OOn Jones’ subgroup of R. Thompson group F Gili Golan, Mark Sapir ∗ April 17, 2015
Abstract
Recently Vaughan Jones showed that the R. Thompson group F encodes in a natural wayall knots and links in R , and a certain subgroup −→ F of F encodes all oriented knots and links.We answer several questions of Jones about −→ F . In particular we prove that the subgroup −→ F is generated by x x , x x , x x (where x i , i ∈ N are the standard generators of F ) and isisomorphic to F , the analog of F where all slopes are powers of and break points are -adicrationals. We also show that −→ F coincides with its commensurator. Hence the linearizationof the permutational representation of F on F/ −→ F is irreducible. We show how to replace in the above results by an arbitrary n , and to construct a series of irreducible representationsof F defined in a similar way. Finally we analyze Jones’ construction and deduce that theThompson index of a link is linearly bounded in terms of the number of crossings in a linkdiagram. Contents F F as a group of homeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 F as a diagram group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.3 A normal form of elements of F . . . . . . . . . . . . . . . . . . . . . . . . . . 62.4 From diagrams to homeomorphisms . . . . . . . . . . . . . . . . . . . . . . . 72.5 From diagrams to pairs of binary trees . . . . . . . . . . . . . . . . . . . . . . 8 −→ F and its properties 10 −→ F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104.2 The subgroup −→ F is isomorphic to F . . . . . . . . . . . . . . . . . . . . . . . 114.3 −→ F is the stabilizer of the set of dyadic fractions with odd sums of digits . . . 12 ∗ The research was supported in part by the NSF grants DMS 1418506, DMS 1318716 and by the BSF grant2010295. a r X i v : . [ m a t h . G R ] A p r The subgroup −→ F n −→ F n − is isomorphic to F n . . . . . . . . . . . . . . . . . . . . . 155.3 The embeddings of F n into F are natural . . . . . . . . . . . . . . . . . . . . 19 -page book embeddable graphs to Thompson graphs . . . . . . 266.6 Step 4: From Thompson graphs to elements of F . . . . . . . . . . . . . . . . 306.7 The Thompson index of a link . . . . . . . . . . . . . . . . . . . . . . . . . . . 306.8 Some open questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326.8.A Unlinked elements of F . . . . . . . . . . . . . . . . . . . . . . . . . . 326.8.B Positive links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326.8.C Random links . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 A recent result of Vaughan Jones [11] shows that Thompson group F encodes in a naturalway all links (this construction is presented in Section 6 below). A subgroup of F , calledby Jones the directed Thompson group −→ F , encodes all oriented links. In order to define −→ F ,Jones associated with every element g of F a graph T ( g ) using the description of elements of F as pairs of binary trees (see Section 3 for details). The group −→ F is the set of all elementsin F for which the associated graph T ( g ) is bipartite. Jones asked for an abstract descriptionof the subgroup −→ F . For example, it is not clear from the definition whether or not −→ F isfinitely generated.We define the graph T ( g ) in a different (but equivalent) way. By [5] F is a diagram group.For every diagram ∆ in F the graph T (∆) is a certain subgraph of ∆ . Then, the subgroup of F composed of all reduced diagrams ∆ in F with T (∆) bipartite is Jones’ subgroup −→ F . Usingthis definition we give several descriptions of −→ F . Recall that for every n ≥ one can definea “brother” F n of F = F as the group of all piecewise linear increasing homeomorphismsof the unit interval where all slopes are powers of n and all breaks of the derivative occurat n -adic fractions, i.e., points of the form an k where a, k are positive integers [1]. It is wellknown that F n is finitely presented for every n (a concrete and easy presentation can befound in [5]). Theorem 1.
Jones’ subgroup −→ F is generated by elements x x , x x , x x where x i , i ∈ N , are the standard generators of F . It is isomorphic to F and coincides with the smallestsubgroup of F which contains x x and is closed under addition (which is a natural binaryoperation on F , see Section 2.2).This theorem implies the following characterization of −→ F which can be found in [11]. Theorem 2.
Jones’ subgroup −→ F is the stabilizer of the set of dyadic fractions from the unitinterval [0 , with odd sums of digits, under the standard action of F on the interval [0 , .As a corollary from Theorem 2 we get the following statement answering a question byVaughan Jones. Corollary 3.
Jones’ subgroup −→ F coincides with its commensurator in F . s noted in [11], this implies that the linearization of the permutational representationof F on F/ −→ F is irreducible (see [14]).In [11], Jones introduced the Thompson index of a link which can be defined as thesmallest number of leaves of a tree diagram (= the number of vertices minus one in thesemigroup diagram) representing that link. The construction in [11] does not give an estimateof the Thompson index. But analyzing and slightly modifying that construction (using someresults from Theoretical Computer Science) we prove that the Thompson index of a link doesnot exceed 12 times the number of crossings in any link diagram.The paper is organized as follows. In Section 2 we give some preliminaries on Thompsongroup F . In Section 3 we give Jones’ definition of the Thompson graph associated withan element of F , we also give the definition in terms of semigroup diagrams [5] and provethe equivalence of two definitions. In Section 4 we define Jones subgroup −→ F and proveTheorems 1 and 2 and Corollary 3. Section 5 contains generalizations of the previous resultsfor arbitrary n . In particular, we show that for every n ≥ , the smallest subgroup of F containing the element x · · · x n − and closed under addition, is isomorphic to F n , can becharacterized in terms of the graphs T (∆) , is the intersection of stabilizers of certain sets ofbinary fractions, and coincides with its commensurator in F . Although Theorems 1 and 2are special cases of the results of Section 5, the direct proofs of these results are much lesstechnical while ideologically similar, and the original questions of Jones concerned the case n = 3 only. Thus we decided to keep the proofs of Theorems 1 and 2. In Section 5 we showhow to adapt these proofs to the general case. In Section 6, we analyze Jones’ constructionand prove the linear upper bound for the Thompson index of a link. Acknowledgments.
We are grateful to Vaughan Jones for asking questions about −→ F .We are also grateful to Victor Guba and Michael Kaufmann for helpful conversations. F F as a group of homeomorphisms The most well known definition of the R. Thompson group F is this (see [3]): F consists of allpiecewise-linear increasing self-homeomorphisms of the unit interval with all slopes powersof and all break points of the derivative dyadic fractions. The group F is generated by twofunctions x and x defined as follows. x ( t ) = t : 0 ≤ t ≤ t + : ≤ t ≤ t + : ≤ t ≤ x ( t ) = t : 0 ≤ t ≤ t − : ≤ t ≤ t + : ≤ t ≤ t + : ≤ t ≤ One can see that x is the identity on [0 , ] and a shrank by the factor of 2 copy of x on [ , . The composition in F is from left to right.Equivalently, the group F can be defined using dyadic subdivisions [1]. We call a subdi-vision of [0 , a dyadic subdivision if it is obtained by repeatedly cutting intervals in half.If S , S are dyadic subdivisions with the same number of pieces, we can define a piecewiselinear map taking each segment of the subdivision S linearly to the corresponding segmentof S . We call such a map a dyadic rearrangement . The group F consists of all dyadicrearrangements. .2 F as a diagram group It was shown in [5, Example 6.4] that the R. Thompson group F is a diagram group overthe semigroup presentation (cid:104) x | x = x (cid:105) .Let us recall the definition of a diagram group (see [5, 6] for more formal definitions). A(semigroup) diagram is a plane directed labeled graph tesselated by cells, defined up to anisotopy of the plane. Each diagram ∆ has the top path top (∆) , the bottom path bot (∆) ,the initial and terminal vertices ι (∆) and τ (∆) . These are common vertices of top (∆) and bot (∆) . The whole diagram is situated between the top and the bottom paths, and everyedge of ∆ belongs to a (directed) path in ∆ between ι (∆) and τ (∆) . More formally, let X be an alphabet. For every x ∈ X we define the trivial diagram ε ( x ) which is just an edgelabeled by x . The top and bottom paths of ε ( x ) are equal to ε ( x ) , the vertices ι ( ε ( x )) and τ ( ε ( x )) are the initial and terminal vertices of the edge. If u and v are words in X , a cell ( u → v ) is a plane graph consisting of two directed labeled paths, the top path labeled by u and the bottom path labeled by v , connecting the same points ι ( u → v ) and τ ( u → v ) .There are three operations that can be applied to diagrams in order to obtain new diagrams.1. Addition.
Given two diagrams ∆ and ∆ , one can identify τ (∆ ) with ι (∆ ) . Theresulting plane graph is again a diagram denoted by ∆ + ∆ , whose top (bottom) path isthe concatenation of the top (bottom) paths of ∆ and ∆ . If u = x x . . . x n is a word in X , then we denote ε ( x ) + ε ( x ) + · · · + ε ( x n ) ( i.e. a simple path labeled by u ) by ε ( u ) andcall this diagram also trivial .2. Multiplication.
If the label of the bottom path of ∆ coincides with the label of thetop path of ∆ , then we can multiply ∆ and ∆ , identifying bot (∆ ) with top (∆ ) . Thenew diagram is denoted by ∆ ◦ ∆ . The vertices ι (∆ ◦ ∆ ) and τ (∆ ◦ ∆ ) coincide withthe corresponding vertices of ∆ , ∆ , top (∆ ◦ ∆ ) = top (∆ ) , bot (∆ ◦ ∆ ) = bot (∆ ) .3. Inversion.
Given a diagram ∆ , we can flip it about a horizontal line obtaining a newdiagram ∆ − whose top (bottom) path coincides with the bottom (top) path of ∆ . (cid:117) (cid:117) ∆ ∆ ∆ ◦ ∆ (cid:117) (cid:117) (cid:117) ∆ ∆ ∆ + ∆ Figure 2.1: The multiplication and addition of diagrams.
Definition 2.1.
A diagram over a collection of cells (i.e., a semigroup presentation) P is anyplane graph obtained from the trivial diagrams and cells of P by the operations of addition,multiplication and inversion. If the top path of a diagram ∆ is labeled by a word u and thebottom path is labeled by a word v , then we call ∆ a ( u, v ) -diagram over P .Two cells in a diagram form a dipole if the bottom part of the first cell coincides withthe top part of the second cell, and the cells are inverses of each other. Thus a dipole is a ubdiagram of the form π ◦ π − where π is a cell. In this case, we can obtain a new diagramby removing the two cells and replacing them by the top path of the first cell. This operationis called elimination of dipoles . The new diagram is called equivalent to the initial one. Adiagram is called reduced if it does not contain dipoles. It is proved in [5, Theorem 3.17] thatevery diagram is equivalent to a unique reduced diagram.Now let P = { c , c , . . . } be a collection of cells. The diagram group DG( P , u ) corre-sponding to the collection of cells P and a word u consists of all reduced ( u, u ) -diagramsobtained from the cells of P and trivial diagrams by using the three operations mentionedabove. The product ∆ ∆ of two diagrams ∆ and ∆ is the reduced diagram obtained byremoving all dipoles from ∆ ◦ ∆ . The fact that DG( P , u ) is a group is proved in [5]. Lemma 2.2 (See [5]) . If X consists of one letter x and P consists of one cell x → x , thenthe group DG( P , x ) is the R. Thompson group F . Since X consists of one letter x , we shall omit the labels of the edges of diagrams in F .Since all edges are oriented from left to right, we will not indicate the orientation of edgesin the pictures of diagrams from F .Thus in the case of the group F , the set of cells P consists of one cell π of the form (cid:114) (cid:114) (cid:114) Figure 2.2: The cell defining the group F . The role of 1 in the group F is played by the trivial diagram ε ( x ) which will be denoteby .Using the addition of diagrams one can define a useful operation of addition on the group F : ∆ ⊕ ∆ = π ◦ (∆ + ∆ ) ◦ π − . Note that if ∆ , ∆ are reduced, then so is ∆ ⊕ ∆ unless ∆ = ∆ = in which case ⊕ = .The following property of ⊕ is obvious: Lemma 2.3.
For every a, b, c, d ∈ F we have ( a ⊕ b )( c ⊕ d ) = ac ⊕ bd. (2.1) In particular a ⊕ b = ( a ⊕ )( ⊕ b ) . Remark 2.4.
Note that the sum ⊕ is not associative but “almost” associative, that is, thereexists an element g ∈ F such that for every a, b, c ∈ F , we have (( a ⊕ b ) ⊕ c ) g = a ⊕ ( b ⊕ c ) . Figure 2.4 below shows that in fact g = x .Thus F can be considered as an algebra with two binary operations: multiplication andaddition. It is a group under multiplication and satisfies the identity (2.1). Algebras with twobinary operations satisfying these conditions form a variety, which we shall call the varietyof Thompson algebras . Note that every group can be turned into a Thompson algebra in atrivial way by setting a ⊕ b = 1 for every a, b . Our main result shows, in particular, that F has a non-trivial structure as a Thompson algebra.The Thompson group F has an obvious involutary automorphism that flips a diagramabout a vertical line (it is also an anti-automorphism with respect to addition). Thus everystatement about F has its left-right dual. ote that for every n ≥ , the group F n is a diagram group over the semigroup presen-tation (cid:104) x | x = x n (cid:105) [5]. This was used in [5] to find a nice presentation of F n for every n .We will use this presentation below. F Let x , x be the standard generators of F . Recall that x i +1 , i ≥ , denotes x − i x x i . Inthese generators, the group F has the following presentation (cid:104) x i , i ≥ | x x j i = x i +1 for every j < i (cid:105) [3].There exists a clear connection between representation of elements of F by diagrams andthe normal form of elements in F . Recall [3] that every element in F is uniquely representablein the following form: x s i . . . x s m i m x − t n j n . . . x − t j , (2.2)where i ≤ · · · ≤ i m (cid:54) = j n ≥ · · · ≥ j ; s , . . . , s m , t , . . . t n ≥ , and if x i and x − i occur in(2.2) for some i ≥ then either x i +1 or x − i +1 also occurs in (2.2). This form is called the normal form of elements in F .We say that a path in a diagram is positive if all the edges in the path are oriented fromleft to right. Let P be the collection of cells which consists of one cell x → x . It was noticedin [6] that every reduced diagram ∆ over P can be divided by its longest positive path fromits initial vertex to its terminal vertex into two parts, positive and negative , denoted by ∆ + and ∆ − , respectively. So ∆ = ∆ + ◦ ∆ − . It is easy to prove by induction on the number ofcells that ∆ + are ( x, x ) -cells and all cells in ∆ − are ( x , x ) -cells.Let us show how given an ( x, x ) -diagram over P one can get the normal form of theelement represented by this diagram. This is the left-right dual of the procedure describedin [6, Example 2] and after Theorem 5.6.41 in [15]. Lemma 2.5.
Let us number the cells of ∆ + by numbers from to k by taking every timethe “leftmost” cell, that is, the cell which is to the left of any other cell attached to the bottompath of the diagram formed by the previous cells. The first cell is attached to the top path of ∆ + (which is the top path of ∆ ). The i th cell in this sequence of cells corresponds to an edgeof the Squier graph Γ( P ) , which has the form ( x (cid:96) i , x → x , x r i ) , where (cid:96) i ( r i ) is the length ofthe path from the initial (resp. terminal) vertex of the diagram (resp. the cell) to the initial(resp. terminal) vertex of the cell (resp. the diagram), such that the path is contained in thebottom path of the diagram formed by the first i − cells. If r i = 0 then we label this cell by 1.If r i (cid:54) = 0 then we label this cell by the element x (cid:96) i of F . Multiplying the labels of all cells, weget the “positive” part of the normal form. In order to find the “negative” part of the normalform, consider (∆ − ) − , number its cells as above and label them as above. The normal formof ∆ is then the product of the normal form of ∆ + and the inverse of the normal form of (∆ − ) − . For example, applying the procedure from Lemma 2.5 to the diagram on Figure 2.3 weget the normal form x x x ( x x x x ) − .Diagrams for the generators of F , x , x , are on Figure 2.4.Lemma 2.5 immediately implies Lemma 2.6. If u is the normal form of ∆ , then the normal form of ⊕ ∆ is obtained from u by increasing every index by 1. In particular, x = ⊕ x , and, in general, x i +1 = ⊕ x i , i ≥ . Thus we get Proposition 2.7.
As a Thompson algebra, F is generated by one element x . (cid:115) (cid:115) (cid:115) (cid:115) (cid:115) (cid:115) (cid:115) (cid:115) Figure 2.3: Reading the normal form of an element of F off its diagram. (cid:117)(cid:117)(cid:117)(cid:117) (cid:117)(cid:117)(cid:117)(cid:117)(cid:117) x x Figure 2.4: Diagrams representing the generators of the R. Thompson group F There is a natural isomorphism between Thompson group F defined as a diagram groupand F defined as a group of homeomorphisms. Let ∆ be a diagram in F . The positivesubdiagram ∆ + describes a binary subdivision of [0 , in the following way. Every edge of ∆ + corresponds to a dyadic sub-interval of [0 , ; that is, an interval of the form [ k n , k +12 n ] forintegers n ≥ and k = 0 , . . . , n − . The top edge top (∆) corresponds to the interval [0 , .For each cell π of ∆ + (hence an ( x, x ) -cell), if top ( π ) corresponds to an interval [ k n , k +12 n ] then the left bottom edge of π corresponds to the left half of the interval, [ k n , k n + n +1 ] ,and the right bottom edge of π corresponds to the right half of the interval, [ k n + n +1 , k +12 n ] .Thus, if the bottom path of ∆ + consists of n edges, ∆ + describes a binary subdivisioncomposed of n intervals. Similarly, the negative subdiagram ∆ − describes a binary sub-division with n intervals as well. The diagram ∆ corresponds to the dyadic rearrangementmapping the top subdivision (associated with ∆ + ) to the bottom subdivision (associated with ∆ − ).Thinking of ∆ + as a subdivision of [0 , , the inner vertices of ∆ (that is, the verticesother than ι (∆) and τ (∆) ) correspond to the break points of the subdivision. Thus, in thediagram ∆ every inner vertex is associated with two break points; one break point of the topsubdivision and one break point of the bottom subdivision. .5 From diagrams to pairs of binary trees Thompson group F can be defined in terms of reduced pairs of binary trees. Let ∆ be adiagram in F with n + 1 vertices. It is possible to put a vertex in the middle of every edgeof the diagram. Then, for each cell π in ∆ + (hence an ( x, x ) -cell) one can draw edges fromthe vertex on top ( π ) to the vertices on the bottom edges of π . We get a binary tree T + withthe root lying on top (∆) and n leaves lying on bot (∆ + ) = top (∆ − ) . A similar constructionin ∆ − gives a second binary tree T − , lying upside down, such that the leaves of T + and T − coincide. Whenever we speak of a pair of binary trees ( T + , T − ) we assume that they havethe same number of leaves, and that T − is drawn upside down so that the leaves of T + and T − coincide.Let T be a binary tree. We call a vertex with two children a caret in the tree. We saythat a pair of binary trees ( T + , T − ) have a common caret if for some i , the i and i + 1 leaveshave a common father both in T + and in T − . We call a pair of trees ( T + , T − ) reduced if ithas no common carets. The construction above maps every reduced diagram ∆ to a reducedpair of binary trees. The correspondence is one to one and enables to view Thompson group F as a group of reduced pairs of binary trees with the proper multiplication. Jones [11] defined for each element of F , viewed as a reduced pair of binary trees, an asso-ciated graph, which he called the Thompson graph of that element. If ( T + , T − ) is a reducedpair of binary trees, we call an edge t in T + or T − a left edge if it connects a vertex to itsleft child. Definition 3.1 (Jones [11]) . Let g be an element of F and ( T + , T − ) the correspondingreduced pair of binary trees. If T + , T − have n common leaves, enumerated l , . . . , l n fromleft to right, we can assume that all of them lie on the same horizontal line. We define agraph J ( g ) as follows. The graph J ( g ) has n vertices v , . . . , v n − . The vertex v lies to theleft of the first leaf l and for all i = 1 , . . . , n − , the vertex v i lies between l i and l i +1 on thehorizontal line. The edges of J ( g ) are defined as follows. For every left edge t of ( T + , T − ) ,we draw a single edge e in J ( g ) which crosses the edge t and no other edge of ( T + , T − ) . Notethat this property determines the end vertices of the edge e .For example, the graph J ( x ) , associated with x , is depicted in Figure 3.5. Definition 3.2.
Let ∆ be a (not necessarily reduced) diagram over the presentation P = (cid:104) x | x = x (cid:105) . The Thompson graph T (∆) is a “subgraph” of the diagram ∆ , defined as follows.The vertex set of T (∆) is the vertex set of ∆ minus the terminal vertex τ (∆) . For everyinner vertex v of ∆ the only incoming edges of v which belong to T (∆) are the top-most andbottom-most incoming edges of v in ∆ . If the top-most and bottom-most incoming edges of v coincide we shall consider them as two distinct edges (hence the quotation marks on theword subgraph). An edge of T (∆) will be called an upper (lower) edge if it is the top-most(bottom-most) incoming edge of an inner vertex in ∆ .The subgraph T ( x x ) of the diagram x x of Thompson group F is depicted in Figure3.6. The upper edges in the graph are colored red, while the lower edges are colored blue.Note that the graph is bipartite. igure 3.5: The graph J ( x ) . Remark 3.3.
Let ∆ be a reduced diagram in F . Consider the positive subdiagram ∆ + .Every cell π in ∆ + is an ( x, x ) -cell. As such, it has a unique bottom vertex separating theleft bottom edge of π from its right bottom edge (see Figure 2.2). Conversely, every innervertex v of ∆ is the bottom vertex of a unique cell π v in ∆ + . Clearly, the top-most incomingedge of v in ∆ is the left bottom edge of the cell π v . Thus, the upper edges in the graph T (∆) are exactly the left bottom edges of the cells in ∆ + . Similarly, the lower edges of thegraph T (∆) are exactly the left top edges of the cells in ∆ − . Remark 3.4.
In Section 6.6 we shall show how to reconstruct g from T ( g ) . Proposition 3.5.
Let ∆ be a reduced diagram in F . Then, the associated graphs T (∆) (from Definition 3.2) and J (∆) (from Definition 3.1) are isomorphic.Proof. Let ( T + , T − ) be the reduced pair of binary trees associated with ∆ . We can assumethat the pair ( T + , T − ) is drawn inside the diagram ∆ as described in Section 2.5. Let T (∆) be the subgraph associated with ∆ . It is possible to stretch each of the upper edges of T (∆) up as follows. By Remark 3.3, e is an upper edge in T (∆) , if and only if e is a left bottomedge of some cell in ∆ + . Let v be the vertex of T + which lies on e and t the edge of T + connecting v to its father. Clearly, t is a left edge in the tree T + . We stretch the edge e slightly so that instead of crossing the vertex v it crosses the edge t of T + (and no otheredges of the tree). Similarly, we stretch every bottom edge of T (∆) down so that instead ofcrossing a vertex of the tree T − , it crosses the edge of the tree connecting the vertex to itsfather. The process is illustrated in Figure 3.7.If the graph T (cid:48) (∆) results from stretching the edges of T (∆) as described, then thereis a one to one correspondence between the left edges of the pair of trees ( T + , T − ) and theedges of T (cid:48) (∆) . Indeed, every edge of T (cid:48) (∆) crosses a single left edge of ( T + , T − ) and every igure 3.6: The graph T ( x x ) .Figure 3.7: Stretching the edges of T (∆) . left edge of ( T + , T − ) is crossed by an edge of T (cid:48) (∆) . Note, that if ( T + , T − ) have n commonleaves, then T (∆) (hence T (cid:48) (∆) ) has n vertices; one to the left of the left most leaf of T + and one between any pair of consecutive leaves of T + . It follows that the graph T (cid:48) (∆) isisomorphic to the graph J (∆) . Since T (∆) and T (cid:48) (∆) are clearly isomorphic as graphs weget the result. −→ F and its properties −→ F Lemma 4.1.
Suppose that a diagram ∆ is obtained from a diagram ∆ (cid:48) by removing a dipole.Suppose that T (∆ (cid:48) ) is bipartite. Then T (∆) is bipartite.Proof. The dipole can be of type π ◦ π − or of type π − ◦ π where π is the cell on Figure2.2. In the first case to get the graph T (∆) we remove from T (∆ (cid:48) ) a vertex with exactly twoedges connecting it to another vertex of T (∆ (cid:48) ) , and the statement is obvious. In the second ase, since T (∆ (cid:48) ) is bipartite, we can label the vertices of ∆ (cid:48) by "+" and "-", so that everytwo incident vertices have opposite signs. Consider the four vertices of the dipole. In T (∆ (cid:48) ) the top and the bottom vertices of the dipole are incident to the left vertex. Hence the topand the bottom vertices have the same label. Note that the edge of the dipole connectingthe left vertex with the right vertex is not an edge of T (∆ (cid:48) ) . Thus, the effect of removingthe dipole on T (∆ (cid:48) ) amounts to identifying the top and the bottom vertices of the dipoleand erasing the lower edge connecting the left vertex with the top vertex and the upper edgeconnecting the left vertex with the bottom one. Since the top and bottom vertices have thesame label, the Thompson graph T (∆) is bipartite. Definition 4.2.
Jones’ subgroup −→ F is the set of all reduced diagrams ∆ in F for which theassociated graph T (∆) is bipartite. Proposition 4.3.
Jones’ subgroup −→ F is indeed a subgroup of F .Proof. Suppose that ∆ , ∆ belong to −→ F . The Thompson graph T (∆ ◦ ∆ ) is the union of T (∆ ) and T (∆ ) with vertices ι (∆ ) and ι (∆ ) identified. Hence T (∆ ◦ ∆ ) is bipartite.By Lemma 4.1, the graph T (∆ ∆ ) is bipartite as well. −→ F is isomorphic to F Lemma 4.4.
The Jones’ subgroup −→ F coincides with the subgroup H which is the smallestsubgroup of F that contains x x and closed under addition.Proof. Clearly H is inside −→ F . Also from Lemma 2.5 it follows that if we add the triv-ial diagram on the right to the reduced diagram representing x x , we get the dia-gram corresponding to the normal form x x x ( x x x ) − = x x x x − ( x x ) − . Hence x x x x − also belongs to H . If we add to this element the diagram on the right, we get ( x x x x )( x x x x ) − , and so the element x x x − belongs to H . By induction, we seethat all elements x n x x − n +12 belong to H . Now consider an arbitrary reduced diagram ∆ in −→ F . Let us enumerate the vertices of ∆ from left to right: , , ..., s so that ι (∆) = 0 , τ (∆) = s .If the Thompson graph T (∆ ) does not contain top nor bottom edges connecting j > with , then ∆ is a sum of ∆ (cid:48) and the trivial diagram (added on the left). The diagram ∆ (cid:48) also belongs to −→ F (its graph T (∆ (cid:48) ) is bipartite), so by induction ∆ (cid:48) is in H , and ∆ is in H also. So assume that T (∆) contains an edge (0 , j ) , j > . Without loss of generality wecan assume that this is an upper edge, that is, it belongs to the positive part of the diagram.Therefore the positive part of the normal form for ∆ in the generators x , x , x , ... startswith x . Let it start with x n x i , i > . The bottom-most cell corresponding to the prefix x n cannot have both its bottom edges on the maximal positive path of ∆ , i.e. its top edgecannot connect with . Indeed if it connects with , then and are in different partsof the bipartite graph T (∆) . The vertex then belongs to the same part as . Then cannot be connected with by a lower edge from T (∆) , so it has to be connected with ,which means that the diagram ∆ is not reduced, a contradiction. Therefore there is an edgeconnecting with j (cid:48) ≤ j . Hence the normal form corresponding to ∆ starts with x n x . If n = 1 , then we can divide by x x (on the left) and get an element from −→ F with a shorternormal form. Hence ∆ is in H since x x is in H . If n > , then we can replace x n x by x n − , and the resulting element would still be in −→ F . Since its normal form is shorter thanthat of ∆ , it is in H , and so ∆ is in H .The proof of Lemma 4.4 proves the following. emma 4.5. The subgroup −→ F is generated by two sets X = { x i x i +1 , i ≥ } and X (cid:48) = { x n +1 i x i +1 x − ni +2 , i ≥ , n ≥ } .Proof. Indeed, in the proof of Lemma 4.4 we proved that X and X (cid:48) are inside −→ F , and everyelement of −→ F is either a sum of the trivial diagram and some other diagram from −→ F or isa product of an element from X ∪ X (cid:48) and a reduced diagram from −→ F with a shorter normalform or is the inverse of such an element. Since by Lemma 2.6 the subgroup generated by X ∪ X (cid:48) is closed under addition of the trivial diagram on the left, the lemma is proved. Lemma 4.6.
The subgroup −→ F is generated by three elements x x , x x , x x .Proof. Let X and X (cid:48) be the sets from Lemma 4.5. It is obvious that for every j > theelement x j x j +1 is equal to ( x j − x j − ) x x . Thus the set X = { x j x j +1 , j ≥ } is containedin the subgroup (cid:104) x x , x x , x x (cid:105) . It remains to show that X (cid:48) ⊆ (cid:104) X (cid:105) . Note that x x x − = x x x − x x x − = x x x x − x x − = x x x x x − x − = ( x x ) ( x x ) − . Similarly, x n +10 x x − n = ( x x ) n +1 ( x x ) − n for every n ≥ . Adding several trivial diagrams on the left, we get x n +1 j x j +1 x − nj +2 = ( x j x j +1 ) n +1 ( x j +2 x j +3 ) − n for every j ≥ . Lemma 4.7.
The subgroup −→ F is isomorphic to F .Proof. The elements x x , x x , x x satisfy the defining relations of F (see [5, page 54]).All proper homomorphic images of F are Abelian [1, Theorem 4.13]. Since x x , x x donot commute, the natural homomorphism from F onto −→ F is an isomorphism.As an immediate corollary of Theorem 1, we get Proposition 4.8 (Compare with Proposition 2.7) . −→ F is a subalgebra of the Thompsonalgebra F generated by one element x x . −→ F is the stabilizer of the set of dyadic fractions with odd sumsof digits Let ∆ be a reduced diagram in F and T (∆) the associated graph. Let ∆ + be the positivesubdiagram. Recall (Section 2.4) that ∆ + describes a binary subdivision of [0 , . Everyedge e in ∆ + corresponds to a dyadic interval of length m for some integer m ≥ . We willcall this length the weight of the edge e and denote it by ω ( e ) . The inner vertices of ∆ + correspond to the break points of the subdivision. Note that if v is an inner vertex of ∆ + ,the dyadic fraction f is the corresponding break point of the subdivision and p is a positivepath in ∆ + from ι (∆) to v , then the weight of the path p (that is, the sum of weights of itsedges) is equal to the fraction f . Lemma 4.9.
Let v be an inner vertex of ∆ . Let e be the (unique) upper incoming edge of v in T (∆) and e be any outgoing upper edge of v in T (∆) , then ω ( e ) < ω ( e ) . roof. By Remark 3.3, e is the left bottom edge of some cell π in ∆ + . Let e (cid:48) be the rightbottom edge of the cell π . Clearly, the edge e (cid:48) is the top-most outgoing edge of v in ∆ . Since e (cid:48) is not the left bottom edge of any cell in ∆ + , by Remark 3.3, the edge e (cid:48) does not belongto T (∆) . Hence, the edge e (cid:54) = e (cid:48) so that e lies under e (cid:48) in ∆ + . From the construction inSection 2.4 it is obvious that ω ( e ) ≤ ω ( e (cid:48) ) = ω ( e ) < ω ( e ) . Definition 4.10.
Let ∆ be a (not necessarily reduced) diagram. Let v be an inner vertexof ∆ . Since every inner vertex of ∆ has a unique incoming upper edge in T (∆) , there is aunique positive path in T (∆) from ι (∆) to v , composed entirely of upper edges. We call thispath the top path from ι (∆) to v . Lemma 4.11.
Let ∆ be a reduced diagram in F . Let v be an inner vertex of ∆ and f thecorresponding break point of the subdivision associated with ∆ + . Let p be the top path from ι (∆) to v . Then, the length of p is equal to the sum of digits in the binary form of f .Proof. Let e , . . . , e k be the edges of p . For each i = 1 , . . . k , the weight ω ( e i ) = mi forsome positive integer m i . By Lemma 4.9, ω ( e ) > · · · > ω ( e k ) . Thus, the weight of p is ω ( p ) = (cid:80) ki =1 12 mi where m > · · · > mk . Clearly, the sum of digits of ω ( p ) in binary formis k . Therefore, the sum of digits of f = ω ( p ) is equal to the number of edges in p .Viewed as a dyadic rearrangement, a reduced diagram ∆ takes the break points of the topsubdivision (associated with ∆ + ) to the break points of the bottom subdivision (associatedwith ∆ − ). Reflecting the diagram ∆ about a horizontal line shows that the analogue ofLemma 4.11 holds for the bottom subdivision: If v is an inner vertex of ∆ and p is the bottom path from ι (∆) to v , composed entirely of lower edges of T (∆) , then the length of p isequal to the sum of digits of the break point corresponding to v in the bottom subdivision. Corollary 4.12.
Let S be the set of dyadic fractions with odd sums of digits. Let ∆ be areduced diagram which stabilizes S and v be an inner vertex of ∆ . Then, the length of the toppath from ι (∆) to v and the length of the bottom path from ι (∆) to v have the same parity.Proof. Let f + , f − be the break points corresponding to v in the top and bottom subdivisionsrespectively. Since ∆ takes f + to f − , the sum of digits of f + and the sum of digits of f − have the same parity. The result follows from Lemma 4.11 and its stated analogue.Now we are ready to prove Theorem 2.
Jones’ subgroup −→ F is the stabilizer of the set of dyadic fractions from the unitinterval [0 , with odd sums of digits, under the standard action of F on the interval [0 , . Proof.
Let S be the set of dyadic fractions with odd sums of digits. Let ∆ be a reduceddiagram which stabilizes S . By Corollary 4.12, for every inner vertex v , the length of thetop path from ι (∆) to v and the length of the bottom path from ι (∆) to v have the sameparity. It is possible to use this parity to assign to every vertex of T (∆) a label "0" or "1".Since for every vertex there is a unique top path and a unique bottom path from ι (∆) to thevertex, neighbors in T (∆) have different labels and T (∆) is bipartite.The other direction follows from Theorem 1. On finite binary fractions x x acts asfollows: x x ( t ) = t = . α → t = . αt = . α → t = . αt = . α → t = . αt = . α → t = . α n particular, x x stabilizes the set S . If ∆ and ∆ (cid:48) are diagrams in F , then viewed asmaps from [0 , to itself, the sum ∆ ⊕ ∆ (cid:48) is defined as (∆ ⊕ ∆ (cid:48) )( t ) = (cid:40) ∆(2 t )2 : t ∈ [0 , ] ∆ (cid:48) (2 t − + : t ∈ [ , It is easy to see that if ∆ and ∆ (cid:48) stabilize S , then ∆ ⊕ ∆ (cid:48) stabilize S as well. Since byTheorem 1, −→ F is the smallest subgroup of F containing x x and closed under sums, theinclusion −→ F ⊂ Stab( S ) follows.In order to prove Corollary 3 we will need the following observations. Remark 4.13.
Let c = ( x x ) − ∈ F . On finite binary fractions c acts as follows: c ( t ) = t = . α → t = . αt = . α → t = . αt = . α → t = . αt = . α → t = . α In particular, if t is a finite binary fraction and m ∈ N then for any large enough n ∈ N , thefirst m digits in the binary form of c n ( t ) are zeros. That is, c n ( t ) < m . Proof.
If the first digit of t is then each application of c adds another to the leadingsequence of zeros in the binary form of t so for every n ≥ m we get the result. If the firstdigit of t is , then t starts with a sequence of ones followed by a . Let l be the length of thissequence. If l > then each application of c reduces the length of the sequence of ones by . Thus, possibly after several applications of c , we can assume that l = 1 or l = 2 . In bothcases, one application of c yields c ( t ) which starts with and we are done by the previouscase. Lemma 4.14.
Let g be an element of R. Thompson group F . Then, there exists m ∈ N such that for any finite binary fraction t < m the sum of digits of t in binary form is equalto the sum of digits of g ( t ) .Proof. The element g maps some binary subdivision B onto a subdivision B . The firstsegment of B is of the form J = [0 , r ] for some positive integer r . Since is mapped to ,on the segment J the function g is defined as a linear function with slope l for some l ∈ Z and with constant number . That is, for every t ∈ J we have g ( t ) = 2 l t . If l ≤ , then forany binary fraction t ≤ r , the application of g adds l zeros to the beginning of the binaryform of g , which does not affect the sum of digits of t . In that case, taking m = r would do.If l > it is possible to take m = max { r, l } . Since m ≥ r , the binary fraction t ∈ J . Since m ≥ l , the binary form of t starts with at least l zeros. The application of g erases the first l zeros and thus does not affect the sum of digits of t .Corollary 3 follows immediately from the following. Theorem 4.15.
Let h be an element of F which does not belong to −→ F . Then the index [ −→ F : −→ F ∩ h −→ F h − ] is infinite. roof. Let h / ∈ −→ F . If the index [ −→ F : −→ F ∩ h −→ F h − ] is finite then there exists r ∈ N such thatfor every g ∈ −→ F , we have g r ∈ h −→ F h − . That is, h − g r h ∈ −→ F . In particular, for every k ∈ N we have h − g rk h ∈ −→ F . Let g = ( x x ) − ∈ −→ F . We will show that for every n large enough, h − g n h / ∈ −→ F and get the required contradiction.Let S be the set of finite binary fractions with odd sums of digits. Since h − / ∈ −→ F ,there exists t ∈ S such that h − ( t ) / ∈ S . Let t = h − ( t ) . By Lemma 4.14 there exists m for which the sum of digits of every binary fraction < m is preserved by h . By Remark4.13 for every n large enough, g n ( t ) < m . Since g n ∈ −→ F and t / ∈ S , the binary fraction g n ( t ) / ∈ S . Since g n ( t ) < m , the sum of digits of h ( g n ( t )) is equal to the sum of digitsof g n ( t ) . Thus h ( g n ( t )) / ∈ S . Therefore, (recall that the composition in F is from left toright), h − g n h ( t ) = h ( g n ( h − ( t ))) = h ( g n ( t )) / ∈ S . The element t belonging to S impliesthat h − g n h does not stabilize S and in particular h − g n h / ∈ −→ F . −→ F n In this section we generalize results of the previous sections from and to arbitrary n . Itturns out that the generalization is quite natural. The proofs follow the same paths and forsome theorems, the proof for arbitrary n is almost identical to the proof of the particularcase considered before. Definition 5.1.
Let ∆ be a (not necessarily reduced) diagram over the presentation P = (cid:104) x | x = x (cid:105) and let n ∈ N . The diagram ∆ is said to be n -good if for every inner vertex v the lengths of the top and bottom paths from ι (∆) to v are equal modulo n .Note that if n = 2 , then being -good is formally weaker than being bipartite. Lemma5.6 shows that these conditions are in fact equivalent for reduced diagrams.The proof of the following lemma is completely analogous to the proof of Lemma 4.1, sowe leave the proof to the reader. Lemma 5.2.
Suppose that a diagram ∆ is obtained from a diagram ∆ (cid:48) by removing a dipole.Suppose that ∆ (cid:48) is n -good for some n ∈ N . Then T (∆) is n -good as well. Definition 5.3.
Let n ∈ N . Jones’ n -subgroup −→ F n is the set of all reduced n -good diagrams ∆ in F .In particular Jones’ -subgroup is the entire Thompson group F . Jones’ -subgroupcoincides with Jones’ subgroup −→ F .The proof of the following proposition is identical to that of Proposition 4.3. Proposition 5.4.
For every n ∈ N , Jones’ n -subgroup −→ F n is indeed a subgroup of F . −→ F n − is isomorphic to F n Let n ≥ . We denote by H n the Thompson subalgebra of F generated by x · · · x n − , i.e.,the smallest subgroup of F containing x · · · x n − and closed under ⊕ . emma 5.5. Let i, d ∈ N ∪ { } . Let ( m , . . . , m d ) be a sequence of positive integers suchthat m d ≥ if d > . Then, d (cid:89) k =0 x m k i + k n − (cid:89) k =1 x i + d + k (cid:34)(cid:32) d − (cid:89) k =0 x m k i + n − k (cid:33) x m d − i + n − d (cid:35) − ∈ H n . Proof.
We first prove the Lemma for i = 0 by induction on d . For d = 0 , the argumentis similar to the one in the proof of Lemma 4.4. If we add the trivial diagram onthe right to the reduced diagram representing (cid:81) n − k =0 x k , we get the diagram correspond-ing to the normal form x (cid:81) n − k =1 x k ( (cid:81) n − k =0 x k ) − = x ( (cid:81) n − k =1 x k ) x − n − ( (cid:81) n − k =0 x k ) − . Hence x ( (cid:81) n − k =1 x k )( x n − ) − also belongs to H n . If we add to this element the diagram on theright, we get x (cid:81) n − k =1 x k [( (cid:81) n − k =0 x k ) x n − ] − = x ( (cid:81) n − k =1 x k )( x n − ) − ( (cid:81) n − k =0 x k ) − . Multi-plying on the right by (cid:81) n − k =0 x k we get that x ( (cid:81) n − k =1 x k )( x n − ) − belongs to H n .Note the effect of the addition of on the right to the element x ( (cid:81) n − k =1 x k )( x n − ) − . Thenegative part of the normal form got multiplied on the left by (cid:81) n − k =0 x k , when originally itstarted with x n − . Similarly, the positive part of the normal form was multiplied by x , whenoriginally it started with x . In particular the exponents of both x and x n − were increasedby one. By repeating the process of adding on the right and multiplying by (cid:81) n − k =0 x k (onthe right) we get that for every positive integer m , the element x m ( (cid:81) n − k =1 x k )( x m − n − ) − belongs to H n as required.Assume that the lemma holds for every non negative integer ≤ d . Let ( m , . . . , m d +1 ) be a sequence of positive integers such that m d +1 ≥ . Let j = min { , . . . , d + 1 } such that m j ≥ . For all r = 1 , . . . , d + 1 − j let n r = m r + j . Let n = m j − . We use the inductionhypothesis on d + 1 − j with the sequence ( n , . . . , n d +1 − j ) . Thus, we have that d +1 − j (cid:89) k =0 x n k k n − (cid:89) k =1 x d +1 − j + k (cid:34)(cid:32) d +1 − j − (cid:89) k =0 x n k n − k (cid:33) x n d +1 − j − n − d +1 − j (cid:35) − ∈ H n Adding the trivial diagram j times on the left we get by Lemma 2.6 the element d +1 − j (cid:89) k =0 x n k k + j n − (cid:89) k =1 x d +1+ k (cid:34)(cid:32) d − j (cid:89) k =0 x n k n − k + j (cid:33) x n d +1 − j − n + d (cid:35) − . We assume that j < d + 1 , the other case being similar. Then, adding on the right resultsin the following element. Note that as in the case d = 0 the positive part of the normal formgets multiplied by (cid:81) jk =0 x k (it currently starts with x j ). Similarly, the negative part of thenormal form is multiplied by (cid:81) n − jk =0 x k . (cid:32) j − (cid:89) k =0 x k (cid:33) x n +1 j d +1 − j (cid:89) k =1 x n k k + j n − (cid:89) k =1 x d +1+ k (cid:34)(cid:32) n − j (cid:89) k =0 x k (cid:33) x n +1 n − j (cid:32) d − j (cid:89) k =1 x n k n − k + j (cid:33) x n d +1 − j − n + d (cid:35) − Substituting n k by m k + j and n + 1 by m j we get (cid:32) j − (cid:89) k =0 x k (cid:33) x m j j d +1 − j (cid:89) k =1 x m k + j k + j n − (cid:89) k =1 x d +1+ k (cid:34)(cid:32) n − j (cid:89) k =0 x k (cid:33) x m j n − j (cid:32) d − j (cid:89) k =1 x m k + j n − k + j (cid:33) x m d +1 − n + d (cid:35) − Then, shifting the indexes we get that j − (cid:89) k =0 x k d +1 (cid:89) k = j x m k k n − (cid:89) k =1 x d +1+ k n − j (cid:89) k =0 x k d (cid:89) k = j x m k n − k x m d +1 − n + d − ∈ H n ultiplying on the right by (cid:81) n − k =0 x k cancels a prefix of the negative part of the normal formso we have that j − (cid:89) k =0 x k d +1 (cid:89) k = j x m k k n − (cid:89) k =1 x d +1+ k j − (cid:89) k =0 x n − k d (cid:89) k = j x m k n − k x m d +1 − n + d − ∈ H n Since m k = 1 for all k = 1 , . . . , j − we have, x d +1 (cid:89) k =1 x m k k n − (cid:89) k =1 x d +1+ k (cid:34) x n − (cid:32) d (cid:89) k =1 x m k n − k (cid:33) x m d +1 − n + d (cid:35) − ∈ H n To get the result for m ≥ it is possible to increase the exponents of x and x n − simul-taneously by repeatedly adding on the right and multiplying by (cid:81) n − k =0 x k . Thus, we havethat d +1 (cid:89) k =0 x m k k n − (cid:89) k =1 x d +1+ k (cid:34)(cid:32) d (cid:89) k =0 x m k n − k (cid:33) x m d +1 − n + d (cid:35) − ∈ H n as required. If i (cid:54) = 0 then by Lemma 2.6, adding the trivial diagram i times on the left, givesthe result. Lemma 5.6.
Let n ≥ . Then, the subgroup −→ F n − coincides with the Thompson subalgebra H n .Proof. Clearly, H n is inside −→ F n − . Let ∆ be a reduced diagram in −→ F n − . We enumeratethe vertices of ∆ from left to right: , , ..., s so that ι (∆) = 0 , τ (∆) = s . We shall need thefollowing lemma. Lemma 5.7.
Let r be the left-most vertex such that there exists (cid:96) < r − such that (cid:96) and r are connected by an edge in T (∆) . Let (cid:96) be the left-most vertex such that ( (cid:96), r ) is an upperor lower edge in T (∆) . Then r − (cid:96) ≥ n . That is, there are at least n − inner vertices of ∆ between the vertices (cid:96) and r .Proof. We assume that the edge ( (cid:96), r ) is an upper edge in T (∆) . Otherwise, we look at ∆ − instead. Note that a priori, there might also be a lower edge in T (∆) connecting the vertices (cid:96) and r . Let ∆ be the subdiagram of ∆ + bounded from above by the edge ( l, r ) and frombelow by (part of) the path bot (∆ + ) . Since every inner vertex i < r has no incoming edgesin ∆ other than ( i − , i ) , every inner vertex of ∆ is connected by an edge in ∆ to theterminal vertex r . It follows that the vertices (cid:96) and r are not connected by a lower edgein T (∆) . Otherwise the same argument for the subdiagram ∆ , bounded from above by top (∆ − ) = bot (∆ + ) and from below by the lower edge from (cid:96) to r , would show that thediagram ∆ is the inverse of ∆ , by contradiction to the diagram ∆ being reduced.Since the length of the top path from ι (∆) to r is (cid:96) +1 , the length of the bottom path from ι (∆) to r is equal to (cid:96) +1 modulo n − . From the minimality of r , there exists some (cid:96) (cid:48) < r suchthat the bottom path from ι (∆) to r is composed of the lower edges (0 , , (1 , . . . , ( (cid:96) (cid:48) − , (cid:96) (cid:48) ) and ( (cid:96) (cid:48) , r ) . Therefore, its length (cid:96) (cid:48) + 1 ≡ (cid:96) + 1( mod n − . From the minimality of (cid:96) andthe absence of a lower edge from (cid:96) to r , it follows that (cid:96) (cid:48) > (cid:96) . Therefore, r − (cid:96) > (cid:96) (cid:48) − (cid:96) ≡ n − implies that r − l > n − .Now we can finish the proof of Lemma 5.6.Consider the subdiagram ∆ of the diagram ∆ + , bounded from above by the upper edge ( (cid:96), r ) and from below by the bottom path bot (∆ + ) . The diagram ∆ has at least n − nner vertices, the first n − of which are connected by arcs (i.e., edges which do not lie on bot (∆ + ) ) to the terminal vertex r .Let x i be the leading term of the positive part of the normal form of ∆ (since ( (cid:96), r ) is anupper edge, such x i exists). Let x m i x m i +1 · · · x m p i + p be the longest sequence of positive powersof consecutive letters x j which forms a prefix of the normal form of ∆ .Each positive letter x j in the normal form of ∆ has a corresponding edge in ∆ + . Namely,the top edge of the cell “labeled by x j ” in the algorithm described in Lemma 2.5. From Lemma2.5 it follows that if for all j = 0 , . . . , p the exponent m j = 1 , then the edge correspondingto the first letter in the sequence (i.e., to x i ) is in fact the edge ( (cid:96), r ) . In this case, the n − mentioned arcs from vertices below the edge ( (cid:96), r ) to r show that the normal formof ∆ starts with (cid:81) n − k =0 x i + k which belongs to H n . Dividing the normal form of ∆ from theleft by (cid:81) n − k =0 x i + k gives an element of −→ F n − with a shorter normal form and we are done byinduction.Thus we can assume that there exists d = max { j = 0 , . . . , p | m d ≥ } . Again, fromLemma 2.5 it follows that the edge corresponding to the last letter of the power x m d d is theedge ( (cid:96), r ) . The arcs below ( (cid:96), r ) show that the prefix x m i x m i +1 · · · x m d i + d of the normal form of ∆ is followed by at least n − letters forming the product (cid:81) n − k =1 x i + d + k . By Lemma 5.5, it ispossible to replace (cid:81) dk =0 x m k i + k (cid:81) n − k =1 x i + d + k by (cid:16)(cid:81) d − k =0 x m k i + n − k (cid:17) x m d − i + n − d and the resultingelement would still be in −→ F n − . Since its normal form is shorter than that of ∆ we are doneby induction. Lemma 5.8.
For all n ≥ , the Thompson subalgebra H n is generated as a group by the set A = { x j · · · x j + n − | j ≥ } .Proof. Let G be the subgroup of F generated by A . It suffices to prove that G is closedunder sums. Let a and b be elements of G . In particular, a = y · · · y t and b = z · · · z s where y , . . . , y t and z , . . . , z s are elements in A ± . By Lemma 2.3, a ⊕ b = ( y ⊕ ) · · · ( y t ⊕ )( ⊕ z ) · · · (1 ⊕ z s ) . Since for any diagram z , we have (1 ⊕ z − ) = (1 ⊕ z ) − and A is closed underaddition of from the left, ( ⊕ z m ) ∈ G for all m = 1 , . . . , s . Thus it suffices to prove thatfor all j ≥ , the element x j · · · x j + n − ⊕ ∈ G .From Lemma 2.5 it follows that (cid:32) n − (cid:89) k =0 x j + k (cid:33) ⊕ = j (cid:89) k =0 x k n − (cid:89) k =0 x j + k (cid:32) j + n − (cid:89) k =0 x k (cid:33) − . Let r ∈ { , . . . , n − } be the residue of j modulo n − . Inserting (cid:81) j + n − r − k = j +1 x k and itsinverse to the right side of the above equation we get that (cid:32) n − (cid:89) k =0 x j + k (cid:33) ⊕ = j (cid:89) k =0 x k j + n − r − (cid:89) k = j +1 x k j + n − r − (cid:89) k = j +1 x k − n − (cid:89) k =0 x j + k (cid:32) j + n − (cid:89) k =0 x k (cid:33) − Since in Thompson group F for all (cid:96) > r we have x − (cid:96) x r = x r x − (cid:96) +1 , it is possible to replace (cid:16)(cid:81) j + n − r − k = j +1 x k (cid:17) − (cid:81) n − k =0 x j + k by (cid:81) n − k =0 x j + k (cid:16)(cid:81) j + n − r − k = j +1 x k + n − (cid:17) − . Thus, (cid:32) n − (cid:89) k =0 x j + k (cid:33) ⊕ = j (cid:89) k =0 x k j + n − r − (cid:89) k = j +1 x k n − (cid:89) k =0 x j + k j + n − r − (cid:89) k = j +1 x k + n − − (cid:32) j + n − (cid:89) k =0 x k (cid:33) − erging adjacent products, we get that (cid:32) n − (cid:89) k =0 x j + k (cid:33) ⊕ = j + n − r − (cid:89) k =0 x k n − (cid:89) k =0 x j + k (cid:32) j +2 n − r − (cid:89) k =0 x k (cid:33) − . Since the length of each of the products (cid:81) j + n − r − k =0 x k , (cid:81) n − k =0 x j + k and (cid:81) j +2 n − r − k =0 x k isdivisible by n − , it is possible to present each of them as a product of elements of A . Theresult clearly follows. Corollary 5.9.
For all n ≥ , the Thompson subalgebra H n is generated as a group by theset { x j . . . x j + n − | j = 0 , . . . , n − } .Proof. For every j ≥ n , the element x j . . . x j + n − is equal to ( x j − n +1 · · · x j − ) x ··· x n − . Lemma 5.10.
For all n ≥ , the subgroup −→ F n − is isomorphic to F n .Proof. The elements x j · · · x j + n − , for j = 0 , . . . , n − satisfy the defining relations of F n (see [5, page 54]). Since all proper homomorphic images of F n are Abelian [1, Theorem 4.13]and x j · · · x j + n − , for j = 0 , do not commute, we get the result.Finally, the proofs of the following theorems are almost identical to the proofs of Theorem2 and Corollary 3 so we leave them to the reader. Theorem 5.11.
Let n ≥ . For each i = 0 , . . . , n − , let S i be the set of all finite binaryfractions from the unit interval [0 , with sums of digits equal to i modulo n . Then, thesubgroup −→ F n is the intersection of the stabilizers of S i , i = 0 , . . . , n − under the naturalaction of F on [0 , . Theorem 5.12.
For all n ≥ , the subgroup −→ F n coincides with its commensurator in F ,hence the linearization of the permutational representation of F on F/ −→ F n is irreducible. F n into F are natural Suppose that G i = DG( P i , u i ) , i = 1 , , are diagram groups, and we can tessellate everycell from P by cells from P , which turns every cell from P into a diagram over P . Thenwe can define a map from G to G which sends every diagram ∆ from G to a diagramobtained by replacing every cell from ∆ by the corresponding diagram over P . This map isobviously a homomorphism. Such homomorphisms were called natural in [7].Instead of giving a more precise general definition of a natural homomorphism from onediagram group into another, we give an example. Recall that F n = DG( (cid:104) x | x = x n (cid:105) , x ) . Let π n be the cell x → x n . It has n + 1 vertices , . . . , n . Suppose that n ≥ . Connect eachvertex , . . . , n − with the vertex n by an arc. The result is an ( x, x n ) -diagram ∆( π n ) overthe presentation (cid:104) x | x = x (cid:105) . Now for every ( x, x ) -diagram ∆ from F n , consider the diagram φ (∆) obtained by replacing every cell π n (resp. π − n ) by a copy of ∆( π n ) (resp. ∆( π n ) − ).The resulting diagram belongs to F = DG( (cid:104) x | x = x (cid:105) , x ) . It is easy to check that the map φ is a homomorphism.A set of generators of F n is described in [5, Page 54]. Their left-right duals also generate F n . If we apply φ to these generators, we get (using Lemma 2.5) elements x j · · · x j + n − , j = 0 , . . . , n − , which are generators of −→ F n − by Corollary 5.9. Since F n does not havea non-injective homomorphism with a non-Abelian image, φ is an isomorphism between F n (cid:114) (cid:114)(cid:114) (cid:114) (cid:114) (cid:114)(cid:114)(cid:114) (cid:114) (cid:114)(cid:114) (cid:114) (cid:114)(cid:114) (cid:114) (cid:114)(cid:114) (cid:114) (cid:114) Figure 5.8: Generators of F . (cid:114) (cid:114) (cid:114)(cid:114) (cid:114) (cid:114) (cid:114)(cid:114)(cid:114) (cid:114) (cid:114)(cid:114) (cid:114) (cid:114)(cid:114) (cid:114) (cid:114)(cid:114) (cid:114) (cid:114) Figure 5.9: Images of generators of F under φ . Red arcs are added by φ . and −→ F n − . Figure 5.8 below shows the left-right duals of generators of F from [5], andFigure 5.9 shows the images of these generators under φ .Note that φ is also an example of a caret replacement homomorphism from F p to F q studied in [2]. It is proved in [2] that the image of F p under φ is undistorted in F . In this section we analyze the connection between elements of Thompson group F and links,as explained by Jones [11]. Given a link L , our analysis yields a linear bound on the wordlength of an element of Thompson group F , which represents L , in terms of the number ofcrossings and the number of unlinked unknots in L . Our construction essentially differs fromthat in [11] only in one step (step number 2).We will sometimes abuse notation and refer to a link and a link diagram as the sameobject. All link diagrams we consider are either connected (i.e., their underlying graph isconnected) or with connected components "far apart", that is, with no connected componentbounding another. The same is true for all plane graphs considered below. This is basically standard knot theory. Let L be a link represented by some link diagram,also denoted by L . If L is connected, then the underlying graph of L is a -regular planegraph and in particular an Eulerian graph. As such, its dual graph is bipartite. Therefore, t is possible to color the regions of L in gray and white so that the unbounded region of L is white and no two adjacent regions (regions which share a boundary component) havethe same color. Clearly, this is true for disconnected links as well. We define a signed planegraph G ( L ) as follows. We put a vertex inside every gray region and draw one edge betweentwo vertices for every common point of the boundaries of these regions (i.e., a crossing inthe link diagram). We label an edge with "+" or "-" according to the type of the crossingas defined in Figure 6.10. The process is demonstrated in Figure 6.12 for the link L fromFigure 6.11 . Figure 6.10: The types of crossings.Figure 6.11: A link L (a) (b)
Figure 6.12: (a) Gray and white regions. (b) The signed plane graph G ( L ) . Clearly this step is reversible: from any plane graph Γ with edges signed by "+" and "-"one can reconstruct a link diagram L such that G ( L ) = Γ . We denote this link diagram L by L (Γ) . Our goal is to turn a signed plane graph into the Thompson graph of some element from F .For this purpose, Jones introduced three moves. Move of type 1: If v is a vertex of degree one, we eliminate v together with the uniqueedge attached to it. igure 6.13: Type 1 move Move of type 2:
Let v be a vertex of degree such that the edges attached to it haveopposite signs and do not share their other end vertex. We eliminate v together with theedges attached to it and identify the endpoints of said edges. Figure 6.14: Type 2 move
Move of Type 3:
If two edges with opposite signs e and e (cid:48) join two vertices such thatthere are no other vertices and no edges between e and e (cid:48) , we erase the edges e and e (cid:48) . Figure 6.15: Type 3 move
If after the application of a move of type the resulting graph has a connected componentwhich bounds another, we “separate” the components so that none of them would bound theother and consider it to be a part of the move. From now on, we shall refer to a moveinverse to a move of type i as a move of type i as well. We say that two signed plane graphs Γ and Γ are equivalent if it is possible to get from one to the other by a finite series ofapplications of moves of types 1-3 and isotopy of the plane. If L is a link and Γ = G ( L ) isthe corresponding signed plane graph, then applying a move of type 1-3 to Γ is equivalentto applying a Reidemeister move to L (and possibly distancing some unlinked componentsof the link). Thus we have the following. Lemma 6.1.
Let L and L be two link diagrams. Let G ( L ) and G ( L ) be the correspondingsigned plane graphs. If G ( L ) and G ( L ) are equivalent, then L and L are link diagramsof the same link. Remark 6.2.
Let L be a link diagram with n crossings. It is easy to see that by applyingReidemeister moves if necessary, it is possible to get a link diagram L (cid:48) with at most n crossings such that the corresponding signed plane graph G ( L (cid:48) ) has no loops. Therefore,from now on all plane graphs considered here are assumed to have no loops. Definition 6.1.
Let Γ be a plane graph. A -page book embedding of Γ is a planar isotopywhich takes all the vertices of Γ to the x -axis and each edge of Γ to an edge whose interioris entirely above the x -axis or entirely below it. Γ is -page book embeddable if there existssuch a planar isotopy. ur definition of a -page book embedding differs slightly from the standard definition.In the standard definition, the planar isotopy can be replaced by any graph embedding.Whenever we speak of -page book embeddings we mean it in the sense of Definition 6.1.For a plane graph Γ , being 2-page book embeddable is equivalent to the condition thatthere exists an infinite oriented simple curve α which passes through all the vertices of Γ ,does not cross any of its edges, and whose infinite rays are in the outer face of Γ . Indeed,given such a curve, an isotopy which takes α to the x -axis (oriented from left to right) is a -page book embedding of Γ .A plane graph Γ is external Hamiltonian if it has a Hamiltonian cycle with at least oneedge on the outer face of Γ . A plane graph is external subhamiltonian if it is a subgraphof some external Hamiltonian plane graph. It is easy to see that a plane graph Γ is -pagebook embeddable if and only if it is external subhamiltonian.The following is an extension of a theorem of Kaufmann and Wiese [12] for simple planegraphs. A graph Γ is said to be k -connected if for every set S of at most k − vertices, thegraph resulting from Γ after the removal of the vertices in S and the edges incident to them,is connected. Theorem 6.2.
Let Γ be a plane graph with no loops. Let D (Γ) be the subdivision of Γ whereeach edge is divided in two. Then D (Γ) is 2-page book embeddable.Proof. The graph D (Γ) is a simple plane graph. As such, it is possible to triangulate it: tocomplete it to a maximal simple plane graph by adding to it a finite number of edges (seeFigure 6.16a). Let Γ (cid:48) be the resulting graph. The boundary of every face (including theouter face) of Γ (cid:48) is composed of exactly 3 edges.We say that a triangle ∆( a, b, c ) in the graph is a separating triangle if its removal wouldmake the graph disconnected. It is easy to see that every separating triangle is not theboundary of a face. By a result of Whitney [16] (see also [4, 10]), every maximal simple planegraph with no separating triangles is external Hamiltonian. Thus, if Γ (cid:48) has no separatingtriangles, it is -page book embeddable and we are done. (Indeed, D (Γ) embeds into Γ (cid:48) ).Assume that Γ (cid:48) has a separating triangle. We will need the following Lemma. Lemma 6.3.
Let a, b, c be the vertices of a triangle in Γ (cid:48) . Then, at least one of the edges ( a, b ) , ( b, c ) and ( c, a ) is not a sub-edge of any of the edges of Γ .Proof. The vertices of Γ (cid:48) can be divided into two disjoint sets: the vertices of Γ and themidpoints of edged of Γ . If ( a, b ) is a sub-edge of an edge of Γ then without loss of generality,we can assume that a is a vertex of Γ and b is a midpoint of an edge of Γ . Since a and c are joined by an edge in Γ (cid:48) , c must be a midpoint of an edge of Γ . Indeed, in Γ (cid:48) , all theneighbors of a vertex of Γ are midpoints of edges of Γ . Then, both end vertices of the edge ( b, c ) are midpoints of edges of Γ . Therefore, ( b, c ) is not a sub-edge of any edge of Γ .Now we can complete the proof. If ∆( a, b, c ) is a separating triangle we choose an edgeof ∆( a, b, c ) which is not a sub-edge of any edge of Γ . We separate it into two edges andtriangulate the resulting graph. The number of separating triangles in the graph decreasesafter this step, no new separating triangle occurs and Lemma 6.3 still holds for the resultinggraph.After a finite number of iterations (for an illustration see Figure 6.16b), we get a maximalsimple plane graph with no separating triangles and thus a -page book embeddable graph.Note that during the process of eliminating the separating triangles of Γ (cid:48) , we have onlydivided edges which are not sub-edges of edges of Γ . Thus, the subdivision D (Γ) , whereevery edge of Γ is divided in two, is a subgraph of the resulting maximal simple plane graphand in particular it is 2-page book embeddable. igure 6.16: (a) a triangulation of the subdivision D (Γ) of Γ from Figure 6.12b. (b) The graph Γ (cid:48) after two iterations. The edges e and f were divided in two and the graph was triangulated. Thereare no more separating triangles so the graph is -page book embeddable. Let Γ be a signed plane graph with no loops. At first we consider the graph Γ as a nonlabeled plane graph. Let D (Γ) be the subdivision of Γ where every edge is divided in two.By Theorem 6.2, D (Γ) is -page book embeddable. Therefore there exists an oriented simplecurve α such that α passes through all the vertices of D (Γ) , does not cross any of its edgesand the two infinite rays of α are in the outer face of D (Γ) . See Figure 6.17a.Let v be a midpoint of an edge e of Γ , considered as a vertex of D (Γ) . When the curve α passes through v , it either remains on the same side of e or crosses the edge e . In the firstcase, we can eliminate the vertex v from D (Γ) and adjust the curve α accordingly. The resultwould be a 2-page book embeddable graph (with one less vertex) together with a suitablecurve α . Eliminating all the possible midpoints of edges of Γ in this way, results in a planegraph Γ (cid:48) . see Figure 6.17b.(a) (b) Figure 6.17: (a) The subdivision D (Γ) and the curve α . (b) Midpoints of edges not crossed by α are erased and the curve α is adjusted. We shall consider the isotopy of the plane which takes the directed curve α to the x -axisdirected from left to right. In accordance with this isotopy, we say that edges to the right of α , are below it and edges to the left of α are above it.Let v be a vertex of Γ (cid:48) which is a midpoint of some edge e of Γ . Then, up to reflections,the curve α passes through v as in Figure 6.18. Lemma 6.3.
It is possible to add an additional vertex to the edge e (thus, dividing itto sub-edges) and adjust the curve α accordingly, so as to remain with a -page bookembeddable graph and a suitable curve α . Moreover, it can be done, so that of the sub-edges of e would be above (below) α and the other one below (above) it. Proof.
The process is demonstrated in Figure 6.19. igure 6.18: The curve α crosses every edge of Γ divided in two in Γ (cid:48) .Figure 6.19: Adding a vertex to an edge of Γ cut in two in Γ (cid:48) and adjusting the curve α accordingly.In the right top (bottom) figure, two of the sub-edges of e are above (below) α and the other oneis below (above) it. Now we consider the labels of the edges of Γ . For each edge of Γ which is not dividedin two in the transition to Γ (cid:48) , we assign the corresponding edge of Γ (cid:48) its label in Γ . If anedge e of Γ is replaced by two edges in Γ (cid:48) we do the following. If e is labeled by "+" ("-")in Γ we use Lemma 6.3 to replace its sub-edges in Γ (cid:48) by sub-edges and adjust the curve α so that two of the sub-edges are above (below) α and the other one is below (above) it.We label the resulting edges of Γ (cid:48) according to their position with respect to α , where anedge above α is labeled by "+" and an edge below α is labeled by "-". We call the resultinggraph Γ (cid:48)(cid:48) . See Figure 6.20. Figure 6.20: The graph Γ (cid:48)(cid:48) with the curve α . Proposition 6.4.
The graph Γ (cid:48)(cid:48) is equivalent to the graph Γ .Proof. The graph Γ (cid:48)(cid:48) results from the graph Γ by dividing some of its edges to sub-edgesand assigning them labels. If e is an edge of Γ labeled by "+" in Γ and replaced by edgein Γ (cid:48)(cid:48) , then two of the edges replacing it are labeled by "+". Therefore, if e + and e − arethe end vertices of e in Γ , then either the sub-edge of e incident to e − or the sub-edge of incident to e + is labeled by "+". Assume the sub-edge of e incident to e − is labeled by"+". Then, applying a type 2 move on the graph Γ (cid:48)(cid:48) , it is possible to eliminate the othertwo sub-edges of e (which meet at a vertex of degree and have opposite signs) and identifytheir end vertices. We get the original edge e labeled by "+".Thus we have proved the following. Theorem 6.4.
Let Γ be a signed plane graph with no loops. Then, there exists a signedplane graph Γ (cid:48)(cid:48) such that1. Γ (cid:48)(cid:48) is equivalent to Γ .2. Γ (cid:48)(cid:48) is a subdivision of Γ where each edge of Γ is divided to sub-edges or not dividedat all.3. Γ (cid:48)(cid:48) can be -page book embedded, so that for every edge e of Γ which is replaced by sub-edges in Γ (cid:48)(cid:48) , the signs of all sub-edges in Γ (cid:48)(cid:48) are compatible with their embeddingbelow or above the x -axis. -page book embeddable graphs to Thompsongraphs Let Γ be a signed plane -page book embeddable graph with no loops. We use an isotopywhich takes all the vertices of Γ to the x axis and each edge of Γ to an edge above or belowthe x -axis. Following [11] we call the resulting graph standard .The image of Γ (cid:48)(cid:48) from Figure 6.20 under the isotopy which takes α to the x -axis (directedfrom left to right) appears in Figure 6.21. All the unlabeled edges in the figure have labelscompatible with their position above or below the x -axis. The graph in the figure is standard.Given a standard graph we enumerate its vertices from left to right by , , . . . . The innervertices of a standard graph are all the vertices except for vertex number . Figure 6.21: A -page book embedding of the graph Γ (cid:48)(cid:48) . Definition 6.5.
Let Γ be a standard graph. We orient all the edges of Γ from left to right.The graph Γ is called Thompson if every inner vertex of Γ has exactly one upper incomingedge (i.e., an edge above the x -axis) and one lower incoming edge (i.e., an edge below the x -axis) and in addition all the upper edges in Γ are labeled by "+" and all the lower edgesin Γ are labeled by "-". Lemma 6.5. [11, Lemma 5.3.3] Let Γ be a standard graph. Then Γ is equivalent to aThompson graph. Proof.
The proof follows closely the proof of [11, Lemma 5.3.3]. We say that a vertex is good if it has at least one upper and one lower incoming edge. We say that an edge is good if itslabel is compatible with its position with respect to the x -axis and it is not superfluous . An pper (lower) edge is said to be superfluous if it is not the bottom-most (top-most) incomingupper (lower) edge of its right end vertex. If Γ is not Thompson, it has a bad vertex or abad edge (where bad is the opposite of good). Thus, at least one of the following occurs in Γ . 1. There is an inner vertex in Γ with no incoming edges.2. There is an inner vertex in Γ with no upper (lower) incoming edge but with a lower(upper) incoming edge.3. There is a superfluous edge in Γ .4. There is an edge in Γ with a label non-compatible with its position.We turn Γ into a Thompson graph in steps, taking care of each of the above problems inturn. All the edges and vertices added during this process are good. In all the figures inthis proof (namely, Figures 6.22-6.25), all the unlabeled black edges can have arbitrary signs.The non-black edges (i.e., the edges attached during this process) have signs compatible withtheir position.Step 3.1. If an inner vertex i has no incoming edges, we connect the vertices ( i − and i by two edges. One upper edge labeled by "+" and one lower edge labeled by "-" as donein Figure 6.22. We do it using a type move. The vertex i becomes good. Figure 6.22: Turning a vertex with no incoming edges to a good vertex.
Applying this step to all the relevant vertices, we can assume that every inner vertex hasat least one incoming edge.Step 3.2. Let i be an inner vertex with no lower incoming edge (the case where i has noupper incoming edge is similar). To correct the problem we apply a move of type 1 followedby a move of type 3 as demonstrated in Figure 6.23 (the vertices in the left figure are thevertices ( i − and i ). One new vertex is added in the process. Figure 6.23: Adding an incoming lower edge.
After several applications of this step, we can assume that all vertices have incomingupper and lower edges and are thus good.Step 3.3. Let e be a superfluous upper edge in Γ (the case of a superfluous lower edge issimilar). We assume that e is the top-most upper incoming edge of its right end vertex v .We apply a move of type 2 to separate the vertex v to two vertices connected by a path oftwo edges with opposite signs. All the incoming edges of v , other than e , remain connectedto its left copy (which we consider to be the vertex v ). The edge e as well as the outgoingedges of v are connected to the right copy of v (which we count as a new vertex). Thenwe apply a move of type 1 followed by a move of type 3 (as in Step 3.2). The process isdemonstrated in Figure 6.24. All the vertices after this step are good. All the new edges aregood and the edge e is no longer superfluous. new vertices are added in this step. igure 6.24: Fixing a superfluous edge. After several iterations of this step we can assume that all vertices are good and thereare no superfluous edges.Step 3.4. Let e be an edge with a label incompatible with its position. We assumethat e is an upper edge labeld by "-", the other case being similar. We fix the problem asdemonstrated in Figure 6.25. We apply two moves of type 2 (see the second figure from theleft), then apply an isotopy which turns e to a lower edge, followed by several moves of type1 and 3 as in Step 3.2. new vertices are added in this step. Figure 6.25: Moving a "-" edge down.
After applying this step to every edge with a non compatible label, we get a graph whereall the vertices and all the edges are good. This is a Thompson graph and it is clearlyequivalent to the original graph Γ . Example 6.6.
We demonstrate the process of turning a standard graph to an equivalentThompson graph on the graph Γ (cid:48)(cid:48) from Figure 6.21. Since every vertex in the graph has atleast one incoming edge, step 3.1 from the proof of Lemma 6.5 is not necessary. Steps 3.2,3.3 and 3.4 are depicted in Figures 6.26, 6.27 and 6.28. All the unlabeled edges in the figureshave signs compatible with their positions. The graph in Figure 6.28 is Thompson. It isequivalent to Γ (cid:48)(cid:48) . Figure 6.26: The graph after an application of Step 3.2.
Lemma 6.7.
Let L be a link which contains u unlinked unknots. Assume that L is repre-sented by a link diagram with n crossings. Then, there exists a Thompson graph Γ with atmost n + u vertices such that L (Γ) is equal to L . Proof.
We shall assume that the link L consists of a single component which cannot beseparated into several unlinked components (the general case will be addressed later below). igure 6.27: The graph after an application of Step 3.3.Figure 6.28: The graph after an application of Step 3.4. Let L (cid:48) be a link diagram which represents L . By assumption, L (cid:48) is connected. Also, byRemark 6.2 we can assume that Γ = G ( L (cid:48) ) has no loops. Since every edge of Γ correspondsto a crossing in L (cid:48) , we have | E (Γ) | = n . If Γ is a tree, then it is equivalent (by means oftype 1 moves) to a Thompson graph composed of a single vertex. This Thompson graphclearly satisfies the conclusion of the lemma. Thus, we can assume that Γ is not a tree. L (cid:48) being connected implies that Γ is connected, therefore | V (Γ) | ≤ | E (Γ) | = n . Let Γ (cid:48)(cid:48) be the -page book embeddable graph equivalent to Γ , from Theorem 6.4 and consider itsembedding defined by the curve α from the theorem.Let m be the number of edges in Γ which are divided to edges in Γ (cid:48)(cid:48) . Then | E (Γ (cid:48)(cid:48) ) | = n + 2 m whereas | V (Γ (cid:48)(cid:48) ) | ≤ n + 2 m .Now we apply the process explained in Lemma 6.5 to the standard graph Γ (cid:48)(cid:48) and considerthe number of additional vertices at each step. In the analysis we distinguish between theoriginal vertices and edges of Γ (cid:48)(cid:48) and those added during the process.In Step 3.1, no new vertices are added.Assume that x vertices are dealt with in Step 3.2. For each of these vertices one newvertex is added. Thus, x new vertices are created in this step. All of them are good.In Step 3.3 we deal with superfluous edges. Every edge added during the process is good.Hence only the n + 2 m original edges of Γ (cid:48)(cid:48) can be superfluous. Each of the x vertices dealtwith in the second step has an original edge of Γ (cid:48)(cid:48) as an incoming edge (otherwise it wouldhave been dealt with in Step 3.1). If that edge is upper (lower), then all the upper (lower)incoming edges of the vertex are original edges of Γ (cid:48)(cid:48) . The bottom-most (top-most) of themis not superfluous. Hence at least x of the original edges of Γ (cid:48)(cid:48) are not superfluous. Thusdealing with at most ( n + 2 m − x ) edges results in at most n + 2 m − x ) additional vertices.Finally, in Step 3.4 we "move" edges whose labels are incompatible with their positionsabove or below the x -axis. Only the original edges of Γ (cid:48)(cid:48) which are edges of Γ (as opposed tosub-edges of edges of Γ ) can have incompatible labels. Moving each of these edges requiresthe addition of new vertices. Thus, at most n − m ) new vertices are added in this step.The total number of vertices in the resulting Thompson graph is at most n + 2 m + x +3( n + 2 m − x ) + 8( n − m ) ≤ n , as requested. ow, in general, let L be a link composed of two unlinked components L and L . Let L (cid:48) be the link diagram of L . We can assume that L (cid:48) has two connected components L (cid:48) and L (cid:48) corresponding to L and L . Indeed, separating unlinked components of a link diagramcan be done without increasing the number of crossings in the diagram. Let Γ and Γ bethe Thompson graphs of L and L , which satisfy the conclusion of the lemma with respectto the number of crossings and unlinked unknots in L (cid:48) and L (cid:48) . Then, drawing Γ to theright of Γ and attaching the left-most vertex of Γ to the right-most vertex of Γ resultsin a Thompson graph Γ corresponding to the link L . Since no vertices are added in thisconstruction and the upper bound n + u is linear, it holds in the general case as well. F Let Γ be a standard graph which is Thompson. As an unlabeled graph, Γ is the Thompsongraph T (∆) of some ( x, x ) -diagram ∆ over the semigroup presentation (cid:104) x | x = x (cid:105) . Indeed,considering the edges of Γ to be the left edges of ∆ (in accordance with Remark 3.3), it ispossible to complete it to a diagram ∆ by attaching suitable right edges. The diagram ∆ does not have to be reduced but it has no π − ◦ π dipoles.In particular, we can finally conclude that the link L from Figure 6.11 corresponds to thefollowing element of Thompson group F : ( x x x x x x x x x x x x x x x )( x x x x x x x x x x x x x x x x x x x ) − . Given a diagram ∆ over (cid:104) x | x = x (cid:105) , we define the signed plane graph Γ(∆) to be theThompson graph T (∆) with all the upper edges labeled by "+" and all the lower edgeslabeled by "-". We say that ∆ represents the link L if G ( L ) = Γ(∆) . Lemma 6.8.
Let ∆ be an ( x, x ) - diagram over the semigroup presentation (cid:104) x | x = x (cid:105) suchthat ∆ has no dipoles of the form π − ◦ π .1. If ∆ is not reduced then the link it represents L = L (Γ(∆)) has an unlinked unknot asa component.2. If ∆ (cid:48) results from ∆ by reducing a π ◦ π − dipole, then the link it represents L (cid:48) = L (Γ(∆ (cid:48) )) results from L by the removal of one unlinked unknot. Proof.
It suffices to prove part (2). To get the graph T (∆ (cid:48) ) we remove from T (∆) a vertexwith exactly two edges, one upper and one lower, connecting it to another vertex of T (∆) .Thus Γ(∆ (cid:48) ) results from Γ(∆) by an application of a move of type and the subsequentremoval of a vertex which becomes isolated. A move of type does not affect the associatedlink. Since an isolated vertex in a graph corresponds to an unknot in a link diagram, itsremoval implies the removal of one unlinked unknot from the corresponding link.Let ∆ be a reduced nontrivial diagram in F we denote by L (∆) the link represented by ∆ . Lemma 6.9.
The link L ( ⊕ ∆) results from the link L (∆) by an addition of one unlinkedunknot. Proof.
To get from the Thompson graph T (∆) to the Thompson graph T ( ⊕ ∆) one has toadd a vertex to the left of the initial vertex of T (∆) and connect it to the initial vertex of T (∆) by two edges; an upper edge and a lower edge. Therefore, Γ( ⊕ ∆) results from Γ(∆) y the addition of an isolated vertex to the left of Γ(∆) (hence, an addition of an unlinkedunknot to the associated link) followed by a type move which does not affect the associatedlink. Lemma 6.10.
Let L be the link which consists of u ≥ unlinked unknots. Then, L isrepresented by the element x u − of Thompson group F . In particular, it is represented by areduced diagram with u + 3 vertices. Proof.
It is easy to check that x represents the unknot (see Figure 6.29). Since for all i ≥ ,we have, x i +1 = ⊕ x i , the result follows from Lemma 6.9. Figure 6.29: The diagram of x , its Thompson graph and the corresponding shaded link. Lemma 6.11.
Let L be a link which is not an unlink. Assume that L contains u unlinkedunknots and has a link diagram with n crossings. Then, there exists a reduced diagram ∆ in F with at most n + u + 1 vertices which represents L . Proof.
By Lemma 6.7 there exists a Thompson graph Γ with at most n + u vertices suchthat L (Γ) = L . Let ∆ be the ( x, x ) -diagram over (cid:104) x | x = x (cid:105) such that Γ(∆) = Γ . Clearly,the number of vertices of ∆ is at most n + u + 1 . As noted above, ∆ does not contain a π − ◦ π dipole. Let ∆ (cid:48) be the reduced diagram equivalent to ∆ and assume that k dipoles(of the form π ◦ π − ) should be canceled to get from ∆ to ∆ (cid:48) . Then, by Lemma 6.8, thediagram ∆ (cid:48) represents the link resulting from L by the removal of k unlinked unknots. Notethat ∆ (cid:48) is not the trivial diagram since L is not a family of unlinked unknots. Let ∆ (cid:48)(cid:48) be thediagram resulting from ∆ (cid:48) by the addition of the trivial diagram k times on the left. Then, ∆ (cid:48)(cid:48) is a reduced diagram which by Lemma 6.9 represents L . Since k vertices were erased inthe transition from ∆ to ∆ (cid:48) and k vertices were added in the transition from ∆ (cid:48) to ∆ (cid:48)(cid:48) , thenumber of vertices of ∆ (cid:48)(cid:48) is at most n + u + 1 . Lemma 6.12.
Let L be a link represented by a link diagram with at most n crossings.Assume that L does not contain any unlinked unknot. Then, L is represented by an element g of Thompson group F which satisfies the following.1. The normal form of g or its inverse starts with x .2. The number of vertices in the reduced diagram of g is at most n + 1 . Proof.
Let ∆ be the reduced diagram from Lemma 6.11 which represents L . Then, thenumber of vertices of ∆ is at most n + 1 . If the normal form of ∆ does not contain x nor x − , then, ∆ = ⊕ ∆ (cid:48) for some reduced diagram ∆ (cid:48) . By Lemma 6.9, the link L representedby ∆ contains an unlinked unknot, a contradiction. et L be a link. Jones [11] defined the Thompson index of L as the minimal number ofvertices in the diagram representing an element g which represents L . Lemmas 6.10 and 6.11imply the following. Theorem 6.13.
The Thompson index of a link containing u unlinked unknots and repre-sented by a link diagram with n crossings does not exceed n + u + 3 . Remark 6.14.
It is known [2] that the word length of an element g from F in the generators x , x does not exceed 3 times the number of vertices in the diagram of g . Thus Theorem6.13 gives a linear upper bound on the word length of an element g representing L in termsof the number of crossings in a link diagram of L . F Let us call an element g of F unlinked if the link L ( g ) is an unlink. The following problemseems difficult Problem 6.15.
Describe the set of unlinked elements of F .Theorem 6.13 implies that the problem of recognizing an unlinked element of F is polyno-mially equivalent to the problem of recognizing an unlink (here elements of F are representedby diagrams or pairs of trees or words in { x , x } and links are represented by link diagrams).By the famous result of Haken [9], the set of unlinked elements is recursive. By [8], it is inNP and by [13] it is also in coNP (modulo the generalized Riemann hypothesis). One can easily verify that many positive elements of F (i.e., elements whose normal formhas trivial negative part) correspond to unlinks (examples: x i , x i x i +1 , etc.). There are,nevertheless, positive elements of F that correspond to non-trivial links (for example, x x ). Problem 6.16.
Which links correspond to positive elements of F ? When does a positiveelement of F represent an unlink? The Jones’ theorem that elements of F represent all links and Remark 6.14 relating thenumber of crossings in a link diagram with the word length of a corresponding element of F suggests the following Problem 6.17.
Consider a (simple) random walk w ( t ) on F , say, with generators x ± , x ± .What can be said about the link corresponding to w ( t ) ? In particular, what is the probabilitythat the link, corresponding to w ( t ) is an unlink? References [1] Kenneth S. Brown, Finiteness properties of groups. Proceedings of the Northwesternconference on cohomology of groups (Evanston, Ill., 1985). J. Pure Appl. Algebra 44(1987), no. 1–3, 45–75.[2] J. Burillo, S. Cleary and M. I. Stein, Metrics and embeddings of generalizations ofThompson’s group F . Trans. Amer. Math. Soc. 353 (2001), no. 4, 1677–1689.
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