Specific examples of Eilenberg-Maclane spaces?

Solution 1:

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.

Solution 2:

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.