# ‘Naturally occurring’ $K(\pi, n)$ spaces, for $n \geq 2$

[edited!] Given a group $$\pi$$ and an integer $$n>1$$, what are examples of Eilenberg–MacLane spaces $$K(\pi, n)$$ that can be constructed as "known" manifolds? (Or if not a manifold, say some space people had a pre-existing desire to study before $$K(\pi,n)$$ spaces were identified as being of interest.)

Constructing $$K(\mathbf Z, 2)$$ as $$\mathbf{CP}^{\infty}$$ is the only example I know — but there must be more out there.

I'm interested in concrete examples (like the one above) that could, e.g., be given in a Topics grad course for topology students. They seem to be scarse, so it would be nice to know what was known.

Note: I've excluded $$n=1$$ because most people know examples (or can figure them out) in this case.

• In what sense is $\mathbb{C} \mathbb{P}^{\infty}$ a manifold? Oct 29, 2010 at 2:17
• Given a nice inclusive definition of "manifold" that allows some examples, what would be an example of a weak homotopy type that is not represented by a manifold? Oct 29, 2010 at 2:46
• Any countable simplicial complex is homotopy-equivalent to a Hilbert manifold. The idea is to inductively embed the skeleton into the Hilbert cube in such a way that you have a regular neighbourhood, making your simplicial complex homotopy-equivalent to the open regular neighbourhood -- and since it's open in Hilbert space it's a manifold. Oct 29, 2010 at 2:49
• There's a lovely paper of Kodama and Michor (2006) where they show that the component of $Imm(S^1,\mathbb R^2)/Diff^+(S^1)$ corresponding to the the figure-8 immersion has that homotopy-type of a $K(\mathbb Z,2)$. Here $Imm(S^1,\mathbb R^2)$ denotes immersions of $S^1$ in the plane, and we're modding out by orientation-preserving reparametrizations. Oct 29, 2010 at 4:37
• The space of lines in $\ell^2$ is a Hilbert manifold. This is not quite $CP^{\infty}$ and has the homotopy type of $K(Z,2)$. Similarly, you can turn $BG$ into a Hilbert manifold for any compact Lie group $G$. Oct 29, 2010 at 14:07

$$\DeclareMathOperator\B{B}\newcommand\TOP{\mathrm{TOP}}\newcommand\PL{\mathrm{PL}}\newcommand\BTOP{{\B}\TOP}\newcommand\BPL{{\B}\PL}$$Let $$\BTOP$$ and $$\BPL$$ be the classifying spaces of topological/PL-sphere bundles and $$\BTOP/\BPL$$ the homotopy fiber of the map $$\BPL \to \BTOP$$. Then $$\TOP/\PL$$ is a model for a $$K(\mathbb{Z}/2\mathbb{Z},3)$$ by Kirby and Siebenmann. This identifies a third cohomology class as obstruction to get a PL-structure on a topological sphere bundle.

• Ah, I had thought there was some result like this, but had only very vague recollections about it. Are there any proofs known other than Kirby and Siebenmann's? Are there any people working on this kind of thing in this decade? All the results I know in this direction are quite "old". Oct 30, 2010 at 15:26
• I think, in the book by Madsen and Milgram are some results of this sort. And I don't really know, what you mean, but while there's certainly less activity in the PL-world today than a few decades ago, at least topological manifolds are a topic, quite a few people are still working on. And the quoted result is surely essential to compare the topological, PL and smooth world. Oct 31, 2010 at 21:12
• I was just wondering who was carrying on the Kirby-Siebenmann, Ranicki, et all torch in the 21st century. A lot of topology grad students I know these days have never really heard the word "PL"... Nov 4, 2010 at 0:50
• @Lennart: What is a reference for this neat fact? Thanks! May 20, 2013 at 13:06
• @David: I think, it is in the Kirby-Siebenmann book "Foundational Essays on Topological Manifolds...". May 20, 2013 at 17:36

If $$M$$ is a hyperfinite type $$I\!I\!I_1$$ factor, then (at least conjecturally), its group of outer automorphisms is a $$K(\mathbb Z,3)$$.

This is based on the following three properties of that von Neumann algebra:
• The group of unitary central elements of $$M$$ is a circle, and thus a $$K(\mathbb Z,1)$$.
• The group of unitaries in $$M$$ is contractible.
• The automorphism group of $$M$$ is contractible (conjectural).

To see that $$\operatorname{Out}(M)\cong K(\mathbb Z,3)$$, apply the long exact sequence of homotopy groups to the following two fiber sequences: $$\begin{gather*} U(Z(M)) \to U(M) \to \operatorname{Inn}(M) \\ \operatorname{Inn}(M) \to \operatorname{Aut}(M) \to \operatorname{Out}(M). \end{gather*}$$

As a consequence, we also get that $${\operatorname B}{\operatorname{Out}(M)}\cong K(\mathbb Z,4)$$.

I recommend my talk "A K(ℤ,4) in nature" (MSRI, April 2014), for an explanation of how to realize $$\operatorname{Out}(M)$$ as the automorphism group of a naturally occurring mathemtical object.

• I enjoyed that talk you gave at MSRI. Is that stack you build really a $K(\mathbb{Z},4)$ or is that also conjectural as in the first part of the answer? Or is it the stack that is the real deal, and we don't really know about the space? May 28, 2014 at 8:06
• Out(M) is not a space: it's a sheaf on Top (and if you force it to be a space, then it acquires the coarse topology). Any Sheaf on Top has an associated homotopy type (via its singular complex), and that's what I claim to be a K(Z,4). Whether or not I should keep the adjective "conjectural" is debatable: I think that I can prove all the claims that I'm making, but I haven't worked out all the details yet... May 28, 2014 at 8:16
• No progress. Last year, I was thinking a bit about this with a graduate student. But we could not complete the proof that the automorphism group of a hyperfinite $III_1$ factor is contractible (a necessary step). I'm still convinced that this is true, but I have no proof to offer. My proof strategy was to realize the hyperfinite $III_1$ as $\mathcal A([0,3])$ for some conformal net $\mathcal A$, show that every automorphism $\alpha$ of $\mathcal A([0,3])$ can be deformed to one which is the identity on $\mathcal A([1,2])$, in a way that depends continuously on $\alpha$ (that part of the... Oct 2, 2017 at 8:30
• ...proof went ok). Then use the action of $Diff([0,3])$ on $\mathcal A([0,3])$ to push the automorphism "off to infinity", where in this case "infinity" is the points $0$ and $3$. The problem is that it's not clear that this is continuous at the end. Oct 2, 2017 at 8:32
• @AndréHenriques any more progress? Aug 28 at 9:15

Following up on Dai's answer, one can go a step further since $$P U(H)$$ is obviously a group. So if we can find a contractible space on which it acts freely, the quotient will be the next level up (namely, a $$K(\mathbb{Z},3)$$).

Such a space can be constructed as follows: take our favourite (separable, though that's not necessary) Hilbert space, $$H$$, and consider $$\operatorname{HS}(H)$$, the space of Hilbert–Schmidt operators on $$H$$. This is isomorphic to the Hilbert tensor product $$H^* \mathbin{\widehat{\otimes}} H$$ so is a Hilbert space. Its unitary group is thus contractible. The group $$U(H)$$ acts on $$\operatorname{HS}(H)$$ by conjugation, and once we divide out by the centre this becomes free. Thus $$P U(H)$$ acts on $$U(\operatorname{HS}(H))$$ freely and so the quotient is a $$K(\mathbb{Z},3)$$.

However, as $$P U(H)$$ does not act centrally on $$U(\operatorname{HS}(H))$$, the iteration stops here.

• That is very nice! Oct 29, 2010 at 7:37

The following example appears in the definition of twisted $K$-theory.

Let $H$ be an infinite dimensional separable Hilbert space over $\mathbb{C}$. Since the unitary group $U(H)$ is contractible, the projective unitary group $PU(H)= U(H)/S^1$ has the homotopy type of $K(\mathbb{Z},2)$. The fact that $BPU(H)\simeq K(\mathbb{Z},3)$ and the fact that $PU(H)$ acts on the space of Fredholm operators $\mathrm{Fred}(H)$ are essential in the definition of twisted $K$-theory.

• Argh, was too slow in my comment to the question. Oct 29, 2010 at 5:38
• Very nice. Is there a place you recommend for an exposition on this? Oct 29, 2010 at 6:28
• Note, however, the mistaken claims by Atiyah and Segal about the compact-open and the strong operator topologies on $PU(H)$ being distinct, and that $PU(H)$ is not a topological group in the compact-open topology. See eg ncatlab.org/nlab/show/U%28%E2%84%8B%29 and ncatlab.org/nlab/show/PU%28%E2%84%8B%29 Aug 29 at 6:30
There is a very nice model of $$K(\mathbb Z,n)$$ which is given by the free abelian topological group on the pointed space $$(S^n,\star)$$, let us call that $$F(S^n,\star)$$. An element in $$F(S^n,\star)$$ is given by a finite set of points in $$S^n \setminus \lbrace\star\rbrace$$ such that each point in this finite set carries a non-zero integer as a label with the obvious addition. The topology is more subtle to describe and made in such a way that $$F(S^n,\star)$$ is an abelian topological group, the inclusion $$S^n \subset F(S^n,\star)$$ is continuous and $$\star=0$$ in $$F(S^n,\star)$$.
Though, I am not sure whether $$F(S^n,\star)$$ is an infinite-dimensional manifold (I think not), it is still pretty regular being a topological group and a CW-complex at the same time.
• Or you could take symmetric prodcuts of $S^n$ labelled by elements of any abelian group $A$: that produces $K(A,n)$. But is that "geometric"? Oct 29, 2010 at 14:00