In the last three notes, we discussed the Bourgain-Gamburd expansion machine and two of its three ingredients, namely quasirandomness and product theorems, leaving only the non-concentration ingredient to discuss. We can summarise the results of the last three notes, in the case of fields of prime order, as the following theorem.

Theorem 1 (Non-concentration implies expansion in )Let be a prime, let , and let be a symmetric set of elements in of cardinality not containing the identity. Write , and suppose that one has the non-concentration propertyfor some and some even integer . Then is a two-sided -expander for some depending only on .

*Proof:* From (1) we see that is not supported in any proper subgroup of , which implies that generates . The claim now follows from the Bourgain-Gamburd expansion machine (Theorem 2 of Notes 4), the product theorem (Theorem 1 of Notes 5), and quasirandomness (Exercise 8 of Notes 3).

Remark 1The same argument also works if we replace by the field of order for some bounded . However, there is a difficulty in the regime when is unbounded, because the quasirandomness property becomes too weak for the Bourgain-Gamburd expansion machine to be directly applicable. On theother hand, the above type of theorem was generalised to the setting of cyclic groups with square-free by Varju, to arbitrary by Bourgain and Varju, and to more general algebraic groups than and square-free by Salehi Golsefidy and Varju. It may be that some modification of the proof techniques in these papers may also be able to handle the field case with unbounded .

It thus remains to construct tools that can establish the non-concentration property (1). The situation is particularly simple in , as we have a good understanding of the subgroups of that group. Indeed, from Theorem 14 from Notes 5, we obtain the following corollary to Theorem 1:

Corollary 2 (Non-concentration implies expansion in )Let be a prime, and let be a symmetric set of elements in of cardinality not containing the identity. Write , and suppose that one has the non-concentration propertyfor some and some even integer , where ranges over all Borel subgroups of . Then, if is sufficiently large depending on , is a two-sided -expander for some depending only on .

It turns out (2) can be verified in many cases by exploiting the solvable nature of the Borel subgroups . We give two examples of this in these notes. The first result, due to Bourgain and Gamburd (with earlier partial results by Gamburd and by Shalom) generalises Selberg’s expander construction to the case when generates a thin subgroup of :

Theorem 3 (Expansion in thin subgroups)Let be a symmetric subset of not containing the identity, and suppose that the group generated by is not virtually solvable. Then as ranges over all sufficiently large primes, the Cayley graphs form a two-sided expander family, where is the usual projection.

Remark 2One corollary of Theorem 3 (or of the non-concentration estimate (3) below) is that generates for all sufficiently large , if is not virtually solvable. This is a special case of a much more general result, known as the strong approximation theorem, although this is certainly not the most direct way to prove such a theorem. Conversely, the strong approximation property is used in generalisations of this result to higher rank groups than .

Exercise 1In the converse direction, if is virtually solvable, show that for sufficiently large , fails to generate . (Hint:use Theorem 14 from Notes 5 to prevent from having bounded index solvable subgroups.)

Exercise 2 (Lubotzsky’s 1-2-3 problem)Let .

- (i) Show that generates a free subgroup of . (
Hint:use a ping-pong argument, as in Exercise 23 of Notes 2.)- (ii) Show that if are two distinct elements of the sector , then there os no element for which . (
Hint:this is another ping-pong argument.) Conclude that has infinite index in . (Contrast this with the situation in which the coefficients in are replaced by or , in which case is either all of , or a finite index subgroup, as demonstrated in Exercise 23 of Notes 2).- (iii) Show that for sufficiently large primes form a two-sided expander family.

Remark 3Theorem 3 has been generalised to arbitrary linear groups, and with replaced by for square-free ; see this paper of Salehi Golsefidy and Varju. In this more general setting, the condition of virtual solvability must be replaced by the condition that the connected component of the Zariski closure of is perfect. An effective version of Theorem 3 (with completely explicit constants) was recently obtained by Kowalski.

The second example concerns Cayley graphs constructed using random elements of .

Theorem 4 (Random generators expand)Let be a prime, and let be two elements of chosen uniformly at random. Then with probability , is a two-sided -expander for some absolute constant .

Remark 4As with Theorem 3, Theorem 4 has also been extended to a number of other groups, such as the Suzuki groups (in this paper of Breuillard, Green, and Tao), and more generally to finite simple groups of Lie type of bounded rank (in forthcoming work of Breuillard, Green, Guralnick, and Tao). There are a number of other constructions of expanding Cayley graphs in such groups (and in other interesting groups, such as the alternating groups) beyond those discussed in these notes; see this recent survey of Lubotzky for further discussion. It has been conjectured by Lubotzky and Weiss thatanypair of (say) that generates the group, is a two-sided -expander for an absolute constant : in the case of , this has been established for a density one set of primes by Breuillard and Gamburd.

** — 1. Expansion in thin subgroups — **

We now prove Theorem 3. The first observation is that the expansion property is monotone in the group :

Exercise 3Let be symmetric subsets of not containing the identity, such that . Suppose that is a two-sided expander family for sufficiently large primes . Show that is also a two-sided expander family.

As a consequence, Theorem 3 follows from the following two statments:

Theorem 5 (Tits alternative)Let be a group. Then exactly one of the following statements holds:

- (i) is virtually solvable.
- (ii) contains a copy of the free group of two generators as a subgroup.

Theorem 6 (Expansion in free groups)Let be generators of a free subgroup of . Then as ranges over all sufficiently large primes, the Cayley graphs form a two-sided expander family.

Theorem 5 is a special case of the famous Tits alternative, which among other things allows one to replace by for any and any field of characteristic zero (and fields of positive characteristic are also allowed, if one adds the requirement that be finitely generated). We will not prove the full Tits alternative here, but instead just give an *ad hoc* proof of the special case in Theorem 5 in the following exercise.

Exercise 4Given any matrix , the singular values are and , and we can apply the singular value decomposition to decomposewhere and are orthonormal bases. (When , these bases are uniquely determined up to phase rotation.) We let be the projection of to the projective complex plane, and similarly define .

Let be a subgroup of . Call a pair a

limit pointof if there exists a sequence with and .

- (i) Show that if is infinite, then there is at least one limit point.
- (ii) Show that if is a limit point, then so is .
- (iii) Show that if there are two limit points with , then there exist that generate a free group. (
Hint:Choose close to and close to , and consider the action of and on , and specifically on small neighbourhoods of , and set up a ping-pong type situation.)- (iv) Show that if is hyperbolic (i.e. it has an eigenvalue greater than 1), with eigenvectors , then the projectivisations of form a limit point. Similarly, if is regular parabolic (i.e. it has an eigenvalue at 1, but is not the identity) with eigenvector , show that is a limit point.
- (v) Show that if has no free subgroup of two generators, then all hyperbolic and regular parabolic elements of have a common eigenvector. Conclude that all such elements lie in a solvable subgroup of .
- (vi) Show that if an element is neither hyperbolic nor regular parabolic, and is not a multiple of the identity, then is conjugate to a rotation by (in particular, ).
- (vii) Establish Theorem 5. (
Hint:show that two square roots of in cannot multiply to another square root of .)

Now we prove Theorem 6. Let be a free subgroup of generated by two generators . Let be the probability measure generating a random walk on , thus is the corresponding generator on . By Corollary 2, it thus suffices to show that

for all sufficiently large , some absolute constant , and some even (depending on , of course), where ranges over Borel subgroups.

As is a homomorphism, one has and so it suffices to show that

To deal with the supremum here, we will use an argument of Bourgain and Gamburd, taking advantage of the fact that all Borel groups of obey a common group law, the point being that free groups such as obey such laws only very rarely. More precisely, we use the fact that the Borel groups are solvable of derived length two; in particular we have

for all . Now, is supported on matrices in whose coefficients have size (where we allow the implied constants to depend on the choice of generators ), and so is supported on matrices in whose coefficients also have size . If is less than a sufficiently small multiple of , these coefficients are then less than (say). As such, if lie in the support of and their projections obey the word law (4) in , then the original matrices obey the word law (4) in . (This lifting of identities from the characteristic setting of to the characteristic setting of is a simple example of the “Lefschetz principle”.)

To summarise, if we let be the set of all elements of that lie in the support of , then (4) holds for all . This severely limits the size of to only be of polynomial size, rather than exponential size:

Proposition 7Let be a subset of the support of (thus, consists of words in of length ) such that the law (4) holds for all . Then .

The proof of this proposition is laid out in the exercise below.

Exercise 5Let be a free group generated by two generators . Let be the set of all words of length at most in .

- (i) Show that if commute, then lie in the same cyclic group, thus for some and .
- (ii) Show that if , there are at most elements of that commute with .
- (iii) Show that if , there are at most elements of with .
- (iv) Prove Proposition 7.

Now we can conclude the proof of Theorem 3:

Exercise 6Let be a free group generated by two generators .

- (i) Show that for some absolute constant . (For much more precise information on , see this paper of Kesten.)
- (ii) Conclude the proof of Theorem 3.

** — 2. Random generators expand — **

We now prove Theorem 4. Let be the free group on two formal generators , and let be the generator of the random walk. For any word and any in a group , let be the element of formed by substituting for respectively in the word ; thus can be viewed as a map for any group . Observe that if is drawn randomly using the distribution , and , then is distributed according to the law , where . Applying Corollary 2, it suffices to show that whenever is a large prime and are chosen uniformly and independently at random from , that with probability , one has

for some absolute constant , where ranges over all Borel subgroups of and is drawn from the law for some even natural number .

Let denote the words in of length at most . We may use the law (4) to obtain good bound on the supremum in (5) assuming a certain non-degeneracy property of the word evaluations :

Exercise 7Let be a natural number, and suppose that is such that for . Show thatfor some absolute constant , where is drawn from the law . (

Hint:use (4) and the hypothesis to lift the problem up to , at which point one can use Proposition 7 and Exercise 6.)

In view of this exercise, it suffices to show that with probability , one has for all for some comparable to a small multiple of . As has elements, it thus suffices by the union bound to show that

for some absolute constant , and any of length less than for some sufficiently small absolute constant .

Let us now fix a non-identity word of length less than , and consider as a function from to for an arbitrary field . We can identify with the set . A routine induction then shows that the expression is then a polynomial in the eight variables of degree and coefficients which are integers of size . Let us then make the additional restriction to the case , in which case we can write and . Then is now a rational function of whose numerator is a polynomial of degree and coefficients of size , and the denominator is a monomial of of degree .

We then specialise this rational function to the field . It is conceivable that when one does so, the rational function collapses to the constant polynomial , thus for all with . (For instance, this would be the case if , by Lagrange’s theorem, if it were not for the fact that is far too large here.) But suppose that this rational function does not collapse to the constant rational function. Applying the Schwarz-Zippel lemma (Exercise 23 from Notes 5), we then see that the set of pairs with and is at most ; adding in the and cases, one still obtains a bound of , which is acceptable since and . Thus, the only remaining case to consider is when the rational function is identically on with .

Now we perform another “Lefschetz principle” maneuvre to change the underlying field. Recall that the denominator of rational function is monomial in , and the numerator has coefficients of size . If is less than for a sufficiently small , we conclude in particular (for large enough) that the coefficients all have magnitude less than . As such, the only way that this function can be identically on is if it is identically on for all with , and hence for or also by taking Zariski closures.

On the other hand, we know that for some choices of , e.g. , contains a copy of the free group on two generators (see e.g. Exercise 23 of Notes 2). As such, it is not possible for any non-identity word to be identically trivial on . Thus this case cannot actually occur, completing the proof of (6) and hence of Theorem 4.

Remark 5We see from the above argument that the existence of subgroups of an algebraic group with good “independence” properties – such as that of generating a free group – can be useful in studying the expansion properties of that algebraic group, even if the field of interest in the latter is distinct from that of the former. For more complicated algebraic groups than , in which laws such as (4) are not always available, it turns out to be useful to place further properties on the subgroup , for instance by requiring that all non-abelian subgroups of that group be Zariski dense (a property which has been calledstrong density), as this turns out to be useful for preventing random walks from concentrating in proper algebraic subgroups. See this paper of Breuillard, Guralnick, Green and Tao for constructions of strongly dense free subgroups of algebraic groups and further discussion.

## 4 comments

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14 February, 2012 at 8:00 am

Lior SilbermanRegarding remark 1: I think using the lazy random walk you can only prove one-sided expansion (you only see the spectral gap at the top), while using even powers of the standard walk will bound the two-sided gap.

[Oops, that was silly of me. The remark has now been deleted, with appropriate changes to all sidedness references. -T]Regarding Remark 2: While this follows from expansion (since the proper subgroups are virtually solvable and hence don’t expand for any bounded set of generators), in proving strong approximation by this route you might as well stop earlier — it already follows from the large girth (which is basically an ingredient in the proof of expansion). Let generate a free group (obtained from the Tits alternative) and choose in that group so that . Observes with Bourgain-Gamburd that every proper subgroup of satisfies the law . It follows that modulo a prime which does not divide the entries of the chosen double commutator the images of do not generate a proper subgroup, and hence are a set of generators.

On the other hand, I don’t see how to generalize this to other groups, which have as a subgroup (even if you assume expansion, since we have just shown that these groups expand). As far as I know the proofs of expansion in higher rank all use the strong approximation theorem rather than imply it, but I’d be happy to learn otherwise.

[Fair enough, I’ll modify the remark accordingly. -T]19 February, 2012 at 4:43 pm

iamdibakardattaDear Prof. Tao,

I am a PhD student in Solid Mechanics at Brown University, USA. According your point of view, what is the ‘ physical significance of entropy ‘ ?

Thank you in advance for the reply.

Regards,

Dibakar

22 February, 2012 at 10:52 am

kittypo in exercise 2: then there os no element

1 March, 2012 at 5:25 pm

254B, Notes 7: Sieving and expanders « What’s new[…] turns out that to prove Theorem 1, the Cayley expansion results in from the previous set of notes are not quite enough; one needs a more general Cayley expansion result in where is square-free […]