Given a topological group $G$, a $G$-space is a topological space $X$ equipped with an action of $G$, such that the map $(g,x) \mapsto g.x$ is continuous. The action is distal if no non-diagonal orbit of $G$ on $X \times X$ has an accumulation point on the diagonal, and minimal if every orbit is dense.


A theorem of Furstenberg (The Structure of Distal Flows, 1963) says that if $G$ is a locally compact group and $X$ is a compact metrizable $G$-space such that the action of $G$ is minimal and distal, then $X$ is built out of isometric extensions: that is, there is a (potentially transfinite) sequence $(X\_{\beta})\_{\beta \le \alpha}$ of compact $G$-spaces and surjective morphisms $X\_{\beta+1} \rightarrow X\_{\beta}$ so that $X\_0$ is the one-point space, $X\_{\alpha} = X$, $X\_{\beta}$ is the inverse limit of $(X\_{\gamma})\_{\gamma < \beta}$ when $\beta$ is a limit ordinal, and for each successor ordinal $\beta+1$ there is a $G$-invariant real-valued function on $X\_{\beta+1} \times X\_{\beta+1}$ that restricts to a compatible metric on each fibre of $X\_{\beta+1} \rightarrow X\_{\beta}$. (Conversely, any $G$-space built out isometric extensions in this way is distal; this direction is much easier and doesn't need any special hypotheses.)


Subsequent authors have generalized this result by removing various hypotheses and obtaining a similar conclusion about the decomposition of distal minimal $G$-spaces; sometimes the $G$-invariant metrics need to be replaced with some version of equicontinuity, but this is just an inevitable consequence of allowing $G$-spaces that are not first-countable. One constant though seems to be that $X$ needs to be compact. Are there any similar results if $X$ is instead some 'nice' topological space, e.g. a Polish space or a locally compact Hausdorff space, but not necessarily compact? Alternatively, are there some instructive examples of distal minimal $G$-spaces that don't break into equicontinuous pieces?


I am mainly interested in the case when $X = G/K$ for some closed subgroup $K$ of a locally compact or Polish group $G$, if that makes life easier (so in particular, the action is transitive).



Furstenberg decomposition for non-compact spaces
Here is a counter example:
take $G=\text{SL}\_2(\mathbb{R})$ and $X=\mathbb{R}^2-\{0,0\}$.
It is easy to check that the action is distal, and that there is no non-trivial factor carrying an invariant metric (the only proper factor is $\mathbb{P}^1(\mathbb{R})$).




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More generally (I think), taking any simple Lie group $G$ and $X=G/U$ for a unipotent subgroup $U$, you will get a distal action with no isometric factors.
This follows from the fact that $U$ has closed orbits on $G/U$ and Theorem 6.2 in <https://arxiv.org/pdf/1408.4217.pdf>.




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However, the following answers positively the case $X=G/K$ where $G$ is locally compact and $K<G$ a compact subgroup.


*Base:* Fix a non constant function in $L^2(G/K)$ and pull it back to a right-$K$-invariant function in $L^2(G)$. Let $K\_1$ be the stabilizer of this function wrt the right $G$ action. Thus $K<K\_1<G$. By the fact that the function is in $L^2(G)$, $K\_1$ is compact. The Hilbert metric of $L^2(G)$ restrict to a metric on the orbit, thus there is a $G$-invariant metric on $G/K\_1$.


*Step:* Assume having a compact group $K\_\alpha\gneq K$. Fix a non constant function in $L^2(K\_\alpha/K)$ and pull it back to $L^2(K\_\alpha)$. Let $K\_{\alpha+1}$ be the stabilizer of this function wrt the right $K\_\alpha$ action. Thus $K<K\_{\alpha+1}\lneq K\_\alpha$. The Hilbert metric of $L^2(K\_\alpha)$ restrict to a metric on the orbit, thus there is a $K\_\alpha$-invariant metric on $K\_\alpha/K\_{\alpha+1}$. Equivalently, there is a $G$-invariant metric on the fibers of $G/K\_{\alpha+1}\to G/K\_\alpha$.


*Limit:* Having a limit ordinal $\beta$ such that $K\_\alpha$ is constructed for every $\alpha\lneq \beta$, define $K\_\beta$ to be the closure of $\cup\_{\alpha\lneq \beta} K\_\alpha$.


*Termination:* With a lot of patient (depending on the cardinality of $G$) you get $\alpha$ such that $K\_\alpha=K$.


