26
$\begingroup$

Let $X$ be a connected CW complex. The Hurewicz map $\pi_*(X) \to H_*(X)$ is given by sending $\sigma: S^n \to X$ to the pushforward of the homology class $\sigma_*[S^n]$. The Hurewicz theorem says that when $\pi_k = 0$ for $k < n$, this map is an isomorphism in degree $n$.

It seems to me that it's extremely rare that this map is an isomorphism in all degrees. In fact, the only cases I can see are those in which $X = K(G,1)$, where $G$ is abelian and of homological dimension 1. (Equivalently, every finitely generated subgroup of $G$ is either trivial or isomorphic to $\Bbb Z$; the first interesting example, then, is $\Bbb Q$.)

It's not true for $X = K(G,2n)$ for any $n>0$; $H^{4n}(K(G,2n);G)$ is nontrivial, coming from the natural transformation $H^{2n}(Y;G) \to H^{4n}(Y;G)$ given by the cup square (and thus necessarily its integral homology is nonzero in either degree $4n$ or degree $4n-1$.) It's not true for any simply connected non-contractible finite-dimensional CW complexes, which necessarily have infinitely many nontrivial higher homotopy groups, but its homology groups are of bounded degree. So it's not likely to be reasonable to look at spaces which are either homologically or homotopically finite. It's not true for products by a Kunneth argument,so our spaces must be irreducible. On the other hand, the most natural source I can think of of spaces which are both homologically and homotopically infinite and has tractable homotopy groups is the infinite symmetric product $\text{SP}^\infty(X)$; on the other hand, infinite symmetric products are in fact products of Eilenberg-MacLane spaces. (In particular, the first interesting thing I can think of, $\text{SP}^\infty(\Bbb{CP}^\infty)$, is out.)

Are there any nontrivial examples, or is it true that all such spaces are necessarily aspherical?

$\endgroup$
  • 2
    $\begingroup$ Since homology is represented by the infinite symmetric product, it seems that your question is equivalent to when $X = \mathrm{SP}^1(X) \hookrightarrow \mathrm{SP}^\infty(X)$ is a homotopy equivalence. Since $\mathrm{SP}^\infty(X)$ is a generalized Eilenberg-Mac Lane space, it seems by what you have already argued, $\mathrm{SP}^\infty X$, and thus $X$, must in fact be an Eilenberg-Mac Lane space that is also a Moore space. $\endgroup$ – JHF Nov 15 '16 at 20:39
  • $\begingroup$ Shouldn't $K(\mathbb Q, 2n+1)$ work as well? $\endgroup$ – Justin Young Nov 15 '16 at 21:08
  • $\begingroup$ @JustinYoung Yup, based on JHF's comment. I wonder how classifiable EM spaces that are also Moore spaces are. $\endgroup$ – user98602 Nov 15 '16 at 22:14
  • $\begingroup$ @JHF I think your comment along with a brief explanation or reference as to why the induced map on homotopy is the Hurewicz map is probably an answer, though I'd like to know eg for what $G$ $K(G,3)$ is a Moore space. $\endgroup$ – user98602 Nov 15 '16 at 22:18
  • $\begingroup$ @MikeMiller A reference for the induced map is the Hurewicz map is in the original Dold-Thom 1958 paper (Satz 6.10). It's in German, however. The strategy of the proof is to first prove it for spheres by induction on dimension using a quasifibration that they constructed earlier, and then use naturality. Anyway, I'm a little hesitant about writing this up as an answer, since I haven't really addressed the question about which Eilenberg-Mac Lane spaces that are also Moore spaces at all, which is really the crux of your question. $\endgroup$ – JHF Nov 15 '16 at 23:44
16
$\begingroup$

This answer was co-written with Michael Andrews at UCLA; thanks for his help on this.

This is possible if and only if $X$ is a product of factors of the form $K(\Bbb Q, 2n+1)$ ($n \geq 1$), $K(\Bbb Z_{p^\infty}, 2n)$ (as $p$ ranges over odd primes and for $n \geq 1$) and $K(\Bbb Z[1/S], 1)$ ($S$ a set of primes), with at most one of each factor (that is, $K(\Bbb Q, 3)$ and $K(\Bbb Q,5)$ are not both allowed, nor are two copies of $K(\Bbb Q,3)$, but $K(\Bbb Z_{3^\infty}, 2) \times K(\Bbb Z_{5^\infty}, 2)$ is OK), a factor of the first type only allowed if there are no factors of the third type, and a factor of the second type at the prime $p$ only allowed if $p$ is in the set of primes $S$. (Note that the statement in the question that it cannot be $K(G,2n)$ is NOT true - that's only true when $G$ is the additive group of a ring, which the Prufer groups are not!)


As noted in the comments, the infinite symmetric product $\text{SP}^\infty(X)$ has homotopy groups naturally isomorphic to the homology groups of $X$, and the Hurewicz map is induced by the map $X = \text{SP}^1(X) \to \text{SP}^\infty(X)$. So we're asking precisely when this map is a (weak) homotopy equivalence. $\text{SP}^\infty(X) = \prod_{i=1}^\infty K(H_i(X;\Bbb Z), i)$, so necessarily $X$ is a product of Eilenberg-MacLane spaces.

If $X \times Y$ has the desired property, so do the factors: the image of the Hurewicz map to $X \times Y$ lies in $H_k(X) \oplus H_k(Y)$ (with each factor of $\pi_k(X) \oplus \pi_k(Y)$ going to the obvious place in homology). So we should start by figuring out which Eilenberg MacLane spaces are also Moore spaces.

Let's suppose $n>1$ for now. We'll return later to the $n=1$ case.

First, $G$ is divisible. For suppose the multiplication-by-p map was not surjective; then $G/pG$ is a $\Bbb F_p$-vector space, so we can get a map $G \to \Bbb F_p$. This induces a map $K(G,n) \to K(\Bbb F_p, n)$, which one verifies is nonzero on $H^n( -; \Bbb F_p)$. Let $\iota \in H^n(K(\Bbb F_p, n); \Bbb F_p)$ be the fundamental class. Then we know that, for instance, the Steenrod power $P^1\iota$ is nonzero if $p > 2$, and thus by naturality, since we know there is an element $\alpha \in K(G,n)$ that pulls back to $\iota$, we see that $P^1\alpha$ must be nonzero. For $p=2$, $\text{Sq}^2\iota$ suffices to perform the same argument. Thus $K(G,n)$ has cohomology in degree $n+2$ or higher, contradicting the Moore hypothesis.

Therefore, $G = \Bbb Q^n \oplus_p \Bbb Z_{p^\infty}^{n_p}$. Let's calculate the homology of the factors. $H_*(K(\Bbb Q, 2n+1)) = \pi_*(K(\Bbb Q, 2n+1))$, as desired; $H^*K(\Bbb Q,2n)$ is a polynomial algebra on a degree $2n$ generator. Now we calculate for $G = \Bbb Z_{p^\infty}$. We have that $K(G,n) = \text{colim}_{k} K(\Bbb Z/p^k, n)$, and homology commutes with directed colimits. $K(\Bbb Z/p^k, n)$ has integral homology that is a finite $p$-group for all $p$; the homology of $K(G,n)$, then, has to be a colimit of finite $p$-groups, thus being a direct sum of copies of $\Bbb Z/p^k$ and $\Bbb Z_{p^\infty}$. In any case, $\text{Tor}(\Bbb Z/p^k, \Bbb F_p)$ is $\Bbb F_p$, as is $\text{Tor}(\Bbb Z/p^\infty, \Bbb F_p)$ (because $\text{Tor}$ commutes with directed colimits, and the induced map on the tor groups of $\Bbb Z/p^k \to \Bbb Z/p^{k+1}$ is the identity.) Thus $K(G,n)$ is a Moore space iff its $\Bbb F_p$ homology is zero outside of degree $0, n+1$ and is $\Bbb F_p$ in those degrees.

So let's prove that this is true when $p$ and $n+1$ are odd. Again, homology commutes with colimits, so we need to compute the induced map on $\Bbb F_p$ homology of $K(\Bbb Z/p^k, n) \to K(\Bbb Z/p^{k+1}, n)$. The easiest way to do this is to actually work with $\Bbb F_p$ cohomology (since the universal coefficient theorem with field coefficients says that the induced map on $\Bbb F_p$ cohomology is just the transpose of the induced map on $\Bbb F_p$ homology). Then the $\Bbb F_p$ cohomology of these spaces is a free algebra on classes given by cohomology operations applied to the fundamental class corresponding to the "reduction mod $p$" homomorphism $\Bbb Z/p^k \to \Bbb Z/p$. This fundamental class maps to zero (it maps to the composite $\Bbb Z/p^k \to \Bbb Z/p^{k+1} \to \Bbb Z/p$). The cohomology operations involved are all either 1) the Bockstein operation corresponding to $\Bbb Z/p \to \Bbb Z/p^{k+1} \to \Bbb Z/p^k$ or 2) that operation, after first applying one of the standard mod $p$ operations. (See this paper.) Call the first $Q_k$. Then the induced map on cohomology must vanish on $P\iota_n$, since mod $p$ cohomology operations are natural, and the induced map kills $\iota_n$. The only one which can fail to vanish is the map on $Q_{k+1}\iota_n$, which maps to $Q_k \iota_n$ (it's not natural, since it depends on the $K(\Bbb Z/p^k, n)$ we're living in!) When $n+1$ and $p$ are odd, the free subalgebra generated by this element is the exterior algebra and thus gives no new elements; when $p$ or $n+1$ is even, it's a polynomial algebra, thus giving many new nonvanishing elements. Thus when $p$ is odd and $n+1$ is even in every degree but $H^{n+1}$, the maps are all zero; the same is true in homology.

Since the maps in the colimit are all either zero (in degrees other than $n+1$) or an isomorphism $\Bbb F_p \to \Bbb F_p$ (in degree $n+1$), the colimits are zero in every degree other than $k+1$, where the final result is $\Bbb F_p$, as desired. It is otherwise false, since we get a whole polynomial algebra from the element surviving in $H^{n+1}$.

Now for the $n=1$ case. Obviously $G$ must be abelian. One could generalize to the case where we only demand the question in degree greater than 1, but I don't care to. It is apparently an open problem to prove that groups with homological dimension $1$ are locally free (that is, every finitely generated subgroup is free); I do not know if the abelian case is known, but if it is, this would be the same as saying that an abelian group has homological dimension 1 iff every finitely generated subgroup is trivial or $\Bbb Z$. This would then imply that the group is a subgroup of $\Bbb Q$, which are up to isomorphism the same as localizations of $\Bbb Z$ at some set of primes. I'm not going to look much more into this.


Now let's think about products when $n \geq 2$; we'll deal with $n=1$ later. Write $Q_n = K(\Bbb Q, 2n+1)$, and $X_{p,n} = K(\Bbb Z_{p^\infty}, 2n+1)$. Then $X_{p,n} \times X_{q,m}$ is also one of our spaces when $p \neq q$, since the tensor product of Prufer groups is automatically zero, and $\text{Tor}(\Bbb Z_{p^\infty}, \Bbb Z_{q^\infty}) = 0$. On the other hand, $\text{Tor}(\Bbb Z_{p^\infty}, \Bbb Z_{p^\infty}) = \Bbb Z_{p^\infty}$, so the Kunneth theorem implies that a product $X_{p,n} \times X_{p,m}$ is not one of our spaces (it's got extra homology in degree $n+m-1$.) We can multiply in at most one copy of $Q_n$; two violates Kunneth (we get a copy of $\Bbb Q \otimes \Bbb Q = \Bbb Q$ in degree $m+n$), but $\text{Tor}(\Bbb Z_{p^\infty}, \Bbb Q) = 0$, and $\Bbb Q \otimes \Bbb Z_{p^\infty} = 0$.

So a simply connected example is a product of at most one $Q_n$ and one (for each $p$) of the $X_{p,n}$. A non-simply connected example, assuming the conjecture stated above, can only exist when there is no $Q_n$ factor (or we'd get too much homology from Kunneth) - and then, if the fundamental group is $\Bbb Z[1/S]$, where $S$ is a set of primes, then no $X_{p,n}$ factor can appear where $p$ is not in $S$ - again, tensor products cause problems. (In particular, if the fundamental group is $\Bbb Z$, we can only have the circle $S^1$.)

$\endgroup$
  • $\begingroup$ Could you elaborate a bit on the proof that $G$ is divisible? I don't follow why $\mathcal P^1 \alpha$ has to be non-zero. $\endgroup$ – Justin Young Nov 18 '16 at 12:55
  • $\begingroup$ @JustinYoung The argument appears to be invalid because the arrows were pointing the wrong direction in my mental calculations. I still believe the result, though. I'll try and correct it. (As a first note, you can use Kunneth to show that $G/pG$ is either $0$ or $\Bbb F_p$ itself.) $\endgroup$ – user98602 Nov 18 '16 at 15:06
  • $\begingroup$ @JustinYoung This is turning out to be quite easy stably and incredibly painful unstably. $\endgroup$ – user98602 Nov 19 '16 at 0:23
  • $\begingroup$ I wouldn't mind seeing the stable argument if you ever get a chance..I still have no solution for this very interesting problem. $\endgroup$ – Justin Young Oct 27 '17 at 14:48
  • $\begingroup$ @JustinYoung Thanks for reminding me about this. I should at least edit this answer to make it less misleading about its accuracy. I would have to try to reconstruct whatever I had in mind, but if you email me I can try sometime this weekend. $\endgroup$ – user98602 Oct 27 '17 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy