Connection of X(n) spectra to formal group laws - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T13:23:11Z http://mathoverflow.net/feeds/question/116663 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116663/connection-of-xn-spectra-to-formal-group-laws Connection of X(n) spectra to formal group laws Jon Beardsley 2012-12-17T23:26:10Z 2012-12-18T04:00:29Z <p>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. </p> <p>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?</p> <p>Thanks! </p> http://mathoverflow.net/questions/116663/connection-of-xn-spectra-to-formal-group-laws/116678#116678 Answer by Akhil Mathew for Connection of X(n) spectra to formal group laws Akhil Mathew 2012-12-18T04:00:29Z 2012-12-18T04:00:29Z <p>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. </p> <p>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 <a href="http://intlpress.com/HHA/v10/n3/a1/v10n3a1.pdf" rel="nofollow">talk</a> 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. </p> <p>It's important, however, that you <em>don't</em> 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 </p> <p>$$B \mathbb{Z}/2 \to M_{FG}$$</p> <p>(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 <em>too</em> much like algebra. (I wrote <a href="http://people.fas.harvard.edu/~amathew/tmfhomotopy.pdf" rel="nofollow">some notes</a> 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 <em>flat</em> 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 <em>derived</em> category of perfect modules on $M_{FG}$.)</p> <p>Let me try to summarize the key inductive argument in Devinatz-Hopkins-Smith's paper, which is to prove: </p> <p><strong>Theorem:</strong> 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. </p> <p>More generally, you can ask when something like this is true: </p> <p><strong>Question:</strong> If $R \to R'$ is a morphism of ring spectra and $R$ "detects nilpotence," then when does $R'$? </p> <p>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:</p> <p><strong>Axiomatic nilpotence theorem:</strong> Suppose $R'$ is obtained from a filtered colimit of spectra $G_k$ such that: </p> <ul> <li>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 $k \to \infty$. </li> <li>Each $G_k$ is Bousfield equivalent to $R$. </li> </ul> <p>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. </p>