Pentagon and hexagon equations following Furusho

In [F] H. Furusho proves the beautiful result that of the three defining equations for associators, the pentagon implies the two hexagons (see also [W]). In this note we present a simpler proof for this theorem (although our paper is less dense, and hence only slightly shorter). In particular, we package the use of algebraic geometry and Groethendieck-Teichmuller groups into a useful and previously known principle, and, less significantly, we eliminate the use of spherical braids.


Introduction
Associators are useful and intricate gadgets that were first introduced and studied by Drinfel'd in [Dr1] and [Dr2]. The theory was later put in the context of parenthesized (a.k.a. non-associative) braids in [LM] and [BN]. Associators arise in several different areas of mathematics, and thus constructing an associator is a task faced by many. Unfortunately, it is a very difficult task: no closed formulas are known at present.
Associators are essentially the solutions to three equations, called the pentagon and the positive and negative hexagons, which live in complicated diagrammatic spaces. Finding an associator amounts to finding a solution to this system of equations. Furusho's result states that the last two of these three equations are superfluous: a solution to the pentagon will automatically be a solution to both hexagons. This statement is quite surprising, and thus we felt that a simpler proof would be of value.
The paper is organized as follows: we first review some definitions, then present the main tool or "extension principle" for the simplified proof: a theorem from [Dr2] and [BN] and also a simpler standard fact which we call the "linearization principle". We then prove the theorem modulo a "Lie algebraic" version, which we call the "Main Lemma", followed by the proof of the Main Lemma, and finally a side note on how one of the algebraic maps used in the proof arises from topology.
The part of the proof which depends on the extension principle is significantly different from that of [F] (and [W]) and significantly simpler: Furusho's proof uses algebraic geometry and Groethendieck-Teichmüller groups to go from the Lie algebra version to the group statement.
The proof of the Main Lemma is essentially Furusho's proof, except for the elimination of the use of spherical braids. Using spherical braids makes the statement of the Main Lemma look very symmetric, so in some sense prettier. However, in our opinion, it makes the combinatorial argument of the proof less transparent: it is the break of symmetry which helps the reader figure out how someone might have discovered the proof. Also, we save the trouble of having to pass back and forth between the spherical and regular chord-diagram spaces. The elimination of spherical braids is easy: the space of chord diagrams of pure spherical 5-braids and that of regular pure 4-braids only differ in taking a quotient by the center of the latter.

Definitions
Algebraically, the space of chord diagrams of pure n-braids, A n , is defined to be the graded completion of the following non-associative algebra, where generation is understood over a fixed field k: Here, 4T is short for four term relations, and L stands for locality.
One usually thinks of A n in terms of horizontal chord diagrams on n strands, where t ij is represented by a chord between strands i and j, and multiplication is done by stacking diagrams: Let us denote by F 2 the free Lie algebra over a field k of characteristic 0, on two generators X and Y , and by UF 2 its universal enveloping algebra, which is isomorphic to k X, Y , the algebra of non-commutative power series over k. By a superscript (m), as in UF (m) 2 , we mean the space or object modulo degree (m + 1), e.g. non-commutative polynomials of degree up to m.
There is a co-product on UF 2 , defined by Δ( Now let us present the main characters of this story: the pentagon and hexagon equations.
The pentagon equation originates from the fact that in a non-associative algebra, there are five ways to multiply four elements, and these are connected by a pentagon of re-associations. This fact and its parenthesized braid analogue are shown in the following figure: The above equality of parenthesized braids implies an algebraic equation in A 4 called the pentagon equation, The hexagons arise from the following equivalences of parenthesized braids: Here, e(x) := e x . An associator is a group-like element of UF 2 which satisfies the pentagon and hexagon equations.
Note that the spaces A 3 and UF 2 are almost isomorphic: An associator is sometimes defined as an element of A 3 , which is equivalent, since associators are commutator-group-like, meaning their abelianization is 1. (This follows from the pentagon equation.)

Theorem and proof
In [F] Furusho proves the following surprising result: To prove Theorem 3.1 we need two tools, which we will call extension and linearization principles, and a major lemma, Lemma 3.2 below.
The "extension principle" is the assertion that an associator modulo degree m can be lifted to an associator modulo degree (m + 1). This statement is highly non-obvious; it takes up a good part of [Dr2] and the main part of [BN]. Its main application is the fact that non-trivial rational associators exist: indeed, up to degree 2 a direct computation shows that P (Φ) = 1 and H ± (Φ) = 1 have a unique non-trivial solution (namely, Φ (2) = 1 + 1 24 [X, Y ]), and then by extension, one may construct a rational associator inductively.
To prove the theorem we use the following main lemma: Proof of Theorem 3.1 from Lemma 3.2. Let us assume that Φ is as in Theorem 3.1, in particular P (Φ) = 1, and, by contradiction, that one of the hexagons fails to hold, i.e. H ± (Φ) = 1. Let m be the minimal degree in which this happens. By the simple computation in low degrees mentioned before, we know that m ≥ 3. Note that by the minimality of m, Φ satisfies the hexagons modulo degree m and hence it is an associator modulo degree m. By extension, there exists a Φ ∈ UF 2 which agrees with Φ modulo degree m and which satisfies both the pentagon and the hexagon equations modulo degree (m + 1). Let ϕ = (Φ − Φ ) (m) , which is homogeneous of degree m since Φ = Φ modulo degree m.
Let us start by proving two simple but necessary lemmas. Throughout this section, ϕ is assumed to be a primitive, homogeneous element of UF(2) of degree ≥ 3, as in Lemma 3.2.

Lemma 4.2. The linearized hexagon dH(ϕ) = 0 is equivalent to the equation
By an abuse of notation, we shall denote the left side of equation (1) by dH(X, Y ).
For the other direction we apply the map i : UF 2 → A 3 given by i(X) = t 12 and i(Y ) = t 23 : where the second equality is due to the fact that ϕ ∈ [F 2 , F 2 ] (since it is primitive of degree ≥ 2) and the fact that the element t 12 + t 23 + t 13 is central in A 3 .
This means that dH(X, Y ) contains only commutators that involve one X and some number of Y 's, and so writing dH in a linear basis we have: X), we know that a n = 0 for all n except possibly n = 2, and a 2 = 0 by the assumption that ϕ is of degree ≥ 3. Thus, dH(X, Y ) = 0, which is equivalent to the Main Lemma by Lemma 4.2.
It is easy to verify that the rest of the proof goes through the same way using this equality in place of the one above.

A side note
The projection q, used in the proof of Lemma 4.1, has an interesting property and a nice topological interpretation.
If we embed A 3 in A 4 on any three strands (i.e. by any embedding of the index set [3] into [4]) and then apply q, we get the "almost isomorphism" between A 3 and F 2 ; i.e. the composition is factoring out by the center of A 3 : q 3 strands embed on any A 4 mod out by center = t 12 + t 23 + t 13 (almost an isomorphism) This is a braid theoretic analogue of the following fact about the symmetric group S 4 . Since S 4 is isomorphic to the group of symmetries of the tetrahedron and each element of S 4 also permutes the three pairs of opposite edges of the tetrahedron, we obtain a map p : S 4 → S 3 . Pre-composing this map with any embedding of S 3 into S 4 induced by an embedding of the set [3] into [4], we get an automorphism of S 3 , namely, an isomorphism from S 3 to the group of symmetries of a face: S 4 permutes sets of opposite edges ⇒ map p: S 4 → S 3 permute any 3 points S 3 S 4 S 3 p ≅ Topologically, q : A 4 → F 2 is induced by a mapq, defined as follows: pB 4 spB 4 pB 3 / full twist q 2 q 1q = Here, pB i denotes the pure braid group on i strands, and spB 4 denotes the group of pure spherical braids on four strands. Spherical braids live in S 2 × I, as opposed to D 2 × I, which means that one strand wrapping all the way around the others is trivial, as shown for strand 1 on the right, and similarly for all other strands. This defines the quotient mapq 1 above. = Forq 2 , take any spherical braid on four strands, pull the last strand straight, and consider the first three strands as a braid in the complement of strand 4. The target space of this map is the group of regular pure 3-braids factored out by the full twist of the three strands: pull the 4-th strand straight on the left side of the spherical relation shown in the figure on the right, and observe that what we get is a full twist of the first 3 strands, which then has to be trivial in the image.
Note that the chord diagram map induced by factoring out by the full twist in pB 3 is exactly the quotient map π : A 3 → F 2 which sends t 12 + t 23 + t 13 to 0.