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Let $M$ be a $\pi_*(MU)$-module. The Landweber exact functor theorem gives conditions for the functor that sends a space $X$ to $ MU(X) \otimes_{\pi_*(MU)} M$ to define a homology theory on spaces, which thus comes from a spectrum.

It'd be nice, though, if one could construct the spectrum directly, instead of going through the homology theory. For instance, it would be nice if one could construct an actual $MU$-module (possibly under further hypotheses) or an $MU$-algebra when $M$ is an algebra. Is there another version of the exact functor theorem that lets one do this?

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    $\begingroup$ I'm sceptical about a possible positive answer because, if there were a more direct construction, I would expect it to be functorial on $M$, but the spectrum representing a cohomology theory is not functorial. $\endgroup$
    – Fernando Muro
    May 22, 2012 at 22:12
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    $\begingroup$ I'll second Fernando's comment. In particular, there are a lot of Landweber exact elliptic cohomology theories. Constructing them functorially is very difficult. Constructing MU-algebras can be terrifyingly difficult depending on how much structure you want. The problem is that you're fundamentally starting with "up to homotopy" data (a module), and rectifying that into an actual spectrum is very, very unlikely to be a canonical procedure. (This isn't specific to homotopy theory, either. The same problem should show up in the differential-graded world.) $\endgroup$
    – Tyler Lawson
    May 23, 2012 at 3:50

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Here are three methods that I know:

  • In the case $M_*=(MU_*/I)[S^{-1}]$ (where $I$ is generated by a regular sequence) there is a more direct construction by reducing to the cases $M_*=MU_*/a$ and $M_*=MU_*[a^{-1}]$. My paper 'Products on MU-modules' is probably the sharpest version, but there are many earlier versions in a similar spirit.

  • In the case $M_*=MU_*[x_1,\dotsc,x_r]$ with $|x_i|=0$ you can use $MU\wedge\Sigma^\infty_+\mathbb{N}^r$ (and this has an $E_\infty$ structure).

  • In the case $M_*=MU_*[n^{-1}]$ (for some $n\in MU_0=\mathbb{Z}$) you can note that there are natural $E_\infty$ maps $$ MU\xleftarrow{f}\Sigma^\infty_+DS^0\xrightarrow{}\Sigma^\infty_+QS^0,$$ where $f$ has degree $n$ on the bottom cell. The smash product $$ MU\wedge_{\Sigma^\infty_+DS^0}\Sigma^\infty_+QS^0$$ then has the required property.

There are some fairly obvious ways to combine these methods and generalise them slightly.

Under the general conditions of the Landweber theorem, I know of several people including myself who have looked quite hard for a more direct construction, but without success.

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    $\begingroup$ I would be very interested in a universal property for one of these spectra (for instance, my understanding is that Lurie has a universal property for complex K-theory), but that might not be a realistic expectation in general. $\endgroup$
    – Akhil Mathew
    May 25, 2012 at 12:26
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I'm not sure that this is exactly what you are looking for, but I looked a bit at the Landweber exact functor theorem in the context of $MU$-modules at the end of a very short paper: Idempotents and Landweber exactness in brave new algebra. Homology, homotopy, and applications 3(2001), 355--359. Theorem 8 there reads: If $M_*$ is a Landweber exact $MU_*$-module, then there is an $MU$-module $M$ such that $\pi_*(M) = M_*$ and, for any finite cell $MU$-module $X$, $\pi_*(X)\otimes_{MU_*} M_* \cong \pi_*(X\wedge_{MU} M)$.

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  • $\begingroup$ This result is quite interesting; does a full proof appear elsewhere? $\endgroup$
    – Akhil Mathew
    May 24, 2012 at 4:53
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Akhil, short though that paper is, I claim that the proof there is as complete as it needs to be.

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  • $\begingroup$ Perhaps this should be a comment on your original answer, rather than a separate answer? $\endgroup$
    – Steve D
    May 26, 2012 at 23:00
  • $\begingroup$ I'm a little confused here. Shouldn't the condition for all finite $MU$ modules be related to flatness over $\pi_* MU$, because one is asking about $\pi_* X \otimes MU_* M_*$ rather than $MU_*(X) \otimes MU_* M_*$ (i.e., in the usual LEFT a comodule structure is being used which doesn't seem to exist here)? Also, I'm not seeing how this is obvious; could you perhaps clarify? $\endgroup$
    – Akhil Mathew
    May 28, 2012 at 3:58
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    $\begingroup$ Steve, sorry about the etiquette of answers vs comments; can't expect an old guy to notice such distinctions. Akhil, I didn't say this is obvious. The paper is on my web page, [102], and the proof takes under two pages (because the serious math is in the references), but it probably shouldn't be repeated here. I can't answer your question precisely because I don't know what you mean by $MU_*M_*$, but here is the key lemma: If $X$ is an $R$-module, where $R$ is a commutative $S$-algebra such that $R_*R$ is $R_*$-flat, then the Hurewicz map gives $X_*$ a structure of $R_*R$-comodule. $\endgroup$
    – Peter May
    Jun 3, 2012 at 20:16
  • $\begingroup$ Whoops, I omitted a subscript and meant $\pi_* X \otimes_{MU_*} M_*$ (where $M_*$ is a graded module over $MU_*$). I will think some more about the lemma you mentioned, thogh. Thanks. $\endgroup$
    – Akhil Mathew
    Jun 5, 2012 at 2:00

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