Eilenberg–MacLane space $K(\mathbb{Z}_2,n)$

Question

We know that the generalized classifying space / Eilenberg–MacLane space $$ B\mathbb{Z}_2=\mathbb{RP}^{\infty} $$ $$ BU(1)=\mathbb{CP}^{\infty} $$

• How do one construct/derive the (infinite dimensional) space explicitly: $$ B^n \mathbb{Z}_2=K(\mathbb{Z}_2,n)=? $$

• How do one construct/derive, when $p$ is a prime: $$ B^n \mathbb{Z}_p=K(\mathbb{Z}_p,n)=? $$

• Can one explain $B\mathbb{Z}_2$, $B^2 \mathbb{Z}_2=K(\mathbb{Z}_2,2)$, $B^n \mathbb{Z}_2=K(\mathbb{Z}_2,n)$, $B^n \mathbb{Z}_p=K(\mathbb{Z}_p,n)$, in a unified consistent but also intuitive way?

Answer 1

Given a connected topological space $X$ with $\pi_n(X) = G$ and $\pi_i(X) = 0$ for $i = 1, \dots, n - 1$, you can build a $K(G, n)$ by attaching cells of dimension at least $n + 2$ to kill the higher homotopy groups.

Suppose you start with $X = \mathbb{RP}^2$ which has $\pi_1(X) \cong \mathbb{Z}_2$. Then $\pi_2(\mathbb{RP}^2) = \mathbb{Z}$ and is generated by the covering map $S^2 \to \mathbb{RP}^2$. Attaching a three-cell to $\mathbb{RP}^2$ with attaching map given by the covering map $S^2 \to \mathbb{RP}^2$ gives a space $X'$ with $\pi_1(X') = \mathbb{Z}_2$ and $\pi_2(X') = 0$; in fact, $X'$ is homotopy equivalent to $\mathbb{RP}^3$. Now note that $\pi_3(\mathbb{RP}^3) \cong \mathbb{Z}$ generated by the covering map $S^3 \to \mathbb{RP}^3$. Attaching a four-cell to $\mathbb{RP}^3$ with attaching map given by the covering map $S^3 \to \mathbb{RP}^3$ gives a space $X''$ with $\pi_1(X'') = \mathbb{Z}_2$ and $\pi_2(X'') = \pi_3(X'') = 0$; in fact, $X''$ is homotopy equivalent to $\mathbb{RP}^4$. Repeating this procedure ad infinitum, we see that $\mathbb{RP}^{\infty}$ is a $K(\mathbb{Z}_2, 1)$.

Suppose now you start with $X = S^2$ which has $\pi_1(X) = 0$ and $\pi_2(X) \cong \mathbb{Z}$. Then $\pi_3(S^2) \cong \mathbb{Z}$ and is generated by the Hopf map $S^3 \to S^2$. Attaching a four-cell to $S^2$ with attaching map given by the Hopf map $S^3 \to S^2$ gives a space $X'$ with $\pi_1(X') = 0$, $\pi_2(X') \cong \mathbb{Z}$, and $\pi_3(X') = 0$; in fact, $X'$ is homotopy equivalent to $\mathbb{CP}^2$. Now note that $\pi_4(\mathbb{CP}^2) = 0$ so we don't need to glue on any five-cells, but $\pi_5(\mathbb{CP}^2) \cong \mathbb{Z}$ generated by the quotient map $S^5 \to \mathbb{CP}^2$. Attaching a six-cell to $\mathbb{CP}^2$ with attaching map given by the quotient map $S^5 \to \mathbb{CP}^2$ gives a space $X''$ with $\pi_1(X'') = 0$, $\pi_2(X'') \cong \mathbb{Z}$, and $\pi_3(X'') = \pi_4(X'') = \pi_5(X'') = 0$; in fact, $X''$ is homotopy equivalent to $\mathbb{CP}^3$. Repeating this procedure ad infinitum, we see that $\mathbb{CP}^{\infty}$ is a $K(\mathbb{Z}, 2)$.

The case for cyclic groups is similar to the first example. Note that $\mathbb{RP}^2$ can be viewed as a circle with a two-cell attached via the double covering map $S^1 \to S^1$. Likewise, a circle with a two-cell attached via the $k$-sheeted covering map $S^1 \to S^1$ is a topological space $X$ with $\pi_1(X) \cong \mathbb{Z}_k$. The above procedure then shows that the infinite lens space $S^{\infty}/\mathbb{Z}_k$ is a $K(\mathbb{Z}_k, 1)$. In order to obtain a $K(\mathbb{Z}_k, n)$, one can form the space $X$ obtained by attaching an $(n + 1)$-cell to $S^n$ with attaching map given by a degree $k$ map $S^n \to S^n$; equivalently, $X$ is the previous space suspended $n - 1$ times. This space has $\pi_n(X) \cong \mathbb{Z}_k$ and $\pi_i(X) = 0$ for $i = 1, \dots, n - 1$. I don't know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(\mathbb{Z}_k, n)$.

As for forming $K(\mathbb{Z}, n)$'s, there is a simple starting space, namely $X = S^n$ (which is $S^1$ suspended $n - 1$ times). Again, I do not know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(\mathbb{Z}, n)$.

For $n \geq 2$, a $K(G, n)$ only exists if $G$ is abelian. If in addition $G$ is finitely generated, then a $K(G, n)$ can be formed by taking products of the examples above (because $\pi_n(X\times Y) \cong \pi_n(X)\oplus\pi_n(Y)$).

For $n = 1$, if $G$ is abelian and finitely generated, then as before, you can just take a product of spaces constructed above to obtain a $K(G, 1)$. However, $G$ can be non-abelian, and taking products of the spaces above cannot produce a $K(G, 1)$ for $G$ not abelian. However, there is a simple way to construct a space $X$ with $\pi_1(X) = G$ which you can then apply the above procedure to in order obtain a $K(G, 1)$. Take a presentation for $G$. First form a bouquet of circles, one for each generator, and now attaching two-cells, one for each relation, with attaching map determined by the word used in the relation; this is the desired $X$ (which is sometimes called a presentation complex for $G$). Again, I do not know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(G, 1)$.

Answer 2

For an abelian group $A$ with a discrete topology we have

$$K(A,0)\simeq A,\qquad K(A,1)\simeq BA.$$

Note the homotopy equivalences rather than equalities. We can model $BA$ by using a bar construction to get a contractible free $A$-space $EA$, and then set $BA=(EA)/A$. I detailed this in an answer here Classifying space $B$SU(n).

One finds that the bar construction model for $EA$ is a topological monoid (that is, they carry an associative, unital product), and in the case that $A$ is abelian, this product descends to make $BA$ into a topological monoid. Hence the bar construction may be applied again to generate a contractible free $BA$-space $E(BA)$, whose $BA$ quotient is a classifying space $B(BA)=B^2A=E(BA)/BA$.

Since the projection $EBA\rightarrow B^2A$ is fibration with contractible total space and fibre $BA\simeq K(A,1)$ we study the long exact homotopy sequence to get

$$\pi_i(B^2A)\cong \pi_{i-1}(BA)$$

In particular, if $BA$ is a $K(A,1)$, then $B^2A$ is a $K(A,2)$.

Now for the same reasons as before $B^2A$ has a classifying space $B^3A\simeq K(A,3)$, and we can proceed by induction we get

$$K(A,n)=B^nA.$$

In general we have no nice way to model $K(A,n)$ except for basically the few cases you outline in the question. A useful fact is that

$$K(A\oplus B,n)\simeq K(A,n)\times K(B,n)$$

so one can apply the theorem on the classification of abelian groups to decompose things into easier to work with factors. Another useful fact is that if $A\rightarrow B$ is a homomorphism then there is an induced map $K(A,n)\rightarrow K(B,n)$, and in particular if $0\rightarrow A\xrightarrow{f} B\xrightarrow{g} C\rightarrow 0$ is short exact then there is a fibration sequence

$$\dots\rightarrow K(C,n-1)\rightarrow K(A,n)\xrightarrow{Kf} K(B,n)\xrightarrow{Kg} K(C,n)\rightarrow K(A,n+1)\rightarrow\dots$$

And this can be useful. For example $0\rightarrow \mathbb{Z}\xrightarrow{\times 2}\mathbb{Z}\rightarrow \mathbb{Z}_2\rightarrow 0$ gives us

$$\dots\rightarrow S^1\rightarrow \mathbb{R}P^\infty\rightarrow \mathbb{C}P^\infty\xrightarrow{\times 2} \mathbb{C}P^\infty\rightarrow K(\mathbb{Z}_2,2)\rightarrow\dots$$

Answer 3

One way to understand Eilenberg-Maclane spaces is using the fact that they represent cohomology. Namely, over the category of CW-complexes, the functors $H^n(-; G)$ and $[-, K(G, n)]$ are naturally isomorphic. If $G$ is moreover an abelian group, then the multiplication map $G \times G \to G$ is a group homomorphism, therefore induces a multiplication map on $K(G, n)$ that makes it an $H$-space; hence $[X, K(G, n)]$ gets a group structure, and these are isomorphic as functors to $\mathsf{Grp}$.

Then $K(\Bbb Z_2, n)$ can be thought as the (unique upto homotopy equivalence) space which represents $H^n(X; \Bbb Z_2)$. For $n = 1$ there is an explicit way to describe this correspondence, since $K(\Bbb Z_2, 1) \simeq \Bbb{RP}^\infty$ and maps $X \to \Bbb{RP}^\infty$ classify line bundles on $X$ upto isomorphism. This bijection $\mathcal{L}(X) \to [X, \Bbb{RP}^\infty]$ can be established by sending a line bundle $\ell/X$ to the Gauss map of it's fiberwise embedding in some Euclidean bundle $X \times \Bbb{R}^\infty$, thinking of $\Bbb{RP}^\infty$ as a Grassmannian.

As Tyrone described, there is a tower of fibrations $$\cdots \to K(\Bbb Z, n) \to K(\Bbb Z, n) \to K(\Bbb Z_2, n) \to K(\Bbb Z, n+1) \to \cdots$$ induced from the short exact sequence $0 \to \Bbb Z \to \Bbb Z \to \Bbb Z_2 \to 0$, but the dual to this tower (or what this tower of fibrations represent) is the Bockstein long exact sequence $$\cdots \to H^n(X; \Bbb Z) \to H^n(X; \Bbb Z) \to H^n(X; \Bbb Z_2) \to H^{n+1}(X; \Bbb Z) \to \cdots$$ induced from the change-of-coefficients sequence $0 \to C^*(X; \Bbb Z) \to C^*(X; \Bbb Z) \to C^*(X; \Bbb Z_2) \to 0$ The snake map $H^n(X; \Bbb Z_2) \to H^{n+1}(X; \Bbb Z)$ is a natural transformation $H^n(-; \Bbb Z_2)\! \Rightarrow\! H^{n+1}(-; \Bbb Z)$ and by representability, is Yoneda-dual to the map $K(\Bbb Z_2, n) \to K(\Bbb Z, n+1)$ that is the "connecting morphism" in the fiber sequence. An alternative description of this map would be as classifying map $K(\Bbb Z_2, n) \to BK(\Bbb Z, n)$ of the fibration $K(\Bbb Z, n) \to K(\Bbb Z, n) \to K(\Bbb Z_2, n)$.

This is mostly how I think about higher Eilenberg-Maclane spaces, as usually they don't have an easy topological model.

There is a sort of striking geometry to $K(\Bbb Z_2, 3)$, and I'll try to explain my limited understanding of the story (if there are one or several misunderstandings here I'd really appreciate if someone could comment on that): Let's call $\text{TOP}(n)$ the group of germs-at-origin of homeomorphisms of $\Bbb R^n$ fixing the origin. Let $\text{PL}(n) \subset \text{TOP}(n)$ be the subgroup of germs of piecewise linear such homeomorphisms. There is a fibration $$\text{TOP}(n)/\text{PL}(n) \to B\text{PL}(n) \to B\text{TOP}(n)$$ There are natural morphisms $\text{TOP}(n) \to \text{TOP}(n+1)$ restricting to $\text{PL}(n) \to \text{PL}(n+1)$, given by sending a germ of a self-homeomorphism of $\Bbb R^n$ to that of $\Bbb R^{n} \times \Bbb R$, extended by identity on the $\Bbb R$-factor. Call the direct limit of the directed system given by these morphisms to be $\text{TOP}$ and $\text{PL}$ respectively. The aforementioned fibration should induce a fibration $$\text{TOP}/\text{PL} \to B\text{PL} \to B\text{TOP}$$ Kirby and Seibenmann proved that $\text{TOP}/\text{PL}$ is a $K(\Bbb Z_2, 3)$-space. A striking consequence of this seems to be the following: if $M$ is a topological manifold, there is a classifying map $M \to B\text{TOP}$ which lifts to a map $M \to B\text{PL}$ if and only if $M$ admits a PL-manifold structure. The $\text{TOP}/\text{PL}$-bundle over $B\text{TOP}$ is classified by a map $B\text{TOP} \to B(\text{TOP}/\text{PL})$, which has fiber $B\text{PL}$. So there seems to be another bundle $B\text{PL} \to B\text{TOP} \to B(\text{TOP}/\text{PL})$, which should be nothing but the one induced from $\text{PL} \to \text{TOP} \to \text{TOP}/\text{PL}$. Then the lifting condition is equivalent to demanding the pushforward $M \to B\text{TOP} \to B(\text{TOP}/\text{PL})$ is nullhomotopic as it factors through the fibers. But since $\text{TOP}/\text{PL} \cong K(\Bbb Z_2, 3)$, this is equivalent to saying the obstruction to $M$ having a PL-structure is homotopical nontriviality of a map $M \to BK(\Bbb Z_2, 3) \cong K(\Bbb Z_2, 4)$, which is an element of $[M, K(\Bbb Z_2, 4)] \cong H^4(M; \Bbb Z_2)$. If I understand correctly this is the Kirby-Seibenmann class. A consequence of this is that any manifold with $H^4 = 0$ has a PL structure (so for example, topological 3-manifolds automatically always have PL structures).

EDIT: As @MikeMiller points out in the comment below, the argument for this conclusion is going to be more subtle that I described. $\text{TOP}/\text{PL}$ is simply the fiber of the fibration $B\text{PL} \to B\text{TOP}$, so does not have an obvious group structure, which makes $B(\text{TOP}/\text{PL})$ nonsensical. Apparently if the obstruction theory is done, the obstruction classes for the existence of that lifting would belong to $H^{k+1}(M; \pi_k \text{TOP}/\text{PL})$, and using the fact that $\text{TOP}/\text{PL} \cong K(\Bbb Z_2, 3)$, one arrives at the fact that the only obstruction lies in $H^4(M; \Bbb Z_2)$.

I don't know if any other $K(\Bbb Z_2, n)$ has some tractable topological model though.