On the structure of set-theoretic polygon equations

Polygon equations generalize the prominent pentagon equation in very much the same way as simplex equations generalize the famous Yang-Baxter equation . In particular, they appeared as “cocycle equations” in Street’s category theory associated with oriented simplices. Whereas the ( N − 1)-simplex equation can be regarded as a realization of the higher Bruhat order B ( N, N − 2), the N -gon equation is a realization of the higher Tamari order T ( N, N − 2). The latter and its dual ˜ T ( N, N − 2), associated with which is the dual N -gon equation, have been shown to arise as suborders of B ( N, N − 2) via a “three-color decomposition”. There are two different reductions of T ( N, N − 2) and ˜ T ( N, N − 2), to T ( N − 1 , N − 3), respectively ˜ T ( N − 1 , N − 3). In this work, we explore the corresponding reductions of (dual) polygon equations, which lead to relations between solutions of neighboring (dual) polygon equations. We also elaborate (dual) polygon equations in this respect explicitly up to the octagon equation.


Introduction
An infinite family of "polygon equations" has been introduced in [13], based on the combinatorial structure of higher Tamari (partial) orders.They appeared earlier as cocycle equations ("higher cocycloids") in a category-theoretical framework based on oriented simplices ("orientals") [57].The 4-gon equation is the condition of associativity for a binary operation.The dual 4-gon equation expresses coassociativity of a comultiplication.The 5-gon or pentagon equation is the most prominent member of this family and plays a profound role in mathematics and mathematical physics.
Of particular relevance is the pentagon equation in the context of bi-and Hopf algebras.The crucial and fairly simple observation here is the following.Given a unital associative algebra A with identity element 1 A , it carries a trivial (left) comultiplication1 ∆ ℓ : A → A ⊗ A, a → a ⊗ 1 A .Non-trivial comultiplications are then obtained via ∆ := W ∆ ℓ W −1 if W ∈ A ⊗ A is an invertible solution of the pentagon equation2 W 12 W 13 W 23 = W 23 W 12 in the threefold tensor product of A, where the indices determine at which two components of it W acts.In fact, each comultiplication on a finite-dimensional algebra can be expressed in this way [48].But also in the rigorous framework of infinite-dimensional C * -algebras such an expression for comultiplication plays a profound role in surpassing technical problems in the duality theory of locally compact groups and for establishing a rigorous setting for quantum groups.Here W appears as a unitary linear operator under the name "multiplicative unitary" or (generalized) Kac-Takesaki operator (see, in particular, [1,38,54,60,61,62,69]).
The pentagon equation, sometimes called "fusion equation", shows up in conformal field theory as a condition for the fusion matrix, which arises from operator product expansion of chiral vertex operators, see [49] for example.
The pentagon equation plays a crucial role as an associativity constraint for the basic functor of a monoidal category, notably because of Mac Lane's coherence theorem [40] (also see [33]).This in turn has roots in Tamari's work on relaxing associativity and Stasheff's work on homotopy associativity [55,56].
One way to obtain topological invariants of manifolds is via triangulations, i.e., decompositions into simplices.To each simplex one assigns a mathematical object and builds from all of them an object (e.g., a "state sum") assigned to a triangulation.This has to be done in such a way that the latter object is invariant under manipulations of triangulations called Pachner moves or bistellar flips [50] (also see, e.g., [39]).In three dimensions, the simplices are tetrahedra.A Pachner (2, 3)-move splits two adjacent tetrahedra into three and this requires that the objects associated with the tetrahedra, constituting a triangulation, satisfy a pentagon relation.Because of the pentagonal Biedenharn-Elliot identity [4], a Wigner 6-j symbol, or a generalization of it, is thus a choice for the object assigned to a tetrahedron [51,63,64]. 3In [3], it was shown that the Wheeler-DeWitt equation for three-dimensional general relativity reduces to the pentagon relation.There is a comprehensive literature around the idea of representing Pachner moves of a triangulation in three dimensions by a solution of the pentagon equation (or a pentagon relation), see in particular [2,8,24,34,47,59].In four dimensions, the dual hexagon equation plays a corresponding role [29] (also see Remark 6.2 below), in five dimensions it is the dual heptagon equation [35,36,37].
In the latter context, it is also worth to mention that the higher Tamari orders, introduced in [13] and underlying polygon equations, are equivalent [67] to higher Stasheff-Tamari orders [16,23] on the set of triangulations of a cyclic polytope.
A pentagonal relation is also satisfied by the Rogers [52] and quantum dilogarithms4 [18,19].Although it does not have the structure of the pentagon equation as considered in this work, there are certain relations, see, e.g., Example 7.20 below. 5omparatively little is known so far about higher polygon equations.An expression for the dual hexagon equation appeared as a 4-cocycle condition in [57,58].
Polygon equations play a role in the theory of completely solvable models of statistical mechanics.If solutions of the N -gon and the dual N -gon equation satisfy a certain compatibility condition, a special solution of the (N − 1)-simplex equation is obtained, see [13,Section 5].This includes and generalizes a relation between solutions of the pentagon equation, its dual, and the 4-simplex (Bazhanov-Stroganov) equation [31].There is also a relation between the pentagon equation, its dual, and the 3-simplex (tetrahedron or Zamolodchikov) equation [31,32,41,53], for which a corresponding generalization to higher polygon and simplex equations is yet unknown.
A class of solutions of odd polygon equations with maps acting on direct sums has recently been obtained in [10].
In the present work, we explore polygon equations in the set-theoretic setting, i.e., we do not assume any additional structure beyond sets and maps between (Cartesian products of) 3 Essentially, (generalized) 6-j symbols arise as follows.If there is a direct sum decomposition them.Using the combinatorics of higher Tamari orders, underlying polygon equations, we derive relations between the sets of solutions of neighboring (dual) polygon equations.These relations are proved for the whole infinite family of (dual) polygon equations.Additional results are provided for (dual) polygon equations up to the octagon (8-gon) equation.Section 2 briefly recalls the definition of higher Bruhat and higher Tamari orders.The reader is referred to [13] for a detailed treatment.Section 3 presents the definition of simplex and polygon equations in a slightly different way from [13].Section 4 deals with reductions of polygon equations and contains the most general results of this work.Sections 5 and 6 present the first few examples of polygon equations and their duals, respectively.They extend results obtained in [13].
Section 7 deals with the special, but most important case of (dual) polygon equations for a single map, acting between Cartesian products of a set U. Partly from the general results in Section 4, but to some extent also independently and from results in Sections 5 and 6, we derive relations between solutions of a (dual) N -gon and (dual) (N + 1)-gon equation, N > 2.
Section 8 contains some concluding remarks.
2 Basics of higher Bruhat and higher Tamari orders n+1 , the packet P (K) is contained in ρ in lexicographic or in reverse lexicographic order.Let A(N, n) denote the set of admissible linear orders of [N ]  n .An equivalence relation is defined on A(N, n) by ρ ∼ ρ ′ if and only if ρ and ρ ′ only differ by exchange of two neighboring elements, not both contained in some packet.Then A(N, n)/∼, supplied with the partial order given via inversions of lexicographically ordered packets, − → P (K) → ← − P (K), is the higher Bruhat order B(N, n).Next we consider the splitting of a packet, P (K) = P o (K) ∪ P e (K), where P o (K) (P e (K)) is the half-packet consisting of elements with odd (even) position in the lexicographically ordered P (K).
We say an element with respect to all K for which J ∈ P (K) and either − → P (K) or ← − P (K) is a subsequence of ρ.It can happen that J is blue in ρ ∈ A(N, n) with respect to some K and red with respect to another K ′ .In such a case we color it green.By ρ (b) , ρ (r) , ρ (g) we denote the blue, red, respectively green subsequence of ρ.
It has been shown in [13] that there are projections are the higher Tamari orders and T (N, n) := B (r) (N, n) are called dual higher Tamari orders.

The inversion operation in case of
Remark 2.1.Associating with K ∈ [N ]  n+1 an n-simplex, the packet P (K) corresponds to the set of its faces, which are (n − 1)-simplices.The pasting scheme given by the above inversion operation then supplies the faces with an orientation.This results in the orientals (oriented simplices) introduced by Street in 1987 [57].It had been conjectured in [12,13] and proved in [67] that the higher Tamari orders are equivalent to the higher Stasheff-Tamari orders in [16,23].All these works thus deal with essentially the same structure.In [11,12] it has been realized in terms of rooted tree-shaped solutions of the Kadomtsev-Petviashvili (KP) hierarchy.

A brief account of simplex and polygon equations
Let N > 2. With J ∈ [N ]  N −2 we associate a set U J .For ρ ∈ A(N, N − 2), let U ρ be the correspondingly ordered Cartesian product of the U J , J ∈ ρ.With K ∈ [N ]  N −1 = P ([N ]) we associate a map The (N − 1)-simplex equation may then be regarded as a realization of B(N, N − 2).The expressions on both sides are compositions of maps R K , K ∈ P ([N ]), applied on the left-hand side in lexicographic, on the right-hand side in reverse lexicographic order.Writing ) is actually in reverse lexicographic order, but the maps are applied in lexicographic order.We have to add the following rules in order for (3.1) to make sense.
(2) Each of the maps R K acts at consecutive positions in the respective multiple Cartesian product of spaces. ( N −2 are such that they do not both belong to P (K) for any K ∈ imposes an equivalence relation on Cartesian products.
Starting with U α , it may be necessary to use the third rule to arrange that R K 1 , respectively R K N , can be applied, which means that the sets associated with elements of P (K 1 ), respectively P (K N ), have to be in lexicographic order and at neighboring positions in the respective multiple Cartesian product of sets.After an application of some R K , it may again be necessary to use the third rule to arrange a further application of a map R K ′ , or to achieve the final reverse lexicographic order U ω .That this works is a consequence of the underlying structure of higher Bruhat orders [13,42,43].
We have to stress that (3.1) is not the form in which simplex equations usually appear in the literature, see [13] for the relation and references.
With each K ∈ [N ] N −1 , now we associate a map .
The N -gon equation may be regarded as a realization of T (N, N − 2).It is well defined if we require the following rules.
(2) Each of the maps T K acts at consecutive positions in the respective multiple Cartesian product of sets. ( N −2 are such that they do not both belong to P (K) for any K ∈ As in the case of simplex equations, to apply or work out a polygon equation, we have to check at each step whether a map T K can be applied directly or whether we first have to use the third rule above to achieve a reordering of the respective multiple Cartesian product.In any case, we have to keep track of the numbering of the sets, even if they are identical as sets.
It is, therefore, convenient to realize the above equivalence relation by introducing explicitly transposition maps (sometimes called flip or switch maps) in the equations, at the price of ending up with a form of the equation that looks more complicated and apparently lost its universal structure, but it is often better suited for applications.Instead of keeping track of the numbering of sets, we then have to keep track of the first position on which a map acts in a multiple Cartesian product.This has been done in [13].For several polygon equations, we will recall the resulting form in Section 5.In this form, we can best deal with the case of prime interest, where all the sets U J are the same and there is only a single map T .
If N is even, (3.2) can be correspondingly expressed as we also associate a map TK : .
The dual N -gon equation may be regarded as a realization of the dual Tamari order T (N, N − 2).Both sides act on α (r) , which is equal to ω (b) totally reversed, and map to ω (r) , which is equal to α (b) totally reversed.
Simplex equations, and also (dual) polygon equations, are interrelated by a kind of integrability feature, which crucially distinguishes them from similar equations.We refer to [13] for the general structure, but in Section 5 we elaborate this feature for some examples of polygon equations.

Reductions of polygon equations
For any fixed k ∈ [N + 1], there is a projection pr k : A(N + 1, n + 1) → A(N, n), obtained by restricting ρ ∈ A(N + 1, n + 1) to the subsequence consisting only of elements K ∈ [N +1] n+1 with k ∈ K.The set of all these subsequences is in bijection with A(N, n), simply by deleting k in each K and an obvious renumbering.Moreover, the projection is compatible with the equivalence relation ∼ and induces a projection B(N + 1, n + 1) → B(N, n).See [13,Remark 2.5].
But only for k ∈ {1, N + 1}, the projection pr k is compatible with the 3-color decomposition, see [13,Remark 2.17].If k = 1, this yields projections In particular, there are thus projections T (N +1, N −1) → T (N, N −2) and T (N +1, N −1) → T (N, N −2), which then induce reductions of the (N +1)-gon to the N -gon equation, respectively the dual N -gon equation.In the same way, the projections T (N + 1, N − 1) → T (N, N − 2) and T (N + 1, N − 1) → T (N, N − 2) induce reductions of the dual (N + 1)-gon to the dual Ngon equation, respectively the N -gon equation.Since we realize Tamari orders by sets and maps between them, we have to make sure, however, that this procedure indeed leads to a realization of the (dual) N -gon equation.This will be made precise in the following subsections.Throughout we assume N > 2.
By a degenerate map (e.g., a solution of a polygon equation), we mean a map whose values do not depend on (at least) one of its arguments.

Reductions induced by pr 1
The reduction of the (N + 1)-gon equation induced by pr 1 is essentially obtained by dropping the map T K with 1 / ∈ K.But we also have to arrange that the remaining maps T K with 1 ∈ K are reduced to maps between (products of) spaces U J with 1 ∈ J ∈ P (K).(1 N , be maps solving the (N + 1)-gon equation.For K with 1 ∈ K, let T K not depend on the last component of its domain.Let (2) Each solution of the N -gon equation can be extended to a degenerate solution of the (N +1)gon equation.
does not depend on the last component of its domain, it induces a map maps to spaces that are, as a consequence of our assumption, disregarded by all other maps, it can be dropped from the (N + 1)-gon equation which thus reduces to where we can now regard the indices as complementary in [N + 1] \ {1}.By a shift in the numbering, this is turned into the standard form of the odd N -gon equation, where the indices are complementary in [N ].
N , be maps solving the Proof .Let N be even and K such that 1 ∈ K.The packet P (K) has N elements and its last member in the lexicographic order is K \ {1}.Hence, K \ {1} is the first element of ← − P e (K).Then where, for J ∈ P (K ′ ), we set again U J := U {1}∪J .Since the domain of the remaining map T 1 only involves spaces that are excluded from the range of all other maps T K , it splits off from the (N + 1)-gon equation which thus reduces to where the complementary indices now refer to where now the indices are complementary in [N ], so that we have the standard form of the (even) N -gon equation.
N is a solution of an odd (N +1)-gon equation, then each map T K with 1 ∈ K has the form solve the even N -gon equation.Each solution of the odd (N + 1)-gon equation is thus an extension, of the above form, of a solution of the (even) N -gon equation.
Recalling that T 1 does not involve any of the spaces U 1ij , 1 < i < j ≤ 5, the pentagon equation implies that the maps, obtained from {T k} by deleting the first component of their codomain, satisfy the tetragon equation (with indices shifted by 1).
Let us now turn to the dual (N +1)-gon equation and consider the subset of maps TK : N , be maps solving the dual (N + 1)-gon equation.For K with 1 ∈ K, let TK not depend on the last component of its domain.Let TK ′ , K ′ = K \ {1}, be obtained from TK by excluding the last component of its domain.Then TK ′ |K ′ ∈ {2,...,N +1} N −1 solve the dual N -gon equation.
where we set U J := U {1}∪J .With the same argument as in the proof of Theorem 4.1, the dual (N + 1)-gon equation where now the complementary indices refer to [N + 1] \ {1}.A shift in the numbering achieves the standard form (3.6) of the dual even N -gon equation.
(2) The proof is analogous to that of part (2) of Theorem 4.1.
N , be maps solving the dual (N +1)gon equation.For K with 1 ∈ K let TK ′ , K ′ = K \ {1}, be obtained from TK by excluding the first component of its codomain.Then The dual (N + 1)-gon equation then reduces to the dual odd N -gon equation, using an argument as in the proofs of the preceding theorems.■

Reductions induced by pr N +1
The projection T (N + 1, N − 1) → T (N, N − 2) induces a reduction of the (N + 1)-gon to the dual N -gon equation.It is essentially obtained by dropping the map T K with N + 1 / ∈ K.But we have to arrange that the remaining maps T K , N + 1 ∈ K, are reduced to maps between (products of) spaces N , be maps solving the (N + 1)-gon equation.For K with N + 1 ∈ K, let T K not depend on the first component of its domain.Set K ′ = K \ {N + 1} and let TK ′ be given by T K with the first component of its domain excluded.Then TK ′ |K ′ ∈ [N ]   N −1 solve the dual N -gon equation.
(2) Each solution of the dual N -gon equation can be extended to a degenerate solution of the (N + 1)-gon equation.
Proof .(1) Let K be such that N + 1 ∈ K. Then K ′ := K \ {N + 1} is the first element of P (K) in lexicographic order, and thus also of P o (K).Hence, we have maps where P ′ o (K) is P o (K) without its first element (in lexicographic order).If these maps do not depend on the first component of their domain, they project to maps where we set U J := U J∪{N +1} for J ∈ P (K ′ ).Under this assumption, maps to spaces that are disregarded by all other T K appearing in the (N + 1)-gon equation.For odd N , the (N + 1)-gon equation where now the indices are complementary in . This is the dual N -gon equation for odd N .
For even N , the (N + 1)-gon equation where again indices are now complementary in [N ].This is the dual N -gon equation for even N .
(2) The proof is analogous to that in preceding theorems.■ Let us now turn to the corresponding reduction of the dual (N +1)-gon equation and consider the maps TK : N , be maps solving the dual (N + 1)-gon equation.For K with N + 1 ∈ K, set K ′ = K \ {N + 1} and let T K ′ be given by TK with the last component of its codomain excluded.Then , and thus the last element of without the last element.Hence, TK induces a map where we set again U J := U J∪{N +1} for J ∈ P (K ′ ).With the kind of argument used in the proofs of preceding theorems, the dual (N + 1)-gon equation reduces to the N -gon equation.■ Remark 4.9.The last result means that, if N is a solution of a dual (N +1)-gon equation, then each map TK with N + 1 ∈ K has the form → U K ′ , and N −1 } solve the N -gon equation.Each solution of the dual (N + 1)-gon equation is thus an extension, of the above form, of a solution of the Ngon equation.

Examples of polygon equations
In this section, we elaborate polygon equations up to the 8-gon equation.

Trigon equation
This is the equation for maps T ij : U i → U j , i < j.On the left-hand side of (5.1), we mean the composition of two maps.If the sets are the same, U i = U, i = 1, 2, 3, and if there is only a single map T , then the trigon equation means that T is idempotent.

Tetragon equation
For i, j = 1, 2, 3, 4, i < j, let a map L ij : U i → U j carry a parameter from a set U ij .Let us further assume that each of the local trigon equations uniquely determines a map Using associativity of compositions, there is a consistency condition, see Figure 1,6 which requires that the maps T ijk have to satisfy the tetragon equation Both sides of this equation act on the lexicographically ordered Cartesian product U 12 ×U 23 ×U 34 and map to U 14 .
Let us introduce a boldface "position index" that indicates the first of two neighboring sets on which the respective map acts in a Cartesian product of more than two sets, These additional indices are redundant, as long as we keep the combinatorial indices and keep track of the numbered sets.
Using complementary index notation, where k stands for the complement of k in {1, 2, 3, 4}, the tetragon equation reads Writing In the simplest case, where all the basic sets are equal and we are dealing with a single map T , we may drop the combinatorial indices, but retain the boldface "position" indices.The tetragon equation is then From local tetragon equations to the pentagon equation.Here ∼ stands for an equivalence, which corresponds to an application of a transposition map P. Essentially, the left diagram is the pentagonal Tamari lattice, in its original form, as displayed in the second figure .Here the action of a map T corresponds to a right associativity map (a If this is an invertible map in a category with a binary operation •, the pentagon relation means that it is a coherent associativity isomorphism in the sense of [40].

Pentagon equation
For i, j, k = 1, . . ., 5, i < j < k, let a map depend on a parameter from a set U ijk .Let us assume that each of the local tetragon equations uniquely determines a map Then it follows that the maps T ijkl , 1 ≤ i < j < k < l ≤ 5, have to satisfy the pentagon equation, see Figure 2. Using complementary index notation, the pentagon equation is Both sides of this equation act on U 123 × U 134 × U 145 and map to U 345 × U 235 × U 125 .
Representing the equivalence relation ∼ in the diagram in Figure 2 by a transposition map, P(a, b) = (b, a), and reading off on which neighboring positions in a multiple Cartesian product a map acts, we get In case of identical basic sets, then renamed to U, and a single map T , the last equation takes the form where the combinatorial indices have been dropped.Now all the information needed is provided by the "position" indices.The latter is our abbreviation of the following more familiar form of the pentagon equation Writing the last restricted form of the pentagon equation is equivalent to the conditions (cf.[30,31]) for all a, b, c ∈ U. Set-theoretic solutions have been obtained in [1,5,6,7,22,25,26,27,28,31,32,45,46,70].
(1) If a * b = b for all a, b ∈ U, (5.4) reduces to the associativity condition for •.If (U, •) is a group and if T is invertible, then this is the only solution [7,31].It underlies one of the Kac-Takesaki operators on a group (see, e.g., [62]).If U is a subset of a group (G, •), not containing the identity element, there are more solutions [31].
(2) If a • b = a, the above system reduces to (a * b) * (a * c) = b * c.In a group, a solution is given by a * b = a −1 b.This underlies another Kac-Takesaki operator on a group (see, e.g., [5,62]).

An example related to incidence geometry
Let V be a vector space and L(a) : V × V → V, a ∈ U, be such that the "local" tetragon equation Then this map satisfies the pentagon equation (5.2).Writing x• a y := L(a)(x, y), we can express the above equation as the parameter-dependent associativity condition (x• a y) An example is given by V = R n , U = (0, 1) ⊂ R, and x • a y := ax + (1 − a)y.Then we obtain the solution of the pentagon equation (also see [26]).It also shows up as a map of "polarizations" resulting from the evolution of tree-shaped matrix KP solitons [14].
The above binary operation can be interpreted as the collinearity of three points A, B, A• a B: A • a B = aA + (1 − a)B.The generalized associativity condition can then be viewed as a (6 2 , 4 3 ) configuration [21], which consists of 6 points and 4 lines, where each point is incident with exactly two lines, each line with exactly 3 points.See Figure 3.A Desargues configuration (10 3 ) consists of 10 points and 10 lines, each point (line) incident with 3 lines (points).It contains 5 configurations of type (6 2 , 4 3 ), which thus constitute a pentagon.Also see [15] for related considerations.

Hexagon equation
For i, j, k, l = 1, . . ., 6, i < j < k < l, let a map depend on a parameter from a set U ijkl .We assume that each of the local pentagon equations uniquely determines a map Then the maps T ijklm , 1 ≤ i < j < k < l < m ≤ 6, have to satisfy the hexagon equation.See Figure 4. Using complementary index notation, the hexagon equation reads Both sides act on U 1234 × U 1245 × U 1256 × U 2345 × U 2356 × U 3456 and map to U 1456 × U 1346 × U 1236 .
Employing transposition maps, according to the equivalences ∼ appearing in Figure 4, and introducing position indices, the hexagon equation takes the form If all basic sets are equal and we are dealing with a single map T , we may drop the combinatorial indices and only retain the (boldface) position indices: (  Remark 5.4.According to our knowledge, without introduction of an auxiliary structure as discussed in [13], the hexagon equation (5.8), and correspondingly all higher even polygon equations, cannot be rewritten without the explicit appearance of transpositions.This is in contrast to the case of the pentagon equation (see (5.5)) and higher odd polygon equations.
Expressing T in terms of two ternary operations, (5.10)

Heptagon equation
Let maps where 1 ≤ i < j < k < l < m ≤ 7, be subject to local hexagon equations where 1 ≤ i < j < k < l < m < n ≤ 7. If these equations uniquely determine maps in two different ways, using local hexagon equations, we find that the maps T ijklmn have to satisfy the heptagon equation, which is in complementary index notation.
Introducing position indices and transposition maps, we can express it as  18]).If all basic sets are equal and there is only a single map T , all the information we need is in the position indices, so that the combinatorial indices can be dropped, Expressing T in terms of three ternary operations, (5.12)

Octagon equation
We consider maps where i < j < k < l < m < p < q.The octagon equation is It arises as the consistency condition of a system of local heptagon equations.
Using position indices and transposition maps, it can be written as

Examples of dual polygon equations
In this section, we elaborate dual polygon equations up to the dual 8-gon equation.

Dual trigon equation
This is the equation for maps Tij : U j → U i , i < j.

Dual tetragon equation
For i, j = 1, 2, 3, 4, i < j, let L ij : U i → U j carry a parameter from a set U ij .Let each of the local trigon equations uniquely determine a map Tijk : Then the maps Tijk have to satisfy a consistency condition, which is obtained by reversing the arrows in Figure 1.This means that the maps Tijk have to satisfy the dual tetragon equation, T123,2 T134 = T234,1 T124 , ( which acts on U 14 .
Introducing position indices and transposition maps, it takes the form If all the basic sets are the same and there is only a single map T , this is which already appeared as a 4-cocycle condition in [58].Writing ) imposes (5.4) on the first two binary operations, and requires in addition for all a, b, c ∈ U.

Dual heptagon equation
The dual heptagon equation is with maps Tijklmp : for all a, b, c, d, e, f ∈ U.

Relations between solutions of neighboring polygon equations
By a (dual) polygon map we mean a solution of a (dual) polygon equation.In this section, the case when all the basic sets appearing in a multiple Cartesian product, on which a (dual) polygon map acts, are the same set U is considered.First, we formulate a few special cases of theorems in Section 4. Though they are rather corollaries, because of their relevance they also deserve to be called theorems.Let N ∈ N, N > 2.
Theorem 7.1.Let T (N +1) be a dual (N + 1)-gon map and T that map with the last component of its codomain cut off.Then T is an N -gon map.
Theorem 7.2.For even N , let T (N +1) be an (N + 1)-gon map and T that map with the first component of its codomain cut off.Then T is an N -gon map.
Theorem 7.3.For odd N , let T (N +1) be a dual (N + 1)-gon map and T that map with the first component of its codomain cut off.Then T is a dual N -gon map.
Further theorems derived from those in Section 4 will be formulated in the following subsections, where we also provide examples for these results and derive more powerful results for the (dual) polygon equations up to the (dual) octagon equation.Proof .This is quickly verified.■ Proposition 7.5.Let T (5) be a pentagon map.Then T (6)  Proof .Writing T (7) (a, b, c) = (b * c, b • c, b ⋄ c), the first two of equations (5.12) become the first two of (5.4).The last two of (5.10) become the last two of (6.5).The first of equations (5.10) becomes the third of (5.4), the third of (5.12) the first of (6.5).■ Proposition 7.8.Let T (6) be a dual hexagon map.Then T (7) (a, b, c) := T (6) (a, b) is a dual heptagon map.
Proof .the first three of conditions (5.13) reduce to (5.12) and the last three to (5.10).The resulting six conditions are those for T (7) to be a heptagon map.■ These results suggest that, more generally, the following statements hold, which determine extensions of each solution of an odd N -gon equation, or each solution of a dual N -gon equation, to a degenerate solution of a (dual) (N + 1)-gon equation.Indeed, these general results follow from theorems in Section 4.
Proof .This a special case of Theorem 4.5 (2).■ 7.2 Further relations between neighboring polygon maps Proposition 7.14.Let T i , i = 1, 2, be trigon maps.Then is a dual tetragon map if and only if the two trigon maps commute.
Proof .This is easily verified.Also see Section 6.2.■ Proposition 7.15.For a map T (5) : U × U → U × U, let us write with a binary operation * and a map T (4) : U × U → U.The following conditions are equivalent: (1) T (5) is a pentagon map.
(2) T (4) is a tetragon map and, with a • b := T (4) (a, b), the binary operations • and * satisfy for all a, b, c ∈ U.
Proof .(5.4) shows that the conditions for T (5) to be a pentagon map are equivalent to T (4)  being a tetragon map (which means associativity of •) and the two compatibility conditions for the two binary operations.■ Corollary 7.16.Let T (4) be a tetragon map.Then is a pentagon map.
Proof .Setting a * b = b solves the two compatibility equations in condition (2) of the preceding proposition.■ Corollary 7.17.Let T (4) be a tetragon map and u a fixed element of U. Then is a pentagon map if and only if T (4) (u, u) = u.
Proof .Setting a * b = u for all a, b ∈ U, solves the first of the two compatibility equations in condition (2) of the preceding proposition and reduces the second to u • u = u.■ Proposition 7.18.Let T (4) be a tetragon map.
(1) The map is a dual pentagon map.
(2) If u is a fixed element of U, then Proof .This is easily verified.■ Proposition 7.19.For a map T (6) : U × U → U × U × U, let us write with a map T (5) : U × U → U × U and a binary operation ⋄.The following conditions are equivalent: (1) T ( 6) is a dual hexagon map.
Proof .This is easily verified.■ Corollary 7.21.If T (5) is a pentagon map, then is a dual hexagon map.
Proof .Setting a ⋄ b := a solves the three equations (6.5).■ Corollary 7.22.Let T (5) be a pentagon map and u ∈ U a fixed element.Then is a dual hexagon map if and only if T (5) (u, u) = (u, u).
Proof .Setting a ⋄ b := u reduces the three equations (6.5) to u * u = u and u • u = u.■ Proposition 7.23.Let T (5) be a dual pentagon map.
(1) The map is a dual hexagon map.
(2) If u is a fixed element of U, then is a dual hexagon map if and only if T (5) (u, u) = (u, u).
(2) T (6) is a hexagon map and, expressed as in (5.9) (so that (5.10) holds), it satisfies the compatibility conditions (5.12) with the above ternary operation.
Proof .This is an immediate consequence of the last part of Section 5.5.■ Corollary 7.25.If T (6) is a hexagon map, then T (7) (a, b, c) := c, T (6) (a, b, c) is a heptagon map.
(1) The map is a dual heptagon map.
(2) If u is a fixed element of U, then is a dual heptagon map if and only if T (6) (u, u, u) = (u, u).
(1) Setting [a, b, c] = a, the last three equations of (6.7) become identities and the first three are equivalent to (5.10) by a renaming of the ternary operations.
Proof .This immediately follows from results in Section 6.6.■ Corollary 7.29.If T (7) is a heptagon map, then is a dual octagon map.
Proof .Setting |a, b, c| = a turns the four equations (6.8) into identities.■ Corollary 7.30.Let T (7) be a heptagon map and u ∈ U a fixed element.Then is a dual octagon map if and only if T (7) (u, u, u) = (u, u, u).
(1) The map is a dual octagon map.
(2) If u is a fixed element of U, then is a dual octagon map if and only if T (7) (u, u, u) = (u, u, u).
Proof .This can be verified using results of Section 6. ■ Preceding results suggest the following conjectures.

Conclusions
The main results of this work concern the structure of solutions of polygon equations.More precisely, we have shown that a solution of a (dual) N -gon equation is related in very simple ways to solutions of the (dual) (N + 1)-gon and (dual) (N − 1)-gon equation.Each solution of a (dual) polygon equation extends to solutions of the higher equations.For a chosen polygon equation, the most important case is when all basic sets are equal and there is only a single polygon map.Expressing polygon equations with the help of transposition maps, we can quite easily verify the above mentioned features for the simplest equations of the family.But for general proofs, we had to return to the underlying framework of higher Tamari orders, as developed in [13].
Our results reveal a beautiful structure of the family of polygon equations.Other nice aspects are the integrability feature [13], recalled in some examples in Section 5, and relations with polyhedra [13].
In this work, we concentrated on the set-theoretic setting.Many results directly pass over to the framework of vector spaces, tensor products, and linear maps, where the dual tetragon equation becomes the coassociativity condition and the pentagon equation plays one of its most important roles, as mentioned in the introduction.A further exploration of polygon equations, beyond the pentagon equation, in this framework, will be left for a separate work.
Also, concerning set-theoretic solutions, we have only set the stage.It is to be expected that more concrete solutions can be obtained by applying methods that have already been exploited in the case of the pentagon equation.Whereas the (dual) pentagon and dual hexagon equation can be expressed in terms of binary operations, the hexagon and higher polygon equations involve n-ary operations with n > 2. There is a vast literature about such generalizations of products, and this should be helpful in finding solutions of N -gon equations with N > 5.In particular, there are corresponding generalizations of co-, bi-and Hopf algebras, for which higher polygon equations may play a role.Also see [58] ("cocycloids").
The question pops up whether there are higher order counterparts of the important relation between solutions of the pentagon equation and bi-or Hopf algebras, mentioned in the introduction.Since the transposition P solves the pentagon equation and also its dual, Corollaries 7.22 and 7.23 show that P ⊗ u and u ⊗ P, where u is a fixed element, are both dual hexagon maps.If U is an algebra A, choosing u = 1 A , these are algebra homomorphisms.They can be regarded as counterparts of the trivial comultiplications ∆ ℓ and ∆ r mentioned in the introduction.It should be obvious how this extends to higher dual polygon equations, since a special solution of the odd N -gon equation is a combination of P's that achieves a total inversion.For example, since P inv := P 1 P 2 P 1 solves the heptagon equation and its dual, it follows that P inv ⊗ 1 A , as well as 1 A ⊗ P inv , solve the dual octagon equation.A similarity transformation, analogous to the construction of non-trivial comultiplications from a trivial one, as mentioned in the introduction, leads to a new algebra homomorphism.But under what conditions does it satisfy the respective (dual) even polygon equation?
As already mentioned in the introduction, solutions of simplex equations can be obtained from solutions of a polygon equation and its dual [13,31,32,41,53].But an additional compatibility condition has to be solved, which needs further exploration.We plan to study this in a separate work.
We have seen in Section 7.2 that a solution of the dual hexagon equation shows up in a pentagonal relation for the (exponentiated) Rogers dilogarithm.Also higher (dual) polygon equations may play a role in this context [65].Surely there is much more to be revealed.
For a non-empty finite subset M of N, and n ∈ N, 1 ≤ n ≤ |M | (where |M | is the cardinality of M ), let M n denote the set of n-element subsets of M .The packet P (M ) of M is the set of (|M | − 1)-element subsets of M .We write − → P (M ) for P (M ) in lexicographic order, and ← − P (M ) for P (M ) in reverse lexicographic order.Let N ∈ N, N > 1, and [N ] = {1, . . ., N }.A linear order (permutation

( 2 )
Each solution of the dual N -gon equation can be extended to a degenerate solution of the dual (N + 1)-gon equation.Proof .(1) If N is even, then K ′ = K \ {1} is the last element of − → P e (K), so that TK : U− → P ′ e (K) × U K\{1} → U← − P o(K) , where − → P ′ e (K) is − → P e (K) without the last element.If TK does not depend on the last component of its domain, it induces a map

Figure 1 .
Figure 1.From local trigon equations to the tetragon equation.Here, and correspondingly in following figures, we suppress the parameters of the maps L.

Figure 4 .
Figure 4. From local pentagon equations to the hexagon equation.