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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?

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    $\begingroup$ Often for infinite dimensional spaces homotopy equivalence often implies homeomorphism. See the discussion here: mathoverflow.net/questions/293382/… I don't understand this phenomenon well enough to confirm this, but it seems likely that many contractible infinite dimensional manifolds are homeomorphic to $S^\infty$. This would give quite a few examples: e.g. any subgroup of $GL_n$. $\endgroup$
    – Phil Tosteson
    Commented Feb 8, 2021 at 22:44
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    $\begingroup$ (1) $S^7$ is not a group. (2) Any group that can act freely on $S^n$ can also act freely on $S^\infty$. (3) I wonder about the converse. In particular, can a non-cyclic group of order act freely on $S^\infty$. (4) There are papers about which groups can act freely on (some finite-dimensional) spheres, but I don't know the state of the art. $\endgroup$
    – Tom Goodwillie
    Commented Feb 8, 2021 at 23:00
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    $\begingroup$ If I'm not mistaken, $S^\infty$ is homeomorphic to a whole Hilbert space, whose unitary group is probably huge, although contractible. Maybe if you edit the tags you can obtain better answers from people in representation theory, analysis, operator algebras, etc. $\endgroup$
    – Fernando Muro
    Commented Feb 9, 2021 at 0:33
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    $\begingroup$ @FernandoMuro I presume that here $S^\infty=\bigcup_nS^n$ rather than the whole unit sphere in a Hilbert space $H$, so I don't think that $S^\infty$ is homeomorphic to $H$. However, this $S^\infty$ is homeomorphic to $\mathbb{R}^\infty=\bigcup_n\mathbb{R}^n$ and to many other spaces such as such as the Stiefel manifold $V_n\mathbb{R}^\infty$, the complement $\mathbb{R}^\infty\setminus(\text{a finite set})$, the ordered configuration space $F_n\mathbb{R}^\infty$ etc. I am not sure where to find these facts in the literature, though. $\endgroup$
    – Neil Strickland
    Commented Feb 9, 2021 at 12:06
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    $\begingroup$ $S^\infty$ in the CW topology is not metrisable, so certainly not homeomorphic to either a Hilbert space or a Hilbert cube. As a (weakly) contractible $\mathbb{R}^\infty$-manifold, $S^\infty$ is homeomorphic to $\mathbb{R}^\infty$ (everything in CW topology). $\endgroup$
    – Tyrone
    Commented Feb 9, 2021 at 15:51

1 Answer 1

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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

  • when points collide, the labels are multiplied (using orientation of the interval to determine the order);
  • points labeled by the identity element of $G$ can always be added or removed;
  • Points can "disappear" by sliding off either end of the interval.

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.

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