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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?

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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.

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  • $\begingroup$ Dear @DustinClausen. I can't see why is it true that $H\mathbb{Z}/p$ smash with $KU$ is 0. $\endgroup$ – user438991 Apr 15 at 6:52
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    $\begingroup$ Dear user438991, I think this fact has come up a few times on MathOverflow. Anyway, here are two proofs. One is using the theory of complex oriented cohomology theories: HZ/p gives the additive formal group and KU gives the multiplicative formal group, so on HZ/p smash KU you must have a formal group which is isomorphic to both the additive and multiplicative one. But this is impossible over a ring of characteristic p, because the heights are different. $\endgroup$ – Dustin Clausen Apr 15 at 8:23
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    $\begingroup$ Another proof is using Adams' self-map v_1: \Sigma^d S/p --> S/p. You can pin down the minimal d (2p-2 for odd p, 8 for p=2), but the only important thing here is that it's positive. When you smash with KU this v_1 becomes a power of the Bott element and hence is invertible, but when you smash with HZ it becomes null for degree reasons. Hence HZ/p smash KU ( = S/p smash HZ smash KU) carries a self-map which is both invertible and null, therefore it must vanish. $\endgroup$ – Dustin Clausen Apr 15 at 8:26
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    $\begingroup$ I guess there are also more computational proofs. You could brute force calculate HZ/p smash KU: by Bott perodicity, it is (with shifts) a certain N-indexed colimit of Z/p-homologies of the space BU ( = colim n BU(n) ). The latter is a well-known calculation, and then it's a matter of identifying the effect of the Bott map \Sigma^2 BU --> BU on homology to be able to pass to the colimit. I guess the key for the vanishing should be the result (of Bott) that the Hurewicz image of the n^{th} power Bott element in \pi_{2n}BU is divisible by (n-1)!, hence dies (mod p) for large n. $\endgroup$ – Dustin Clausen Apr 15 at 10:11
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    $\begingroup$ You can probably shortcut the computational proof by using Snaith's presentation KU = \Sigma^\infty_+ BU(1)[bott^{-1}]. Then it's enough to invert Bott on the homology of the space BU(1). That this dies with finite coefficients follows from the fact that the pontryagin ring structure on H_*(BU(1)) is a divided power algebra. $\endgroup$ – Dustin Clausen Apr 15 at 10:12
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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 $d=|v_0|+1+|v_1|+1=2p$.

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    $\begingroup$ Ha. We were apparently posting basically the same thing at the same time. $\endgroup$ – Nicholas Kuhn Apr 11 at 16:47
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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.

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    $\begingroup$ Is ku really not harmonic? Here's a back-of-the-envelope calculation: chromatic fracture presents $L_1 ku$ as an extension of $ku$ by $Q / Z \otimes \Sigma^{-1} KU(-\infty, 0)$. $L_{K(< \infty)} ku := L_{\bigvee_{j < \infty} K(j)} ku$ is $L_1 ku$ again, so the harmonic localization $L_{K(\le \infty)} ku$ can be recovered from a fracture square involving $L_1 ku$ and $L_{K(\infty)} ku$. $L_{K(\infty)} ku$ is $ku^\wedge_p$, so $L_{K(< \infty)} L_{K(\infty)} ku$ has a similar coconnective part, and I think the coconnective parts cancel in the fracture square to give $L_{K(\le \infty)} ku = ku$. $\endgroup$ – Eric Peterson Apr 16 at 13:46
  • $\begingroup$ I imagine I'm missing something, not you, but I'd of course like to know what! $\endgroup$ – Eric Peterson Apr 16 at 13:47
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    $\begingroup$ @Eric Peterson The fact that there are nontrivial maps from the "dissonant" spectrum HZ/2 to a suspension of ku proves that ku is not harmonic. $\endgroup$ – Nicholas Kuhn Apr 16 at 19:07
  • $\begingroup$ Oh: harmonic localization is $L_{K(< \infty)}$, not $L_{K(\le \infty)}$, or else HZ/2 itself would be recoverable. Great, thanks! $\endgroup$ – Eric Peterson Apr 16 at 19:11
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(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).

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  • $\begingroup$ Dear @Ekie. Are you proving the claim in the OP question for $ku_{(p)}$? I'm a little confused. $\endgroup$ – Tsk Apr 15 at 6:49
  • $\begingroup$ @Tsk I am proving the claim that $Hom_{A}(H^*(ku; \mathbb{Z}/p), A) = 0$, for normal $ku$. I can understand if my answer seems convoluted, as it has grown/changed a little bit. Let me know which parts are confusing, and I can try to address that. $\endgroup$ – Ekie Apr 16 at 1:39
  • $\begingroup$ Ah, I see, yes I'm definitely using a $p$-local splitting. Hmm I feel as though the module argument still works out when using $p$-completion instead, which would give the right answer for mod $p$ homology, but I'm not sure. In any case, this works for regular $ku$ when $p=2$. $\endgroup$ – Ekie Apr 17 at 22:53

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