Title. For what topological groups $G$ can we take $EG \rightarrow BG$ to be of the form $S^{\infty} \rightarrow BG$?
If $G$ is a subgroup of either $S^0,S^1,S^3$ or $S^7$ this induces a free action on $S^{\infty}$ and thus a $G-$principal bundle $S^{\infty} \rightarrow BG$. Does the reverse direction hold? That is, if $G$ acts freely on $S^{\infty}$ is it contained in some of the $S^0,S^1,...,S^7$ as a subgroup such that the induced action on $S^{\infty}$ is the same?
I like to think of $EG$ and $BG$ in terms of configuration spaces.
The space $BG$ can be identified with the following configuration space. It consists of configurations of finitely many points in the open interval $(0,1)$ with the points having labels in the topological group $G$. It is topologized so that
It is a nice exercise to see that this agrees with the usual definition of $BG$ as the geometric realization of a simplicial space.
$EG$ has a similar description as configurations of points in the half-open interval $[0, 1)$. In this case points cannot slide off the closed end, and can only "disappear" by sliding off the open end.
Sliding everything off the open end gives a contraction onto the empty configuration, whence $EG$ is contractible.
The map $EG \to BG$ is just the restriction of configurations.
The action of $G$ on $EG$ is the following. Each configuration in $EG$ may be view as having the point $0 \in [0,1)$ as part of it - either it is already labeled or we give it the label $e \in G$. The action of $G$ just multiplies the label of the point $0$ on the left.
From these descriptions (or the usual simplicial ones) you can realize $EG$ as a certain colimit of simple spaces which consist of products of intervals (open and half-open) and copies of $G$.
If $G$ is a finite dimensional Lie group with countably many components, then from this colimit description it is possible to see that locally $EG$ is of the form $K \times \mathbb{R}^\infty$ where $K$ is a neighborhood retract of $\mathbb{R}^\infty$ (which might be different at different points - we do not care). If $G$ has finitely many components $K$ will even be a locally finite CW-complex. If there are countably many components $K$ will look like a finite dimensional countable CW-complex, which can still be embedded nicely in $\mathbb{R}^\infty$ as a neighborhood retract.
From the results cited in this excellent MO answer: https://mathoverflow.net/a/293409/184
we deduce the following surprising facts (1) $EG$ is actually locally modeled on $\mathbb{R}^\infty$ and (2) for spaces locally modeled on $\mathbb{R}^\infty$, homotopy equivalence implies homeomorphism.
Since both $EG$ and $S^\infty$ are contractible spaces locally modeled on $\mathbb{R}^\infty$, it follows that we have a homeomorphism $EG \cong S^\infty$.
So in summary: For any finite dimensional Lie group $G$ with countably many components you may take $EG \cong S^\infty$. For example $G$ can be any countable discrete group. However the free action of $G$ on $S^\infty$ is realized through a possibly strange homeomorphism and likely has nothing to do with $G$ acting on finite dimensional spheres.
