In the proof of the Nilpotence Theorem, or at least in Ravenel's account of it in his Orange Book, a sequence of spectra are used, denoted $X(n)$ with $X(0)=\mathbb{S}$ and and $X(\infty)=MU$ such that $\langle X(n)\rangle\geq\langle X(n+1)\rangle$. These are the Thom spectra associated to the map $\Omega SU(n)\to BU$. They are homotopy equivalent to $MU$ up through degree 2n-1. They all have associated Hurewicz maps $h(n):\pi_\ast(R)\to X(n)_\ast(R)$ for $R$ a finite ring spectra. We are interested, for the Nilpotence Theorem, in determining when $h(n)(\alpha)=0$, ultimately for $n=0$. To this end, Ravenel proves that if $h(n+1)(\alpha)=0$ then $h(n)(\alpha)$ is nilpotent.

So, from this theorem we know then that any of the $X(n)$ spectra detect nilpotence just as well as $MU$. The nice thing of course about $MU$ (or one of the many nice things) is that it has at least one other interpretation (i.e. aside from its geometric interpretation as a Thom spectrum). This is of course that $MU_\ast$ determines formal group laws over rings, and you have all of this amazing stuff happen. Do you have any similar such interpretations for these $X(n)$ spectra, or are they ONLY geometric in nature? Or I guess, to rephrase, does anyone KNOW of any other ways of thinking about these things?


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    There's Tyler Lawson's answer to this question:… . – Eric Peterson Dec 17 '12 at 23:49
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    See Ravenel's green book (which I guess is red now), where he talks about these... they sort of classify approximations to complex orientations. I'd be really happy to see something like "The X(n) are the global sections of a derived stack whose underlying classical stack is the stack of n-buds of formal groups." (where the latter stack can be found in, for example, Goerss' notes on the subject.) The trouble is that we don't know the X(n) are anything better than E_2-ring spectra. – Dylan Wilson Dec 17 '12 at 23:55
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    @Sean: I don't necessarily expect it... it just seems that there's some sort of connection between the two notions, and I'd like to see it done more formally. So that's a first guess. – Dylan Wilson Dec 18 '12 at 7:12
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    It's known that the $X(n)$ spectra don't admit $E_4$ structures. To see this, there would have to be an $E_4$-map $X(n)\to H\mathbb{Z}/2$. Look at the associated map on mod-2 homology. The element $\xi^2_1$ is in the image, but it generates the subalgebra $\mathbb{Z}/2[\xi^2_i]$ under the $E_4$-Dyer-Lashof operations. The homology of X(n) is $\mathbb{Z}/2[b_1,\ldots,b_n]$, which is not big enough to surject onto it. – Tyler Lawson Dec 18 '12 at 16:42
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    Me-4-years-ago had an indexing error: the above is valid with $E_3$ instead of $E_4$. – Tyler Lawson Dec 29 '16 at 0:12
up vote 18 down vote accepted

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.

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    This is a very nice answer. I'd make one additional remark: nevermind explaining everything away using formal groups, it's very difficult even to phrase the algebraic component of D-H-S's second half in that language. Neil Strickland has put substantial effort into this, and I've tried my hand a few times, but it's extremely stubborn. There are a lot of things in and around this proof's mechanics that I think are super interesting exactly because of this stubbornness. The proof of the nilpotence theorems is a real gem of thought that, decades later, doesn't feel completely demystified. – Eric Peterson Dec 18 '12 at 4:44
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    For instance, here's Nishida's proof of nilpotence in the sphere spectrum, for elements of order $p$. Let $\alpha \in \pi_n (S^0)$ have order $p$. We'd like to prove $\alpha$ is nilpotent. It's equivalent to say that $S^n \to S^0 \to S^0[1/\alpha]$ is the zero map, where we can form the localization in the world of $E_\infty$-algebras. But $S^0[1/\alpha]$ is an $E_2$-algebra with $p=0$, any any such is a wedge of Eilenberg-MacLane spectra, so the composite is zero (as $\alpha$ is zero in homology). As I understand, this was the inspiration for D-H-S's proof, and I think that it fits ... – Akhil Mathew Dec 18 '12 at 14:25
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    ...into the same pattern, except you have to work with these more complicated $\Sigma^\infty_+ J_{p-1} S^{2m}_(p)$, and also $\Sigma^\infty_+ \Omega^2 S^{2m+1}$. Anyway, since you're apparently going to be at MIT in the spring, we should discuss this in person sometime (hopefully I can get this completely sorted out in my head by then). – Akhil Mathew Dec 18 '12 at 14:29
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    @Jon: I don't know the details, but I believe that John Palmieri has proved analogs of this for modules over the Steenrod algebra. There are also algebraic "nilpotence theorems," for instance the Quillen-Venkov theorem in group cohomology and probably others that I don't know about. Anyway, I too hope that I'm wrong. It's a really amazing fact that $MU$ turns out to be such a powerful tool for studying stable homotopy (with formal groups on the one hand and nilpotence on the other). As far as I know, a satisfying explanation for this isn't known. – Akhil Mathew Dec 18 '12 at 14:51
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    @Akhil: Yes, good idea, let's chat about this in person. The comodule-spectra-for-$\Sigma^\infty_+ \Omega^2 S^{2m+1}$ stuff is precisely what I find most resistant --- I think the core of it is that I have no idea what geometric object should be attached to $H_* \Omega^2 S^{2m+1}$. Ravenel has a paper about this at (along with various others in the citations), which is intriguing but not so enlightening. – Eric Peterson Dec 18 '12 at 18:12

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