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Let $MU$ be the complex bordism spectrum and let $H\mathbb{Z}$ be the Eilenberg-Maclane spectrum.

Is it know what the structure of the complex cobordism cohomology $MU^{*}(H\mathbb{Z})$ is?

EDIT: What if instead $H\mathbb{Z}$, one consider $H\mathbb{Z}/(p)$ for a prime $p$?

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  • $\begingroup$ Hrmm.. we know $MU_*H\mathbb{Z}=H\mathbb{Z}_*MU=\mathbb{Z}[b_1,b_2,...]$, so we might hope to run the universal coefficient spectral sequence, although the structure as $MU_*$-module is somewhat complicated. $\endgroup$ Commented Nov 14, 2018 at 19:23
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    $\begingroup$ If I remember correctly, it is $0$, but I don't remember a reference off the head. $\endgroup$
    – user43326
    Commented Nov 14, 2018 at 19:24

1 Answer 1

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One can prove that $\mathrm{Map}(H\mathbf{F}_p,MU)$ is contractible. We know that $H\mathbf{F}_p$ is dissonant (Theorem 4.7 of Ravenel's "Localization with Respect to Certain Periodic Homology Theories"), but $MU$ is harmonic (Theorem 4.2 of that paper). Since dissonant spectra (resp. harmonic spectra) are by definition $E$-acyclic (resp. local) for the spectrum $E = \bigvee_p E_p$, where $E_p = \bigvee_{0\leq n<\infty} K(n)$, the claim follows. (I just realized that this is Corollary 4.10 of Ravenel's paper.)

It is, however, not the case that $\mathrm{Map}(H\mathbf{Z},MU)$ is contractible. (What I wrote previously was incorrect.) The spectrum $H\mathbf{Z}$ is not dissonant, so we cannot immediately apply the above argument. Since $MU$ is harmonic, there is, however, an equivalence between $\mathrm{Map}(H\mathbf{Z},MU)$ and $\mathrm{Map}(L_E H\mathbf{Z},MU)$. We therefore need to understand $L_E H\mathbf{Z}$. By the discussion at this question, we can conclude that $L_{E_p} H\mathbf{Z} \simeq H\mathbf{Q}_p$. It therefore suffices to understand $MU^\ast(H\mathbf{Q})$. But $H\mathbf{Q}$ is the colimit of multiplication by $2,3,5,7,\cdots$ on the sphere, so $MU^\ast(H\mathbf{Q})$ admits a description in terms of $\lim^0$ and $\lim^1$ of multiplication by $2,3,5,7,\cdots$ on $\pi_\ast MU$. In particular, the $\lim^1$ term is $\mathrm{Ext}^1_\mathbf{Z}(\mathbf{Q,Z}) \cong \widehat{\mathbf{Z}}/\mathbf{Z}$.

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    $\begingroup$ I don't think that the last bit of this is right. $MU^0(H\mathbb{Q}_p)$ is not a ring. We can write $H\mathbb{Q}=S\mathbb{Q}$ as the telescope of multiplication by $2,3,4,5,\dotsc$ on $H$ or on $S$. From the first description together with $F(H/n,MU)=0$ we get $F(H,MU)=F(H\mathbb{Q},MU)=F(S\mathbb{Q},MU)$. The second description relates $[S\mathbb{Q},MU]_*$ to $\lim^0$ and $\lim^1$ of multiplication by $2,3,4,\dotsc$ on $\pi_*(MU)$. Here $\lim^0=0$ but $\lim^1$ involves $\text{Ext}(\mathbb{Q},\mathbb{Z})=\widehat{\mathbb{Z}}/\mathbb{Z}$. $\endgroup$ Commented Nov 14, 2018 at 20:59
  • $\begingroup$ @NeilStrickland you're right, thanks! I'll edit my answer. $\endgroup$
    – skd
    Commented Nov 14, 2018 at 21:04
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    $\begingroup$ Just a general comment that this isn't so "chromatic": It's a theorem of Margolis that maps out of $H\mathbb{F}_p$ to a bounded below spectrum of finite type are the same as maps of modules over the Steenrod algebra on cohomology into $\mathcal{A}^*$; this is already enough to show that $MU^*(H\mathbb{F}_p) = 0$. Then the sequence $\mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z}$ tells you that $Y^*\mathrm{H}\mathbb{Z}$ will always be a humongous sum of $\mathrm{Ext}(\mathbb{Q}, ?)$'s if $Y^*(H\mathbb{F}_p)$ vanishes for all $p$. TY Lin's paper on duality and EM spectra has more on this. $\endgroup$ Commented Nov 14, 2018 at 23:21

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