summary:The centralizer of a semisimple isometric extension of a minimal flow is described

Siemaszko, Artur

Institute of Mathematics AS CR, v. v. i.

Commentationes Mathematicae Universitatis CarolinaeArtur SiemaszkoA note on the centralizer of topological isometric extensionsCommentationes Mathematicae Universitatis Carolinae, Vol. 46 (2005), No. 1, 137--143Persistent URL: http://dml.cz/dmlcz/119515Terms of use:© Charles University in Prague, Faculty of Mathematics and Physics, 2005Institute of Mathematics of the Academy of Sciences of the Czech Republic provides access todigitized documents strictly for personal use. Each copy of any part of this document mustcontain these Terms of use.This paper has been digitized, optimized for electronic delivery and stampedwith digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://project.dml.czComment.Math.Univ.Carolin. 46,1 (2005)137–143 137A note on the centralizerof topological isometric extensionsArtur SiemaszkoAbstract. The centralizer of a semisimple isometric extension of a minimal flow is de-scribed.Keywords: topological dynamics, isometric extensions, centralizerClassification: Primary 37B05; Secondary 54H20If S : X −→ X is a homeomorphism of a compact metric space X then a pair(X,S) is called a compact metric flow . Such a homeomorphism “generates” acontinuous action of the group Z on X :X × Z ∋ (x, n) 7−→ Sn(x) ∈ X.A flow (X,S) is called minimal if for every closed subset F ⊂ X with S(F ) = Feither F = ∅ or F = X . Equivalently, (X,S) is minimal iff every orbit is densein X : cl{Sn(x) : n ∈ Z} = X for each x ∈ X .Let us consider the following commutative diagram(1)Xψϕ@@@@@@@@Y π// Zwhere X = (X,S), Y = (Y, T ), Z = (Z,U) are compact metric flows and ϕ is agroup (say G-) extension. In a case of Z being minimal, Y is called an isometricextension of Z. As a straightforward corollary of Theorem 3.1 of [3] we have thefollowing.Proposition 1. Assume that in (1) Z is minimal. Then every minimal subflowof Y is an isometric extension of Z.Proof: Let N ⊂ Y be a minimal subset. Take a minimal subset M ⊂ ψ−1(N).By Theorem 3.1 of [3], there exists a closed subgroupH < G and a homomorphismResearch partly supported by KBN Grant 5 PO3A 027 21.138 A. Siemaszkoφ : (M,S) −→ Z such that φ is an H-extension. Since for every x ∈ M , φ(x) =ϕ(x) and obviously ψ(M) = N , we have the following commutative diagram(M,S)ψ|Mφ""EEEEEEEEE(N,T )π|N// Zthus (N,T ) is an isometric extension of Z. Remark 1. In Theorem 3.1 of [3] Z is assumed to be a minimal rotation. Never-theless, the arguments are valid for an arbitrary minimal flow.Remark 2. If we assume that in (1) Y is semisimple, the above proposition“allows” us to consider only isometric extensions of Z.It is well-known that if in (1) Y is a minimal isometric extension of Z, then Xcan be assumed to be minimal and one may easily verify that X is a K-extensionof Y, where K = {g ∈ G : ∀x∈X ψ(gx) = ψ(x)} and that π−1(z) is homeomorphicto a homogeneous space G/K for every z ∈ Z.We will be interested in special kinds of isometric extensions, which are factorsof “cocycle” extensions of minimal flows. Let Z = (Z,U) be a minimal flow andα : Z −→ G be a continuous function (called a topological cocycle) with values ina compact topological group. We define a homeomorphism Uα : Z×G −→ Z×Gby Uα(z, g) = (Uz, α(z)g). Of course (Z ×G,Uα) is a G-extension of Z. Now letK < G be a closed subgroup. We define a homeomorphism UG,K,α : Z×G/K −→Z×G/K by UG,K,α(z, gK) = (Uz, α(z)gK). Then, obviously, (Z×G/K,UG,K,α)is an isometric extension of Z.Remark 3. We can realize a group extension ϕ : X −→ Z as a topological cocycleif and only if there is a continuous selector of the homomorphism ϕ (i.e. thereexists a continuous map j : Z −→ X such that ϕ ◦ j = IdZ).Let us now consider the situation described by the following commutative di-agram:(Z ×G′, Uα′)ψ′ϕ′))RRRRRRRRRRRRRRR(Z ×G,Uα)ϕvvmmmmmmmmmmmmmmψ(Z ×G′/K ′, UG′,K ′,α′)π′// (Z,U) (Z ×G/K,UG,K,α)πoowhere ϕ, ϕ′, π and π′ are projections onto the first coordinate, while ψ and ψ′are natural projections along the second coordinate.The centralizer of isometric extensions 139Our first aim is to describe those homomorphismst : (Z ×G′/K ′, UG′,K ′,α′) −→ (Z ×G/K,UG,K,α),which are lifting of some element ofAut(Z) = {S : Z −→ Z| S is a homeomorphism and S ◦ U = U ◦ S}.The methods used in the proof of the next proposition, as well as in the restof the paper, are adapted from the proof of Theorem 1.4 and other parts of [2].The main difficulty is that, in contrary to the measure theoretical case, we do nothave at our disposal continuous selectors for a continuous map, in general. So forinstance not every group extension is a cocycle extension and we are not able touse transform functions (which are constructed with use of measurable selectors)to define some maps. Moreover, on various groups we have different topologicalstructures (in this paper, on the centralizer we consider the uniform topology,while in the measure theoretical case the centralizer is furnished with the weaktopology).Proposition 2. Assume thatt : (Z ×G′/K ′, UG′,K ′,α′) −→ (Z ×G/K,UG,K,α)is a homomorphism between two minimal isometric extensions of Z for whichthere exists s ∈ Aut(Z) such that s ◦ π′ = π ◦ t. Let K < G be irreducible (i.e. Kdoes not contain nontrivial normal subgroups of G). Then(i) there exist a continuous group epimorphism l : G′ −→ G and a continuousfunction f : Z −→ G such that(2) t(z, g′K ′) = (s(z), f(z)l(g′)K),l(K ′) ⊂ K and(3) l(α′(z)) = f(Uz)−1α(sz)f(z);(ii) l−1(K) = K ′ if and only if t is an isomorphism;(iii) if l|K ′ is one-to-one then K′ < G′ is irreducible;(iv) if K ′ < G′ is irreducible and t is an isomorphism then l is one-to-one andl(K ′) = K.Proof: (i) First observe that (z, g′, sz, g)J7−→ (z, g′, g) establishes an isomor-phism between ((ϕ′ × ϕ)−1Graph(s), Uα′ × Uα) and (Z × G′ × G,Uα′×α◦s),where (α′ × α ◦ s)(z) = (α′(z), α ◦ s(z)). Let us take a Uα′×α◦s-minimal sub-set M ⊂ J((ψ′ × ψ)−1Graph(t)) ⊂ J((ϕ′ × ϕ)−1Graph(s)).140 A. SiemaszkoLet ρ : Z ×G′/K ′ −→ G/K be a continuous map determined by(4) t(z, g′K ′) = (s(z), ρ(z, g′K ′)).Since t is a homomorphism of flows, ρ satisfies(5) ρ(Uz, α′(z)g′K ′) = α ◦ s(z)ρ(z, g′K ′).From (5) it follows that a continuous mapZ ×G′ ×G ∋ (z, g′, g)F7−→ g−1ρ(z, g′K ′) ∈ G/Kis Uα′×α◦s-invariant. Therefore F |M is constant, hence there is k ∈ G such that(6) ρ(z, g′K ′) = gkKfor every (z, g′, g) ∈ M . Without loss of generality we may assume that k = 1G,for if k 6= 1G, replace α, M and t by α1(z) = kα(z)k−1, M1 = {(z, g′, kgk−1) :(z, g′, g) ∈M} and t1(z, g′K ′) = (s(z), kgk−1K) if t(z, g′K ′) = (z, gkK), respec-tively. Then all corresponding objects are isomorphic and ρ1 : Z×G′/K ′ −→ G/K(defined as above for t1) satisfies ρ1(z, g′K ′) = gK for all (z, g′, g) ∈M1.Put H = {(g′, g)|M(g′, g) =M}. Then H is a closed subgroup of G′ ×G. Theorem 3 ([3, Theorem 3.1]). The flow (M,Uα′×α◦s) is an H-extension of Z.Now let NG′ = {g′ ∈ G′ : (g′, 1G) ∈ H} and NG = {g ∈ G : (1G′ , g) ∈ H}.Both NG′ and NG are closed normal subgroups of H (by the assumption ofminimality, H has full projections onto both coordinates).Lemma 4 ([3, Lemma 3.4]). There exists a continuous group homomorphismξ : G′/NG′ −→ G/NG such thatH =⋃g′∈G′τ ′−1(g′NG′)× τ−1(ξ(g′NG′)),where τ ′ : G′ −→ G′/NG′ and τ : G −→ G/NG are natural homomorphisms.Proof: Let us take k ∈ NG. For each (z, g′, g) ∈ M , (z, g′, gk) ∈ M since(1G′ , k) ∈ H . By the choice of M , we have t(z, g′K ′) = (s(z), gK) = (s(z), gkK),hence k ∈ K. Thus NG < K, hence by irreducibility of K < G, NG = {1G}. Itfollows that if l = ξ ◦ τ ′ then H = {(g′, l(g′)) : g′ ∈ G′}. Now put f : Z −→ G,(7) f(z) = gl(g′)−1,The centralizer of isometric extensions 141if (z, g′, g) ∈ M . Observe that if (z, g′, g), (z, g′1, g1) ∈ M then (g′−1g′1, g−1g1) ∈H and since l is a group homomorphism, f is well-defined. Obviously f is con-tinuous. Therefore (4), (6) (with k = 1G) and (7) give (2). Also, by (7) and theinvariance of M , we have (3).Now take k′ ∈ K ′. Find z ∈ Z and k ∈ K such that (z, k′, k) ∈M . Then(s(z), f(z)K) = t(z,K ′) = t(z, k′K ′) = (s(z), f(z)l(k′)K)and thus l(k′) ∈ K. Therefore l(K ′) ⊂ K.(ii) If t is an isomorphism and g′ ∈ l−1(K) then we havet(z, g′K ′) = (s(z), f(z)l(g′)K) = (s(z), f(z)K) = t(z,K ′),hence g′ ∈ K ′.Now if l−1(K) = K ′ and t(z, g′K ′) = t(z1, g′1K′) then z = z1 and l(g′−1g′1) ∈K, hence g′−1g′1 ∈ K′.Now (iii) and (iv) are obvious. Given a flow (Z,U), by C(Z,U) (C(U) for short, if the phase space Z isestablished) let us denote the centralizer of (Z,U), i.e. the set of all continuoustransformations from Z onto Z which commute with U . If we endow the subsetAut(Z) = Aut(U) ⊂ C(Z,U) of all invertible elements with the topology ofuniform convergence (we denote the closure operation in this topology by “cl”),it becomes a Polish group.As an immediate consequence of the above proposition we have the following.Proposition 5. Let (Z ×G/K,UG,K,α) be a minimal isometric extension of Z.Then every lift s̃ ∈ C(UG,K,α) of s ∈ Aut(Z) is of the forms̃(z, gK) = sf,l(z, gK) = (s(z), f(z)l(g)K),where f : Z −→ G is a continuous function and l : G −→ G is a continuous groupepimorphism such that l(K) ⊂ K and(8) l(α(z)) = f(Uz)−1α(sz)f(z).Moreover sf,l is invertible if and only if so is l. In such a case l(K) = K.Denote Adk : G −→ G, Adk(g) = k−1gk and NG(K) = {g ∈ G : g−1Kg =K}.Corollary 6. (i) Every s ∈ Aut(Z) which can be lifted to C(UG,K,α) can alsobe lifted to C(Uα).(ii) If Z is a minimal rotation then every element of C(UG,K,α) is a lift ofsome s ∈ Aut(Z).142 A. Siemaszko(iii) If Z is a minimal rotation and all elements of C(Uα) are invertible thenso are all elements of C(UG,K,α).(iv) If Ĩd ∈ C(UG,K,α) is a lift of IdZ thenĨd(z, gK) = (z, gkK) = (IdZ)k,Adk(z, gK),where k ∈ NG(K).(v) All lifts of the identity are invertible.(vi) If s, s−1 ∈ Aut(Z) and s, s−1 can both be lifted to C(UG,K,α) then alllifts of s are invertible.Proof: One can easily check that (i), (ii) and (iii) follow. To show (iv) putĨd = (IdZ)f,l and observe that, by (8) (with s = IdZ), a continuous map Z×G ∋(z, g) 7−→ g−1f(z)l(g) ∈ G is Uα-invariant, hence constant. Thus there existsk ∈ G such that for every (z, g) ∈ Z ×G(9) f(z)l(g) = gk.Putting g = 1G in (9), we get f(z) = k for every z ∈ Z and then l(g) = k−1gkfor each g ∈ G, so that l = Adk. Since Adk is invertible, we immediately havethat k ∈ NG(K).Now (v) and (vi) follow from (iv) and Proposition 5. Remark 4. Proposition 5 generalizes some results of Section 3 of [3] and Sec-tion 2 of [4] where the case of group extensions (instead of isometric extensions)is considered.Now putLK(U,α) = {s ∈ Aut(Z) : s, s−1 can be lifted to C(UG,K,α)},C̃(UG,K,α) = {s̃ ∈ C(UG,K,α) : s̃ is a lift of some s ∈ LK(U,α)}.From Corollary 6(vi) it follows that the above sets are groups. Notice also thatC̃(UG,K,α) is a closed subgroup of Aut(UG,K,α) (from Proposition 5 it followsthat s̃ ∈ C̃(UG,K,α) iff the factor Z is invariant under s̃).Let us define σ : NG(K)/K −→ C̃(UG,K,α) by σ(kK) = Idk,Adk . We easilysee that σ is an injective group homomorphism.Using Proposition 5 define a map πK : C̃(UG,K,α) −→ Aut(Z) by πK(sf,l) = s,which is a continuous group homomorphism. Moreover, by Corollary 6(iv),Im(σ) = Ker(πK), hence the image of σ is a closed subgroup. If we endowNG(K)/K with the quotient topology, σ becomes a continuous injective homo-morphism between Polish groups. Therefore σ is open.On LK(U,α) = C̃(UG,K,α)/Ker(πK) consider the quotient topology and call itThe centralizer of isometric extensions 143the LK -topology. By [1], LK(U,α) furnished with the LK -topology becomes aPolish group. Notice also that the LK -topology is stronger than the uniformtopology on Aut(Z) restricted to LK(U,α).We have constructed the following short exact sequence of Polish topologicalgroups.(10) 1 −→ NG(K)/Kσ−→ C̃(UG,K,α)πK−→ LK(U,α) −→ 1.Proposition 7. If Aut(Z) = cl{Un : n ∈ Z} = LK(U,α), thenC̃(UG,K,α) = cl{σ(kK) ◦ (UG,K,α)n : n ∈ Z, k ∈ NG(K)}.Proof: Since Aut(Z) = LK(U,α) the LK -topology and uniform topology areequal. To end the proof notice that, by Corollary 6(iv),π−1K ({Un : n ∈ Z}) = {σ(kK) ◦ (UG,K,α)n : n ∈ Z, k ∈ NG(K)}.Since the uniform topology on Aut(Z) is equal to the quotient topology (withrespect to Ker(πk)), the pre-image of the dense in Aut(Z) set {Un : n ∈ Z} isdense in C̃(UG,K,α). Corollary 8. Let Z be a minimal rotation. If K 6= {1G} and K < G is irre-ducible, then LK(U,α) 6= C(U).Proof: Recall that an arbitrary minimal flow Z is a rotation iff Aut(Z) iscompact, notice that NG(K)/K is compact and use (10) to finish the proof. References[1] Brown L.G., Topologically complete groups, Proc. Amer. Math. Soc. 35 (1972), 593–600.[2] Danilenko A., Lemańczyk M., Isometric extensions, 2-cocycles and ergodicity of skew pro-ducts, Studia Math. 137 (1999), 123–142.[3] Glasner E., Mentzen M.K., Siemaszko A., A natural family of factors for minimal flows,Contemp. Math. 215 (1998), 19–42.[4] Siemaszko A., Ergodic and Topological Properties of Distal Dynamical Systems, PhD The-sis, Toruń, 1996.Faculty of Mathematics and Computer Science, University of Warmia and Mazuryin Olsztyn, ul. Żo lnierska 14A, 10-561 Olsztyn, PolandE-mail : artur@uwm.edu.pl(Received May 12, 2004, revised October 5, 2004)

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A note on the centralizer of topological isometric extensions

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A note on the centralizer of topological isometric extensions

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Institute of Mathematics AS CR, v. v. i.

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