Ergodicity of a ne automorphismssimply connected nilpotent Lie group and is a lattice in G. De...

Chapter 3

Ergodicity of affineautomorphisms

3.1 Linear automorphisms of nilmanifolds

Definition 3.1.1. An automorphism of a Lie group G is a diffeomor-phism Ψ : G → G that is also a group isomorphism. An automorphismof a Lie algebra g is a linear isomorphism ψ : g→ g that also a Lie algebramorphism.

Definition 3.1.2. An automorphism (or linear automorphism, whenit is necessary to distinguish from affine automorphisms, which will definedlater) of a compact nimanifold M = G/Γ is a diffeomorphism f : M → Mthat lifts to an automorphism of the universal cover G of M . Here G is asimply connected nilpotent Lie group and Γ is a lattice in G.

Definition 3.1.3. An automorphism of a discrete group Γ is a groupisomorphism Ψ : Γ→ Γ.

Let G, g, Γ and M be as above. Their groups of automorphisms arerespectively denoted by Aut(G), Aut(g), Aut(Γ), Aut(M).

We know from Corollary 1.1.13 that the derivative DeΨ of a Lie groupautomorphism is an automorphism of g = Lie(G). By the same corollary andthe bijectivity of the exponential map, we also know that Ψ is determinedby DeΨ. In addition, Corollary 1.1.16 shows that every ψ ∈ Aut(g) is thederivative of some Ψ. To summarize, we have

Lemma 3.1.4. De : Aut(G)→ Aut(g) is bijective.

47

CHAPTER 3. ERGODICITY OF AFFINE AUTOMORPHISMS 48

We now give a characterization of Aut(G), or more generally, all Liegroup morphisms from G to itself.

Proposition 3.1.5. Suppose Ψ : G → G is a Lie group morphism of asimply connected nilpotent Lie group G, and X is a Mal’cev basis of g adaptedto a filtration {gi}s+1

i=1 . Then in the coordinate system (1.16),(1) Ψ is represented by the matrix ψ = DeΨ, which has the block form

ψ1 ∗ · · · ∗ψ2 ∗

. . ....ψr

, (3.1)

where the i-th block corresponds to the ui component of a vector u and isspanned by Xm−mi+1, · · · , Xm−mi+1.

(2) ψ2, · · · , ψr are determined by ψ1.(3) Every eigenvalue of ψi is the product of i eigenvalues of ψ1, possibly

with repetition.

Proof. (1) This is because of the fact thatDeΨ is a Lie algebra morphism andpreserves each g(i) in the lower central series. (Recall thatXm−mi+1, · · · , Xm

is a basis of g(i).)(2) Denote by πi the natural projection from g(i) to g(i)/g(i+1). Then

claim is equivalent to that: for all 1 ≤ i ≤ s and any given W ∈ g(i), theprojection πi(ψW ) ∈ g(i)/g(i+1) is determined by ψ1. Because g(i) is spannedby vectors of the form [Y1, [Y2, · · · , [Yi−1, Yi] · · · ]], it suffices to prove for avector W of this form. In this case ψW = [ψY1, [ψY2, · · · , [ψYi−1, ψYi] · · · ]].

Observe that if any of the Yj ’s is replaced by a translate Yj + V whereV ∈ g(2), then the bracket vector [Y1, [Y2, · · · , [Yi−1, Yi] · · · ]] is translated by[Y1, [Y2, · · · , [V, · · · [Yi−1, Yi] · · · ]], which is in g(i+1) by Lemma 1.4.3. There-fore πiW depends only on π1Y1, · · · , π1Yi. Similarly, since ψ preserves g(2),πi(ψW ) depends only on π1(ψY1), · · · , π1(ψYi). Note that by construction,π1(ψYj) is in turn determined by ψ1 and ψ1Yj . Thus when the Yj ’s aregiven, πi(ψW ) depends only on ψ1. This proves part (2).

(3) Remark that ψi is isomorphic to the morphism ψ induces on g(i)/g(i+1).If λ is an eigenvalue of ψi, if and only if the subspace

Eλi := {V ∈ g(i) : (ψ − λId)nV ∈ g(i+1) for sufficiently large n},

which contains g(i+1), is strictly larger than g(i+1). Here the exponent ncan be replaced with any positive integer greater than or equal to di =


dim(g(i)/g(i+1)), but using an unspecified large exponent will be more con-venient for the calculation below. Moreover, with Λ1 denoting the set ofeigenvalues of ψ1,

g(i)/g(i+1) =⊕λ∈Λ1

(Eλi /g(i+1)).

Because g(i) is spanned by vectors of the form [Y1, [Y2, · · · , [Yi−1, Yi] · · · ]],it is spanned by the subspaces

Eλ1,··· ,λii = span {[Y1, [Y2, · · · , [Yi−1, Yi] · · · ]] : Yj ∈ E

λj1 },

where the (λ1, · · · , λi) ∈ Λi1. Since ψ preserves the filtration g(i), all the

Eλi ’s, and the Eλ1,··· ,λii ’s, are ψ-invariant.

To conclude the proof, it suffices to prove that

Eλ1,··· ,λii ⊆ Eλ1+···+λi

i . (3.2)

This will in turn follow inductively from the claim that

[Eλ1 , Eµj ] ⊆ Eλµj+1. (3.3)

To show (3.3), let X ∈ Eλ1 , Y ∈ Eλj . Remark that

(ψ − λµId)[X,Y ] =ψ[X,Y ]− λµ[X,Y ] = [ψX,ψY ]− [λX, µY ]

=[(ψ − λId)X,ψY ] + [λX, (ψ − µId)Y ].(3.4)

The vectors (ψ − λId)X and λX are still in Eλ1 , and ψY and (ψ − µId)Yremain in Eλj . Furthermore ψ commutes with (ψ − λId). Therefore, byiterating (3.4) n times, we will see that

(ψ − λµId)n[X,Y ] =

n∑k=0

(n

k

)[λk(ψ − λId)n−kX,ψn−k(ψ − λId)kY ].

When n is sufficiently large, either k or n−k is sufficiently large. So the k-thterm belongs to either [g(2), g(j)] or [g(1), g(j+1)]. By Lemma 1.4.3, both ofthese are in g(j+2) and therefore (ψ−λµId)n[X,Y ] ∈ g(j+2) for all sufficiently

large n. In other words, [X,Y ] ∈ Eλµj+1. This establishes (3.3), and (3.2)then follows.

We will show below that Aut(Γ) is the same as Aut(M).

Lemma 3.1.6. Aut(Γ) ∼= Aut(M) ∼= {Ψ ∈ Aut(G) : Ψ(Γ) = Γ}.


Proof. We first prove Aut(Γ) ∼= {Ψ ∈ Aut(G) : Ψ(Γ) = Γ}. Clearly, if Ψ ∈Aut(G) satisfies Ψ(Γ) = Γ then the restriction of Ψ on Γ is an automorphism.We now show every element of Aut(Γ) can be uniquely realized in this way.

We can follow the proof of the “only if” part of Theorem 2.3.1 to con-struct a Mal’cev basis X of g adapted to the lower central series {g(i)},that satisfies Claim 2.3.2. In particular, Γ is generated by γj = expXj ,1 ≤ j ≤ m. By Porposition 1.6.6, the map φ in (1.19) is a diffeomorphismfrom Rm to G. Suppose A is an automorphism of Γ, define Ψ : G→ G by

Ψ(φ(u)) = exp(u1 exp−1A(γ1)) · · · exp(um exp−1A(γm)).

Then, by Proposition 1.6.6 again, Ψ is a polynomial map in terms of thecoordinates (1.16) or (1.19).

By Claim 2.3.2.(2), Γ = φ(Zd). And because A is an automorphism ofΓ, Ψ coincides with A on Γ. In particular Ψ(Γ) = Γ.

The map Ψ(g)Ψ(h)Ψ(gh)−1 is polynomial and equals identity when g, h ∈Γ. Since Γ is Zariski dense in G, so is Γ × Γ in G × G. It follows thatΨ(g)Ψ(h) = Ψ(gh) for all gh. Hence Ψ is a Lie group morphism extendingA. One can similarly define a Lie group morphism Ψ that is polynomial inthe coordinates (1.16) and extends A−1 to G. Then Ψ◦Ψ′ and Ψ◦Ψ are Liegroup morphisms from G to itself and coincide with the identity map on Γ.Because both Ψ ◦ Ψ and Ψ ◦Ψ are polynomial, this implies they agree withthe identity map on G. Therefore Ψ is invertible and belong to Aut(G).

Finally, suppose Ψ,Ψ′ ∈ Aut(G) both preserve Γ and coincide with Aon Γ. Because they are polynomial maps by Lemma 1.6.4 and Γ is Zariksidense in G, Ψ = Ψ′. This shows the uniqueness of the realization of everyA ∈ Aut(Γ) in Aut(G). The claim is established.

We then show Aut(M) ∼= {Ψ ∈ Aut(G) : Ψ(Γ) = Γ}. This is becauseevery element f of Aut(M) descends from an automorphism Ψ of G. Itsuffices to show what Ψ preserves Γ. Note that for two points gγ and g inG that descends to the same point gΓ ∈ M , Ψ(gΓ) = Ψ(g)Ψ(γ) and Ψ(g)should project to the same point in M in order for f to be well-defined.Therefore Ψ(γ) ∈ Γ for all γ ∈ Γ.

From now on, we will always identify Aut(Γ) and Aut(M) with the abovesubgroup of Aut(G).

Example 3.1.7. For Td = Rd/Zd,

Aut(Td) ∼= Aut(Zd) ∼= {A ∈ Aut(Rd) : A.Zd = Zd}={A ∈ GL(d,R) : A.Zd = Zd} = GL(d,Z).


Since every Lie group morphism Ψ : G → G preserves the lower cen-tral series {G(i)}. Hence for all indices j < i, Ψ induces an autormorphism

Ψ(i)(j)of the quotient group G(j)/G(i). It follows from the lemma above that

all elements Ψ of Aut(M) preserves the subgroups Γ(i) ⊆ Γ. Hence for allindices j < i, Ψ induces an autormorphism of the quotient group G(j)/G(i),that preserves the lattice Γ(j)/Γ(i), and hence an automorphism of the cor-responding quotient, which is the compact nilmanifold G(j)/G(i)Γ(j).

Definition 3.1.8. A number λ ∈ C is an algebraic integer if it is theroot of an integer polynomial p(x) = xk + ak−1x

k−1 + · · · + a0 ∈ Z[x] withleading coefficient 1. It is an algebraic unit if in addition ak ∈ {±1}.

Corollary 3.1.9. If Ψ ∈ Aut(M), then all the eigenvalues of ψ = DeΨ arealgebraic units.

Proof. By Proposition 3.1.5, it suffices to write DeΨ in the form (3.1) andprove that the eigenvalues of each ψi are algebraic units. In this case,ψi is isomorphic to the isomorphism of the abelian Lie algebra g(i)/g(i+1),which is identified with the abelian Lie group G(i)/G(i+1) via the exponen-tial map. The lattice Γ(i)/Γ(i+1), which is the Z-span of the basis vectorsXm−mi+1, · · · , Xm−mi+1 of the i-th block of the current coordinate system,is ψi-invariant. Thus ψi ∈ GL(di,Z), which guarantees that it has onlyalgebraic units as eigenvalues.

A related fact to Corollary 3.1.9 is:

Lemma 3.1.10. If Ψ ∈ Aut(M), then DeΨ is a matrix with rational entrieswith respect to the Q-structure on g determined by Γ.

Proof. It suffices to show that DeΨ preserves the set of rational points ofg. By Corollary 1.1.13, this is equivalent to showing that Ψ sends rationalelements of G to rational elements. If g is rational, then gn ∈ Γ for somen ∈ N by Corollary 2.3.8. So Ψ(g) is also rational because Ψ(g)n = Ψ(gn)is also in Γ.

In addition to linear automorphisms, we define a larger automorphismgroup.

Definition 3.1.11. An affine automorphism of a compact nilmanifold M =G/Γ is a diffeomorphism T : M →M of the form Tx = gΨ(x) where g ∈ Gand Ψ ∈ Aut(M).


It is clear that affine automorphisms form a group, denoted by Aff(M),generated by linear automorphisms and translations, i.e. diffeomorphismsTg(x) = gx where g ∈ G.

Proposition 3.1.12. Every affine automorphism of a compact manifold Mpreserves the Haar measure mM on M .

Proof. Suppose first Ψ ∈ Aut(M) and ψ = DeΨi. From the proof of Corol-lary 3.1.9 above, we know that every ψi in the form (3.1) has determinant±1. In particular, ψ sends the volume form dvol on g to either dvol or−dvol. In consequence, ψ preserves the Haar measure mg, which then isidentified with mG and projects to mM .

Because every left translation Lg preserves mG, Tg preserves mM for allg ∈ G. The porposition follows as every affine automorphism has the formTg ◦Ψ.

Exercises

Exercise 3.1.1. Suppose A ∈ SL(d,Q) does not have roots of unity amongits eigenvalues. By Exercise 1.5.4, the Lie group G = {g ∈ SL(d,R) :limn→∞A

ngA−n = Id} is nilpotent.(1) Show that G is simply connected.(2) Show that Γ = G ∩ SL(d,Z) is a lattice in G.(3) Suppose B ∈ SL(d,Z) commutes with A. Show that g → BgB−1

induces an automorphism of the compact nilmanifold G/Γ.

Exercise 3.1.2. Show that the automorphism group Aut(H3) of the 3-dimensional Heisenberg Lie group is isomorphic to GL(2,R) nR2.

Exercise 3.1.3. Show that the automorphism group Aut(M) of the 3-dimensional Heisenberg nilmanifold is isomorphic to GL(2,Z) n Z2.

Exercise 3.1.4. Show that: Aff(M) ∼= Aut(M) n(G/(Γ ∩ C(G))

). Here

C(G) denotes the center of G.

3.2 Basic facts from ergodic theory

We aim to study the ergodic theory of affine automorphisms of a compactnilmanifold M . For this purpose, wet first define some basic notions ofergodic theory.


Definition 3.2.1. A topological dynamical system is a pair (X,T )where X is a compact metric space and T : X → X is a continuous map.A measure preserving dynamical system (X,T,B, µ) is a topologicaldynamical system (X,T ) equipped with a T -invariant σ-algebra B and a T -invariant B-measurable probability measure µ.

Example 3.2.2. By Proposition 3.1.12, if M = G/Γ is a compact nilman-ifold, T ∈ Aff(M), BM is the Borel σ-algebra of M , and mM is the uniqueleft-invariant probability measure on M , then (M,T,BM ,mM ) is a measurepreserving dynamical system.

Here B is T -invariant if T−1B ∈ B for all B ∈ B. A set B ∈ B is T -invariant modulo µ if T−1B = B up to a null set, i.e. µ((T−1B)4B) = 0.

Definition 3.2.3. A measure preserving dynamical system (X,T,B, µ) is:(1) ergodic if for all T -invariant (modulo µ) subset A ∈ B, µ(A) = 0

or 1.(2) weakly mixing if for all measurable subsets A,B ∈ B,

limN→∞

1

N

N−1∑n=0

∣∣µ(T−nA ∩B)− µ(A)µ(B)∣∣ = 0.

(3) mixing if for all measurable subsets A,B ∈ B,

limn→∞

µ(T−nA ∩B) = µ(A)µ(B).

(4) Kolmogrov mixing (K-mixing) if for every T -invariant σ-algebraA of B, the Kolmogrov-Sinai entropy of (X,A, T, µ) is positive unless µ(A) ∈{0, 1} for all A ∈ A.

The definition of K-mixing uses the definition of the Kolmogrov-Sinaientropy. We will give the precise definition in future sections. 1

For now, it suffices for the reader to know the following hierarchy:

Theorem 3.2.4. ergodic ⇐ weakly mixing ⇐ mixing ⇐Kolmogrov mixing

These are standard facts from ergodic theory. The first two implicationsare easy to prove. The last one can be found in e.g. [Par81, §3.3].

1The definition of K-mixing given here is actually called completely positive en-tropy (CPE). The original definition of K-mixing is a different one. Two theorems,respectively due to Pinsker and Rokhlin-Sinai, together assert that CPE is equivalent toK-mixing.


For a measure preserving dynamical system (X,T,B, µ), T induces anoperator on functions on X, which we still denote by T , by Tf = f ◦T . Theoperator is isometric on every Lp(µ) because T preserves µ.

Denote the inner production on L2(µ) by

〈f1, f2〉 =

∫f1f2dµ.

Because T is a unitary operator on L2(µ), two eigenfunctions with distincteigenvalues must be orthogonal.

The constant functions form a one dimensional (over R or C dependingon the context) subspace of eigenfunctions for T , with 1 being the eigenvalue.Write L2

0(µ) for the orthogonal complement of this subspace in L2(µ).

Theorem 3.2.5. The following are equivalent for a measure preserving dy-namical system (X,T,B, µ):

(1) (X,T,B, µ) is ergodic;(2) If f is measurable and Tf = f µ-a.e., then f = constant µ-a.e.(3) If f ∈ L2(µ) and Tf = f µ-a.e., then f = constant µ-a.e.(4) 1 is not an eigenvalue of T on L2

0(µ).(5) (von Neumann’s L2 Ergodic Theorem) 1

N

∑N−1n=0 T

nf converges to theconstant

∫fdµ in L2-norm for all f ∈ L2(µ).

Proof. (1)⇒(2) It suffices to prove for real valued functions. Suppose f :X → R satsifies Tf = f a.e. but is not constant a.e.. Then there exists Ysuch that A = {x ∈ X : f(x) ≤ R} has measure µ(A) ∈ (0, 1). However bythe T -invariance of f , µ((T−1A)4A) = 0. Hence A is T -invariant moduloµ. This shows T is not ergodic.

(2)⇒(3) is evident.(3)⇒(1) Note that if A is T -invariant modulo µ, then f = 1A satifies

Tf = f µ-a.e. So assuming (3), one will have either 1A = 0 or 1A = 1 a.e.,or equivalently µ(A) = 0 or µ(A) = 1.

(3)⇔(4) is obvious, because it is already known that constant functionsare eigenfunctions with eigenvalue 1.

(5)⇒(4) Suppose for contradiction that f is a non-constant functionfrom L2

0(µ) that is an eigenfunction for eigenvalue 1. That is, Tf = f a.e.Then 1

N

∑N−1n=0 T

nf = f . So by the claim (5), f =∫fdµ is a constant a.e,

contradicting the hypothesis on f .(3)⇒(5) Let V ⊆ L2(µ) be the closure of the subspace {Th − h : h ∈

L2(µ)}, and let V ⊥ be its orthogonal complement. Then since L2(µ) is aHilbert space, L2(µ) = V ⊕ V ⊥.


We first show that for g ∈ V , 1N

∑N−1n=0 〈Tnf, g〉 → 0. It suffices to show

this for g = Th− h where h ∈ L2(µ). In this case

1

N

N−1∑n=0

〈Tnf, g〉 =1

N

N−1∑n=0

(〈Tnf, Th〉 − 〈Tnf, h〉)

=1

N

N−1∑n=0

(〈Tnf, Th〉 − 〈Tn+1f, Th〉)

=1

N(〈f, Th〉 − 〈TN−1f, h〉 → 0.

We then show that for g ∈ V ⊥, 1N

∑N−1n=0 〈Tnf, g〉 = 〈f, g〉. Because

〈Tn+1f −Tnf, g〉 = 0, we have 〈Tnf, g〉 = 〈f, g〉 for all n. The claim follows.Combining the two previous facts, we know that 1

N

∑N−1n=0 〈Tnf, g〉 →

〈πV ⊥f, g〉 where πV ⊥ is the orthogonal projection from L2(µ) to V ⊥. Inother words, 1

N

∑N−1n=0 T

nf weakly converges to πV ⊥f in L2. Since∥∥∥∥∥T (1

N

N−1∑n=0

Tnf)− 1

N

N−1∑n=0

Tnf

∥∥∥∥∥L2(µ)

=

∥∥∥∥ 1

N(TNf − f)

∥∥∥∥L2(µ)

→ 0,

this implies πV ⊥f is T -invariant, and hence by property (3) a constant func-tion a.e. for all f ∈ L2. In particular, this shows V ⊥ = {constant functions}and thus V = L2

0(µ).For a function of the form f = Th− h, 1

N

∑N−1n=0 T

nf = 1N (TNf − f)→

0. Because such functions are dense in V = L20(µ), for every f ∈ L2

0(µ),1N

∑N−1n=0 T

nf → 0 in L2 norm.Finally, for all f ∈ L2(µ), f −

∫fdµ ∈ L2

0(µ). So

1

N

N−1∑n=0

Tnf =

∫fdµ+

1

N

N−1∑n=0

Tn(f −∫fdµ)→

∫fdµ+ 0 =

∫fdµ

in L2(µ).

The most important theorem in ergodic theory is Birkhoff’s ergodic the-orem, which strengthens L2 ergodic theorem.

Theorem 3.2.6 (Birkhoff’s Pointwise Ergodic Theorem). If (X,T,B, µ) isergodic and f ∈ L1(µ), then 1

N

∑N−1n=0 T

nf pointwisely converges to∫fdµ

at µ-almost every point.


All T -invariant B-measurable probability measures form a convex set ofwhich the ergodic measures are the extremal points. When this set degen-erates into a single point, we have the following definition

Definition 3.2.7. Given (X,B), a continuous map T : X → X is uniquelyergodic if there is only one T -invariant B-measurable probability measureon X.

The next theorem says every point in this convex set can be uniquelywritten as a convex combination of the extremal points.

Theorem 3.2.8 (Ergodic decomposition). In a measure preserving dynam-ical system (X,T,B, µ), there is a measurable map that assignes to µ-almostevery x ∈ X an ergodic T -invariant probability measure µEx, called the er-godic component of µ , such that µ =

∫µExdµ.

The construction of µEx is not hard. First let A ⊆ B be the σ-subalgebraof all subsets that are T -invariant modulo µ. Then for every f ∈ L2(B, µ)there is a conditional expectation function E(f |A) : X → C such that:

1. E(f |A) is A-measurable;

2.∫B E(f |A)(x)dµ(x) =

∫B f(x)dµ(x);

3. E(f |A)(x) ≥ 0 if f ≥ 0.

The function E(f |A) is in fact the orthogonal projection of f from the Hilbertspace L2(B, µ) to the closed subspace L2(A, µ). For each given x, the cor-respondence IAx (f) = E(f |A)(x) from L2(B, µ) to C is a linear functionaland one can show that |E(f |A)(x)| ≤ ‖f‖L∞ . In particular, one can restrictIAx to C(X) ⊆ L2(B, µ) and get a bounded linear functional on C(X). Be-cause X is compact, by Riesz representation theorem, there is a measureµAx such that E(f |A)(x) = IAx (f) =

∫fdµAx for all f ∈ C(X). The ergodic

component is given by µEx = µAx . In general, for all Lebesgue probabil-ity measure spaces (X,B, µ) and countably generated σ-subalgebra A ⊆ B,one can construct conditional measures µAx in this way (See, for example,[?EW11, Theorem 5.14]).

We next state some equivalence definitions of weak mixing and mixingwithout proofs.

Theorem 3.2.9. The following are equivalent for a measure preserving dy-namical system (X,T,B, µ):

(1) (X,T,B, µ) is weakly mixing;


(2) For all functions f, g ∈ L20(µ), limN→∞

1N

∑N−1n=0

∣∣〈Tnf, g〉∣∣ = 0.(3) (X ×X,T × T,B × B, µ× µ) is ergodic;(4) (X ×X,T × T,B × B, µ× µ) is weakly mixing;(5) T does not have eigenvalues on L2

0(µ).

Theorem 3.2.10. The following are equivalent for a measure preservingdynamical system (X,T,B, µ):

(1) (X,T,B, µ) is mixing;(2) For all functions f, g ∈ L2

0(µ), limN→∞∣∣〈Tnf, g〉∣∣ = 0.

Theorems 3.2.8-3.2.10 can be found in, for instance, [Par81, Ch. 1 & 3].

Definition 3.2.11. A measure preserving dynamical system (Y, S,A, ν) isa factor of another measure preserving dyamical system (X,T,B, µ) if thereis a measurable surjective map π : (X,B)→ (Y,A) satisfying π ◦ T = S ◦ πand π∗µ = ν.

Notice that each factor system can be realized within X by restrictingto a σ-subalgebra: (Y, S,A, ν) is isomorphic to (X,T, π−1A, µ). It followseasily that if T is ergodic (resp. weakly mixing, mixing) then so is S.

Lemma 3.2.12. If π : (X,B) → (Y,A) is a measurable surjective mapbetween two measurable spaces where X and Y are compact metric spaces.If homeomorphisms T : X → X and S : Y → Y satsify π ◦ T = S ◦ π. Thenfor every ergodic S-invariant probability measure ν, there is an ergodic T -invariant probability measure µ such that π∗µ = ν.

Proof. By Birkhoff’s Ergodic Theorem, ν-a.e. y is generic for ν in the sensethat 1

N

∑N−1n=0 δSny converges to ν in weak-∗ topology. For every x that

lifts y, a subsequence of 1N

∑N−1n=0 δSny weak-∗ converges because the set of

probability measures on a compact metric space is compact in the weak-∗

topology. Any such sequential limit is a T -invariant probability measurethat converges to ν, and so are all its ergodic components because of theergodicity of ν. Let µ be such an ergodic component.

Exercises

Exercise 3.2.1. Show that by Definition 3.2.3, weak mixing implies ergod-icity.

Exercise 3.2.2. Show that the translation Tx = x + b of the torus Td,where b = (b1, · · · , bd) ∈ Td, is ergodic with respect to the Haar measuremTd if and only if 1, b1, · · · , bd are linearly independent over Q. Moreover,show that T is ergodic if and only if it is uniquely ergodic.


Exercise 3.2.3. Show that the translation in Exercise 3.2.3 is not weaklymixing.

3.3 Ergodicity of affine automorphisms of tori

Notation 3.3.1. In the future, when we say an affine automorphism T ofa nilmanifold M is ergodic, it will be by defaut with respect to the probabilityHaar measure mM . For simplicity, the measures L2(mM ) and L2

0(mM ) willbe denoted by L2(M) and L2

0(M).

In this section, we will give necessary and sufficient conditions for anaffine automorphism f of Td to be ergodic.

Recall that T ∈ Aff(Td) has the form Tx = Ax+b where A is an elementof Aut(Td), which by Example 3.1.7 can be identified with GL(d,Z), andb ∈ Rd. Note that b can be chosen from Td instead of Rd.

As detA = 1, all eigenvalues of A are algebraic units. We first show ansufficient condition:

Lemma 3.3.2. If A ∈ GL(d,Z) does not have roots of unity among itseigenvalues, then for all b ∈ Rd, the affine automprhism Tx = Ax+ b of Tdis ergodic.

Proof. By Theorem 3.2.5 it suffices to show that 1 is not among the eigen-values of f on L2

0(Td).The Fourier modes {e(ξ · x) : ξ ∈ Zd} form an orthonormal basis of

L2(Td).2 Those with ξ 6= 0 form an orthonormal basis of L20(Td). Remark

thate(ξ · Tx) = e(ξb)e(ξ ·Ax) = e(ATξ · x). (3.5)

Assume f(x) ∈ L20(Td) is a non-zero eigenfunction with eigenvalue 1;

that is Tf = f . Take the Fourier series f(x) =∑

ξ∈Zd\{0} f(ξ)e(ξ · x) in L2,then

Tf(x) =∑

ξ∈Zd\{0}

f(ξ)e(ξb)e(ATξ · x)

=∑

ξ∈Zd\{0}

f((AT)−1ξ)e((AT)−1ξ · b)e(ξ · x).(3.6)

Comparing Tf with f , we know that

f(ξ) = f((AT)−1ξ)e((AT )−1ξ · b) (3.7)

2Here and in the remainder of these notes, e(z) will denote the function e2πiz, whichcan be regarded as a function from either R or T1 to the unit circle in C.


for all ξ ∈ Zd\{0}. In particular, |f(ξ)| = |f((AT)−1ξ)|, and thus |f(ξ)| =|f((AT)−nξ)| for all n ∈ N by iteration.

Since f is non-trivial, we can fix a ξ∗ 6= 0 such that f(ξ∗) 6= 0. Because‖f‖L2 =

∑ξ |f(ξ)|2, there are only finitely many ξ with |f(ξ)| ≥ |f(ξ∗)|.

In particular, (AT)−nξ∗ is in this finite set for all n ∈ N. So (AT)−nξ∗ =(AT)−mξ∗ for some m,n with m > n. This shows ξ∗ is a fixed vectorof (AT)m−n. So 1 is an eigenvalue of (AT)m−n, and thus of Am−n. Itfollows that among the eigenvalues of A, there must be an (m− n)-th rootof unity.

The following is a necessary condition, weaker than the sufficient condi-tion above.

Lemma 3.3.3. If an affine automprhism Tx = Ax + b of Td is ergodic,then A ∈ GL(d,Z) does not have any roots of unity other than 1 among itseigenvalues.

Proof. Assume for contradiction that u 6= 1 is a primitive k-th roots of unity,where k > 1, and at the same time an eigenvalue of A. Then u is also aneigenvalue of AT.

Let V =⊕

λ 6=1 ker(AT − λId)d be the direct sum of all generalizedeigenspaces whose corresponding eigenvalue is not 1. Then V is a rationalsubspace of Rd. Moreover, the eigenspace ker(AT−uId) of AT correspondingto the eigenvalue u is a subset of

V ∩(

ker((AT)k − Id

)\k−1⋃j=1

ker((AT)j − Id

)). (3.8)

So (3.8) is non-empty. Because AT ∈ GL(d,Z), every ker((AT)j − Id) is arational vector subspace, and thus one can find a non-zero rational vectorξ∗ in (3.8) above. By rescaling, one may assume ξ∗ ∈ Zd\{0}.

Then (AT)kξ∗ = ξ∗, but ξ∗, ATξ∗, · · · , (AT)k−1ξ∗ are all distinct. Thevector

∑k−1j=0(AT)jξ∗ is a fixed vector of AT in V . Since 1 is not an eigen-

vector of AT on V by construction,∑k−1

j=0(AT)jξ∗ = 0.

We will define Fourier coefficients at fequencies ξ∗, ATξ∗, · · · , (AT)k−1ξ∗

by f((AT)jξ∗) = e(∑j−1

l=0 (AT)lξ ·b). Note that this definition holds for j = k

as well, in which case f((AT)kξ∗) = f(ξ∗) = 1 = e(∑k−1

l=0 (AT)lξ · b). By thisdefinition, (3.7) holds at the frequency ξ = (AT)jξ∗ for j = 0, · · · , k − 1.Therefore the function f(x) =

∑k−1j=0 f((AT)jξ∗)e((AT)jξ∗ · b) satisfies Tf =

f . Moreover, since these k frequecies are distinct and |f((AT)jξ∗)| = 1 for


each j, f is a non-trivial function in L20(Td). Thus 1 is an eigenvalue for T

in L20(Td), and by Theorem 3.2.5, T is not ergodic.

Lemma 3.3.2 and Lemma 3.3.3 leave to us the case when 1 is the onlyroot of unity among the eigenvalues of A. We will focus on this case below.

Lemma 3.3.4. If the eigenvalues of A include 1 but no other roots of unity,then an affine automorphism Tx+ b of Td is not ergodic if and only if thereexists ξ ∈ Zd\{0} such that ATξ = ξ and ξ · b ∈ Z.

Proof. By proof of , T fails to be ergodic if and only if there exists an L2

function f satisfying (3.7) for all ξ ∈ Zd\{0}. This can happen only if foreach ξ with non-vanishing f(ξ), (AT)kξ = ξ for some k ∈ N. Because theeigenvalues of AT, which are the same as those of A, does not have rootsof unity of higher orders, we must have k = 1 and ATξ = ξ for all ξ withf(ξ) 6= 0. For these ξ, (3.7) becomes f(ξ) = f(ξ)e(ξ ·b), and thus e(ξ ·b) = 1.Equivalently ξ · b ∈ Z.

Conversely, if ATξ = ξ and ξ · b, then f(x) = e(ξ · x) is a non-trivialsolution to (3.7).

Summarizing the lemmas above, we have the following criterion for er-godicity:

Theorem 3.3.5. Suppose Tx = Ax + b is an affine automorphism of Tdwhere A ∈ GL(d,Z) and b ∈ R. Then T is ergodic if and only if:

A does not have roots of unity other than 1 among its eigenvalues, andfor every ξ ∈ Zd\{0} such that ATξ = ξ, ξ · b /∈ Z.

Example 3.3.6. When A =

(1 01 1

)and b = (α, 0), T is the affine

automorphism of T2 given by T (x, y) = (x + α, y + x) where x, y ∈ R/Z.

The only eigenvalue of AT =

(1 10 1

)is 1, and the eigenvectors in Zd\{0}

are ξ = (q, 0). By Theorem 3.3.5, T is ergodic if and only if for everyq ∈ Z\{0}, qα /∈ Z, i.e. α is irrational.

In this example, T projects to the rotation Sx = x+ α of the circle T1.By Exercise 3.2.3, S is ergodic if and only if α is irrational. In other words,the ergodicity of T is equivalent to that of S.

In general, suppose 1, but no other roots of unity, is among the eigenval-ues of A. The corresponding generalized eigenspace is V 1 = ker(A − Id)d.Using Jordan canonical form, we know that there exists 1 ≤ k ≤ d such that

{0} ( ker(A− Id) ( ker(A− Id)2 · · · ( ker(A− Id)k = V 1. (3.9)


Recall that Cd =⊕

ζ VζC where ζ ranges over all eigenvalues of A and

V ζC = kerCd(A − ζId)d is the generalized eigenspace. The subspace V 6=1

C =⊕ζ 6=1 V

ζC = kerCd

∏ζ 6=1(A − ζId)d is defined over Q as 1 is rational. Let

V 6=1 = Rd ∩ V 6=1C , which is an A-invariant rational subspace of Rd. The

subspace V = V 6=1⊕ker(A−Id)k−1 is again an A-invariant rational subspaceof Rd, and W1 = Rd/V = V 1/ ker(A− Id)k−1. Because (A− Id)v ∈ ker(A−Id)k−1 for all v ∈ V 1, A projects to the identity map on W1. In fact V isthe largest A-invariant subspace for which A induces the identity map onthe quotient.

The affine transform x → Ax + b projects to a translation x → x + b1on W1. Where b1 is the projection of V in W1. Because V is a rationalsubspace, Λ1 = Zd/(Zd ∩ V ) is a lattice in W1, and Y1 = W1/Λ1 is a torus,on which T induces the rotation S1(x) = x+ b1. The torus X1 is the largestquotient torus of Td on which T induces a rotation. (Y1, S1,mY1) is a factordynamical system of (Td, T,mTd), which we call the maximal rotationfactor.

Corollary 3.3.7. An affine toral automorphism Tx = Ax + b is ergodic ifand only if the eigenvalues of A include no non-trivial roots of unity and themaximal rotation factor of T is ergodic.

Proof. The “only if” direction follows from Theorem 3.3.5 and the ergodicityof factors. For the opposite statement, notice that under the assumptions,by Theorem 3.3.5 the map T can be non-ergodic only if for some ξ ∈ Zd\{0},ATξ = ξ and ξ · b ∈ Z.

We adapt the notations above. For v ∈ V ζC where ζ 6= 1, ξ ·v = ATξ ·v =

ξ ·Av = λ(ξ · v), and thus ξ · v = 0.On the other hand, one can see using Jordan canonical forms that for v ∈

ker(A− Id)k−1 there exists w ∈ ker(A− Id)k = V 1 such that v = (A− Id)w.Thus ξ · v = (AT − Id)ξ · w = 0 · w = 0.

It follows that ξ, as an element of (Rd)∗, vanishes on V and descends toa non-trivial linear functional ξ1 on W1. Moreover, ξ assumes integer valueson Λ1. Thus ξ1 is a Fourier frequency from the torus Y1 and ξ · b = ξ1 · b1.Thus ξ1 · b1 6= 0. Again by Theorem 3.3.5 , S1 is not ergodic, which yieldsthe desired contradiction.

Exercises

Exercise 3.3.1. If T (x1, x2, · · · , xd) equals

(x1 + α1, x2 + x1 + α2, x3 + x2 + α3, · · · , xd + xd−1 + αd)


on Td, show that T is ergodic if and only if α1 is irrational.

3.4 Ergodicity and unique ergodicity on principalbundles

In this section, suppose (X0, T0,B0, µ0) is an ergodic measure preservingdynamical system, K is a second countable compact abelian group, and Xis a left principal K-bundle over X0 equipped with the product σ-algebraB = B0 × BK , where BK is the Borel σ-algebra of K. We remark that theexpression B0×BK is indeed an abuse of notation asX is not a direct productbetween X0 and K. To define the product σ-algebra, we actually should lookat local charts U of X0, the preimage π−1(U) of which is homeomorphicto U ×K. On each of these preimages, we can define a product σ-algebraBπ−1(U) which is identified with B0|U×BK through the homeomorphism. Theglobal product σ-algebra B is the σ-algebra generated by all the Bπ−1(U)’s.

Let T : X → X be a measurable extension of T0, i.e. a measurablemap such that π ◦ T = T0 ◦ π, where π : X → X0 is the bundle projection.In addition, assume that T acts on the fibers by an automorphism A of K,i.e.

T (zx) = A(z)Tx,∀z ∈ K,µ-a.e. x ∈ X. (3.10)

Note π−1(B0) is the σ-algebra on X consisting of all unions of K-fibers.By abusing notation, we denote it indifferently by B0. Then for every proba-bility measure ν and ν-almost all x, νB0

x is supported on a single fiber, whichis the K-orbit of x.

The group K acts freely on the principal bundle X, write Lz(x) = zxfor z ∈ K and x ∈ X. For every x, the Haar measure mK on K pushesforward to a K-invariant measure on the orbit K.x. By invariance of mK ,this measure depends only on the orbit K.x but not on the choice of x withinthe orbit, and hence is a function on the underlying quotient X0. For eachx0 ∈ X0, which represents a K-orbit in X, denote this measure by mK,x0 .Then the measure

µ =

∫X0

mK,x0dµ0(x0) (3.11)

is equal to the direct product µ0×mK on local charts of the principal bundlestructure.

For every z ∈ K, as each mK,x0 is translation invariant, (Lz)∗µ = µ. Forf ∈ L2(µ), define Lzf = f ◦ Lz. Then Lz is an isometry of L2(µ)

Lemma 3.4.1. The measure µ in (3.11) is T -invariant.


Proof. T∗µ =∫X0T∗mK,x0dµ0(x0). As (T0)∗µ0 = µ0, it suffices to show

T∗mK,x0 = mK,T0x0 . Consider the fiber π−1(x0) above x0 in X, which isthe K-orbit that x0 represent, then T (π−1(x0)) is the fiber π−1(T0x0) asπ ◦ T = T0 ◦ π, and T∗mK,x0 is a probability measure on this fiber. Itremains to show T∗mK,x0 is the unique K-invariant probability measureon TF0. To see this, note that for z ∈ K, T ◦ Lz = LA(z) ◦ T . Hence

(Lz)∗T∗mK,x0 = T∗(L−1A (z))∗mK,x0 = T∗mK,x0 . The lemma is proved.

Write K for the Pontryagin dual of K, which is the group of all contin-uous characters ξ : K → T1. Then K is a countable discrete abelian groupand A induces a transpose AT on K by AT(ξ) · z = ξ ·A(z)

For any function f ∈ L2(µ), because f is L2-integrable with respect tomK,x0 along µ0-almost every fiber π−1(x0), one can perform a Fourier seriesdecomposition along the fibers by letting

f(ξ, x) =

∫z∈K

f(zx)e(−ξ · z)dz,∀ξ ∈ K. (3.12)

Then, in L2(µ)-sense, for x ∈M and z ∈ K,

f(zx) =∑ξ∈K

e(ξ · z)f(ξ, x). (3.13)

In particular,

f(x) =∑ξ∈K

f(ξ, x). (3.14)

Because the decomposition into eigenfunctions is unique, x → f(ξ, x) is aneigenfunction for the K-action in the sense that

f(ξ, zx) = e(ξ · z)f(ξ, x). (3.15)

Lemma 3.4.2. Assuming (3.10), then in the fiberwise Fourier series above,

T f(ξ, x) = f((AT)−1ξ, Tx).

Proof. Notice that Tf(x) =∑

ξ∈K f(ξ, Tx). By (3.15),

f(ξ, T (zx)) =f(ξ, A(z)Tx) = e(ξ ·A(z))f(ξ, Tx) = e(ATξ · z))f(ξ, Tx).

So f(ATξ, x) = T f(ξ, Tx). The lemma follows by a coordinate change inξ.


We now introduce an important property, essentially due to Furstenberg[Fur61], and extended by Parry [Par69b].

Theorem 3.4.3. In the above setting, assume that for all k ∈ N and ξ ∈ Ksuch that (AT)kξ = ξ, ATξ = ξ.

Then the implications (1)⇒ (2)⇒ (3) hold for the following properties:(1) µ is the unique T -invariant measure projecting to µ0;(2) There do not exist a non-trivial continuous character ξ : K → T1

and a measurable assignment to µ0-a.e. x0 ∈ X0 of a (ker ξ)-orbit Ex0 ∈π−1(x0)/ ker ξ inside the K-orbit π−1(x0), such that ATξ = ξ and TEx0 =ET0x0 for µ0-a.e. x0;

(3) µ is ergodic.Moreover, if in addition the automorphism A in (3.10) is unipotent, i.e.

(A− Id)n = 0 for some n, then the implication (3)⇒ (1) is also true.

When using the notation A − Id, we are thinking of K as an additiveabelian group.

Proof. (1)⇒(2): Suppose there exist ξ and the measurable assignment x0 →Ex0 as in (2). The probability Haar measure mker ξ on the compact abeliansubgroup ker ξ pushes forward to a ker ξ-invariant measure mker ξ,x0 on Ex0 ⊂π−1(x0).

Since ATξ = ξ, A ker ξ = ker ξ and A−1 ker ξ = ker ξ. In other words, Ais an automorphism of the compact abelian group ker ξ and hence preservesmker ξ.

It follows that, because for each z ∈ ker ξ and x ∈ π−1(x0), T (zx) =A(z)Tx where A(z) ∈ ker ξ, the pushforward T∗mker ξ,x0 is translation invari-ant under elements of ker ξ. Since it is supported on the orbit TEx0 = ET0x0 .Thus T∗mker ξ,x0 = mker ξ,T0x0 .

Define µ′ =∫X0

mker ξ,x0dµ0(x0). Then µ′ projects to µ0 and

T∗µ′ =

∫X0

T∗mker ξ,x0dµ0(x0) =

∫X0

mker ξ,T0x0dµ0(x0)

=

∫X0

mker ξ,x0d(T0)∗µ0(x0) =

∫X0

mker ξ,x0dµ0(x0) = µ′.

However, (µ′)B0x = mker ξ,x0 6= mK,x0 = µB0

x . So µ is not the unique T -invariant measure projecting to µ0, contradicting statement (1).

(2)⇒(3): Suppose T is µ is not ergodic and there is a non-constant func-tion f ∈ L2(µ) such that Tf = f almost everywhere3. Decompose f as in

3The argument here still works if f is a non-constant eigenfunction of T .


(3.12), then there is a non-zero ξ such that f(ξ, x) 6= 0 in L2(µ). By Lemma3.4.2, f(ATξ, x) = h(ξ, Tx) has the same norm in L2(µ). Because the Fouriermodes f(ξ, x) are orthogonal for different ξ’s, and f is L2-integrable, ξ musthave a finite orbit under AT. By assumption AT ξ = ξ.

By (3.15), |f(ξ, x)| is constant along each K-orbit, and thus factorsthrough a function a on X0. Moreover, a is T0-invariant modulo µ0 as f(ξ, x)is T -invariant in L2(µ). So by ergodicity of (T0, µ0), a is µ0-almost every-where constant. Thus without loss of generality, we may assume |f(ξ, x)| = 1for µ-almost all x ∈ X.

For each x0, define Ex0 = {x ∈ π−1(x0) : f(ξ, x) = 1} ⊂ π−1(x). Thanksto (3.15), Ex0 is a ker ξ-orbit. Moreover, TEx0 = ET0x0 by the T -invariance

of the function f(ξ, x). This contradicts claim (2) and therefore we haveshowed the implication (2)⇒(3).

(3)⇒(1) when A is unipotent: We first prove this in the special casewhere A = Id. Suppose a T -invariant probability measure ν projects to µ0

on X0. Decompose ν =∫νB0x dν(x) into conditional measures along fibers.

Because x → νB0x is F-measurable, νB0

x depends only on the fiber, i.e. theprojection x0 of x in X0. In this case we write νB0

x0= νB0

x . Then

ν =

∫XνB0

π(x)dν(x) =

∫XνB0x0

dπ∗ν(x0) =

∫X0

νB0x0

dµ0(x0).

Consider the average

ν =

∫K

(Lz)∗νdmK(z) =

∫x0∈X0

∫z∈K

(Lz)∗νB0x0

dmK(z)dµ0(x0).

Because∫z∈K(Lz)∗ν

B0x0

dmK(z) is K-invariant on the K-orbit over x0, itis equal to mK,x0 , thus ν = µ. Because A = Id, T commutes with thetranslation Lz by z ∈ K. Hence (Lz)∗ν is also T -invariant for all z ∈ K.So µ is an average of T -invariant probability measures (Lz)∗ν. Since µ isergodic, these measures coincide for mK-almost all z. That is, (Lz)∗ν = νfor almost all z, and µ = ν = ν. The unique ergodicity is established.

Now assume (A− Id)n = 0 instead. When n = 1 this was proved above.For induction, suppose the claim is known when (A − Id)n−1 = 0. ThenA preserves the close subgroup ker(A − Id) and induces an automorphismA1 on the quotient group K1 = K/ ker(A − Id). Because for all z ∈ K,(A− Id)n−1(z) ∈ ker(A− Id), (A1 − Id)n−1 = 0.

Let X1 = X/ ker(A− Id) be the space of ker(A− Id)-orbits on X. ThenX1 is an intermediate K1-principal bundle over X0. Because T (zx) =


A(z).Tx = z.Tx for all z ∈ ker(A − Id), T sends ker(A − Id)-orbits toker(A − Id)-orbits, and thus descends to a map T1 on X1. The map T1

further projects to T0. Let π1 : X → X1 and π0 : X1 → X0 be the bundleprojections and µ1 = (π1)∗µ. Then µ is an ergodic T1-invariant measureand projects to µ0.

For x1 ∈ X1 and z1 ∈ K1, choose any lifts x and z respectively in Xand K. The equality T (zx) = A(z).Tx yields T1(z1x1) = A1(z1).T1(x1)after projection. So by inductive assumption, µ1 is the unique T1-invariantmeasure projecting to µ0.

Now X is a ker(A−Id)-principal bundle over X1. Since T (zx) = z.Tx forall z ∈ ker(A− Id), by the special case A = Id, µ is the unqiue T1-invariantmeasure projecting to µ1.

For a T -invariant measure µ′ projecting to µ0, let µ′1 = (π1)∗µ′ be its

projection on the intermediate bundle X1. Because (π0)∗µ′1 = π∗µ

′ = µ0,µ′1 must be µ1 and it in turn follows that µ′ = µ. This establishes the proofof statement (1).

Corollary 3.4.4. An affine toral automorphism Tx = Ax + b on Td isuniquely ergodic if and only if A ∈ GL(d,Z) is unipotent and T is ergodic.

Proof. Suppose first A is unipotent and T is ergodic, i.e. its only eigenvalueis 1. Define the maximal rotation factor Y1 as in the last paragraphs of §3.3.Let k be the length of the flag as in (3.9). Then Rd = V 1 = ker(A − Id)k.For each 1 ≤ j ≤ k, ker(A − Id)k−j is a rational subspace. More generally,we can define a torus Yj = Wj/Λj , where Wj = Rd/ ker(A − Id)k−j isa rational subspace, and Λj = Zd/(Zd ∩ ker(A − Id)k−1). T projects toan affine automorphism Tj on Yj , which is ergodic because T is. Note thatYk = Td. Then each Yj+1 is a torus principal bundle over Yj where the fiber isparametrized by a torus quotient of the vector space ker(A−Id)k−j/ ker(A−Id)k−j−1. Note that the map Aj+1 induced by A on Rd/ ker(A − Id)k−j−1

restricts to the identity map on ker(A − Id)k−j/ ker(A − Id)k−j−1. Thusfor all z ∈ ker(A − Id)k−j/ ker(A − Id)k−j−1 and x ∈ Yj+1, Tj+1(x + z) =Aj+1(x+z)+b = Tj+1x+z. So by the implication (3)⇒(1) in Theorem 3.4.3,if Tj is uniquely ergodic, then Tj+1 is uniquely ergodic as Tj+1 is ergodic.Becaue T1 is an ergodic rotation of Y1, it is uniquely ergodic (Exercise 3.2.3).It follows by induction that all the Tj ’s, including T = Tk, are uniquelyergodic.

Conversely, suppose T is uniquely ergodic. Then it clearly must alsobe ergodic with respect to mTd (otherwise mTd splits into distinct ergodiccomponents). Assume A is not unipotent, then the subspace V 1 = ker(A−


Id)k in (3.9) is a proper rational A-invariant subspace of Rd. The affinetranform x→ Ax+ b projects from Rd to an affine transform x→ A′x+ b′

on W = R/V 1 where A′ has no eigenvalue 1. Moreover Λ = Zd/(Zd ∩ V 1)is a lattice in W , and x → A′x + b′ induces an affine automorphism S ofthe torus Y = W/Λ. Thanks to Lemma 3.2.12, it suffices to prove that S isnot uniquely ergodic. Equivalently, we need to show: if A ∈ GL(d,Z) doesnot have eigenvalue 1, then the affine automorphism Tx = Ax + b is notuniquely ergodic.

In this case, note that A − Id is an invertible matrix. For v = −(A −Id)−1b ∈ Rd, v = Av + b and therefore v projects to a fixed point of T inTd. The point mass at this point is a T -invariant probability measure otherthan mTd . So T is not uniquely ergodic.

Exercises

Exercise 3.4.1. Let T : T2 → T2 be the continuous map

T (x, y) = (x+ α, y + x2) mod Z2,

where α is irrational and the torus T2 is parametrized by [0, 1)2. Show thatT is ergodic.

Exercise 3.4.2. Prove Corollary 3.3.7 using Theorem 3.4.3.

3.5 Ergodicity and unique ergodicity of affine nil-manifold automorphisms

Suppose M = G/Γ is a compact manifold. Let M1 = G/G(2)Γ be thehorizontal torus of M , which is a quotient of M . Let T ∈ Aff(M) be anaffine automorphism of M , then Tx = bA(x) where b ∈ G and A ∈ Aut(M).By Lemma 3.1.6, A can be viewed as an automorphism of G that preservesΓ. Since A preserves both G(2) and Γ, it projects to a linear automorphismA1 ∈ Aut(M1). Thus T projects to the affine transform T1x = g1A1(x)where g1 is the projection of g in G/G(2). Moregenerally, for each 1 ≤ j ≤ s,T projects to Ti ∈ Aut(Mi), where Mi = G/G(i+1)Γ, and Tix = giAi(x),where gi and Ai are respectively the projections of g and Ai to G/G(i+1).

The main result of this section is:

Theorem 3.5.1 (Parry [Par69a]). T is ergodic if and only if T1 is.


Because T1 is a factor of T , the “only if” part is a general property offactor maps. So only the “if” direction needs to be proved. We now assumeT1 is ergodic. Then no eigenvalue of T1 on L2

0(M1) is equal to 1. Thetheorem would follow from the following fact:

Proposition 3.5.2. If T1 is ergodic, then every eigenfunction of T onL2

0(M) has the form f1 ◦ π where f1 is an eigenfunction of T1 on L20(M1)

and π : M →M1 is the projection.

We now reduce the proposition to a simpler setting.Recall that M = Ms is a principal Tds-bundle over Ms−1 where ds =

dimG(s). More precisely, the torus group G(s)/Γ(s)∼= Tds freely acts on Ms

with the quotient space modulo the orbit equivalence relation being Ms−1,where Γ(s) = G(s) ∩ Γ. Below, identify G(s) with Rds and Γ(s) with Zds .

Any function f ∈ L2(M) can be decomposed into a Fourier series along

the Tds-fibers by (3.12), where the Fourier frequencies are from Zds = Tds .Note that f ∈ L2

0(M) if and only if f(0, x) = 0.For z ∈ G(s),

T (zx) = bA(zx) = bA(z).A(x) = A(z)b.A(x) = A(z).Tx (3.16)

because A preserves G(s) and A(z) ∈ G(s) commutes with b. Thus Lemma3.4.2 applies in this seeting.

For g ∈ G, let (Lgf)(x) = f(gx). Because f is bounded, by Luzin’sTheorem, the map g → Lgf from G to L2

0(M) is continuous for a given f .

Proof of Proposition 3.5.2. First of all, let n ∈ N be such that the order ofevery root of unity among the eigenvalues of DeA ∈ Aut(g) divides n. ThenDeA

n has no root of unity other than 1 among its eigenvalues. Supposef ∈ L2

0(M) is an eigenfunction of T , then it is also an eigenvalue of Tn. Ifthe proposition holds for Tn, f factors through an eigenfunction f1 of Tn1on M1. Because f is an eigenfunction of T and T projects to T1 on M1,f1 must be an eigenfunction of T1. Hence the proposition also holds for T .Therefore, one may assume DeA has no root of unity other than 1.

The statement holds trivially if G is abelian. Suppose for induction thatG is s-step nilpotent and the proposition is true for all cases where the stepof nilpotency is bounded by s−1. In particular, every eigenfunction of Ts−1

in L20(Ms−1) factors through an eigenfunction of T1 in on L2

0(M1). Hencethe problem reduces to show that every eigenfunction f of T factors throughan eigenfunction of Ts−1.

Suppose for the sake of contradiction that f is not a constant and let θ bethe eigenvalue such that Tf = θf . Then it follows the proof of implication


(2)⇒(3) in Theorem 3.4.3 that ATξ = ξ and f(ξ, x) 6= 0 in L2 for someξ ∈ Zds\{0}. So by Lemma3.4.2, f(ξ, Tx) = λf(ξ, x). In other words,x→ f(ξ, x) is a non-constant eigenfunction for T .

Let kerTds ξ be the kernel of ξ on Tds , which is a closed subgroup ofdimension ds − 1 and the projection of ker ξ ∈ Rds ∼= G(s). Let G′ =G/ ker ξ and M ′ = M/ kerTds ξ, then G′ is a simply connected nilpotent Liegroup by Corollary 1.5.14, and M ′ = G′/Γ′ is a compact nilmanifold whereΓ′ = Γ/(Γ ∩ ker ξ). The lower central series of G′ is G′(i) = G(i)/ ker ξ, in

particular, G′(s)∼= R. By (3.15), x → f(ξ, x) = 0 is constant along orbits

of kerTds ξ, thus f(ξ, x) factors through a function f ′(x) in L2(M ′). M ′ is aprincipal T1-bundle over Ms−1 where the fiber T1 is given by Tds/ kerTds ξ.So by (3.15), for every z′ ∈ T1 and almost all x′ ∈M ′, f ′(z′x′) = e(z′)f ′(x′).

Therefore, after replacing M and f respectively by M ′ and f ′, we mayassume that ds = 1 and a non-zero T -invariant function f ∈ L2

0(M) satisfies

f(zx) = e(z)f(x), ∀z ∈ G(s)∼= R. (3.17)

It remains to show in this case that f factors through a function on Ms−1.Observe that under these conditions, |f | is T -invariant and is constant alongG(s)/Γ(s)-fibers. Thus |f | factors through a Ts−1-invariant function on Ms−1,and therefore by inductive hypothesis, through a T1-invariant function onM1. As T1 is ergodic, |f | is constant. By replacing f with f

|f | , it may also

be assumed that |f | = 1.Let H = {g ∈ G(s−1) : f is an eigenfunction of Lg} and let λ(h) be the

corresponding eigenvalue for h ∈ H. Then H is obviously a subgroup, andλ is a group morphism from H to the unit circle in C. Moreover, H is closedthanks to the continuity above. By (3.17), G(s) ⊆ H and λ(z) = e(z) forz ∈ G(s)

∼= R, where Z ⊆ R is identified with Γ(s)∼= Z. In addition, H is

a normal subgroup of G(s−1). To see this, suppose g ∈ G(s−1) and h ∈ H.Then because h−1ghg−1 ∈ G(s),

Lghg−1f =(Lh−1ghg−1 ◦ Lh)f = Lh−1ghg−1(λ(h)f) = e(h−1ghg−1)λ(h)f,

f is an eigenfunction of Lghg−1 with eigenvalue e(h−1ghg−1)λ(h).Remark that the group morphism λ can be extended to a function on

G by λ(g) = 〈Lgf, f〉. A priori, the resulting function does not need to begroup morphism any more. However, for g ∈ G and h ∈ H, we still have

λ(hg) = 〈LgLhf, f〉 = 〈λ(h)Lgf, f〉 = λ(h)λ(g). (3.18)


Furthermore, as Lg is unitary, λ(g−1) = 〈Lg−1f, f〉 = 〈f, Lg(f)〉 = λ(g).Therefore

λ(gh) = λ(h−1g−1) = λ(h−1)λ(g−1) = λ(h)λ(g). (3.19)

These also show that, if g, k ∈ G are such that gk ∈ H, then

λ(gk) = λ(g−1)−1λ(k). (3.20)

In particular, we can apply (3.20) to gh and g−1h−1 where g ∈ G and h ∈ H,because ghg−1h−1 ∈ H in this case. This would yield

λ(ghg−1h−1) =λ(h−1g−1)−1λ(g−1h−1)

=λ(h−1)−1λ(g−1)−1λ(g−1)λ(h−1) = 1.(3.21)

In other words, [G,H] ⊆ H ∩G(s) is in contained in kerλ and

f(gx) = f(x), ∀g ∈ [G,H]. (3.22)

Comparing (3.17) to (3.22), we see that [G,H] ( G(s). Since G(s)∼= R, it

follows that [G,H] is discrete. In particular, if H0 denotes the connectedcomponent of H (which is a Lie subgroup as H is closed), then [G,H0] istrivial. Therefore, H0 is in the center Z of G.

Decompose the complexified abelian Lie algebra (g(s−1)/g(s)) ⊗ C as a

direct sum⊕

ζ VζC of generalized eigenspaces of the linear tranform induced

by DeA, where ζ is the corresponding eigenvalue. Then g(s−1)/g(s) =⊕

ζ Vζ

where V ζ = (V ζC⊕V

ζC )∩(g(s−1)/g(s)) and the sum runs over a subset of indices

ζ that contains exactly one value from each imaginary comjugate pairs ofeigenvalues. Let V ζ be the preimage of V ζ in g(s−1). Then g(s) ⊆ V ζ . Take

Y ∈ V ζ and g = expY .Suppose first |ζ| < 1, then as n → ∞, (DeA)nY → g(s). So there are

Wn ∈ g and Yn ∈ g(s) such that Wn → 0 and (DeA)nY = Wn + Yn. Since[Wn, Yn] = 0, An(g) = exp(Wn + Yn) = gnwn where wn = expWn → 0, andyn = expYn ∈ G(s).

For almost every x ∈M ,

Tn(gx) =b ·A(b) · · · ·An−1(b)An(gx)

=Qn(b)An(g)An(x)

=Qn(b)An(g)Qn(b)−1Tn(x)

=Qn(b)An(g)Qn(b)−1An(g)−1An(g)Tn(x)

=Qn(b)An(g)Qn(b)−1An(g)−1gnwnTn(x)


whereQn(b) = b·A(b) · · ··An−1(b). BecauseQn(b)An(g)Qn(b)−1An(g)−1gn ∈G(s) ⊆ H and f is assumed to be T -invariant, we have

f(x) = f(Tnx);

Lgf(x) = f(Tn(gx)) = λ(Qn(b)An(g)Qn(b)−1An(g)−1gn

)f(wnT

nx).

In consequence,

|〈f, Lgf〉| =∣∣∣∣∫ f(Tnx)f(wnTnx)dmM (x)

∣∣∣∣=

∣∣∣∣∫ f(x)f(wnx)dmM (x)

∣∣∣∣ = |〈f, Lwnf〉|.

Since Lwnf is continuous in wn, as n→∞, Lwnf tends to f in L2(M).Soby taking limit we obtain that |〈f, Lgf〉| = ‖f‖2L2(M) = ‖Lgf‖2L2(M).This canhappen if and only if Lgf = ηf for some η from the unit circle in C, i.e.when f is an eigenvector of Lg. Thus g ∈ H. Because this argument appliesto exp tY for all t ∈ R, we know that Y ∈ h.

When |ζ| > 1, |ζ−1| < 1 and ζ−1 is an eigenvalue of the linear mapinduced by DeA

−1 on g(s−1)/g(s), with Y being the eigenvector. So one canwork with T−1 in stead of T , and prove that g ∈ H.

Finally, assume ζ = 1, then (ζ − 1)nπ(Y ) = 0 for some n ∈ N whereπ(Y ) ∈ V ζ is the projection of Y in g(s−1)/g(s). So (DeA− 1)nY ∈ g(s) ⊆ h.We claim in this case that Y ∈ h and hence g ∈ H. For this, it suffices toshow that if a vector W ∈ h satisfies (DeA)W −W ∈ h, then W ∈ h.

For such a vector W , let w = expW ∈ G(s−1), then

A(w) · w−1 = exp((DeA)W )w−1 = exp(W + ((DeA)W −W )

)w−1

= exp((DeA)W −W )w · w−1 = exp((DeA)W −W ) ∈ H.

Here we used that (DeA)W −W ∈ h ⊆ z and hence [W, (DeA)W −W ] = 0.In addition, because A(w) ∈ G(s−1), bA(w)b−1A(w)−1 ∈ G(s) ⊆ H. In thiscase, notice

Lwf(x) =(LwTf)(x) = Tf(wx) = f(bA(w)A(x))

=f(bA(w)b−1A(w)−1A(w)w−1w · bA(x))

=λ(bA(w)b−1A(w)−1)λ(A(w)w−1)f(w · Tx)

=λ(bA(w)b−1A(w)−1)λ(A(w)w−1) · (TLwf)(x).


Therefore, Lwf is also an eigenfunction of T , and so is (Lwf)f . Moreover,for all z ∈ G(s),(

(Lwf)f)(zx) = f(wzx)f(zx) = f(zwx)f(zx) = λ(z)λ(z)f(wx)f(x)

=((Lwf)f

)(x).

Thus (Lwf)f factors through an invariant function of Ts−1 on Ms−1, andtherefore by inductive hypothesis, factors through an eigenfunction φ of T1

of on M1.If the eigenvalue for φ is not equal to 1, then

〈Lwf, f〉 =

∫(Lwf)fdmM =

∫φdM1 = 〈φ, 1〉 = 0.

Otherwise, because T1 is ergodic, φ is a constant c and Lwf = c(f)−1 = cf .In paricular, |c| = 1 as Lw is an isometry. In this case, 〈Lwf, f〉 = c.To summarize, |〈Lwf, f〉| is either 0 or 1. Remark that this is not onlytrue for w but for all wt = exp tW . Because |〈Lwf, f〉| is continuous in w,|〈Lwf, f〉| = |〈Lw0f, f〉| = |〈f, f〉|. This happens only when φ is a constantc, in which case Lwf = cf and w ∈ H.

In summary, we have proved that h ⊆ g(s−1) contains the subspace⊕|ζ|6=1

or ζ=1

V ζ . Recall that H0 ⊆ Z, thus h ⊆ z ∩ g(s−1) and the auto-

morphism φ of g(s−1)/(z ∩ g(s−1)) induced by DeA has only eigenvaluesthat are not equal to 1 but of modulus 1. Observe that s(s) ⊆ z, henceg(s−1)/(z∩ g(s−1)) is abelian and can be identified with G(s−1)/(Z ∩G(s−1)).With this identification, A induces a map, which acts as φ, on the vectorspace G(s−1)/(Z ∩ G(s−1)), while preserving the Γ(s−1)/(Z ∩ Γ(s−1)), whichis a lattice by Proposition 2.3.10. This shows that all eigenvalues of φ arealgebraic units. It is a basic fact from number theory that if all Galoisconjugates of an algebraic unit ζ has modulus 1, then ζ is a root of unity.4

However, all eigenvalues of φ are eigenvalues of DeA, which earlier simpli-cification include no non-trivial roots of unity. Therefore g(s−1)/(z ∩ g(s−1))is trivial and g(s−1) ⊂ z. This makes the step of nilpotency s− 1 instead ofs, a contradiction. The proof is completed.

Since we have a complete criterion for ergodicity of affine nilmanifoldautomorphisms, it is natural to ask about unique ergodicity.

4To see this, prove first that such algebraic units in every given number field form afinite group.


Theorem 3.5.3 (Parry [Par69a]). Let Tx = bA(x) be an affine automor-phism of a compact nilmanifold M = G/Γ, then the following are equivalent:

(1) T is uniquely ergodic;(2) The induced affine automorphism T1 on the horizontal torus is unique

ergodic;(3) DeA is unipotent and T is ergodic.

Proof. (1)⇒(1): Notice that by Lemma 3.2.12 for every ergodic T1-invariantprobability measure there is at least one ergodic T -invariant probability mea-sure that lifts it. Therefore, if T is uniquely ergodic, then so is T1.

(2)⇒(3): First, if T1 is unique ergodic then it is ergodic by Corollary3.4.4. It then follows T is ergodic by Theorem 3.5.1.

Again by Corollary 3.4.4, the automorphismA1 induced byA onG/G(2)∼=

Rd1 is unipotent, i.e. only has eigenvalue 1. As A1 can be identified withDeA1, by Proposition 3.1.5, DeA also only has eigenvalue 1.

(3)⇒(1): We inductively prove this in terms of the step of nilpotency.When s = 1, T = T1 and the claim is empty. Suppose the claim is knownfor (s− 1)-step nilmanifolds and T1 is uniquely ergodic, then the projectedmap Ts−1 on Ms−1 = G/G(s)Γ is uniquely ergodic.

Because A acts on G(s) by the matrix DeA|g(s), which only has eigenvalue1 and is unipotent, it follows from (3.16) and the implication (3)⇒(1) inTheorem 3.4.3 that T is uniquely ergodic.

Exercises

Exercise 3.5.1. For all automorphisms T of the three dimensional Heisen-berg nilmanifold (classified in Exercise 3.1.3), either show T is ergodic ordescribe the ergodic decomposition of mM .

Exercise 3.5.2. Let h : R2 → R be a continuous function, and T be the dif-feomorphism of the 3-dimensional Heisenberg group defined by T ((x, y, z)Γ) =(α, β, h(x, y))((x, y, z)Γ). Show that T preserves mM and is ergodic with re-spect to mM if and only if T1(x, y) = (x+ α, y + β) is ergodic on T2.

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