I'm learning about Bousfield localizations. For a triangulated category satisfying some axioms, a Bousfield localizations can be described as an idempotent functor $L:D \to D$.

I thought there is a bijection between Bousfield localizations, and localizing subcategories (the kernel of $L$ determines $L$).

But there is a set of localizing subcategories of $D$, and a category of idempotent functors $D \to D$. I guess what I mean is, the idempotent functor can have automorphisms (even the identity functor can have automorphisms)

Is there a way to fix that, in the definition of the localization functor $L:D \to D$? To "rigidify" i.e. add a little more data so that it has no automorphisms?

The next thing I want to understand is, why do Bousfield localizations form a poset, but I got hung up on this. If I am thinking about it the wrong way, let me have it!

  • $\begingroup$ You might be interested to look into Bousfield lattices, in the context of the stable homotopy category. I believe they have also been studied for the derived category of a ring and for the stable module category. A nice reference is Hovey-Palmieri-Strickland's Axiomatic Stable Homotopy Theory $\endgroup$ – David White Jan 31 '16 at 17:36

There's an underived version of the story that might be worth working through first or instead. Namely, for a category $C$, there's a natural correspondence between reflective subcategories of $C$ and idempotent monads on $C$. Now one might similarly be concerned in this case that reflective subcategories don't seem to have automorphisms while idempotent monads do; for example the subcategory $C$ corresponds to the identity $\text{id}_C$ which can have automorphisms.

But the sense in which $\text{id}_C$ has automorphisms is as a functor; this doesn't imply that it has automorphisms as an idempotent monad (although I haven't checked either way). So it may still be the case that for the right definition of Bousfield localization, these still form a set (probably in fact a preorder, which is what reflective subcategories form) and in particular have no interesting automorphisms.

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    $\begingroup$ As you guessed, an idempotent monad only has trivial automorphisms. (This is forced by compatibility with the unit.) $\endgroup$ – Zhen Lin Jan 31 '16 at 13:33
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    $\begingroup$ In fact, it is not hard to show that if $F$ and $G$ are idempotent monads then there exists a map of monads $T: F \to G$ if and only if the reflective subcategory of $G$-local objects is contained in the reflective subcategory of $F$-local objects, in which case there is exactly one such $T$ (uniquely determined by the condition that T agrees with the unit of $G$ on $F$-local objects). $\endgroup$ – Yonatan Harpaz Jan 31 '16 at 14:31

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