# Specific examples of Eilenberg-Maclane spaces?

Given an integer $$n$$ and a group $$G$$ (abelian if $$n \geq 2$$), it's always possible to construct a $$K(G,n)$$ as a cell complex. The standard procedure is to choose a presentation $$\langle S | R \rangle$$ of $$G$$, construct a wedge sum of $$n$$-spheres (one for each generator in $$S$$), then attach an $$(n+1)$$-cell for each relation in $$R$$ via an attaching map determined by that relation. This gives you an $$(n-1)$$-connected space with $$\pi_n = G$$, and you continue attaching higher dimensional cells to kill off the higher homotopy groups.

This approach works, at least in theory, but relies on knowing what the next highest homotopy group at each stage is. The details get cumbersome as you attach more cells, and it's not clear to me that you can hope to get an explicit cell structure in this way. However, for some cases like $$K(\mathbb{Z},2)=\mathbb{C} P^\infty$$, the cell structure is rather simple to describe: $$\mathbb{C} P^\infty$$ has exactly one cell in every even dimension, and the attaching map for the $$2k$$-cell corresponds to the fiber bundle $$S^{2k-1} \rightarrow \mathbb{C} P^{k-1}$$.

My question is this: are there any other examples of $$K(G,n)$$'s that turn out to be familiar spaces like $$S^1$$, $$\mathbb{C} P^\infty$$, and $$\mathbb{R} P^\infty$$? Or just ones with easy to understand cell structures at all?

• I recall seeing the fact that the only $K(G,1)$ that can have finite skeletons are those with $G$ a free (abelian) group. – Connor Malin Jun 22 '19 at 17:20
• @ConnorMalin you might be misremembering something. See my answer for other examples. – Lee Mosher Jun 23 '19 at 2:54
• @ConnorMalin The correct set of adjectives would be that $G$ must not have finite-order elements. Otherwise, it would have a $\Bbb Z_n$ subgroup, so $K(G, 1)$ would be covered by a $K(\Bbb Z_n, 1)$ which has homology in infinitely many dimensions, so does not admit any model by a finite dimensional CW complex. But if $K(G, 1)$ had one such model, one could lift it upstairs. – Balarka Sen Jun 23 '19 at 20:10

For $$K(G,1)$$ spaces there are some geometric methods. The key thing about the $$K(G,1)$$ property is that a connected CW complex possesses that property if and only if the universal covering space of that complex is contractible. And there are several examples of theorems of the form "such-and-such hypotheses imply that the universal covering space is contractible".

Here's one such class of examples, coming from Riemannian geometry. Let $$M$$ be a smooth connected $$m$$-manifold, and let $$g$$ be a complete Riemannian metric on $$M$$ (for what it's worth, all smooth compact manifolds have finite CW-complex structures). If all sectional curvatures of $$g$$ are non-positive, then $$M$$ is a $$K(G,1)$$ space. The reason this is true is because the universal cover $$\widetilde M$$ is simply connected $$m$$-manifold, the lift $$\tilde g$$ is a complete Riemannian metric with nonpositive sectional curvatures, and now one applies the Cartan-Hadamard theorem to conclude that $$\widetilde M$$ is diffeomorphic to $$\mathbb R^m$$ and is therefore contractible, and so $$M$$ is a $$K(G,1)$$ space.

For some very specific examples, if $$m=2$$ and if $$M$$ is any connected 2-manifold that is not homeomorphic to $$\mathbb S^2$$ or $$\mathbb R P^2$$ then $$M$$ is a $$K(G,1)$$ space (because all such surfaces have a complete Riemannian metric of constant curvature $$0$$ or $$1$$, by an application of the Riemann mapping theorem). So, for example, for each $$g \ge 2$$ the closed, oriented surface of genus $$g$$ is a $$K(G,1)$$ space.

I'll add two more examples to the mix: Let $$K \subset S^3$$ be a knot, then $$M = S^3 \setminus K$$ is a $$K(G, 1)$$ space. This following from Papakyriakopoulos's sphere theorem: if $$\pi_2M \neq 0$$ then there's an embedded PL sphere $$S^2 \to M \subset S^3$$ which is not nullhomotopic. By Alexander's theorem both components of the complement of this sphere in $$S^3$$ must be balls, and $$K$$ being a connected subspace is contained entirely in one of them. The other component is an embedded ball in $$M$$ bounding the sphere, which is a contradiction. This implies $$\pi_2M = 0$$; pass to the universal cover $$\tilde{M}$$ which is $$2$$-connected so by Hurewicz's theorem $$\pi_3\tilde{M} = H_3\tilde{M}$$ which is also zero as $$\tilde{M}$$ is noncompact. Subsequently by Hurewicz's theorem again $$\pi_n \tilde{M} = H_n \tilde{M} = 0$$ for all $$n > 3$$, which means $$\tilde{M}$$ therefore $$\pi_n M = 0$$ for all $$n \geq 2$$. Therefore $$M$$ is a $$K(\pi_1 M, 1)$$.

I think it's interesting to understand which knot complements admit non-positively curved metrics in the context of Lee Mosher's answer; the knots whose complements admit complete constant curvature $$-1$$ metrics are called hyperbolic knots. My understanding is that it's a theorem of Thurston that every knot is either a torus knot, a satellite knot, or a hyperbolic knot. Examples of satellite knots include those of the form $$K = K_1 \# K_2$$, so one can take an embedded torus by taking a $$\varepsilon$$-sphere around the tube $$S^0 \times I$$ joining $$K_1$$ and $$K_2$$ in the connected sum construction, and replacing $$K_1$$ and $$K_2$$ by big spheres. This is an essential $$T^2 \subset M$$ which is not boundary-parallel. If we assume $$M$$ has a complete hyperbolic metric, this gives rise to a $$\Bbb Z^2$$ subgroup of $$\pi_1 M$$. The generators should correspond to deck transformations $$\alpha, \beta$$ of $$\Bbb H^3 = \tilde{M}$$ which are parabolic isometries fixing common point at the boundary $$\partial \Bbb H^3$$, so take a small horosphere around that point, and the quotient of the horosphere by $$\langle \alpha, \beta \rangle$$ should be the original torus $$T^2$$. The horoball bounding that horosphere would then be a cusp coning off the $$T^2$$, which forces $$T^2$$ to be boundary-parallel, a contradiction. I think one can do similar arguments to show torus knots aren't hyperbolic either, but with the embedded annulus given by cutting the torus along the knot, although I am not quite able to carry out this argument.

On a completely different flavor, one class of Eilenberg-Maclane spaces which have a startlingly simple description are $$K(\Bbb Q, n)$$'s for odd $$n$$. The following result is true: $$H^*(K(\Bbb Q, n); \Bbb Q)$$ is isomorphic to $$\Bbb Q[x]$$, $$|x| = n$$ for even $$n$$ and $$\Bbb Q[x]/(x^2)$$, $$|x| = n$$ for odd $$n$$, and all coefficients in the cohomology groups would be $$\Bbb Q$$ from this point. Let's try to sketch a proof of this; our strategy would be to run the cohomology Serre spectral sequence on the pathspace fibration $$K(\Bbb Q, n) \to PK(\Bbb Q, n+1) \to K(\Bbb Q, n+1)$$; for convenience we will denote this as $$K(n) \to P \to K(n+1)$$. Right off the bat, note that $$H^*(P) = 0$$ except in degree $$0$$, so the $$E^\infty$$ page keeping track of various factor groups in a filtration for $$H^*P$$ has to be all zero except in the $$(0, 0)$$ position. This means everything in the spectral sequence must die in the next page, eventually.

Let's prove the result for $$n+1$$ assuming it's true for $$n = 2k$$. The $$E^2$$ page consists of the rows $$q = 0, n, 2n, \cdots$$ in which $$H^*K(n+1)$$ are arranged. No differential hits the $$E^2_{0, i}$$ terms for $$0 < i < n+1$$ so they must be zero as it would survive to the $$E^\infty$$ page otherwise. Since all the intermediate rows are zero, $$E^2 = E^{n+1}$$ and first nontrivial differential is $$d_{n+1} : E^{n+1}_{0, n} \to E^{n+1}_{n+1, 0}$$. Since no other differential comes out of the $$(0, n)$$ position, it must die in $$E^{n+2}$$ from the discussion in the previous paragraph. That is to say, $$d_{n+1}$$ is injective. Moreover no other differential enters the $$(n+1, 0)$$ position so $$d_{n+1}$$ has to be an isomorphism as well. But $$E^{n+1}_{0,n} = H^0(K(n+1), H^nK(n)) = H^nK(n) = \Bbb Q$$ so $$H_{n+1} K(n+1) = E^{n+1}_{n+1, 0} = \Bbb Q$$. Let's call $$x$$ to be the cohomology class generating $$H^nK(n)$$ (which therefore generates $$H^*K(n) \cong \Bbb Q[x]$$) and $$y$$ be the cohomology class generating $$H^{n+1}K(n+1)$$. At the level of generators, $$d_{n+1}(1 \otimes x) = y \otimes 1$$. By the multiplicative structure of the spectral sequence, the differential $$d_{n+1} : \Bbb Q = E^{n+1}_{0, kn} \to E^{n+1}_{n+1, k(n-1)}$$ satisfies $$d_{n+1}(1 \otimes x^k) = ky \otimes x^{k-1}$$ therefore injective and also an isomorphism since $$k$$ is invertible in $$\Bbb Q$$. $$d_{n+1} : E^{n+1}_{n+1, k(n-1)} \to E^{n+1}_{2(n+1), k(n-2)}$$ must be zero since kernel of this map contains image of the previous $$d_{n+1}$$, which is the full group $$E^{n+1}_{n+1, k(n-1)}$$, therefore $$E^{n+2}_{n+1, k(n-1)} = 0$$. To sum it all up, the first $$n+1$$ rows of the $$E^{n+2}$$ page are zero. By arguing inductively now, the groups $$H^pK(n+1)$$ in the $$(p, 0)$$ position are all zero for $$p > n+1$$ since no nonzero differential hits them, making it survive to $$E^\infty$$. So we have the desired isomorphism $$H^*K(n+1) \cong \Bbb Q[y]/(y^2)$$.

So we know that for $$n$$ odd, $$H^*(K(\Bbb Q, n); \Bbb Q)$$ is $$\Bbb Q$$ in degree $$n$$ and $$0$$ otherwise. Consider the class $$C$$ of finitely generated $$\Bbb Q$$-modules amongst all abelian groups and note that this is a Serre class (i.e., closed under taking subgroups, quotients, and extensions) so since $$\pi_* K(\Bbb Q, n)$$ are in $$C$$, $$H_*(K(\Bbb Q, n); \Bbb Z)$$ are also in $$C$$ by Hurewicz theorem (mod $$C$$). Since $$H_*(K(\Bbb Q, n); \Bbb Z)$$ are $$\Bbb Q$$-modules, by universal coefficient theorem (Ext term vanishes) $$H^*(K(\Bbb Q, n); \Bbb Q) =\text{Hom}(H_*(K(\Bbb Q, n); \Bbb Z), \Bbb Q)= H_*(K(\Bbb Q, n); \Bbb Z)$$.

This means $$K(\Bbb Q, n)$$ is (for $$n$$ odd) the Moore space $$M(\Bbb Q, n) = S^n_\Bbb Q$$, also called a rational sphere, which can be constructed by taking homotopy colimit of $$S^n \stackrel{\times 2}{\to} S^n \stackrel{\times 3}{\to} S^n \to \cdots$$, or explicitly, taking a bunch of cylinders $$S^n \times I$$ and gluing the top of the $$k$$-th cylinder to the bottom of the $$(k+1)$$-th by a degree $$k$$ map. To me it is not at all obvious that all the higher homotopy of $$S^n$$ gets killed off just by rationalizing $$\pi_n S^n$$ in this fashion! The general result along these lines is that of Sullivan, which says that $$X_\Bbb Q$$ is a rationalization of $$X$$, i.e., $$\pi_* X_\Bbb Q$$ is a $$\Bbb Q$$-module and there is a map $$X \to X_\Bbb Q$$ which is an isomorphism on $$\pi_*(-) \otimes \Bbb Q$$ if and only if $$H_* X_\Bbb Q$$ is a $$\Bbb Q$$-module and $$X \to X_\Bbb Q$$ an isomorphism on $$H_*(-) \otimes \Bbb Q$$ (you need some hypothesis on $$X$$ here, the simplest being $$X$$ is simply connected but being a simple space is enough), i.e., a homology-rationalization is a homotopy-rationalization.