# What are explicit obstructions to realizability of formal group laws as complex-oriented ring spectra?

Recall that a complex-oriented spectrum is a ring spectrum E with a map $MU \to E$.

Analogously, a ring with a (1-d commutative) formal group law is (represented by) a ring $R$ with a map $L \to R$ (where $L$ is the Lazard ring).

In the cases where $L \to R$ is Landweber exact, this can be made explicit: However, the explicit lifting condition can't just be Landweber exactness. For example: the additive formal group law, though not Landweber-exact, corresponds to $H\mathbb{Z}$; the $n$th Honda formal group law, though not Landweber-exact, corresponds to $K(n)$.

My question is not "are there rings with (1-d commutative) formal group laws for which there is no corresponding map between ring spectra," but instead along the lines of "how do I build an explicit example."

Edit: A stupid explicit example of a formal group law that doesn't lift to a complex-orientable spectrum is any formal group law over $\mathbb{F}_p$ that is not isomorphic to the additive formal group law over $\mathbb{F}_p$. The only complex orientable spectra associated to formal group laws over $\mathbb{F}_p$ is $H\mathbb{F}_p$ (everything is concentrated in one degree). However, we can of course have a formal group law over $\mathbb{F}_p[[u_n, u_n^{-1}]]$ which lifts to a periodic ring spectrum.

What are explicit obstructions to realizability of formal group laws as complex-oriented ring spectra?

A vague guess is that the difference might come from the algebra of the ring $MU^*$ behaving differently from the 'homotopical' algebra of the ring spectrum $MU$, but I'm not sure how to proceed from this.

• This is a good question. There are examples of explicit obstructions when you ask for more structure on $E$ than simply being a ring spectrum. I do not know an example where realizability is impossible. – Tyler Lawson Apr 18 '15 at 5:54
• Just a suggestion - start from some "subtle" $E$ and either take arbitrary subring $R\subset\pi_*(E)$ containing the image of $L\to\pi_*(E)$ or, in the opposite direction, take arbitrary ring homomorphism $\pi_*(E)\to R$ to some other ring. It is intuitively clear that $R$ might be "arbitrarily bad". I realize this is very imprecise but somehow it gives rise to a feeling that it must not be that hard to stumble on counterexamples... – მამუკა ჯიბლაძე Apr 18 '15 at 9:02
• I do not know the answer. In particular, I do not know whether realization is possible if I kill a non-regular sequence from $MU_*$ or related spectra, e.g. if I do something like $BP\langle 2 \rangle_* /(v_1^2, v_1v_2, v_2^2)$. – Lennart Meier Apr 19 '15 at 22:38
• Regarding your edit, if you impose the constraint that your formal group law comes from a graded ring $R$ with a map $MU^* \to R$ of graded rings, the additive formal group is indeed the only possible formal group law on the graded ring $\Bbb F_p$. – Tyler Lawson May 5 '15 at 14:41

The other major class of cases where one can construct $E$ is when $R$ is a localised regular quotient of $L$, or in other words $R=L[S^{-1}]/I$, where the ideal $I$ can be generated by a regular sequence. There is a long history of general results of this type. I think that the sharpest versions are in my papers "Products on $MU$-modules" and "Realising formal groups" (although the methods used are not so different from earlier work). Note that the latter paper works with the periodic spectrum $MP=\bigvee_{n\in\mathbb{Z}}\Sigma^{2n}MU$ rather than $MU$ itself; this is more natural for many purposes. Note that in this context we focus on $\pi_0$ and note that $\pi_0(MP)=L$.

There is one more useful construction which is less often discussed in the literature. Choose generators $a_1,a_2,\dotsc$ for $L=\pi_0(MP)$. Let $A$ be the monoid (under multiplication) generated by these elements, and put $T=\Sigma^\infty_+A$, or equivalently $T=\bigvee_{a\in A}S^0$. Now $\pi_*(T)=\pi_*(S)\otimes L$, and there is an evident map $f\colon T\to MP$ of naive ring spectra, which is an isomorphism on $\pi_0$. If we want to make an $MP$-algebra $E$, we could hope to start by making a $T$-algebra $E'$, and then put $E=E'\wedge_TMP$. In particular, suppose that $I$ is an ideal in $L$ that is generated by some subset of monomials in the generators $a_i$. Then it is easy to construct $T/I$ as $\bigvee_{b\in B}S^0$ for a suitable subset $B\subseteq A$, and we can hope to construct $MP/I$ as $T/I\wedge_TMP$. In particular, we can choose lifts in $\pi_0(MP)$ of the chromatic generators $v_k$, and use these as some of the integral generators $a_i$; then we can form $MP/(v_1^2,v_1v_2,v_2^2)$, which is similar but not identical to the thing that @LennartMeier mentioned in his comment.

However, to make this work, we need more highly structured versions of $T$, $MP$ and $f$ (because there is no general construction of smash products of modules over unstructured ring spectra). We can construct strictly commutative versions of $T$ and $MP$, and also of $T/I$ when $I$ is generated by monomials, but unfortunately $f\colon T\to MP$ cannot be a map of strictly commutative rings, because $\pi_0(MP)$ has interesting power operations and $\pi_0(T)$ does not. There are some pitfalls with model category structures that can cause trouble here, and I have not checked all the details, but I think one can do the following, for example in the category of EKMM spectra.

• In the EKMM context, the $0$-sphere spectrum $S$ is not cofibrant. We write $C$ for the cofibrant replacement, and $C^m$ for the $m$-fold smash power.
• Let $A_i$ denote the submonoid of $A$ generated by $a_i$ and put $T_i=\Sigma^\infty_+A_i=\bigvee_mS$. The spectrum $T=\Sigma^\infty_+A$ is then the smash product of all the $T_i$ (by which I mean, the colimit over $n$ of $\bigwedge_{i=1}^nT_i$).
• Let $T_i^c$ denote the cofibrant replacement of $T_i$, in the category of strictly commutative ring spectra. This has no simple description. Let $T^c$ denote the smash product of the $T_i^c$; I think this is the cofibrant replacement of $T$ in the commutative category.
• Put $T_i^a=\bigvee_mC^m$. This is the free associative ring generated by $C$, and is the cofibrant replacement of $T_i$ in the associative category. There is a natural weak equivalence $T_i^a\to T_i^c$ of associative rings.
• Let $T^a$ be the smash product of all the objects $T^a_i$. This is an associative ring, but is not cofibrant as such. I think that does not matter.
• Now take a strictly commutative model of $MP$. Using the freeness property of $T_i^a$ we get a map $f^a_i\colon T^a_i\to MP$ carrying the generator to $a_i$. As $MP$ is commutative, we can smash these together and pass to a colimit to get a map $T^a\to MP$ of strictly associative ring spectra. This gives us an associative $T^a$-algebra. To be on the safe side, we should probably take the cofibrant replacement of this in the category of $T^a$-algebras, to get an object $M^a$. We can then take $M^c=T^c\wedge_{T^a}M^a$. This is an associative $T^c$-algebra, and I think we have done enough cofibrant replacement to ensure that the underlying homotopy type is still $MP$.
• Now let $I$ be an ideal in $L$ that is generated by monomials, and let $B$ denote the set of monomials that do not lie in $I$. There is a fairly direct way to make $\Sigma^\infty_+B$ into a strictly commutative ring spectrum, and we let $T^c/I$ denote the cofibrant replacement. This has a natural map from $T^c$. Now we can define $MP/I=T^c/I\wedge_TM^c$.
• Is this the only other class of cases? – Catherine Ray Aug 13 '15 at 22:29
• Can we construct Morava K-theory (in such a way that we explicitly know that the Eilenberg-Steenrod exactness axiom is satisfied) via the method you describe above? – Catherine Ray Oct 14 '15 at 21:08