Given a (graded) ring $R$, to define a formal group law it is equivalent to define a ring homomorphism $\phi:L\to R$ where $L$ is Lazard's ring. Is there any notion of defining a formal group law on ring spectrum $E$, i.e. a map $f:\mathbb{H}L\to E$? Would this in any way extend the notion of formal group laws on algebraic rings or the Landweber Exact Functor Theorem?
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asked Dec 31, 2011 at 15:24
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3$\begingroup$ Classically, the analog would be where you replace $L$ by $MU$ and ask for a map $MU \rightarrow E$. This is equivalent to having a complex oriented ring spectrum. I'm not exactly sure what the "Landweber Exact Functor Theorem" would say here: as far as I know the Landweber exact functor theorem is already a statement involving spectra, so to have an "analog" you would need to fill in the analogy "(rings):(ring spectra):: (ring spectra): (???)"... and I don't know how to do that. $\endgroup$– Dylan WilsonDec 31, 2011 at 16:37
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$\begingroup$ I agree with Dylan here, from what I've heard, the analog of a formal group law on ring spectra is $f:MU\rightarrow E$, i.e. a complex orientation. I think the Landweber Exact Functor Theorem is used to prove this is the ``right'' analog. Have you looked at Jacob Lurie's notes on Chromatic Homotopy Theory? math.harvard.edu/~lurie/252x.html $\endgroup$– David WhiteDec 31, 2011 at 19:19
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$\begingroup$ Yeah I guess you're right. The analogy already exists as MU. But I wonder about HL nonetheless. $\endgroup$– Jonathan BeardsleyDec 31, 2011 at 19:21
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$\begingroup$ Yeah actually I think what I'm asking is significantly different from the correspondence you mention, though it may not be of any relevance anyway. $\endgroup$– Jonathan BeardsleyDec 31, 2011 at 19:37
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1 Answer
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I think the right correspondence is the one mentioned in the comments. The main reason is that if you have a map of ring spectra $HL \to E$ then $E$ is an Eilenberg-Maclane spectrum. So in particular, you are only getting information about formal group laws over EM spectra and not any other kinds of spectra. However, it might be interesting to compute this. It is easy not easy to do or it is trivial. The main sources for computing this type of mapping space (maps of commutative ring spectra) that I know of is the obstruction theory of Goerss and Hopkins. There could be others though.
answered Dec 31, 2011 at 20:33
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$\begingroup$ The existence of a map from HL to E is sufficient to say that E is an EM spectrum? $\endgroup$ Dec 31, 2011 at 21:50
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3$\begingroup$ @JonB: Yes, the existence of such a ring map is sufficient to show that E is homotopy equivalent to a coproduct of suspensions of Eilenberg-Mac Lane spectra (i.e. a generalized E-M spectrum). $\endgroup$ Dec 31, 2011 at 23:25
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$\begingroup$ I think really what I'm interested in, and this is something I asked Tyler about in Boston, is whether or not you can just define a formal group law ON a general ring spectrum, i.e. without elements. I don't really know enough yet to be able to talk about this, but I'm not talking about complex orientations. $\endgroup$ Jan 24, 2012 at 20:40