**Theorem** (Hilbert). Suppose that is an algebraically closed field (say, the field of complex numbers). Let be an ideal of the ring of polynomials generated by polynomials , be a polynomial from the same ring. Suppose that there are no tuples of elements from which satisfy the sequence of equalities and inequality . Then in is for some natural number .

**Proof. **The idea is to first show that

**(Goal 1)** () belongs to a bigger ideal generated by for some polynomial ,

and then show that

**(Goal 2)** for some natural .

Then from the first of these statements, it follows that for some polynomial , and from the second statement, it follows that is in .

So let

(a) be our system of equalities and inequality that does not have a solution in .

The proof proceeds by induction on three parameters:

- – the number of variables occurring in . Let be the smallest subscript of all these variables.
- – the number of which actually contain the variable .
- – the smallest non-zero number which is the degree of viewed as a polynomial in .

*Note that* does not depend on .

We order the triples lexicographically and assume (by induction) that for smaller triples the theorem holds.

- Without loss of generality has degree in . Then we can write where is a polynomial not containing (that will be the required polynomial from goals 1 and 2) and is a polynomial of degree in .Then neither of the following systems of equalities and inequalities has a solution in (obvious):

(b)(c)

- First we will deal with the system (b). This system is equivalent to the system(b*)
The triple for (b*) is smaller than the triple for (a): the number of variables is at most the same but either does not contain , in which case, the number of polynomials containing (the second coordinate of the triple) is smaller for (b*), or the degree in of is smaller than and so the first two coordinates are the same but the third is smaller for (b*) .Thus we can apply the Nullstellensatz for (b*) and get

(d) .

The first goal is achieved.

- To achieve the second goal we will use (c) and consider two cases.3.1. Suppose that . Then appears in one of the . WLOG . View as a polynomial in with coefficients in the field of rational fractions . Then we can divide by with a remainder. It is easy to see that when we do the division, the only denominators we get are powers of (the highest coefficient of ). Therefore we have an equality where is a natural number, are polynomials, and the -degree of is less than . It is easy to see that the system
(c*)

does not have a solution in because it is equivalent to (c). The triple of (c*) is smaller than the triple for (a) (the first two coordinates are the same, but the third coordinate is smaller for (c*). Therefore by induction we get

and we have the second goal. Note that we have not even used the fact that is algebraically closed. It will be used in the second case.

3.2. Suppose that . Divide by

(here is just a natural number bigger than ). As before we get

(e)where (we can always achieve the strict inequality by multiplying both sides by , recall that -degree of is 0). Now consider the system

(f)

We wish to show that (f) does not have a solution in . Suppose it has a solution , that is if we substitute each by we get true equalities and inequality. Replace back by the variable to obtain a system in one variable . Note that since do not contain , these polynomials are still equal to 0 after our substitution. But becomes where is from and is a polynomial in which is not 0. The result of our substitution in , the polynomial , is of degree in , and so it has roots in (since the field is algebraically closed!). Combining each of these roots with our values for we get a solution of the system . That cannot be a solution of the system (a). Hence each is a root of (the result of our substitution applied to ). Moreover the exponent in (f) is so large that the multiplicity of each root of is at least the same as in . Hence has at least roots in which is a contradiction since the degree of is smaller than .

Thus, indeed, (f) does not have a solution. The triple for (f) is smaller than the triple for (a). Hence we have (by the induction assumption):

for some natural . Hence

and the second goal is achieved.

The theorem is proved.

**Remark.**One also needs to consider the case in 3.2 which is trivial (since the field is algebraically closed, it is infinite and no non-zero polynomial can vanish for every -tuple of elements of ).

]]>

First we show that Jones’ subgroup of is maximal in a certain subgroup of index of which itself is isomorphic to . (Recall that is isomorphic to the Thompson-Brown group and is 3-generated.) Moreover, we prove that there are exactly three subgroups of containing . The preimage of under the natural injective endomorphism of is then a maximal subgroup of and it is not difficult to show that this subgroup does not fix a point in the open unit interval. The fact that there are only three subgroups of containing Jones’ subgroup is very counter-intuitive and answers a question by Saul Schleimer.

The second solution gives many implicit examples (which may or may not be finitely generated). Note that the generator of does not fix any point in the open unit interval. Now take any element such is a proper subgroup of and not inside any proper subgroup of finite index. Then any maximal subgroup containing is an example we need. The fact that is not inside any proper subgroup of finite index is easy to establish. Indeed, every non-Abelian homomorphic image of is itself. So it is enough to show that the images of in generate the whole . For example, one can take .

The non-trivial thing is to prove that is proper. Indeed, the only previously known (more precisely – previously published) way to show that a subgroup of is proper is to show that it is inside some proper subgroup of finite index of or fixes a point on the open unit interval. We employ a 2-dimensional analog of Stallings’ solution of the membership problem for subgroups of free groups. For every subgroup of viewed as the diagram group over the Dunce hat, we construct what we called a Stallings 2-core which is a directed 2-complex with a distinguished 1-path and a map into the Dunce hat (which is also a directed 2-complex). In the case of the free group, the Stallings core is an automaton which accepts a reduced word if and only if it belongs to the subgroup. In the case of , the Stallings 2-core is a 2-automaton which “accepts” reduced diagrams from (but not only these diagrams). In the case of we construct the Stallings 2-core and show that is not “accepted” by it. So , and is indeed a proper subgroup of . Although the method of Stallings 2-cores was never published before, we discussed it with Victor Guba long ago (in 1998 or even earlier). We wanted to write a paper about it but never had time to do that.

Note that the question of when the Stallings 2-core of a subgroup of a diagram group accepts only the diagrams from is very interesting because in that case, is a diagram group itself. This is how, together with Victor Guba, we represented the derived subgroup as a diagram group (see Theorem 26 in this paper).

If is a finite set of finite binary fractions, then it is easy to see that the subgroup of all elements of that fix each number from is just the direct product of copies of . Savchuk studied these subgroups too, proving that their Schreier graphs are also amenable. We prove in our paper that for every there are only finitely many subgroups of containing (moreover the lattice of subgroups of containing is anti-isomorphic to the Boolean lattice of subsets of ). This implies that is quasi-residually finite (in our terminology), that is it has a decreasing sequence of finitely generated subgroups such that the intersection of these subgroups is trivial and for each there are only finitely many subgroups of containing . That is also a counter-intuitive property of .

]]>Then some boundary label of , read counterclockwise, is a relator of (all relators here and below are cyclic words) and has most 2-letter subwords from the following set called *regular words*. It follows (see Theorem 3 in the paper), that every amenable 2-generated group as above has a relator such that, say, more than 99 percent of all 2-letter subwords of are regular. The authors claim (Theorem 8) that does not have such relators. For this, the authors introduce weights of irregular 2-letter words. Say, weights 24 while weights 23, and weights sometimes 17 sometimes 15 depending on the neighbor letters. The weight is the sum of weights of irregular 2-letter subwords and the number of maximal subwords containing only irregular 2-letter subwords. Let be the length of . If 99 percent of 2-letter subwords of are regular, then , and the authors claim that it is impossible that a relator of satisfies this property.

The authors use the semigroup diagram representation of elements of as in our papers with Victor Guba and the action of on binary trees with a marked leaf. It is very easy to describe the action of each of the two standard generators (the authors denote them by ) on trees: basically moves the marker one leaf to the left, and does some changes to the right of the marker. Analyzing the proof, one can quickly come to the conclusion that the only properties of the authors are using are the following:

(1) The group is torsion-free and non-commutative.

(2) Every relator contains a (cyclic) subword of the form where , and commutes with in (this immediately follows from Lemma 13 and is explained in the proof of Lemma 14).

Victor Guba noticed that the wreath product acting in the standard way on the bi-infinite sequences of integers with a marked place ( moves the marker by 1 to the left, increases the value of the marked integer by 1) has the same properties. To prove that, one just has to repeat the proof of Lemma 13 of the paper. It works for the action of on marked bi-infinite sequences of numbers just as well as it works for the action of on trees. What is used there is that moves the marker to the left while does not change the configuration to the left of the marker. Since is amenable, the proof in the paper cannot possibly work. The concrete “fatal” mistake in the proof is in Case 3 on page 20 (the proof of Lemma 14). The authors did not notice that after they replace the subword by the word (which can be done since commutes with ), some subwords of the form have new neighbors, and so their weight may increase (by 2, see above). That leads to increasing the total weight , and so the inequality is no longer true after the rewriting.

So the paper contains some nice ideas (Theorem 3, in particular), but the proof is wrong. It is interesting whether the next “proof” will claim amenability or nonamenability. We’ll probably know it next year.

**Update 1:** The authors have put a new version of their paper on the arXiv (see the link above).

The general idea remains the same but the weight system has changed, it has become more complicated. The authors said that the new proof does not work for . Of course there are many other 2-generated amenable groups. But in any case the new proof needs to be checked.

**Update 2:** Gili Golan (a student in Bar Ilan University) found a mistake in the new version of the paper. Here is her message:

In Lemma 18, second to last line on page 18 of the paper, it says that if then .

The proof of this claim (under some assumptions) is given in the third paragraph on page 20, but I don’t understand why the argument works.If we have a red maximal edge from to some vertex to the right of it is possible that removing it by applying results in a maximal red edge from to (and from to ). Then multiplying by would get us to as required.

**Update 3: **Note that so far nobody could find a counterexample to the statement that does not have a relator with the number of regular 2-letter subwords . Note also that there are no counterexamples to Lemma 18 either. One can construct a 2-generated non-amenable group where for every , there exists a relator with the number of regular 2-letter subwords . Just take a group where the words satisfy a small cancelation condition and .

**Update 4: **A counterexample to the proof of Lemma 18 has been found by Victor Guba. The word where is a relator of but the subword constructed as in the proof of Lemma 18 does not satisfy all the conditions of that lemma. Matt Zaremsky found a subword of that word, , which does satisfy all conditions of Lemma 18, so this is not a counterexample to Lemma 18.

**Update 5: **Lemma 18 is wrong. Victor Guba found a counterexample. Here is his message:

]]>Let . The only thing one needs to check here is the fact that is a relation in . This is quite easy.

Now, has only two subwords of the form . For each of them, we see before such a subword.

Let be two elements in , all elements in the support of have length . Then RD is equivalent to the existence of a universal polynomial such that where is the norm. It is easy to see that is a function such that for every . Since the norm squared is the sum of squares of coefficients, it is clear that the problem is about triangles with side labels in the Cayley graph of : the “more” triangles with a given base label, the harder it is to get RD.

Even though the known classes of groups with RD are quite diverse, the methods of proving RD are “asymptotically similar”. The reason why the free groups have RD is that every geodesic triangle on a tree has a center which belongs to every side of the triangle. For Cayley graphs of hyperbolic groups, a center of a triangle may not belong to all three sides, but it is at bounded distance from all three sides (this is Rips’ definition of hyperbolic groups). For triangles in the Cayley graphs of relatively hyperbolic groups (as in our paper with Cornelia Drutu) and in symmetric spaces of Lie groups such as every triangle has an “inscribed” nice and relatively small triangle from a certain family of triangles. Indira Chatterji and Kim Ruane used clouds of centers, and Laura Ciobanu, Derek Holt and Sarah Rees used a condition which can be interpreted as a center-like condition.

I just wrote a paper “The Rapid Decay property and centroids in groups”, where an easy to formulate and check “centroid” condition is given. It follows from the centroid-type conditions used before and implies property RD. A not quite successful attempt to do that was made at the end of our paper with Cornelia Drutu. There we formulated our property “(**)-relative hyperbolicity”. But that property is not general enough and the definition of (**) in our paper contains errors.

The centroid property is the following.

Let be a countable group acting almost freely (i.e., the stabilizers of points have uniformly bounded finite orders) by isometries on a metric space , . We assume that is the length function defined by . Let be a map from the set of pairs to . We can view as embedded into (by the map ), a pair as the vertices of triangle , and as a “center” of that triangle. We say that and satisfies the *centroid property* if for some polynomial we have

For every and every the number of elements does not exceed

For every in the number of elements , in does not exceed .

For every the number of elements in the set does not exceed

In this case is called the space of centroids of and is called the centroid map.

The meaning is clear: if we fix the label (in ) of one side of the triangle with vertices and vary the third vertex keeping the length of one side , then the number of possible centroids of these triangles is not large (at most a polynomial in ).

In the paper, I show that almost all groups that are known to have RD such as hyperbolic groups, many uniform lattices in direct products of semisimple Lie groups of ranks 1 and 2, the mapping class groups, infinitely presented small cancelation groups, Artin groups of large type, have the centroid property or its relative version. The only class for which it is not known (to me) are graph products of groups. An undergraduate student Mitch Kleban is now working on that, it is his Summer REU problem. I think that the centroid property is interesting in its own right as a very weak form of hyperbolicity.

Another thing which I included in the paper is an easy non-amenability-like corollary from RD. If in the definition of RD one restricts oneself to indicator functions of finite sets, then it is easy (one line) to prove that for every group with RD there is a polynomial such that for every two finite sets of elements where all elements of have length , we have

In particular, it immediately implies that does not have amenable subgroups of superpolynomial growth. It would be interesting to find out if (1) (which I call “superpolynomial expansion property) implies RD. Using (1), it is easy to observe that the free product of all free Abelian groups of finite ranks with an appropriate length function is an example of a countable group without RD and without amenable subgroups of superpolynomial growth. Alexander Olshanskii proved that one of the standard embeddings of countable groups into 2-generated groups does not add extra amenable subgroups (i.e. every amenable subgroup of the bigger group is either cyclic of is conjugated to a subgroup of the smaller group), and cyclic subgroups in the bigger group that are not conjugated to subgroups of the smaller group are undistorted. Therefore there exists a 2-generated group without RD and without amenable subgroups of superpolynomial growth. Another example was constructed by Denis Osin. It would be interesting to find a finitely presented example. It should not be that difficult using S-machines and one of the numerous versions of the Higman embedding theorem.

A very interestiong problem (obviously related to a problem of Alain Valette whether uniform lattices in higher rank semi-simple Lie groups always have RD) is whether uniform lattices in satisfy the (relative) centroid property or the superpolynomial expansion property.

]]>THEOREM There exists a constant such that for every natural number we have

PROOF. The inequality is obvious, so we should only prove the inequality For every natural number consider the middle binomial coefficient in the Newton expansion of .

EXERCISE 1. Prove that . HINT: Use the Pascal triangle formula to prove that the middle binomial coefficient is the biggest one. Then use the fact that the sum of the binomial coefficients is , and the number of them is .

For a prime and a number we define as the largest exponent of that divides . Note that and if divides , then .

EXERCISE 2. For every natural number and prime we have where is such that . HINT: There are numbers that are divisible by , numbers that are divisible by , etc.

LEMMA. If , i.e., divides , then .

PROOF. Let be the natural number such that . Then by Exercise 2.

because for every positive real number we have . Therefore Now let us deduce the theorem from this lemma. Since for every , we can assume that is even. Let . Every prime that divides does not exceed . Therefore , where all primes are different and , so . Hence By Lemma, each does not exceed . Hence by Exercise 1

for some Thus .

]]>Here is an abstract:

The Tarski number of a non-amenable group is the minimal number of pieces

in a paradoxical decomposition of . In this paper we investigate how

Tarski numbers may change under various group-theoretic operations. Using these

estimates and the theory of Golod-Shafarevich groups, we show that

the Tarski numbers of -generated non-amenable groups can be arbitrarily large, and that the Tarski numbers of finite index subgroups of a given finitely generated non-amenable group can be arbitrarily large. In particular, for some number , the property

of having Tarski number is not invariant under quasi-isometry.

We also use -Betti numbers to show that there exist groups with Tarski number .

These provide the first examples of non-amenable groups without free subgroups

whose Tarski number has been computed precisely.

]]>

Lemma 1. Let us have a homogeneous system of linear equations with variables with integer coefficients, the abs. value of each coefficient is at most . Then there exists a non-trivial integer solution with provided .

Proof (half a page) see in Lang “Algebra”.

Consider the function with rational coefficients. Set (choose so that it is an integer). By Lemma 1, we can find integers such that all derivatives , , moreover absolute values of are bounded by something like for very large by Lemma 1. Since are algebraically independent (here we use that is irrational), the function is not identically 0. Let be the smallest integer such that all derivatives up to the order of are 0 at each , but the derivative for some (for simplicity let , the proof is the same for any ). Then belongs to the field , that is where are rational numbers. Moreover the least common denominator of is bounded by for some constant . The conjugate of (i.e.,) is bounded by for some constant . Then the norm of in , i.e. the product of and its conjugate, which is an integer , is bounded by . Now consider the function . Then differs from by some factor bounded by (which is “small”). By the maximum modulus principle, the value of is bounded above by the value of where is a complex number on a circle of large enough radius (we need only that is bigger than each ). Taking large enough, we get that is bounded from above by a number of the form . For ( is an appropriate constant) this gives a contradiction with the previous inequality for (the denominator grows much faster than the numerator, so the fraction cannot be always greater than 1).

]]>