Polish topologies on groups of non-singular transformations

In this paper, we prove several results concerning Polish group topologies on groups of non-singular transformation. We ﬁrst prove that the group of measure-preserving transformations of the real line whose support has ﬁnite measure carries no Polish group topology. We then characterize the Borel σ -ﬁnite measures λ on a standard Borel space for which the group of λ -preserving transformations has the automatic continuity property. We ﬁnally show that the natural Polish topology on the group of all non-singular transformations is actually its only Polish group topology.


Introduction
The study of measure-preserving (or more generally non-singular) transformations on a standard measured space (Y, λ) is broadened once one realises that such transformations form a Polish group.Indeed, the Baire Category Theorem is then available and so the question of generic properties of such transformations arises naturally.
As a somewhat degenerate case, one may first look at the case where the measure λ is completely atomic.Then Aut(Y, λ) only acts by permuting atoms of the same measure and thus splits as a direct product of permutation groups.In the case where all the atoms have the same measure and λ is infinite, we get the Polish group S ∞ of permutations of the integers.In this group, the generic permutation has only finite orbits and infinitely many orbits of size n for every n ∈ N.Such permutations thus form a comeager conjugacy class.
Actually a much stronger property called ample generics holds for the Polish group S ∞ .This has several nice consequences as was shown by Kechris and Rosendal [9], among which the automatic continuity property, which in turn implies that its Polish group topology is unique.Franc ¸ois Le Maître Definition 1.1 A Polish group G has the automatic continuity property if whenever π : G → H is a group homomorphism taking values in a separable topological group H , the homomorphism π has to be continuous.
It is well-known that as soon as λ has a non-atomic part, the group Aut(Y, λ) fails to have ample generics.However, it was shown by Ben Yaacov, Berenstein and Melleray that when λ is a non-atomic finite measure, Aut(X, λ) still has the automatic continuity property [2] (see also [12,Section 2] for a more direct proof).Later on Sabok developed a framework to show automatic continuity for automorphism groups of metric structures [15].In particular, he got another proof of automatic continuity for Aut(Y, λ), and then Malicki simplified his approach [13] .We first observe that this framework can also be applied when λ is infinite.
Theorem 1.2 Let (Y, λ) be a standard Borel space equipped with a non-atomic σ -finite infinite measure λ.Then Aut(Y, λ) has the automatic continuity property.
Note that as a concrete example for the above result, one can take Y to be the reals equipped with the Lebesgue measure.In general, we can actually characterize when Aut(Y, η) has the automatic continuity as follows, where the η -atomic multiplicity of a real r > 0 is the number of atoms whose measure is equal to r.
The group Aut(Y, η at ) is a permutation group, so it has the automatic continuity and thus we focus on Aut(Y, η cont ), assuming that η cont is non-trivial.Observe that η cont is then equivalent to an atomless probability measure, so we may as well assume η cont is a probability measure.We are thus led to ask: Question Let (X, µ) be a standard probability space.Does the group Aut * (X, µ) of all non-singular transformations of (X, µ) have the automatic continuity property ?
The main difficulty with this question is that the framework of Sabok is not available for Aut * (X, µ) because it cannot be the automorphism group of a complete homogeneous metric structure as was recently shown by Ben Yaacov [1].While we cannot answer this question, we still manage to obtain a basic consequence of automatic continuity, namely having a unique Polish group topology.
Theorem 1.4 Let (X, µ) be a standard probability space.The group Aut * (X, µ) has a unique Polish group topology, namely the strong topology.
The techniques we use to prove the above theorem are quite standard, except for the fact that we use the automatic continuity for Aut(X, µ) so as to know that Aut(X, µ) is a Borel subgroup of Aut * (X, µ) for any Polish group topology on Aut * (X, µ).This trick may be of use for other Polish groups.
Finally, we prove a result in the line of research started by Rosendal [14] by showing that the group of all measure-preserving transformations of the real line which have finite support cannot carry any Polish group topology.
The paper is organized in two independent sections.Section 2 deals with groups of measure-preserving transformations over a σ -finite space.After a preliminary section, we start with the above-mentioned absence of Polish group topology on the group of finite support transformations in Section 2.2.We then check that Aut(Y, ν) has the automatic continuity property in Section 2.3, and we prove Theorem 1.3 in Section 2.4.Section 3 is finally devoted to the proof of the uniqueness of the Polish group topology of Aut * (X, µ) (Theorem 1.4).
Remark Throughout the paper, we will often neglect what happens on null sets without explicit mention.
Franc ¸ois Le Maître 2 Groups of transformations preserving a σ-finite measure

Preliminaries
A standard σ -finite space is a standard Borel space equipped with a Borel nonatomic σ -finite infinite measure.All such spaces are isomorphic to R equipped with the Lebesgue measure, and we fix from now on such a standard σ -finite space (Y, λ).
The first group we are interested in is denoted by Aut(Y, λ) and consists of all Borel bijections T : Y → Y which preserve the measure λ: for all Borel A ⊆ Y , we have λ(A) = λ(T −1 (A)).As usual, two such bijections are identified if they coincide on a conull set.
Consider the space MAlg f (Y, λ) of finite measure Borel subsets of Y where we identify would only be a pseudo-metric if we had not identified sets up to measure zero.)We have the following well-known lemma.
Proof We first prove completeness.Let (A n ) be a Cauchy sequence, up to taking a subsequence we may assume that for all n ∈ N, λ(A n △ A n+1 ) < 2 −n .It then follows from the Borel-Cantelli Lemma that for almost every y ∈ Y , we have some N ∈ N such that y ̸ ∈ A n △ A n+1 for all n ⩾ N .Now denote by A the set of all y ∈ Y such that there is N ∈ N such that for all n ⩾ N , y ∈ A n .It follows from the second-to-last sentence that if y ̸ ∈ A then there is N ∈ N such that for all n ⩾ N , y ̸ ∈ A n .We thus have that for all N ∈ N, ).Since the latter has measure at most n⩾N 2 −n = 2 −N+1 which tends to zero, we conclude that A n → A as wanted.
The separability is then obtained by noting that we may as well assume Y = R endowed with the Lebesgue measure, and then finite unions of rational endpoints intervals are dense in MAlg(Y, λ).Now, if (X, µ) is a standard probability space then every Borel subset has finite measure, and by definition the measure algebra (MAlg(X, µ), d µ ) is defined as its set of Borel subsets up to measure zero, equipped with the metric For T ∈ Aut(Y, λ), we define its support to be the following Borel set, which is only well-defined up to measure zero: 2.2 Absence of Polish group topology on Aut f (Y, λ)

Non-Polishability
Our first lemma is well-known, we provide a proof for the reader's convenience.
Lemma 2.4 For all R > 0, the set of T ∈ Aut(Y, λ) such that λ(supp T) ⩽ R is closed.
Proof Take T such that λ(supp T) > R, then there exists a partition of supp T in countably many sets of positive measure (A i ) i∈N such that for all i ∈ N, we have λ(T(A i ) ∩ A i ) = 0.By our hypothesis, we may then find n ∈ N such that and up to shrinking each A i we may furthermore assume so each B i is contained in the support of T ′ , and since they are disjoint we conclude that the support of T ′ has measure greater than the Baire Category Theorem yields that there is n ∈ N such that F n has nonempty interior.Since τ is second-countable, we deduce that Aut f (Y, λ) is covered by countably many F n -translates.This means that Aut f (Y, λ) contains a countable set which is n-dense1 for the metric d λ given by: Let us explain why this cannot happen.
Fix a Borel set A ⊆ Y of measure 3n, and identify A with the circle S 1 equipped with the finite measure 3nλ, where λ is the Haar measure on S 1 .Take z ∈ S 1 and consider T z the translation by z in S 1 , which we can see through our identification as a measure preserving transformation of (Y, λ) supported on A. Observe that for z ̸ = z ′ , we have there is an uncountable subgroup all whose distinct elements are 3n-apart for the metric d λ , contradicting the fact that Aut f (Y, λ) contains a countable set which is n-dense for the metric d λ by the pigeonhole principle.

Non-existence of a Polish group topology
We now upgrade the previous theorem to see that G := Aut f (Y, λ) cannot carry a Polish group topology.Fortunately, the arguments we need were carried out by Kallman in [6] to prove the uniqueness of the Polish topology of Aut(Y, λ).We reproduce them here for the convenience of the reader.
For a Borel subset A ⊆ Y we let: Take T ̸ ∈ G Y\A .By definition the support of T intersects A, so there is By the previous lemma and the fact that centralizers are always closed in topological groups, whenever τ is a Hausdorff group topology on Aut f (Y, λ) we have that the set G Y\A is τ -closed.Also note that for all T ∈ Aut f (Y, λ) and all A ⊆ Y , we have: (1) • △ and ∩ are the usual set-theoretic operations, thus making (MAlg f (Y, λ), △, ∩) a Boolean ring (without unit).
It is well-known that MAlg f (Y, λ) is homogeneous and complete as a metric structure.
Remark Note that Malicki and Sabok make a slightly different definition by naming the empty set and keeping only the operation △, in line with Proposition 2.2.Here we prefer to provide a definition of the structure on MAlg f (Y, λ) which makes finitely generated substructures easier to understand.
Lemma 2.11 Finite tuples of disjoint finite measure subsets of (Y, λ) are ample and relevant.
Lemma 2.12 MAlg f (Y, λ) locally has finite automorphisms and has the extension property.
Proof Every finitely generated substructure of MAlg f (Y, λ) has a unit X so that we may see it as a substructure of the measure algebra MAlg(X, λ X ), which up to rescaling is the measure algebra over a standard probability space.We now prove (ii)⇒(i).Let (r i ) n i=1 be the reals whose η -atomic multiplicity belongs to [2, +∞[ and let A i be the set of atoms of measure r i .Let (s j ) j∈J denote the reals whose η -atomic multiplicity is infinite and let B j be the set of atoms of measure s j .Finally, let η n.a.be the non-atomic part of η .We then have a decomposition where S(B j ) is equipped with the topology of pointwise convergence, viewing B j as a discrete set.
Let us show that Aut(Y, η n.a. ), n i=1 S(A i ) and j∈J S(B j ) have automatic continuity.Since η is σ -finite, η n.a. also is.We then have three cases to check.Since any finite product of groups with automatic continuity has automatic continuity, we conclude from (3) that Aut(Y, η) has the automatic continuity property.
3 The group of non-singular transformations

Preliminaries
A standard probability space is a standard Borel space equipped with a Borel nonatomic probability measure.All such spaces are isomorphic, and we fix from now on such a standard probability space (X, µ).
A Borel bijection T of (X, µ) is called non-singular if the pushforward measure T * µ is equivalent to µ, that is, if for all Borel A ⊆ X , we have µ(A) = 0 if and only if µ(T −1 (A)) = 0. Denote by Aut * (X, µ) the group of non-singular Borel bijections of (X, µ), two such bijections being identified if they coincide up to measure zero.
The strong topology on Aut * (X, µ) is a metrizable group topology defined by declaring that a sequence (T n ) of elements of Aut * (X, µ) strongly converges to T ∈ Aut * (X, µ) if for all Borel A ⊆ X , one has µ(T n (A) △ T(A)) → 0 and We refer the reader to Danilenko and Silva [4] for more on this topology, which is actually a Polish group topology.Our purpose here will be to show that it is the unique Polish group topology one can put on Aut * (X, µ).
For T ∈ Aut * (X, µ), we define as before its support to be the Borel set: supp T := {x ∈ X : T(x) ̸ = x} Note that we have again the following relation: for all S, T ∈ Aut * (X, µ), supp(STS −1 ) = S(supp T).

Theorem 1 . 3
Let (Y, η) be a standard Borel space equipped with a Borel σ -finite measure η .Then the following are equivalent: (i) Aut(Y, η) has the automatic continuity property.(ii) There are only finitely many positive reals whose η -atomic multiplicity belongs to [2, +∞[.Let us now consider the group Aut * (Y, η) of non-singular transformations of (Y, η), ie the group of Borel bijections which preserve η -null sets.If η at denotes the atomic part of λ and η cont denotes the atomless part, we see that Aut * (Y, η) splits as a direct product: Aut

Lemma 2 . 8
Let τ be a Hausdorff group topology onG = Aut f (Y, λ).For all A, B ⊆ Y , the set G(A, B) is τ -closed.Proof Observe first that A ⊆ B if and only if G A ⩽ G B .The direct implication is clear; conversely, if A is not a subset of Bthen we find a transformation supported on A \ B, thus witnessing that G A ̸ ⩽ G B .By equality (1) we then have G(A, B) = {T ∈ Aut f (Y, λ) : TG A T −1 ⊆ G B }. So, by the previous lemma G(A, B) is the set of all T ∈ Aut f (Y, λ) such that for all U ∈ G A , TUT −1 commutes with every element of G X\B .This is clearly a τ -closed condition.Now take A ⊆ Y , let ϵ > 0, and pick The above proposition implies that Aut(Y, λ) is a Polish group as it is a closed subgroup of the isometry group of a separable complete metric space.The corresponding topology is called the weak topology; it is thus defined by T n → T if and only if for all A ⊆ Y of finite measure, one has: ν) is another standard probability space, any isometry between (MAlg(X, µ), d µ ) and (MAlg(Z, ν), d ν ) sending ∅ to ∅ comes from a measure-preserving bijection which is unique up to a null set (see Kechris [8, Section 1 (B)]).Using the σ -finiteness of (Y, ν) and the above fact, we easily get the following proposition.Proposition 2.2 Aut(Y, λ) is equal to the group of isometries of (MAlg f (Y, λ), d λ ) which fix ∅.
Franc ¸ois Le Maître Definition 2.5 A subgroup H of a Polish group G is called Polishable if it admits a Polish group topology which refines the topology of G. Remark By a direct application of the Lusin-Suslin theorem (see, eg, Kechris [7, Theorem 15.1]), every Polishable subgroup of a Polish group G is a Borel subset of G.Here it follows from the above lemma that Aut f (Y, λ) is F σ in Aut(Y, λ) (in particular Borel).Nevertheless, we have the following result.
Theorem 2.6 The subgroup Aut f (Y, λ) ⩽ Aut(Y, λ) is not Polishable.Proof Suppose that Aut f (Y, λ) is Polishable.Then by definition its Polish group topology τ refines the weak topology.For each n ∈ N, let: 13eorem 2.13The group Aut(Y, λ) of measure-preserving transformation of an infinite σ -finite standard measured space has the automatic continuity property.Remark Let MAlg 1 (Y, λ) denote the closed set of all A ∈ MAlg f (Y, λ) whose measure is at most 1.It is easy to check that the Aut(Y, λ)-action on MAlg 1 (Y, λ) is approximately oligomorphic and that Aut(Y, λ) is a closed subgroup of the isometry group of MAlg 1 (Y, λ).By Ben Yaacov and Tsankov [3, Theorem 2.4], we conclude that Aut(Y, λ) is a Roelcke-precompact Polish group.We finally use the previous results to characterize automatic continuity for Aut(Y, η), where (Y, η) be a standard Borel space equipped with a Borel σ -finite measure η , possibly with atoms.Recall that for such a measure there are only countably many atoms and they have finite measure (by σ -finiteness), and that each atom is a singleton (because Y is standard).Let us say that the η -atomic multiplicity of a positive real r is the (possibly infinite) number of atoms in Y whose measure is equal to r. Theorem 2.15 Let (Y, η) be a standard Borel space equipped with a Borel σ -finite measure η .Then the following are equivalent: (i) Aut(Y, η) has the automatic continuity property.(ii)Thereare only finitely many positive reals whose η -atomic multiplicity belongs to [2, +∞[.ProofWe first prove the contrapositive of (i)⇒(ii).Suppose there are infinitely many positive reals whose η -atomic multiplicity belongs to [2, +∞[ and enumerate them as (r n ) n∈N .Then if A n is the set of atoms of measure r n , we see that each A n is Aut(Y, η)-invariant and we thus get natural surjection: The result then follows from [15, Lemma 8.1 and Lemma 8.2].As a consequence of Malicki's theorem [13, Theorem 3.4], we thus have the following result.nS(A n )For each n, let σ n be the signature map S(A n ) ↠ {±1}.By composing our previous homomorphism with (σ n ) n∈N we get a continuous surjection Aut(Y, η) ↠ {±1} N .Since the latter has 2 2 ℵ 0 distinct homomorphisms onto {±1} (indeed each ultrafilter on N provides such a homomorphism) and there are at most 2 ℵ 0 continuous homomorphisms Aut(Y, η) → {±1}, we conclude that Aut(Y, η) does not have the automatic continuity property.
Journal of Logic & Analysis 14:4 (2022) Polish topologies on groups of non-singular transformations 11 • If η n.a. is trivial, Aut(Y, η n.a. ) also is and hence has automatic continuity.• If η n.a. is finite, Aut(Y, η n.a. ) has automatic continuity by Ben Yaacov, Berenstein and Melleray [2, Theorem 6.2].• If η n.a. is infinite, Aut(Y, η n.a. ) has automatic continuity by Theorem 2.13.The group n i=1 S(A i ) is finite and thus has automatic continuity.Finally the group j∈J S(B j ) is a countable product of groups with ample generics and hence has ample generics.By Kechris and Christian Rosendal [9, Theorem 1.10] it has automatic continuity.