Here by stable homotopy category I mean the homotopy category of spectra, or maybe just some monogenic, Brown, algebraic, etc. stable homotopy category (in the language of Hovey, Palmieri and Strickland).

It is my understanding that it is still open as to whether or not a given localizing subcategory of the stable homotopy category can in fact be realized as the set of acyclics of some homology theory. Is this indeed the correct framing of the question? It seems to me that a localizing subcategory necessarily induces a localization functor (the left adjoint of the inclusion), but it not known whether or not this functor corresponds to a homology theory. Perhaps this is not correct, please let me know.

Assuming the correct understanding of the above situation, how does this fit into the framework of (co)reflective subcategories? We have Casacuberta et al.'s proof that Vopenka's principle implies that every localizing subcategory is co-reflective. Does that answer the above question (which is discussed in various papers of Hovey, Palmieri, Strickland and others) modulo the necessary set-theoretic assumption, or is this notion of coreflectivity unrelated?

And lastly, does the above paper, in answering questions about colocalizing subcategories, make strong statements regarding cohomological Bousfield classes?


2 Answers 2


I'm not sure if I constitute an expert or this constitutes a real answer but let me try.

If I understand correctly your first question is whether it is open that every localizing subcategory of an algebraic stable homotopy category $\mathcal{S}$ (so it should be triangulated, symmetric monoidal in a way compatible with the triangulation, be compactly generated and the compact objects should be rigid, and monogenic say) is a homological Bousfield class. This is, as far as I know, open - it is not even known in general that there are a set of localizing subcategories rather than a proper class. On the other hand it is known, under more general hypotheses, that there is a set of homological Bousfield classes (see Theorem 3.1 here).

It is possible to produce counterexamples i.e., to build rather simpleminded tensor triangulated categories where one can write down both the localizing subcategories and homological Bousfield classes and check that they don't agree but I don't know of any monogenic counterexample.

It is not true, without some further hypotheses, that the inclusion of every localizing subcategory $\mathcal{L} \to \mathcal{S}$ has a right adjoint (i.e. that every localizing subcategory is coreflective). If $\mathcal{L}$ is generated by a set, and hence by a single object, Brown representability guarantees us a right adjoint, but there is no reason a priori that $\mathcal{L}$ should have a generating set if it is just handed to us. This is where the work of Casacuberta, Gutiérrez, and Rosický comes in. They prove that, assuming Vopěnka's principle, every localizing subcategory of the homotopy category of a combinatorial stable model category is generated by a set of objects and hence is coreflective. This doesn't settle the question of whether all such subcategories are Bousfield classes though (or if there is a set of them) and as Tom points out this question seems to be of a different nature.

For your final question the CGR paper does not make strong statements about cohomological Bousfield classes that I am aware of. Under Vopěnka (with the same combinatorial model hypothesis) they prove that there is a bijection between localizing and colocalizing subcategories given by taking orthogonals. Thus if every colocalizing subcategory were singly generated this would prove that every localizing subcategory is a cohomological Bousfield class. However, they say they were not able to prove this, even under Vopěnka, and in the final part of the paper discuss the analogy with torsion classes where the analogous generation result is known not to hold (already in abelian groups).

  • $\begingroup$ Wow thanks very much Greg. Do you recall meeting in Barcelona? Your answer clears up a lot of my confusion. $\endgroup$ Feb 16, 2012 at 15:41
  • $\begingroup$ Pleasure to be able to help. I do remember meeting you in Barcelona - it's cool that it seems you've been thinking about Bousfield classes since then. $\endgroup$ Feb 17, 2012 at 6:55
  • $\begingroup$ Yeah absolutely. I'm working a little minor project with them right now actually. Hopefully I can actually figure something out! $\endgroup$ Feb 18, 2012 at 16:55

I was thinking of asking a similar question. This won't be an answer, but it might be a slightly useful reframing. Presumably an expert will give us a real answer.

For spectra $X$ and $Y$, let's write $X\perp Y$ if the mapping spectrum $Hom(X,Y)$ is contractible, i.e. if the cohomology $Y^\star X$ is trivial, i.e. if in the stable homotopy category there is no nontrivial map $X\to \Sigma^nY$ for any $n\in\mathbb Z$.

Consider pairs $(\mathcal C,\mathcal L)$ of classes of stable homotopy types such that ($X\in \mathcal C$ iff $X\perp Y$ for all $Y\in \mathcal L$) and ($Y\in \mathcal L$ iff $X\perp Y$ for all $X\in \mathcal C$). Let's call such a thing a matched pair, just to have a term for it.

The best matched pairs are those which have localization functors. This means that for every spectrum $X$ there is a universal example of a map $X\to L$ from $X$ to an element of $\mathcal L$.

Here are some properties that a class $C$ of stable homotopy types may or may not have:

  1. $\mathcal C$ is the first member of a localization pair $(\mathcal C,\mathcal L)$.

  2. $\mathcal C$ is the first member of a matched pair $(\mathcal C,\mathcal L)$.

  3. There is a class $\mathcal E$ of spectra such that $X\in \mathcal C$ iff for every $E\in \mathcal E$ we have $E_\star X=0$.

  4. $\mathcal C$ is closed under homotopy colimits and desuspension.

Of course 1 implies 2. The converse seems like a good question.

The question of whether a given matched pair $(\mathcal C,\mathcal L)$ is a localization pair can be reformulated in a couple of ways. Given a fibration sequence $\star\to C\to X\to L\to \star$, the following three conditions are equivalent:

(a) $L\in \mathcal L$ and $X\to L$ is universal among maps from $X$ to objects in $\mathcal L$.

(b) $C\in \mathcal C$ and $L\in \mathcal L$.

(c) $C\in \mathcal C$ and $C\to X$ is universal among maps to $X$ from objects in $\mathcal C$.

Clearly 2 implies 4 and 3 implies 4.

3 is like saying that $\mathcal C$ is a Bousfield acyclicity class, except that we are allowing $\mathcal E$ to be a proper class instead of a set.

(In the same sense 2 says that $\mathcal C$ is what might be called a cohomological Bousfield class.)

If we take a "What, me worry?" attitude toward set theory, then 3 implies 1: every Bousfield class has a localization functor.

In fact it seems that within the framework of this "Alfred E. von Neumann set theory" 4 implies 1, by the usual argument (carried out rigorously by Bousfield with appropriate hypotheses): To create a localization of $X$, start throwing elements of $\mathcal C$ at it and take cofibers, and keep doing this transfinitely until you are done. The resulting spectrum $L$ should belong the class $\mathcal L$ of all spectra that satisfy $C\perp L$ for all $C\in\mathcal C$. The (co)fiber of the map $X\to L$ is made from $\mathcal C$ by repeated colimits, and so is in $\mathcal C$. This makes $X\to L$ universal. Applying that same construction when $X$ satisfies $X\perp L$ for $L\in \mathcal L$, we find that such $X$ is a retract of an object in $\mathcal C$, therefore is in $\mathcal C$.

So modulo set theory 1=2=4.

But as far as I know the question of whether 3 is strictly stronger is of a different nature.

About (co)reflexivity: 1 implies that $\mathcal C$ is coreflexive. To get a converse you have to assume more, like that the class is closed under desuspension. For example, if $\mathcal C$ consists of all spectra with $\pi_n=0$ for $n<0$ and $\mathcal L$ consists of all those with $\pi_n=0$ for $n\ge 0$ then this pair is not a matched pair, although every $X$ has a universal map $X\to L$ and a universal map $C\to X$ and these form a fibration sequence.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.