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 $p$?

  • 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. – Denis Nardin Nov 14 at 19:23
  • 2
    If I remember correctly, it is $0$, but I don't remember a reference off the head. – user43326 Nov 14 at 19:24
up vote 14 down vote accepted

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 $\mathrm{Ext}^1_\mathbf{Z}(\mathbf{Q,Z}) \cong \widehat{\mathbf{Z}}/\mathbf{Z}$.

  • 1
    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 $\text{Ext}(\mathbb{Q},\mathbb{Z})=\widehat{\mathbb{Z}}/\mathbb{Z}$. – Neil Strickland Nov 14 at 20:59
  • @NeilStrickland you're right, thanks! I'll edit my answer. – skd Nov 14 at 21:04
  • 5
    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. – Dylan Wilson Nov 14 at 23:21
  • Good to know, thanks @DylanWilson! – skd Nov 15 at 1:03

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.