. Step 4: A curve with whiskers is thinly homotopic to one without.
Assume the curve has whiskers, and consider the subwords which whenreduced would be trivial. By concatenation and inclusion, we can assumethat any such word is maximal (i.e. is not contained in a larger subwordwhich is trivial when reduced). It follows that any two such words areseparated by a nontrivial letter. Take the closure of the preimage of eachsuch word. It should be clear that any two such preimages are disjoint(otherwise the two words could be concatenated and would form a largerreducible one). Each one of these preimages is a closed interval, but theset of all such intervals is not necessarily closed. However, since every treecontains at least one tip, and thus a point where the derivative vanishes,it follows that any limit point of these intervals which is not in one ofthem is in C .Now, for any whisker consider its root. Take its preimage under γ and note that any interval in I − C will contain at most two such points(belonging to two different whiskers, as there is always a point in C in the‘interior’ of every whisker). Performing a similar procedure to the one instep 2 above (but simpler since S now has at most two points), we canhave a thin homotopy between the given curve to the one which has thederivative vanish at the root of every whisker.Now, we can just apply the previous three steps to each one of thewhiskers, performing all the homotopies simultaneously. The only subtlepoint is that one should take the edges of the entire curve and not ofthe individual whiskers separately when defining the sequence { l ′ n } ∞ n =1 instep 3. Clearly, the two partials exist everywhere and are continuous awayfrom the limit points of the set of roots of the whiskers. Using the twoinequalities on partials at the end of the previous step it is easy to show hat the partials are continuous at such limit points as well. Thus the re-sulting ‘total’ homotopy is C . And since the images of those homotopiesare always contained in the image of the curve the total homotopy is ofrank at most one.The fact that the resulting curve has no whiskers is obvious and theproof is complete.We can now prove the following Theorem 3.
Assume γ : I → M is a loop with a vanishing derivative atthe endpoints. Then the following are equivalent:a- γ is a whisker.b- γ is thinly homotopic to the constant loop.c- If G is a semi-simple Lie group, then the holonomy of any smooth G -connection around γ is trivial.d- γ factors through a tree via continuous maps.Proof. ( a ) = ⇒ ( b ) is nothing but a special case of theorem 2 above.( b ) = ⇒ ( c ) was proven in [5].To see that ( c ) = ⇒ ( a ), assume γ is not a whisker. Thus, its reducedword is nontrivial. Therefore, one of the truncations to finitely manyletters is not reducible. Let us denote the letters in this truncation by a , . . . , a n and the corresponding word by ω . Consider the images ofthe intervals corresponding to these letters. They are, when restrictedto their interiors, disjoint embeddings. It follows at once that given anyLie group G , and an n -tuple of group elements g , . . . , g n , that there is asmooth G -connection whose parallel transport along the loop is the imageof ( g , . . . , g n ) under the word map G n → G defined by the (truncated)word. However, it was shown in [10], that the image of a word map G n → G defined by a nontrivial word, would contain a nonempty opensubset of the identity if G is compact and semi-simple. In particular itwould contain nontrivial elements. It is not difficult to see that even if G is not compact the image of the word map will be nontrivial. Thereason for this is that since G is semi-simple then, if K is its analyticsubgroup (from the Cartan decomposition) and Z is its center, then K/Z is a compact semi-simple group. It is clear that the word map descendsto the quotient (
K/Z ) n . The image of this descended map is non-trivial,and thus its image before quotienting must have been non-trivial as well.And thus, there is a smooth G -connection whose holonomy around γ isnontrivial.( a ) = ⇒ ( d ) is steps 1 and 3 of theorem 2. It remains to prove that( d ) = ⇒ ( a ). Suppose that the curve factorizes through a tree. Weshall shift the tree to put its root at the origin. We will be done if wecan show that the word associated to the curve is reducible. Suppose γ = f ◦ ˜ γ with ˜ γ : I → T . It is easy to see that if a and b are twodifferent letters in the word, then their images (i.e. the images of theopen intervals corresponding to them) under ˜ γ are disjoint. Additionally,it follows, using theorem 1 and simple connectivity of the tree, that theimages of the same letter under ˜ γ are either identical or disjoint. Weshall treat different occurrences of the same letter which have differentimages under ˜ γ as different letters. For any letter, pick an edge whoseinterior is contained in the image of the letter under ˜ γ . Due to the waywe have assumed our trees to be embedded in l , it follows that | ˜ γ ( t ) | isa monotone function of t when restricted to map to an edge. Call a letter ositive if this function is increasing and negative if it is decreasing. It iseasy to see that any letter must appear an even number of times, beingalternatingly positive and negative, with the first occurrence in the wordbeing positive. Pair a positive occurence of the letter with the first nextnegative one. It is straightforward to check that this pairing reduces theword to a trivial one and the proof is complete.We now have the following immediate corollary: Corollary : If γ and γ are two curves with coinciding endpointsat which their derivatives vanish then the following are equivalent:a- The reduced word associated to γ is the same as that for γ .b- γ is thinly homotopic to γ .c- γ and γ are holonomically equivalent for a semi-simple G .d- γ · γ factors through a tree via continuous maps. Appendix
We sketch here the proof of the modification of the results of [9] that wasused above.
Lemma.
Assume S ⊂ I is a closed set, of measure 0 which contains both { } and { } . Then there is a monotone increasing surjective function ψ : I → I whose set of critical values is precisely S . Moreover, one canassume that A = sup x ∈ I | ψ ′ ( x ) | ≥ is independent of S .Proof. The complement of S is a countable collection of disjoint openintervals. Note that it is naturally a linearly ordered set which we shalldenote by I . Assume the lengths of these intervals are { l n } ∞ n =1 , andhave been ordered to be non-increasing. Since | S | = 0, if follows that P ∞ n =1 l n = 1. It is easy to see that there is a non-increasing sequence { l ′ n } ∞ n =1 which satisfies: • l ′ n ≥ l n for all n ∈ N . • P ∞ n =1 l ′ n = 2. • lim n →∞ l n l ′ n = 0.Let f : I → I be a C , monotone increasing, onto function whosederivative vanishes only at { } and { } , such that sup x ∈ I | f ′ ( x ) | = 2.Suppose x ∈ S , let g ( x ) = P i ∈I ,i 2] is a continuous, monotone injection. Therefore S ′ = g ( S )is a closed set. Thus, its complement is a countable collection of openintervals. Suppose, ( y , y ) = ( g ( x ) , g ( x )) is one such interval. Notethat if x − x = l n then y − y = l ′ n . It follows that we have a naturalbijection between the elements of I and the complement of S ′ , such thatan interval of length l n corresponds to an interval of length l ′ n .Define a function b ψ to be simply an extension of g − from S ′ to allof [0 , 2] using a scaled, translated version of f to map the complementof S ′ onto the complement of S . More precisely, if a ′ is the endpoint ofan interval forming the complement of S ′ , and a is the endpoint of thecorresponding interval forming the complement of S , then b ψ restricted to[ a ′ , a ′ + l ′ ] is given by: b ψ ( x ) = a + lf (cid:16) x − a ′ l ′ (cid:17) t is obvious that b ψ is monotone, that it is C on the complement ofthe set of limit points of S ′ , that it is differentiable everywhere, and thatits set of critical values is precisely S . That b ψ is C everywhere followseasily from the fact that lim n →∞ l n l ′ n = 0. Finally, it is trivial to verifythat sup x ∈ [0 , | b ψ ′ ( x ) | ≤ 2. Setting ψ ( x ) = b ψ ( x ) we obtain the functionwe want. References [1] S. Kobayashi, ‘La connexion des varietes fibrees. II’, C. R. Acad. Sci.Paris , (1954), 318–319.[2] K. Chen, ‘Integration of paths - a faithful representation of paths bynon-commutative formal power series’, Trans. Amer. Math. Soc. ,(1958), 395–407.[3] R. Hain, ‘The geometry of the mixed Hodge structure on on thefundamental group’, Algebraic geometry, Bowdoin, 1985 (Brunswick,Maine, 1985), 247–282, Proc. Sympos. Pure Math., , Part 2, Amer.Math. Soc., Providence, RI, 1987.[4] A. Ashtekar, J. Lewandowski, ‘Representation theory of analyticholonomy C ∗ -algebras’, in Knots and quantum gravity (Riverside,CA, 1993), 21–61, Oxford Lecture Ser. Math. Appl., , Oxford Univ.Press, New York, (1994).[5] A. Caetano, R. F. Picken, ‘An Axiomatic Definition of Holonomy’,Internat. J. Math. , (1994), no. 6, 835–848.[6] B. Hambly, T. Lyons, ‘Uniqueness of the signature of a path ofbounded variation and the reduced path group’, Ann. of Math. (2) , (2010), no.1, 109–167.[7] A. Sard, ‘Hausdorff Measure of Critical Images on Banach Manifolds’,Amer. J. Math. , (1965), 158–174.[8] J. W. Cannon, G. R. Conner, ‘The Combinatorial Structure of theHawaiian Earring Group’, Topology Appl. , (2000), no.3, 225–271.[9] S. M. Bates, A. Norton, ‘On Sets of Critical Values in the Real Line’,Duke Math. J. , (1996), no. 2, 399 – 413.[10] A. Borel, ‘On free subgroups of semi simple groups’, Enseign. Math.(2) , (1983), no. 1-2, 151–164. Department of Mathematics, American University of Beirut, Beirut, Lebanon.Email : [email protected]@aub.edu.lb