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

Question

[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.

Answer 1

$\DeclareMathOperator\B{B}\newcommand\TOP{\mathrm{TOP}}\newcommand\PL{\mathrm{PL}}\newcommand\BTOP{{\B}\TOP}\newcommand\BPL{{\B}\PL}$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.

Answer 2

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*}

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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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

This is all very classical and was studied in detail in

• Dold, Albrecht; Thom, René, Quasifaserungen und unendliche symmetrische Produkte, Ann. of Math. (2) 67 1958 239–281. https://doi.org/10.2307/1970005, https://www.jstor.org/stable/1970005