Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
2  
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. –  Fernando Muro May 22 '12 at 22:12
2  
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.) –  Tyler Lawson May 23 '12 at 3:50
add comment

3 Answers 3

up vote 14 down vote accepted

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.

share|improve this answer
    
Thanks! This is pretty helpful. –  Akhil Mathew May 25 '12 at 12:24
1  
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. –  Akhil Mathew May 25 '12 at 12:26
add comment

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 $X$, $\pi_*(X)\otimes_{MU_*} M_* \cong \pi_*(X\wedge_{MU} M)$.

share|improve this answer
    
This result is quite interesting; does a full proof appear elsewhere? –  Akhil Mathew May 24 '12 at 4:53
add comment

Akhil, short though that paper is, I claim that the proof there is as complete as it needs to be.

share|improve this answer
    
Perhaps this should be a comment on your original answer, rather than a separate answer? –  Steve D May 26 '12 at 23:00
    
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? –  Akhil Mathew May 28 '12 at 3:58
    
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. –  Peter May Jun 3 '12 at 20:16
    
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. –  Akhil Mathew Jun 5 '12 at 2:00
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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