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 property
for 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 1 The 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 property
for 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 2 One 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 1 In 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 3 Theorem 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 4 As 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 that any pair
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 3 Let
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 4 Given any matrix
, the singular values are
and
, and we can apply the singular value decomposition to decompose
where
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 point of
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
, 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
. 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 7 Let
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 5 Let
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 6 Let
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
, 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 7 Let
be a natural number, and suppose that
is such that
for
. Show that
for 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
, 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 5 We 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 called strong 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
Comments feed for this article
14 February, 2012 at 8:00 am
Lior Silberman
Regarding 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
iamdibakardatta
Dear 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
kit
typo 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 [...]