Chromotopy

Devinatz--Hopkins--Smith III

In the first post of this series we set out to investigate the ability of complex cobordism to detect nilpotence in stable homotopy, and we reduced this question to an inductive argument with sort of major and minor stepping stones, and we formulated the two theorems we’d need to pass between the minor stepping stones and the major ones. In the next post, we showed that our problem does land somewhere in this tower of reductions, and now we have to show the inductive part that lets us reduce to the base case of $X(1) = \mathbb{S}$ – namely,

Second half: $F_{p^{k+1}-1}$ is Bousfield-equivalent to $F_{p^k-1}$, powering the secondary induction down to $F_0 = X(n)$.

The exact theorem we intend to prove will actually be more general than this, and we’ll specialize to the situation above at the very end.

To start, let E be a space with a designated map $\xi: E \to BU$ classifying a stable complex vector bundle and a designated fibration $p: E \to J_r S^{2m}$. The James filtration on $J_r S^{2m}$ gives rise to a filtration on E by pullback; let $E_j$ be the pullback of $E \xrightarrow{p} J_r S^{2m}$ along the inclusion $J_j S^{2m} \hookrightarrow J_r S^{2m}$. In particular, $E_0$ is the (homotopy) fiber of p. Our goal will be to construct a certain self-map b of $E_0^\xi$ such that we can relate the Bousfield classes $\langle E_0^\xi \rangle$, $\langle E^\xi \rangle$, and $\langle b^{-1} E_0^\xi \rangle$.

As a first step toward this end, we can construct a map between Thom spectra by starting with

$$E \xrightarrow{p \times \mathrm{id}} J_r S^{2m} \times E \xrightarrow{\xi \circ \pi_2} BU,$$

and Thomifying to get a map

$$E^\xi \to J_r S^{2m}_+ \wedge E^\xi,$$

describing the interaction of the map p with the vector bundle. Once again we make use of the stable, multiplicative Snaith splitting

$$J_r S^{2m}_+ \simeq \bigvee_{j=0}^r S^{2mj}$$

to build a sequence of maps

$$\theta_i: E^\xi \to J_r S^{2m}_+ \wedge E^\xi \simeq \left( \bigvee_{j=0}^r S^{2mj} \right) \wedge E^\xi \xrightarrow{\mathrm{collapse}} S^{2mi} \wedge E^\xi.$$

In fact, these maps are compatible with the lifted James filtration on E, in the sense that they induce maps

$$\theta_i: E^\xi_q / E^\xi_{q-h} \to S^{2mi} \wedge E^\xi_{q-i} / E^\xi_{q-h-i}$$

on the quotients.

Due to the multiplicativity of the splitting, we have an equality of the two composites

$$J_r S^{2m}_+ \xrightarrow{\Delta} J_r S^{2m}_+ \wedge J_r S^{2m}_+ \xrightarrow{\mathrm{collapse}} S^{2mi} \wedge S^{2m(t-i)},$$ $$J_r S^{2m}_+ \to S^{2mt} \xrightarrow{\binom{t}{i}} S^{2mt} \xrightarrow{\simeq} S^{2mi} \wedge S^{2m(t-i)}.$$

Using the fact that

$$E \xrightarrow{p \times p \times \mathrm{id}} J_r S^{2m} \times J_r S^{2m} \times E$$

factors in the following two ways

$$E \xrightarrow{p \times \mathrm{id}} J_r S^{2m} \times E \xrightarrow{\Delta \times \mathrm{id}} J_r S^{2m} \times J_r S^{2m} \times E,$$ $$E \xrightarrow{p \times \mathrm{id}} J_r S^{2m} \times E \xrightarrow{\mathrm{id} \times (p \times \mathrm{id})} J_r S^{2m} \times J_r S^{2m} \times E,$$

we get that this fact about multiplicativity turns into the following relation among the thetas:

$$\theta_{i+j} = \binom{i+j}{i} \theta_i \circ \theta_j.$$

In fact, if we’re careful enough, we can also show that this relation hnews among the filtered version of the thetas as well.

These thetas are what we’ll use to build our self-map – but they’re going the wrong direction, so we’ll need to invert a few of them. To understand what that might mean, we first note that the filtered map $\theta_j: E_j^\xi / E_{j-1}^\xi \to \Sigma^{2mj} E_0^\xi$ is already an equivalence (the reasoning for this is technical but straightforward, so I’ll omit it). Taking that for granted, we can build the commuting rectangle

$$\begin{array}{ccccc} E^\xi_j / E^\xi_0 & \to & E^\xi_{j+1} / E^\xi_0 & \to & E^\xi_{j+1} \\ \downarrow & & \downarrow & & \downarrow \\ \Sigma^{2m} E^\xi_{j-1} & \to & \Sigma^{2m} E^\xi_j & \to & \Sigma^{2m} E^\xi_j / E^\xi_{j-1},\end{array}$$

where the vertical arrows are all $\theta_1$. In the case that r! is invertible, we recall that the filtered $\theta_1$ is invertible and induct to show that $\theta_r E^\xi / E_0^\xi \to \Sigma^{2m} E^\xi_{r-1}$ is invertible too. Finally, our map b is constructed as the fourfnew horizontal composite in

$$\begin{array}{ccccccccc}\Sigma^{2m(r+1)-2} E^\xi_0 & \xrightarrow{\theta_r^{-1}} & \Sigma^{2m-2} E^\xi / E^\xi_{r-1} & \to & \Sigma^{2m-1} E^\xi_{r-1} & \xrightarrow{\theta_1^{-1}} & \Sigma^{-1} E^\xi / E^\xi_0 & \to & E_0^\xi \\ & & \uparrow & & & & \uparrow \\ & & \Sigma^{2m-2} E^\xi & & & & \Sigma^{-1} E^\xi \\ & & \uparrow & & & & \uparrow \\ & & \Sigma^{2m-2} E^\xi_{r-1} & & & & \Sigma^{-1} E^\xi_0,\end{array}$$

where the consecutive unlabeled maps form cofiber sequences.

We can now show

Main theorem: On Bousfield classes, we have $\langle E^\xi_0 \rangle = \langle E^\xi \rangle \vee \langle b^{-1} E_0^\xi \rangle$.

To show that the inequality $\langle E^\xi_0 \rangle \le \langle E^\xi \rangle \vee \langle b^{-1}E_0^\xi \rangle$, we just have to use the cofibers in the definition of b; fix a map $f: X \to Y$ and recall that a map $t: T \to Y$ has null composite with the cofiber of f, $Y \to C(f)$, if and only if t lifts to a map to X. To show the other direction, recall that $b^{-1} E_0^\xi$ is formed as the colimit of the sequence of iterates of b and that smashing commutes with colimits, so we can induct up.

At this point, our two remaining goals are:

Subproject 1: $\langle b^{-1} E^\xi_0 \rangle$ is trivial, under certain conditions. Subproject 2: Our original situation involving $F_k$ and so forth fit these conditions.

We’ll address at least one of those next time.

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