# Are Eilenberg-MacLane spaces limits of manifolds?

In this answer, mme shows that for any compact Lie group $$G$$, there is a model for its classifying space $$BG$$ which is the direct limit of closed manifolds. If $$G$$ is discrete (i.e. $$\dim G = 0$$), then a model for $$BG$$ is also a model for $$K(G, 1)$$ since $$\pi_i(BG) \cong \pi_{i-1}(G)$$.

Let $$G$$ be a countable abelian group and $$n > 1$$. Does there exist a model for $$K(G, n)$$ which is a direct limit of closed manifolds?

Recall, a space $$X$$ is a model for $$K(G, n)$$ if $$\pi_n(X) \cong G$$ and $$\pi_i(X) = 0$$ for $$i \neq n$$.

There is such a model for $$K(\mathbb{Z}, 2)$$, namely $$\mathbb{CP}^{\infty}$$ which arises as the direct limit of finite-dimensional complex projective spaces $$\{\mathbb{CP}^m\}_{m\geq 1}$$. This can also be viewed in the context of the linked answer as a model for $$BU(1)$$ is a model for $$K(\mathbb{Z}, 2)$$.

The above question may be too difficult to answer in general. I would be interested to know what happens for the next simplest cases, namely $$K(\mathbb{Z}/m\mathbb{Z}, 2)$$ and $$K(\mathbb{Z}, 3)$$.

• I believe that any homotopy type that is represented by a countable CW complex is also represented by an increasing union of closed manifolds. Commented May 22 at 1:29
• @Tom Goodwillie: just out of curiosity, how would you represent a bouquet of circles? Commented May 22 at 8:55
• @Nandor: one way would be to take a connect-sum of $k$ copies of $S^1 \times S^n$, and then take the natural inclusions as $n$ varies. Commented May 22 at 16:43
• @Ryan Budney: is the connected sum a manifold? Commented May 22 at 20:43
• @Nandor: yes it is. I don't think I know any category other than connected, oriented manifolds where it is defined. Commented May 23 at 3:07

First consider a finite CW complex $$X$$. Let $$M\sim X$$ be a compact manifold (with boundary), let $$M_n$$ be $$M\times D^n$$ and use the inclusions $$\dots \subset \partial M_n\subset M_n\subset \partial M_{n+1}\subset M_{n+1}\subset \dots$$ where we are viewing $$\partial M_{n+1}$$ as the double of $$M_n$$. The union as $$n\to\infty$$ is both homotopy equivalent to $$X$$ and an increasing union of closed manifolds.
For a countably infinite complex, do something similar: Write it as an increasing sequence of finite complexes $$X(k)$$. For each $$k$$ choose a compact manifold $$M(k)\sim X(k)$$, and arrange for $$M(k)$$ to be embedded in $$M(k+1)$$ (by a map in the homotopy class of the inclusion $$X_k\to X_{k+1}$$). We can arrange for $$M(k)$$ to be in the boundary of $$M(k+1)$$. Then $$X$$ is equivalent to the union of $$\dots \subset M(k)\subset M(k+1)\subset \dots,$$ which is also the union of the closed manifolds $$\dots \subset \partial M(k)\subset \partial M(k+1)\subset \dots,$$
• Given a finite CW complex, I know you can embed it into Euclidean space of large enough dimension, and then take a thickening to obtain a manifold with boundary with the same homotopy type. From this perspective, I can see how we can arrange for $M(k)$ to be embedded in $M(k+1)$, but I don't see how we can arrange for $M(k)$ to be in the boundary of $M(k+1)$. Can this be seen from this thickening idea or should I be constructing these manifolds differently? Commented May 22 at 13:16
• @MichaelAlbanese $M(k+1)=M(k+1)\times 0$ is contained in the boundary of $M(k+1)\times [0,1]$, so replace $M(k+1)$ by $M(k+1)\times [0,1]$. Commented May 22 at 16:11
• Thanks. Is it clear that there is a countable CW complex which is a model for $K(G, n)$? I guess it follows from here, but I was hoping for something more direct. My idea was to take a presentation for $G$, build the analogue of a presentation complex using $n$ and $(n+1)$-cells, then kill higher homotopy groups by attaching more cells. However, its not clear to me if this will only require countably many cells. It does if $G$ is finitely presented (the first CW complex will be finite and hence have finitely generated homotopy groups by Serre). Commented May 22 at 20:39
• Here's a simplicial construction. Let $S^n$ be a simplicial model for a based $n$-sphere, such as $\Delta[n]/\partial\Delta^n$. Let $S^n[G]$ be the simplicial abelian group obtained from $S^n$ by applying the functor from based sets to abelian groups (left adjoint to the forgetful functor). By Dold-Kan, the homotopy groups of (the realization of) $S^n[G]$ are the reduced homology groups of $S^n$ with coefficients in $G$; this CW complex is a $K(G,n)$. If $G$ is countable, then in each simplicial degree this simplicial abelian group is countable, so that the CW complex has countably many cells. Commented May 22 at 21:16