The Koopman Operator

Spectral Theoryof a dynamical system refers to the properties of eigenvalues and eigenfunctions of the Koopman operator of a dynamical system.

We recall: Given(X,B, µ,T), we can take the space of complex-valued square-integrable observablesL²(µ). This is a Hilbert space, equipped with inner product

hf,gi= Z

X

f(x)·g(x) dµ.

TheKoopman operatoris dened as

UT :L²(µ)→L²(µ), UTf =f ◦T.

The Koopman Operator

ByT-invariance ofµ, it is a unitary operator (it preserves the inner product). Indeed

hU_Tf,U_Tgi = Z

X

f ◦T(x)·g◦T(x) dµ

= Z

X

(f ·g)◦T(x) dµ= Z

X

f ·g dµ=hf,gi, and thereforeU_T^∗U_T =U_TU_T^∗ =I.

Theorem: The eigenvalues of U_T form a multiplicative group subgroup of the unit circle. Eigenfunctions to dierent eigenvalues are orthogonal, and ifµis ergodic, then the eigenfunctions have constant modulus and each eigenspace is one-dimensional.

Examples: A transformation is weakly mixing if 1 is the only eigenvalue (and it has multiplicity 1, so constant functions are the only eigenfunctions). This is not so easy to see from our earlier denition of weak mixing, but we will not give the proof for now.

For rational circle rotations(S¹,B,Leb,R_p/q) the eigenvalues are {e^2πir/q :r ∈ {0,1, . . . ,q−1}}, and each eigenvalue has innite multiplicity. Note: no ergodicity!

For irrational circle rotations(S¹,B,Leb,Rα)the eigenvalues are {e^2πinα:n ∈Z}, and each eigenfunction of e^2πinα isx 7→e^2πinx, up to a multiplicative constant.

The Koopman Operator

Proof: Ifλis the eigenvalue of eigenfunctionv, then

hv,vi=hU_Tv,UTvi=hλv, λvi=|λ|²hv,vi,

soλlies on the unit circle. Assuming thatλ, µare eigenvalues with eigenfunctionsv andw respectively, we have

UT(vw) = (vw)◦T = (v◦T)·(w ◦T) =UTv·UTw =λµ(vw), soλµ is an eigenvalue. Also

U_T(¯v) = ¯v◦T =v◦T =U_Tv = ¯λ¯v =λ⁻¹v¯,

so the eigenvalues form a multiplicative group of the unit circle. If v andw are eigenfunctions to dierent eigenvaluesλandµ, then

hv,wi=hU_Tv,U_Twi=hλv, µwi=λ¯µhv,wi, and this can only be true ofhv,wi=0.

Assume now thatµ is ergodic, so the only eigenvectors of eigenvalue 1 are constantµ-a.e. Ifv is the eigenfunction of eigenvalueλ, then|v|is an eigenfunction of eigenvalue |λ|=1, so

|v|is constant; we can scale|v|=1. Ifw is another eigenfunction ofλ, scaled so that|w|=1 and independent ofv, thenv/w is an eigenfunction of 1, sov =w µ-a.e.

The Koopman Operator

LemmaIf(Y,S, ν) is a measure-theoretical factor of(X,T, µ) (with factor mapπ andν=µ◦π⁻¹), then every eigenvalue of (Y,S, ν) is also an eigenvalues of(X,T, µ).

In particular, the spectrum of(Y,S, ν) is contained in the spectrum of(X,T, µ), and isomorphic systems have the same eigenvalues and spectrum.

Proof: Letg be an eigenvalue of(Y,S, ν), with eigenvalue e^2πiα. Thenf :=g◦π is an eigenvector of(X,T, µ), because

f ◦T =g ◦π◦T =g ◦S◦π=e^2πiαg ◦π =e^2πiαf µ-a.e.

Hencef is an eigenfunction of (X,T, µ) with the same eigenvalue e^2πiα.

Given a non-negative measureν ∈ M(T) on the circle, the Fourier coecients ofν are dened as

ˆ ν(n) =

Z

T

zⁿdν= Z ₁

0 e^2πinxdν(e^2πix).

For every sequence(z_j)j∈N⊂CandN ∈N, we have

N

X

j,k=1

z_jz¯_kν(jˆ −k) =

N

X

j,k=1

Z ₁

0 z_je^2πijxz_ke^2πikxdν

= Z ₁

0 N

X

j=1

z_je^2πijx

N

X

k=1

z_ke^2πikxdν

= Z ₁

0

k

N

X

j=1

z_je^2πijxk²dν≥0.

Spectral Measures

This property of(ˆν(n))n∈Z is calledpositive deniteness.

Conversely, the Bochner-Herglotz Theorem states that for every positive denite sequence(an)n∈Z ⊂C, there is a unique

non-negative measureν∈ M(T) such thatνˆ(n) =an for eachn, andν(T) =pP

n|a_n|².

Let(X,B, µ,T)be an invertible dynamical system. Given a functionf ∈L²(µ), the sequence a_n:=hU_Tⁿf,fi=R

X f ◦Tⁿf dµ is positive denite because

N

X

j,k=1

z_jz¯_kaj−k =

N

X

j,k=1

z_jz_khU_T^j−kf,fi=

N

X

j,k=1

hz_jU_T^j f,z_kU_T^kfi

=

* _N X

j=1

z_jU_T^j f,

N

X

k=1

z_kU_T^kf +

=k

N

X

j=1

z_jU_T^j fk²≥0.

Therefore the Bochner-Herglotz Theorem associates a

non-negative measureν_f ∈ M(T) tof, which is called thespectral measure off.

Spectral Measures

Remarks

I If U is an invertible unitary operator, then ˆ

νf(−n) = hU⁻ⁿf,fi=hU⁻ⁿf,U⁻ⁿUⁿfi

= hf,Uⁿfi=hUⁿf,fi

= νˆ_f(n),

for every n∈N. Therefore it makes sense to dene ˆ

ν_f(−n) := ˆν_f(n)also for non-invertible unitary operators.

Most of the theory remains valid.

Remarks

I For U =U_T, the Koopman operator of an invertible dynamical system (X,B, µ,T), the Fourier coecients

ˆ

ν_f(n) =hU_Tⁿf,fi=R

Xf ◦Tⁿf dµ¯ are the autocorrelation coecients of the observablef ∈L²(µ).

If µis mixing, then ˆν_f(n)→0 for everyf ∈L²(µ) with R

X f dµ=0.

In fact, the correlation coecients hU_Tⁿf,gi=R

X f ◦Tⁿg d¯ µ of two observables f,g ∈L²(µ) are the Fourier coecients of acomplex measure σ_f,g; this is an application of a somewhat more general version of the Bochner-Herglotz Theorem.

Spectral Measures

Suppose the unitary operatorU acts on a the Hilbert space H. We can decomposeHinto subspaces that are the linear spans of U-orbits of well-chosen functions inH:

Theorem: Let U be an invertible unitary operator acting on a separable Hilbert spaceH. Then there is a (possibly nite) sequence of functionsh_j ∈Hsuch that

H=⊕_jSpan(Uⁿhj :n∈Z)

and if j 6=k, then (1)

Span(Uⁿh_j :n∈Z)⊥Span(Uⁿh_k :n∈Z).

The corresponding spectral measures satises ν_h1 ν_h2 ν_h3 . . . Moreover, if(h_j⁰)satisfy (1), thenνhj ∼ν_h⁰

j for eachj.

Denition: The spectral measure νh₁ of the leading function h₁ in (1) is called themaximal spectral type. If U =U_T is the Koopman operator of an invertible dynamical system, then we callν_h1 the spectral measure ofT and we will denote it as ν_T.

Example: Iff is an eigenfunction of U_T to eigenvalueλscaled so thatkfk₂=1, thenν_f =δ_λ is the Dirac measure at the

eigenvalue. Indeed, δˆ_λ(n) =

Z

T

zⁿdδ_λ =λⁿ=hλⁿf,fi=hU_Tⁿf,fi.

For each eigenfunctionf, Span(U_Tⁿf :n∈Z) =:Span(f) is only a one-dimensional subspace. However, closure of the span of all eigenvalues Span(f :UTf =λf), called theKronecker factor, can be as large as the whole Hilbert spaceL²(µ).

Pure Point Spectrum

The spectral measure ofT decomposes as ν_T =ν_pp+ν_ac+ν_sing where

I νpp is thediscreteor pure point part ofν_T. It is an at most countable linear combination of Dirac measures, namely at every eigenvalue, so in particular at λ=1. For weak mixing transformations νpp =cδ₀ for some c ∈(0,1].

I νac is absolutely continuous w.r.t. Lebesgue measure.

I νsing is non-atomic but singular w.r.t. Lebesgue measure.

Then partsν_ac+ν_sing =ν_cont together are called the continuous partof the spectral measure.

Denition: A measure-preserving dynamical system(X,B, µ,T) is said to have apure point spectrum (also calleddiscrete spectrum if the collection of eigenfunctions of the Koopman operatorU_T spans L²(µ). That is: the Kronecker factor isL²(µ).

Equivalently, the spectral measureν_T =νpp is a countable linear combination of Dirac measures.

Pure Point Spectrum

We quote (without proof) two structure theorems due to Halmos &

von Neumann, that illustrate the use of pure point spectrum transformations.

TheoremTwo measure-preserving dynamical systems with pure point spectra are isomorphic if and only if their eigenvalues are the same.

TheoremAn ergodic probability measure preserving system

(X,T, µ)on compact metric space has pure point spectrum if and only if it is isomorphic to a rotation on a compact metrizable Abelian groupG with Haar measureµ_G, so there is g₀ ∈G such thatTx =φ⁻¹(φ(x) +g₀), where φ:X →G is the isomorphism.

We give some examples to illustrate the second theorem. Assume that(X,T, µ) has pure point spectrum.

Example: Letα be irrational suppose that the set of eigenvalues of U_T is {e^2πinα :n∈Z}.

Then the system is isomorphic to(S¹,R_α,Leb), with eigenfunctions v_n=e^2πinx

These are the usual Fourier modes, and they spanL²(S¹,Leb).

Pure Point Spectrum

Example: Letα andβ two irrationals that are rationally independent, i.e.,xα+yβ+z =0 forx,y,z ∈Q implies x=y =z =0.

Then the group rotation

R_α,β :T² →T², (x,y)7→(x+α,y+β).

has spectrum{e^{2πi(mα+nβ)} :m,n ∈Z}. The eigenfunctions are

vm,n=e^2πi(mx+ny).

These are the two-dimensional Fourier modes, and they span L²(T²,Leb).

The following example, calleddyadic odometeror dyadic adding machinehas spectrum{e^2πim2−n :m,n∈N}.

It is a mapa:{0,1}^N→ {0,1}^Ndened by add-and-carry.

x = 01110001101101010100. . . +1 = 10000000000000000000. . . a(x) = 11110001101101010100. . . +1 = 10000000000000000000. . . a²(x) = 00001001101101010100. . .

The dyadic odometer

Let us write the mapa down as a computer algorithm:

c :=1 ; k :=1 Repeat

s :=x_k +c;

If s ≥2 then c :=1 elsec :=0 x_k :=s mod2 ; k:=k+1 Until c =0

Check that this algorithm indeed gives

a(11111111111. . .) =00000000000. . .

Let us also generalize this to show that the odometer is a topological group under addition.

The additionz =x+y of two sequencesx,y ∈X with add-and-carry goes according to the algorithm:

c :=0 ; k :=1 Repeat for allk ∈N

s :=xk +yk+c;

If s ≥2 then c :=1 elsec :=0 z_k :=s mod2; k :=k+1 One can check that this is continuous inx andy.

The dyadic odometer

As an interval map, the odometer has the form

a(x) =x−(1−3·2¹⁻ⁿ) ifx ∈[1−2¹⁻ⁿ,1−2⁻ⁿ), n≥1, It preserves Lebesgue measure.

Figure:The dyadic odometer represented as a mapaon the interval (von Neumann-Kakutani map).

The mapa permutes then-cylinders cyclically, so if we dene v_m,n:{0,1}^N→Cas

v_m,n(x) =e^2πimk/2n ifx∈[a₁. . .a_n], k =

n

X

j=1

a_j2^j−1,

then

vm,n◦a=e^2πim/2nvm,n. This shows thate^2πim/2n are indeed all eigenvalues.

Exercise: Verify that these eigenfunctions form a orthonormal system. Show that the dyadic adding machine is uniquely ergodic.

The dyadic odometer

TheoremThe dyadic odometer has pure point spectrum.

Proof: Let(X,a) be the odometer, with a-invariant measure µ.

Then-cylinder Z_[0ⁿ_]= [0. . .0](n zeros) is periodic with period 2ⁿ under the mapa. AbbreviateZ_jⁿ=a^j(Z_[0ⁿ_]). Now

λⁿ_m:=e^2πim/2n for 0≤m<2ⁿ

is an eigenvalue, because we can construct a corresponding eigenfunctionvm,n of the Koopman operator as

v_m,n|_Zⁿ

j =e^−2πijm/2n. In particular,

(v_m,n)n∈N,0≤m<2ⁿ forms an orthogonal system,

and it is also easy to check that theL²(µ)-norms kv_m,nk₂=1 for alln ∈N,0≤m<2ⁿ.

Proof continued: To show that it is a complete orthonormal system, i.e.,

Span({v_m,n:n∈N,0≤m<2ⁿ}) is dense in L²(µ), it suces to show that ifg ∈L²(µ) is such that R

X g vm,ndµ=0 for alln∈N,0≤m<2ⁿ, theng ≡0 µ-a.e. Since C(X)is dense inL²(µ), we can assume thatg is continuous.

Assume that there is a cylinder setZ_jⁿ such that R

Z_jⁿg dµ6=0. Let wε=ε+ (1−ε)e^−2πij/2nv₁,n,

The dyadic odometer

Proof continued: Thenwε|_Zⁿ

j =1 and|w_ε(x)| ≤1−ε² <1 for x∈/ Z_jⁿfor ε >0 suciently small. Clearlyw_ε is a linear

combination of eigenfunctions, so Z

X

g wεdµ=0.

The algebraic powerw_ε^r is a linear combination of eigenfunctions too, and hence also Z

X

g w_ε^rdµ=0.

But since|w_ε^r(x)|<(1−ε²)^r →0 for eachx∈/ Z_jⁿ, we get limr→∞R

Xg w_ε^rdµ=R

Z_jⁿg dµ6=0. This contradiction shows that (v_m,n)n∈N,0≤m<2ⁿ is indeed a complete orthonormal system,