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Landweber exactness gives a criterion for when a complex oriented cohomology theory $E$ can be recovered from the formal group law over $E_{*}$ determined by the complex orientation. That is it gives a criterion for the following to hold (see for example here)

$$E_{*}(X)\cong MU_{*}(X)\otimes_{MU_{*}}E_{*}$$

Since the additive formal group law over $\mathbb{Z}$ is NOT Landweber exact, when $E=H\mathbb{Z}$ this isomorphism cannot hold for all spaces $X$.

My question is: what is an example of such a space?

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Firstly, you ask for a space $X$. I will instead talk about finite spectra, but they become spaces if you suspend them enough times, so that does not really make a difference.

There is a kind of tautological answer to your question as follows. By the nilpotence technology of Hopkins, Devinatz and Smith, for suitable sequences of natural numbers $i_k$ there are generalised Moore spectra $S/(v_0^{i_0},\dotsc,v_n^{i_n})$ such that $$ MU_*(S/(v_0^{i_0},\dotsc,v_n^{i_n})) = MU_*/(v_0^{i_0},\dotsc,v_n^{i_n}) $$ and there are cofibre sequences $$ \Sigma^{|v_n^{i_n}|} S/(v_0^{i_0},\dotsc,v_{n-1}^{i_{n-1}}) \xrightarrow{} S/(v_0^{i_0},\dotsc,v_{n-1}^{i_{n-1}}) \to S/(v_0^{i_0},\dotsc,v_n^{i_n}) $$ with the obvious effect in $MU$-homology. Any spectrum of the form $S/(v_0^{i_0},v_1^{i_1})$ will be an answer for your question. Examples for this $n=1$ case were already constructed by Adams before the nilpotence theory was available (and this was a big part of the motivation for the nilpotence programme).

However, one might ask for a more elementary example. I think that one can proceed as follows. The Hopf map $\eta\colon S^1\to S^0$ has order $2$ and so extends to give a map $\eta'\colon S^1/2\to S^0$. Let $Q$ be the third suspension of the Spanier-Whitehead dual of the cofibre of $\eta'$. This has cells in dimensions $0$, $1$ and $3$. In mod $2$ homology, the bottom two cells are connected by $\text{Sq}^1$ and the top two are connected by $\text{Sq}^2$ so the cell diagram looks like a question mark and the complex is sometimes called the question mark complex. I think it works out that $MU_*Q=MU_*/(2,v_1)x\oplus MU_*y$ with $|x|=0$ and $|y|=3$. On the other hand, we have $H_*Q=\mathbb{Z}/2x\oplus\mathbb{Z}z$ with $x$ mapping to $x$ and $y$ mapping to $2z$, so the map $\mathbb{Z}\otimes_{MU_*}MU_*Q\to H_*Q$ is not surjective. However, the argument is a bit intricate.

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  • $\begingroup$ This is great, thank you! Is there away to get the "2" out of there, and just have a map of spaces (finite spectra) such that, for its cofiber, one of the maps in the long exact sequence in MU-homology is multiplication by v1, which then becomes zero in the LES for ordinary homology? Because then the groups will differ, if I am not being dumb. $\endgroup$ – John Greenwood Nov 20 at 21:44
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Take $X=H\mathbb{F}_p$ for $p$ an odd prime. Then $MU_*(X)\cong\mathbb{F}_p[b_1,b_2,\ldots]$ where $|b_i|=2i$. In particular, the right hand side is concentrated entirely in even degrees. On the other hand, the left hand side may be computed as $\mathbb{F}_p[\xi_1,\xi_2,\ldots]\otimes E(\bar{\tau_1},\bar{\tau_2},\ldots)$, which has classes in odd degrees, e.g. $|\bar{\tau_1}|=2p-1$.

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  • $\begingroup$ Technically $X$ here is a spectrum and not a space (I don't know if the OP is satisfied with an example with a spectrum) $\endgroup$ – Denis Nardin Nov 20 at 8:55
  • $\begingroup$ Thanks for this, I decided to accept the other answer since I did technically ask for a space. $\endgroup$ – John Greenwood Nov 20 at 21:45

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