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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.

In the previous set of notes we saw how a representation-theoretic property of groups, namely Kazhdan’s property (T), could be used to demonstrate expansion in Cayley graphs. In this set of notes we discuss a different representation-theoretic property of groups, namely *quasirandomness*, which is also useful for demonstrating expansion in Cayley graphs, though in a somewhat different way to property (T). For instance, whereas property (T), being qualitative in nature, is only interesting for infinite groups such as or , and only creates Cayley graphs after passing to a finite quotient, quasirandomness is a quantitative property which is directly applicable to finite groups, and is able to deduce expansion in a Cayley graph, provided that random walks in that graph are known to become sufficiently “flat” in a certain sense.

The definition of quasirandomness is easy enough to state:

Definition 1 (Quasirandom groups)Let be a finite group, and let . We say that is-quasirandomif all non-trivial unitary representations of have dimension at least . (Recall a representation istrivialif is the identity for all .)

Exercise 1Let be a finite group, and let . A unitary representation is said to beirreducibleif has no -invariant subspaces other than and . Show that is -quasirandom if and only if every non-trivial irreducible representation of has dimension at least .

Remark 1The terminology “quasirandom group” was introduced explicitly (though with slightly different notational conventions) by Gowers in 2008 in his detailed study of the concept; the name arises because dense Cayley graphs in quasirandom groups are quasirandom graphs in the sense of Chung, Graham, and Wilson, as we shall see below. This property had already been used implicitly to construct expander graphs by Sarnak and Xue in 1991, and more recently by Gamburd in 2002 and by Bourgain and Gamburd in 2008. One can of course define quasirandomness for more general locally compact groups than the finite ones, but we will only need this concept in the finite case. (A paper of Kunze and Stein from 1960, for instance, exploits the quasirandomness properties of the locally compact group to obtain mixing estimates in that group.)

Quasirandomness behaves fairly well with respect to quotients and short exact sequences:

Exercise 2Let be a short exact sequence of finite groups .

- (i) If is -quasirandom, show that is -quasirandom also. (Equivalently: any quotient of a -quasirandom finite group is again a -quasirandom finite group.)
- (ii) Conversely, if and are both -quasirandom, show that is -quasirandom also. (In particular, the direct or semidirect product of two -quasirandom finite groups is again a -quasirandom finite group.)

Informally, we will call *quasirandom* if it is -quasirandom for some “large” , though the precise meaning of “large” will depend on context. For applications to expansion in Cayley graphs, “large” will mean “ for some constant independent of the size of “, but other regimes of are certainly of interest.

The way we have set things up, the trivial group is infinitely quasirandom (i.e. it is -quasirandom for every ). This is however a degenerate case and will not be discussed further here. In the non-trivial case, a finite group can only be quasirandom if it is large and has no large subgroups:

Exercise 3Let , and let be a finite -quasirandom group.

- (i) Show that if is non-trivial, then . (
Hint:use the mean zero component of the regular representation .) In particular, non-trivial finite groups cannot be infinitely quasirandom.- (ii) Show that any proper subgroup of has index . (
Hint:use the mean zero component of the quasiregular representation.)

The following exercise shows that quasirandom groups have to be quite non-abelian, and in particular perfect:

Exercise 4 (Quasirandomness, abelianness, and perfection)Let be a finite group.

- (i) If is abelian and non-trivial, show that is not -quasirandom. (
Hint:use Fourier analysis or the classification of finite abelian groups.)- (ii) Show that is -quasirandom if and only if it is perfect, i.e. the commutator group is equal to . (Equivalently, is -quasirandom if and only if it has no non-trivial abelian quotients.)

Later on we shall see that there is a converse to the above two exercises; any non-trivial perfect finite group with no large subgroups will be quasirandom.

Exercise 5Let be a finite -quasirandom group. Show that for any subgroup of , is -quasirandom, where is the index of in . (Hint:use induced representations.)

Now we give an example of a more quasirandom group.

Lemma 2 (Frobenius lemma)If is a field of some prime order , then is -quasirandom.

This should be compared with the cardinality of the special linear group, which is easily computed to be .

*Proof:* We may of course take to be odd. Suppose for contradiction that we have a non-trivial representation on a unitary group of some dimension with . Set to be the group element

and suppose first that is non-trivial. Since , we have ; thus all the eigenvalues of are roots of unity. On the other hand, by conjugating by diagonal matrices in , we see that is conjugate to (and hence conjugate to ) whenever is a quadratic residue mod . As such, the eigenvalues of must be permuted by the operation for any quadratic residue mod . Since has at least one non-trivial eigenvalue, and there are distinct quadratic residues, we conclude that has at least distinct eigenvalues. But is a matrix with , a contradiction. Thus lies in the kernel of . By conjugation, we then see that this kernel contains all unipotent matrices. But these matrices generate (see exercise below), and so is trivial, a contradiction.

Exercise 6Show that for any prime , the unipotent matricesfor ranging over generate as a group.

Exercise 7Let be a finite group, and let . If is generated by a collection of -quasirandom subgroups, show that is itself -quasirandom.

Exercise 8Show that is -quasirandom for any and any prime . (This is not sharp; the optimal bound here is , which follows from the results of Landazuri and Seitz.)

As a corollary of the above results and Exercise 2, we see that the projective special linear group is also -quasirandom.

Remark 2One can ask whether the bound in Lemma 2 is sharp, assuming of course that is odd. Noting that acts linearly on the plane , we see that it also acts projectively on the projective line , which has elements. Thus acts via the quasiregular representation on the -dimensional space , and also on the -dimensional subspace ; this latter representation (known as the Steinberg representation) is irreducible. This shows that the bound cannot be improved beyond . More generally, given any character , acts on the -dimensional space of functions that obey the twisted dilation invariance for all and ; these are known as the principal series representations. When is the trivial character, this is the quasiregular representation discussed earlier. For most other characters, this is an irreducible representation, but it turns out that when is the quadratic representation (thus taking values in while being non-trivial), the principal series representation splits into the direct sum of two -dimensional representations, which comes very close to matching the bound in Lemma 2. There is a parallel series of representations to the principal series (known as the discrete series) which is more complicated to describe (roughly speaking, one has to embed in a quadratic extension and then use a rotated version of the above construction, to change a split torus into a non-split torus), but can generate irreducible representations of dimension , showing that the bound in Lemma 2 is in fact exactly sharp. These constructions can be generalised to arbitrary finite groups of Lie type using Deligne-Luzstig theory, but this is beyond the scope of this course (and of my own knowledge in the subject).

Exercise 9Let be an odd prime. Show that for any , the alternating group is -quasirandom. (Hint:show that all cycles of order in are conjugate to each other in (and not just in ); in particular, a cycle is conjugate to its power for all . Also, as , is simple, and so the cycles of order generate the entire group.)

Remark 3By using more precise information on the representations of the alternating group (using the theory of Specht modules and Young tableaux), one can show the slightly sharper statement that is -quasirandom for (but is only -quasirandom for due to icosahedral symmetry, and -quasirandom for due to lack of perfectness). Using Exercise 3 with the index subgroup , we see that the bound cannot be improved. Thus, (for large ) is not as quasirandom as the special linear groups (for large and bounded), because in the latter case the quasirandomness is as strong as a power of the size of the group, whereas in the former case it is only logarithmic in size.If one replaces the alternating group with the slightly larger symmetric group , then quasirandomness is destroyed (since , having the abelian quotient , is not perfect); indeed, is -quasirandom and no better.

Remark 4Thanks to the monumental achievement of the classification of finite simple groups, we know that apart from a finite number (26, to be precise) of sporadic exceptions, all finite simple groups (up to isomorphism) are either a cyclic group , an alternating group , or is a finite simple group of Lie type such as . (We will define the concept of a finite simple group of Lie type more precisely in later notes, but suffice to say for now that such groups are constructed from reductive algebraic groups, for instance is constructed from in characteristic .) In the case of finite simple groups of Lie type with bounded rank , it is known from the work of Landazuri and Seitz that such groups are -quasirandom for some depending only on the rank. On the other hand, by the previous remark, the large alternating groups do not have this property, and one can show that the finite simple groups of Lie type with large rank also do not have this property. Thus, we see using the classification that if a finite simple group is -quasirandom for some and is sufficiently large depending on , then is a finite simple group of Lie type with rank . It would be of interest to see if there was an alternate way to establish this fact that did not rely on the classification, as it may lead to an alternate approach to proving the classification (or perhaps a weakened version thereof).

A key reason why quasirandomness is desirable for the purposes of demonstrating expansion is that quasirandom groups happen to be rapidly mixing at large scales, as we shall see below the fold. As such, quasirandomness is an important tool for demonstrating expansion in Cayley graphs, though because expansion is a phenomenon that must hold at all scales, one needs to supplement quasirandomness with some additional input that creates mixing at small or medium scales also before one can deduce expansion. As an example of this technique of combining quasirandomness with mixing at small and medium scales, we present a proof (due to Sarnak-Xue, and simplified by Gamburd) of a weak version of the famous “3/16 theorem” of Selberg on the least non-trivial eigenvalue of the Laplacian on a modular curve, which among other things can be used to construct a family of expander Cayley graphs in (compare this with the property (T)-based methods in the previous notes, which could construct expander Cayley graphs in for any fixed ).

Van Vu and I have just uploaded to the arXiv our preprint “A sharp inverse Littlewood-Offord theorem“, which we have submitted to Random Structures and Algorithms. This paper gives a solution to the (inverse) Littlewood-Offord problem of understanding when random walks are concentrated in the case when the concentration is of polynomial size in the length of the walk; our description is sharp up to epsilon powers of . The theory of inverse Littlewood-Offord problems and related topics has been of importance in recent developments in the spectral theory of discrete random matrices (e.g. a “robust” variant of these theorems was crucial in our work on the circular law).

For simplicity I will restrict attention to the Bernoulli random walk. Given real numbers , one can form the random variable

where are iid random signs (with either sign +1, -1 chosen with probability 1/2). This is a discrete random variable which typically takes values. However, if there are various arithmetic relations between the step sizes , then many of the possible sums collide, and certain values may then arise with much higher probability. To measure this, define the *concentration probability* by the formula

.

Intuitively, this probability measures the amount of additive structure present between the . There are two (opposing) problems in the subject:

- (Forward Littlewood-Offord problem) Given some structural assumptions on , what bounds can one place on ?
- (Inverse Littlewood-Offord problem) Given some bounds on , what structural assumptions can one then conclude about ?

Ideally one would like answers to both of these problems which come close to inverting each other, and this is the guiding motivation for our paper.

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