On the Ore condition for the group ring of R.\,Thompson's group F
aa r X i v : . [ m a t h . G R ] J a n On the Ore condition for the group ring ofR. Thompson’s group F V. S. Guba ∗ Vologda State University,15 Lenin Street,VologdaRussia160600E-mail: [email protected]
Abstract
Let R = K [ G ] be a group ring of a group G over a field K . The Ore conditionsays that for any a, b ∈ R there exist u, v ∈ R such that au = bv , where u = 0 or v = 0 . It always holds whenever G is amenable. Recently it was shown thatfor R. Thompson’s group F the converse is also true. So the famous amenabilityproblem for F is equivalent to the question on the Ore condition for the group ringof the same group.It is easy to see that the problem on the Ore condition for K [ F ] is equivalent tothe same property for the monoid ring K [ M ] , where M is the monoid of positiveelements of F . In this paper we reduce the problem to the case when a , b arehomogeneous elements of the same degree in the monoid ring. We study the case ofdegree and find solutions of the Ore equation. For the case of degree , we study thecase of linear combinations of monomials from S = { x , x x , x x , x , x x } . Thisset is not doubling, that is, there are nonempty finite subsets X ⊂ M ⊂ F such that | SX | < | X | . As a consequence, the Ore condition holds for linear combinations ofthese monomials. We give an estimate for the degree of u , v in the above equation.The case of monomials of higher degree is open as well as the case of degree for monomials on x , x , ..., x m , where m ≥ . Recall that negative answer to anyof these questions will immediately imply non-amenability of F . Acknowledgements
The author is grateful to A. Yu. Ol’shanskii, Matt Brin, Mark Sapir, Rostislav Grigorchuk,and David Kielak for helpful discussions of the results of this paper. ∗ This work is supported by the Russian Foundation for Basic Research, project no. 20-01-00465. ntroduction Let M be a monoid given by the following infinite presentation h x , x , x , . . . | x j x i = x i x j +1 (0 ≤ i < j ) i . (1)It is easy to see that any word in these generators can be reduced to a word of theform x i . . . x i k , where k ≥ and ≤ i ≤ · · · ≤ i k . The rewrite system x j x i → x i x j +1 ( j > i ≥ ) turns out to be terminating and confluent so the above normal form is unique.For any elements a, b ∈ M there exists a least common right multiple of a , b . Also itis easy to check that M is a cancellative monoid. A classical Ore theorem states that fora cancellative monoid M with common right multiples, there exists a natural embeddingof M into its group of quotients (see [17] for details). Any element of this group belongsto M M − , and the group is given be the same presentation. We denote it by F .This group was found by Richard J. Thompson in the 60s. We refer to the survey [6] fordetails. (See also [3, 4, 5].) It is easy to see that for any n ≥ , one has x n = x − ( n − x x n − so the group is generated by x , x . It can be given by the following presentation withtwo defining relations h x , x | x x = x x x , x x = x x x i , (2)where a b ⇌ b − ab .Each element of F can be uniquely represented by a normal form , that is, an expres-sion of the form x i x i · · · x i k x − j l · · · x − j x − j , (3)where k, l ≥ , ≤ i ≤ i ≤ · · · ≤ i k , ≤ j ≤ j ≤ · · · ≤ j l and the following istrue: if (3) contains both x i and x − i for some i ≥ , then it also contains x i +1 or x − i +1 (inparticular, i k = j l ).An equivalent definition of F can be given in the following way. Let us considerall strictly increasing continuous piecewise-linear functions from the closed unit intervalonto itself. Take only those of them that are differentiable except at finitely many dyadicrational numbers and such that all slopes (derivatives) are integer powers of . Thesefunctions form a group under composition. This group is isomorphic to F . Another usefulrepresentation of F by piecewise-linear functions can be obtained if we replace [0 , by [0 , ∞ ) in the previous definition and impose the restriction that near infinity all functionshave the form t t + c , where c is an integer.It is known that F is a diagram group over the simplest semigroup presentation P = h x = x i . See [14] for the theory of these groups. In our paper, we will use diagrampresentation for the elements of F . This is based on non-spherical diagrams over P . Adetailed description can be found in [15]. We will mostly use diagrams that representelements of the monoid M . Such objects are called positive diagrams over x = x . Let usgive a brief illustration. 2iven a normal form of an element in F , it is easy to draw the corresponding non-spherical diagram, and vice versa. The following example illustrates the diagram thatcorresponds to the element g = x x x x − x − represented by its normal form: r r r r r r r r r x x x x x − x − x − Here we assume that each egde of the diagram is labelled by a letter x . The horizontalpath in the picture cuts the diagram into two parts, positive and negative . The positivepart represents an element of M , namely, x x x . The top label of this positive diagramhas label x , the bottom part of it has label x . So we have a positive ( x , x ) -diagramover P . The set of all positive diagrams with top label x m and bottom label x n , where m ≤ n , will be denoted by S m,n . Given a positive ( x m , x n ) -diagram ∆ and a positive ( x n , x k ) -diagram ∆ , one can concatenate them obtaining an ( x m , x k ) -diagram denotedby ∆ ◦ ∆ . This operation is very natural: we identify the bottom path of ∆ with thetop path of ∆ .We also say product instead of concatenation, and can multiply sets of diagrams inthis way. Clearly, S m,n S n,k = S m,k . All sets S m,n are finite, and the cardinalities of themare given by numbers from Catalan triangle; see Section 1.Recall that a group G is called amenable whenever there exists a finitely additivenormalized invariant mean on G , that is, a mapping µ : P ( G ) → [0 , such that• µ ( A ∪ B ) = µ ( A ) + µ ( B ) for any disjoint subsets A, B ⊆ G ,• µ ( G ) = 1 ,• µ ( Ag ) = µ ( gA ) = µ ( A ) for any A ⊆ G , g ∈ G .One gets an equivalent definition of amenability if only one-sided invariance of the meanis assumed, say, the condition µ ( Ag ) = µ ( A ) ( A ⊆ G , g ∈ G ) . The proof can be found in[13].The class of amenable groups includes all finite groups and all abelian groups. Itis invariant under taking subgroups, quotient groups, group extensions, and ascendingunions of groups. The closure of the class of finite and abelian groups under theseoperations is the class EA of elementary amenable groups. A free group of rank > isnot amenable. There are many useful criteria for (non)amenability. Here we would liketo mention Folner criterion from [11]. For our reasons, it is convenient to formulate it asfollows. A group G is amenable if and only if for any finite set A ⊆ G and for any ε > , thereexists a finite set S such that | AS | < (1 + ε ) | S | . A contains the identity element. Also one can extend A to thegenerating set of G provided the group is finitely generated. In this case we see that theset S from the above statement is almost invariant under (left) multiplication by elementsin A .It was proved in [3] that the group F has no free subgroups of rank > . It is alsoknown that F is not elementary amenable (see [7]). However, the famous problem aboutamenability of F is still open. The question whether F is amenable was asked by RossGeoghegan in 1979; see [12]. There were many attempts of various authors to solve thisproblem in both directions. We will not review a detailed history of the problem; thisinformation can be found in many references. However, to emphasize the difficulty ofthe question, we mention the paper [18], where it was shown that if F is amenable, thenFolner sets for it have a very fast growth. Besides, we would like to refer to the paper [2]where the authors obtained an estimate of the isoperimetric constant of the group F inits standard set of generators { x , x } . This estimate has not been improved so far.Now we are going to describe a new approach to the problem in terms of equationsin the group ring of F . Tamari [19] shows that if a group G is amenable, then the group ring R = K [ G ] satsfiesOre condition for any field K . This means that for any a, b ∈ R there exist u, v ∈ R suchthat au = bv , where u = 0 or v = 0 .Let us generalize this statement. Suppose that instead of one linear equation au = bv with coefficients in R we have a system of them, where the number of variables exceedsthe number of equations: a u + · · · + a n u n = 0 · · · · · · · · · a m u + · · · + a mn u n = 0 where n > m , a ij ∈ R for all ≤ i ≤ m , ≤ j ≤ n . We are interested in solutions ( u , ..., u n ) ∈ R n .We claim that for amenable group G , this system always has a nonzero solution.Indeed, let A ⊆ G be the union of supports of all the coefficients of the form a ij . ByF ø lner criterion, for any ε > there exists a finite set X such that | AX | < (1 + ε ) | X | . Let u j ( ≤ j ≤ n ) be linear combinations of elements in X with indefinite coefficients fromthe field. This gives n | X | variables taking their values in K . For any of m equations inthe system, we collect all terms on any element in AX and impose the condition that thesum of them is zero. This leads to an ordinary system of m | AX | linear equations. Sucha system has a nonzero solution whenever the number of variables exceeds the number ofequations. This holds if m (1 + ε ) < n , so it suffices to claim ε < n − mm .In a recent paper [1], Bartholdi shows that the converse to the above statement is true.This gives a new criterion for amenabilty of groups. Although Thorem 1.1 in [1] concerns4he so-called GOE and MEP properties of automata (Gardens of Eden and MutuallyErasable Patterns), the proof of it allows one to extract the following statement. Theorem 1 (Bartholdi)
For any group G , the following two properties are equaivalent.(i) G is amenable(ii) For any field K and for any system of m linear equations over R = K [ G ] in n > m variables, there exists a nonzero solution. In the Appendix to the same paper, Kielak shows that if the group ring K [ G ] hasno zero divisors, both properties are equivalent to the Ore condition. In particular, thisholds for R. Thompson’s group F . It is orderable, so there are no zero divisors in a groupring over a field. So we quote the following Theorem 2 (Kielak)
The group F is amenable if and only if the group ring K [ F ] overany field satisfies Ore condition. The following elementary property of the group F is well known. Lemma 1
For any g , ..., g n ∈ F there exist g ∈ F such that g g, . . . , g n g ∈ M . Using this fact, one can easily show that the Ore condition for K [ F ] is equivalent tothe Ore condition for K [ M ] . Lemma 2
For any field K , the group ring K [ F ] satisfies Ore condition if and only ifthe monoid ring K [ M ] satisfies Ore condition. Proof.
Suppose that any equation of the form au = bv , where a, b ∈ K [ F ] , has anonzero solution in the same ring. Let a, b ∈ K [ M ] . Then there exist u, v ∈ K [ F ] , where u = 0 or v = 0 , such that au = bv . Let X be the union of supports of u and v . Thisis a finite subset in F . According to Lemma 1, there exists g ∈ F such that Xg ⊂ M .Therefore, aug = bvg , where ( ug, vg ) is a non-zero solution of the equation in K [ M ] .Converesly, suppose that the Ore condition holds in K [ M ] . Take any a, b ∈ K [ F ] .Taking Y as a union of their supports, find g ∈ G such that Y g ⊂ M . Then ag , bg belongto K [ M ] and so there exists a nonzero solution of the equation ag · u = bg · v in K [ M ] .Here ( gu, gv ) brings a nonzero solution of the equation in K [ F ] with coefficients a , b .The role of this easy lemma is that the monoid ring K [ M ] has a simpler structurethan the group ring of F . This is in fact a graded ring of skew polynomials where thenon-commutative variables are multiplied using the rule x j x i = x i x j +1 ( j > i ).Now we are going to reduce the general problem of solving equations to the one forhomogeneous polynomials a , b . Lemma 3
Suppose that any equation of the form au = bv has a nonzero solutionin K [ M ] provided a , b are homogeneous polynomials of the same degree. Then K [ M ] satisfies Ore condition. roof. Notice that if both a , b are homogeneous but not of the same degree then onecan multiply one of them on the right by a monomial such that the degrees will coincide.Say, if m = deg a < deg b = n then we solve the equation ax n − m u = bv for monomials ofthe same degree, and this gets us the solution for the pair a, b .Also it is easy to see that if au = bv , then one can replace u , v by their homogeneouscomponents of minimal degree. This will give another solution to the same equationwhere u , v are also homogeneous.Any nonzero element a ∈ K [ M ] we can written as a sum of homogeneous components: a = a + · · · + a k , where k ≥ , deg a < · · · < deg a k . By the width of a we mean thedifference of highest and lowest degrees: α = deg a k − deg a . Similarly, let b = b + · · · + b l be a sum of its homogeneous components, where l ≥ , deg b < · · · < deg b l , and thewidth of b is β = deg b l − deg b .We proceed by induction on α + β . The cases a = 0 and b = 0 are obvious so we willassume that both polynomials a , b are nonzero. The situation α + β = 0 means that a , b are homogeneous. In this case the equation has a nonzero solution.Let the sum of widths be greater than zero, and assume that α ≤ β without loss ofgenerality. Here β > so l ≥ . Let us solve the equation a u = b v for homogeneous a , b , where u , v are homogeneous and nonzero.Let b ′ = bv − au = b v + · · · + b l v − a u −· · ·− a k u . The terms a u and b v cancelin this difference. The case b ′ = 0 already gives us a nonzero solution of the equation for a , b . So assume that b ′ = 0 .Let us compare the homogeneous components in b ′ of highest and lowest degree.The highest one does not exceed max(deg( b l v ) , deg( a k u )) . We see that deg( b l v ) =deg b l + deg v = β + deg b v and deg( a k u ) = deg a k + deg u = α + deg( a u ) . Since a u and b v are equal, we have deg( b l v ) − deg( a k u ) = β − α ≥ . Therefore, thehighest degree of homogeneous components in b ′ does not exceed deg b l + deg v .On the other hand, all monomials in b ′ have degree strictly greater that deg b +deg v = deg a + deg u since degrees of the a i s and the b j s strictly increase. So thedifference between the highest and the lowest degree of monomials in b ′ is strictly lessthan (deg b l + deg v ) − (deg b + deg v ) = β . Hence the sum of widths of a and b ′ isstrictly less than α + β , and we can use the inductive assumption.Solving the equation au = b ′ v , we find its nonzero solution (both u , v are nonzero).Now au = ( bv − au ) v so we have a ( u + u v ) = bv v . This solution is nonzero because v , v = 0 , and the group ring K [ F ] ⊃ K [ M ] has no zero divisors. This completes theproof.Now we have a bunch of equations in K [ M ] indexed by two parameters. The fist oneis d , the degree of homogeneous polynomials a and b . The second one is m , where m isthe highest subscript in variables we involve. The general strategy can be the following:we try to solve as much equations in K [ M ] as we can, using this classification. For apair of numbers d ≥ , m ≥ , we can take a , b as linear combinations of monomialsof degree d in variables x , x , ... , x m with indefinite coefficients. We can think aboutthese coefficients as elements of the field of rational functions over K with a number ofvariables. 6ore precisely, we can state the general problem as follows. Any finite system ofmonomials of degree d ≥ is contained in a set of the form S m +1 ,m + d +1 for some m ≥ .This set consists of all elements in the monoid M with normal forms x i . . . x i d , where i ≤ · · · ≤ i d and i ≤ m , i ≤ m + 1 , ... , i d ≤ m + d − . Let K [ S ] denote the set of alllinear combinations of elements of S ⊂ M with coefficients in K . Problem P d,m : Given two elements a, b ∈ K [ S m +1 ,m + d +1 ] , find a nonzero solution ofthe equation au = bv , where u, v ∈ K [ M ] , or prove that it does not exist. According to Theorem 2 by Kielak, and Lemma 3 on homogeneous equations, we havethe following alternative. If the Problem P d,m has positive solution for any d, m ≥ (thatis, we can find nonzero solutions), then the group F is amenable. If this Problem hasnegative solution for at least one case, then F is not amenable. The first case we are going to start with, is the case of polynomials of degree d = 1 forarbitrary m . Let us show that equations of the form ( α x + α x + · · · + α m x m ) u = ( β x + β x + · · · + β m x m ) v (4)have nonzero solutions in K [ M ] for arbitrary coefficients α i , β i ∈ K ( ≤ i ≤ m ). Ourfirst approach will be based on cardinality reasons; later we are going to improve it.All elements of the set X m = { x , x , ..., x m } are viewed as positive diagrams over x = x with top label x m +1 and bottom label x m +2 . Each of these diagrams consists ofone cell; the diagram for x i ( ≤ i ≤ m ) looks in the following way: r r rr r xx x x i x m − i We will need some more notation. An edge labelled by x i will be denoted ε ( x i ) . Also ifwe have two diagrams ∆ and ∆ , then ∆ + ∆ denotes the diagram where the rightmostpoint of ∆ is identified with the leftmost point of ∆ . The diagram in the picture canthus be denoted as ε ( x i ) + ( x = x ) + ε ( x m − i ) , where ( x = x ) is a positive cell.All these diagrams belong to the set S m +1 ,m +2 . Let n ≫ be a sufficiently largeinteger. We are going to multiple the set X m that we identify with S m +1 ,m +2 , by the set Y = S m +2 ,n . Our aim is to show that | X m Y | < | Y | . This will imply that equation (4)has nonzero solutions.Notice that the set X m has cardinality m + 1 which can be arbitrarily large. The listof elements of X m Y taken with repetitions will have size ( m + 1) | Y | . Nevertheless, it hasless than | Y | distinct elements. 7t is known from standard combinatorics that the cardinality of S k,n is given by anelement of the Catalan triangle: b nk = k (2 n − k − n !( n − k )! . (5)So in our case we have | X m Y || Y | = |S m +1 ,n ||S m +2 ,n | = b n,m +1 b n,m +2 = m + 1 m + 2 · n − m − n − m − . The desired inequality holds whenever n > ( m +1)( m +2)2 . Here the cardinality approachgives us a quadratic estimate to the degree of u and v in a potential solution. In fact, wecan improve this estimate finding solutions of degree just m in variables x , x , ... , x m ,which we are going to present in the next Section finding solutions of potentially minimaldegree for (4). Now let us finish our proof.We need to repeat the same argument that was used in the proof of Tamari’s theoremin the beginning. Namely, we take u , v as linear combinations of the elements in Y withfree coefficients (variables). The total number of variables is | Y | . Making a substitutionof these expressions into (4), we get a linear combination of | X m Y | elements of M .Claiming that all of them have equal coefficients in the left-hand side and the right-handside, we get | X m Y | < | Y | linear equations over K . Since the system with this propertyhas a nonzero solution in the field, we get the desired solution for our equation (4).So let us state a weak form of our result on the equations for the case of polynomialsof degree as follows. Theorem 3 a) For any m ≥ , the set of elements X m = { x , x , ..., x m } is notdoubling, that is, there exists a finite subset Y ⊂ M such that | X m Y | < | Y | .b) If a, b ∈ K [ M ] are linear combinations of monomials x , x , ... , x m of degree ,then the equation au = bv in k [ M ] has a nonzero solution, where deg u = deg v ≤ m ( m +1)2 . Here the degrees of u and v do not exceed n − ( m + 2) , so the above estimate holdsif we take n = ( m +1)( m +2)2 + 1 .In other words, Problem P ,m has positive solution for any m ≥ .Now let d = 2 , that is, we deal with linear combinations of monomials of degreetwo. The first interesting case is m = 1 . This means that a , b belong to K [ S ] , where S = S m +1 ,m + d +1 = S , = { x , x x , x x , x , x } . The above is just the set of all positive ( x , x ) -diagrams over x = x (each of them has two cells). We are going to show thatfor this case the Problem P d,m = P , has positive solution.We will use the construction from [9]. Notice that the set S in the notation of thatpaper is what we call S , , if to replace forests by diagrams. As a set of elements of M ,this is exactly x { x , x x , x x , x , x } . The first x is ignored in our case. What is called ruinous in Donnelly’s paper, is called doubling in our terminilogy.So let S = { x , x x , x x , x , x } = S , . If one takes Y = S ,n , then the condition | SY | < | Y | never holds. The idea of the proof of [9, Theorem 1] is as follows (using our8otation). Take n ≫ and exclude from S ,n all diagrams of the form ε ( x )+∆+ ε ( x )+ ε ( x ) and ε ( x ) + ε ( x ) + ∆ + ε ( x ) . Here ∆ is any ( x, x n − ) -diagram. Let Y be the correspondingset difference; its cardinality is | Y | = b n − c n − , where c k is the k th Catalan number,and b n = n − n !( n − is the number from Catalan triangle.The product SY is contained in S ,n . It is not hard to describe the elements thatdo not appear in the difference of these sets. To do that, we introduce a few technicalconcepts.Let Ξ be a positive semigroup diagram over x = x . We call it simple whenever it hasexactly one top cell (that is, the cell whose top boundary is contained in the top pathof Ξ ). In the language of rooted binary forests, this is equivalent to the property thatexactly one tree of the forest is nontrivial. Suppose that we removed the top cell of Ξ . Ifthe result is again a simple diagram, then we call Ξ a diagram.Now look at those 2-simple diagrams from S ,n such that if we remove their top cellstwo times, we get a diagram of the form ε ( x )+∆+ ε ( x )+ ε ( x ) or ε ( x )+ ε ( x )+∆+ ε ( x ) . Nowit is easy to check that there are 3 ways to add top cells twice for any of these diagramsin order to get a 2-simple preimage. We hope the reader can check this geometric factdrawing a few pictures. Hence the set SY differs from S by at least c n − elements.Therefore we get the following estimate: | SY || Y | ≤ b n − c n − b n − c n − = 13 · n − n + 562 n − n − n + 142 n − . Clearly, this quotient is less than 2 for n ≫ . In fact, the inequality | SY | < | Y | holdsfor all n ≥ . We proved Theorem 4
The equation of the form ( α x + α x x + α x x + α x + α x x ) u = ( β x + β x x + β x x + β x + β x x ) v in the monoid ring K [ M ] has a nonzero solution with the property deg u, v ≤ . Here α ij , β ij were arbitrary coefficients from K . The equation has a solution in K [ Y ] ,where Y ⊂ S n consist of monomials of degree n − ≤ . Notice that we do not knowwhat is the minimum degree of u , v for solutions of this equation. A rough computersearch can show that n > , but we do not even know whether the minimum value of n is close to or .We think it is interesting to find the value of n by the following reasons. If wego towards amenability of F , we need to be able to solve more and more complicatedequations in K [ M ] . We can do that for d = 1 and arbitrary m ; we also know the answerfor the case d = 2 , m = 1 . The cases that come after that are already unknown. This is d = 2 , m = 2 , where a , b are linear combinations of 9 monomials of degree 2: x , x x , x x , x x , x , x x , x x , x , x x . This is one possible candidate to obtain the negative answer. If, nevertheless, this Prob-lem P , has positive answer (that is, there exists nonzero solutions), then one can tryProblem P , , where a , b are linear combinations of 14 monomials of degree 3: 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 . au = bv , where a = α x + α x , b = β x + β x ,then the description of all its solutions is easy. Namely, u = ( β x + β x ) w , v =( α x + α x ) w for any w ∈ K [ M ] . This means that the intersection of two principalright ideals aR ∩ bR is a principal right ideal ( R = K [ M ] ).We also know how to describe all solutions of the equation ( α x + α x + α x ) u =( β x + β x + β x ) v . This will be done in the next Section. For this equation, thedescription has more complicated form; the intersection aR ∩ bR is no longer a principalright ideal.If instead of one equation we have a system of equations of the form ( α x + β x ) u = ( α x + β x ) u = · · · = ( α k x + β k x ) u k for any k ≥ , then it also has a nonzero solution. This obviously follows from cardinalityreasons: one can take a set of the form Y = { , x − x , ( x − x ) , . . . , ( x − x ) n } for n ≫ and notice that the cardinality of Y almost coincides with the cardinality of { x , x } Y .This construction is quite trivial, so we will offer a more explicit form of the solution.The product ( α x + β x )( α x + β x ) · · · ( α k x + β k x k +1 ) is left divisible by α i x + β i x for any ≤ i ≤ k , which can be checked directly.A much more interesting example of a system of equations looks as follows. Let usstate it as a separate problem. Problem Q k : Given k + 1 linear combinations of elements x , x , x , consider asystem of k equations with k + 1 unknowns: ( α x + β x + γ x ) u = ( α x + β x + γ x ) u = · · · = ( α k x + β k x + γ k x ) u k . Find a nonzero solution of this system, where u , u , . . . , u k ∈ K [ M ] , or prove that itdoes not exist. Notice that Q has been already considered. To solve Q in positive, it suffices to finda finite set Y with the property | AY | < | Y | , where A = { x , x , x } . This can be doneby cardinality reasons simliar to the above proof of Theorem 4. The estimate there willbe also n ≥ . We do not give details here since we are able to prove a much strongerfact. Namely, using the result of [16], we can construct a finite set Y with the property | AY | < | Y | . The size of Y is really huge, it does not have transparent description. Thisimmediately implies that Problem Q has a positive solution. We announce this resulthere; details will appear in forthcoming papers.Donnelly shows in [10] that F is non-amenable if and only if there exists ε > suchthat for any finite set Y ⊂ F , one has | AY | ≥ (1 + ε ) | Y | , where A = { x , x , x } (seealso [8]). For the set Y here, one can assume without loss of generality that Y is containedin S ,n for some n . This gives some evidence that the amenability problem for F has very10lose relationship with the family of Problems Q k . The case k = 4 looks as a possiblecandidiate to a negative solution (that is, all solutions are zero). If true, this will implythat the constant ε = fits into the above condition.Descriptions of solutions for equations or their systems seem also be important becauseif we try to prove that some equation or system has only zero solutions, then we are basedon the description of simpler equations for which we know all solutions. This situation isvaguely similar to the classical case when the descrition of all Pythagorian triples impliesFermat’s proof that X + Y = Z has no solutions in positive integers. Here we canexpect a similar effect. In this Section we go back to the equation (4) finding its solutions of possibly minimaldegree. Namely, instead of quadratic power of deg u , deg v with respect to m , we will findsolutions with deg u = deg v = m . It looks truthful that the value m here is minimal.We do not offer the proof of this fact although we give a complete description to the setof solutions of our equations for the cases m = 1 and m = 2 .For our needs it will be convenient to assume that the coefficients in the equation areindependent variables, that is, they belong to the field of rational functions over somefield L . By K we denote the field of rational functions with coefficients in L over anumber of variables of the form α i , β i ( i ≥ ).So let R = K [ M ] and let a = α x + α x + · · · + α m x m , b = β x + β x + · · · + β m x m . We are interested in finding nonzero solution of the equation au = bv , where u , v arehomogeneous polynomials in R of possibly minimal degree. Theorem 5
The equation ( α x + α x + · · · + α m x m ) u = ( β x + β x + · · · + β m x m ) v has a nonzero solution, where u , v are homogeneous polynomials of degree m in variables x , x , . . . , x m . Proof.
We keep the notaion a , b for the coefficients of the above equation. Weintoduce a sequence of elements σ i = a · β i + b · ( − α i ) ∈ aR + bR for all ≤ i ≤ m , wherethe coefficient on x i in σ i is zero. So each σ i has the form γ i x + · · · + γ i,i − x i − + γ i,i +1 x i +1 + · · · + γ im x m for any ≤ i ≤ m , where all γ ij = α j β i − α i β j are nonzero for j = i .We introduce polynomials of the form f k ( x k , . . . , x k + m − ) of degree k for each ≤ k ≤ m . Each of these polynomials will be equal to au k − + bv k − for some homogeneous11olynomials of degree k − . This will be done by induction. At each step we will checkthat monomial x k − occurs in both u k − , v k − with nonzero coefficient.For the case k = 1 , we take f ( x , . . . , x m ) = σ . By definition, this is a linearcombination of x , ... , x m with nonzero coefficients. It also equals au + bv , where u = β , v = − α . Notice that coefficients on x here are nonzero.Let k < m . Multiplying f k ( x k , . . . , x k + m − ) by k − P i =0 γ ki x i on the right and using definingrelations of the monoid M , we get the element f k ( x k , . . . , x k + m − ) · k − X i =0 γ ki x i = k − X i =0 γ ki x i · f k ( x k +1 , . . . , x k + m ) . We used the fact that for any i < k and for any product w of x k , x k +1 , ... , it holds theequality wx i = x i w ′ , where w ′ is obtained from w by increasing all subscripts by . Thisfollows directly from the defining relations of M .Notice that σ k = k − P i =0 γ ki x i + m P i = k +1 γ ki x i . Therefore, we can rewrite the above equalityas follows: ( au k − + bv k − ) · k − X i =0 γ ki x i = ( σ k − m X i = k +1 γ ki x i ) · f k ( x k +1 , . . . , x k + m ) . Now let us take into account that σ k = a · β k + b · ( − α k ) . Also let us define the polynomial f k +1 ( x k +1 , . . . , x k + m ) = m X i = k +1 γ ki x i · f k ( x k +1 , . . . , x k + m ) . This is a homogeneous polynomial of degree k + 1 in variables x k +1 , ... , x k + m . Sendingthis polynomial to the left-hand side of the above equality, we see that it is equal to − ( au k − + bv k − ) · k − X i =0 γ ki x i + ( a · β k + b · ( − α k )) · f k ( x k +1 , . . . , x k + m ) . This means that f k +1 ( x k +1 , . . . , x k + m ) = au k + bv k , where u k = − u k − k − X i =0 γ ki x i + β k f k ( x k +1 , . . . , x k + m ) ,v k = − v k − k − X i =0 γ ki x i − α k f k ( x k +1 , . . . , x k + m ) . These are homogeneous polynomials of degree k . Notice that the term x k occurs in bothexpressions with nonzero coefficient. Indeed, x k − had nonzero coefficients in u k − , v k − ;here it is multiplied by − γ k x , and the rest of the sum in both expressions does notcontain x at all. 12ontinuing in this way, at the last step we get f m ( x m , ..., x m − ) as a homogeneouspolynomial of degree m . It has the form au m − + bv m − . Now we multiply it on the rightby σ m = γ m x + · · · + γ m,m − x m − . Aplying defining relations of M , we get to f m ( x m , ..., x m − )( γ m x + · · · + γ m,m − x m − ) = ( γ m x + · · · + γ m,m − x m − ) f m ( x m +1 , ..., x m ) . Now f m σ m becomes au m − σ m + bv m − σ m . On the other hand, it equals σ m f m ( x m +1 , ..., x m ) ,where σ m ∈ aR + bR is expressed according to its defintion. So the same element alsoequals ( aβ m − bα m ) f m ( x m +1 , ..., x m ) .Comparing both things, we have an equality of the form au = bv , where u = − u m − σ m + β m f m ( x m +1 , ..., x m ) , v = v m − σ m + α m f m ( x m +1 , ..., x m ) . By the same reasons as above, both u and v have monomial x m with nonzero coeffi-cient (the monomials x m − with nonzero coefficients from u m − , v m − are multiplied by ± γ m x ).It is also clear that u , v are homogeneous polynomials of degree m in variables x , x ,... , x m , as it was stated. This completes the proof.Now let us observe the case m = 1 in order to describe the set of all solutions. Here a = α x + α x , b = β x + β x will be linear combinations of x , x with arbitrarycoefficients from any field. We exclude trivial case when the elements a , b are proportional.The equation au = bv now can be rewritten as x ( α u − β v ) = x ( β v − α u ) . By ourassumption, both expressions here are nonzero. All monomials in the right-hand side areleft divisible by x . Thus we can write β v − α u = x w for some polynomial w ∈ K [ M ] .Using the fact that x x = x x and cancelling by x on the left, we get α u − β v = x w .Solving the above system, one can write u = ( β x + β x ) w , v = ( α x + α x ) w withoutloss of generality, where w was replaced by ( α β − α β ) w up to a nonzero coefficient.Now let us go to a more interesting case m = 2 in order to give a complete descriptionof the set of solutions for au = bv , where a = α x + α x + α x , b = β x + β x + β x .To avoid unnecessary work with many coefficients, let us apply a linear transformation,reducing the equation to the following one: ( x + αx ) u = ( x + βx ) v, (6)where α, β are some coefficients. We will assume both of them are nonzero: otherwisethe description turns out to be trivial.First of all, we take a solution of this equation extracted from the proof of Theorem 5.One can check directly that the following polynomials satisfy (6): u = βx x + β x x − αx x − αβx x − αβx − αβ x x ,v = βx − αx x − α x − α βx x . We say that ( u , v ) is a basic solution. Now we are going to show how to extract allsolutions from it.By M we denote the submonoid of M generated by x , x , ... .13 emma 4 For any v ∈ K [ M ] there exist w ∈ K [ M ] , w , w ∈ K [ M ] such that v = v w + x w + w . Proof.
The idea of the proof is analogous to The Remainder Theorem. We are goingto show that v = x x + w modulo the principal right ideal v R , where R = K [ M ] . Allequalities below will be done modulo this ideal. Each of them can be multiplied on theright.We know that x = β − ( αx x + α x + α βx x ) . That is, x = x ξ + η for some ξ , η ∈ K [ M ] . Let us prove by induction on k ≥ that x k = x ξ k + η k for some ξ k , η k ∈ K [ M ] . We know that for k = 2 . Let this equality hold for some k ≥ . Then x k +10 = ( x ξ k + η k ) x = x φ ( ξ k ) + x φ ( η k ) , where φ is an endomorphism that takes each x i to x i +1 ( i ≥ ). Therefore, x k +10 = ( x ξ + η ) φ ( ξ k ) + x φ ( η k ) = x ξ k +1 + η k +1 , where ξ k +1 = ξ φ ( ξ k ) + φ ( η k ) , η k +1 = η φ ( ξ k ) .Now we can decompose v by powers of x , that is, v = ζ + x ζ + x ζ + · · · + x d ζ d forsome d , where ζ , . . . , ζ d ∈ K [ M ] . Replacing here all x k by x ξ k + η k for k ≥ , we get thedesrired equality of the form v = x w + w modulo the ideal v R , where w , w ∈ K [ M ] .Hence there exists w ∈ R such that v = v w + x w + w .The proof is complete.Let ( u, v ) be any solution of (6). We have au = bv = b ( v w + x w + w ) accordingto Lemma 4, together with au = bv multiplied by w on the right. This implies a ( u − u w ) = b ( x w + w ) . Recall that b = x + βx , so the right-hand side belongs to R + x R , where R = K [ M ] . Since a = x + αx , the term w = u − u w in its normalform cannot involve x . So it belongs to R .We have ( x + αx ) w = ( x + βx )( x w + w ) , that is, x w + αx w = x ( x + βx ) w +( x + βx ) w . Comparing coefficients, we get w = ( x + βx ) w and αx w = ( x + βx ) w .Hence αx ( x + βx ) w = ( x + βx ) w . This implies that x w is left divisible by x .This is also true for all monomials involved in x w . From elementary properties of themonoid M it follows that w is left divisible by x , so we can put w = x w for some w ∈ K [ M ] . Now x w = x x w = x x w , and we can cancel the equality by x onthe left.Now we get α ( x + βx ) w = ( x + βx ) w . Notice that u , v are homogeneous of somedegree d . So deg x w = deg v = d . Therefore deg w = d − . Also deg w = deg w − v − d − . So we are trying to solve the equation ( x + βx ) w = ( x + βx )( αw ) in K [ M ] in polynomials of degree d − . One can apply φ − decreasing all subscripts by1. This gives us equation ( x + βx ) φ − ( w ) = ( x + βx ) φ − ( αw ) in K [ M ] for whichwe eventually know all solutions of degree < d by induction. Any such solution brings usa solution ( u, v ) of the original equation of degree d .Let Sol ( d, α, β ) denote the set of solutions of (6) with parameters α , β , where deg u =deg v = d . This set is zero for d = 0 and d = 1 . Let ( u ′ , v ′ ) be arbitrtary solutionfrom Sol ( d − , β, β ) . We assume by induction this set is known for us. Now we canexpress w = φ ( u ′ ) and αw = φ ( v ′ ) . From αx w = ( x + βx ) w = ( x + βx ) x w = x ( x + βx ) w , we get w = α − ( x + βx ) φ ( u ′ ) . Therefore, u = u w + w = u w + α − ( x + βx ) φ ( u ′ ) and v = v w + x w + w = v w + α − x φ ( v ′ ) + x φ ( u ′ ) . These14qualities (cid:26) u = u w + α − ( x + βx ) φ ( u ′ ) v = v w + α − x φ ( v ′ ) + x φ ( u ′ ) (7)allow us to describe all solutions of degree d for parameters α , β provided all solutions ofdegree d − for parameters β , β are known.We see that the general case of parameters is reduced to the case α = β . Also if weknow the first component u of the solution, then the second component v can be uniquelyexpress from it. To present information in a compact from, we formulate the followingstatement. Theorem 6