18
$\begingroup$

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 $K(\mathbb{Z}, 3)$.

$\endgroup$
5
  • 2
    $\begingroup$ I believe that any homotopy type that is represented by a countable CW complex is also represented by an increasing union of closed manifolds. $\endgroup$ Commented May 22 at 1:29
  • 2
    $\begingroup$ @Tom Goodwillie: just out of curiosity, how would you represent a bouquet of circles? $\endgroup$
    – Nandor
    Commented May 22 at 8:55
  • $\begingroup$ @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. $\endgroup$ Commented May 22 at 16:43
  • $\begingroup$ @Ryan Budney: is the connected sum a manifold? $\endgroup$
    – Nandor
    Commented May 22 at 20:43
  • $\begingroup$ @Nandor: yes it is. I don't think I know any category other than connected, oriented manifolds where it is defined. $\endgroup$ Commented May 23 at 3:07

1 Answer 1

21
$\begingroup$

Any homotopy type that is represented by a countable CW complex is also represented by an increasing union of closed manifolds:

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 $$ \dots \subset \partial M(k)\subset \partial M(k+1)\subset \dots, $$

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$ Commented May 22 at 13:16
  • $\begingroup$ @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 $M(k+1)\times [0,1]$. $\endgroup$ Commented May 22 at 16:11
  • $\begingroup$ 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). $\endgroup$ Commented May 22 at 20:39
  • 2
    $\begingroup$ 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. $\endgroup$ Commented May 22 at 21:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.