$\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ has two interesting properties: on one hand it is non-compact, but on the other hand it admits a unique $\text{SL}(d,\mathbb R)$-invariant *finite* measure on $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$.


I wonder if there exists a bounded measurable subset $F$ of $\text{SL}(d,\mathbb R)$ (with subspace topology, of course) that is Borel isomorphic to $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$?


By "Borel isomorphic" I mean: There exists a bijection $f:F\subset \text{SL}(d,\mathbb R) \to \text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ such that both $f$ and $f^{-1}$ are Borel measurable, with $F$ inheriting the subspace topology from $\text{SL}(d,\mathbb R)$. Of course, "up to a subset of measure zero" is always allowed.


If this is true, then it is good to know a constructive proof but the proof for the existence of such an $F$ will also be greatly appreciated!




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Here I am not assuming $F$ is a "fundamental domain" for $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$, whose definition itself has many interpretations and I asked $F$ to be bounded and thus *may not be homeomorphic to* $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ (but as a possible approach one may want to construct such an $F$ whose closure is homeomorphic to the one-point compactification of $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ modulo a set of measure zero). I only want this $F$ to be *measure theoretically isomorphic to* $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$ but not topologically. Disproving this claim seems also very hard.



Existence of a bounded measurable subset of $\text{SL}(d,\mathbb R)$ that is Borel isomorphic to $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$?
By the Iwasawa decomposition, $\text{SL}(d,\mathbb R)$ is homeomorphic to $\mathbb{R}^n\times\text{SO}(d,\mathbb R)$, where $n=(d^2+d-2)/2$. As $\mathbb{R}^n$ is homeomorphic to any open ball in $\mathbb{R}^n$, it follows that $\text{SL}(d,\mathbb R)$ is homeomorphic to a bounded Borel subset of $\text{SL}(d,\mathbb R)$; this homeomorphism can be constructed explicitly. Now take a nice fundamental domain of $\text{SL}(d,\mathbb R)/\text{SL}(d,\mathbb Z)$; e.g. take a suitable Borel subset. We see that this fundamental domain is also homeomorphic to a bounded Borel subset of $\text{SL}(d,\mathbb R)$, and we are done.


