As far as I know, there is still no such interpretation. The closest I've heard is some rumored (but unpublished) work in derived algebraic geometry interpreting MU as some kind of representing object.
Such a construction of MU in terms of formal group data be very welcome (probably even more now than when Ravenel wrote the green book).
EDIT: Some elaboration.
We do know a lot about MU. We know that it has an orientation (Chern classes for vector bundles), and in it's universal for this property. It's not then extremely suprising that we get a formal group law from the tensor product for line bundles, but the fact that MU carries a universal formal group law, and that MU ^ MU carries a universal pair of isomorphic formal group laws, is surprising. At this point it's something we observe algebraically. Even Lurie's definition of derived formal group laws, assuming I understand correctly, is geared to construct formal group laws objects in derived algebraic geometry carrying a connection to the formal group law data that we already know is there on the spectrum level, and hence ties it to the story we already knew for MU implicitly.
Some reasons these days we might want to know how to construct MU from formal group law data:
- Selfish, ordinary homotopy-theoretic reasons. It's very useful to be able to construct other spectra with specific connections to formal group law data (like K-theory, TMF, etc) and constructing them is generally very difficult. Things like the Landweber exact functor theorem, the Hopkins-Miller theorem, and Lurie's recent work give us a lot of progress in this direction, but they only apply to restricted circumstances. None of these general methods will construct ordinary integral cohomology, corresponding to the additive formal group law (only rational cohomology). If we understood how to build MU, we might understand how to generalize.
- Equivariant homotopy theory. I would tentatively say that we don't have nearly as good computational and "qualitative" pictures of the equivariant stable categories, because we don't have something like the startling MU-picture that relates it all to some stack like the moduli stack of 1-dimensional formal group laws. If we found MU by _accident_ then we don't really know how the analogue should play out in other, more general, stable categories.
- Motivic homotopy theory. Hopkins and Morel found that there is some data to formal group laws appearing in motivic stable homotopy theory via the motivic bordism spectrum MGL. I'm not up with the state of the art here but a better understanding of this connection would be very important too - for understanding MGL itself, but also hopefully for understanding the analogues of chromatic data in these categories related to algebraic geometry.
- (space reserved for connections to other subjects that I've forgotten)
Although I'm not aware an interpretation of $X(n)$ in terms of formal group laws (aside from
the ones in Eric's and Dylan's comments), I would like to point out that the nilpotence theorem is fundamentally a geometric fact and is proved that way. The nilpotence theorem has a number of corollaries to the effect that the formal group perspective gives a very good description of the global structure of the stable homotopy category, but its proof requires explicit, computational facts about very specific spectra.
For example, one way to phrase the nilpotence theorem is that, for any connective spectrum $X$, the $E_\infty$-page of the Adams-Novikov spectral sequence (drawn with $t-s$ horizontally and $s$ vertically) has a "vanishing curve" which is asymptotically flat: that is, has slope tending to zero as $t-s \to \infty$. That is, the maximum possible Adams-Novikov filtration of an element in $\pi_k X$ grows in $k$ by some function which is $o(k)$. (Actually determining what this function is seems to be an open problem; see Hopkins's enjoyable talk about Ravenel's work for some discussion.) If you know that there is such a vanishing curve, you can get the nilpotence theorem in its first form (about ring spectra) at once by noting that an element in $\pi_* R$ for any ring spectrum which is killed by the $MU$-Hurewicz is detected in the ANSS in positive filtration and its powers live on a line of positive slope, which has to overtake the asymptotically flat vanishing curve. This is something that's very much specific for $MU$. With mod $2$ (or even integral) homology and with the sphere spectrum, the image of $J$ elements already rule out such a vanishing line in the classical ASS.
It's important, however, that you don't get the vanishing curve at $E_2$, which is the part that comes from algebra. The $E_2$-term (which is the cohomology of $M_{FG}$) has lots of non-nilpotent elements, for instance the element corresponding to $\eta$. A quick way to see this is to consider the map
$$B \mathbb{Z}/2 \to M_{FG}$$
(which corresponds geometrically to the map
$S^0 \to KO$). The element $\eta$ is not nilpotent for $B \mathbb{Z}/2$, and therefore it can't be nilpotent in $H^*(M_{FG})$, although $\eta^3$ is killed by a $d_3$ differential in the spectral sequence for $KO$. The nilpotence theorem is somehow saying something global about the structure of such spectral sequences, that there have to be a lot of differentials, which create this vanishing line. So at some level, it's saying what appears to be the opposite: that homotopy can't look too much like algebra.
(I wrote some notes on the spectral sequence for $KO$, and ultimately for $TMF$ at the prime $3$, which I mention because working through these spectral sequences helped me appreciate some of this technology. In fact, it's even better: you get flat vanishing lines at finite stages; this is related to the Hopkins-Ravenel smashing theorem which states that this happens for any $E(n)$-local spectrum.) So there's no nilpotence theorem for $M_{FG}$. (As a related note: there's no thick subcategory theorem for the derived category of perfect modules on $M_{FG}$.)
Let me try to summarize the key inductive argument in Devinatz-Hopkins-Smith's paper, which is to prove:
Theorem: If $R$ is a connective, associative ring spectrum and $\alpha \in \pi_* R$ is such that $X(n+1)_* \alpha =0$, then $X(n)_* \alpha $ is nilpotent.
More generally, you can ask when something like this is true:
Question: If $R \to R'$ is a morphism of ring spectra and $R$ "detects nilpotence," then when does $R'$?
For example, if $R'$ and $R$ are Bousfield equivalent, then it's easy to get the result, but the $X(n)$ are not Bousfield equivalent. However, they do turn out to Bousfield equivalent on "telescopes of connective spectra," which turns out to be all you need.
In particular, if $R'$ annihilates a localization $\alpha^{-1} T$ for $T$ a connective ring spectrum (which is to say that $\alpha$ is nilpotent in $R'$-homology), then so does $R$.
That's what D-H-S show, and their argument is summarized in:
Axiomatic nilpotence theorem:
Suppose $R'$ is obtained from a filtered colimit of spectra $G_k$ such that:
- The $G_k$ have good $R'$-based Adams spectral sequences: that is, in the $E_\infty$-page of the $R'$-based ASS for $G_k \wedge X$ (for any connective $X$), there is a vanishing line of slope $\epsilon_k$ which tends to zero as $k \to \infty$.
- Each $G_k$ is Bousfield equivalent to $R$.
D-H-S check these conditions for $X(n) \to X(n+1)$. The first step is something that they do purely algebraically (at $E_2$) using a series of May-type spectral sequences, and it's the piece of the argument which you might be able to get using facts about formal groups. But the second part -- which is way, way harder -- seems to require some geometry and facts about concrete things like $E_2$-algebras and Thom spectra and partial James constructions. I suspect that, at some level, the use of such geometry is an inescapable feature of the whole business.
Best Answer
The following is a communal answer from the algebraic topology Discord [1], primarily put forward by Irakli Patchkoria (correcting previous half-answers by Tyler Lawson and me). Kiran suggested it be recorded here to ease future reference.
The idea is to produce two topological realizations $M$, $N$ of a single $MU_*$–module by finding two distinct resolutions whose effect on homotopy is the same. The two associated square-zero extensions then give a counterexample. We'll reduce complexity first by considering $ku$–modules rather than $MU$–modules, and second by aiming for a $ku$–module whose homotopy cleaves into small even and odd parts, forcing its $ku_*$–module structure to trivialize.
$\DeclareMathOperator{\Sq}{Sq} \newcommand{\F}{\mathbb{F}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\HFtwo}{H\F_2} \newcommand{\Susp}{\Sigma} \newcommand{\co}{\colon\thinspace}$ The nice homotopy groups of complex $K$–theory, $ku_* = \Z[u]$, can be used to show that its bottom $k$–invariant $\kappa_{ku}$ is $ku$–linear: in the diagram $$ \begin{array}{ccccc} & & \Susp^4 ku \\ & & u \downarrow \\ \Susp^{-1} H\Z & \to & \Susp^2 ku & \to & \Susp^2 H\Z \\ & & u \downarrow \\ & & ku & \to & H\Z, \end{array} $$ the vertical maps are multiplication by homotopy elements, hence are $ku$–linear; in turn the horizontal co/fibers are also $ku$–linear; and, finally, the $k$–invariant appears as the middle composite, hence is also $ku$–linear. Similarly, we can show the $ku$–linearity of the bottom $k$–invariant of $ku/2$ and of the Bockstein map $\beta\co \HFtwo \to \Susp H\Z$ (relying on the $ku$–linearity of $2\co H\Z \to H\Z$).
Stringing some of these together gives a $ku$–linear composite $$\HFtwo \xrightarrow{\kappa_{ku/2}} \Susp^3 \HFtwo \xrightarrow{\beta} \Susp^4 H\Z \to \Susp^4 \HFtwo.$$ The bottom $k$–invariant $\kappa_{ku/2}$ of $ku/2$ is given as the Milnor primitive $Q_2 = \Sq^3 + \Sq^2 \Sq^1$, the composite of the latter two maps is given as $\Sq^1$, and hence the whole composite is the nontrivial Steenrod operation $$\Sq^1 Q_2 = \Sq^1(\Sq^3 + \Sq^2 \Sq^1) = \Sq^3 \Sq^1.$$ Meanwhile, the homotopy groups of the cofiber $M$ of this composite are $\Susp \F_2 \oplus \Susp^4 \F_2$, which splits as a $ku_*$–module — hence this $ku_*$–module could alternatively be modeled by $N = \Susp H\F_2 \oplus \Susp^4 \HFtwo$ (i.e., the cofiber of the zero map). To finish, set $E = ku \oplus M$ and $F = ku \oplus N$.