Subgroups of abelian Polish groups

Greg Hjorth

April 4, 2005

1 Introduction

Ilijas Farah and Slawek Solecki showed in [2] that every Polish group contains arbitrarily complicated Borel subgroups and go on to ask whether the same can be shown for Polishable subgroups. As a partial answer, this notes shows that an uncountable abelian Polish group contains Polishable subgroups of unbounded Borel complexity.

Acknowledgements: These results were largely proved during the course of the conference at CRM during the year dedicated to Set Theory, and came about after listening to Farah and Solecki describe their recent work. I am very grateful for a number of conversations with both of them, including one lengthy discussion carried out in the taxi on the way out to the airport for the flight home.

I am also grateful to CRM for making this meeting possible, and I further thank Joan Bagaria and the other members of the organizing committee for inviting me to participate.

I am more than indebted to the anonymous referee for a careful and exacting reading of the earlier version of this paper.

2 Notation

In what follows, G is an uncountable abelian Polish group and d(·, ·) a compatible, complete, two sidedinvariant metric. We write the group operations on G with reference to it being abelian – thus + is the group operation and n · g stands for

g + g + ...(n times).. + g.

0 is the group identity in G.

We will find it convenient to use the notation of an infinite sum. Thus we say that

g = ∑_{n=0}^{∞} g_{n}

if

d(∑_{n=0}^{N} g_{n}, g) → 0.

Since d is two sided-invariant metric we obtain that

∑_{n=0}^{∞} g_{n} = ∑_{n=0}^{∞} h_{n}

if and only if

∑_{n=0}^{∞}(g_{n} − h_{n}) = 0.

For g a group element we use o(g) to denote the order – that is to say, the least ℓ with ℓ · g = 0 if such an ℓ exists, and infinity otherwise.

3 The subgroups

We choose a sequence of non-zero elements

(g_{i})_{i}

in G such that:

(I) for any k ∈ ℕ and (n_{ℓ})_{ℓ≤k} chosen from ℤ, with each

|n_{ℓ}| ≤ 4ℓk

and

h = n_{0} · g_{0} + n_{1} · g_{1}... + n_{k} · g_{k} ∈ G
h ≠ 0

we have that for any choice of (n̂_{ℓ})_{ℓ>k} with each |n̂_{ℓ}| < 4ℓk that

h ≠ ∑_{ℓ>k} n̂_{ℓ} · g_{ℓ};

(II) for any ℓ and n with |n| ≤ ℓ^{2} we have

d(0, n · g_{ℓ}) < 2^{−ℓ};

(III) either every g_{i} has infinite order, or the g_{i}’s all have the same finite order, or o(g_{i}) → ∞ as i → ∞.

We obtain (I) and (II) just by taking any sequence which converges to 0 sufficiently fast. (III) then follows by refining the sequence as needed.

Definition If h ∈ G then a k-representation of h is an expression of the form

∑_{ℓ∈ℕ} n_{ℓ} · g_{ℓ}

with

lim_{M→∞} ∑_{ℓ=0}^{M} n_{ℓ} · g_{ℓ} = h

and each |n_{ℓ}| ≤ ℓ · k. A representation is a k-representation for some k; a group element is representable if it has a representation.

Lemma 3.1 If

∑_{ℓ∈ℕ} n_{ℓ} · g_{ℓ},

∑_{ℓ∈ℕ} n̂_{ℓ} · g_{ℓ},

are two 2k-representations of h then for all ℓ > k

n_{ℓ} · g_{ℓ} = n̂_{ℓ} · g_{ℓ}.

Proof By the definition of 2k-representation we have |n_{ℓ}|, |n̂_{ℓ}| ≤ 2ℓk each ℓ, and then by our assumption (I) on the (g_{i}) sequence we have that at all M ≥ k

∑_{ℓ≤M}(n_{ℓ} − n̂_{ℓ}) · g_{ℓ} = 0.

From this the lemma follows. □ Definition We let H be the subgroup of all h which have some representation

∑_{ℓ∈ℕ} n_{ℓ} · g_{ℓ}

with

\frac{|n_{ℓ}|}{ℓ} → 0.

Definition If

∑_{ℓ∈ℕ} n_{ℓ} · g_{ℓ}

is a representation of h, then we let

φ(∑_{ℓ∈ℕ} n_{ℓ} · g_{ℓ}) = sup_{ℓ} \frac{|n_{ℓ}|}{ℓ}.

If h ∈ H we let ψ(h) be the infinum of

φ(∑_{ℓ∈ℕ} n_{ℓ} · g_{ℓ})

as ∑ n_{ℓ} · g_{ℓ} ranges over representations of h. Lemma 3.2 In the metric

d_{H}(h, h′) = ψ(h − h′) + d(h, h′)

we have that H is a Polish group. Proof First we need to check that this is indeed a metric, and that follows from the observation that given any two representations

∑ n_{ℓ} · g_{ℓ},
∑ n̂_{ℓ} · g_{ℓ},

for h − h′ and h′ − h′′ respectively, we have φ(∑(n_{ℓ} + n̂_{ℓ}) · g_{ℓ}) ≤ φ(∑ n_{ℓ} · g_{ℓ}) + φ(∑ n̂_{ℓ} · g_{ℓ}).

The metric is separable since the representations with finite support (∀^{∞}ℓ(n_{ℓ} = 0)) give rise to a dense subgroup. Continuity of the group operations follows since

ψ(h) + ψ(h′) ≥ ψ(h + h′).

This only leaves us with a campaign to show that the metric is complete.

For this purpose, consider a Cauchy sequence (h_{i})_{i} in H, with each h_{i} having

∑_{ℓ∈ℕ} n_{ℓ}^{i} · g_{ℓ}

as a representation. Since

|ψ(h_{i}) − ψ(h_{j})| ≤ ψ(h_{i} − h_{j})

we get that (ψ(h_{i}))_{i∈ℕ} is Cauchy and so there is a fixed k such that each is a k-representation. Since the space of k-representations is precompact, we may assume there exist (n_{ℓ}^{∞})_{ℓ∈ℕ} such that

∀^{∞}j(n_{ℓ}^{j} = n_{ℓ}^{∞}).

Claim(1): As j → ∞ and then M → ∞

d_{H}(h_{j}, ∑_{ℓ=0}^{M} n_{ℓ}^{∞} · g_{ℓ}) → 0.

Proof of Claim: Fix ϵ > 0. We need to show that for all sufficiently large j we have

∀^{∞}M(d_{H}(h_{j}, ∑_{ℓ=0}^{M} n_{ℓ}^{∞} · g_{ℓ}) < ϵ).

Choose j large enough that d_{H}(h_{j}, h_{j′}) < ϵ all j′ ≥ j and n_{ℓ}^{j} = n_{ℓ}^{∞} all ℓ ≤ k. We may assume without loss of generality that ϵ < \frac{1}{k} . Thus at each j′ ≥ j we have that if

∑_{ℓ∈ℕ} m_{ℓ} · g_{ℓ}

is a 2k-representation of h_{j} − h′_{j} with φ value less than ϵ then

m_{ℓ} = 0

at all ℓ ≤ k. Since by 3.1 we have

∑(n_{ℓ}^{j} − n_{ℓ}^{j′}) · g_{ℓ}

as the unique 2k-representation of h_{j} − h_{j′} with coefficient 0 at the ℓ^{th} coordinate all ℓ ≤ k, and thus

∑(n_{ℓ}^{j} − n_{ℓ}^{j′}) · g_{ℓ}

is the unique 2k-representation of h_{j} − h_{j′} with φ value less than ϵ.

Therefore if at arbitrarily large M we have

d_{H}(h_{j}, ∑_{ℓ=0}^{M} n_{ℓ}^{∞} · g_{ℓ}) > ϵ

then for arbitrarily large M we have

d_{H}(∑_{ℓ=0}^{M} n_{ℓ}^{j} · g_{ℓ}, ∑_{ℓ=0}^{M} n_{ℓ}^{∞} · g_{ℓ}) > ϵ,

but then going to some j′ with n_{ℓ}^{j′} = n_{ℓ}^{∞} for all ℓ ≤ M we have

φ(∑(n_{ℓ}^{j} − n_{ℓ}^{j′}) · g_{ℓ}) > ϵ,

with a contradiction. (□Claim) Claim(2): As N → ∞

φ(∑_{ℓ=N}^{∞} n_{ℓ}^{∞} · g_{ℓ}) → 0.

Proof of Claim: Consider ϵ > 0. Choose M_{0} such that at all j ≥ M_{0} and ℓ ≤ k we have n_{ℓ}^{j} = n_{ℓ}^{∞}. Choose M_{1} > M_{0} such that at all j, j′ ≥ M_{1} we have ψ(h_{j} − h_{j′}) < ϵ. Choose N_{0} such that at all ℓ ≥ N_{0}

\frac{|n_{ℓ}^{M_{1}}|}{ℓ} ≤ ϵ,

∴ φ( ∑_{ℓ=N_{0}}^{∞} n_{ℓ}^{M_{1}} · g_{ℓ}) ≤ ϵ.

As in the proof of claim(1) we may assume that ϵ < \frac{1}{k}, and again as in that proof we have that for all sufficiently large j′

∑_{0}^{∞}(n_{ℓ}^{j′} − n_{ℓ}^{M_{1}}) · g_{ℓ}

is the only possible representation of h_{j′} − h_{M_{1}} with

φ(∑_{0}^{∞}(n_{ℓ}^{j′} − n_{ℓ}^{M_{1}}) · g_{ℓ}) < ϵ,

and hence by 3.1 for j′ sufficiently large we have ∑_{0}^{∞}(n_{ℓ}^{j′} − n_{ℓ}^{M_{1}}) · g_{ℓ} is the representation of h_{j′} − h_{j} with φ(∑_{0}^{∞}(n_{ℓ}^{j′} − n_{ℓ}^{M_{1}}) · g_{ℓ}) < ϵ. Hence at each N > N_{0}

φ( ∑_{ℓ=N_{0}}^{N} n_{ℓ}^{∞} · g_{ℓ}) ≤ φ( ∑_{ℓ=N_{0}}^{N}(n_{ℓ}^{j′} − n_{ℓ}^{M_{1}}) · g_{ℓ}) + φ( ∑_{ℓ=N_{0}}^{N} n_{ℓ}^{M_{1}} · g_{ℓ})

≤ φ(∑_{0}^{∞}(n_{ℓ}^{j′} − n_{ℓ}^{M_{1}}) · g_{ℓ}) + ϵ = ψ(h_{j′} − h_{M_{1}}) + ϵ < 2ϵ.

Letting ϵ → 0 this proves the claim. (□Claim)

These two claims between them demonstrate that

∑ n_{ℓ}^{∞} · g_{ℓ}

represents some h_{∞} ∈ H and moreover

ψ(h_{j} − h_{∞}) → 0

as j → ∞.

□

Definition We define the notion of α-calibration by induction on the countable ordinal α.

A 0-calibration is an infinite A ⊂ ℕ. An α + 1-calibration is an infinite A ⊂ ℕ equipped with a partition

A = ˙⋃B_{n},

where each B_{n} is an α-calibration. For λ a limit, with a specified^{1}

α_{n} ↗ λ,

we say that A ⊂ ℕ is λ-calibration if it is partitioned

A = ˙⋃B_{n},

with each B_{n} an α_{n}-calibration.

Definition For A a 0-calibration, a representation

∑ n_{ℓ} · g_{ℓ}

is A-0-good if there is an n_{∞} such that

∀^{∞}ℓ ∈ A(n_{ℓ} = n_{∞});

that is to say, the n_{ℓ}’s, for ℓ ∈ A, converge to n_{∞}. We then let π_{A,0}(∑ n_{ℓ} · g_{ℓ}) be this indicated n_{∞} and

φ_{A,0}(∑ n_{ℓ} · g_{ℓ}) = φ(∑ n_{ℓ} · g_{ℓ}) + |π_{A,0}(∑ n_{ℓ} · g_{ℓ})| + |{ℓ ∈ A : n_{ℓ} ≠ π_{A,0}(∑ n_{ℓ} · g_{ℓ})}|;

in other words, we add into the norm on representations for H information about the eventual value n_{∞} along with a counter for the number of times this value is missed.

For A an α + 1-calibration, A = ˙⋃B_{n}, we say that

∑ n_{ℓ} · g_{ℓ}

is A-α + 1-good if it is B_{n}-α-good at every n and there is an m_{∞} such that

∀^{∞}n(π_{B_{n},α}(∑ n_{ℓ} · g_{ℓ}) = m_{∞}).

We then let π_{A,α+1}(∑ n_{ℓ} · g_{ℓ}) be this eventual m_{∞} value and let

φ_{A,α+1}(∑ n_{ℓ} · g_{ℓ}) = φ(∑ n_{ℓ} · g_{ℓ}) + ∑_{n} 2^{−n} \frac{φ_{B_{n},α}(∑ n_{ℓ} · g_{ℓ})}{1 + φ_{B_{n},α}(∑ n_{ℓ} · g_{ℓ})}

+|π_{A,α+1}(∑ n_{ℓ} · g_{ℓ})| + |{n : π_{B_{n},α}(∑ n_{ℓ} · g_{ℓ}) ≠ π_{A,α+1}(∑ n_{ℓ} · g_{ℓ})}|;

that is to say, we merge into a single norm information about the eventual value m_{∞}, the set of exceptional n’s at which this value is not yet reached, along with a bounded version of norms for the various B_{n} sets.

For λ a limit ordinal, α_{n} ↗ λ, A a λ-calibration, we can define A-λ-goodness in the exactly parallel manner, replacing each mention of B_{n}-α-goodness by B_{n}-α_{n}-goodness.

1For the purpose of this definition, imagine that all our limit ordinals come with a previously decided cofinal sequence.

We want to now define π_{A,α}(h) in a way that is independent of the representation. Here there is a split in cases, depending on the situation dictated at (III).

Observe first of all by 3.1 we have that if ∑ n_{ℓ} · g_{ℓ} and ∑ n′_{ℓ} · g_{ℓ} are two A-α-representations of h ∈ H then the terms

n_{ℓ} · g_{ℓ}, n′_{ℓ} · g_{ℓ}

differ at only finitely many places.

Therefore in the case of every g_{ℓ} having infinite order we get

π_{A,α}(∑ n_{ℓ} · g_{ℓ}) = π_{A,α}(∑ n′_{ℓ} · g_{ℓ})

for any two distinct representations. Therefore we just do the natural thing and let π_{A,α}(h) be the common value realized on all the A-α-good representations.

The next case is when the g_{ℓ}’s all have the same finite order. Use M to denote this common value. We then let π_{A,α}(h) be the unique m ∈ {0, 1, ..., M − 1} for which there is some A-α-good representation having

π_{A,α}(∑ n_{ℓ} · g_{ℓ}) = m.

In the case that the g_{ℓ}’s have strictly increasing but finite orders, we let π_{A,α}(h) be the unique m ∈ ℤ for which there is some A-α-good representation with

π_{A,α}(∑ n_{ℓ} · g_{ℓ}) = m.

Note also that the collection of A-α-good elements is closed under the group operations – a critical fact underpinning the next definition.

Definition For A an α-calibration we let

H_{A,α}

be the subgroup of H consisting of all elements which admit an A-α-good representation. For h such an element, we let

ψ_{A,α}(h)

be the infinum of

φ_{A,α}(∑_{ℓ∈ℕ} n_{ℓ} · g_{ℓ})

as ∑ n_{ℓ} · g_{ℓ} ranges over all possible representations. We then let d_{A,α}(·, ·) be defined by

d_{A,α}(h, h′) = ψ_{A,α}(h − h′) + d_{H}(h, h′).

Lemma 3.3 d_{A,α} defines a complete, separable metric on H_{A,α} for any α-calibration A.

Proof That this is a metric follows as in the proof for H. Separability follows by induction on α. For α = 0 we observe that the set of elements with a representation

∑ n_{ℓ} · g_{ℓ}

for which there is some m_{∞} with

∀^{∞}ℓ(n_{ℓ} = m_{∞})

forms a countable dense subset. For A = ˙⋃B_{n} with each B_{n} a β_{n}-calibration, we assume that there are countable dense subsets D_{n} of each H_{B_{n},β_{n}} and we take as our countable dense subset those representations which on finitely many B_{n}’s are an element of a corresponding D_{n}, and elsewhere are eventually constant.

Again, these arguments are quickly dismissed, and the main battle is for completeness. So we suppose that we have some d_{A,α}-Cauchy sequence (h_{j})_{j∈ℕ}, each h_{j} having

∑_{ℓ∈ℕ} n_{ℓ}^{j} · g_{ℓ}

as a representation. Again by the compactness of the space of k-representations, as in 3.2, we may assume there are (n_{ℓ}^{∞})_{ℓ>k} such that

∀^{∞}j(n_{ℓ}^{j} = n_{ℓ}^{∞}).

Note that for all sufficiently large j, j′ we have

ψ_{A,α}(h_{j} − h_{j′}) < 1,

and hence

π_{A,α}(h_{j} − h_{j′}) = 0

by definition of ψ_{A,α}(·).

Thus we may find a single m_{∞} such that

∀^{∞}j(π_{A,α}(h_{j}) = m_{∞}).

Let h_{∞} be the limit in the d_{H}(·, ·) metric of the (h_{j})_{j} sequence, with representation

∑_{ℓ} n_{ℓ}^{∞} · g_{ℓ}.

Note that each |n_{ℓ}^{∞}| ≤ ℓ · k since ∀^{∞}j(n_{ℓ}^{j} = n_{ℓ}^{∞}) and each |n_{ℓ}^{j}| ≤ ℓ · k. Thus ∑ n_{ℓ}^{∞} · g_{ℓ} is a k-representation.

In the case that α = 0 we have the following claim:

Claim: There are only finitely many a ∈ A with

n_{a}^{∞} ≠ m_{∞}.

Proof of Claim: We again use the fact that 2k-representations are unique on coordinates bigger than k. Thus we obtain that if for some c ≥ 2k we have 2c-many a’s with n_{a}^{∞} ≠ m_{∞}, then for all sufficiently large j we have all 2k representations of h_{j} have at least c many coordinates which present a value other than

m_{∞} = π_{A,0}(h_{j}),

and hence we are left with ψ_{A,0}(h_{j}) > c.

Thus to recap the argument, we have seen that if at every c there exist more than c many a′s with n_{a}^{∞} ≠ m_{∞} then ψ_{A,0}(h_{j}) → ∞ as j → ∞, contradicting the assumption that (ψ_{j})_{j∈ℕ} is Cauchy. (□Claim)

Thus we have h_{∞} ∈ H_{A,0}, π_{A,0}(h_{∞}) = m_{∞}; the function

j ↦ {a ∈ A : n_{a}^{j} ≠ m_{∞}}

is eventually constant by the definition of ψ_{A,0}(·) and uniqueness of the k-representation on coordinates bigger than k. Therefore with respect to d_{A,0}(·, ·) we have

h_{j} → h_{∞}.

If α > 0, A = ˙⋃B_{n}, each B_{n} a β_{n}-calibration, we may assume inductively that there are (p_{n}) such that at every n

∀^{∞}j(π_{B_{n},β_{n}}(h_{j}) = p_{n}),
h_{∞} ∈ H_{B_{n},β_{n}},

and

h_{j} → h_{∞}

in d_{B_{n},β_{n}}(·, ·). By Cauchyness of the (h_{j}) sequence we must eventually have that for all sufficiently large j, j′

∀n(π_{B_{n},β_{n}}(h_{j}) = π_{B_{n},β_{n}}(h_{j′})),

and hence the (p_{n})-sequence is eventually constant, with value m_{∞}.

□

Lemma 3.4 Given any Δ ∼_{α}^{0} set D ⊂ ω^{ω} and α-calibration A = ˙⋃B_{n}, each B_{n} a β_{n}-calibration, there is continuous

f_{α}^{A,D} : ω^{ω} → G

such that (0) for every y there is a representation

∑ n_{ℓ} · g_{ℓ}

of f_{α}^{A,D}(y) with each n_{ℓ} ∈ {0, 1}.

(i) if y ∈ D then ∀^{∞}n(π_{B_{n},β_{n}}(f_{α}^{A,D}(y)) = 1); (ii) if y ∉ D then ∀^{∞}n(π_{B_{n},β_{n}}(f_{α}^{A,D}(y)) = 0); (iii) the representation of each f_{α}^{A,D}(y) has support in A and each f_{α}^{A,D}(y) ∈ H_{A,α}.

Proof This is proved by induction on α, in a manner entirely parallel to the similar claims from [3]. First of all one sees that for any clopen D ⊂ ω^{ω} and A a 0-calibration we can certainly find continuous f^{A,D} : ω^{ω} → G with

f^{A,D}(x) = 0

if x ∉ D, but

f^{A,D}(x) = lim_{N→∞} ∑_{ℓ∈A,ℓ≤N} g_{ℓ},

if x ∈ D. This forms the base of our induction.

Now we do the inductive step, and suppose that D is Δ ∼_{α}^{0}, A = ˙⋃B_{n} is an α-calibration, each B_{n} a β_{n}-calibration. We recall the classical fact (compare [3]) that there must be Δ ∼_{β_{n}}^{0} sets (E_{n})_{n} such that

(a) if y ∈ D then ∀^{∞}n(y ∈ E_{n});

(b) if y ∉ D then ∀^{∞}n(y ∉ E_{n}). Appealing to our inductive hypothesis we may find

f_{n} : ω^{ω} → H_{B_{n},β_{n}},

continuous with respect to the Polish G-topology, each f_{n}(x) having support inside B_{n}, with π_{B_{n},β_{n}}(f_{n}(x)) = 1 if x ∈ E_{n}, π_{B_{n},β_{n}}(f_{n}(x)) = 0 if x ∉ E_{n}, and every f_{n}(x) having a representation with coordinates taken solely from {0, 1}. We can then simply let

f^{A,D}(x) = lim_{N→∞}f_{0}(x) + f_{1}(x) + f_{2}(x) + ... + f_{N}(x).

Our inductive assumptions on the f_{n}’s suffice to give that this is in H, and then we obtain it will be in H_{A,α} with either (a) or (b) above by the choice of the E_{n} sets. It is continuous by condition (II) on the g_{ℓ}-sequence and the fact that each f_{n}(x) has support inside B_{n}. □

Corollary 3.5 If A = ˙⋃B_{n} is an α + 1-calibration, then H_{A,α+1} is not Π ∼_{α}^{0}.

Proof Let C ∈ Σ ∼_{α}^{0} be of the form

C = ⋃ D_{n},

where each D_{n} ∈ Δ ∼_{α}^{0}. Following the last lemma we may find continuous

f_{n} : ω^{ω} → H_{B_{n},β_{n}}

such that (0) for every y there is a representation

∑_{ℓ∈B_{n}} n_{ℓ} · g_{ℓ}

of f_{n}(y) with each n_{ℓ} ∈ {0, 1}.

(i) if y ∈ ⋃_{i≤n} D_{i} then π_{B_{2n},β_{2n}}(f_{2n}(y)) = 1; (ii) if y ∉ ⋃_{i≤n} D_{i} then π_{B_{2n},β_{2n}}(f_{2n}(y)) = 0; (iii) the representation of each f_{n}(y) has its support in B_{n}; (iv) for m odd at every y we have π_{B_{m},β_{m}}(f_{m}(y)) = 1.

After this we can simply take

f^{C}(y) = lim_{N→∞}f_{0}(y) + f_{1}(y) + ...f_{N}(y).

Then in order for the values of the π_{B_{n},β_{n}}(f^{C}(y)) to settle down into a limit we must have that y ∈ C, and thus membership C is a necessary condition for f(y) to be in H_{A,α+1}. But conversely membership in C is also a sufficient condition for f(y) to be in this subgroup, and thus we have continuously reduced membership in an arbitrary Σ ∼_{α}^{0} set to membership in H_{A,α+1}.

□

4 After thoughts

As observed by Slawek Solecki, this result for uncountable abelian Polish groups implies the same for uncountable locally compact Polish groups.

The main point is that we can use approximation by Lie groups – in particular the theorem that every connected locally compact group has arbitrarily small normal subgroups for which the resulting quotient is Lie. (See [5].)

One case is that the group is totally disconnected, and hence, by local compactness (see [5]) it will have a neighborhood basis of clopen compact subgroups. Appealing to [6] we obtain an uncountable abelian compact subgroup, and finish by the results of the last section.

The other case is that there exists a closed connected subgroup, which in turn will have an uncountable Lie group as a quotient. But any Lie group contains one-parametrized subgroups, arrived at via the exponential map from the tangent bundle at the identity, and hence a continuous image of ℝ or 𝕋, and again we fall back into the cases covered by the last section.

The most optimistic conjecture, then, would be that every uncountable Polish group contains an uncountable abelian subgroup. From this and the results above we could obtain arbitrarily complicated Polishable subgroups of any uncountable Polish group.

However the spirit of fearless scientific research demands that we should never take a position of optimism when pessimism is unrefuted. Thus:

Conjecture 4.1 There is an uncountable Polish group all of whose abelian subgroups are discrete.

It may even be that some minor variation of the the surjectively universal two-sided invariant group (see [4], [1]) provides a counterexample. Accordingly it seems unlikely that the construction above will answer the question raised in [2].

References

[1] H. Becker and A. S. Kechris, The descriptive set theory of Polish group actions, London Mathematical Society Lecture Note Series, 232, Cambridge University Press, Cambridge, 1996.

[2] I.  Farah and S.  Solecki, Borel subgroups of Polish groups, preprint,

   http://www.math.yorku.ca/~ifarah/preprints.html.

[3] G. Hjorth, A. S. Kechris, and A. Louveau, The Borel equivalence relations induced actions of the infinite symmetric group, Annals of Pure and Applied Logic, vol. 92(1998), pp. 63–112.

[4] A. S. Kechris, Definable equivalence relations and Polish group actions, manuscript, Caltech, 1993.

[5] D. Montgomery and L. Zippin, Topological transformation groups, reprint of the 1955 original, Robert E. Krieger Publishing Co., Huntington, N.Y., 1974. xi+289 pp.

[6] E. I. Zelmanov, On periodic compact groups, Israel Journal of Mathematics, vol. 77(1992), pp. 83–95.

greg@math.ucla.edu

MSB 6363 UCLA CA 90095-1555