Polish Bitopological Groups and Polish Bitopological Groupoids

1. Introduction

Polish spaces and Polish groups play a fundamental role in descriptive set theory, topological dynamics, and the study of classification problems. A Polish space is a separable topological space whose topology can be induced by a complete metric. A Polish group is a topological group whose underlying topological space is Polish [1,3].

The theory of Polish group actions provides a natural setting for studying orbit equivalence relations and the classification of mathematical structures. In particular, Polish groups frequently arise as automorphism groups of countable structures, while their actions generate important examples of Borel equivalence relations [1].

A related direction is the study of spaces endowed with two topologies. Such spaces allow two distinct notions of convergence, continuity, and openness to coexist on a common underlying set. The purpose of this paper is to formulate a Polish version of this two-topology framework for groups and groupoids.

The basic idea is to equip an algebraic group or groupoid with two Polish topologies and require its structural operations to be continuous with respect to both topologies. The two topologies need not coincide. Moreover, they need not generate the same Borel sigma-algebra unless an additional compatibility condition is imposed.

We emphasize that the definitions below specify a framework rather than asserting that every result is new. The standard theory of Polish groups and topological groupoids provides the underlying tools; our purpose is to organize these tools in a simultaneous two-topology setting.

The main contributions of the paper are as follows.

label=(0)

We introduce Polish bitopological groups and distinguish them from Borel-compatible examples.

lbbel=(0)

We establish basic results concerning closed subgroups, products, and continuous homomorphisms.

lcbel=(0)

We formulate a corresponding definition for Polish bitopological groupoids and examine their spaces of composable arrows.

ldbel=(0)

We discuss the relationship between the two Borel structures and identify several questions for further investigation.

2. Preliminaries

We begin by recalling the principal definitions.

Definition 1.

A topological space $(X,\mathcal{T})$ is called a Polish space if it is separable and its topology is induced by a complete metric.

A Polish topology is necessarily metrizable and second countable. We refer to [3] for the standard properties of Polish spaces.

Definition 2.

A bitopological space is a triple

$$(X,{\mathcal{T}}_1,{\mathcal{T}}_2),$$

where ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ are topologies on the same set X. It is called a Polish bitopological space if both $(X,{\mathcal{T}}_1)$ and $(X,{\mathcal{T}}_2)$ are Polish spaces.

For $i\in \{1,2\}$, write

$${\mathcal{B}}_i=\mathcal{B}(X,{\mathcal{T}}_i)$$

for the Borel sigma-algebra generated by ${\mathcal{T}}_i$.

Definition 3.

A Polish bitopological space $(X,{\mathcal{T}}_1,{\mathcal{T}}_2)$ is called Borel-compatible if

$$\mathcal{B}(X,{\mathcal{T}}_1)=\mathcal{B}(X,{\mathcal{T}}_2).$$

Borel compatibility is an additional condition. It does not follow merely from the fact that both topologies are Polish.

Definition 4.

Let X and Y be bitopological spaces, with topologies $({\mathcal{T}}_1,{\mathcal{T}}_2)$ and $({\mathcal{S}}_1,{\mathcal{S}}_2)$, respectively. A map $f:X\to Y$ is called bitopologically continuous if

$$f:(X,{\mathcal{T}}_i)⟶(Y,{\mathcal{S}}_i)$$

is continuous for each $i\in \{1,2\}$.

We use the standard convention that a topological group has jointly continuous multiplication and continuous inversion. The analogous convention will be used for topological groupoids.

3. Polish Bitopological Groups

3.1. Definition and Basic Properties

Definition 5.

Let G be a group, and let ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ be topologies on G. The triple

$$(G,{\mathcal{T}}_1,{\mathcal{T}}_2)$$

is called a Polish bitopological group if, for every $i\in \{1,2\}$, the triple $(G,{\mathcal{T}}_i,·)$ is a Polish topological group.

Equivalently, both topologies are Polish and the maps

$$m:G\times G\to G,m(x,y)=xy,$$

and

$$inv:G\to G,inv\left(x\right)=x^{-1},$$

are continuous with respect to each topology.

The requirement is that each topology separately makes the group into a Polish topological group. We do not require ${\mathcal{T}}_1\subseteq {\mathcal{T}}_2$ or ${\mathcal{T}}_2\subseteq {\mathcal{T}}_1$.

Remark 1.

A Polish bitopological group is not necessarily a Polish group with respect to a single preferred topology. It is a group carrying two possibly different Polish group topologies.

Proposition 1.

Let $(G,{\mathcal{T}}_1,{\mathcal{T}}_2)$ be a Polish bitopological group. Then the identity element e is closed in both topologies, and the diagonal subgroup

$${\Delta }_G=\{(g,g):g\in G\}$$

is closed in

$$(G,{\mathcal{T}}_i)\times (G,{\mathcal{T}}_i)$$

for each $i\in \{1,2\}$.

Proof. 

Every Polish space is Hausdorff. Hence each singleton $\left\{e\right\}$ is closed in $(G,{\mathcal{T}}_i)$.

For each i, the diagonal of a Hausdorff space is closed in its Cartesian square. Applying this fact to $(G,{\mathcal{T}}_i)$ proves the assertion. □

3.2. Closed Subgroups

Theorem 1.

Let $(G,{\mathcal{T}}_1,{\mathcal{T}}_2)$ be a Polish bitopological group, and let $H\le G$ be a subgroup that is closed in both ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$. Then

$$(H,{\mathcal{T}}_1{|}_H,{\mathcal{T}}_2{|}_H)$$

is a Polish bitopological group.

Proof. 

For each $i\in \{1,2\}$, the subspace $(H,{\mathcal{T}}_i{|}_H)$ is closed in the Polish space $(G,{\mathcal{T}}_i)$. Consequently, it is Polish.

Since H is a subgroup, multiplication and inversion restrict to maps

$$m_H:H\times H\to H,{inv}_H:H\to H.$$

These maps are continuous in the respective subspace topologies because they are restrictions of the continuous multiplication and inversion maps on G.

Thus each $(H,{\mathcal{T}}_i{|}_H)$ is a Polish topological group. □

Corollary 1.

If $H\le G$ is closed in both topologies, then the inclusion

$$\iota :H\hookrightarrow G$$

is continuous for each of the two topologies.

Proof. 

The inclusion of a subspace into its ambient space is continuous. Apply this observation to each subspace topology. □

Remark 2.

Closedness in only one of the two topologies guarantees that the corresponding subspace is Polish, but it does not by itself guarantee that the subspace is Polish for the other topology.

3.3. Products

Theorem 2.

Let

$$(G,{\mathcal{T}}_1,{\mathcal{T}}_2)\mathrm{and}(H,{\mathcal{S}}_1,{\mathcal{S}}_2)$$

be Polish bitopological groups. Define

$${\mathcal{P}}_i={\mathcal{T}}_i\times {\mathcal{S}}_i,i=1,2.$$

Then

$$(G\times H,{\mathcal{P}}_1,{\mathcal{P}}_2)$$

is a Polish bitopological group under the product group operation.

Proof. 

For each i, the product of the Polish spaces $(G,{\mathcal{T}}_i)$ and $(H,{\mathcal{S}}_i)$ is Polish.

The multiplication map on $G\times H$ is

$$((g_1,h_1),(g_2,h_2))⟼(g_1{g}_2,h_1{h}_2).$$

It is continuous for ${\mathcal{P}}_i$ because its coordinate maps are continuous.

Similarly, inversion is given by

$$(g,h)⟼(g^{-1},h^{-1}),$$

which is continuous for each product topology. Therefore both product topologies define Polish group structures. □

3.4. Homomorphisms

Definition 6.

Let $(G,{\mathcal{T}}_1,{\mathcal{T}}_2)$ and $(H,{\mathcal{S}}_1,{\mathcal{S}}_2)$ be Polish bitopological groups. A group homomorphism $f:G\to H$ is called a bitopological homomorphism if

$$f:(G,{\mathcal{T}}_i)\to (H,{\mathcal{S}}_i)$$

is continuous for $i=1,2$.

Proposition 2.

Let $f:G\to H$ be a bitopological homomorphism. Then f is Borel measurable with respect to each of the two Borel structures:

$$f^{-1}\left(B\right)\in \mathcal{B}(G,{\mathcal{T}}_i)\mathrm{whenever}B\in \mathcal{B}(H,{\mathcal{S}}_i).$$

Proof. 

For each i, the map

$$f:(G,{\mathcal{T}}_i)\to (H,{\mathcal{S}}_i)$$

is continuous. The inverse image of every open set is therefore open. Since inverse images commute with complements and countable unions, the same assertion holds for all Borel sets. □

Remark 3.

If the two topologies on each group generate the same Borel sigma-algebra, the preceding proposition yields measurability relative to a common Borel structure. Without this compatibility, the two measurability statements concern potentially different sigma-algebras.

4. Examples and Borel Compatibility

4.1. The Diagonal Example

Example 1.

Let G be any Polish group with topology $\mathcal{T}$. Then

$$(G,\mathcal{T},\mathcal{T})$$

is a Polish bitopological group.

This example shows that the class is nonempty and contains every ordinary Polish group as a special case.

4.2. Two Distinct Polish Group Topologies

Example 2.

Let $G=(\mathbb{R},+)$, regarded as an abstract additive group. Choose a discontinuous additive bijection

$$\phi :\mathbb{R}\to \mathbb{R}.$$

Such a map can be obtained from a nontrivial automorphism of the vector space $\mathbb{R}$ over $\mathbb{Q}$.

Let ${\mathcal{T}}_1$ be the usual topology on $\mathbb{R}$, and define

$${\mathcal{T}}_2=\{{\phi }^{-1}\left(U\right):U\mathrm{is}\mathrm{usual}-\mathrm{open}\}.$$

Then $(\mathbb{R},{\mathcal{T}}_2)$ is a Polish topological group, because φ is a group isomorphism and a homeomorphism from $(\mathbb{R},{\mathcal{T}}_2)$ onto the usual real line.

Thus

$$(\mathbb{R},{\mathcal{T}}_1,{\mathcal{T}}_2)$$

is a Polish bitopological group. Since φ is discontinuous in the usual topology, ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ are distinct.

4.3. Borel Structures

The descriptive-set-theoretic significance of Polish spaces is closely connected with their Borel sets [3]. This motivates the following observation.

Proposition 3.

Suppose that

$$(G,{\mathcal{T}}_1,{\mathcal{T}}_2)$$

is a Polish bitopological group and

$$\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2).$$

Then the identity map

$$id:(G,{\mathcal{T}}_1)\to (G,{\mathcal{T}}_2)$$

is a Borel isomorphism.

Proof. 

For every ${\mathcal{T}}_2$-open set U, the set U belongs to $\mathcal{B}(G,{\mathcal{T}}_2)$. By the assumed equality, it also belongs to $\mathcal{B}(G,{\mathcal{T}}_1)$. Hence the identity map from $(G,{\mathcal{T}}_1)$ to $(G,{\mathcal{T}}_2)$ is Borel measurable.

The same argument with the two topologies interchanged shows that the inverse identity map is Borel measurable. □

Remark 4.

Borel compatibility does not imply that the two topologies are equal. In particular, Borel isomorphism is weaker than homeomorphism. The distinction is important in descriptive set theory.

5. Polish Bitopological Groupoids

5.1. Topological Groupoids

We recall the basic groupoid notation. A groupoid consists of a space of arrows G, a space of objects $G^{\left(0\right)}$, source and target maps

$$s,t:G\to G^{\left(0\right)},$$

an identity map

$$u:G^{\left(0\right)}\to G,$$

an inversion map

$$inv:G\to G,$$

and a composition map defined on the space of composable pairs

$$G^{\left(2\right)}=\{(g,h)\in G\times G:s\left(g\right)=t\left(h\right)\}.$$

We use the convention that $gh$ is defined when $s\left(g\right)=t\left(h\right)$, with

$$s\left(gh\right)=s\left(h\right),t\left(gh\right)=t\left(g\right).$$

A topological groupoid is a groupoid for which the structural maps are continuous, with $G^{\left(2\right)}$ equipped with the subspace topology inherited from $G\times G$. Standard background on topological groupoids can be found in [2].

5.2. Definition

Definition 7.

A Polish bitopological groupoid is a groupoid

$$\mathcal{G}=(G,G^{\left(0\right)},s,t,u,inv,m)$$

equipped with two topologies ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ on G and two topologies ${\mathcal{T}}_1^{\left(0\right)}$ and ${\mathcal{T}}_2^{\left(0\right)}$ on $G^{\left(0\right)}$, such that the following conditions hold for each $i\in \{1,2\}$:

• $(G,{\mathcal{T}}_i)$ and $(G^{\left(0\right)},{\mathcal{T}}_i^{\left(0\right)})$ are Polish spaces.
• The source and target maps

  $$s,t:(G,{\mathcal{T}}_i)\to (G^{\left(0\right)},{\mathcal{T}}_i^{\left(0\right)})$$

  are continuous.
• The unit map

  $$u:(G^{\left(0\right)},{\mathcal{T}}_i^{\left(0\right)})\to (G,{\mathcal{T}}_i)$$

  and inversion map

  $$inv:(G,{\mathcal{T}}_i)\to (G,{\mathcal{T}}_i)$$

  are continuous.
• The composition map

  $$m:G^{\left(2\right)}\to G,m(g,h)=gh,$$

  is continuous when $G^{\left(2\right)}$ is equipped with the subspace topology inherited from $(G,{\mathcal{T}}_i)\times (G,{\mathcal{T}}_i)$.

Remark 5.

The two topologies on the arrow space need not coincide, and neither do the two topologies on the object space. The definition requires the groupoid operations to be continuous at each level separately.

5.3. The Space of Composable Arrows

The following elementary fact explains why the composable-pair space has a useful Polish structure.

Lemma 1.

Let $(G,\mathcal{T})$ and $(G^{\left(0\right)},{\mathcal{T}}^{\left(0\right)})$ be Polish spaces, where $(G^{\left(0\right)},{\mathcal{T}}^{\left(0\right)})$ is Hausdorff. Suppose that

$$s,t:(G,\mathcal{T})\to (G^{\left(0\right)},{\mathcal{T}}^{\left(0\right)})$$

are continuous. Then

$$G^{\left(2\right)}=\{(g,h):s\left(g\right)=t\left(h\right)\}$$

is closed in $G\times G$. In particular, $G^{\left(2\right)}$ is Polish.

Proof. 

Consider the continuous map

$$F:G\times G\to G^{\left(0\right)}\times G^{\left(0\right)},F(g,h)=(s\left(g\right),t\left(h\right)).$$

Since $G^{\left(0\right)}$ is Hausdorff, its diagonal

$${\Delta }_{G^{\left(0\right)}}=\{(x,x):x\in G^{\left(0\right)}\}$$

is closed.

We have

$$G^{\left(2\right)}=F^{-1}\left({\Delta }_{G^{\left(0\right)}}\right).$$

Therefore $G^{\left(2\right)}$ is closed in $G\times G$. The product of two Polish spaces is Polish, and a closed subspace of a Polish space is Polish. Hence $G^{\left(2\right)}$ is Polish. □

Corollary 2.

If $\mathcal{G}$ is a Polish bitopological groupoid, then $G^{\left(2\right)}$ is Polish with respect to each of the two subspace topologies induced by ${\mathcal{T}}_1\times {\mathcal{T}}_1$ and ${\mathcal{T}}_2\times {\mathcal{T}}_2$.

Proof. 

Apply the lemma separately to each topology. The object space is Polish and therefore Hausdorff for both topologies, and the source and target maps are continuous for both. □

5.4. Examples

Example 3.

Every Polish bitopological group determines a Polish bitopological groupoid with one object.

Indeed, regard G as the arrow space and let

$$G^{\left(0\right)}=\left\{e\right\}.$$

The source and target maps are constant, the unit map sends the unique object to e, and the groupoid composition and inversion are the group operations. All required maps are continuous for both topologies.

Example 4.

Let X be a Polish bitopological space. Its pair groupoid is

$$\mathcal{G}=X\times X,G^{\left(0\right)}=X.$$

An arrow $(x,y)$ is regarded as an arrow from y to x. The structural maps are

$$s(x,y)=y,t(x,y)=x,$$

$$u\left(x\right)=(x,x),inv(x,y)=(y,x).$$

Composition is given by

$$(x,y)(y,z)=(x,z).$$

Equip $X\times X$ with the product topology ${\mathcal{T}}_i\times {\mathcal{T}}_i$ for each $i\in \{1,2\}$. Then the pair groupoid is a Polish bitopological groupoid.

The composable-pair space is naturally identified with $X^3$, and the composition map is continuous for each of the two product structures.

6. Morphisms and Subgroupoids

Definition 8.

Let $\mathcal{G}$ and $\mathcal{H}$ be Polish bitopological groupoids. A bitopological groupoid morphism consists of maps

$$F:G\to H,F^{\left(0\right)}:G^{\left(0\right)}\to H^{\left(0\right)}$$

that preserve source, target, units, inversion, and composition, and are continuous with respect to both corresponding topologies.

Proposition 4.

The composition of two bitopological groupoid morphisms is again a bitopological groupoid morphism.

Proof. 

Let

$$(F,F^{\left(0\right)}):\mathcal{G}\to \mathcal{H}$$

and

$$(K,K^{\left(0\right)}):\mathcal{H}\to \mathcal{K}$$

be such morphisms.

The composite maps are

$$K\circ F:G\to K_1,K^{\left(0\right)}\circ F^{\left(0\right)}:G^{\left(0\right)}\to K_1^{\left(0\right)},$$

where $K_1$ denotes the arrow space of the target groupoid. They are continuous for each topology because compositions of continuous maps are continuous.

Since both original morphisms preserve the groupoid structure, their composite preserves source, target, units, inversion, and composition. Hence it is a bitopological groupoid morphism. □

Definition 9.

A subgroupoid $\mathcal{H}$ of a groupoid $\mathcal{G}$ is called closed in both topologies if its arrow space is closed in $(G,{\mathcal{T}}_i)$ and its object space is closed in $(G^{\left(0\right)},{\mathcal{T}}_i^{\left(0\right)})$ for $i=1,2$.

Theorem 3.

Let $\mathcal{G}$ be a Polish bitopological groupoid, and let $\mathcal{H}$ be a subgroupoid closed in both topologies. With the induced subspace topologies, $\mathcal{H}$ is a Polish bitopological groupoid.

Proof. 

For each i, the arrow space of $\mathcal{H}$ is a closed subspace of the Polish space $(G,{\mathcal{T}}_i)$. Likewise, its object space is closed in $(G^{\left(0\right)},{\mathcal{T}}_i^{\left(0\right)})$. Both spaces are therefore Polish.

The structural maps of $\mathcal{H}$ are restrictions of the corresponding structural maps of $\mathcal{G}$. Because $\mathcal{H}$ is a subgroupoid, these restrictions take values in the appropriate subspaces. They are continuous in the induced topologies.

The composition map is the restriction of the composition map of $\mathcal{G}$ to the composable pairs in $\mathcal{H}$. The subspace topology on these pairs is the one inherited from the product of the arrow-space topologies. Thus composition is continuous for each i.

All the defining conditions are satisfied. □

7. Orbit Relations and Descriptive Set Theory

Let $\mathcal{G}$ be a topological groupoid with object space X. Its orbit relation is

$$x{E}_{\mathcal{G}}y⟺\mathrm{there}\mathrm{exists}g\in G\mathrm{with}s\left(g\right)=y,t\left(g\right)=x,$$

where $x{E}_{\mathcal{G}}y$ means that x and y belong to the same groupoid orbit. Orbit equivalence relations associated with Polish group actions form a central part of descriptive set theory [1]. The groupoid setting extends this viewpoint and includes equivalence relations and transformation groupoids.

For a Polish bitopological groupoid, we obtain two topological descriptions of the same underlying algebraic orbit relation.

Proposition 5.

Let $\mathcal{G}$ be a Polish bitopological groupoid. The orbit relation $E_{\mathcal{G}}$ is the same set-theoretic relation whether it is considered using the first or the second topology.

Proof. 

The orbit relation is defined by the existence of an arrow with prescribed source and target. This condition depends only on the underlying groupoid structure, not on the choice of topology. Therefore the two topologies produce the same underlying relation. □

Remark 6.

The preceding proposition does not assert that the orbit relation is Borel in either product topology. Borelness is a further descriptive property that requires additional hypotheses.

Proposition 6.

Suppose that the two topologies on the object space of a Polish bitopological groupoid generate the same Borel sigma-algebra. If its orbit relation is Borel with respect to the first product Borel structure, then it is Borel with respect to the second product Borel structure.

Proof. 

Write the two object-space Borel structures as ${\mathcal{B}}_1$ and ${\mathcal{B}}_2$, and suppose that ${\mathcal{B}}_1={\mathcal{B}}_2$.

The Borel sigma-algebra on the product of two second-countable spaces is generated by measurable rectangles. Consequently, the product Borel structures agree:

$${\mathcal{B}}_1\otimes {\mathcal{B}}_1={\mathcal{B}}_2\otimes {\mathcal{B}}_2.$$

Thus a subset of $X\times X$ is Borel for one product Borel structure if and only if it is Borel for the other. □

8. Further Structural Questions

The framework suggests several directions for additional research.

8.1. Compatibility of the Two Topologies

Let $(G,{\mathcal{T}}_1,{\mathcal{T}}_2)$ be a Polish bitopological group. Since ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ are both Polish group topologies on the same abstract group G, there are several natural levels of compatibility between them. These levels should be distinguished carefully. In particular, equality of the two Borel $\sigma$-algebras is substantially weaker than equality of the topologies themselves.

For $i=1,2$, let ${\mathcal{B}}_i=\mathcal{B}(G,{\mathcal{T}}_i)$ denote the Borel $\sigma$-algebra associated with ${\mathcal{T}}_i$. We say that the two Polish group topologies are Borel-compatible if ${\mathcal{B}}_1={\mathcal{B}}_2$. This condition means that the two topological structures have the same measurable sets, although they may have different open sets and different convergent sequences.

Now, we have the following implications:

$${\mathcal{T}}_1={\mathcal{T}}_2⟹id:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)\mathrm{is}\mathrm{a}\mathrm{homeomorphism}$$

and

$$id:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)\mathrm{is}\mathrm{a}\mathrm{homeomorphism}⟹{\mathcal{B}}_1={\mathcal{B}}_2.$$

The converse implications do not hold in general.

Proposition 7.

Let $(G,{\mathcal{T}}_1,{\mathcal{T}}_2)$ be a Polish bitopological group. The following conditions are equivalent:

label=()

$\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2)$;

lbbel=()

the identity map

$$id:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)$$

is a Borel isomorphism;

lcbel=()

the identity map

$$id:(G,{\mathcal{T}}_2)⟶(G,{\mathcal{T}}_1)$$

is a Borel isomorphism.

Proof. 

Suppose first that $\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2)$.

For every ${\mathcal{T}}_2$-open set U, we have $U\in \mathcal{B}(G,{\mathcal{T}}_2)=\mathcal{B}(G,{\mathcal{T}}_1)$.

Hence the identity map $id:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)$ is Borel measurable. Interchanging ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ shows that the inverse identity map is also Borel measurable. Thus the identity is a Borel isomorphism.

Conversely, suppose that the identity map in (ii) is a Borel isomorphism. If $A\in \mathcal{B}(G,{\mathcal{T}}_2)$, then $A={id}^{-1}\left(A\right)$ belongs to $\mathcal{B}(G,{\mathcal{T}}_1)$. Hence $\mathcal{B}(G,{\mathcal{T}}_2)\subseteq \mathcal{B}(G,{\mathcal{T}}_1)$.

Since the inverse identity is also Borel measurable, the reverse inclusion holds. Therefore $\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2)$.

The equivalence with (iii) follows symmetrically. □

Thus Borel compatibility can be interpreted as saying that the two Polish groups have the same underlying measurable space, even though their topological structures may differ.

8.1.1. Continuity of the Identity Maps

A stronger form of compatibility is obtained by considering the identity maps ${id}_{12}:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)$ and ${id}_{21}:(G,{\mathcal{T}}_2)⟶(G,{\mathcal{T}}_1)$.

The map ${id}_{12}$ is continuous if and only if ${\mathcal{T}}_2\subseteq {\mathcal{T}}_1$.

Indeed, continuity of the identity means that every ${\mathcal{T}}_2$-open set is ${\mathcal{T}}_1$-open. Similarly,

$${id}_{21}\mathrm{is}\mathrm{continuous}⟺{\mathcal{T}}_1\subseteq {\mathcal{T}}_2.$$

Therefore, both identity maps are continuous precisely when ${\mathcal{T}}_1={\mathcal{T}}_2$.

Proposition 8.

Let $(G,{\mathcal{T}}_1,{\mathcal{T}}_2)$ be a Polish bitopological group. Then ${\mathcal{T}}_1={\mathcal{T}}_2$ if and only if the identity map $id:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)$ is a homeomorphism.

Proof. 

If ${\mathcal{T}}_1={\mathcal{T}}_2$, the identity map is clearly a homeomorphism.

Conversely, suppose that the identity map is a homeomorphism. Then it is continuous, so ${\mathcal{T}}_2\subseteq {\mathcal{T}}_1$. Its inverse is also continuous, giving ${\mathcal{T}}_1\subseteq {\mathcal{T}}_2$.

Therefore ${\mathcal{T}}_1={\mathcal{T}}_2$. □

It follows that the three conditions

$${\mathcal{B}}_1={\mathcal{B}}_2,{id}_{12}\mathrm{is}\mathrm{Borel},{id}_{12}\mathrm{is}\mathrm{a}\mathrm{homeomorphism}$$

should not be confused. Borel measurability is much weaker than continuity.

8.1.2. A Hierarchy of Compatibility

The preceding observations give the following hierarchy:

$${\mathcal{T}}_1={\mathcal{T}}_2⟹{id}_{12}\mathrm{is}\mathrm{a}\mathrm{homeomorphism}⟹{id}_{12}\mathrm{is}\mathrm{Borel}.$$

In the present setting, the first implication is actually an equivalence, because a homeomorphism whose underlying map is the identity forces equality of the two topologies. Moreover,

$${id}_{12}\mathrm{is}\mathrm{Borel}⟺{\mathcal{B}}_2\subseteq {\mathcal{B}}_1,$$

while

$${id}_{12}\mathrm{and}{id}_{21}\mathrm{are}\mathrm{both}\mathrm{Borel}⟺{\mathcal{B}}_1={\mathcal{B}}_2.$$

Thus one may distinguish at least four situations:

label=(0)

${\mathcal{T}}_1={\mathcal{T}}_2$;

lbbel=(0)

${\mathcal{T}}_1\ne {\mathcal{T}}_2$, but one topology is finer than the other;

lcbel=(0)

${\mathcal{T}}_1\ne {\mathcal{T}}_2$ and the two topologies are incomparable, but their Borel $\sigma$-algebras coincide;

ldbel=(0)

the two topologies generate different Borel $\sigma$-algebras.

The third situation is particularly interesting from the descriptive-set-theoretic point of view. The topology can change while the measurable structure remains unchanged.

8.1.3. The Role of the Borel Structures

Suppose that ${\mathcal{B}}_1={\mathcal{B}}_2=:\mathcal{B}$.

Then $(G,\mathcal{B})$ is a common standard Borel space for both Polish topologies. In this case, any subset of G that is Borel with respect to one topology is automatically Borel with respect to the other.

This observation is useful for comparing measurable maps. Let X be another standard Borel space and let

$$f:G⟶X$$

be a map. If f is Borel measurable with respect to ${\mathcal{T}}_1$ and

$$\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2),$$

then f is also Borel measurable with respect to ${\mathcal{T}}_2$. Indeed, for every Borel set $B\subseteq X$,

$$f^{-1}\left(B\right)\in \mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2).$$

It is automatically Borel measurable with respect to ${\mathcal{T}}_2$. Indeed, for every Borel set $B\subseteq X$, $f^{-1}\left(B\right)\in {\mathcal{B}}_1={\mathcal{B}}_2$.

Thus, under Borel compatibility, the two topologies induce the same class of Borel measurable functions into standard Borel spaces.

Proposition 9.

Suppose that $\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2)$.

Let X be a standard Borel space. Then a map $f:G\to X$ is Borel measurable with respect to $(G,{\mathcal{T}}_1)$ if and only if it is Borel measurable with respect to $(G,{\mathcal{T}}_2)$.

Proof. 

If f is Borel measurable with respect to ${\mathcal{T}}_1$, then for every Borel set $B\subseteq X$, $f^{-1}\left(B\right)\in \mathcal{B}(G,{\mathcal{T}}_1)$.

By the assumed equality of Borel structures, $f^{-1}\left(B\right)\in \mathcal{B}(G,{\mathcal{T}}_2$). Hence f is Borel measurable with respect to ${\mathcal{T}}_2$. The converse follows by symmetry. □

8.1.4. When One Topology Is Finer than the Other

Suppose that ${\mathcal{T}}_2\subseteq {\mathcal{T}}_1$. Then the identity map $id:(G,{\mathcal{T}}_1)\to (G,{\mathcal{T}}_2)$ is continuous. Since every ${\mathcal{T}}_2$-open set is also ${\mathcal{T}}_1$-Borel, we obtain ${\mathcal{B}}_2\subseteq {\mathcal{B}}_1$.

Hence a refinement of the topology produces a refinement of the corresponding Borel $\sigma$-algebra at the level of inclusion. However, the inclusion ${\mathcal{B}}_2\subseteq {\mathcal{B}}_1$ does not by itself imply ${\mathcal{T}}_2\subseteq {\mathcal{T}}_1$.

Borel information is weaker than topological information.

If, in addition, ${\mathcal{B}}_1={\mathcal{B}}_2$, then the identity map is a Borel isomorphism, but it need not be continuous in either direction.

8.1.5. The Borel Identity Map and Automatic Continuity

Since both $(G,{\mathcal{T}}_1)$ and $(G,{\mathcal{T}}_2)$ are Polish groups, a natural question is whether a Borel identity homomorphism must automatically be continuous.

This is an important point where the group structure becomes stronger than the structure of arbitrary Polish spaces. The classical automatic continuity theorem for Polish groups implies that a Borel measurable homomorphism between Polish groups is continuous. Consequently, if $id:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)$ is Borel measurable, then it is continuous.

In particular, for two Polish group topologies on the same abstract group, Borel compatibility is much stronger than merely saying that the two spaces are Borel isomorphic: the identity map is a group homomorphism, and its Borel regularity can force topological regularity.

Thus, in the Polish group setting, one should distinguish the general descriptive-set-theoretic situation from the more rigid group-theoretic situation.

Remark 7.

The automatic continuity statement requires the hypotheses of the relevant automatic continuity theorem. It should therefore not be applied to arbitrary Borel maps between Polish spaces. The fact that the identity map is a group homomorphism is essential.

8.1.6. A Useful Comparison Principle

The preceding discussion motivates the following comparison principle.

Proposition 10.

Let $(G,{\mathcal{T}}_1,{\mathcal{T}}_2)$ be a Polish bitopological group. Assume that the identity map $id:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)$ is Borel measurable and that the corresponding automatic continuity theorem applies to Borel homomorphisms between the two Polish groups. Then $id:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)$ is continuous, and hence ${\mathcal{T}}_2\subseteq {\mathcal{T}}_1$.

If the inverse identity is also Borel measurable, then ${\mathcal{T}}_1={\mathcal{T}}_2$.

Proof. 

The identity map is a group homomorphism. By the assumed automatic continuity theorem, its Borel measurability implies continuity. Therefore ${\mathcal{T}}_2\subseteq {\mathcal{T}}_1$.

If the inverse identity is also Borel measurable, the same argument gives ${\mathcal{T}}_1\subseteq {\mathcal{T}}_2$. Hence, ${\mathcal{T}}_1={\mathcal{T}}_2$. □

This observation shows that, under suitable automatic continuity hypotheses, two Polish group topologies on the same abstract group cannot behave like completely unrelated Polish structures if their identity maps are Borel measurable.

8.2. Compatibility Questions for Polish Bitopological Groups

Let $(G,{\mathcal{T}}_1,{\mathcal{T}}_2)$ be a Polish bitopological group. For $i=1,2$, write ${\mathcal{B}}_i=\mathcal{B}(G,{\mathcal{T}}_i)$ for the Borel $\sigma$-algebra generated by ${\mathcal{T}}_i$. The comparison of ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ can be studied at several different levels. One may compare the topologies themselves, the corresponding Borel structures, or the continuity and measurability of the identity homomorphism ${id}_G⟶G$.

For arbitrary Polish spaces, equality of Borel structures is strictly weaker than equality of topologies. In the category of Polish groups, however, the algebraic structure imposes strong additional rigidity. In particular, the Banach–Pettis automatic continuity theorem implies that a Borel measurable homomorphism between Polish groups is continuous. Consequently, Borel compatibility of two Polish group topologies on the same abstract group forces the identity map to be continuous in both directions, and hence forces equality of the two topologies. This is one of the main structural differences between Polish spaces and Polish groups. See, for example, the discussion of automatic continuity in [3] and [1].

We now address the questions listed above.

8.2.1. Algebraic Conditions Implying Borel Compatibility

Given two Polish group topologies ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ on the same abstract group, what algebraic conditions imply $\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2$)?

A useful sufficient condition is that the underlying abstract group have a strong uniqueness property for Polish group topologies. One formulation is the following.

Definition 10.

An abstract group G is said to have theunique Polish group topology propertyif whenever ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ are Polish group topologies on G, one has ${\mathcal{T}}_1={\mathcal{T}}_2$.

Clearly, this property immediately implies Borel compatibility. Indeed,

$${\mathcal{T}}_1={\mathcal{T}}_2⟹\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2).$$

A stronger and very useful sufficient condition is automatic continuity.

Definition 11.

A Polish group G is said to have theautomatic continuity propertyif every continuous-group homomorphism from G into a separable topological group is continuous whenever the homomorphism is considered with respect to the relevant measurable or Baire structure; in particular, one frequently uses the formulation that every Baire measurable homomorphism from G into a Polish group is continuous.

For the present problem, the following consequence is more directly relevant.

Proposition 11.

Suppose that G has the property that every Borel measurable homomorphism from G into a Polish group is continuous. If ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ are Polish group topologies on G and $\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2)$, then ${\mathcal{T}}_1={\mathcal{T}}_2$.

Proof. 

The identity map ${id}_{12}:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)$ is a group homomorphism. Since the two Borel structures coincide, ${id}_{12}$ is Borel measurable. By the assumed automatic continuity property, ${id}_{12}$ is continuous. Therefore ${\mathcal{T}}_2\subseteq {\mathcal{T}}_1$.

The inverse identity ${id}_{21}:(G,{\mathcal{T}}_2)⟶(G,{\mathcal{T}}_1)$ is also Borel measurable. Applying the same argument gives ${\mathcal{T}}_1\subseteq {\mathcal{T}}_2$. Hence ${\mathcal{T}}_1={\mathcal{T}}_2$. □

For arbitrary Polish groups, the required automatic continuity statement follows from the classical Banach–Pettis theorem: a Borel measurable homomorphism between Polish groups is continuous. Thus we obtain the following important conclusion.

Theorem 4.

Let ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ be Polish group topologies on the same abstract group G. If $\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2)$, then ${\mathcal{T}}_1={\mathcal{T}}_2$.

Proof. 

The identity map ${id}_{12}:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)$ is a Borel measurable group homomorphism. By the Banach–Pettis automatic continuity theorem, it is continuous. The inverse identity is similarly a Borel measurable group homomorphism and is therefore continuous. Hence the identity is a homeomorphism, and hence ${\mathcal{T}}_1={\mathcal{T}}_2$. □

This result shows that, for Polish groups, the Borel structure together with the abstract group operation contains enough information to recover the Polish group topology.

Consequently, algebraic conditions that imply uniqueness of the Polish group topology automatically imply Borel compatibility. Examples of such uniqueness phenomena occur for several important groups with strong automatic-continuity properties. For instance, certain automorphism groups of standard measured structures have unique Polish group topologies as a consequence of automatic continuity results [1].

It is important, however, not to claim that properties such as local compactness or non-Archimedeanness alone always imply uniqueness. Additional hypotheses are generally required.

8.2.2. When Does Borel Compatibility Imply Continuity?

Under what hypotheses does Borel compatibility imply that the identity map is continuous?

For Polish groups, the answer is particularly strong.

Suppose that $\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2)$.

Then the identity map ${id}_{12}:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)$ is Borel measurable. Since it is also a group homomorphism, the Banach–Pettis theorem gives continuity:

$${id}_{12}:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)\mathrm{is}\mathrm{continuous}.$$

Applying the same argument to the inverse identity gives

$${id}_{21}:(G,{\mathcal{T}}_2)⟶(G,{\mathcal{T}}_1)\mathrm{continuous}.$$

Therefore ${\mathcal{T}}_1={\mathcal{T}}_2$.

Thus we have the equivalence

Theorem 5.

For two Polish group topologies ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ on the same abstract group G, the following conditions are equivalent:

$$\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2);$$

the identity map

$$id:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)$$

is Borel measurable and its inverse is Borel measurable; the identity map

$$id:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)$$

is a homeomorphism;

$${\mathcal{T}}_1={\mathcal{T}}_2.$$

Proof. 

The equivalence of (i) and (ii) follows directly from the definition of the Borel structures.

Clearly, (iii) and (iv) are equivalent.

Assume (i). Then the identity map in (ii) is a Borel measurable group homomorphism. By the Banach–Pettis theorem it is continuous. The same applies to its inverse. Hence (iii) holds.

Finally, (iii) implies (i), because a homeomorphism maps Borel sets to Borel sets. □

This theorem is particularly useful for the present theory. It shows that two genuinely distinct Polish group topologies on the same abstract group must necessarily have different Borel structures.

8.2.3. Comparable Polish Group Topologies

If one topology is contained in the other, when must the two Borel $\sigma$-algebras coincide?

Suppose that ${\mathcal{T}}_2\subseteq {\mathcal{T}}_1$.

Then $id:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)$ is automatically continuous. Hence $\mathcal{B}(G,{\mathcal{T}}_2)\subseteq \mathcal{B}(G,{\mathcal{T}}_1)$.

For arbitrary topological spaces, equality of the Borel $\sigma$-algebras does not follow from this inclusion. In the Polish group setting, however, the situation is much more rigid.

Theorem 6.

Let ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ be Polish group topologies on the same abstract group G. If ${\mathcal{T}}_2\subseteq {\mathcal{T}}_1$, then ${\mathcal{T}}_1={\mathcal{T}}_2$. Therefore, $\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2)$.

Proof. 

The inclusion ${\mathcal{T}}_2\subseteq {\mathcal{T}}_1$ means that $id:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)$ is a continuous bijective group homomorphism.

A continuous bijective homomorphism between Polish groups is a topological isomorphism. Equivalently, its inverse is continuous. Therefore $id:(G,{\mathcal{T}}_2)⟶(G,{\mathcal{T}}_1)$ is continuous. Thus ${\mathcal{T}}_1\subseteq {\mathcal{T}}_2$.

Together with the assumed inclusion ${\mathcal{T}}_2\subseteq {\mathcal{T}}_1$

we obtain ${\mathcal{T}}_1={\mathcal{T}}_2$.

The equality of the Borel structures follows immediately. □

Hence two distinct Polish group topologies on the same abstract group cannot be comparable by inclusion. They must be incomparable as topologies.

This gives a useful structural observation:

$$\begin{array}{|c|}\hline {\mathcal{T}}_1\ne {\mathcal{T}}_2⟹{\mathcal{T}}_1⊈{\mathcal{T}}_2\mathrm{and}{\mathcal{T}}_2⊈{\mathcal{T}}_1.\\ \hline\end{array}$$

Thus the two-topology theory becomes genuinely interesting only when the two Polish group topologies are incomparable.

8.2.4. Algebraic Restrictions on Pairs of Polish Group Topologies

8.2.4.1. Automatic continuity.

Suppose G has the property that every Borel measurable homomorphism from G into a Polish group is continuous. If ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ are Polish group topologies with $\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2)$, then the identity maps are continuous in both directions, and therefore ${\mathcal{T}}_1={\mathcal{T}}_2$.

More generally, if G has the stronger property that every homomorphism from G into a separable topological group is continuous, then G admits at most one separable group topology of the relevant type. This is the mechanism behind many uniqueness-of-topology results.

8.2.4.2. Local compactness.

Local compactness by itself should not be stated as an automatic uniqueness condition for arbitrary abstract groups. Nevertheless, it becomes powerful when combined with additional algebraic hypotheses.

For example, if an abstract group admits a locally compact Polish topology and the Borel structure can be characterized intrinsically from a countable algebraic family of subsets, then any other locally compact Polish group topology may be forced to have the same Borel structure. Automatic continuity can then yield equality of the two topologies.

Thus a useful strategy is:

$$\mathrm{algebraic}\mathrm{description}\mathrm{of}\mathrm{Borel}\mathrm{sets}⟹\mathrm{Borel}\mathrm{compatibility}⟹\mathrm{automatic}\mathrm{continuity}⟹\mathrm{uniqueness}\mathrm{of}\mathrm{topology}.$$

Results of this type are known for important classes of locally compact groups and for groups with sufficiently rigid algebraic structures.

8.2.4.3. Non-Archimedean groups.

A Polish group is called non-Archimedean if it has a neighborhood basis at the identity consisting of open subgroups. Equivalently, its topology is determined by a suitable countable family of open subgroups.

If ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ are two non-Archimedean Polish group topologies on G, the comparison of the two structures can therefore be reduced, in many cases, to comparison of the corresponding families of open subgroups.

For example, suppose there are countable families

$${\mathcal{U}}_1=U_n\in \mathbb{N},{\mathcal{U}}_2=V_n\in \mathbb{N},$$

of open subgroups forming neighborhood bases at the identity for ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$, respectively. If the two families generate the same Borel structure and the identity maps are Borel measurable, then automatic continuity again gives ${\mathcal{T}}_1={\mathcal{T}}_2$.

Non-Archimedeanness alone, however, does not imply uniqueness. Additional algebraic or dynamical hypotheses are required.

8.2.4.4. Groups with strong uniqueness properties.

There are Polish groups for which automatic continuity is known from the structure of the group. Classical examples arise among certain automorphism groups and groups of measure-preserving transformations. For such groups, automatic continuity can imply that a natural Polish topology is the only Polish topology on the underlying abstract group [1].

This suggests the following general principle.

Proposition 12.

Suppose that G has a property implying that every Borel measurable group homomorphism $G⟶H$ into a Polish group H is continuous. Then G has at most one Polish group topology.

Proof. 

Let ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$ be Polish group topologies on G. The identity map $id:(G,{\mathcal{T}}_1)⟶(G,{\mathcal{T}}_2)$ is a group homomorphism between Polish groups. If the relevant automatic-continuity property applies to this homomorphism, it is continuous. The same argument applies to the inverse identity. Hence ${\mathcal{T}}_1={\mathcal{T}}_2.$ □

8.2.5. Closed Subgroups, Quotients, Products, and Extensions

How does Borel compatibility behave under closed subgroups, quotients, products, and extensions?

The answer is particularly clean for closed subgroups and products.

8.2.5.1. Closed subgroups.

Let $H\le G$ be closed in both ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$. The induced topologies ${\mathcal{T}}_1{{|}_H,{\mathcal{T}}_2|}_H$ are Polish. Moreover, $\mathcal{B}(H,{\mathcal{T}}_i{|}_H)$ is the trace of $\mathcal{B}(G,{\mathcal{T}}_i)$ on H.

If $\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2)$, then $\mathcal{B}(H,{\mathcal{T}}_1{|}_H)=\mathcal{B}(H,{\mathcal{T}}_2{|}_H)$.

Proposition 13.

Let H be closed in both topologies. If the two Borel structures on G coincide, then the two induced Borel structures on H coincide.

Proof. 

Observer $B\cap H\in \mathcal{B}(G,{\mathcal{T}}_2)$. We have, $\mathcal{B}(H,{\mathcal{T}}_1{|}_H)=\mathcal{B}(H,{\mathcal{T}}_2{|}_H)$.

Thus the two induced Borel structures on H coincide. □

Since H is itself a Polish group for both induced topologies, the automatic continuity argument can be applied to H.

8.2.5.2. Products.

Let $G=G_1\times G_2$ and suppose each factor carries two Polish group topologies

$${\mathcal{T}}_1^{\left(j\right)},{\mathcal{T}}_2^{\left(j\right)},j=1,2.$$

Consider the product topologies

$${\mathcal{T}}_i^{\left(1\right)}\times {\mathcal{T}}_i^{\left(2\right)},i=1,2.$$

$$\mathcal{B}\left(G_1\times G_2,{\mathcal{T}}_2^{\left(1\right)}\times {\mathcal{T}}_2^{\left(2\right)}\right).$$

This follows from the fact that the Borel structure of a product of standard Borel spaces is generated by measurable rectangles. Consequently, Borel compatibility is stable under finite products.

8.2.5.3. Quotients.

Let $N⊴G$ be closed in both topologies. For $i=1,2$, let

$$q_i:(G,{\mathcal{T}}_i)⟶G/N$$

be the quotient homomorphism, and equip $G/N$ with the quotient topology induced by ${\mathcal{T}}_i$.

Since N is closed, each quotient is a Polish group. Denote the corresponding quotient topologies by

$${\overline{\mathcal{T}}}_1,{\overline{\mathcal{T}}}_2.$$

If $\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2)$, then the two quotient Borel structures agree under the natural identification of the two abstract quotient groups, provided the quotient Borel structures are understood through the quotient maps.

Indeed, a subset $A\subseteq G/N$ is naturally associated with the N-saturated subset $q^{-1}\left(A\right)\subseteq G$.

Thus the comparison of quotient Borel structures can be reduced to the comparison of saturated Borel subsets of G.

Remark 8.

For quotients it is important to distinguish the quotient topology from the abstract quotient group. Equality of Borel structures on G gives a strong compatibility statement, but the precise identification of quotient Borel structures should be formulated through the quotient maps.

8.2.5.4. Extensions.

Consider a short exact sequence of abstract groups

$$1⟶N⟶G\stackrel{q}{⟶}Q⟶1.$$

Suppose that G carries two Polish group topologies ${\mathcal{T}}_1$ and ${\mathcal{T}}_2$, and that N is closed in both. Assume also that the induced topologies on N and Q have compatible Borel structures.

In favorable situations, these assumptions allow one to compare the Borel structures on G through the extension

$$N\hookrightarrow G\twoheadrightarrow Q.$$

For a direct product $G\cong N\times Q$, the result follows immediately from the product argument. For a semidirect product $G\cong N⋊Q$, one additionally has to compare the two topologies of the conjugation action

$$Q⟶Aut\left(N\right).$$

Thus extensions introduce an additional compatibility problem: not only the Borel structures of N and Q, but also the continuity of the action defining the extension must be compared.

8.2.6. Actions and Orbit Equivalence Relations

If a Polish bitopological group acts continuously with respect to both topologies, under what conditions do the two actions determine the same Borel orbit equivalence relation?

Let $(G,{\mathcal{T}}_1,{\mathcal{T}}_2)$ be a Polish bitopological group and let $(X,{\mathcal{S}}_1,{\mathcal{S}}_2)$ be a Polish bitopological space. Suppose that the same abstract group action $\alpha \times X⟶X$ is continuous for both pairs $(G,{\mathcal{T}}_1)\times (X,{\mathcal{S}}_1)$ and $(G,{\mathcal{T}}_2)\times (X,{\mathcal{S}}_2)$.

The underlying orbit relation is

$$x,E_G,y⟺\exists g\in G,\alpha (g,x)=y.$$

Because the action is the same algebraic action, the set-theoretic relation $E_G$ is independent of the choice of topology. What may differ is its descriptive complexity.

A sufficient condition for the two Borel orbit structures to coincide is $\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2)$ and $\mathcal{B}(X,{\mathcal{S}}_1)=\mathcal{B}(X,{\mathcal{S}}_2)$. Observe $\mathcal{B}(G\times X,{\mathcal{T}}_2\times {\mathcal{S}}_2)$.

Therefore any Borel construction involving the same algebraic action has the same ambient Borel structure in the two descriptions.

In particular, if $E_G$ is Borel for one of the product structures, it is Borel for the other.

Proposition 14.

Suppose that $\mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2)$ and $\mathcal{B}(X,{\mathcal{S}}_1)=\mathcal{B}(X,{\mathcal{S}}_2)$. If the orbit relation $E_G\subseteq X\times X$ is Borel with respect to $\mathcal{B}(X,{\mathcal{S}}_1)\otimes \mathcal{B}(X,{\mathcal{S}}_1)$, then it is also Borel with respect to $\mathcal{B}(X,{\mathcal{S}}_2)\otimes \mathcal{B}(X,{\mathcal{S}}_2)$. In particular,

$$E_G\in \mathcal{B}(X,{\mathcal{S}}_1)\otimes \mathcal{B}(X,{\mathcal{S}}_1)⟺E_G\in \mathcal{B}(X,{\mathcal{S}}_2)\otimes \mathcal{B}(X,{\mathcal{S}}_2).$$

Proof. 

Consider that $\mathcal{B}(X,{\mathcal{S}}_2)\otimes \mathcal{B}(X,{\mathcal{S}}_2)$. Indeed, the product $\sigma$-algebra is determined entirely by the underlying $\sigma$-algebras on the two factors. Since $E_G$ is Borel with respect to $\mathcal{B}(X,{\mathcal{S}}_1)\otimes \mathcal{B}(X,{\mathcal{S}}_1)$, and the latter $\sigma$-algebra coincides with $\mathcal{B}(X,{\mathcal{S}}_2)\otimes \mathcal{B}(X,{\mathcal{S}}_2)$, we obtain

$$E_G\in \mathcal{B}(X,{\mathcal{S}}_2)\otimes \mathcal{B}(X,{\mathcal{S}}_2).$$

Thus a subset of $X\times X$, and in particular the orbit relation $E_G$, is Borel with respect to one product Borel structure if and only if it is Borel with respect to the other. □

Notice that the equality of the two Borel structures on the group alone is not sufficient to guarantee equality of the Borel structures on the orbit space. The Borel structure of the phase space X must also be controlled.

8.2.7. Polish Bitopological Groupoids

For a Polish bitopological groupoid, what additional conditions guarantee that the two induced Borel structures on the orbit space or on the orbit relation coincide?

Proposition 15.

Suppose $\mathcal{B}(G^{\left(0\right)},{\mathcal{T}}_i^{\left(0\right)})$ for the Borel structures on the object space.

$$(x,y)\in G^{\left(0\right)}\times G^{\left(0\right)}:\exists g\in G,s\left(g\right)=y,t\left(g\right)=x.$$

As a set, $E_{\mathcal{G}}$ is independent of the choice of topology. The question is whether its Borel status is also independent of the topology.

A first sufficient condition is equality of the Borel structures on the object space.

$$\mathcal{B}(G^{\left(0\right)},{\mathcal{T}}_2^{\left(0\right)})\otimes \mathcal{B}(G^{\left(0\right)},{\mathcal{T}}_2^{\left(0\right)}).$$

Consequently, $E_{\mathcal{G}}$ is Borel in one object-space product structure if and only if it is Borel in the other.

Proof. 

The assertion follows directly from equality of the two Borel $\sigma$-algebras. The product of standard Borel spaces carries the product Borel structure generated by measurable rectangles. Thus equality of the factors implies equality of their product Borel structures. □

There is, however, a distinction between the Borel structure of the orbit relation and the Borel structure of the orbit space.

Let ${\pi }^{\left(0\right)}⟶G^{\left(0\right)}/E_{\mathcal{G}}$ be the quotient map. The quotient set

$$G^{\left(0\right)}/E_{\mathcal{G}}$$

consists of the groupoid orbits.

In general, the orbit space need not be a standard Borel space. Therefore it is not enough merely to know that the relation $E_{\mathcal{G}}$ is Borel.

A useful sufficient condition is that the orbit equivalence relation be smooth.

Definition 12.

The orbit relation $E_{\mathcal{G}}$ is calledsmoothif there exists a standard Borel space Y and a Borel map $f:G^{\left(0\right)}⟶Y$

such that

$$x,E_{\mathcal{G}},y⟺f\left(x\right)=f\left(y\right).$$

Theorem 7.

If $E_{\mathcal{G}}$ is smooth, the orbit space can be represented by a standard Borel quotient. In this situation, if the two topologies on $G^{\left(0\right)}$ generate the same Borel structure, then the same Borel classification map can be used for both descriptions.

$\mathcal{B}(G^{\left(0\right)},{\mathcal{T}}_2^{\left(0\right)})$

and that the orbit relation $E_{\mathcal{G}}$ is smooth. Then the two topologies induce the same standard Borel orbit structure. In particular, the quotient can be represented by the same standard Borel set in both descriptions.

Proof. 

By Borel compatibility, the identity map on $G^{\left(0\right)}$ is a Borel isomorphism between the two object-space structures.

Let $f:G^{\left(0\right)}⟶Y$ be a Borel reduction of $E_{\mathcal{G}}$ to equality on a standard Borel space Y. Since the Borel structures on $G^{\left(0\right)}$ coincide, f is Borel measurable with respect to both topologies.

Moreover,

$$x,E_{\mathcal{G}},y⟺f\left(x\right)=f\left(y\right)$$

is independent of the topology. Hence the same map f gives the same standard Borel parametrization of the orbit classes in both descriptions. □

A stronger situation occurs when the orbit relation is closed.

Proposition 16.

Suppose that $E_{\mathcal{G}}$ is closed in

$$(G^{\left(0\right)},{\mathcal{T}}_i^{\left(0\right)})\times (G^{\left(0\right)},{\mathcal{T}}_i^{\left(0\right)})$$

for both $i=1,2$. Then $E_{\mathcal{G}}$ is Borel in both product structures.

If, in addition, the two object-space Borel structures coincide, then the two descriptions determine exactly the same Borel orbit relation.

Proof. 

A closed subset of a Polish space is Borel. Therefore $E_{\mathcal{G}}$ is Borel in both product structures. If the object-space Borel structures coincide, the two product Borel structures coincide as well. □

One can also formulate a groupoid analogue of the compatibility condition at the level of arrows.

$$(g,h)\left(g\right)=t\left(h\right)$$

has the same underlying Borel description in both structures.

This provides a natural measurable framework for comparing the two versions of the groupoid composition map.

8.2.8. A Compatibility Theorem

The preceding discussion can be summarized in the following statement.

Theorem 8.

$$\mathcal{B}(G^{\left(0\right)},{\mathcal{T}}_2^{\left(0\right)}).$$

Then the following properties are independent of the choice of $i\in 1,2$:

• Borel measurability of the source and target maps;
• the Borel structure of the composable-pair space $G^{\left(2\right)}$;
• Borelness of the orbit relation $E_{\mathcal{G}}$;
• smoothness of the orbit relation;
• Borel measurability of maps defined purely in terms of the underlying groupoid structure.

Proof. 

The equality of the Borel structures on G and $G^{\left(0\right)}$ implies that the source and target maps have the same Borel measurability properties in both descriptions.

$${(s,t)}^{-1}\left({\Delta }_{G^{\left(0\right)}}\right),$$

its Borel structure is determined by the Borel structure of G, the Borel structure of $G^{\left(0\right)}$, and the Borel maps s and t. Thus the composable-pair construction is compatible with both descriptions.

The orbit relation is a subset of $G^{\left(0\right)}\times G^{\left(0\right)}$ defined entirely by the underlying groupoid structure. Equality of the object-space Borel structures gives equality of the corresponding product Borel structures. Therefore Borelness of $E_{\mathcal{G}}$ is the same for both structures.

Finally, smoothness is defined by the existence of a Borel reduction to equality on a standard Borel space. Since the Borel structure on $G^{\left(0\right)}$ is the same in both descriptions, the existence of such a reduction is independent of the chosen topology. □

8.2.9. Summary of the Compatibility Hierarchy

The preceding results show that the compatibility hierarchy for arbitrary Polish spaces must be modified when a group structure is present.

For arbitrary Polish spaces one has the implications

$${\mathcal{T}}_1={\mathcal{T}}_2⟹id\mathrm{is}\mathrm{a}\mathrm{homeomorphism}⟹id\mathrm{is}\mathrm{a}\mathrm{Borel}\mathrm{isomorphism},$$

but the converse implication from Borel isomorphism to homeomorphism can fail.

For Polish group topologies on the same abstract group, automatic continuity gives the stronger conclusion

$$\begin{array}{|c|}\hline \mathcal{B}(G,{\mathcal{T}}_1)=\mathcal{B}(G,{\mathcal{T}}_2)⟺id:(G,{\mathcal{T}}_1)\to (G,{\mathcal{T}}_2)\mathrm{is}\mathrm{a}\mathrm{homeomorphism}⟺{\mathcal{T}}_1={\mathcal{T}}_2.\\ \hline\end{array}$$

Thus, for Polish groups, the abstract group operation together with the Borel structure determines the Polish group topology.

On the other hand, if

$${\mathcal{T}}_1\ne {\mathcal{T}}_2,$$

then necessarily

$$\mathcal{B}(G,{\mathcal{T}}_1)\ne \mathcal{B}(G,{\mathcal{T}}_2).$$

In particular, two distinct Polish group topologies on the same abstract group cannot be Borel-compatible.

This observation changes the interpretation of the notion of “Borel-compatible Polish bitopological group.” Such an object, under the present definition, necessarily has

$${\mathcal{T}}_1={\mathcal{T}}_2.$$

Therefore, if the goal is to study genuinely distinct Polish topologies with a common measurable structure, one should consider a weaker category in which the two structures are not both required to be Polish group topologies, or one should replace equality of the Borel structures by a weaker notion of Borel comparison.

For the present theory, this distinction is fundamental:

$$\begin{array}{|c|}\hline \begin{array}{c}\mathrm{distinct}\mathrm{Polish}\mathrm{group}\mathrm{topologies}\\ \Downarrow \\ \mathrm{different}\mathrm{Borel}\mathrm{structures},\end{array}\begin{array}{c}\mathrm{same}\mathrm{Borel}\mathrm{structure}\\ \Downarrow \\ \mathrm{same}\mathrm{Polish}\mathrm{group}\mathrm{topology}.\end{array}\\ \hline\end{array}$$

Consequently, the study of genuinely nontrivial Polish bitopological groups should focus on pairs

$$({\mathcal{T}}_1,{\mathcal{T}}_2)\mathrm{with}{\mathcal{T}}_1\ne {\mathcal{T}}_2$$

and hence

$$\mathcal{B}(G,{\mathcal{T}}_1)\ne \mathcal{B}(G,{\mathcal{T}}_2),$$

while the comparison of the two Borel structures becomes a question of inclusion, mutual Borel reducibility, or other measurable relationships rather than equality.

This also suggests a natural reformulation for Polish bitopological groupoids: one may investigate when the two object and arrow Borel structures are equal, included in one another, or Borel isomorphic, while allowing the two underlying Polish topologies to remain genuinely distinct. Such a formulation preserves the descriptive-set-theoretic content of the bitopological framework without collapsing the group case to the diagonal topology.

8.3. Actions of Polish Bitopological Groups

Suppose that $(G,{\mathcal{T}}_1,{\mathcal{T}}_2)$ is a Polish bitopological group and $(X,{\mathcal{S}}_1,{\mathcal{S}}_2)$ is a Polish bitopological space. A natural definition of a bitopological action is a group action

$$\alpha :G\times X\to X$$

such that

$$\alpha :(G,{\mathcal{T}}_i)\times (X,{\mathcal{S}}_i)⟶(X,{\mathcal{S}}_i)$$

is continuous for each i.

One can then investigate the orbit relation

$$x{E}_{G}y⟺\exists g\in G,\alpha (g,x)=y,$$

and compare its descriptive complexity in the two product Borel structures.

The classical theory of Polish group actions provides the natural point of departure [1].

These questions connect the present framework with the broader theory of topological groupoids [2].

9. Conclusions

We have introduced Polish bitopological groups and Polish bitopological groupoids by requiring each of two topologies to define a Polish structure compatible with the relevant algebraic operations.

For groups, we established permanence under closed subgroups and products and recorded the behavior of continuous homomorphisms. For groupoids, we proved that the space of composable arrows is Polish under each of the two induced product topologies, and we established a closed-subgroupoid permanence result.

We also distinguished the existence of two Polish topologies from the stronger requirement that they generate the same Borel structure. This distinction is particularly relevant when studying orbit equivalence relations and measurable actions.

The framework provides a starting point for a systematic study of two-topology structures in descriptive set theory and topological algebra. Further work should investigate Borel compatibility, the classification of bitopological group actions, and appropriate notions of equivalence for Polish bitopological groupoids.