5
$\begingroup$

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?

$\endgroup$

2 Answers 2

8
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ Nov 20, 2019 at 21:44
5
$\begingroup$

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. $|\bar{\tau_1}|=2p-1$.

$\endgroup$
2
  • $\begingroup$ 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) $\endgroup$ Nov 20, 2019 at 8:55
  • $\begingroup$ Thanks for this, I decided to accept the other answer since I did technically ask for a space. $\endgroup$ Nov 20, 2019 at 21:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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