Does $\mathbb C\mathbb P^\infty$ have a (commutative) group structure? More specifically, is it homeomorphic to $FS^2$, (the connected component of) the free commutative group on $S^2$?

$\mathbb C\mathbb P^\infty$ is well known to have the homotopy type of the classifying space of a commutative group, $\mathbb C^\times$. One model for the classifying space is the bar construction, which is a functor from groups to spaces and is product preserving, so the classifying space of a commutative group is again a commutative group (since the group operation is a homomorphism iff the group is a commutative). The Dold-Thom theorem says that the component of the free commutative group on $S^n$ is a $K(\mathbb Z,n)$. So those are two fairly nice group models. But is either actually homeomorphic to the space we started with, $\mathbb C\mathbb P^\infty$?

By $\mathbb C\mathbb P^\infty$ I mean the colimit of the $\mathbb C\mathbb P^n$ under the closed inclusions. Similarly, I topologize the free commutative group on a space $X$ as the colimit of the inclusions of the subset of words of length $n$, which is topologized as a quotient of $(X\times\{\pm1\}\cup\{*\})^n$, that is, sequences of $n$ letters, each of which may be a generator, the inverse of a generator, or the identity element. So $\mathbb C\mathbb P^\infty$, $BS^1$, and $FS^2$ are CW complexes with finite skeleta. Are they all locally modeled on $\mathbb R^\infty$? Do such infinite dimensional manifolds have homeomorphism types determined by their homotopy types?

Appendix: Motivation

In the question, I invoked the Dold-Thom theorem to motivate the particular model $FS^2$ under consideration. But this case is easy and should motivate the general theorem, rather than vice versa. So I will suggest a proof. Moreover, I will motivate the choice of space, how we get from $\mathbb C\mathbb P^\infty$ to $FS^2=F\mathbb C\mathbb P^1$. We want to make a model for $B\mathbb C^\times$ that is as nice as possible. If we identify $\mathbb C^\infty=\mathbb C[x]$, it is a ring, so its multiplication makes $\mathbb C\mathbb P^\infty=\mathbb P(\mathbb C[x])$ into an topological monoid. That’s a pretty nice structure; it just lacks inverses. So we should consider $\mathbb P(\mathbb C(x))$. For any reasonable topology on $\mathbb C(x)$, the vector space is contractible, as is the complement of the origin; and the multiplication on the ring is continuous. Thus $\mathbb P(\mathbb C(x))$ is a topological monoid homotopy equivalent to $\mathbb C\mathbb P^\infty$. This is better than the previous model because every element has an inverse. However, the inversion map is continuous only for some topologies. For counterexample, the finest possible topology is given by making the topological vector space the colimit of its finite dimensional subspaces (a colimit indexed by an uncountable filtering poset). But then the inverse is not continuous: the copy of $\mathbb C$ given by $y\mapsto x-y$ has inverses a set of linearly independent vectors, thus discrete.

Turning away from topological vector spaces, how else can we put a topology on $P(\mathbb C(x))$? A polynomial is determined (up to scale) by its zeros and a rational function by its zeros and poles. Then we can think of $\mathbb P(\mathbb C[x])$ as something like the free commutative monoid on $S^2=\mathbb C\mathbb P^1$ (specifically, the free monoid modulo the point at infinity). Similarly, $\mathbb P(\mathbb C(x))$ is the component of the free commutative group on $\mathbb C\mathbb P^1$, that is, $F\mathbb C\mathbb P^1$. To put the zeros and poles on the same footing, we need the topology described from the beginning.

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    $\begingroup$ There's also the minor issue that $\Bbb C(x)$ is a vector space of uncountable dimension over $\Bbb C$ and so its projective space is slightly different than $\Bbb{CP}^\infty$. $\endgroup$ Commented Mar 31, 2016 at 3:07
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    $\begingroup$ I think that the tangent bundle to $\Bbb{CP}^\infty$ is not parallelizable, hence this space cannot be a Lie group (modulo tecnicalities caused by infinite-dimensionality). $\endgroup$ Commented Mar 31, 2016 at 3:31
  • $\begingroup$ @TylerLawson Yes, but. Cardinality shows that there cannot be a linear isomorphism. If you use the finest topology, the colimit of the finite dimensional subspaces, then the basis is discrete, so $\mathbb C(x)$ admits an uncountable discrete subspace, while $\mathbb C[x]$ does not. So neither is there a non-linear homeomorphism. But my questions is whether $\mathbb C[x]$ under the finest topology is (equivariantly) homeomorphic to $\mathbb C(x)$ under a different topology that I implicitly specify. $\endgroup$ Commented Mar 31, 2016 at 3:37
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    $\begingroup$ @VictorProtsak Do you have a reference that carefully considers infinite dimensional tangent bundles? I expect these are trivial because of the Eilenberg swindle. $\endgroup$ Commented Mar 31, 2016 at 3:45
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    $\begingroup$ No, otherwise I would have posted an answer :) $\endgroup$ Commented Mar 31, 2016 at 4:23

6 Answers 6


I noticed that this question still has no accepted answer and all existing answers are rather long. It seems that the answer can be easily obtained using some results of infinite-dimensional topology, namely, the theory of manifolds modeled on the direct limit $\mathbb R^\infty$ of Euclidean spaces (see Chapter 5 of Sakai's book).

Two results of this theory will be important for our purposes:

Characterization Theorem 5.4.1. A topological space $X$ is is homeomorphic to an open subspace of $\mathbb R^\infty$ if and only if any embedding $f:B\to X$ of a closed subspace $B$ of a finite-dimensional compact metrizable space $A$ can be extended to an embedding $\bar f:A\to X$.

Classification Theorem 5.5.1. Two $\mathbb R^\infty$-manifolds are homeomorphic if and only if they are homotopically equivalent.

Now using the Characterization Theorem, it can be shown that $\mathbb{CP^\infty}$ is an $\mathbb R^\infty$-manifold.

Next, in his question Ben Wieland writes that $\mathbb{CP}^\infty$ is homotopically equivalent to the connected component $G_0$ of the free topological Abelian group $G$ over the sphere. Using the Characterization Theorem of Sakai once more, one can show (and this was done by Zarichnyi in 1982) that $G_0$ is an $\mathbb R^\infty$-manifold. Since $\mathbb{CP}^\infty$ and $G_0$ are two homotopically equivalent $\mathbb R^\infty$-manifolds, the Classification Theorem ensures that $\mathbb{CP}^\infty$ is homeomorphic to $G_0$ and hence has a compatible topogical group structure.

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    $\begingroup$ I am so glad someone finished this off cleanly! $\endgroup$ Commented Feb 21, 2018 at 16:21

I believe that the answer to the first question is yes, but I don't know about the second one.

It is well-known that ${\mathbb C}P^\infty$ is homeomorphic to the bar construction of the circle group $S^1$. You can see this from Milnor's model for the bar construction. The n-th skeleton of the bar construction is homeomorphic to the quotient $(S^1)^{*n+1}/S^1=S^{2n+1}/S^1={\mathbb C}P^n$.

It follows that ${\mathbb C}P^\infty$ has a structure of a topological abelian group. I don't know an explicit geometric description of this group structure.

{EDIT: As Tom pointed out, Milnor's model for $BG$ is, in most cases, homotopy equivalent but not homeomorphic to the bar construction, so this argument is inadequate. By inspecting low dimensional skeleta, it seems likely that in our case the two spaces are homeomorphic, by a non-obvious map. A later edit See also Ben Wieland's answer. }

It also is well-known that ${\mathbb C}P^n$ is homeomorphic to $Sp^n({\mathbb C}P^1)$, the n-fold symmetric product of ${\mathbb C}P^1$. You can construct a map $SP^n({\mathbb C}P^1)\to {\mathbb C}P^n$ as follows: let $[[u_1; v_1],\ldots, [u_n; v_n]]$ be a point of $SP^n({\mathbb C}P^1)$. Consider the the homogeneous polynomial of two variables $\prod_{i=1}^n(u_ix-v_iy)$. The coefficients of this polynomial define a point in ${\mathbb C}P^n$. The injectivity of this map follows from uniqueness of factorization of polynomials, and surjectivity follows from the fundamental theorem of algebra. Taking limit as $n$ goes to infinity, we obtain that ${\mathbb C}P^\infty$ is homeomorphic to $SP^\infty({\mathbb C}P^1)$ - the pointed free commutative monoid generated by ${\mathbb C}P^1$.

Note that ${\mathbb R}P^\infty$ is homeomorphic to the bar construction on the group ${\mathbb Z}/2$ by a similar argument (EDIT: With the same caveat as above). Thus ${\mathbb R}P^\infty$ also has an abelian group structure. However, as far as I know ${\mathbb R}P^\infty$ is not homeomorphic to a free commutative monoid.

  • $\begingroup$ Gregory, how can one see that the bar construction has a structure of abelian topological group? $\endgroup$ Commented Mar 31, 2016 at 19:37
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    $\begingroup$ If G is abelian then the multiplication $G\times G \to G$ and the inverse map $G\to G$ are group homomorphisms. So you get a multiplication map $BG\times BG\cong B(G\times G) \to BG$ and similarly an inverse map. $\endgroup$ Commented Mar 31, 2016 at 20:06
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    $\begingroup$ Curiously, if $G$ has order $2$ then the topological abelian group $BG$ inherits the property that $2x=0$ for all $x$. So if it really is homeomorphic to $\mathbb R P^\infty$ then it's very unlike a Lie group. $\endgroup$ Commented Apr 1, 2016 at 1:42
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    $\begingroup$ If Milnor's $BG$ is the one that has the infinite join $EG$ as a principal $G$-bundle, then this in effect uses faces but not degeneracies in its construction, so it is not obviously homeomorphic to the one that satisfies $B(G\times H)\cong BG\times BH$. $\endgroup$ Commented Apr 1, 2016 at 1:45
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    $\begingroup$ @Tom You are right, in general Milnor's join and the bar construction are only homotopy equivalent. I think in our two cases they are homeomorphic. For each $n$, there is a natural map from the $n$-th stage of Milnor's $BG$ to the $n$-th stage of the bar construction. In cases we care about, the source this map is $CP^n$ or $RP^n$. In particular it is a closed PL manifold, and the map is a quotient map with contractible preimages. So its source and target ought to be homeomorphic. I will try to write a proper proof later. If I am wrong, someone will point it out in the meantime... $\endgroup$ Commented Apr 1, 2016 at 8:57

$\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}\def\PP{\mathbb{P}}$This is a report of partial progress, not a full answer. Probably the most interesting thing, at the end, is that I show Segal's model of $B(\mathbb{Z}/2)$ really is $\mathbb{RP}^{\infty}$ and really does have an abelian group structure, but I wind up having to cite a pretty difficult paper to do it.

Let $G$ be a commutative metrizable group. According to these notes by Somnath Basu, Segal constructs $BG$ in the following manner. (I think that Basu has added a lot of insight of his own.) Let $EG_n$ be the set of all step functions $[0,1] \to G$ with at most $n$ discontinuities, where two step functions are equated if they disagree at finitely many points, metrized by (for example) $d_{EG_n}(\alpha, \beta) = \int_{[0,1]} d_G(\alpha(x), \beta(x)) dx$. For example, let $C^2$ be the cyclic group of order $2$. Then $EC^2_1$ consists of two closed line segments glued at their endpoints. Explicitly, the point $t$ in the first copy of $[0,1]$ corresponds to $$f_t(x) := \begin{cases} 0 & x<t \\ 1 & x>t \end{cases}$$ and the $t$ in the second copy corresponds to $$g_t(x) := \begin{cases} 1 & x<t \\ 0 & x>t. \end{cases}$$ The end points are glued because $f_0=g_1$ and $f_1=g_0$.

Similarly, $ES^1_1$ is $S^1 \times [0,1]$ with $\left( S^1 \times \{ 0,1 \} \right) \cup \left( \{0 \} \times [0,1] \right)$ contracted to a point, which is clearly $S^2$.

Segal defines $EG$ as the direct limit $\bigcup EG_n$, and $BG_n$ and $BG$ as the obvious quotients $EG_n/G$ and $EG/G$. Pointwise multiplication gives a map $EG_m \times EG_n \to EG_{m+n}$ and, if $G$ is commutative, it descends to the quotient $BG_m \times BG_n \to BG_{m+n}$. On the infinite unions $BG$ and $EG$, we get groups.

Remark: There is a natural surjection $F(BG_1) \to BG$, since $BG$ is generated by $BG_1$ as a group, but the restriction to the identity component of $F(BG_1)$ isn't an isomorphism. Let $$\phi(t,g)(x) = \begin{cases} 1 & x < t \\ g & x>t \end{cases}.$$ Then $\phi(t_1,g_1) + \phi(t_1,-g_1) - \phi(t_2, g_2) - \phi(t_2, -g_2)$ is in the connected component of the identity and in the kernel of the map. So this doesn't seem like a good way to approach the free group issue.

So we would like to know, for this particular construction, are $(EC^2)_n$, $(BC^2)_n$, $(ES^1)_n$ and $(BS^1)_n$ homeomorphic to $S^n$, $\mathbb{RP}^n$, $S^{2n+1}$ and $\mathbb{CP}^n$ respectively.

Set $EG^{\circ}_n := EG_n \setminus EG_{n-1}$ and $BG^{\circ}_n := BG_n \setminus BG_{n-1}$. Then $EG_n^{\circ}$ clearly isomorphic to $G \times (G \setminus \{ 0 \})^{n-1} \times \{ 0 < x_1 < x_2 < \cdots < x_n < 1 \} \cong G \times (G \setminus \{ 0 \})^{n-1} \times \RR^n$, and $BG_n^{\circ} \cong (G \setminus \{ 0 \})^{n-1} \times \RR^{n-1}$.

We see that $(EC^2)_n^{\circ} = \RR^n \sqcup \RR^n$, so $EC_n$ has a cell structure with two cells of each dimension. If we knew it was a regular cell decomposition, then we would know $(EC^2)_n \cong S^n$. Similarly, $(BS^1)_n^{\circ} \cong \RR^n \times \RR^n \cong \CC^n$, but it seems hard to understand the attachment maps.

I have seen $(EC^2)_n$ before! Consider $n=3$ first. $(EC^2)_3$ is divided into two maximal cells. Let $\Delta_3 = \{ (x_1, x_2, x_3) \in \RR_{\geq 0}^3 : \sum x_i = 1 \}$. Each maximal cell is of the form $(z,0,1-z) \cong (z',0,1-z')$ for any $z$, $z'$. We can explicitly build the quotient map $\Delta^3 \to (EC^2)^3$ as a map from $\Delta^3$ to $3 \times 3$ matrices by $$ (x,y,z) \mapsto \begin{pmatrix} 1 & x & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & y \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & z & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} . $$ The image is the matrices $\left( \begin{smallmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{smallmatrix} \right)$ obeying $a$, $b$, $c$, $\det \left( \begin{smallmatrix} a & b \\ 1 & c \end{smallmatrix} \right) \geq 1$, $a+c=1$. Think of this as related to the reduced word $s_1 s_2 s_1$ in the Weyl group of $GL_3$. Fomin and Shapiro build similar maps $\Delta^n \to U$ where $U$ is the unipotent radical of any reductive group $G$, corresponding to reduced words of length $n$ in the Weyl group of $G$. They build a corresponding cell complex and conjecture it is regular.

Hersh proved Fomin and Shapiro's conjecture! Theorem 6.13 of her paper shows that the cell complex for $(EC^2)_n$ is regular. (The CW complex she is discussing the closure of one of the two cells in our cellular structure.) I would love to see someone write out a simpler proof of this result, though. I went through Hersh's proof line by line and convinced myself it worked, but I never felt like I understood it conceptually.

It is also worth noting there is a different set of models: I'll call them $\widehat{EG}_n$ and $\widehat{BG}_n$ which are clearly isomorphic to $\CC \PP^n$ and $\RR \PP^n$.

A point of $\widehat{EG}_n$ is a pair $( (x_1, x_2, \ldots, x_n), f )$ where $0 \leq x_1 \leq x_2 \leq \cdots \leq x_n \leq 1$ and $f$ is a function in $EG_n$ whose discontinuity set is contained in $(x_1, x_2, \ldots, x_n)$. For example, $(EC^2)_1$ is made of four line segments glued into a square: Two of them are $\{ (t,f_t) \}$ and $\{ (t, g_t) \}$ for $f_t$ and $g_t$ as above; the other two are $\{ (t,0) \}$ and $\{ (t,1) \}$, where $0$ and $1$ are the constant functions. Again, $\widehat{BG}_n = \widehat{EG}_n/G$.

There are obvious maps $\widehat{EG}_n \to EG_n$ and $\widehat{BG}_n \to BG_n$. If $f$ has $k$ discontinuities, its preimage in $\widehat{EG}_n$ is a simplex of dimension $n-k$ (just specify the locations for the $n-k$ points not forced by $f$).

It is also easy to see that $\widehat{ES^1}_n \cong S^{2n+1}$ (and likewise for the other three cases). Given $((x_1, \ldots, x_n), f)$, formally set $x_0=0$ and $x_{n+1}=1$, and let $\theta_k$ be the value of $f$ on $(x_{k-1}, x_k)$. Set $r_k = \sqrt{x_{k}-x_{k-1}}$. Identify $((x_1, \ldots, x_n), f)$ with $$(r_1 \cos \theta_1, r_1 \sin \theta_1, r_2 \cos \theta_2, r_2 \sin \theta_2, \ldots, r_n \cos \theta_n, r_n \sin \theta_n).$$ Notice that $\theta_k$ is undefined precisely when $r_k=0$, so this makes sense, and it is easy to check that it gives a homeomorphism $\widehat{ES^1}_n \cong S^{2n+1}$. But there doesn't seem to be a group structure on $\widehat{BG}$.

  • $\begingroup$ If I am not mistaken, the attaching maps in this complex are quotient maps, with equivalence classes being contractible. There are results that say that for "nice enough" quotient maps where the source is a manifold (in our case a closed ball), the quotient space is homeomorphic to the source. I was wondering if one could use a result like this to prove that the complex is regular. $\endgroup$ Commented Apr 1, 2016 at 19:30
  • $\begingroup$ I'd love citations to a result like this! $\endgroup$ Commented Apr 1, 2016 at 20:46
  • $\begingroup$ The book of Daverman and Venema calvin.edu/~venema/embeddingsbook/embeddings-noprint.pdf gives some criteria. See in particular section 2.3 $\endgroup$ Commented Apr 1, 2016 at 22:11
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    $\begingroup$ Here's one way to describe $BG$ for $G=S^1$ (or any other abelian Lie group, for example $\mathbb Z/2$): Form a direct sum of copies of $G$, one for each point in $\Delta^1$. Divide by the sum of the two copies corresponding to the endpoints. Now give this group the minimum topology (as a topological abelian group) such that the evident map $\Delta^1\times G\to BG$ is continuous. In other words, $BG$ is the universal example of a topological abelian group equipped with a based loop of continuous homomorphisms from $G$. $\endgroup$ Commented Apr 2, 2016 at 17:31
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    $\begingroup$ Yes, I know. I just like this way of saying it, and thought somebody else might find it useful. $\endgroup$ Commented Apr 3, 2016 at 3:26

Putting it all together...

A nice description of $EG$ is given by the space of piecewise constant functions from the interval $[0,1]$ to $G$. That is, the choice of a finite subset of the interval and a locally constant function defined away from the subset (in particular, we don’t care about the values at discontinuities). If a function has a removable discontinuity, we identify it with the resolution; in particular, discontinuities at the endpoints are not needed. This space is contractible: comb the discontinuities to the right, while introducing a discontinuity from the left with the identity to the left of it. When all the discontinuities, including the new one, reach the right, we are left with the constant function with value the identity. This set is a group under pointwise multiplication. It contains the constant functions as a subgroup, so $BG=EG/G$ is a classifying space for $G$. If $G$ is abelian, so is $EG$ and thus $BG$.

This space is naturally identified with the bar construction, although we don't need this; indeed, it is a good way to prove several properties of the bar construction. Specifically, both $EG$ and $BG$ are product preserving. Thus they take abelian groups to abelian groups, although we already saw that. Both are filtered by the number of discontinuities. $E_nG$ is parameterized by the choice of $n$ discontinuities and $n+1$ choices of values between them: $G^{n+1}\times \Delta^n$, which gives a precise definition of the topology. The face identifications correspond to ignoring the value of the function on an interval of length zero. The degeneracy identifications correspond to ignoring removable discontinuities. Milnor’s iterated join construction $E^M_nG=G^{*n+1}$ can be thought of as a multiset of cardinality $n$ in the interval and with values in $G$ on the intervals that remain, ie, ignoring the intervals of length zero, but keeping track of removable discontinuities, even with multiplicity.* So the fiber of $E^M_nG\to E_nG$ over a function that requires $k$ discontinuities is $\Delta^{n-k}$, the choice of $n-k$ points to add to the multiset of formal discontinuities. Since $G$ acts freely on these simplices, the fibers of $B^M_nG\to B_nG$ are also simplices.

$B^M_nG$ and $B_nG$ are not homeomorphic in general. In particular, $E^M_1\mathbb Z/3$ is a graph with 6 vertices (a $K_{3,3}$), while $E_1\mathbb Z/3$ is a graph with 3 vertices (a $K_3$ with multiplicity 2). But in a few special cases, they are homeomorphic. If $G$ is a sphere, namely, if $G=\mathbb Z/2,S^1,SU(2)$, then $E_n^MG=G^{*n+1}$ is a sphere and $B_n^MG$ is a manifold. $B_nG$ is a quotient of $B_n^MG$ whose fibers are disks. It is well-known that such a quotient is again a manifold, indeed, homeomorphic to the original manifold. If the source is a PL manifold, the map simplicial, and the fibers PL isomorphic to disks, then the isomorphism is PL, as I will sketch at the end. If we were careful about constructing the homeomorphisms, they would probably be compatible across $n$, showing that $B^MG$ is homeomorphic to $BG$. Alternately, we can apply surgery theory to show that the embeddings of projective spaces (over $\mathbb R$, $\mathbb C$, or $\mathbb H$) in the standard homology classes are unique,** to produce isomorphisms of towers and show that the limits are homeomorphic.

Thus $\mathbb C\mathbb P^\infty$ and $\mathbb R\mathbb P^\infty$ are homeomorphic to the abelian groups $BS^1$ and $B\mathbb Z/2$. Also, $\mathbb H\mathbb P^\infty$ is homeomorphic to $BSU(2)$. Taking an appropriate circle bundle over $\mathbb C\mathbb P^\infty$ gives an abelian group structure on a lens space. Left open are various other homeomorphisms, such as with other group models, such as the lens space with $B\mathbb Z/n$ or projective space with the free abelian group on the sphere and the lens space with the free abelian group on the Moore space.

Finally, we need to prove that if $M$ is a PL manifold without boundary and $X$ is a quotient simplicial complex such that the fibers are all disks, then $X$ is a PL manifold, indeed, isomorphic to $M$, by an unspecified map homotopic to the given non-injective quotient map $M\to X$. The proof is to factor the quotient into a sequence of basic collapses $M\to X_1\to X_2\cdots\to X$. Specifically, the collapses are indexed by the simplicies of $X$, ordered by increasing dimension, for a triangulation with the properties that all simplices are embedded (we must subdivide for $X=B_nG$) and that the fibers are constant over the interior of each simplex. We define $X_r$ to perform the collapses over the first $r$ simplices. That is, $\pi\colon X_r\to X$ should be an isomorphism over the interiors of the first $r$ simplicies and over later simplices should have the same fibers as $M\to X$. At each step $X_r$ should be a quotient of $M$ and admit $X$ as a quotient $\pi\colon X_r\to X$. Construct it by induction: take $X_r$ and consider the next simplex $\Delta_{r+1}\subset X$. Since the boundary $\partial\Delta_{r+1}$ of $\Delta_{r+1}$ is lower dimensional, each of its simplices have already been collapsed, so $\pi^{-1}(\partial\Delta_{r+1})\to\partial\Delta_{r+1}$ is an isomorphism. Thus we can take $X_r$, remove the preimage of the interior $\pi^{-1}(\mathring\Delta_{r+1})$ and glue in $\Delta_{r+1}$ along its boundary to define $X_{r+1}$.

So we factor our collapse into a sequence of such collapses, thus defining basic collapses. They are ones such that the fibers are all either a single point or a $D^p$ and the points with big fibers form a $D^q$. By induction, $X_r$ is a manifold, indeed isomorphic to $M$. I claim that the set of points in big fibers $\mathring D^q\times D^p$ has closure contained in a chart. Thus the theorem reduces to proving that the quotient of $\mathbb R^n$ by a basic collapse is again $\mathbb R^n$. I claim that a basic collapse is isomorphic to a standard example, as follows. Take the $\ell^1$ metric on $\mathbb R^n=\mathbb R^p\times\mathbb R^q\times\mathbb R^{n-p-q}$. The closure of the set of big fibers will be the unit ball times a point $B_{p+q}\times \{0\}\subset \mathbb R^{p+1}\times\mathbb R^{n-p-q}$. The equivalence relation is that two points are identified if they have the same projection to $B_q\subset\mathbb R^q$. The fibers are balls of various radii in $\mathbb R^p$. Over the boundary of $B_q$, they are balls of radius zero, single points, while over the interior of $B_q$ they are balls of positive radius, $D^p$. It is easy to prove that the standard example works. The key is proving that any map with the same fibers is locally isomorphic to it.

* Milnor’s construction is supposed to correspond to applying the face identifications, but I am confused about that. I think that one should identify each face of a $1$-simplex with a $0$-simplex, but there aren’t very many $0$-simplices, so it seems like the two ends must be glued together. Specifically, I think that should identify functions with the formal discontinuity at $0$ with functions with no breakpoints, and similarly with one discontinuity at $1$. But this means identifying configurations with breakpoint at $0$ with configurations with breakpoint at $1$. Compared to $G*G$, it should glue together the two polar copies of $G$. (This identification will eventually be forced by the degeneracies.) But this does not have the same homotopy type as $E_1G$, so it cannot be correct.

** The case of greatest interest $\mathbb C\mathbb P^n\subset \mathbb C\mathbb P^{n+1}$ is a shocking result. This is codimension 2, the inaccessible realm of knot theory. There are lots of ways of knotting $S^{2n}$ in $S^{2n+2}$, but take the connected sum with complex projective space and they all unknot.

Like David, I found the description of $BG$ in the notes of Somnath Basu. He attributes it to Segal 1968, but it appears original to me. A similar construction, with the advantage of generality, but also with the disadvantages of generality, appears in Drinfeld 2003 following Besser 1998.

  • $\begingroup$ Could you point me to sources regarding "It is well-known that the quotient of a manifold by a contractible equivalence relation is homeomorphic to the original manifold." I spent a good deal of last night reading maths.ed.ac.uk/~aar/papers/daverman.pdf looking for theorems about when a quotient of a manifold is homeomorphic to the original manifold. I was left with two competing feelings (more) $\endgroup$ Commented Apr 4, 2016 at 1:26
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    $\begingroup$ (1) "wow, this is much more complicated than I though it was" (2) "all of these examples involve contracting wild, fractal objects. Surely, it isn't so hard if all I want to deal with is contracting polyhedral subcomplexes of a simplicial complex." But I couldn't find any theorems which said contracting nice things is straightforward. This literature is very foreign to me, so I'd appreciate any pointers. $\endgroup$ Commented Apr 4, 2016 at 1:28
  • $\begingroup$ $E_nG$ (or the $n$-skeleton of the realization of any simplicial space) is obtained from the category of nonempty ordered sets $\lbrace 0,\dots ,k\rbrace$, $0\le k\le n$, with monotone maps as morphisms. If you use the subcategory having the same objects but just the injective monotone maps, you don't get $E^M_nG$. You get $E_n^MG$ if you use the poset of nonempty subsets of $\lbrace 0,\dots ,n\rbrace$. $\endgroup$ Commented Apr 4, 2016 at 4:29
  • $\begingroup$ @DavidSpeyer for the case of collapsing a contractible complex to a point, first do the case where the complex is collapsible (like here). In general, you have to expand it, which takes some care to make sure that it is still embedded. But I think more general equivalence relations are an exercise. It's locally just crossing with a disk. Take a triangulation downstairs so that the relation is constant on the interiors of simplices. Do the collapses over the vertices. Then do the collapses over the edges. Only interesting over the interiors of the edges, so all independent and products. $\endgroup$ Commented Apr 4, 2016 at 18:29
  • $\begingroup$ @DavidSpeyer Actually, the claim is false. If we have a contractible $n$ manifold with boundary a non-simply-connected homology sphere, then it embeds in a manifold by adding a collar, but collapsing it to a point yields the cone on the homology sphere, which is not homeomorphic to a manifold. Similarly, if there were a counterexample to the 4D PL/smooth Poincaré conjecture it would bound a contractible manifold, that would yield an example that is not a PL manifold. Moreover, I think it would be PL-cellular. $\endgroup$ Commented Apr 4, 2016 at 21:14

Let $G$ be a set. Let $EG$ be the realization of the nerve of the category that has $G$ as object set, with one morphism for each source and target. This is contractible if $G$ is not empty. If $G$ is a group then $G$ acts freely on $EG$ by translation and the orbit space is the bar construction $BG$.

I just realized that topologically $EG$ is the same for all finite sets $G$ having at least two elements: it is an infinite simplex (the direct limit of $\Delta^n$ as $n\to \infty$, the map $\Delta^n\to \Delta^{n+1}$ being embedding as a face). This is also the same as $S^\infty$, the direct limit of $S^n$.

Here's how I'm thinking: $EG$ is the direct limit of its skeleta. The $n$-skeleton is the realization of the simplicial set (with only finitely many nondegenerate simplices) whose $p$-simplices are all sequences $(g_0,\dots ,g_p)$ such that there are at most $n$ jumps, a jump being an instance of $g_{i+i}\neq g_{i}$. Choose an ordering of the finite set $G$, and consider the simplicial set $X_n$ whose $p$-simplices are all sequences $(g_0,\dots ,g_p)$ such that there are at most $n$ backwards jumps, instances of $g_{i+1}<g_i$. This contains the $n$-skeleton and is contained in some skeleton (in fact, in the $(mn+m+n)$-skeleton where $G$ has $m+1$ elements). Therefore $EG$ is the direct limit of the $X_n$. And I claim that $X_n$ is topologically a simplex of dimension $mn+m+n$, and that the inclusion of $X_n$ into $X_{n+1}$ corresponds to the inclusion of something topologically equivalent to a face, namely the union of $n+1$ codimension one faces of a face of a certain dimension.

Another way to describe $EG$ is the closed unit ball in $\mathbb R^\infty$ (direct limit of disks rather than simplices). This is also homeomorphic to $S^{\infty}$, because if $D^n$ is embedded in $S^n$ as a hemisphere and $S^n$ is embedded in $D^{n+1}$ as the boundary then the union of $$ \dots \subset D^n\subset S^n\subset D^{n+1}\subset S^{n+1}\subset \dots $$ is both $D^\infty$ and $S^\infty$. (Maybe it's also homeomorphic to $\mathbb R^\infty$.)

Anyway, this shows that all of the spaces $BG$ for nontrivial finite $G$ (in particular all of the topological abelian groups $BG$ for nontrivial finite abelian $G$) have the same universal covering space; they are all locally homeomorphic to $\mathbb R^\infty$.

But these topological abelian groups $EG$ and $BG$ are not very reminiscent of Lie groups, as you can see from the fact that as an abstract group $EG$ or its quotient $BG$ has exponent $N$ if $G$ has exponent $N$.

To prove the claim above, that $X_n$ is topologically a simplex, take the step function viewpoint: a point in $EG$ is a way of cutting up $\Delta^1$ into finitely many segments and labeling these by elements of $G$, with the rule that a cut between two equal labels can be erased and that when two cuts come together the label between disappears. $X_n$ is the part where there is the constraint that at most $n$ cuts can involve a backwards jump. But this is precisely the simplex of all possible ways of cutting the interval into $(m+1)(n+1)$ subintervals (some of which might have length zero) and labeling them by listing all of the elements of the ordered set $G$ in order, $n+1$ times over.

  • $\begingroup$ The strategy suggested in my question was to prove (1) that everything is homogeneous, (2) that everything is locally modeled on $\mathbb R^\infty$, and (3) that for such manifolds homotopy equivalence implies homeomorphism. But since we pretty much managed to answer the question by explicit construction, I posted it as a separate question. $\endgroup$ Commented Apr 6, 2016 at 4:00

This is obviously only a partial answer, concerning $B\mathbb{Z}/2 = \mathbb{RP}^\infty$. In particular, I think that there is a much simpler way to show that it has the structure of an Abelian group.

First of all, a definition.

Definition Let $X$ be a topological space. We can (formally) write elements of the $n$-th symmetric power of $X, Sym^n(X)$ as sums $x_1 + \cdots + x_n$. Then the $n$-th truncated product of $X$, denoted $TP^n(X)$, is the quotient of $Sym^n(X)$ by the relation $2x = 0$. More precisely, we say that $$ \sum k_i x_i \sim \sum k_i' x_i \iff k_i \equiv k_i' \pmod 2 $$

As with symmetric powers, we have inclusions $TP^n(X) \hookrightarrow TP^{n+1}(X)$ obtained by specifying a basepoint.

Theorem (Mostovoy, http://www.math.cinvestav.mx/~mostovoy/Papers/roots.pdf) The space $TP^n(\mathbb{RP}^1)$ is homeomorphic to $\mathbb{RP}^n$.

Taking the colimit for these inclusions, we obtain $\mathbb{RP}^\infty$, which is the collection of all finite sums of elements $\sum_i x_i$ with $x_i \in \mathbb{RP}^1$ and no term repeating. Moreover, this has an obvious structure as a commutative monoid by combining sums; every element is its own inverse, and so it is an Abelian group.

Remark While looking at Mostovoy's paper, I realize this may be a little more subtle than I'd thought when starting to write this up. The question is as to whether or not the inclusions $\mathbb{RP}^n \hookrightarrow \mathbb{RP}^{n+1}$ are the standard inclusions or are simply homotopic to the standard inclusions. It is a little discomfiting that we need to specify a basepoint in order to obtain $TP^n(X) \hookrightarrow TP^{n+1}(X)$ in our case, which should be natural. However, we don't need to specify one to obtain $TP^n(X) \hookrightarrow TP^{n+2}(X)$, and so perhaps one could look instead at the colimit of $$ TP^1(\mathbb{RP}^1) \hookrightarrow TP^3(\mathbb{RP}^1) \hookrightarrow TP^5(\mathbb{RP}^1) \hookrightarrow \cdots $$ which is natural and should give the same limit...

  • $\begingroup$ In the last display, do you mean $TP^3(\mathbb{RP}^1$ instead of $TP^1(\mathbb{RP}^3)$? $\endgroup$ Commented Apr 6, 2016 at 11:56
  • $\begingroup$ Oh yes, silly me. Thanks for catching that. $\endgroup$
    – Simon Rose
    Commented Apr 6, 2016 at 11:58
  • $\begingroup$ The uniqueness of $\mathbb R\mathbb P^n$ in $\mathbb R\mathbb P^{n+1}$ is a hard theorem (needs the h-cobordism theorem) but it is overkill. For both your answer and my answer we can prove that the complement is a $\mathbb R^{n+1}$ and then it becomes fairly easy. (Same for $\mathbb C\mathbb P^n$. $\mathbb H\mathbb P^n$ probably does require h-cobordism, but not surgery.) $\endgroup$ Commented Apr 6, 2016 at 20:14
  • $\begingroup$ Another way to see this is via Gavin Wraith's characterization of $S^\infty$ as the topological Boolean algebra freely generated from $[0,1]$ viewed as a topological distributive lattice (see his "Using the generic interval"). Since $S^\infty$ is a topological Boolean algebra, in particular it is a topological $\mathbb F_2$-vector space; and $\mathbb R\mathbb P^\infty$ is the quotient of it by a free involution which, in fact, can be taken the Boolean negation, so this is quotient by the subgroup $\{0,1\}$. $\endgroup$ Commented Feb 21, 2018 at 8:27

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