Chromatic Homotopy and Telescopic Localization

3 Chromatic Localizations

The moduli stack of formal groups is filtered by the open substacks of formal groups of height $\leq n$. Chromatic localizations are a way to turn this algebraic filtration into a topological one, and their study was pioneered by Doug Ravenel. To talk about them, we will briefly review Bousfield localizations of the category $\Sp $.

Given a spectrum $X$, there is an adjunction

\[\adjunction {L_X}{\Sp }{\Sp _X}{i}\]

such that

• $L_X$ inverts $X$-equivalences: that is morphisms $f$ such that $f\otimes X$ is an equivalence.
• $L_X$ kills (sends to $0$) the $X$-acyclic objects, i.e those objects $Y$ such that $Y\otimes X = 0$.
• $i$ is fully faithful, so $\Sp _X$ is a reflective subcategory of $\Sp $.
• The essential image of $i$ consists of $X$-local spectra, that is objects $Z$ such that there are no nonzero maps from $X$-acyclic objects to $Z$.

The composite $i\circ L_X$ will often be shortened to $L_X$. The unit of the adjunction gives a natural map $Y \to L_XY$, characterized by the fact that it is an $X$-equivalence to an $X$-local object.

The construction $L_X$ doesn’t depend on all of $X$ but rather on the Bousfield class, that is $\langle X\rangle = \{X\text {-acyclic objects}\}$.

We can often break up a Bousfield localization into smaller pieces, and glue them back together.

Lemma 3.1. Suppose $L_E$ preserves $F$-acyclic objects. Then

is a pullback square.

Proof. Let $P = L_EX\times _{L_EL_FX}L_FX$.
- • $P$ is $E\oplus F$ local. Indeed, if $Z$ is $E\oplus F$ acyclic, $P^Z = 0 \times _0 0 = 0$.
- • $X \to P$ is an $E\oplus F$ equivalence. To see it is an $E$-equivalence, after tensoring with $E$ it becomes
  
  \[X\otimes E \xrightarrow {\sim } X\otimes E\times _{E\otimes L_FX}E\otimes L_FX\]
  
  To see it is an $F$-equivalence, by the hypothesis on $L_E$, we learn that the natural transformation $Y \to L_EY$ is an $F$-equivalence. Thus after tensoring with $F$, we get
  
  \[X\otimes F \xrightarrow {\sim } X\otimes F\times _{X\otimes F}X\otimes F\]
□

If $X$ is a type $n$ spectrum, We can invert a $v_n$-self map to get $X[v_n^{-1}]$, which is called the telescope of $X$ and denoted $T(n)$. By the almost uniqueness of $v_n$-self maps, $T(n)$ only depends on $X$. Essentially by the thick subcategory theorem, $\langle T(n)\rangle $ only depends on $n$.

There are two flavors of chromatic localizations that are studied. The first are the telescopic and finite localizations $L_{T(n)}$ and $L_{n}^f:=L_{\oplus _{0}^nT(i)}$. The second are the $K(n)$ and $E_n$ localizations $L_{K(n)}$ and $L_n:=L_{E(n)} = L_{\oplus _0^n K(n)}$. The hope is that we can understand stable homotopy via the towers of localizations

\[ X \to \dots \to L_nX \to L_{n-1}X \to \dots L_1X \to L_0X\]

(and similarly for $L_n^f$ in place of $L_n$).

The two flavors of localizations are related to each other. If $Y\otimes T(n) = 0$, then $X\otimes T(n) \otimes K(n) = 0$, but $T(n)\otimes K(n)$ is a nonzero sum of copies of $K(n)$, so $X\otimes K(n) = 0$. Thus we get factorizations of the natural maps

(-tikz- diagram) ,

An important property of $L_nX$ is that it is colimit preserving:

Theorem 3.2 (Smashing Theorem). $L_nX = L_n\SP \otimes X$

The same is true of $L_n^f$, but it is easier to prove, as will now be explained.

Lemma 3.3. The $L_n^f$-acyclic spectra coincide with $\Ind (\Sp _{\geq n+1})$: that is they are filtered colimits of type $\geq n+1$ spectra.

Proof. It is easy to see that $\Ind (\Sp _{\geq n+1})$ consists of $T(n)$-acyclic spectra; we will show the reverse inclusion. First let $n = 0$, and suppose $X$ is $T(0)$-acyclic. Then there is a cofibre sequence

\[X \to p^{-1}X = X\otimes T(0) \to X\otimes \SP /p^\infty \]

where $\SP /p^\infty $ is the colimit of $\SP /p^n$ over all $n$. Since $X\otimes T(0)$ vanishes, we learn that $X = \Sigma ^{-1}X\otimes \SP /p^\infty $. $X$ is a filtered colimit of finite spectra, and after tensoring with $\SP /p^n$, this becomes a filtered colimit of type $1$ spectra.

Now we can induct on $n$. For example, let $n=1$, and assume that in addition, $X$ is $T(1)$-acyclic. Then there is a cofibre sequence

\[X\otimes \SP /p^n \to X\otimes v_1^{-1}\SP /p^n = X\otimes T(1) \to X\otimes \SP /p^n,v_1^\infty \]

, so since $X\otimes T(1) = 0$, we learn that $X = \Sigma ^{-2}X\otimes \SP /p^\infty ,v_1^{\infty }$, which is in $\Ind (\Sp _{\geq 2})$. □

Remark 3.4. The argument in the above lemma shows that there is a cofibre sequence

\[\Sigma ^{-1-n}\SP /v_0^\infty ,\dots v_{n}^\infty \to \SP \to L_n^f\SP \]

Since $L_n^f$ kills a category that is generated by compact objects, it preserves filtered colimits. It also preserves finite colimits, so $L_n^f$ preserves all colimits. The only colimits preserving endomorphisms of $\Sp $ are given by tensoring, so we learn

Corollary 3.5. $L_n^fX = L_n^f\SP \otimes X$.

The corollary above is one way to see that $L_m^f$ preserves $\oplus _{m+1}^nT(i)$-acyclic objects. Thus we learn from Lemma 3.1:

Corollary 3.6. There is a pullback diagram

Note that the same is true with $K(n)$ replacing $T(n)$ and $L_n$ replacing $L_n^f$ by the smashing theorem. These pullback squares allow one to reduce the study of $L_n^f$ to the study of $L_{T(n)}$.

The exact relation between $L_{T(n)}$ and $L_{K(n)}$ is not known. It was conjectured by Ravenel that there is no difference between the two.

Conjecture 3.7 (Telescope conjecture). The map $L_{T(n)}X \to L_{K(n)}X$ is an equivalence.

This conjecture is known to be true for $n=1,0$, and many believe it to be false otherwise. Nevertheless, so long as we are concerned with rings or finite spectra, the nilpotence theorem implies that $T(n)$ and $K(n)$ behave similarly.

Lemma 3.8. If $R$ is a ring spectrum, $R\otimes T(n) = 0 \iff R\otimes K(n) = 0$.

Proof. Let $V_n$ be a type $n$ spectrum that is an $\EE _1$-ring. For example, one can start with any type $n$ spectrum $X$ and replace it with its endomorphism ring $X\otimes DX$. Let $v_n$ be a central $v_n$-self map, so that $T(n) = V_n[v_n^{-1}]$ is a ring. Then
$\seteqnumber{0}{}{0}$
\begin{align*} & R\otimes T(n) = 0\\ \iff & \text {the unit of $R\otimes T(n)$ is nilpotent}\\ \iff & \text {the unit of $R\otimes T(n)\otimes K(m)$ is nilpotent for all $m$}\\ \iff & \text {the unit of $R\otimes T(n)\otimes K(n)$ is nilpotent}\\ \iff & R\otimes T(n)\otimes K(n) = 0\\ \iff & R\otimes K(n) = 0 \end{align*} Where in the second step, we use the nilpotence theorem, and in the last step we use the fact that $T(n)\otimes K(n)$ is a nonzero free $K(n)$-module. □