Good reference for homology of $K(\mathbb{Z}, 2n)$?

Question

The homology algebra $H_*( K(\mathbb{Z},2n); \mathbb{Z})$ contains a divided polynomial algebra on a generator $x$ of dimension $2n$.
I suppose I could read through the Cartan seminar for a proof, but I'm hoping someone knows of a nice simple argument for this fact.

Answer 1

We need to do three things: (1) show that the homology contains a polynomial algebra, (2) show that powers of the generator are sufficiently divisible, and (3) show that torsion doesn't interfere.

Let me repeat Hatcher's path through (1), since we need the particular generator to show that it is divisible. According to Dold-Thom (Hatcher 4.K; cf Dold-Kan¹), a model for $K(Z,m)$ is the free commutative monoid on pointed² $S^m$. According to James (Hatcher 4.J), the homology of the free associative monoid $JS^m$ on $S^m$ is a polynomial algebra on one generator in degree $m$. Thus there is a map of monoids $JS^m\to K(Z,m)$ so its map on homology is compatible with the ring structure. It is easy to see (eg, by the Serre spectral sequence) that the rational homology of $K(Z,m)$ is the signed-commutative algebra free on a generator in degree $m$, so this map is a rational isomorphism if $m=2n$.

Now for step (2), where we divide the James class by $k!$. If we restrict to words of length $k$ in these monoids, the affect of imposing commutativity is to quotient by the $k$th symmetric group. This piece of the James construction is a pseudo-manifold representing the class in degree $nk$ which is the $k$th power of the generator. The action of the symmetric group is generically free, so this cycle has multiplicity $k!$ when it lands in $K(Z,2n)$. Thus it is divisible. In other words, choose a fundamental domain for the action of the symmetric group on the $k$th filtration of the James construction. This yields is an $kn$-chain which when symmetrized is the whole of the $k$th filtration, and thus a cycle. Projecting it to the quotient is an alternative way of symmetrizing, so it is a cycle for $K(Z,2n)$.

I will leave step (3) as an exercise. What this shows so far is that the integral homology hits the divided power algebra in the rational homology, but the canonical choice of the divided element should make it plausible that these divisions are the right choices, ie, $x^{(k)}x^{(l)}=\binom {k+l}k x^{(k+l)}$.

¹If one accepts the fragment of Dold-Kan of the agreement of the two things one can do to a simplicial abelian group (1) homology of (either) associated chain complex and (2) the homotopy groups of the geometric realization; then a version of Dold-Thom follows trivially. Indeed, the very definition of simplicial homology is the free abelian group (ie, simplicial group) on the simplicial set.

² by "the free (commutative) monoid on a pointed set," I mean that we impose on the free object the relation that the basepoint is the identity for the monoid.

Answer 2

Do you want to know the integral homology?

If you are happy with homology with coefficients in $\mathbb{F}_p$, the best way to compute (and describe) the homology of Eilenberg-MacLane spaces is the technique developed in a paper by Ravenel and Wilson (MathSciNet). See also Wilson's "sampler" (MathSciNet). They used the structure called Hopf ring (a collection of Hopf algebras equipped with other operation) to describe $\{H_*(K(\mathbb{Z},n);\mathbb{F}_p)\}_{n\ge 0}$ as a whole.

The point is they worked in homology instead of cohomology so that we can use $$ K(\mathbb{Z},n)\times K(\mathbb{Z},m) \longrightarrow K(\mathbb{Z},m+n)$$ to "generate" $H_*(K(\mathbb{Z},n);\mathbb{F}_p)$ from $H_*(K(\mathbb{Z},m);\mathbb{F}_p)$ for $m<n$.

It turns out the Hopf ring structure is compatible with the Eilenberg-Moore spectral sequence and we obtain the Hopf algebra structure on the homology of Eilenberg-MacLane spaces easily. It is also important their technique works for generalized homology theories. In fact, their motivation was to compute the Morava $K$-theory of Eilenberg-MacLane spaces.

Answer 3

I am not sure if this qualifies as a simple argument, but to me it is very nice. I did not get the reference in the comments above - so sorry if this is close to that. I dont have an exact reference since this is pieced together from several places.

Let $A$ be any abelian group. Let $X=X(A)$ be the simplicial set defined by:

$X_p = \{$colorings of the $n$ faces of the standard $p$ simplex by elements in $A$ such that when restricted to any $n+1$ face the sum with alternating sign of the colorings of the faces are $0\}$

The face and degeneracy maps are defined by restriction and pull-back (the latter introduces 0's when an $n$ face is degenerate.

with this $X_p$ is a single point when $p < n$ (a single empty coloring), $X_{n}=A$, $X_{n+1}$ is "relations", and $X_p$ for $p>n+1$ is given by its restrictions to the $n+1$ faces. I.e. it is $n+(1 or 2?)$ co-skeletal. This means that its geometric realization $\lvert X \rvert$ is a $K(A,n)$.

In fact this is a simplicial group by adding the colorings, and $X(A)\times X(A)$ is isomorphic as simplicial groups to $X(A\oplus A)$ by definition.

To simplify we now use $A=\mathbb{Z}$ and to understand this product on the $\mathbb{Z}$ span of the simplices, I will assume some familiarity with the Eilenberg-Zilber operator, which appears naturally in the product of simplicial sets. Using this we describe the power map

(1) $H_n(X(A),A)^{\otimes k} \to H_{kn}(X(A)\times\cdots\times X(A),A)=H_{kn}(X(A^{\oplus k}),A) \to H_{kn}(X(A),A)$

$H_n(X(A),A)$ is a single $A$ and it is generated by the $n$ simplex $\alpha$ colored by $1$, the tensor product $\alpha^{\otimes k}$ can when mapped to the middle term of (1) be written as the sum

$\sum_{\sigma \in S(k,n)}$ sgn$(\sigma) \beta_{\sigma}$

where $S(k,N)$ is the permutations on $\{1,\dots,kn\}$ which preserves the order of the $k$ sequences SEQ$\strut_i=\{in+1,\dots,(i+1)n\}, i=0,k-1$ (a generalized shuffle), and $\beta_{\sigma}$ is the associated product of degenerations defined by

(2) $\beta_{\sigma}=(\sigma_1^* \alpha)\times \cdots \times (\sigma_k^* \alpha)$,

where $\sigma_i$ is the order preserving surjective map from $\{0,\dots,kn\}$ to $\{0,\dots,n\}$ defined by increasing one precisely when a number is in the image of $\sigma$ restricted to SEQ$\strut_i$.

In the case of $n$ even it is clear that the sign of the permutations $\sigma$ which simply permutes the sequences SEQ$\strut_i$ (without intertwining them) has sign 1 and thus we may act on the $\sigma$'s in the sum by these with out changing the sign, this action, however, permutes the factors in (2), but when mapped to the last factor in (1), they are the same, so the image of the sum in (1) is divisible by $k!$.

To see injectivity of the product one can see that the power map is injective on rational homology by using a Hopf-algebra argument as in section 3.C of Hatchers book.

Answer 4

For a reference, you might see this paper of Birgit Richter. A rough outline follows:

Since $X = K(Z,n)$ can be made a commutative topological monoid per Ben Wieland's answer, its singular chain complex $C_*(X)$ is made into a commutative and associative differential graded algebra via the commutative associative Eilenberg-Zilber shuffle product $C_*(X) \otimes C_*(X) \to C_*(X \times X)$ composed with the multiplication on $X$.

The formula for the shuffle product is slightly involved when you iterate it, but essentially: to multiply $\alpha_1 \cdots \alpha_n$ with $\alpha_i$ of degree $k_i$, you sum over all ways to divide a set of size $\sum k_i$ into subsets of size $k_1, \cdots, k_n$ of a product of certain degeneracy operators, depending on the subdivision, applied to the chains $\alpha_i$. (With signs.)

If all the $\alpha_i$ are equal and in positive even degree $k$, then the signs don't interfere and we are summing over all ways to divide $nk$ into $n$ equal pieces of size $k$. However, because the chains are now equal we get the same term if we simply permute the pieces of size $k$, and so each term appears in the sum at least $n!$ times. (Note that $k > 0$ is necessary here in order for these permutations to actually give different subivisions.) As a result, the n-fold product chain is divisible by $n!$.

Answer 5

Nice question. Unfortunately the answer I posted an hour ago is wrong because I switched the structures of homology and cohomology for the James reduced product. It is the cohomology that is a divided polynomial ring while the homology is a polynomial ring.