Regularity and inverse theorems for uniformity norms on compact abelian groups and nilmanifolds

Theorem A.1.

Let X be a k-step cfr nilspace such that the factor Xk-1 is toral. Let G denote the Lie group Θ⁢(X), let G∙ denote the degree-k filtration (Θi⁢(X))i≥0, and for an arbitrary fixed x∈X let Γ=StabG⁡(x). Then X is isomorphic as a compact nilspace to the coset space G/Γ with cube sets Cn⁢(X)=(Cn⁢(G∙)⋅Γ〚n〛)/Γ〚n〛, n≥0.

Proof.

Fix x∈X and let Γ=StabG⁡(x).

We first claim that Γ is discrete. Indeed, letting h:Θ⁢(X)→Θ⁢(Xk-1) be the natural continuous homomorphism defined by h⁢(α)⁢(y)=πk-1⁢(α⁢(x)) (see [3, Lemma 2.9.3]), note that h⁢(Γ) is a subgroup of the stabilizer of πk-1⁢(x) in Θ⁢(Xk-1), and since Xk-1 is toral, this stabilizer is discrete (see the proof of [3, Theorem 2.9.17]), so h⁢(Γ) is discrete. Then, since h-1⁢(h⁢(Γ)) is a union of cosets of ker⁡(h), it suffices to show that Γ∩ker⁡(h) is discrete. This follows from [3, Lemma 2.9.9], since no non-trivial element of τ⁢(Zk) stabilizes x.

By [3, Corollary 2.9.12] the Lie group Θ⁢(X)0 acts transitively on the connected components of X, and since Xk-1 is toral, it follows that 〈Θ⁢(X)0,Zk〉 acts transitively on X. Indeed, if x,y∈X are in different components, there is g′∈Θ⁢(Xk-1)0 such that g′⁢πk-1⁢(x)=πk-1⁢(y). Then there is g∈Θ⁢(X)0 such that h⁢(g)=g′, and since g is path-connected to the identity in G, it follows that gx is in the same component as x. Moreover, by definition of h we have πk-1⁢(g⁢x)=g′⁢πk-1⁢(x)=πk-1⁢(y). There is therefore z∈Zk such that z⁢g⁢x=y, which proves the claimed transitivity. Now since G⊃〈Θ⁢(X)0,Zk〉, we have that G also acts transitively on X, whence X is homeomorphic to the coset space G/Γ (see [25, Chapter II, Theorem 3.2]). In particular, since X is compact, we have that Γ is cocompact.

Recall from [4, Definition 3.2.38] that two cubes c1,c2∈Cn⁢(X) are said to be translation equivalent if there is an element c∈Cn⁢(G∙) such that c2⁢(v)=c⁢(v)⋅c1⁢(v). We now show that Cn⁢(X)=πΓ〚n〛⁢(Cn⁢(G∙)), i.e. that every cube on X is translation equivalent to the constant x cube. First we claim that for every cube c∈Cn⁢(X) there is a cube c′∈Cn⁢(X) that is translation equivalent to the constant x cube and such that πk-1∘c=πk-1∘c′. Indeed, given c∈Cn⁢(X), we have

and since X is toral the latter cube is translation equivalent to the cube with constant value x′=πk-1⁢(x), i.e. πk-1∘c=c~⋅x′ for some cube c~ on the group Θ⁢(Xk-1)0 with the filtration (Θi⁢(Xk-1)0)i≥0. By the unique factorization result for these cubes [4, Lemma 2.2.5], we have

where g~j∈Θcodim⁡(Fj)⁢(Xk-1)0. By [3, Theorem 2.9.10 (ii)], for each j∈[0,2n) there is gj∈Θcodim⁡(Fj)⁢(X)0 such that h⁢(gj)=g~j. Let c* be the cube in Cn⁢(Θ⁢(X)0) defined by

Let c′=c*⋅x. This is in Cn⁢(X), and is translation equivalent to the constant x cube. Moreover, by construction πk-1∘c′ equals

This proves our claim.

It follows from [4, Theorem 3.2.19] and the definition of degree-k bundles (in particular [4, (3.5)]) that c-c′∈Cn⁢(𝒟k⁢(Zk)). But then, using translations from τ⁢(Zk)=Θk⁢(X), we can correct c′ further to obtain c, thus showing that c is itself a translation cube with translations from Θ⁢(X). (Such a correction procedure has been used in previous arguments, see for instance the proof of [4, Lemma 3.2.25].)

We have thus shown that Cn⁢(X)⊂πΓ〚n〛⁢(Cn⁢(G∙)). The opposite inclusion is clear, by definition of the groups Θi⁢(X). ∎

Theorem A.2.

Let X be a k-step cfr nilspace. If Ck⁢(X) is connected, then X is toral.

Proof.

We argue by induction on k. For k=1 the statement is clear. For k>1, first note that Ck⁢(Xk-1) is connected (by continuity of πk-1), and so (since projection to a k-1 face of a k cube is a continuous map) we have also that Ck-1⁢(Xk-1) is connected, so by induction we have that Xk-1 is toral. Now suppose for a contradiction that X is not toral. Then the last structure group Zk must be a disconnected compact abelian Lie group. By quotienting out the torus factor of Zk if necessary, we can assume that X now has k-th structure group Zk being a finite abelian group of cardinality greater than 1. We shall now deduce that Ck⁢(X) must be disconnected, a contradiction.

By Theorem A.1 we have that X is isomorphic to the coset nilspace (G/Γ,G∙), where G=Θ⁢(X) and Γ=StabG⁡(x) for some fixed point x∈X. Hence Ck⁢(X)=Ck⁢(G∙)/Γ〚k〛. Let σk be the Gray code map on G〚k〛 (see [4, Definition 2.2.22]), and recall that restricted to Ck⁢(G∙) this map takes values in Gk (see [3, Proposition 2.2.25]) and that Gk≅Zk (see [4, Lemma 3.2.37]). We know that shifting any value c⁢(v) of a cube c∈Ck⁢(G∙) by any element of Zk still gives a cube in Ck⁢(G∙) (see [4, Remark 3.2.12]). It follows that σk maps Ck⁢(G∙) onto Zk. On the other hand, the map σk only takes the value idG on Γ〚k〛, since Γ∩Gk={idG} (as the action of Gk≅Zk is free). Now let C denote the identity component of Cn⁢(G∙). It is standard that C is normal in Cn⁢(G∙). We also have σk⁢(C⋅Γ〚k〛)={idG}. Indeed, since σk is continuous and Zk is discrete, for every element c⋅γ∈C⋅Γ〚k〛 we have σk⁢(γ)=0, and c⋅γ is in the same component as γ, so we must also have σk⁢(c⋅γ)=0. But then the product set C⋅Γ〚k〛 must be a proper subgroup of Cn⁢(G∙) (otherwise its image under σk would be Gk). Thus we have shown that Cn⁢(G∙)/C⋅Γ〚k〛 is not the one point space. Hence there are at least two disjoint cosets of C⋅Γ〚k〛 forming a cover of Cn⁢(G∙). Since the latter group is a Lie group, C is open, and therefore these covering cosets of C⋅Γ〚k〛 are open sets. But then the quotient map q:Cn⁢(G∙)→Cn⁢(G∙)/Γ〚k〛 (which is open) sends these cosets to disjoint open sets covering Ck⁢(G∙)/Γ〚k〛, so Ck⁢(X) is disconnected. ∎

Lemma A.3.

Let X be a k-step cfr nilspace such that Xk-1 is toral. Then for every integer n≥0 the connected components of Cn⁢(X) have equal positive Haar measure.

Proof.

Recall that Cn⁢(X) is a compact abelian bundle with base Cn⁢(Xk-1), bundle projection π:=πk-1〚n〛, and structure group Z~k:=Cn⁢(𝒟k⁢(Zk)), where Zk is the k-th structure group of X (see[3, Lemma 2.2.12]). The Haar measure μ on Cn⁢(X) is invariant under the continuous action of Z~k, by construction (see [3, Proposition 2.2.5]). Assuming that there is more than one component of Cn⁢(X), let c1,c2 be any points in distinct components C1, C2, respectively. Then, since Xk-1 is toral, by [3, Theorem 2.9.17] there is a cube c∈Cn⁢(Θ⁢(Xk-1)∙0) such that c⋅π⁢(c1)=π⁢(c2). By [3, Theorem 2.9.10] there is a cube c~∈Cn⁢(Θ⁢(X)∙0) such that π⁢(c~⋅c1)=π⁢(c2). There is therefore z∈Z~k such that c~⋅c1+z=c2. We also have c~⋅c1 still in C1, because the map c1↦c~⋅c1 is a composition of multiplications by face-group elements of the form gF where F is a face in 〚n〛 and g is in the connected Lie group Θcodim⁡(F)⁢(X)0. Hence (C1+z)∩C2 is non-empty (containing c2), so C1+z⊂C2 (since C1+z is connected and C2 is a maximal connected set), whence μ⁢(C1)=μ⁢(C1+z)≤μ⁢(C2). Similarly μ⁢(C2)≤μ⁢(C1). ∎

Lemma A.4.

For every k-step compact nilspace X and every d≥2, the function defined by f↦∥f∥Ud is a seminorm on L∞⁢(X).

Proof of Lemma A.4.

The lemma follows from results in [7], namely from [7, Proposition 3.6], which shows that the Haar measures μ〚n〛 on Cn⁢(X) form a cubic coupling, and [7, Corollary 3.17], which yields the seminorm properties for a general cubic coupling. ∎

Lemma A.5 (Taylor expansion).

Let g∈poly⁡(Z,G∙), where G∙ has degree at most s. Then there are unique Taylor coefficientsgi∈Gi such that for all n∈Z we have

Conversely, every such expression defines a map g∈poly⁡(Z,G∙). Moreover, if H≤G and g is H-valued, then gi∈H for each i.

Proof of Proposition 6.3.

Since ϕ∘β is a morphism ℤ→G/Γ, it suffices to prove the following statement: for every morphism ϕ:ℤ→G/Γ, there is a morphism ψ:ℤ→G (whence ψ∈poly⁡(ℤ,G∙)) such that πΓ∘ψ=ϕ. We prove this by descending induction on j∈[k+1], showing that the statement holds for maps ϕ taking values in (Gj⁢Γ)/Γ. For j=k+1, since Gk+1={idG}, the map ϕ is constant and the statement is trivially verified letting ψ be a constant Γ-valued map. For j<k+1, suppose that the statement holds for j+1 and that ϕ takes values in (Gj⁢Γ)/Γ. It follows from the filtration property that Gj+1⁢Γ is a normal subgroup of Gj⁢Γ and that the quotient Gj⁢Γ/(Gj+1⁢Γ) is an abelian group. Denoting this abelian group by Aj, let qj:(Gj⁢Γ)/Γ→Aj be the quotient map for the action of Gj+1 on (Gj⁢Γ)/Γ. Note that qj is a nilspace morphism. More precisely, for every cube c⁢Γ〚n〛 on (Gj⁢Γ)/Γ (where c∈Gj〚n〛∩Cn⁢(G∙)), we have qj∘(c⁢Γ〚n〛)=(q~j∘c)⁢Γ〚n〛, where q~j is the quotient homomorphism Gj→Gj/Gj+1; this implies that every (j+1)-face of qj∘(c⁢Γ〚n〛) has value 0 under the Gray-code map σj+1, so qj is a morphism into 𝒟j⁢(Aj). It follows that qj∘ϕ is a morphism ℤ→𝒟j⁢(Aj), and is in particular a polynomial map of degree at most k, so by Lemma A.5 we have qj∘ϕ⁢(x)=∑ℓ=0kaℓ⁢(xℓ) for x∈ℤ, for some aℓ∈Aj, and binomial coefficients (xℓ). Since qj is surjective, there exist elements b0,b1,…,bk in Gj such that qj⁢(bℓ⁢Γ)=aℓ for each ℓ. Let α:ℤ→G be the polynomial map

and note that qj⁢(α⁢(x)⁢Γ)=qj∘ϕ⁢(x) for all x. It follows that the map α-1⁢ϕ is a morphism ℤ→(Gj+1⁢Γ)/Γ, so by induction there is a map ψ′∈poly⁡(ℤ,G∙) such that

Then ψ⁢(x):=α⁢(x)⁢ψ′⁢(x) is a map in poly⁡(ℤ,G∙) with the required property. ∎

Lemma B.1.

Let (Ω,A,λ) be a probability space, let B be a sub-σ-algebra of A, and suppose that S∈A satisfies ∥1S-E(1S|B)∥L2≤ϵ. Then S′={x∈Ω:E(1S|B)(x)>ϵ12} satisfies λ⁢(S⁢Δ⁢S′)<5⁢ϵ12.

Proof.

We first observe that

Moreover, from the assumption and the triangle inequality we have

whence ∥𝔼(1S|ℬ)∥L22≥∥1S∥L22-2ϵ=λ(S)-2ϵ. Therefore λ⁢(S′∖S)<2⁢ϵ12.

On the other hand, we have

so λ⁢(S′∩S)≥λ⁢(S)-3⁢ϵ12, whence λ⁢(S∖S′)≤3⁢ϵ12.

Combining the main two inequalities above, the result follows. ∎

Lemma B.2.

Let (Ω,A,λ) be a probability space, let B0,B1 be sub-σ-algebras of A such that B0⁢⊥⊥λ⁡B1, let Si∈Bi, i=0,1, and suppose that λ⁢(S0⁢Δ⁢S1)≤ϵ. Then there exists C∈B0∧B1 such that λ⁢(C⁢Δ⁢Si)≤10⁢ϵ14 for i=0,1.

Proof.

The assumption ∥1S0-1S1∥L22≤ϵ implies

The assumption ℬ0⁢⊥⊥λ⁡ℬ1 implies that 𝔼(1S0|ℬ1) is ℬ0∧ℬ1-measurable (in particular, we have 𝔼(1S0|ℬ1)=𝔼(1S0|ℬ0∧ℬ1)). By Lemma B.1 with ℬ=ℬ0∧ℬ1 and 𝒜=ℬ0, the set

is in ℬ0∧ℬ1 and satisfies λ⁢(C⁢Δ⁢S0)≤5⁢(2⁢ϵ12)12≤10⁢ϵ14. Similarly, by Lemma B.1 with 𝒜=ℬ1 instead of 𝒜=ℬ0, this set C satisfies λ⁢(C⁢Δ⁢S1)≤10⁢ϵ14. ∎

Lemma B.3.

Let (X,A,λ) be the ultraproduct of probability spaces (Xi,Ai,λi). For each i let Bi,0,Bi,1 be sub-σ-algebras of Ai such that Bi,0⁢⊥⊥λi⁡Bi,1. For j=0,1 let Bj be the Loeb σ-algebra corresponding to the sequence (Bi,j)i∈N, and let C be the Loeb σ-algebra corresponding to (Bi,0∧λiBi,1)i∈N. Then B0∧λB1=λC and B0⁢⊥⊥λ⁡B1.

Proof.

Note that the inclusion ℬ0∧λℬ1⊃λ𝒞 is clear, for if A∈𝒞, then there are sets Ai∈ℬi,0∧λiℬi,1 such that A=λ∏i→ωAi, so ∏i→ωAi is in ℬj up to a null set, j=0,1, whence A∈ℬ0∧λℬ1. For the opposite inclusion, let Q be in ℬ0∧λℬ1, so for j=0,1 there are sets Qi,j∈ℬi,j for each i∈ℕ such that Q=λ∏i→ωQi,j. Then

so letting ϵi=λi⁢(Qi,0⁢Δ⁢Qi,1), we have limω⁡ϵi=0. By Lemma B.2, for each i there is a set Ci∈ℬi,0∧λiℬi,1 such that

Let R=∏i→ωCi. By construction R∈𝒞, and by the last inequality we have R=λQ, so the required inclusion holds. Finally, the desired conclusion ℬ0⁢⊥⊥λ⁡ℬ1 can be seen to follow from ℬi,0⁢⊥⊥λi⁡ℬi,1, i∈ℕ, using the definition of conditional independence [7, Definition 2.9] and basic facts about Loeb probability spaces. More precisely, by [7, Theorem 2.4 and Remark 2.5] it suffices to show that every function f in L∞⁢(ℬ1) satisfies 𝔼(f|ℬ0)=λ𝔼(f|ℬ0∧λℬ1). To show this, we use first that f is λ-almost-surely equal to a measurable function of the form f′=limω⁡fi′ (see [35, Corollary 5.1]), and then we prove the equality

by deducing it from the fact that, by the assumption ℬi,0⁢⊥⊥λi⁡ℬi,1, the analogous equality holds for the fi′. This last deduction is enabled by the fact that 𝔼(⋅|ℬ0)=limω𝔼(⋅|ℬi,0), a fact which is confirmed in a straightforward way by checking that for any function of the form g=limω⁡gi∈L1⁢(𝒜) (with each gi measurable) we have that limω𝔼(gi|ℬi,0) satisfies the defining property of the conditional expectation 𝔼(g|ℬ0), i.e. that for every h∈L1⁢(ℬ0) we have ∫𝐗hgdλ=∫𝐗hlimω𝔼(gi|ℬi,0)dλ. This last equality is seen using an S-integrable lifting of h (see [35, Theorem 6.4]), commuting ultralimit and integrals as afforded by [35, Theorem 6.2, part 4], and basic properties of ultralimits. ∎

Lemma B.4.

Let G be an amenable group acting on a Borel probability space (Ω,A,λ) by measure-preserving transformations, and let S∈A be such that for some ϵ>0 we have λ⁢(S⁢Δ⁢(g⋅S))≤ϵ for every g∈G. Then there exists a set S′∈A such that g⋅S′=λS′ for all g∈G and λ⁢(S⁢Δ⁢S′)≤5⁢ϵ14.

Proof.

We first suppose that G is countable. Let (Fj)j∈ℕ be a Følner sequence in G and for each j let hj=𝔼g∈Fj⁢1g⋅S. By the mean ergodic theorem for amenable groups [43, Theorem 2.1], letting ℬ be the σ-algebra of G-invariant sets in 𝒜, and f be a version of 𝔼(1S|ℬ), we have ∥f-hj∥L2→0 as j→∞. Note that for every j we have

so letting j→∞ yields ∥1S-f∥L2≤ϵ12. By Lemma B.1, the set S′={x∈Ω:f⁢(x)>ϵ14} satisfies λ⁢(S⁢Δ⁢S′)≤5⁢ϵ14, and since f is G-invariant, we have g⋅S′=λS′ for every g∈G.

We now reduce the general case to the countable case. It suffices to prove that if G is a group acting on a separable metric space (X,d) by isometries, then there is a countable group G0≤G such that if x∈X is a fixed point for G0, then it is a fixed point for G (we then apply this with X the measure algebra of 𝒜). Let (xi)i be a dense sequence in X. For each i, the orbit G⋅xi is itself separable, so there is a countable set Si⊂G such that Si⋅xi is dense in this orbit. Let G0 be the subgroup of G generated by ⋃iSi. Observe that for every i∈ℕ, g∈G and ϵ>0, there is g′∈Si⊂G0 such that d⁢(g⋅xi,g′⋅xi)<ϵ. Suppose for a contradiction that there is x∈X that is G0-invariant but not G-invariant, so d⁢(g⋅x,x)=ϵ>0. Then by the density of (xi)i there is i such that d⁢(x,xi)<ϵ100, so

which by the isometry property equals d⁢(g⋅x,x)-2⁢d⁢(x,xi)≥98⁢ϵ100. Hence

By the earlier observation, there is g′∈G0 such that d⁢(g⋅xi,g′⋅xi)<ϵ100, so

Combining this last inequality with d⁢(x,xi)<ϵ100 and the triangle inequality and isometry property, we deduce that d⁢(g′⋅x,x)≥d⁢(g′⋅xi,xi)-2⁢d⁢(x,xi)≥95⁢ϵ100, which contradicts that x is G0-invariant. ∎

Lemma B.5.

Let Y be a compact Polish space, let d be a metric compatible with the weak topology on P⁢(Y), and let (Xi,λi)i∈N be a sequence of Borel probability spaces. For each i∈N let fi:Xi→Y be a Borel function, and let ω be a non-principal ultrafilter on N. Then, letting f=limω⁡fi, we have limω⁡d⁢(λi∘fi-1,λ∘f-1)=0.

Proof.

As shown in [29, Theorem (17.19)], one can always metrize this space of probability measures with a metric of the form d′⁢(μ,ν)=∑r∈ℕ12r⁢|∫hr⁢dμ-∫hr⁢dν|, for a sequence of continuous functions hr:Y→ℂ with ∥hr∥∞≤1, r∈ℕ. Since d and d′ metrize the same topology, it suffices to prove that limω⁡d′⁢(λi∘fi-1,λ∘f-1)=0.

Suppose for a contradiction that for some b∈(0,1) and some set S∈ω, for every i∈S we have d′⁢(λi∘fi-1,λ∘f-1)>b. Then, for each i∈S, a short argument by contradiction shows that there exists r=r⁢(i)∈[1,2⁢⌈log2⁡(2b)⌉] such that

Using the ultrafilter properties, we then deduce that for some fixed integer r there is a set S′⊂S with S′∈ω such that for all i∈S′ we have

Now we have two exhaustive possibilities. The first one is that some S′′⊂S′ with S′′∈ω satisfies ∫Xihr∘fi⁢dλi≥∫Xhr∘f⁢dλ+b2 for all i∈S′′; but then, commuting ultralimit and integrals (as in the proof of Lemma B.3), we obtain

a contradiction. The other option is that some S′′⊂S′ with S′′∈ω satisfies

then we deduce similarly that

obtaining again a contradiction. ∎

Lemma B.6.

Let (Xi)i∈N, (Yi)i∈N be sequences of Polish spaces, and for each i∈N let μi be a Borel probability measure on B⁢(Xi) and let νi be a Borel probability measure on B⁢(Xi)⊗B⁢(Yi). Let (X,LX,μ), (X×Y,LX×Y,ν) be the corresponding Loeb probability spaces. Suppose that the projection πi:Xi×Yi→Xi, (x,y)↦x is measure preserving for every i∈N. Then the projection π:X×Y→X, (x,y)↦x is measurable with respect to LX, LX×Y, and is measure-preserving with respect to μ,ν.

Proof.

The preimage under π of any internal measurable set in 𝐗 is an internal measurable set in 𝐗×𝐘, and it is also clear that if A is an internal measurable subset of 𝐗, then ν∘π-1⁢(A)=μ⁢(A). (These claims follow from the fact the projections πi are measure-preserving maps and that taking ultraproducts commutes with taking preimages under the projections.) Now ℒ𝐗 consists precisely of sets S such that for every ϵ>0 there exist internal measurable sets Ai,Ao⊂𝐗 with Ai⊂S⊂Ao and μ⁢(Ao∖Ai)<ϵ (see [35, Section 2.1]). This combined with the properties already established for π for internal sets implies that π-1⁢(ℒ𝐗)⊂ℒ𝐗×𝐘 and μ∘π-1=ν, as required. ∎