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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 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
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 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
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 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
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 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.
In the previous set of notes, we saw that one could derive expansion of Cayley graphs from three ingredients: non-concentration, product theorems, and quasirandomness. Quasirandomness was discussed in Notes 3. In the current set of notes, we discuss product theorems. Roughly speaking, these theorems assert that in certain circumstances, a finite subset of a group
either exhibits expansion (in the sense that
, say, is significantly larger than
), or is somehow “close to” or “trapped” by a genuine group.
Theorem 1 (Product theorem in
) Let
, let
be a finite field, and let
be a finite subset of
. Let
be sufficiently small depending on
. Then at least one of the following statements holds:
- (Expansion) One has
.
- (Close to
) One has
.
- (Trapping)
is contained in a proper subgroup of
.
We will prove this theorem (which was proven first in the cases for fields
of prime order by Helfgott, and then for
and general
by Dinai, and finally to general
and
independently by Pyber-Szabo and by Breuillard-Green-Tao) later in this notes. A more qualitative version of this proposition was also previously obtained by Hrushovski. There are also generalisations of the product theorem of importance to number theory, in which the field
is replaced by a cyclic ring
(with
not necessarily prime); this was achieved first for
and
square-free by Bourgain, Gamburd, and Sarnak, by Varju for general
and
square-free, and finally by this paper of Bourgain and Varju for arbitrary
and
.
Exercise 1 (Diameter bound) Assuming Theorem 1, show that whenever
is a symmetric set of generators of
for some finite field
and some
, then any element of
can be expressed as the product of
elements from
. (Equivalently, if we add the identity element to
, then
for some
.) This is a special case of a conjecture of Babai and Seress, who conjectured that the bound should hold uniformly for all finite simple groups (in particular, the implied constants here should not actually depend on
. The methods used to handle the
case can handle other finite groups of Lie type of bounded rank, but at present we do not have bounds that are independent of the rank. On the other hand, a recent paper of Helfgott and Seress has almost resolved the conjecture for the permutation groups
.
A key tool to establish product theorems is an argument which is sometimes referred to as the pivot argument. To illustrate this argument, let us first discuss a much simpler (and older) theorem, essentially due to Freiman, which has a much weaker conclusion but is valid in any group :
Theorem 2 (Baby product theorem) Let
be a group, and let
be a finite non-empty subset of
. Then one of the following statements hold:
- (Expansion) One has
.
- (Close to a subgroup)
is contained in a left-coset of a group
with
.
To prove this theorem, we suppose that the first conclusion does not hold, thus . Our task is then to place
inside the left-coset of a fairly small group
.
To do this, we take a group element , and consider the intersection
. A priori, the size of this set could range from anywhere from
to
. However, we can use the hypothesis
to obtain an important dichotomy, reminiscent of the classical fact that two cosets
of a subgroup
of
are either identical or disjoint:
Proposition 3 (Dichotomy) If
, then exactly one of the following occurs:
- (Non-involved case)
is empty.
- (Involved case)
.
Proof: Suppose we are not in the pivot case, so that is non-empty. Let
be an element of
, then
and
both lie in
. The sets
and
then both lie in
. As these sets have cardinality
and lie in
, which has cardinality less than
, we conclude from the inclusion-exclusion formula that
But the left-hand side is equal to , and the claim follows.
The above proposition provides a clear separation between two types of elements : the “non-involved” elements, which have nothing to do with
(in the sense that
, and the “involved” elements, which have a lot to do with
(in the sense that
. The key point is that there is a significant “gap” between the non-involved and involved elements; there are no elements that are only “slightly involved”, in that
and
intersect a little but not a lot. It is this gap that will allow us to upgrade approximate structure to exact structure. Namely,
Proposition 4 The set
of involved elements is a finite group, and is equal to
.
Proof: It is clear that the identity element is involved, and that if
is involved then so is
(since
. Now suppose that
are both involved. Then
and
have cardinality greater than
and are both subsets of
, and so have non-empty intersection. In particular,
is non-empty, and so
is non-empty. By Proposition 3, this makes
involved. It is then clear that
is a group.
If , then
is non-empty, and so from Proposition 3
is involved. Conversely, if
is involved, then
. Thus we have
as claimed. In particular,
is finite.
Now we can quickly wrap up the proof of Theorem 2. By construction, for all
,which by double counting shows that
. As
, we see that
is contained in a right coset
of
; setting
, we conclude that
is contained in a left coset
of
.
is a conjugate of
, and so
. If
, then
and
both lie in
and have cardinality
, so must overlap; and so
. Thus
, and so
, and Theorem 2 follows.
Exercise 2 Show that the constant
in Theorem 2 cannot be replaced by any larger constant.
Exercise 3 Let
be a finite non-empty set such that
. Show that
. (Hint: If
, show that
for some
.)
Exercise 4 Let
be a finite non-empty set such that
. Show that there is a finite group
with
and a group element
such that
and
.
Below the fold, we give further examples of the pivot argument in other group-like situations, including Theorem 2 and also the “sum-product theorem” of Bourgain-Katz-Tao and Bourgain-Glibichuk-Konyagin.
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
-quasirandom if all non-trivial unitary representations
of
have dimension at least
. (Recall a representation is trivial if
is the identity for all
.)
Exercise 1 Let
be a finite group, and let
. A unitary representation
is said to be irreducible if
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 1 The 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 2 Let
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 3 Let
, 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 5 Let
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 6 Show that for any prime
, the unipotent matrices
for
ranging over
generate
as a group.
Exercise 7 Let
be a finite group, and let
. If
is generated by a collection
of
-quasirandom subgroups, show that
is itself
-quasirandom.
Exercise 8 Show 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 2 One 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 9 Let
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 3 By 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 4 Thanks 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
).
In the previous set of notes we introduced the notion of expansion in arbitrary -regular graphs. For the rest of the course, we will now focus attention primarily to a special type of
-regular graph, namely a Cayley graph.
Definition 1 (Cayley graph) Let
be a group, and let
be a finite subset of
. We assume that
is symmetric (thus
whenever
) and does not contain the identity
(this is to avoid loops). Then the (right-invariant) Cayley graph
is defined to be the graph with vertex set
and edge set
, thus each vertex
is connected to the
elements
for
, and so
is a
-regular graph.
Example 2 The graph in Exercise 3 of Notes 1 is the Cayley graph on
with generators
.
Remark 3 We call the above Cayley graphs right-invariant because every right translation
on
is a graph automorphism of
. This group of automorphisms acts transitively on the vertex set of the Cayley graph. One can thus view a Cayley graph as a homogeneous space of
, as it “looks the same” from every vertex. One could of course also consider left-invariant Cayley graphs, in which
is connected to
rather than
. However, the two such graphs are isomorphic using the inverse map
, so we may without loss of generality restrict our attention throughout to left Cayley graphs.
Remark 4 For minor technical reasons, it will be convenient later on to allow
to contain the identity and to come with multiplicity (i.e. it will be a multiset rather than a set). If one does so, of course, the resulting Cayley graph will now contain some loops and multiple edges.
For the purposes of building expander families, we would of course want the underlying groupto be finite. However, it will be convenient at various times to “lift” a finite Cayley graph up to an infinite one, and so we permit
to be infinite in our definition of a Cayley graph.
We will also sometimes consider a generalisation of a Cayley graph, known as a Schreier graph:
Definition 5 (Schreier graph) Let
be a finite group that acts (on the left) on a space
, thus there is a map
from
to
such that
and
for all
and
. Let
be a symmetric subset of
which acts freely on
in the sense that
for all
and
, and
for all distinct
and
. Then the Schreier graph
is defined to be the graph with vertex set
and edge set
.
Example 6 Every Cayley graph
is also a Schreier graph
, using the obvious left-action of
on itself. The
-regular graphs formed from
permutations
that were studied in the previous set of notes is also a Schreier graph provided that
for all distinct
, with the underlying group being the permutation group
(which acts on the vertex set
in the obvious manner), and
.
Exercise 7 If
is an even integer, show that every
-regular graph is a Schreier graph involving a set
of generators of cardinality
. (Hint: you may assume without proof Petersen’s 2-factor theorem, which asserts that every
-regular graph with
even can be decomposed into
edge-disjoint
-regular graphs. Now use the previous example.)
We return now to Cayley graphs. It is easy to characterise qualitative expansion properties of Cayley graphs:
Exercise 8 (Qualitative expansion) Let
be a finite Cayley graph.
- (i) Show that
is a one-sided
-expander for
for some
if and only if
generates
.
- (ii) Show that
is a two-sided
-expander for
for some
if and only if
generates
, and furthermore
intersects each index
subgroup of
.
We will however be interested in more quantitative expansion properties, in which the expansion constant is independent of the size of the Cayley graph, so that one can construct non-trivial expander families
of Cayley graphs.
One can analyse the expansion of Cayley graphs in a number of ways. For instance, by taking the edge expansion viewpoint, one can study Cayley graphs combinatorially, using the product set operation
of subsets of .
Exercise 9 (Combinatorial description of expansion) Let
be a family of finite
-regular Cayley graphs. Show that
is a one-sided expander family if and only if there is a constant
independent of
such that
for all sufficiently large
and all subsets
of
with
.
One can also give a combinatorial description of two-sided expansion, but it is more complicated and we will not use it here.
Exercise 10 (Abelian groups do not expand) Let
be a family of finite
-regular Cayley graphs, with the
all abelian, and the
generating
. Show that
are a one-sided expander family if and only if the Cayley graphs have bounded cardinality (i.e.
). (Hint: assume for contradiction that
is a one-sided expander family with
, and show by two different arguments that
grows at least exponentially in
and also at most polynomially in
, giving the desired contradiction.)
The left-invariant nature of Cayley graphs also suggests that such graphs can be profitably analysed using some sort of Fourier analysis; as the underlying symmetry group is not necessarily abelian, one should use the Fourier analysis of non-abelian groups, which is better known as (unitary) representation theory. The Fourier-analytic nature of Cayley graphs can be highlighted by recalling the operation of convolution of two functions , defined by the formula
This convolution operation is bilinear and associative (at least when one imposes a suitable decay condition on the functions, such as compact support), but is not commutative unless is abelian. (If one is more algebraically minded, one can also identify
(when
is finite, at least) with the group algebra
, in which case convolution is simply the multiplication operation in this algebra.) The adjacency operator
on a Cayley graph
can then be viewed as a convolution
where is the probability density
where is the Kronecker delta function on
. Using the spectral definition of expansion, we thus see that
is a one-sided expander if and only if
whenever is orthogonal to the constant function
, and is a two-sided expander if
whenever is orthogonal to the constant function
.
We remark that the above spectral definition of expansion can be easily extended to symmetric sets which contain the identity or have multiplicity (i.e. are multisets). (We retain symmetry, though, in order to keep the operation of convolution by
self-adjoint.) In particular, one can say (with some slight abuse of notation) that a set of elements
of
(possibly with repetition, and possibly with some elements equalling the identity) generates a one-sided or two-sided
-expander if the associated symmetric probability density
obeys either (2) or (3).
We saw in the last set of notes that expansion can be characterised in terms of random walks. One can of course specialise this characterisation to the Cayley graph case:
Exercise 11 (Random walk description of expansion) Let
be a family of finite
-regular Cayley graphs, and let
be the associated probability density functions. Let
be a constant.
- Show that the
are a two-sided expander family if and only if there exists a
such that for all sufficiently large
, one has
for some
, where
denotes the convolution of
copies of
.
- Show that the
are a one-sided expander family if and only if there exists a
such that for all sufficiently large
, one has
for some
.
In this set of notes, we will connect expansion of Cayley graphs to an important property of certain infinite groups, known as Kazhdan’s property (T) (or property (T) for short). In 1973, Margulis exploited this property to create the first known explicit and deterministic examples of expanding Cayley graphs. As it turns out, property (T) is somewhat overpowered for this purpose; in particular, we now know that there are many families of Cayley graphs for which the associated infinite group does not obey property (T) (or weaker variants of this property, such as property ). In later notes we will therefore turn to other methods of creating Cayley graphs that do not rely on property (T). Nevertheless, property (T) is of substantial intrinsic interest, and also has many connections to other parts of mathematics than the theory of expander graphs, so it is worth spending some time to discuss it here.
The material here is based in part on this recent text on property (T) by Bekka, de la Harpe, and Valette (available online here).
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In the Winter quarter (starting on January 9), I will be teaching a graduate course on expansion in groups of Lie type. This course will focus on constructions of expanding Cayley graphs on finite groups of Lie type (such as the special linear groups , or their simple quotients
, but also including more exotic “twisted” groups of Lie type, such as the Steinberg or Suzuki-Ree groups), including the “classical” constructions of Margulis and of Selberg, but also the more recent constructions of Bourgain-Gamburd and later authors (including some very recent work of Ben Green, Emmanuel Breuillard, Rob Guralnick, and myself which is nearing completion and which I plan to post about shortly). As usual, I plan to start posting lecture notes on this blog before the course begins.
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