Farah and Solecki showed the following in Borel subgroups of Polish groups:

Theorem: Every Polish group $G$ admits Borel subgroups of arbitrarily high Borel rank.

However, the construction is far from uniform. I want to know if the procedure can be made uniform, i.e. if we have a reasonably absolute transfinite operation which, on countable ordinal $\alpha$, spits out a Borel subgroup of $G$ which is not $\Pi_\alpha$. There are various morally equivalent ways to make this precise, for example:

Question: Work in $\text{ZF} + \text{AD}$ (axiom of determinacy). Suppose $G$ is a Polish group. Is there is an injection from $\omega_1$ to the set of Borel subgroups of $G$?

If such an injection existed, then after pruning, we can arrange the Borel ranks of subgroups to be strictly increasing. This is because under AD, Hjorth proved in An absoluteness principle for Borel sets there is no injection from $\omega_1$ to $\Pi_\alpha$ subsets of $\mathbb{R}$, for any fixed $\alpha < \omega_1$.

Some remarks:

• The answer might depend on $G$. I would accept a positive or negative answer for any specific (uncountable) $G$.
• $\omega_1$ is reducible to the set of Borel subsets of $\mathbb{R}$: e.g. let $X_\alpha$ be the set of well-founded linear orders of rank less than $\alpha$.
• If we looked at subgroups up to conjugacy, rather than just subgroups, the answer is "Yes" for $S_\infty$ (in fact we just need closed subgroups). But otherwise it may depend on properties of the group $G$; for instance if $G$ is abelian then conjugacy is trivial, so we reduce to the original question.
• We can ask the same question for various adjectives for Borel subgroups; for instance we can restrict attention to Polishable groups, or to non-Archimedian Polishable groups.

Farah and Solecki showed the following in Borel subgroups of Polish groups:

Theorem: Every Polish group $G$ admits Borel subgroups of arbitrarily high Borel rank.

However, the construction is far from uniform. I want to know if the procedure can be made uniform, i.e. if we have a reasonably absolute transfinite operation which, on countable ordinal $\alpha$, spits out a Borel subgroup of $G$ which is not $\Pi_\alpha$. There are various morally equivalent ways to make this precise, for example:

Question: Work in $\text{ZF} + \text{AD}$ (axiom of determinacy). Suppose $G$ is a Polish group. Is there is an injection from $\omega_1$ to the set of Borel subgroups of $G$?

Edit. Actually I was misreading the Farah Solecki proof, it actually seems to be uniform. Instead, restrict to Polishable subgroups. End edit

If such an injection existed, then after pruning, we can arrange the Borel ranks of subgroups to be strictly increasing. This is because under AD, Hjorth proved in An absoluteness principle for Borel sets there is no injection from $\omega_1$ to $\Pi_\alpha$ subsets of $\mathbb{R}$, for any fixed $\alpha < \omega_1$.

Some remarks:

• The answer might depend on $G$. I would accept a positive or negative answer for any specific (uncountable) $G$.
• $\omega_1$ is reducible to the set of Borel subsets of $\mathbb{R}$: e.g. let $X_\alpha$ be the set of well-founded linear orders of rank less than $\alpha$.
• If we looked at subgroups up to conjugacy, rather than just subgroups, the answer is "Yes" for $S_\infty$ (in fact we just need closed subgroups). But otherwise it may depend on properties of the group $G$; for instance if $G$ is abelian then conjugacy is trivial, so we reduce to the original question.
• We can ask the same question for various adjectives for Borel subgroups; for instance we can restrict attention to Polishable groups, or to non-Archimedian Polishable groups.

Farah and Solecki showed the following:

Theorem: Every Polish group $G$ admits Borel subgroups of arbitrarily high Borel rank.

However, the construction is far from uniform. I want to know if the procedure can be made uniform, i.e. if we have a reasonably absolute transfinite operation which, on countable ordinal $\alpha$, spits out a Borel subgroup of $G$ which is not $\Pi_\alpha$. There are various morally equivalent ways to make this precise, for example:

Question: Work in $ZF + AD$ (axiom of determinacy). Suppose $G$ is a Polish group. Is there is an injection from $\omega_1$ to the set of Borel subgroups of $G$?

If such an injection existed, then after pruning, we can arrange the Borel ranks of subgroups to be strictly increasing. This is because under AD, Hjorth proved there is no injection from $\omega_1$ to $\Pi_\alpha$ subsets of $\mathbb{R}$, for any fixed $\alpha < \omega_1$.

Some remarks:

• The answer might depend on $G$. I would accept a positive or negative answer for any specific (uncountable) $G$.
• $\omega_1$ is reducible to the set of Borel subsets of $\mathbb{R}$: e.g. let $X_\alpha$ be the set of well-founded linear orders of rank less than $\alpha$.
• If we looked at subgroups up to conjugacy, rather than just subgroups, the answer is "Yes" for $S_\infty$ (in fact we just need closed subgroups). But otherwise it may depend on properties of the group $G$; for instance if $G$ is abelian then conjugacy is trivial, so we reduce to the original question.
• We can ask the same question for various adjectives for Borel subgroups; for instance we can restrict attention to Polishable groups, or to non-Archimedian Polishable groups.

Farah and Solecki showed the following in Borel subgroups of Polish groups:

Theorem: Every Polish group $G$ admits Borel subgroups of arbitrarily high Borel rank.

However, the construction is far from uniform. I want to know if the procedure can be made uniform, i.e. if we have a reasonably absolute transfinite operation which, on countable ordinal $\alpha$, spits out a Borel subgroup of $G$ which is not $\Pi_\alpha$. There are various morally equivalent ways to make this precise, for example:

Question: Work in $\text{ZF} + \text{AD}$ (axiom of determinacy). Suppose $G$ is a Polish group. Is there is an injection from $\omega_1$ to the set of Borel subgroups of $G$?

If such an injection existed, then after pruning, we can arrange the Borel ranks of subgroups to be strictly increasing. This is because under AD, Hjorth proved in An absoluteness principle for Borel sets there is no injection from $\omega_1$ to $\Pi_\alpha$ subsets of $\mathbb{R}$, for any fixed $\alpha < \omega_1$.

Some remarks:

• The answer might depend on $G$. I would accept a positive or negative answer for any specific (uncountable) $G$.
• $\omega_1$ is reducible to the set of Borel subsets of $\mathbb{R}$: e.g. let $X_\alpha$ be the set of well-founded linear orders of rank less than $\alpha$.
• If we looked at subgroups up to conjugacy, rather than just subgroups, the answer is "Yes" for $S_\infty$ (in fact we just need closed subgroups). But otherwise it may depend on properties of the group $G$; for instance if $G$ is abelian then conjugacy is trivial, so we reduce to the original question.
• We can ask the same question for various adjectives for Borel subgroups; for instance we can restrict attention to Polishable groups, or to non-Archimedian Polishable groups.