Maps from mod-$p$ Eilenberg-MacLane spectrum to connective $K$-theory spectrum

Question

Let $ku$ be the connective cover of the complex $K$-theory spectrum $KU$. Consider the mod-$p$ Eilenberg-MacLane spectrum $H\mathbb{Z}/p$.

I want to see that $[H\mathbb{Z}/p,ku]=0$.

Since $H\mathbb{Z}/p$ is dissonant (Theorem 4.7 of Ravenel's "Localization with Respect to Certain Periodic Homology Theories") the result will follow once we know that $ku$ is harmonic (Corollary 4.11 in loc. cit.)

Is it true that the spectrum $ku$ is harmonic?

If not, how can I prove the claim?

Certainly, an idea is to use directly the theorem of Margolis (“Eilenberg-MacLane Spectra”): For any spectrum $Y$ of finite type, there is an isomorphism $$[H\mathbb{Z}/p,Y]\to \text{Hom}_{\mathcal{A}}(H^{*}(Y,\mathbb{Z}/p),\mathcal{A})$$ where $\mathcal{A}$ is the Steenrod algebra.

Is it true that $\text{Hom}_{\mathcal{A}}(H^{*}(ku,\mathbb{Z}/p),\mathcal{A})=0$?

If the answer is yes, why is that?

Answer 1

You can also cheat: since HZ/p is connective,$$ [HZ/p,ku] = [HZ/p,KU] = [HZ/p \wedge KU,KU]_{KU-modules} = 0$$ since $HZ/p \wedge KU = 0$.

Edit: Let me stress that the other answers give more information, namely the calculation of maps from HZ/p to ku in all degrees.

Answer 2

There are cofibrations $$\Sigma^2 kU\xrightarrow{v} kU\to H\xrightarrow{\alpha}\Sigma^3 kU$$ and $$ H\xrightarrow{p}H\to H/p\xrightarrow{\beta} \Sigma H. $$ The composite $\alpha\beta\colon H/p\to\Sigma^4 kU$ is nontrivial. In fact, I am fairly sure that the map $(\alpha\beta)^*$ induces an isomorphism $(kU^*kU)/(p,v)\to kU^*(H/p)$. To prove this (using the above cofibrations), one would only need to check that the sequence $(v,p)$ is regular on $kU^*kU$, and I think that that is known.

We get essentially the same picture if we replace $kU$ by the Adams summand $BP\langle 1\rangle$ as in Ekie's answer. Then you can use the Adams spectral sequence which starts with $\text{Ext}^{**}_{E(1)}(\mathbb{F}_p,\mathcal{A})$. Here $\mathcal{A}$ has the form $E(1)\otimes R$ for some $R$, and $E(1)$ is self-dual up to a shift so we just get $\text{Ext}^{**}=\text{Ext}^{0*}=\Sigma^d R$, where $d=|v_0|+1+|v_1|+1=2p$.

Answer 3

Lets just work with $p=2$. $ku$ is certainly not harmonic, but $[H\mathbb Z/2, ku]=0$. The Adams spectral sequence works very nicely to show all of this: The $E_2$--term is $Ext^{s,t}_{A}(H^*(ku), H^*(H\mathbb Z/2)) = Ext^{s,t}_{A}(A//E(1), A) = Ext^{s,t}_{E(1)}(\mathbb Z/2, A)$. Since $A$ is a free module over the Hopf algebra $E(1)$, and we conclude that $A$ is an injective $E(1)$-module. Thus the Adams $E_2$--term is just $Hom$, and we learn that there is an isomorphism $[H\mathbb Z/2, \Sigma^tku] = Hom_{E(1)}(\Sigma^t\mathbb Z/2, A)$. The smallest $t$ for which this is nonzero is $t=4$, corresponding to the element $Sq^1Sq^2Sq^1 \in A$. The associated map $H\mathbb Z/2 \rightarrow \Sigma^4 ku$ can be taken to be the composite of the `Bockstein' map $H\mathbb Z/2 \rightarrow \Sigma H\mathbb Z$ with one suspension of the map $H\mathbb Z \rightarrow \Sigma^3 ku$ arising from the cofibration sequence $$ \Sigma^2 ku \rightarrow ku \rightarrow H\mathbb Z \rightarrow \Sigma^3 ku.$$

I am sure the odd prime version is similar.

Answer 4

(Edited to include the case when $p$ is odd. Hopefully there are no mistakes as I don't usually work with $p$ odd.)

This is true, assuming you mean degree 0 maps. We have $ku\simeq l \vee \Sigma^{2} l \vee \dots \vee \Sigma^{2(p-2)} l$ where $l$ is called the Adams summand (usually denoted $BP\langle 1 \rangle$). Notice when $p=2$ that $ku=l$. The cohomology of $l$ is $$H^*(l; \mathbb{F}_p) \cong \mathcal{A}//E(1) = \mathcal{A} \otimes_{E(1)} \mathbb{F}_p$$ where $E(1)$ is the exterior algebra generated by the Milnor primitives $Q_0$ and $Q_1$ in degrees 1 and $2p-1$, respectively. For example, if $p=2$, then $Q_0 = Sq^1$ and $Q_1 = Sq^2 Sq^1 + Sq^3$. Thus $$H^*(ku) \cong \mathcal{A}//E(1) \oplus \Sigma^2 \mathcal{A}//E(1) \oplus \dots \oplus \Sigma^{2p-4} \mathcal{A}//E(1).$$

Now look at each factor: by change of rings, $$\text{Hom}_{\mathcal{A}}(\Sigma^{2n} H^{*}(l;\mathbb{F}_p),\mathcal{A}) \cong \text{Hom}_{E(1)}(\Sigma^{2n} \mathbb{F}_p, \mathcal{A}|_{E(1)})$$ for $0\leq 2n \leq 2p-4$, where $\mathcal{A}|_{E(1)}$ means restriction to $E(1)$.

There are no nonzero $E(1)$-maps from $\mathbb{F}_p$ to $\mathcal{A}|_{E(1)}$ because the bottom class of $\mathcal{A}$ supports nontrivial operations (e.g. $Q_0$). This shows the group is 0 for $n=0$, which is the only case to consider if $p=2$. If $p$ is odd, then the odd primary Steenrod algebra is generated by the Bockstein $\beta$ in degree 1 and the reduced powers $P^i$ in degree $2i(p-1)$. The first $P^i$ is in degree $2p-2 > 2p-4$, hence the generator of $\Sigma^{2n} \mathbb{F}_2$ can never hit a nonzero class when $n>0$, so $\text{Hom}_{E(1)}(\Sigma^{2n} \mathbb{F}_p, \mathcal{A}|_{E(1)}) = 0$ for all $n$ with $0\leq 2n \leq 2p-4$.

As noted in other answers, this statement is not true if we look at suspensions of $ku$ instead, because e.g. there is a nontrivial map $\Sigma^4 \mathbb{F}_2 \to \mathcal{A}$ where the generator of $\Sigma^4 \mathbb{F}_2$ hits $Q_0Q_1$ (since $Q_0Q_1$ supports no operation when $\mathcal{A}$ is restricted to $E(1)$, there is no longer a contradiction here).