(Linear) Fourier analysis can be viewed as a tool to study an arbitrary function on (say) the integers
, by looking at how such a function correlates with linear phases such as
, where
is the fundamental character, and
is a frequency. These correlations control a number of expressions relating to
, such as the expected behaviour of
on arithmetic progressions
of length three.
In this course we will be studying higher-order correlations, such as the correlation of with quadratic phases such as
, as these will control the expected behaviour of
on more complex patterns, such as arithmetic progressions
of length four. In order to do this, we must first understand the behaviour of exponential sums such as
Such sums are closely related to the distribution of expressions such as in the unit circle
, as
varies from
to
. More generally, one is interested in the distribution of polynomials
of one or more variables taking values in a torus
; for instance, one might be interested in the distribution of the quadruplet
as
both vary from
to
. Roughly speaking, once we understand these types of distributions, then the general machinery of quadratic Fourier analysis will then allow us to understand the distribution of the quadruplet
for more general classes of functions
; this can lead for instance to an understanding of the distribution of arithmetic progressions of length
in the primes, if
is somehow related to the primes.
More generally, to find arithmetic progressions such as in a set
, it would suffice to understand the equidistribution of the quadruplet
in
as
and
vary. This is the starting point for the fundamental connection between combinatorics (and more specifically, the task of finding patterns inside sets) and dynamics (and more specifically, the theory of equidistribution and recurrence in measure-preserving dynamical systems, which is a subfield of ergodic theory). This connection was explored in one of my previous classes; it will also be important in this course (particularly as a source of motivation), but the primary focus will be on finitary, and Fourier-based, methods.
The theory of equidistribution of polynomial orbits was developed in the linear case by Dirichlet and Kronecker, and in the polynomial case by Weyl. There are two regimes of interest; the (qualitative) asymptotic regime in which the scale parameter is sent to infinity, and the (quantitative) single-scale regime in which
is kept fixed (but large). Traditionally, it is the asymptotic regime which is studied, which connects the subject to other asymptotic fields of mathematics, such as dynamical systems and ergodic theory. However, for many applications (such as the study of the primes), it is the single-scale regime which is of greater importance. The two regimes are not directly equivalent, but are closely related: the single-scale theory can be usually used to derive analogous results in the asymptotic regime, and conversely the arguments in the asymptotic regime can serve as a simplified model to show the way to proceed in the single-scale regime. The analogy between the two can be made tighter by introducing the (qualitative) ultralimit regime, which is formally equivalent to the single-scale regime (except for the fact that explicitly quantitative bounds are abandoned in the ultralimit), but resembles the asymptotic regime quite closely.
We will view the equidistribution theory of polynomial orbits as a special case of Ratner’s theorem, which we will study in more generality later in this course.
For the finitary portion of the course, we will be using asymptotic notation: ,
, or
denotes the bound
for some absolute constant
, and if we need
to depend on additional parameters then we will indicate this by subscripts, e.g.
means that
for some
depending only on
. In the ultralimit theory we will use an analogue of asymptotic notation, which we will review later in these notes.
— 1. Asymptotic equidistribution theory —
Before we look at the single-scale equidistribution theory (both in its finitary form, and its ultralimit form), we will first study the slightly simpler, and much more classical, asymptotic equidistribution theory.
Suppose we have a sequence of points in a compact metric space
. For any finite
, we can define the probability measure
which is the average of the Dirac point masses on each of the points , where we use
as shorthand for
(with
). Asymptotic equidistribution theory is concerned with the limiting behaviour of these probability measures
in the limit
, for various sequences
of interest. In particular, we say that the sequence
is asymptotically equidistributed on
with respect to a reference Borel probability measure
on
if the
converge in the vague topology to
, or in other words that
for all continuous scalar-valued functions . Note (from the Riesz representation theorem) that any sequence is asymptotically equidistributed with respect to at most one Borel probability measure
.
It is also useful to have a slightly stronger notion of equidistribution: we say that a sequence is totally asymptotically equidistributed if it is asymptotically equidistributed on every infinite arithmetic progression, i.e. that the sequence
is asymptotically equidistributed for all integers
and
.
A doubly infinite sequence , indexed by the integers rather than the natural numbers, is said to be asymptotically equidistributed relative to
if both halves of the sequence
and
are asymptotically equidistributed relative to
. (This omits
entirely, but it is easy to see that any individual element of the sequence has no impact on the asymptotic equidistribution.) Similarly, one can define the notion of a doubly infinite sequence being totally asymptotically equidistributed relative to
.
Example 1 If
, and
whenever
for some natural number
and
otherwise, show that the sequence
is not asymptotically equidistributed with respect to any measure. Thus we see that asymptotic equidistribution requires all scales to behave “the same” in the limit.
Exercise 1 If
is a sequence into a compact metric space
, and
is a probability measure on
, show that
is asymptotically equidistributed with respect to
if and only if one has
for all open sets
in
whose boundary
has measure zero. (Hint: for the “only if” part, use Urysohn’s lemma. For the “if” part, reduce (1) to functions
taking values between
and
, and observe that almost all of the level sets
have a boundary of measure zero.) What happens if the requirement that
have measure zero is omitted?
Exercise 2 Let
be a sequence in a compact metric space
which is equidistributed relative to some probability measure
. Show that for any open set
in
with
, the set
is infinite, and furthermore has positive lower density in the sense that
In particular, if the support of
is equal to
, show that the set
is dense in
.
Exercise 3 Let
be a sequence into a compact metric space
which is equidistributed relative to some probability measure
. Let
be a compactly supported, piecewise continuous function with only finitely many pieces. Show that for any
one has
and for any open
whose boundary has measure zero, one has
In this set of notes, will be a torus (i.e. a compact connected abelian Lie group), which from the theory of Lie groups is isomorphic to the standard torus
, where
is the dimension of the torus. This torus is then equipped with Haar measure, which is the unique Borel probability measure on the torus which is translation-invariant. One can identify the standard torus
with the standard fundamental domain
, in which case the Haar measure is equated with the usual Lebesgue measure. We shall call a sequence
in
(asymptotically) equidistributed if it is (asymptotically) equidistributed with respect to Haar measure.
We have a simple criterion for when a sequence is asymptotically equidistributed, that reduces the problem to that of estimating exponential sums:
Proposition 1 (Weyl equidistribution criterion) Let
. Then
is asymptotically equidistributed if and only if
for all
, where
. Here we use the dot product
which maps
to
.
Proof: The “only if” part is immediate from (1). For the “if” part, we see from (2) that (1) holds whenever is a plane wave
for some
(checking the
case separately), and thus by linearity whenever
is a trigonometric polynomial. But by Fourier analysis (or from the Stone-Weierstrass theorem), the trigonometric polynomials are dense in
in the uniform topology. The claim now follows from a standard limiting argument.
As one consequence of this proposition, one can reduce multidimensional equidistribution to single-dimensional equidistribution:
Corollary 2 Let
. Then
is asymptotically equidistributed in
if and only if, for each
, the sequence
is asymptotically equidistributed in
.
Exercise 4 Show that a sequence
is totally asymptotically equidistributed if and only if one has
This quickly gives a test for equidistribution for linear sequences, sometimes known as the equidistribution theorem:
Exercise 5 Let
. By using the geometric series formula, show that the following are equivalent:
- The sequence
is asymptotically equidistributed on
.
- The sequence
is totally asymptotically equidistributed on
.
- The sequence
is totally asymptotically equidistributed on
.
is irrational, in the sense that
for any non-zero
.
Remark 1 One can view Exercise 5 as an assertion that a linear sequence
will equidistribute itself unless there is an “obvious” algebraic obstruction to it doing so, such as
being constant for some non-zero
. This theme of algebraic obstructions being the “only” obstructions to uniform distribution will be present throughout the course.
Exercise 5 shows that linear sequences with irrational shift are equidistributed. At the other extreme, if
is rational in the sense that
for some positive integer
, then the sequence
is clearly periodic of period
, and definitely not equidistributed.
In the one-dimensional case , these are the only two possibilities. But in higher dimensions, one can have a mixture of the two extremes, that exhibits irrational behaviour in some directions and periodic behaviour in others. Consider for instance the two-dimensional sequence
. The first coordinate is totally asymptotically equidistributed in
, while the second coordinate is periodic; the shift
is neither irrational nor rational, but is a mixture of both. As such, we see that the two-dimensional sequence is equidistributed with respect to Haar measure on the group
.
This phenomenon generalises:
Proposition 3 (Ratner’s theorem for abelian linear sequences) Let
be a torus, and let
for some
. Then there exists a decomposition
, where
is totally asymptotically equidistributed on
in a subtorus
of
(with
, of course), and
is periodic (or equivalently, that
is rational).
Proof: We induct on the dimension of the torus
. The claim is vacuous for
, so suppose that
and that the claim has already been proven for tori of smaller dimension. Without loss of generality we may identify
with
.
If is irrational, then we are done by Exercise 5, so we may assume that
is not irrational; thus
for some non-zero
. We then write
, where
is a positive integer and
is irreducible (i.e.
is not a proper multiple of any other element of
); thus
is rational. We may thus write
, where
is rational, and
. Thus, we can split
, where
and
. Clearly
is periodic, while
takes values in the subtorus
of
. The claim now follows by applying the induction hypothesis to
(and noting that the sum of two periodic sequences is again periodic).
As a corollary of the above proposition, we see that any linear sequence in a torus
is equidistributed in some union of finite cosets of a subtorus
. It is easy to see that this torus
is uniquely determined by
, although there is a slight ambiguity in the decomposition
because one can add or subtract a periodic linear sequence taking values in
from
and add it to
(or vice versa).
Having discussed the linear case, we now consider the more general situation of polynomial sequences in tori. To get from the linear case to the polynomial case, the fundamental tool is
Lemma 4 (van der Corput inequality) Let
be a sequence of complex numbers of magnitude at most
. Then for every
, we have
Proof: For each , we have
and hence on averaging
Applying Cauchy-Schwarz, we conclude
We expand out the left-hand side as
The diagonal contribution is
. By symmetry, the off-diagonal contribution can be dominated by the contribution when
. Making the change of variables
,
(accepting a further error of
), we obtain the claim.
Corollary 5 (van der Corput lemma) Let
be such that the derivative sequence
is asymptotically equidistributed on
for all positive integers
. Then
is asymptotically equidistributed on
. Similarly with
replaced by
.
Proof: We just prove the claim for , as the claim for
is analogous (and can in any case be deduced from the
case.)
By Proposition 1, we need to show that for each non-zero , the exponential sum
goes to zero as . Fix an
. By Lemma 4, this expression is bounded by
On the other hand, for each fixed positive integer , we have from hypothesis and Proposition 1 that
goes to zero as
. Taking limit superior as
, we conclude that
Since is arbitrary, the claim follows.
Remark 2 There is another famous lemma by van der Corput concerning oscillatory integrals, but it is not directly related to the material discussed here.
Corollary 5 has the following immediate corollary:
Corollary 6 (Weyl equidistribution theorem for polynomials) Let
be an integer, and let
be a polynomial of degree
with
. If
is irrational, then
is asymptotically equidistributed on
.
Proof: We induct on . For
this follows from Exercise 5. Now suppose that
, and that the claim has already been proven for smaller values of
. For any positive integer
, we observe that
is a polynomial of degree
in
with leading coefficient
. As
is irrational,
is irrational also, and so by the induction hypothesis,
is asymptotically equidistributed. The claim now follows from Corollary 5.
Exercise 6 Let
be a polynomial of degree
in
. Show that the following are equivalent:
is asymptotically equidistributed on
.
is totally asymptotically equidistributed on
.
is totally asymptotically equidistributed on
.
- There does not exist a non-zero
such that
.
(Hint: it is convenient to first use Corollary 2 to reduce to the one-dimensional case.)
This gives a polynomial variant of Ratner’s theorem:
Exercise 7 (Ratner’s theorem for abelian polynomial sequences) Let
be a torus, and let
be a polynomial map from
to
of some degree
. Show that there exists a decomposition
, where
are polynomials of degree
,
is totally asymptotically equidistributed in a subtorus
of
on
, and
is periodic (or equivalently, that all non-constant coefficients of
are rational).
In particular, we see that polynomial sequences in a torus are equidistributed with respect to a finite combination of Haar measures of cosets of a subtorus. Note that this finite combination can have multiplicity; for instance, when considering the polynomial map , it is not hard to see that this map is equidistributed with respect to
times the Haar probability measure on
, plus
times the Haar probability measure on
.
Exercise 7 gives a satisfactory description of the asymptotic equidistribution of arbitrary polynomial sequences in tori. We give just one example of how such a description can be useful:
Exercise 8 (Recurrence) Let
be a torus, let
be a polynomial map from
to
, and let
be an integer. Show that there exists a sequence
of positive integers going to infinity such that
.
We discussed recurrence for one-dimensional sequences . It is also of interest to establish an analogous theory for multi-dimensional sequences, as follows.
Definition 7 A multidimensional sequence
is asymptotically equidistributed relative to a probability measure
if, for every continuous, compactly supported function
and every function
, one has
as
. The sequence is totally asymptotically equidistributed relative to
if the sequence
is asymptotically equidistributed relative to
for all positive integers
and all
.
Exercise 9 Show that this definition of equidistribution on
coincides with the preceding definition of equidistribution on
in the one-dimensional case
.
Exercise 10 (Multidimensional Weyl criterion) Let
be a multidimensional sequence. Show that
is asymptotically equidistributed if and only if
for all
and all rectangular boxes
in
. Then show that
is totally asymptotically equidistributed if and only if
Exercise 11 Let
, and let
be the linear sequence
. Show that the following are equivalent:
- The sequence
is asymptotically equidistributed on
.
- The sequence
is totally asymptotically equidistributed on
.
- We have
for any non-zero
.
Exercise 12 (Multidimensional van der Corput lemma) Let
be such that the sequence
is asymptotically equidistributed on
for all
outside of a hyperplane in
. Show that
is asymptotically equidistributed on
.
Exercise 13 Let
be a polynomial map from
to
of degree
, where
are coefficients. Show that the following are equivalent:
is asymptotically equidistributed on
.
is totally asymptotically equidistributed on
.
- There does not exist a non-zero
such that
for all
.
Exercise 14 (Ratner’s theorem for abelian multidimensional polynomial sequences) Let
be a torus, and let
be a polynomial map from
to
of some degree
. Show that there exists a decomposition
, where
are polynomials of degree
,
is totally asymptotically equidistributed in a subtorus
of
on
, and
is periodic with respect to some finite index sublattice of
(or equivalently, that all non-constant coefficients of
are rational).
We give just one application of this multidimensional theory, that gives a hint as to why the theory of equidistribution of polynomials may be relevant:
Exercise 15 (Szemerédi’s theorem for polynomial Bohr sets) Let
be a torus, let
be a polynomial map from
to
, and let
be an open ball of
. Let
be the set
. Show that if the set
has positive upper density in the sense that
, then
contains arbitrarily long arithmetic progressions. (Hint: consider the polynomial map from
to
that maps
to
. One can also use the one-dimensional theory by freezing
and only looking at the equidistribution in
.)
— 2. Single-scale equidistribution theory —
We now turn from the asymptotic equidistribution theory to the equidistribution theory at a single scale . Thus, instead of analysing the qualitative distribution of infinite sequence
, we consider instead the quantitative distribution of a finite sequence
, where
is a (large) natural number and
. To make everything quantitative, we will replace the notion of a continuous function by that of a Lipschitz function. Recall that the (inhomogeneous) Lipschitz norm
of a function
on a metric space
is defined by the formula
We also define the homogeneous Lipschitz seminorm
Definition 8 Let
be a compact metric space, let
, let
be a probability measure on
. A finite sequence
is said to be
-equidistributed relative to
if one has
We say that the sequence
is totally
-equidistributed relative to
if one has
for all Lipschitz functions
and all arithmetic progressions
in
of length at least
.
In these notes, we will only apply this concept to the torus with the Haar measure
and the metric inherited from the Euclidean metric. However, in subsequent notes we will also consider equidistribution in other spaces, most notably on nilmanifolds.
Exercise 16 Let
be a sequence in a metric space
, and let
be a probability measure on
. Show that the sequence
is asymptotically equidistributed relative to
if and only if, for every
,
is
-equidistributed relative to
whenever
is sufficiently large depending on
, or equivalently if
is
-equidistributed relative to
for all
, where
as
. (Hint: You will need the Arzelá-Ascoli theorem.)
Similarly, show that
is totally asymptotically equidistributed relative to
if and only if, for every
,
is totally
-equidistributed relative to
whenever
is sufficiently large depending on
, or equivalently if
is totally
-equidistributed relative to
for all
, where
as
.
Remark 3 More succinctly, (total) asymptotic equidistribution of
is equivalent to (total)
-equidistribution of
as
, where
denotes a quantity that goes to zero as
. (Thus we see that asymptotic notation such as
can efficiently conceal a surprisingly large number of quantifiers.)
Exercise 17 Let
be a large integer, and let
be a sequence in the standard torus
with Haar measure. Show that whenever
is a positive multiple of
, then the sequence
is
-equidistributed. What happens if
is not a multiple of
?
If furthermore
, show that
is
-equidistributed. Why is a condition such as
necessary?
Note that the above exercise does not specify the exact relationship between and
when one is given an asymptotically equidistributed sequence
; this relationship is the additional piece of information provided by single-scale equidistribution that is not present in asymptotic equidistribution.
It turns out that much of the asymptotic equidistribution theory has a counterpart for single-scale equidistribution. We begin with the Weyl criterion.
Proposition 9 (Single-scale Weyl equidistribution criterion) Let
be a sequence in
, and let
.
- If
is
-equidistributed, and
has magnitude
, then one has
if
is a small enough absolute constant.
- Conversely, if
is not
-equidistributed, then there exists
with magnitude
, such that
for some
depending on
.
Proof: The first claim is immediate as the function has mean zero and Lipschitz constant
, so we turn to the second claim. By hypothesis, (6) fails for some Lipschitz
. We may subtract off the mean and assume that
; we can then normalise the Lipschitz norm to be one; thus we now have
We introduce a summation parameter , and consider the Fejér partial Fourier series
where are the Fourier coefficients
and is the Fourier multiplier
Standard Fourier analysis shows that we have the convolution representation
where is the Fejér kernel
Using the kernel bounds
and
where is the distance from
to the nearest integer, and the Lipschitz nature of
, we see that
Thus, if we choose to be a sufficiently small multiple of
(depending on
), one has
and thus by the pigeonhole principle (and the trivial bound and
) we have
for some non-zero of magnitude
, and the claim follows.
There is an analogue for total equidistribution:
Exercise 18 Let
be a sequence in
, and let
.
- If
is totally
-equidistributed,
has magnitude
, and
is a rational of height at most
, then one has
if
is a small enough constant depending only on
.
- Conversely, if
is not totally
-equidistributed, then there exists
with magnitude
, and a rational
of height
, such that
for some
depending on
.
This gives a version of Exercise 5:
Exercise 19 Let
, let
, and let
. Suppose that the linear sequence
is not totally
-equidistributed. Show that there exists a non-zero
with
such that
.
Next, we give an analogue of Corollary 5:
Exercise 20 (Single-scale van der Corput lemma) Let
be a sequence which is not totally
-equidistributed for some
. Let
for some sufficiently large
depending only on
. Then there exists at least
integers
such that the sequence
is not totally
-equidistributed (where we extend
by zero outside of
). (Hint: apply Lemma 4.)
Just as in the asymptotic setting, we can use the van der Corput lemma to extend the linear equidistribution theory to polynomial sequences. To get satisfactory results, though, we will need an additional input, namely the following classical lemma:
Lemma 10 (Vinogradov lemma) Let
,
,
, and
. Suppose that
for at least
values of
. Then there exists a positive integer
such that
.
The key point here is that one starts with many multiples of being somewhat close (
) to an integer, but concludes that there is a single multiple of
which is very close (
, ignoring factors of
) to an integer.
Proof: By the pigeonhole principle, we can find two distinct integers with
such that
. Setting
, we thus have
. We may assume that
since we are done otherwise. Since
, we have
(say).
Now partition into
arithmetic progressions
for some
. By the pigeonhole principle, there must exist an
for which the set
has cardinality at least . On the other hand, since
, we see that this set consists of intervals of length at most
, punctuated by gaps of length at least
(say). Since the gaps are at least
times as large as the intervals, we see that if two or more these intervals appear in the set, then the cardinality of the set is at most
, a contradiction. Thus at most one interval appears in the set, which implies that
, and the claim follows.
Remark 4 The numerical constants can of course be improved, but this is not our focus here.
Exercise 21 Let
be a polynomial sequence
, let
, and let
. Suppose that the polynomial sequence
is not totally
-equidistributed on
. Show that there exists a non-zero
with
such that
. (Hint: Induct on
starting with Exercise 19 for the base case, and then using Exercise 20 and Lemma 10 to continue the induction.)
Note the denominator; the higher-degree coefficients of a polynomial need to be very rational in order not to cause equidistribution.
The above exercise only controls the top degree coefficient, but we can in fact control all coefficients this way:
Lemma 11 With the hypotheses of Exercise 21, we can in fact find a non-zero
with
such that
for all
.
Proof: We shall just establish the one-dimensional case , as the general dimensional case then follows from Exercise 18.
The case follows from Exercise 19, so assume inductively that
and that the claim has already been proven for smaller values of
. We allow all implied constants to depend on
. From Exercise 21, we already can find a positive
with
such that
. We now partition
into arithmetic progressions of spacing
and length
for some sufficiently large
; then by the pigeonhole principle, we see that
fails to be totally
-equidistributed on one of these progressions. But on one such progression (which can be identified with
) the degree
component of
is essentially constant (up to errors much smaller than
) if
is large enough; if one then applies the induction hypothesis to the remaining portion of
on this progression, we can obtain the claim.
This gives us the following analogue of Exercise 7. We say that a subtorus of some dimension
of a standard torus
has complexity at most
if there exists an invertible linear transformation
with integer coefficients (which can thus be viewed as a homeomorphism of
that maps
to the standard torus
, and such that all coefficients have magnitude at most
.
Exercise 22 Show that every subtorus (i.e. compact connected Lie subgroup)
of
has finite complexity. (Hint: Let
be the Lie algebra of
, then identify
with a subspace of
and
with
. Show that
is a full rank sublattice of
, and is thus generated by
independent generators.)
Proposition 12 (Single-scale Ratner’s theorem for abelian polynomial sequences) Let
be a polynomial map from
to
of some degree
, and let
be an increasing function. Then there exists an integer
and a decomposition
into polynomials of degree
, where
- (
is smooth) The
coefficient
of
has size
. In particular, on the interval
,
is Lipschitz with homogeneous norm
.
- (
is equidistributed) There exists a subtorus
of
of complexity at most
and some dimension
, such that
takes values in
and is totally
-equidistributed on
in this torus (after identifying this torus with
using an invertible linear transformation of complexity at most
).
- (
is rational) The coefficients
of
are such that
for some
and all
. In particular,
and
is periodic with period
.
If furthermore
is of polynomial growth, and more precisely
for some
, then one can take
.
Example 2 Consider the linear flow
in
on
. This flow can be decomposed into a smooth flow
with a homogeneous Lipschitz norm of
, an equidistributed flow
which will be
-equidistributed on the subtorus
for a reasonably small
(in fact one can take
as small as
for some small absolute constant
), and a rational flow
, which is periodic with period
. This example illustrates how all three components of this decomposition arise naturally in the single-scale case.
Remark 5 Comparing this result with the asymptotically equidistributed analogue in Example 7, we notice several differences. Firstly, we now have the smooth component
, which did not previously make an appearance (except implicitly, as the constant term in
). Secondly, the equidistribution of the component
is not infinite, but is the next best thing, namely it is given by an arbitrary function
of the quantity
, which controls the other components of the decomposition.
Proof: The case is trivial, so suppose inductively that
, and that the claim has already been proven for lower degrees. Then for fixed degree, the case
is vacuously true, so we make a further inductive assumption
and the claim has already been proven for smaller dimensions (keeping
fixed).
If is already totally
-equidistributed then we are done (setting
and
and
), so suppose that this is not the case. Applying Exercise 21, we conclude that there is some non-zero
with
such that
for all . We split
where
is irreducible and
is a positive integer. We can therefore split
where
,
for some positive integer
, and
. This then gives a decomposition
, with
taking values in the subtorus
, which can be identified with
after an invertible linear transformation with integer coefficients of size
. If one applies the induction hypothesis to
(with
replaced by a suitably larger function
) one then obtains the claim.
The final claim about polynomial bounds can be verified by a closer inspection of the argument (noting that all intermediate steps are polynomially quantitative, and that the length of the induction is bounded by ).
Remark 6 It is instructive to see how this smooth-equidistributed-rational decomposition evolves as
increases. Roughly speaking, the torus
that the
component is equidistributed on is stable at most scales, but there will be a finite number of times in which a “growth spurt” occurs and
jumps up in dimension. For instance, consider the linear flow
on the two-dimensional torus. At scales
(and with
fixed, and
assumed to be sufficiently large depending on
),
consists entirely of the smooth component. But as
increases past
, the first component of
no longer qualifies as smooth, and becomes equidistributed instead; thus in the range
, we have
and
(with
remaining trivial), with the torus
increasing from the trivial torus
to
. A second transition occurs when
exceeds
, at which point
encompasses all of
. Evolving things in a somewhat different direction, if one then increases
so that
is much larger than
, then
will now entirely consist of a rational component
. These sorts of dynamics are not directly seen if one only looks at the asymptotic theory, which roughly speaking is concerned with the limit after taking
, and then taking a second limit by making the growth function
go to infinity.
There is a multidimensional version of Proposition 12, but we will not describe it here; see this paper of Ben Green and myself for a statement (and also see the next section for the ultralimit counterpart of this statement).
Remark 7 These single-scale abelian Ratner theorems are a special case of a more general single-scale nilpotent Ratner theorem, which will play an important role in later aspects of the theory, and which was the main result of the aforementioned paper of Ben Green and myself.
As an example of this theorem in action, we give a single-scale strengthening of Exercise 8 (and Exercise 15):
Exercise 23 (Recurrence) Let
be a polynomial map from
to
of degree
, and let
be an integer. Show that for every
and
, and every integer
, we have
Exercise 24 (Multiple recurrence) With the notation of Exercise 23, establish that
for any
.
Exercise 25 (Syndeticity) A set of integers is syndetic if it has bounded gaps (or equivalently, if a finite number of translates of this set can cover all of
). Let
be a polynomial and let
. Show that the set
is syndetic. (Hint: first reduce to the case when
is (totally) asymptotically equidistributed. Then, if
is large enough, show (by inspection of the proof of Exercise 21) that the translates
are
-equidistributed on
uniformly for all
, for any fixed
. Note how the asymptotic theory and the single-scale theory need to work together to obtain this result.)
— 3. Ultralimit equidistribution theory —
The single-scale theory was somewhat more complicated than the asymptotic theory, in part because one had to juggle parameters such as , and (for the Ratner-type theorems)
as well. However, one can clean up this theory somewhat (especially if one does not wish to quantify the dependence of bounds on the equidistribution parameter
) by using an ultralimit, which causes the
and
parameters to disappear, at the cost of converting the finitary theory to an infinitary one. Ultralimit analysis was discussed in this previous blog post; we give a quick review here.
We first fix a non-principal ultrafilter . A property
pertaining to a natural number
is said to hold for all
sufficiently close to
if the set of
for which
holds lies in the ultrafilter
. Two sequences
of objects are equivalent if one has
for all
sufficiently close to
, and we define the ultralimit
to be the equivalence class of all sequences equivalent to
, with the convention that
is identified with its own ultralimit
. Given any sequence
of sets, the ultraproduct
is the space of all ultralimits
, where
for all
sufficiently close to
. The ultraproduct
of a single set
is the ultrapower of
and is denoted
.
Ultralimits of real numbers (i.e. elements of ) will be called limit real numbers; similarly one defines limit natural numbers, limit complex numbers, etc. Ordinary numbers will be called standard numbers to distinguish them from limit numbers, thus for instance a limit real number is an ultralimit of standard real numbers. All the usual arithmetic operations and relations on standard numbers are inherited by their limit analogues; for instance, a limit real number
is larger than another
if one has
for all
sufficiently close to
. The axioms of a non-principal ultrafilter ensure that these relations and operations on limit numbers obey the same axioms as their standard counterparts (the formalisation of this principle is Los’s theorem).
Ultraproducts of sets will be called limit sets; they are roughly analogous to “measurable sets” in measure theory. Ultraproducts of finite sets will be called limit finite sets. Thus, for instance, if is a limit natural number, then
is a limit finite set, and can be identified with the set of limit natural numbers between
and
.
Given a sequence of functions , we can form the ultralimit
by the formula
one easily verifies that this is a well-defined function between the two ultraproducts. We refer to ultralimits of functions as limit functions; they are roughly analogous to “measurable functions” in measurable theory. We identify every standard function with its ultralimit
, which extends the original function
.
Given two limit numbers , we write
,
, or
if we have
for some standard
. We also write
if we have
for every standard
; thus for any limit numbers
with
, exactly one of
and
is true. A limit real is said to be bounded if it is of the form
, and infinitesimal if it is of the form
; similarly for limit complex numbers. Note that the bounded limit reals are a subring of the limit reals, and the infinitesimal limit reals are an ideal of the bounded limit reals.
Exercise 26 Show that every bounded limit real number
has a unique decomposition
, where
is a standard real (called the standard part of
) and
is infinitesimal.
We now give the analogue of single-scale equidistribution in the ultralimit setting.
Definition 13 (Ultralimit equidistribution) Let
be a standard compact metric space, let
be an unbounded limit natural number, and let
be a limit function. We say that
is equidistributed with respect to a (standard) Borel probability measure
on
if one has
for all standard continuous functions
. Here, we define the expectation of a limit function in the obvious limit manner, thus
if
and
.
We say that
is totally equidistributed relative to
if the sequence
is equidistributed on
for every standard
and
(extending
arbitrarily outside
if necessary).
Remark 8 One could just as easily replace the space of continuous functions by any dense subclass in the uniform topology, such as the space of Lipschitz functions.
The ultralimit notion of equidistribution is closely related to that of both asymptotic equidistribution and single-scale equidistribution, as the following exercises indicate:
Exercise 27 (Asymptotic equidistribution vs. ultralimit equidistribution) Let
be a sequence into a standard compact metric space (which can then be extended from a map from
to
as usual), let
be a Borel probability measure on
. Show that
is asymptotically equidistributed on
with respect to
if and only if
is equidistributed on
for every unbounded natural number
and every choice of non-principal ultrafilter
.
Exercise 28 (Single-scale equidistribution vs. ultralimit equidistribution) For every
, let
be a natural number that goes to infinity as
, let
be a map to a standard compact metric space. Let
be a Borel probability measure on
. Write
and
for the ultralimits. Show that
is equidistributed with respect to
if and only if, for every standard
,
is
-equidistributed with respect to
for all
sufficiently close to
.
In view of these correspondences, it is thus not surprising that one has ultralimit analogues of the asymptotic and single-scale theory. These analogues tend to be logically equivalent to the single-scale counterparts (once one concedes all quantitative bounds), but are formally similar (though not identical) to the asymptotic counterparts, thus providing a bridge between the two theories, which we can summarise by the following three statements:
- Asymptotic theory is analogous to ultralimit theory (in particular, the statements and proofs are formally similar);
- ultralimit theory is logically equivalent to qualitative finitary theory; and
- quantitative finitary theory is a strengthening of qualitative finitary theory.
For instance, here is the ultralimit version of the Weyl criterion:
Exercise 29 (Ultralimit Weyl equidistribution criterion) Let
be a limit function for some unbounded
and standard
. Then
is equidistributed if and only if
for all standard
. {Hint: mimic the proof of Proposition 1.}
Exercise 30 Use Exercise 29 to recover a weak version of Proposition 9, in which the quantities
,
are replaced by (ineffective) functions of
that decay to zero as
. Conversely, use this weak version to recover Exercise 29. (Hint: see this previous blog post for further examples of this type of logical equivalence between ultralimit statements and qualitative finitary statements.)
Exercise 31 With the notation of Exercise 29, show that
is totally equidistributed if and only if
for all standard
and standard rational
.
Exercise 32 With the notation of Exercise 29, show that
is equidistributed in
on
if and only if
is equidistributed in
on
for every non-zero standard
.
Now we establish the ultralimit version of the linear equidistribution criterion:
Exercise 33 Let
, and let
be an unbounded integer. Show that the following are equivalent:
- The sequence
is equidistributed on
.
- The sequence
is totally equidistributed on
.
is irrational to scale
, in the sense that
for any non-zero standard
.
Note that in the ultralimit setting, assertions such as make perfectly rigorous sense (it means that
for every standard
), but when using finitary asymptotic big-O notation
Next, we establish the analogue of the van der Corput lemma:
Exercise 34 (van der Corput lemma, ultralimit version) Let
be an unbounded integer, and let
be a limit sequence. Let
be unbounded, and suppose that the derivative sequence
is equidistributed on
for
values of
(extending
by arbitrarily outside of
). Show that
is equidistributed on
. Similarly “equidistributed” replaced by “totally equidistributed”.
Here is the analogue of the Vinogradov lemma:
Exercise 35 (Vinogradov lemma, ultralimit version) Let
,
be unbounded, and
be infinitesimal. Suppose that
for
values of
. Show that there exists a positive standard integer
such that
.
These two lemmas allow us to establish the ultralimit polynomial equidistribution theory:
Exercise 36 Let
be a polynomial sequence
with
standard, and
. Let
be an unbounded natural number. Suppose that
is not totally equidistributed on
. Show that there exists a non-zero standard
with
.
Exercise 37 With the hypotheses of Exercise 36, show in fact that there exists a non-zero standard
such that
for all
.
Exercise 38 (Ultralimit Ratner’s theorem for abelian polynomial sequences) Let
be a polynomial map from
to
of some standard degree
. Let
be an unbounded natural number. Then there exists a decomposition
into polynomials of degree
, where
- (
is smooth) The
coefficient
of
has size
. In particular, on the interval
,
is Lipschitz with homogeneous norm
.
- (
is equidistributed) There exists a standard subtorus
of
, such that
takes values in
and is totally equidistributed on
in this torus.
- (
is rational) The coefficients
of
are standard rational elements of
. In particular, there is a standard positive integer
such that
and
is periodic with period
.
Exercise 39 Show that the torus
is uniquely determined by
, and decomposition
in Exercise 38 is unique up to expressions taking values in
(i.e. if one is given another decomposition
, then
and
differ by expressions taking values in
).
Exercise 40 (Recurrence) Let
be a polynomial map from
to
of some standard degree
, and let
be an unbounded natural number. Show that for every standard
and every
, we have
and more generally
for any standard
.
As before, there are also multidimensional analogues of this theory. We shall just state the main results without proof:
Definition 14 (Multidimensional equidistribution) Let
be a standard compact metric space, let
be an unbounded limit natural number, let
be standard, and let
be a limit function. We say that
is equidistributed with respect to a (standard) Borel probability measure
on
if one has
for every standard box
and for all standard continuous functions
.
We say that
is totally equidistributed relative to
if the sequence
is equidistributed on
for every standard
and
(extending
arbitrarily outside
if necessary).
Remark 9 One can replace the indicators
by many other classes, such as indicators of standard convex sets, or standard open sets whose boundary has measure zero, or continuous or Lipschitz functions.
Theorem 15 (Multidimensional ultralimit Ratner’s theorem for abelian polynomial sequences) Let
be standard integers, and let
be a polynomial map from
to
of degree
. Let
be an unbounded natural number. Then there exists a decomposition
into polynomials of degree
, where
- (
is smooth) The
coefficient
of
has size
for every multi-index
. In particular, on the interval
,
is Lipschitz with homogeneous norm
.
- (
is equidistributed) There exists a standard subtorus
of
, such that
takes values in
and is totally equidistributed on
in this torus.
- (
is rational) The coefficients
of
are standard rational elements of
. In particular, there is a standard positive integer
such that
and
is periodic with period
.
Proof: This is implicitly in this paper of Ben Green and myself; the result is phrased using the language of single-scale equidistribution, but this easily implies the ultralimit version.
Update, Apr 5: Another exercise added; some corrections.
44 comments
Comments feed for this article
29 March, 2010 at 10:53 am
Diego
Dear Terry,
Will there be applications to, say, dispersive PDEs in this course, or only number-theoretic oriented applications?
Thanks!
29 March, 2010 at 12:48 pm
Terence Tao
An overview of the course can be found at
https://terrytao.wordpress.com/2010/03/09/course-announcement-254b-higher-order-fourier-analysis/
The focus will be primarily on applications in additive combinatorics. There are tantalising hints that higher order Fourier analysis could also be useful in some aspects of harmonic analysis, such as the theory of the trilinear Hilbert transform, or the Lambda(p) problem for sets such as the square numbers, but there has been little concrete progress in this direction so far.
29 March, 2010 at 4:04 pm
Diego
Thanks Terry, I’ll try to keep up.
30 March, 2010 at 4:10 am
Vincent
Thank you for the interesting notes! A stupid notational question though: what is the meaning of the dot in the Weyl Criterion for equidistribution? Does it take values in the ordinary torus T?
[Corrected, thanks – T.]
30 March, 2010 at 7:25 am
Sujit
Small typo. \Phi should be U in Exercise 3.
[Corrected, thanks – T.]
1 April, 2010 at 12:14 pm
Josh
“…it is not hard to see that this map is equidistributed with respect to 1/3 times the Haar probability measure on (T)\times {0 mod Z}, plus 2/3 times the Haar probability measure on (T)\times {1 mod Z}”
Perhaps the last statement should be “(T)\times {1/3 mod Z}” ?
[Corrected, thanks – T.]
5 April, 2010 at 9:10 am
Eckhard
First, thank you for the notes and the whole blog, which I find very interesting to read.
Second, I think the plural of ‘torus’ is ‘tori’, not ‘torii’.
[Corrected, thanks – T.]
5 April, 2010 at 9:50 am
Jason
In the definition of totally delta-equidistributed, I think f(n) should read f(x(n)).
Thanks for the great notes–Jason
[Corrected, thanks – T.]
5 April, 2010 at 10:20 am
brad
Perhaps I’m missing something, but for the Fejer kernel estimates, I think one has only 1/(R \|x_j\|^2) rather than 1/(R\|x_j\|)^2; and to get the O(1/R) bound I think one has to resort to the full force of a \min(R,1/(R\|x_j\|^2)) estimate (at any rate I did).
More pedantically, the expression involving a product of sines should have x_j multiplied by 2\pi, I think.
[Corrected, thanks – T.]
9 April, 2010 at 3:00 am
Sungjin Kim
One minor thing, the error term H/N in the proof of corollary 5 should be changed to sqrt(H/N) too.
[Corrected, thanks – T.]
9 April, 2010 at 9:50 am
Jason
Small detail: In Exercise 21, should “the linear sequence P” read “the polynomial sequence P” ? [Corrected, thanks – T.]
12 April, 2010 at 10:41 am
mmailliw/william
In the beginning of Section 3, you say
“thus for any limit numbers {X, Y} with {Y>0}, exactly one of {X=O(Y)} and {X=o(Y)} is true”.
Doesn’t X=o(Y) imply X=O(Y) by your definition? Maybe it should be changed to “exactly one of {Y=O(|X|)} and {X=o(Y)} is true”.
[Corrected, thanks -T.]
21 April, 2010 at 3:13 pm
brad
There’s a classical metric theory of asymptotic equidistribution; for example, for any increasing sequence of integers
, the sequence
is equidistributed mod 1 for almost all
. Is their any analogue of these results in the single scale case? For instance, one might make the following guess:
For a fixed sequence of increasing integers
and any
, for almost all
, there is a sequence
(depending on
) so that
and
is
-equidistributed for all N.
If I'm not mistaken this is true in the case that
, but there might be other obstructions for less canonical sequences. (There is also another motivation, by analogy to Carleson's theorem, using partial summation.)
21 April, 2010 at 8:56 pm
Terence Tao
It seems quite likely that most of the classical results of asymptotic equidistribution have more quantitative single-scale counterparts, but the literature of the latter is much sparser currently than that of the former. A very rough rule of thumb is that as long as one avoids the use of the ergodic theorem (which is perhaps the first genuinely asymptotic result that utilises multiple scales), or the use of infinitary topological tools such as closures and compactifications, then asymptotic equidistribution arguments are usually fairly easy to make quantitative (and the bounds are then usually “polynomial” in nature in some sense).
Note though that the exact value of
here will be somewhat sensitive to the precise definition of equidistribution. In my notes here, I used a notion based on the Lipschitz norm, but this is not the only possible choice, and if one uses a different quantitative measure of equidistribution, then one might get a different power of
showing up. But I would guess that a polynomially quantitative version of the above theorem should probably exist.
23 April, 2010 at 12:19 pm
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6 November, 2013 at 8:26 am
Joel Moreira
Why is Exercise 15 called Szemeredi’s theorem for polynomials? Does it imply Szemeredi’s theorem (or polynomial Szemeredi) in some sense? Thank you.
[Exercise reworded to make the connection to Szemeredi’s theorem clearer – T.]
25 December, 2014 at 4:09 pm
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16 April, 2015 at 3:05 pm
MrCactu5 (@MonsieurCactus)
Can you go through the second-to-last sentence in the proof of Lemma 10 more slowly. Why do you get
?
I reasoned set has size at most
and we need to fit two intervals + a gap which is about 50 times larger.
Then I reasons the set is at most
but that might be false.
16 April, 2015 at 3:25 pm
Terence Tao
The length of each gap is at least
as long as either of the neighbouring intervals, each of which have length at least 1, so the number of integers in each gap is at least
times the number of integers in the neighbouring intervals (one can tighten the factor of 10 a bit if needed, but this is certainly enough of a safety margin to deal with roundoff error). Summing, this shows that the total number of integers in the gaps is at least
times the total number of integers in the intervals, which gives the claim since there are only about
integers in all.
12 June, 2017 at 12:55 pm
Anonymous
I’m having a little trouble with exercise 4: total asymptotic equidistribution seems to be equivalent to
for any non-zero
, and any
. Not sure how to connect this to the statement
for rational
without some sort of structural assumption on the sequence
…
12 June, 2017 at 1:00 pm
Anonymous
Below remark 1:
“…if {\alpha} is rational in the sense that {m\alpha = 0} for some positive integer {m}…”
Do you mean that {m\alpha = 0 \mod 1}?
22 June, 2017 at 5:10 pm
Terence Tao
Write
. Each natural number
can be uniquely expressed as
for some
. This can be used to break up sums of the second type into sums of the first type. To go in the reverse direction, one needs to perform a Fourier expansion in
.
13 June, 2017 at 10:03 am
Anonymous
In the second half of exercise 17, should the sequence be
equidistributed rather that
equidistributed?
[Corrected, thanks – T.]
15 June, 2017 at 8:50 am
Anonymous
In exercise 20, should the condition
$$ 1 \leq H \leq \delta^{-C_d} N $$
be replaced with
$$ 1 \leq H \leq \delta^{C_d} N $$
instead?
[Corrected, thanks – T.]
2 October, 2017 at 5:46 am
farlabb
I am trying to understand the proof of proposition 1 (Weyls equidistribution criterion) and I don’t understand the “if” part. I think I am missing something silly, but here it goes: If
then we have that
for
, so assuming equation (2) we indeed have that (1) holds for such functions. In general, if
is any function, then by linearity I get that
. So I get that integral of any function on the torus is just 0, which seems weird to me. Could you explain in more detail the limiting argument to me? A lot of thanks in advance!
2 October, 2017 at 9:23 am
Terence Tao
The contribution of
has to be treated separately. If (1) holds for a sequence of functions
that tend uniformly to a limit
, then it is not hard to show that (1) also holds for
.
13 October, 2017 at 4:54 am
farlabb
Dear prof. Tao, I am having trouble with exercise 4. I do not understand how to show that if
is totally asymptotically equidistributed, then
. Using your hint I could prove the converse to this, but I do not understand how to use your hint (fourier expansion in
) to prove this case.
17 October, 2017 at 10:51 am
Terence Tao
Fourier expansion is not needed for the converse direction. If
, split up the
summation into
sums over arithmetic progressions modulo
by making a change of variables
.
27 March, 2019 at 4:26 am
Maths student
I believe that the uniqueness of the asymptotic equidistribution needs only the easy implication of Riesz’ representation theorem.
29 August, 2019 at 11:51 pm
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12 October, 2021 at 5:59 am
Anonymous
In Lemma 11, I’m having a hard time figuring out how to identify the pigeon-holed arithmetic progression
, which lives inside
, with the progression
, where
is the length of
. My concern is that we are translating
by a constant on the order of
,

with
seems to discount the effect of the shift by
, which would change the coefficients of the relevant polynomial
has different coefficients that
.
but identifying
20 October, 2021 at 1:19 pm
Terence Tao
The stated conclusion of Lemma 11 is invariant with respect to these shifts (the powers of
that show up cancel each other out).
26 November, 2021 at 8:51 pm
Novice
Dear Prof Tao,
In the statement of the Proposition 3, we have that
is totally asymptotically equidistributed on
in a **subtorus**
of
. How is this **subtorus** defined?
In the proof of proposition we have defined the subtorus
and then we apply induction hypothesis, but no where it is described how subtorus
is defined?
Can you please elaborate on this? Thankyou!
28 November, 2021 at 10:51 am
Terence Tao
The subtorus
is constructed recursively. After identifying
with
, there are two cases. If
is irrational, then Exercise 5 lets us take
. Otherwise, we take
to be whatever subtorus was recursively assigned to
.
29 November, 2021 at 9:01 pm
Novice
Dear Prof Tao,
Also, can you please explain how are we able to apply induction hypothesis to the subtorus
?
To apply induction hypothesis we need
to have dimension less than the dimension of
since we assumed that such a decomposition exists for tori of smaller dimension.
But I don’t see why the subtorus
of
has dimension less than d (=dimension of
)
4 December, 2021 at 11:45 am
Terence Tao
The dimension of a torus is equal to the dimension of its tangent space at the origin. The tangent space of
at the origin is a linear subspace of the tangent space
of
that cannot be all of
(since the exponential map would then make
equal to
), so it has dimension at most
.
15 December, 2021 at 11:22 pm
Anonymous
Can I get a hint with Exercises 23/24? I can approximate the set by a Fejer kernel argument, so it looks like I’m interested in analyzing quantities like
(or take the expectation of arithmetic progressions). I can see an upper bound in terms of a maximum of
where
is the minimal “stopping time” that
using a polynomial growth function
But, I’m having trouble with the lower bound. Any hints?
16 December, 2021 at 1:54 am
Anonymous
Ah, a single AP does the job
3 January, 2022 at 8:57 pm
Novice
Dear Prof. Tao,
Can you please give some explaination on how to proceed for exercise 17? I am unable to bound the difference by
for the case when
is a multiple of
(from the definition of
-equidistribution) . Similarly I am not unable to bound the difference for the case when
.
[Try the case
first. Think about why a Riemann sum of a Lipschitz function is close to the Riemann integral. -T]