I want to show that $H\mathbb{Z}$ is not a retract of $ku$, where $ku$ is the connective cover of the complex $K$-theory and $H\mathbb{Z}$ is the Eilenberg-MacLane spectrum.
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2$\begingroup$ Well you have to use something that distinguishes ku from HZ[t]. For example, take mod 2 homology everywhere. We have $H_3 H\mathbb{Z}\ne 0$ but $H_3 ku=0$, so we’re done. $\endgroup$– Dylan WilsonMay 8, 2019 at 2:31
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$\begingroup$ @DylanWilson. Thank you for your answer. Could you please elaborate why is it true that $H_{3}H\mathbb{Z}\neq 0$ and $H_{3}ku=0$? $\endgroup$– TskMay 8, 2019 at 5:55
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$\begingroup$ As a matter of fact $H_3ku\neq 0$ ($\tau _2$ lives there). $\endgroup$– user43326May 8, 2019 at 11:19
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$\begingroup$ I don't know what $\tau_2$ is, but the mod $2$ homology of $ku$ is definitely $0$. For example, it is known that its homology is, as a comodule algebra over the dual Steenrod algebra, given by $\mathbb{F}_2[\xi_1^2, \xi_2^2,\xi_3,\xi_4,\ldots]$, which doesn't have anything in degree $3$. Maybe you're thinking about the $\mathbb{Z}$-homology of $ku$? $\endgroup$– Achim KrauseMay 8, 2019 at 12:01
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2$\begingroup$ @Tsk you can use the result that Achim writes or you can just look at the homology of the first few spaces in the spectra ku and HZ. $\endgroup$– Dylan WilsonMay 8, 2019 at 14:02
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2 Answers
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If you had a map $H\mathbb Z\to ku$ inducing an isomorphism on $\pi_0$ then you could compose with the usual map $ku\to KU$ to get a map $H\mathbb Z\to KU$ inducing an isomorphism on $\pi_0$. This can be ruled out in a number of ways, I imagine, but here's one: Using the periodicity equivalence $ \Sigma^{2j}KU\sim KU$ you would then get maps $\Sigma^{2j}H\mathbb Z\to KU$ for all integers $j$ such that all together they give an equivalence $\coprod_j \Sigma^{2j}H\mathbb Z\sim KU$.
ADDED in response to the reasonable complaint that I merely reduced the question to a similar question:
To me, and perhaps to many of us, $KU$ is a more familiar object than $ku$. I thought it was worth pointing out that if one could split off that one $\mathbb Z$ from $ku$ then one could split off all the $\mathbb Z$'s from $KU$. And "everybody knows that" $KU$ is not a product of Eilenberg-Maclane spectra, i.e. that periodic $K$-theory is truly an extraordinary cohomology theory.
One quick way of verifying this is to observe that while the reduced integral cohomology groups of $\mathbb RP^{2n}$ are killed by $2$ the group $\tilde KU(\mathbb RP^{2n})$ is not, if $n\ge 2$. In fact, let $L$ be the complexification of the nontrivial line bundle; $c_1(L)$ is the nontrivial element of $H^2$, so that $c_2(L\oplus L)=c_1(L)\cup c_1(L)$ is the nontrivial element of $H^4$ and $L\oplus L$ is not stably trivial.
answered May 9, 2019 at 1:47
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$\begingroup$ But isn't the whole point that we need a way to distinguish $KU$ from $\bigvee \Sigma^{2j}H\mathbb{Z}$? I think justifying that claim leads back to one of the arguments already discussed, which, in the end, all boil down to computing some cohomology group of something like SU or BSU. I think that's unavoidable. $\endgroup$ May 9, 2019 at 8:49
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$\begingroup$ It's probably obvious to everyone reading this, but it might be worth mentioning why $KU$ cannot possibly be equivalent to a wedge of Eilenberg-MacLane spaces (e.g. by looking at $KU^*BC_2$ and seeing that the Atiyah-Hirzebruch spectral sequence does not collapse, or the fact that $KU\wedge H\mathbb{F}_p=0$) $\endgroup$ May 9, 2019 at 9:25
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$\begingroup$ Dear @DenisNardin Why $KU\wedge H\mathbb{F}_p =0$? $\endgroup$– TskMay 10, 2019 at 8:40
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2$\begingroup$ @ Tsk Note that this is equivalent to saying that the integral homology groups of $KU$ are $\mathbb Q$ vector spaces. You can compute that $H_0(KU)\cong\mathbb Q$ by working out what the Bott map $\Sigma^2BU\to BU$ looks like on integral (co)homology. On cohomology it must take all decomposables to $0$, and it takes $c_j$ to $\Sigma^2(n_jc_{j-1}+decomposable)$ where the integer $n_j$ turns out to be $\pm j$. $\endgroup$ May 10, 2019 at 19:33
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If it were the case, then $K(Z,n)$ would be a retract of $\underline{ku}_n$ where $\underline{ku}_n$ is the $n$-th infinite loop space associated to the spectrum $ku$. However, $ku$ is what is called $BP\langle 1 \rangle$ (at the prime 2), and we know that the (2-local) homology of $BP\langle 1 \rangle _j$ is torsion free by work of Wilson {\it The Ω-spectrum for Brown-Peterson cohomology. Part II}, American Journal of Mathematics, {\bf 97} 1975, 101-123 whereas the homology of $K(Z,3)$ is not.
Edited after comments by Dylan Wilson and Achim Krause
Here is a simple argument following the original idea of the OP (deleted from the post later).
Consider the cofibration $\Sigma^2ku \to ku \to HZ$. Take the mod 2 homology to get a long exact sequence $$HZ/2 _{*-2}(ku) \to HZ/2 _*(ku) \to HZ/2 _*(HZ)\to HZ/2 _{*-3}(ku).$$ By induction on *, we see that the connecting homomorphism is epi, so that as graded vector spaces $$HZ/2 _*(HZ) \cong HZ/2 _*(ku) \oplus HZ/2 _{*-3}(ku).$$
This implies that $HZ/2 _*(HZ)$ can't be a summand of $HZ/2 _*(ku)$. It also shows, even without any knowledge of $HZ/2 _*(HZ)$, $HZ/2_3(ku)$ is larger than $HZ/2 _3(HZ)$. Thus contradicting retraction.
To get the claim in Dylan Wilson's comment for $HZ/2_3HZ$ it suffices to play the same game with the cofibration $HZ \to HZ \to HZ/2$.
Incidentally, by analyzing the long exact sequence which splits as short exact sequences, we end up with the description of $HZ/2_*HZ$ in my comment and that of $HZ/2_*(ku)$ in Achim Krause.
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5$\begingroup$ You can avoid $2$-localizing and quoting Wilson by observing that the third space in the spectrum for $ku$ is $SU$, and that $K(\mathbb{Z},3)$ can't be a retract of that because, as you said, there's torsion in the homology of $K(\mathbb{Z},3)$ but not in the one of $SU$. $\endgroup$ May 8, 2019 at 12:08
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3$\begingroup$ (not to mention the assertion that ku is BP<1>... which, if you knew, would also imply the result with a homology calculation at the spectrum level.) $\endgroup$ May 8, 2019 at 14:04
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1$\begingroup$ @AchimKrause. Are you consider mod $2$-homology? For example, if we consider integral cohomology $H^{6}(K(\mathbb{Z},3),\mathbb{Z})\neq 0$ (the $k$-invariant lives there) but $H^{6}(SU,\mathbb{Z})=0$. Right? $\endgroup$ May 8, 2019 at 18:20
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$\begingroup$ @user438991 In any case, $H^*(SU,Z)$ is free, whereas $H^*(K(Z,3),Z)$ has lots of torsion. $\endgroup$ May 8, 2019 at 18:37
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$\begingroup$ In the new argument, where do you use any fact that distinguishes $ku$ from $H\mathbb{Z}[t]$? For example, I don't understand how you prove that $H_3H\mathbb{Z} \to H_0ku$ is surjective 'by induction' without knowing the fact that I keep coming back to about the homology of $ku$ (I think these facts are equivalent anyway). You must use something because, after all, we also have a cofiber sequence $\Sigma^2(H\mathbb{Z}[t]) \to H\mathbb{Z}[t] \to H\mathbb{Z}$ where $|t|=2$. $\endgroup$ May 9, 2019 at 8:47