In this lecture, we move away from recurrence, and instead focus on the *structure* of topological dynamical systems. One remarkable feature of this subject is that starting from fairly “soft” notions of structure, such as topological structure, one can extract much more “hard” or “rigid” notions of structure, such as *geometric* or *algebraic* structure. The key concept needed to capture this structure is that of an *isometric system*, or more generally an *isometric extension*, which we shall discuss in this lecture. As an application of this theory we characterise the distribution of polynomial sequences in torii (a baby case of a variant of Ratner’s theorem due to (Leon) Green, which we will cover later in this course).

We begin with a key definition.

Definition 1(Equicontinuous and isometric systems). Let be a topological dynamical system.

- We say that the system is
isometricif there exists a metric d on X such that the shift maps are all isometries, thus for all n and all x,y. (Of course, once T is an isometry, all powers are automatically isometries also, so it suffices to check the n=1 case.)- We say that the system is
equicontinuousif there exists a metric d on X such that the shift maps form a uniformly equicontinuous family, thus for every there exists such that whenever n, x, y are such that . (As X is compact, equicontinuity and uniform equicontinuity are equivalent concepts.)

**Example 1. **The circle shift on is both isometric and equicontinuous. On the other hand, the Bernoulli shift on is neither isometric nor equicontinuous (why?).

**Example 2.** Any finite dynamical system is both isometric and equicontinuous (as one can see by using the discrete metric).

Since all metrics are essentially equivalent, we see that the choice of metric is not actually important when checking equicontinuity, but it seems to be more important when checking for isometry. Nevertheless, there is actually no distinction between the two properties:

**Exercise 1**. Show that a topological dynamical system is isometric if and only if it is equicontinuous. (Hint: one direction is obvious. For the other, if is a uniformly equicontinuous family with respect to a metric d, consider the modified metric .)

**Remark 1.** From this exercise we see that we can upgrade *topological* structure (equicontinuity) to *geometric* structure (isometry). The motif of studying topology through geometry pervades modern topology; witness for instance Perelman’s proof of the Poincaré conjecture.

**Exercise 2**. (Ultrafilter characterisation of equicontinuity) Let be a topological dynamical system. Show that X is equicontinuous if and only if the maps are homeomorphisms for every .

Now we upgrade the geometric structure of isometry to the *algebraic *structure of being a compact abelian group action.

Definition 2.(Kronecker system) A topological dynamical system is said to be aKronecker systemif it is isomorphic to a system of the form , where is a compact abelian metrisable topological group, and is a group rotation for some .

**Example 3.** The circle rotation system is a Kronecker system, as is the standard shift on a cyclic group . Any product of Kronecker systems is again a Kronecker system.

Let us first observe that a Kronecker system is equicontinuous (and hence isometric). Indeed, the compactness of the topological group K (and the joint continuity of the addition law ) easily ensures that the group rotations are uniformly equicontinuous as varies. Since the shifts are all group rotations, the claim follows.

On the other hand, not every equicontinuous or isometric system is Kronecker. Consider for instance a finite dynamical system which is the disjoint union of two cyclic shifts of distinct order; it is not hard to see that this is not a Kronecker system. Nevertheless, it clearly contains Kronecker systems within it. Indeed, we have

Proposition 1.Everyminimalequicontinuous (or isometric) system is a Kronecker system, i.e. isomorphic to an abelian group rotation . Furthermore, the orbit is dense in K.

**Proof.** By Exercise 1, we may assume that the system is isometric, thus we can find a metric d such that all the shift maps are isometries. We view the as lying inside the space of continuous maps from X to itself, endowed with the uniform topology. Let be the closure of the maps . One easily verifies that G is a closed metrisable topological group of isometries in ; from the Arzelà-Ascoli theorem we see that G is compact. Also, since and commute for every n and m, we see upon taking limits that G is abelian.

Now let be an arbitrary point. Then we see that the image of G under the evaluation map is a compact non-empty invariant subset of X, and thus equal to all of X by minimality. If we then define the stabiliser , we see that is a closed (hence compact) subgroup of the abelian group G. Since , we thus see that there is a continuous bijection from the quotient group (with the quotient topology) to X. Since both spaces here are compact Hausdorff, this map is a homeomorphism. This map is thus an isomorphism of topological dynamical systems between the Kronecker system K (with the group rotation given by ) and X. Since K is a compact metrisable (thanks to Hausdorff distance) topological group, the claim follows (relabeling the group operation as +). Note that the density of in K is clear from construction.

**Remark 2. ** Once one knows that X is homeomorphic to a Kronecker system with dense, one can *a posteriori* return to the proof and conclude that the stabiliser is trivial. But I do not see a way to establish that fact directly. In any case, when we move to isometric extensions below, the analogue of the stabiliser can certainly be non-trivial.

To get from minimal isometric systems to non-minimal isometric systems, we have

Proposition 2.Any isometric system can be partitioned as the union of disjoint minimal isometric systems.

**Proof.** Since minimal systems are automatically disjoint, it suffices to show that every point is contained in a minimal dynamical system, or equivalently that the orbit closure is minimal. If this is not the case, then there exists such that x does not lie in the orbit closure of y. But by definition of orbit closure, we can find a sequence such that converges to y. By the isometry property, this implies that converges to x, and so x is indeed in the orbit closure of y, a contradiction.

Thus every equicontinuous or isometric system can be expressed as a union of disjoint Kronecker systems.

We can use the algebraic structure of isometric systems to obtain much quicker (and slightly stronger) proofs of various recurrence theorems. For instance, we can give a short proof of (a slight strengthening of) the multiple Birkhoff recurrence theorem (Theorem 3 from Lecture 4) as follows:

Proposition 3. (Multiple Birkhoff for isometric systems) Let be an isometric system. Then for every there exists a sequence such that for every integer k.

**Proof.** By Proposition 2 followed by Proposition 1, it suffices to check this for Kronecker systems in which is dense in K. But then we can find a sequence such that in K, and thus (since K is a topological group) in K for all k. The claim follows.

The above argument illustrates one of the reasons why it is desirable to have an algebraic structural theory of various types of dynamical systems; it makes it much easier to answer many interesting questions regarding such systems, such as those involving recurrence.

— The Kronecker factor —

We have seen isometric systems are basically Kronecker systems (or unions thereof). Of course, not all systems are isometric. However, it turns out that every system contains a maximal isometric *factor*. Recall that a factor of a topological dynamical system is a surjective morphism from X to another topological dynamical system . (We shall sometimes abuse notation and refer to as the factor, when it is really the quadruplet .) We say that one factor *refines* or is *finer than* another factor if we can factorise for some continuous map . (Note from surjectivity that this map, if it exists, is unique.) We say that two factors are *equivalent* if they refine each other. Observe that modulo equivalence, refinement is a partial ordering on factors.

**Example 4.** The identity factor is finer than any other factor of X, which in turn is finer than the trivial factor that maps to a point.

**Exercise 3.** Show that any factor of a minimal topological dynamical system is again minimal.

We note two useful operations on factors. Firstly, given two factors and , one can define their join , where is the compact subspace of the product system , and is the surjective morphism . One can verify that is the least common refinement of and , hence the name.

Secondly, given a chain of factors (thus refines for all ), one can form their inverse limit by first letting be the factoring maps for all , observing that for all , and then defining to be the compact subspace of the product system defined as

(1)

and then setting . One easily verifies that is indeed a factor of X, and it is the least upper bound of the .

Now we observe that these operations interact well with the isometry property:

**Exercise 4.** Let and be two factors such that Y and Y’ are both isometric. Then is also isometric.

Lemma 1.Let be a totally ordered set of factors with isometric. Then the inverse limit of the is such that Y is also isometric.

**Proof.** Observe that we have factor maps which are surjective morphisms, which themselves factor as for and some surjective morphisms . Let us fix some metric d on Y. For each , consider the compact subset of . These sets decrease as increases, and their intersection is the diagonal (why?). Applying the finite intersection property in the compact sets , we conclude that for every there exists such that whenever .

Now suppose for contradiction that Y is not isometric, and hence not uniformly equicontinuous. Then there exists a sequences with , an , and a sequence of integers such that . By compactness we may assume that both converge to the same point. But by the preceding discussion, we can find such that whenever . In other words, for any z in , the fibre has diameter at most .

Now let and . Then and converge to the same point z in , and so by equicontinuity of , goes to zero for some metric on (the exact choice of metric is not important, as remarked earlier). By compactness and passing to a subsequence we can assume that and both converge to some point in . On the other hand, from the preceding discussion and the triangle inequality, we see that the fibres and are separated by a distance at least in Y. On the other hand, the distance between and must go to zero as (as a simple sequential compactness argument shows), and similarly the distance between and goes to zero. Since has diameter at most , we obtain a contradiction. The claim follows.

Combining Exercise 2 and Lemma 1 with Zorn’s lemma (and noting that with the trivial factor , the image is clearly isometric) we obtain

Corollary 1. (Existence of maximal isometric factor) For every topological dynamical system there is a factor with Y isometric, and which is maximal with respect to refinement among all such factors with this property. This factor is unique up to equivalence.

By Proposition 1 and Exercise 3, the maximal isometric factor of a minimal system is a Kronecker system, and we refer to it as the* Kronecker factor* of that minimal system X.

**Exercise 5.** (Explicit description of Kronecker factor) Let be a minimal topological dynamical system, and let be the set

(2)

where V ranges over all open neighbourhoods of the diagonal of , and is the product shift. Let be the finest equivalence relation on X such that the set is closed and contains Q. (The existence and uniqueness of can be established by intersecting over all candidates together.) Show that the projection map to the equivalence classes of (with the quotient topology) is (up to isomorphism) the Kronecker factor of X.

The Kronecker factor is also closely related to the concept of an eigenfunction. We say that a continuous function is an *eigenfunction* of a topological dynamical system if it is not identically zero and we have for some , which we refer to as an *eigenvalue* for T.

**Exercise 6**. Let be a minimal topological dynamical system.

- Show that if is an eigenvalue for T, then lies in the unit circle , and furthermore there exists a unimodular eigenfunction with this eigenvalue. (
*Hint*: the zero set of an eigenfunction is a closed shift-invariant subset of X.) - Show that for every eigenvalue , the eigenspace is one-dimensional, i.e. all eigenvalues have geometric multiplicity 1. (
*Hint*: first establish this in the case .) - If is a unimodular eigenfunction with non-trivial eigenvalue , show that is an isometric factor of X, where is given the shift . Conclude in particular that , where is the Kronecker factor, is a character of K, and c is a constant. Conversely, show that all functions of the form are eigenfunctions. (From this, it is possible to reconstruct the Kronecker factor canonically from the eigenfunctions of X; we leave the details to the reader.)

We will see eigenfunctions (and various generalisations of the eigenfunction concept) playing a decisive role in the structure theory of measure-preserving systems, which we will get to in a few lectures.

— Isometric extensions —

To cover more general systems than just the isometric systems, we need the more flexible concept of an *isometric extension*.

Definition 3(Extensions). If is a factor of , we say that is anextensionof , and refer to as theprojection maporfactor map. We refer to the (compact) spaces for as thefibresof this extension.

**Example 5**. The skew shift is an extension of the circle shift, with the fibres being the “vertical” circles. All systems are extensions of a point, and (somewhat trivially) are also extensions of themselves.

Definition 4(Isometric extensions). Let be an extension of a topological dynamical system with projection map . We say that this extension isisometricif there exists a metric on each fiber with the following properties:

- (Isometry) For every and , we have .
- (Continuity) The function formed by gluing together all the is continuous (where we view the domain as a compact subspace of ).
- (Isometry, again) For any , the metric spaces and are isometric.

**Example 6**. The skew shift is an isometric extension of the circle shift, where we give each fibre the standard metric.

**Example 7.** A topological dynamical system is an isometric extension of a point if and only if it is isometric.

**Exercise 7**. If X is minimal, show that properties 1 and 2 in the above definition automatically imply property 3. Furthermore, in this case show that the isometry group of any fibre acts transitively on that fibre. Show however that property 3 can fail even when properties 1 and 2 hold if X is not assumed to be minimal.

**Exercise 7′. ** (Topological characterisation of isometric extensions) Let be a extension of a minimal topological dynamical system with factor map , and let d be a metric on X. Show that X is an isometric extension if and only if the shift maps are uniformly equicontinuous relative to in the sense that for every there exists such that every with and , we have for all n.

An important subclass of isometric extensions are the *group extensions*. Recall that an *automorphism* of a topological dynamical system is an isomorphism of that system to itself, i.e. a homeomorphism that commutes with the shift.

Definition 5. (Group extensions) Let be a topological dynamical system. Suppose that we have a compact group G of automorphisms of X (where we endow G with the uniform topology). Then the quotient space is also a compact metrisable space, and one easily sees that the projection map is a factor map. We refer to X as agroup extensionof Y (or of any other system isomorphic to Y). We refer to G as thestructure groupof the extension. We say that the group extension is anabelian group extensionif G is abelian.

**Example 8.** (Cocycle extensions) If G is a compact topological metrisable group, is a topological dynamical system, and a continuous map , then we define the *cocycle extension* to be the product space with the shift , and with the factor map . One easily verifies that X is a group extension of Y with structure group G. The converse is not quite true for topological reasons; not every G-bundle can be globally trivialised, although one can still describe general group extensions by patching together cocycle extensions on local trivalisations.

**Example 9.** The skew shift is a cocycle extension (and hence group extension) of the circle shift , with being the identity map. Any Kronecker system is an abelian group extension of a point.

**Exercise 8**. Show that every group extension is an isometric extension. (Hint: the group G acts equicontinuously on itself, and thus isometrically on itself by choosing the right metric, as in Exercise 1.)

**Exercise 9.** Let be a topological dynamical system, and G a compact topological metrisable group. We say that two cocycles are *cohomologous* if we have for some continuous map . Show that if are cohomologous, then the cocycle extensions and are isomorphic. Understanding exactly which cocycles are cohomologous to each other is a major topic of study in dynamical systems (though not one which we will pursue here).

In view of Proposition 1 and Exercise 14, it is reasonable to ask whether every minimal isometric extension is a group extension. The answer is no (though actually constructing a counterexample is a little tricky). The reason is that we can form intermediate systems between a system Y=G\X and a group extension X of that system by quotienting out a subgroup. Indeed, if H is a closed subgroup of the structure group G, then H\X is a factor of X and an isometric extension of G\X, but need not be a group extension of G\X (basically because G/H need not be a group). But this is the only obstruction to obtaining an analogue of Proposition 1:

Lemma 2. Suppose that X is an isometric extension of another topological dynamical system Y with projection map . Suppose also that X is minimal. Then there exists a group extension Z of Y with structure group G (thus ) and a closed subgroup H of G such that X is isomorphic to H\Z, and is (after applying the isomorphisms) the projection map from H\Z to G\Z; thus we have the commutative diagram. (4)

**Proof**. For each , let be the metric space with the metric given by Definition 5. Thus for any integer n and any , is an isometry from to ; taking limits, we see for any that is an isometry from to . Also, the clearly commute with the shift T.

Fix a point , and set .

Let W be the space of all pairs where and f is an isometry from to . This is a compact metrisable space with a shift and an action of G that commutes with U. We let Z be the orbit closure in W of the G-orbit under the shift U. If we fix a point , then Z projects onto X by the map , and onto Y by the map ; these maps of course commute with the projection from X to Y. Because X is minimal (and thus equal to all of its orbit closures), one sees that all of these projections are surjective morphisms, thus Z extends both Y and X. Also, one verifies that Z is a group extension over Y with structure group G, and a group extension over X with structure group given by the stabiliser . The claim follows.

**Exercise 10.** Show that if an minimal extension is finite, then it is automatically an abelian group extension. (*Hint*: recall from Lecture 2 that minimal finite systems are equivalent to shifts on a cyclic group.)

An important feature of isometric or group extensions is that they tend to preserve recurrence properties of the system. We will see this phenomenon prominently when we turn to the ergodic theory analogue of isometric extensions, but for now let us give a simple illustrative result in this direction:

Proposition 4.Let be an isometric extension of with factor map , and let y be a recurrent point of Y (see Lecture 3 for a definition). Then every point x in the fibre is a recurrent point in X.

**Proof. **It will be convenient to use ultrafilters. In view of Lemma 2, it suffices to prove the claim for group extensions (note that recurrence is preserved under morphisms). Since y is recurrent, there exists such that (see Exercise 9 from Lecture 3). Thus . Since Y = G\X, this implies that for some . We can iterate this (recalling that G commutes with T) to conclude that for all positive integers n. But by considering the action of g on G, we know (from the Birkhoff recurrence theorem from Lecture 3) that we have for some and ; canceling the h, and then applying to x, we conclude that , and thus . If we write for some , we conclude that and so x is recurrent as desired.

— Application: distribution of polynomial sequences in torii —

Now we apply the above theory to the following specific problem:

Problem 1.Let be a polynomial sequence in a d-dimensional torus, thus for some . Compute the orbit closure .

(We will be vague here about what “compute” means.)

**Example 10.** Is the orbit dense in the two-dimensional torus ?

The answer should of course depend on the polynomial P; for instance if P is constant then the orbit closure is clearly a point. Similarly, if the polynomial P has a constraint of the form for some non-zero and , then the orbit closure is clearly going to be contained inside the proper subset of the torus. For instance, is clearly not dense in the two-dimensional torus, as it is contained in the closed one-dimensional subtorus .

In the above example, it is clear that the problem of computing the orbit closure of reduces to computing the orbit closure of . More generally, if a polynomial obeys a constraint for some non-zero *irreducible* (i.e. m does not factor as m = q m’ for some and , or equivalently that the greatest common divisor of the coefficients of m is 1), then some elementary number theory shows that the set is isomorphic (after an invertible affine transformation with integer coefficients on the torus) to the standard subtorus ).

**Exercise 11**. Prove the above claim. (Hint: the Euclidean algorithm may come in handy.)

Because of this, we see that whenever we have a constraint of the form with m irreducible, we can reduce Problem 1 to an instance of Problem 1 with one lower dimension. What about if m is not irreducible? A typical example of this would be when (I’m going to drop the “mod 1” terms to remove clutter). Here, we have the constraint , which constrains P to the union of two one-dimensional torii, rather than a single one-dimensional torus. But we can eliminate this multiplicity by the trick of working with the odd and even components and respectively. One observes that each component obeys an irreducible constraint, namely and respectively, and so by the preceding discussion, the problem of computing the orbit closures for each of these components reduces to that of computing an orbit closure in a torus of one lower dimension.

**Exercise 12. **More generally, show that whenever P obeys a constraint with m not necessarily irreducible, then there exists an integer such that the orbits obey a constraint with m’ irreducible.

From Exercises 11 and 12, we see that every time we have a constraint of the form for some non-zero m, we can reduce Problem 1 to one or more copies of Problem 1 in one lower dimension. So, without loss of generality (and by inducting on dimension) we may assume that no such constraint exists. (We will see this “induction on dimension” type of argument much later in this course, when we study Ratner-type theorems in more detail.)

Now that all the “obvious” restrictions on the orbit have been removed, one might now expect P(n) to be uniformly distributed throughout the torus. Happily, this is indeed the case (at least at the topological level):

Theorem 1.(Equidistribution theorem) Let be a polynomial sequence which does not obey any constraint of the form with non-zero. Then the orbit is dense in (i.e. the orbit closure is the whole torus).

**Remark 3.** The recurrence theorems we have already encountered (e.g. Corollary 1 from Lecture 4, or Theorem 1 from Lecture 5) do not seem to directly establish this result, instead giving the weaker result that every element in is a limit point.

**Exercise 13.** Assuming Theorem 1, show that the answer to Problem 1 is always “a finite union of subtorii”, regardless of what the coefficients of P are.

Theorem 1 can be proven using Weyl’s theory of equidistribution, which is based on bounds on exponential sums; but we shall instead use a topological dynamics argument based on some ideas of Furstenberg. Amusingly, this argument will use some *global* topology (specifically, winding numbers) and not just *local* (point-set) topology.

To begin proving this theorem, let us first consider the linear one-dimensional case, in which one considers the orbit closure of for some . The constant term only affects this closure by a translation and we can ignore it. One then easily checks that the orbit closure is a closed subgroup of . Fortunately, we have a classification of these objects:

Lemma 3.Let H be a closed subgroup of . Then either , or H is a cyclic group of the form for some .

**Proof.** If H is not all of , then its complement, being a non-empty open set, is the union of disjoint open intervals. Let x be the boundary of one of these intervals, then x lies in the closed set H. Translating the group H by x, we conclude that 0 is also the boundary of one of these intervals. Since H = -H, we thus see that 0 is an isolated point in H. If we then let y be the closest non-zero element of H to the origin (the case when H={0} can of course be checked separately), we check (using the Euclidean algorithm) that y generates H, and the claim easily follows.

**Exercise 14.** Using the above lemma, prove Theorem 1 in the case when d=1 and P is linear.

**Exercise 15. **Obtain another proof of Lemma 3 using Fourier analysis and the fact that the only non-trivial subgroups of (the Pontryagin dual of are the groups for .

Now we consider the linear case in higher dimensions. The key lemma is

Lemma 4.Let H be a closed subgroup of for some such that , where is the canonical projection. Then either or for some with final coefficient non-zero.

**Proof. **The fibre is isomorphic to a closed subgroup of , so we can apply Lemma 3. If this subgroup is full, then it is not hard to see that , so suppose instead that is isomorphic to the cyclic group of order N. We then apply the homomorphism , and observe that is a closed subgroup of whose fibres are a point, i.e. is a graph for some . Observe that the projection map is a continuous bijection from the compact Hausdorff space to the compact Hausdorff space , and is thus a homeomorphism; in particular, is continuous. Also, since is a group, must be a homomorphism. It is then a standard exercise to conclude that is linear, and therefore takes the form for some integers . The claim then follows by some routine algebra.

**Exercise 16.** Using the above lemma, prove Theorem 1 in the case when d is arbitrary and P is linear.

We now turn to the polynomial case. The basic idea is to re-express P(n) in terms of the orbit of some topological dynamical system on a torus. We have already seen this happen with the skew shift , where the orbits exhibit quadratic behaviour in n. More generally, an iterated skew shift such as

(5)

generates orbits whose final coefficient contains degree d terms such as . What we would like to do is find criteria under which we could demonstrate that systems such as (5) are *minimal*; this would mean that every orbit closure in that system is dense, which would clearly be relevant for proving results such as Theorem 1.

To do this, we will exploit the fact that systems such as (5) can be built as towers of isometric extensions; for instance, the system (5) is an isometric extension over the same system (5) associated to d-1 (which, in the case d=1, is simply a point). Now, isometric extensions don’t always preserve minimality; for instance, if one takes a trivial cocycle extension then the system is certainly non-minimal, as every horizontal slice of that system is a subsystem. More generally, any cocycle extension which is cohomologous to the trivial cocycle (see Exercise 9) will not be minimal. However, it turns out that if one has a topological obstruction to triviality, then minimality is preserved. We will formulate this fact using the machinery of *winding numbers*. Recall that every continuous map has a winding number , which can be defined as the unique integer such that f is homotopic to the linear map . Note that the map is linear, and also that is unchanged if one continuously deforms f.

We now give a variant of a lemma of Furstenberg.

Lemma 5.Let be a minimal topological dynamical system. Let be a cocycle such that for every non-zero there exists a loop such that is homotopic to and . Then is also minimal.

**Proof. **We induct on d. The case d=0 is trivial, so suppose and the claim has already been proven for d-1. Suppose for contradiction that contains a proper minimal subsystem Z. Then is a subsystem of Y, and must therefore equal all of Y, by minimality of Y. Now we use the action of on , which commutes with the shift . For every , we see that is also a minimal subsystem, and so is either equal to Z or disjoint from Z. If we let , we conclude that H is a closed subgroup of .

We now claim that the projection of H to must be all of . For if this were not the case, we could project Z down to , where is the projection of , and obtain a proper subsystem of that extension. But by induction hypothesis we see that is minimal, a contradiction, thus proving the claim.

We can now apply Lemma 4. If H is all of then Z is all of , a contradiction. Thus we have for some non-zero , and thus Z must take the form

(6)

for some . Arguing as in the proof of Lemma 4 we can show that Y is homeomorphic to the image of Z under the map and so must be continuous. Since Z is shift-invariant, we must have the equation

. (7)

We apply this for y in the loop associated to m by hypothesis, and take degrees to conclude

. (8)

But as is homotopic to , we have and thus , contradicting the hypothesis.

**Exercise 17.**Using the above lemma and an induction on d, show that the system (5) is minimal whenever is irrational. [The key, of course, is to make a good choice for the loop that makes all computations easy.]

**Exercise 18. **More generally, show that the product of any finite number of systems of the form (5) remains minimal, as long as the numbers that generate each factor system are linearly independent with respect to each other and to 1 over the rationals .

It is now possible to deduce Theorem 1 from Exercise 18 and a little bit of linear algebra. We sketch the ideas as follows. Firstly we take all the non-constant coefficients that appear in P and look at the space they span, together with 1, over the rationals . This is a finite-dimensional space, and so has a basis containing 1 which is linearly independent over . The non-constant coefficients of P are rational linear combinations of elements of this basis; by dividing the basis elements by some suitable integer (and using the trick of passing from P(n) to P(qn+r) if necessary) we can ensure that the coefficients of P are in fact integer linear combinations of basis elements. This allows us to write P as an affine-linear combination (with integer coefficients) of the coefficients of an orbit in the type of product system considered in Exercise 18. If this affine transformation has full rank, then we are done; otherwise, the affine transformation maps to some subspace of the torus of the form , contradicting the hypothesis on P. Theorem 1 follows.

[*Update*, Feb 9: Proof of Lemma 2 corrected.]

[*Update*, Feb 11: Exercise 6 corrected.]

[*Update*, Feb 12: Proof of Lemma 1 corrected.]

[*Update*, Feb 18: Exercise 6.3 corrected.]

[*Update*, Feb 19: Exercise 6.3 corrected again.]

[*Update*, Mar 24: Exercise 7′ corrected.]

[*Update*, Mar 31: Some minor corrections.]

[*Update*, Sep 27: More corrections.]

## 52 comments

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26 January, 2008 at 12:16 pm

NilaySir

1. In the definitions of isometric and equicontinous, where does n belong to – set of integers or set of naturals?

2. In example 2, I think if we use the discreet metric we will not necessarily obtain an isometric system. Consider the two point system {x,y} such that Tx = x, Ty = x. Then d(x,y) = 1 but d(Tx,Ty) = 0.

26 January, 2008 at 6:50 pm

Terence TaoDear Nilay,

In these lectures, the shift T is understood to be invertible unless otherwise specified (see Lecture 1, paragraph 5).

Cute exercise: show that a non-invertible shift cannot generate an isometric (non-invertible) topological dynamical system. (It’s clear that non-injectivity causes an obstruction; what is more tricky is to show that non-surjectivity does also.)

28 January, 2008 at 10:48 am

254A, Lecture 7: Structural theory of topological dynamical systems « What’s new[…] Exercise 3. Show that an inverse limit of a totally ordered set of distal factors is still distal. (This turns out to be slightly easier than Lemma 1 from the previous lecture.) […]

11 February, 2008 at 1:31 pm

mmailliw/williamIn Question 6.3, I think we need to suppose that g = 1 at some point x to get the desired decomposition g = \chi o \phi.

Otherwise, we have the following counterexample:

X is the two-point set {a, b}, T is the shift sending a to b and vice versa, and

g: X –> S^1 is an eigenfunction for the eigenvalue \lambda = -1 with g(a) = i and g(b) = -i (clearly this is unimodular).

As X is isometric, it is its own Kronecker factor (and any representation of this Kronecker factor would be as a group of order two).

However, if \chi is a character from a group of order two to S^1, its image must be contained in the subset of S^1 consisting of order-two elements, i.e. {-1, 1}. This means that g = \chi o \pi is impossible as the former has range {i, -i} and the latter must have range contained in {-1, 1}.

11 February, 2008 at 2:43 pm

Terence TaoDear mmailliw/william,

Thanks for pointing that out. I’ll add a multiplicative constant to the decomposition to fix the problem.

11 February, 2008 at 9:14 pm

254A, Lecture 11: Compact systems « What’s new[…] are analogous to the equicontinuous or isometric systems in topological dynamics discussed in Lecture 6, and as with those systems, we will be able to characterise such systems (or more precisely, the […]

11 February, 2008 at 10:31 pm

ZaherDear Prof.;

In the proof of lemma 1 (last paragraph), it seems to me that we should consider the fibres and and use that is uniformly equicontinuous.

11 February, 2008 at 11:06 pm

AnonymousDear Prof Tao:

This is an off topic suggestion. Would it make sense to create a “graveyard” thread so that people who are not familiar with laTex/Wordpress can test their formatting before submitting the real posts?

12 February, 2008 at 11:57 am

Terence TaoDear Zaher,

Thanks for the correction!

Dear Anonymous:

What I have been doing so far is correcting any LaTeX errors I see manually. I think if someone wanted to do extensive testing of LaTeX then doing so in their own blog (where they can delete or re-edit the content at will, and also have previewing options) would make more sense.

18 February, 2008 at 9:10 am

ZaherDear Prof.;

Concerning Exercise 6.3, the unimodular eigenfunction may not be surjective and hence not a factor. mmailliw/william (above commenter) already gave an example. Yet, I think that the decomposition g= c \pi o \chi may still true even if g is not onto.

18 February, 2008 at 11:33 am

Terence TaoGood point; I’ve amended the exercise accordingly.

21 February, 2008 at 10:24 am

254A, Lecture 9: Ergodicity « What’s new[…] Exercise 5. Show that the circle shift (with the usual Lebesgue measure) is ergodic if and only if is irrational. (Hint: analyse the equation for (say) using Fourier analysis. Added, Feb 21: As pointed out to me in class, another way to proceed is to use the Lebesgue density theorem (or Lebesgue differentiation theorem) combined with Exercise 14 from Lecture 6.) […]

23 February, 2008 at 2:36 pm

254A, Lecture 12: Weakly mixing systems « What’s new[…] Exercise 7. Using Exercise 6, give another proof of Theorem 1 from Lecture 6. […]

2 March, 2008 at 9:42 am

254A, Lecture 13: Compact extensions « What’s new[…] be described as (inverse limits of) group quotient extensions, somewhat analogously to Lemma 2 from Lecture 6. Roughly speaking, the idea is to first use some spectral theory to approximate conditionally […]

24 March, 2008 at 7:45 am

sugatasir,

i have tried to prove the ‘topological characterisation of isometric extensions’,but could not figure it out.can i just ask you what is your definition of isometric spaces,i.e.when are two spaces called isometric.

24 March, 2008 at 8:12 am

Terence TaoDear Sugata,

Oops! For exercise 7′, it’s only possible to obtain the isometry of fibres in the case when Y is minimal. I’ve added that hypothesis to the question.

24 March, 2008 at 7:51 pm

Sugatasir,

i tried to solve the problem just after group extensions.does your definition of isometry imply that ‘the isometry’ is onto,then there is a problem again.

consider X=a discrete topological dynamical system containing n points.T=identity map on X.name the points of X 1,2,…now consider S^(n-1) acting on first (n-1) points of X.then Y=[S^(n-1).{1},{n}].then pie inverse{S^(n-1).{1}}={1,2…,n-1},pie inverse{n}={n}.then these two spaces can’t be isometric.

24 March, 2008 at 7:58 pm

Terence TaoDear Sugata,

As I mentioned in my previous comment, one has to make the additional assumption that Y is minimal in order to mix the fibres together properly. With your example, Y is not minimal (the point {n}, for instance, is a proper subsystem of Y, since the shift here is just the identity).

30 March, 2008 at 5:13 am

liuxiaochuanDear professor 陶:

(1)The last line of definition 2 I think it should be .

(2)At the end of the second paragraph of the proof of lemma 1, it is at most

p.s. I realized I can input Chinese characters. :-)

刘

31 March, 2008 at 8:02 pm

Terence TaoDear 刘: Thanks for the correction!

1 April, 2008 at 5:44 pm

liuxiaochuanDear professor 陶：

About the eigenfunction T. Should Tf(x) mean f(T(x)) or or something else?

刘

2 April, 2008 at 10:02 am

Terence TaoDear 刘: Tf is always understood to be the function (see equation (1) of Lecture 1).

10 April, 2008 at 12:58 am

liuxiaochuanDear Professor 陶:

(1)In exercise 8, should the condition “X is minimal ” be added or not? I wonder if a system is isometric extension of itself if and only if it is minimal.

(2)At the second paragraph of the proof of lemma 2, I think the definition of G should be ; at the third paragraph, the onto X map should be

(3)The proof of proposition 4, “…it suffices to prove the claim for “group” extensions”.

(4)After exercise 11, the definition of should be , where “n” is missing.

刘

11 April, 2008 at 7:02 am

liuxiaochuanDear Professor 陶:

I don’t quite follow the last paragraph. How can one see that the system (5) could still be minimal under a affine transformation?

yours，

刘

11 April, 2008 at 10:26 am

Terence TaoDear 刘:

When applying the affine transformation, we don’t try to preserve minimality, but we can preserve the property of being dense. If for instance is an affine map of full rank from one torus to another (with ), and E is any set which is dense in the domain of f, then f(E) will be dense in the range. In particular, given any minimal dynamical system on , the image of any dense orbit under f will be dense in .

11 April, 2008 at 4:24 pm

liuxiaochuanDear Professor 陶:

I’ve got several corrections above, although it may be my misunderstanding and they may be wrong.

刘

27 September, 2008 at 9:00 am

Terence TaoDear 刘:

Thanks (very belatedly) for the corrections!

27 September, 2008 at 12:51 pm

What is a gauge? « What’s new[…] There was nothing terribly special about circles in this example; one can also define group extensions, or more generally homogeneous space extensions, of dynamical systems, and have a similar theory, although one has to take a little care with the order of operations when the structure group is non-abelian; see e.g. these lecture notes. […]

19 January, 2009 at 11:34 am

An inverse theorem for the uniformity seminorms associated with the action of F^infty_p « What’s new[…] some standard theory of Mackey and Furstenberg on isometric extensions and group extensions (cf. my lectures on the topological dynamics counterpart to these topics), and of Furstenberg-Weiss on […]

26 July, 2009 at 6:01 am

liuxiaochuanDear Professor Tao:

In Definition 5, why is a quotient space?

26 July, 2009 at 6:02 am

liuxiaochuanDear Professor Tao:

In Definition 5, why is a quotient space?Is there any typoes there?

26 July, 2009 at 7:50 am

Terence TaoIf one defines the equivalence relation on X by declaring iff for some , then Y becomes the quotient space of that relation, see

http://en.wikipedia.org/wiki/Quotient_space

26 July, 2009 at 8:05 am

陶哲轩遍历论习题解答：第六讲 « Liu Xiaochuan’s Weblog[…] (注：遍历论为陶哲轩教授于08年年初的一门课程，我尝试将所有习题做出来，这是第五讲的十八个习题。这里是这门课程的主页，这里是本讲的链接。关于拓扑动力系统两种扩张的定义，我用中文写过一个帖子。 […]

22 October, 2009 at 12:23 am

liuxiaochuanDear Professor Tao:

I am going over the proof of lemma 1 when I find a problem (I think). The metric d is only defined on The space Y. But metric d is also used on . The metric used there should be derived from d and thus being compatible with each other. Maybe it is worth mentioning to avoid misunderstanding.

22 October, 2009 at 7:53 am

Terence TaoAh, yes, one needs to pick another metric for . It doesn’t really matter which one, though; from compactness, all metrics are essentially equivalent to each other for the purposes of understanding equicontinuity. (If one wanted to fix a metric, one could use the Hausdorff metric on the fibres with respect to d, but this is not necessary.)

18 June, 2010 at 2:53 am

Solutions to Ergodic Theory：Lecture twelve « Xiaochuan Liu's Weblog[…] 7. Using Exercise 6, give another proof of Theorem 1 from Lecture 6: (Equidistribution theorem) Let be a polynomial sequence which does not obey any constraint of […]

20 June, 2010 at 5:41 pm

Solutions to Ergodic Theory：Lecture Seven « Xiaochuan Liu's Weblog[…] Exercise 4: Show that every topological dynamical system has a maximal distal factor. (Hint: repeat the proof of Corollary 1 from the lecture six.) […]

14 August, 2011 at 2:18 am

pavel zorinDear Prof. Tao,

the claim in the second paragraph of the proof of Lemma 5 should rely on the hypothesis on , since for we obtain . However, I think that it would be easier to use the full characterization of closed subgroups of torii, replacing the condition

by

for a finite family of and continuing the argument as stated.

This way the induction on is eliminated (or rather hidden). Also, there is no need to argue by a double contradiction.

Note that the same lemma and proof appear as Lemma 2.6.30 in the online preprint of your “year two…” book, which I would like to thank you for on this occasion: the introduction to the ultrafilter approach to topological dynamics has been very illuminating for me.

Finally, I have a question: how does the converse implication in Exercise 2 work? (i.e. “ are homeomorphisms” implies “ are equicontinuous”)

best regards,

pavel

15 August, 2011 at 8:05 am

Terence TaoThe hypothesis on is used when the induction hypothesis is invoked (because if obeys the hypothesis in d dimensions, then the projection does too). In particular, will not work for this step (except when d=1) because will vanish also.

It is true that one can handle the full torus in one jump, but I think I preferred the inductive approach for pedagogical reasons, to illustrate the more general technique of proving things about dynamical systems by climbing up towers of extensions that are as simple as possible.

For Exercise 2: I must have had a slick ultrafilter-based proof of this at some point, but I can’t reconstruct it. Here is a rather messy (and partially ultrafilter-based) proof: put a metric on X. Suppose equicontinuity failed, then there is some such that one can find arbitrarily close x,y such that for some n. By shifting back and forth, this implies furthermore that for any syndetic set A, one can find arbitrarily close x,y such that for some n in A.

On the other hand, the are homeomorphisms, and are in particular injective (this is equivalent to distality). Specialising to p a minimal idempotent, we thus have for all x. Among other things, this implies that every point in X is almost periodic, and more generally every k-tuple in is almost periodic.

Using these two observations, one can now recursively find a sequence of pairs of increasingly close points, such that for each M, the set of n for which for all is non-empty (and hence syndetic, by joint almost periodicity). One can then construct an ultrafilter p for which for all m, contradicting the homeomorphism property.

16 August, 2011 at 8:50 am

pavel zorinThank you for the explanations. I have now figured out what confused me in Lemma 5: it was not clear to me why the projection of to must be a proper subsystem if the projection of to is not surjective.

To get it straight, here is the argument which I have filled in: by the induction hypothesis we know that the projection of to is surjective. This means that, given fixed, for every there exists a such that . But the orbit of the latter point under is just the orbit of the former point shifted by . Since both orbits are dense in by minimality, we see that is invariant under translation by , so that is contained in the projection of .

I have also found the following modification of your argument for Exercise 2:

consider with the uniform topology and the (continuous) action of thereon via . For every minimal , the element is almost periodic in . This means that the set

is syndetic for every positive . But since is surjective, the latter set is equal to

.

Uniform continuity of for every finite concludes the argument.

best regards,

pavel

16 August, 2011 at 9:50 am

Terence TaoDear Pavel,

This is a nice argument, although I’m a little worried about whether you are assuming implicitly that Y is compact in the uniform topology (which, by Arzela-Ascoli, is very close to the equicontinuity that one is trying to prove!). It is easy to show that Y is compact in the weak topology and closed in the uniform topology, but to get compactness requires some argument (and without compactness, one has to be careful regarding the different formulations of almost periodicity).

16 August, 2011 at 3:55 pm

pavel zorinThis is trickier than I believed it would be and I only have a partial result for the moment. is compact and right topological for the pointwise topology, but also left topological since it consists of homeomorphisms. Moreover, algebraically is a group, since whenever is a minimal idempotent and , we have that . By Ellis’ Theorem, is a topological group, i.e. the multiplication is jointly continuous. But contains a dense commutative subset, hence is commutative.

Assume now that is minimal. Then the action of is transitive and we see that is a Kronecker system (more precisely, the quotient of by the stabilizer of any point).

I hope that the commutativity will help to show that is compact for the uniform topology too. Unfortunately, it is no longer evident that is dense, so that structure theory of monothetic groups is not immediately available.

best regards,

pavel

24 July, 2012 at 12:08 pm

RexFor Exercise 6, what does it mean for an eigenfunction to be unimodular? I didn’t see the definition anywhere…

24 July, 2012 at 10:11 pm

Terence TaoIn this context, “unimodular” means that it takes values in S^1 (i.e. it has unit modulus).

25 July, 2012 at 9:22 am

RexWhat is an example of an isometric extension where is not an isometric system?

25 July, 2012 at 2:19 pm

Terence TaoSee Example 5.

For a less trivial example, let be a system, let be a compact metric space, and for each , let be an isometry that varies continuously in y (using the uniform topology in K). Then the skew-product system with shift is an isometric extension of Y.

9 June, 2015 at 11:59 am

Clemens MDear Prof. Tao

I don’t see why the inverse limit factor space is metrisable. Could you provide some insight on that?

10 June, 2015 at 12:57 am

Terence TaoIn this set of notes, topological dynamical systems are always understood to be metrisable (see Notes 1). The space Y is a factor of X, which is assumed metrisable, and thus inherits a metric (Hausdorff distance between the fibres in X).

11 June, 2015 at 12:10 am

Clemens MMore specifically, I am trying to understand why the space as defined in equation (1) is metrisable. From your answer I understand that the Hausdorff distance could be used as a metric for the space. But I am probably still misunderstanding something, because I don’t see how this metric could generate the correct topology.

Consider for example with $T = id$ and $Y$ the quotient space obtained by identifying all points in the interval with each other. Then this quotient space is clearly metrisable, but the Hausdorff metric does not seem to be the correct choice. Each one-point set has Hausdorff distance at least from , so with respect to the Hausdorff distance, would be an open set whereas in the quotient topology it isn’t.

12 June, 2015 at 12:21 am

Terence TaoHmm, fair enough – but one can still show that any Hausdorff quotient of a compact metric space is metrisable by appealing to Urysohn’s metrisation theorem (take a countable dense subset of the original space, and consider -neighbourhoods of the fibres they lie in, to obtain a countable base for the quotient).

12 June, 2015 at 5:16 am

Clemens MThank you very much for answering my questions!

To be honest, this approach with the -neighbourhoods of fibres did not work for me because I was not able to show that for any open set in the quotient space and every there is a fibre from the countable collection of fibres such that for some .

However, I found a similar proof in a book in the library yesterday (Willard – General Topology), so I see now that the space is metrisable.

12 March, 2019 at 3:58 am

maths studentIt felt like compact Hausdorff uniform spaces should have been metrisable (ie. countable entourage system), but sadly with the unit interval seems to be a counterexample… Too bad, would’ve simplified the def. of Kronecker systems…