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Let $\mathrm{Homeo}(M)$ and $\mathrm{Borel}(M)$ be the groups of homeomorphic and Borel automorphisms of a space $M$, respectively.

Question: Are $\mathrm{Homeo}(M)$ and $\mathrm{Borel}(M)$ isomorphic as abstract groups for any reasonable non-discrete space $M$? Say, $M=\mathbb{R}$ or $M=\mathbb{S}^1$.

Obviously $\mathrm{Homeo}(M)\subsetneq\mathrm{Borel}(M)$ if $M$ is non-discrete, but with infinite cardinality this does not prohibit an algebraic isomorphism. The literature on $\mathrm{Homeo}(\mathbb{R})$ and $\mathrm{Borel}(\mathbb{R})$ that I have seen so far has not been explicit enough for me to figure out the answer in this case.

Thank you.

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    $\begingroup$ Random thoughts: (1) for, say, $M = [0,1]$, the Borel automorphism group contains continuum many pairwise commuting elements of order 2 (take continuum many disjoint pairs of points, and consider the maps that swap the pair while fixing everything else). Is that true of the homeomorphism group? Seems unlikely. (2) All uncountable Polish spaces have isomorphic Borel automorphism groups. But their homeomorphism groups can't all be isomorphic, right? $\endgroup$ May 13, 2017 at 1:59
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    $\begingroup$ @NateEldredge Let $M=\mathbb R$ for notational simplicity. For each $n\in \mathbb N$ let $f_n$ be the identity outside $(n,n+1)$ an autohomeomorphism of $[n,n+1]$ fixing $n$ and $n+1$, such as $x\mapsto n+\sqrt{x-n}$. Then for any set $B\subseteq \mathbb N$ there is a naturally defined $f_B = $ the composition of all $f_n$, $n\in B$, and the continuum many maps $f_B$ all commute with each other. $\endgroup$
    – Goldstern
    May 13, 2017 at 8:41
  • $\begingroup$ Continuing Nate Eldredge's second remark, it is known that $\mathrm{Homeo}(\mathbb{S}^2)$ (2-sphere) is simple but $\mathrm{Homeo}(\mathbb{R})$ is not (the subgroup of orientation preserving homeomorphisms is normal). $\endgroup$
    – Bedovlat
    May 13, 2017 at 11:44
  • $\begingroup$ There are many natural follow-up questions, about what would be reasons that for "most" spaces, these groups should be non-isomorphic. $\endgroup$
    – YCor
    Feb 10, 2019 at 2:31
  • $\begingroup$ The answer is clear for $M=\mathbf{R}$, because $\mathrm{Homeo}(\mathbf{R})$ has a torsion-free subgroup of index 2 (and hence no element of finite order $\ge 3$) while $\mathrm{Borel}(\mathbf{R})$ has element of all possible orders. Similarly $\mathrm{Homeo}(S^1)$ has all its finite subgroups cyclic or dihedral, while $\mathrm{Borel}(S^1)\simeq\mathrm{Borel}(\mathbf{R})$ contains copies of all finite groups. $\endgroup$
    – YCor
    Feb 18, 2020 at 9:48

2 Answers 2

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Let $\bar X = X\cup \{\infty\}$ be the 1-point compactification of a discrete space $X$. Then the autohomeomorphisms of $\bar X$ are all permutations $p$ of $X$ (extended by $p(\infty)=\infty$), whereas the Borel automorphisms are all permutations of $\bar X$ (since every subset of $\bar X$ is Borel, even open$\cup$closed). Algebraically, these two groups are isomorphic.

(But is this space "reasonable"? Felix Hausdorff might disagree...)

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    $\begingroup$ Seems reasonable to me. Why would Hausdorff reject it? $\endgroup$
    – Todd Trimble
    May 13, 2017 at 0:46
  • $\begingroup$ Sorry. It was late in the night, and the space looked non-Hausdorff... $\endgroup$
    – Goldstern
    May 13, 2017 at 8:32
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    $\begingroup$ This is indeed an answer. Equivalently, take $\{\frac1n\}_{n=1}^\infty$. And I think the space is Polish, since it is complete in the metric induced from $\mathbb{R}$, and countable, right? $\endgroup$
    – Bedovlat
    May 13, 2017 at 8:33
  • $\begingroup$ Definitely Polish. $\endgroup$
    – Todd Trimble
    May 13, 2017 at 10:23
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    $\begingroup$ Polish iff $X$ is countable. Also to continue with trivial details, $X$ has to be infinite for the groups to be isomorphic. $\endgroup$
    – YCor
    May 13, 2017 at 11:18
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For every uncountable Polish space $M$ (hence every non-discrete metrizable manifold with countably many components), the groups $\mathrm{Homeo}(M)$ and $\mathrm{Borel}(M)$ are non-isomorphic. Furthermore, $\mathrm{Borel}(M)$ is not isomorphic to any subgroup of $\mathrm{Homeo}(M)$.

Indeed, $\mathrm{Homeo}(M)$ is a separable metrizable group, and I claim that the subgroup $\mathfrak{S}_{\mathrm{fin}}(M)$ of finitely supported permutations of $M$ (which is a subgroup of $\mathrm{Borel}(M)$) does not embed into any separable metrizable topological group $G$. Indeed, let $(M_i)_{i\in I}$ be pairwise disjoint finite subsets of $M$, each of cardinal $\ge 3$, with $I$ uncountable, and let $\Gamma_i$ be a non-abelian subgroup of permutations of $M_i$, extended to the identity outside $M_i$.

By contradiction, let $f$ be an embedding of $\mathfrak{S}_{\mathrm{fin}}(M)$ into $G$. Since every subset of a separable metrizable space is separable, there exists a countable subset $J$ of $I$ such that $f\Big(\bigcup_{j\in J}f(M_j)\Big)$ is dense in $f\Big(\bigcup_{i\in I}f(M_i)\Big)$. Choose $i\in I-J$. Then $[M_i,M_j]=\{1\}$ for all $j\in J$, so $[f(M_i),f(M_j)]=\{1\}$ . By density, we deduce $[f(M_i),f(M_i)]=\{1\}$. But the latter equals $f([M_i,M_i])$, which is not trivial since $M_i$ is non-abelian and $f$ is injective. Contradiction.


For countable Polish spaces, one can give a full answer too: basically Goldstern's example is the only one along with discrete ones.

Let $M$ be a countable Polish space. Then $\mathrm{Homeo}(M)$ is non-isomorphic to $\mathrm{Borel}(M)$ (which equals $\mathrm{S}(M)$, the whole permutation group), with the only exceptions when $X$ is finite, infinite discrete or the 1-point compactification of an infinite countable set. Namely, beyond these exceptions, $\mathrm{Homeo}(M)$ is infinite and has at least 5 normal subgroups.

For $M$ infinite, by the Onofri-Schreier-Ulam theorem, $\mathrm{S}(M)$ has exactly 4 normal subgroups (trivial, whole, finite support, even finite support). In $\mathrm{Homeo}(M)$ we already have such a chain, where the intermediate subgroups are the group $\mathfrak{S}_{\mathrm{fin}}(M_{\mathrm{iso}})$ of finitely supported permutations of the subset $M_{\mathrm{iso}}$ of isolated points of $M$ ($M_{\mathrm{iso}}$ is dense in $M$, hence infinite) and its index 2 subgroup. Hence, if $\mathrm{Homeo}(M)$ is homeomorphic to $\mathrm{S}(M)$, then every normal subgroup is one of these.

Let $M_{\mathrm{acc}}$ be the set of accumulation points. If $M_{\mathrm{acc}}$ is empty then $M$ is discrete and this case is clear. Next we assume $M_{\mathrm{acc}}$ non-empty.

Let $H_M$ be the group of those self-homeomorphisms of $M$ that are identity on $M_{\mathrm{acc}}$, so $\mathfrak{S}_{\mathrm{fin}}(M_{\mathrm{iso}})\subset H_M$. Hence we either have (a) $H_M=\mathrm{Homeo}(M)$ or (b) $H_M=\mathfrak{S}_{\mathrm{fin}}(M_{\mathrm{iso}})$. Also denote $G_x$ as the subgroup of $\mathrm{Homeo}(M)$ consisting of those $g$ that are identity at the neighborhood of $x\in M$; if $x$ is $\mathrm{Homeo}(M)$-invariant then $G_x$ is a normal subgroup.

Lemma: for every topological metrizable space and point $x_0$ which is isolated among accumulation point and neighborhood $V$ of $x_0$, there exists a self-homeomorphism fixing $x_0$, not with finite support, and identity outside $V$.

Indeed, one can suppose that $V\smallsetminus\{x_0\}$ consists of isolated points of $X$; choose an injective sequence $(y_n)$ tending to $x_0$. Then permuting $y_{2n-1}$ and $y_{2n}$ and fixing all the remainder yields the desired permutation.$\Box$

Choose $x_0$ an isolated point in $M_{\mathrm{acc}}$. Applying the lemma to $x_0$, we see that $H_M$ is not reduced to $\mathfrak{S}_{\mathrm{fin}}(M_{\mathrm{iso}})$, which excludes (b). So assume (a): $\mathrm{Homeo}(M)$ acts trivially on $M_{\mathrm{acc}}$.

In particular, $G_{x_0}$ is a normal subgroup; by the lemma it is a proper subgroup of $\mathrm{Homeo}(M)$. If we assume that $X$ is not the 1-point compactification of a discrete set, we either (a') have another point in $M_{\mathrm{acc}}$, and hence another isolated point in $M_{\mathrm{acc}}$ or (b') there exists an open infinite discrete subset in $M$. In both cases, we deduce (using the lemma in Case (a')) that $G_{x_0}$ is not reduced to $\mathfrak{S}_{\mathrm{fin}}(M_{\mathrm{iso}})$.

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  • $\begingroup$ PS since they appear in the proof, let me mention the following fact (which I never heard, but is easy): let $X$ be a metrizable space with a single non-isolated point $x$. Then $X$ is homeomorphic to exactly one of the following three spaces: (a) 1-point compactification of an infinite discrete countable set (b) disjoint union of the latter with a infinite discrete countable set (c) $\{2^{-n}e_m,n,m\ge 0\}$ where $(e_m)$ is a Hilbert basis. (a) is compact, (b) is locally compact but not compact, and (c) is Polish but not locally compact. $\endgroup$
    – YCor
    Feb 18, 2020 at 13:52

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