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CW -> Neighborhhood retract and clarification.
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Chris Schommer-Pries
Chris Schommer-Pries
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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 not too hardpossible to see that locally $EG$ is of the form $K \times \mathbb{R}^\infty$ for a locally finite CW-complexwhere $K$ 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.

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 not too hard to see that locally $EG$ is of the form $K \times \mathbb{R}^\infty$ for a locally finite CW-complex $K$ (which might be different at different points - we do not care).

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.

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.

Source Link
Chris Schommer-Pries
Chris Schommer-Pries
  • 27.5k
  • 3
  • 91
  • 171

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 not too hard to see that locally $EG$ is of the form $K \times \mathbb{R}^\infty$ for a locally finite CW-complex $K$ (which might be different at different points - we do not care).

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.