# Example of a space X exhibiting the Landweber non-exactness of the additive formal group over the integers?

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?

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.
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. $$|\bar{\tau_1}|=2p-1$$.
• 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) – Denis Nardin Nov 20 '19 at 8:55