The complex topological $K$-theory spectrum is not an $H\mathbb{Z}$-module.

Question

Why is it true that the complex topological $K$-theory spectrum is not an $H\mathbb{Z}$-module? I mean, why the nontriviality of the first $k$-invariant implies the claim above?

Answer 1

Let $E$ be a $H\mathbb{Z}$-module spectrum and consider the smash $H\mathbb{Z}\wedge E$. I claim in this case that it is the wedge

$$H\mathbb{Z}\wedge E\simeq \bigvee_{k\in\mathbb{Z}} \Sigma^kHG_k$$

where $G_k=H_k(E)$. To see this, for each $k\in\mathbb{Z}$ take the Moore spectrum $\Sigma^kMG_k$ and a map $\alpha_k:\Sigma^kMG_k\rightarrow H\mathbb{Z}\wedge E$ which induces an isomorphism

$$\alpha_k:\pi_k(HG_k)\cong H_k(E)\xrightarrow\cong\pi_k(H\mathbb{Z}\wedge E)=H_k(E).$$

It is a standard result that such a map should exist, and it is indeed easy to construct one explicitly starting with a given presentation of $MG_k$ as a cofiber of a map between wedges of spheres.

Now fix $k$ and consider the map

$$\beta_k:H\mathbb{Z}\wedge\Sigma^kMG_k\xrightarrow{1\wedge\alpha_k}H\mathbb{Z}\wedge H\mathbb{Z}\wedge E\xrightarrow{\mu\wedge 1}H\mathbb{Z}\wedge E$$

where $\mu$ is the ring-spectrum product on $H\mathbb{Z}$. It is clear from the definitions that this map induces an isomorphism on $\pi_k$, and the trivial homomorphism on $\pi_l$ for $l\neq k$. Therefore, when we define

$$\bigvee_{k\in\mathbb{Z}} \Sigma^kHG_k\xrightarrow{\bigvee\beta_k}H\mathbb{Z}\wedge E$$

we easily see that it is a weak equivalence, and we get our claim.

The point is that an $H\mathbb{Z}$-module spectrum is a GES (generalised Eilenberg-Mac Lane Spectrum, i.e. a wedge of suspensions of EM-spectra) and in particular has no non-trivial Postnikov invariants. On the other hand, the K-theory spectrum $KU$ indeed does have non-trivial Postnikov invariants. Hence it is not a GES, and so not an $H\mathbb{Z}$-module.