I am looking for a result along the lines of the following statement: If $D$ is a triangulated category, and $E_i$ is a set of generators (every object can be obtained up to isomorphism by shifts and cones of objects in $E_i$), then $D$ is equivalent to $D(R)$, for $R = \operatorname{End}(\bigoplus E_i))$. I guess I want to assume also that $D$ is enriched in $k$-vector spaces, and each $Hom$ space is finite dimensional. (I am really thinking about $D^b(\operatorname{Coh}(P^n_k))$.)

I've heard something like this before, but I can't find a precise result. Can someone suggest a reference, or tell me what exactly is true?

If true, I think it is reasonable to send $R$ to $\bigoplus E_i$, and use the existence of a finite free resolution for $R$-mod to build a functor on other objects... but there are a lot of details here, and I could easily miss some crucial hypothesis...



In $D(R)$, $\operatorname{Hom}(R,R[t])=0$ for $t\neq0$, so you definitely need the condition that $\operatorname{Hom}(E_i,E_j[t])=0$ for $t\neq0$.

In my paper "Morita theory for derived categories" (J. London Math. Soc. (2) 39 (1989), no. 3, 436-456) I proved a result for derived categories of rings with a proof along the lines of the idea in your final paragraph, although that paper's very old, and there are much more elegant ways of doing it now. The proof doesn't work for arbitrary triangulated categories, but it can be adapted for $D^b(\operatorname{Coh}(P^n_k))$.

Key words you might like to search for are "tilting complex" or "tilting object".

  • $\begingroup$ Thanks! When you say that there are more elegant modern ways of approaching this, what do you mean? $\endgroup$ – Lorenzo Jun 10 '16 at 0:49
  • $\begingroup$ @AreaMan Later authors, particularly Keller, realized that it's more enlightening to work in more general contexts than derived categories of module categories, such as differentially graded algebras ["Deriving DG categories", Bernhard Keller, Ann. Sci. Ecole Norm. Sup. (4) 27 (1994), no. 1, 63-102] or $A_\infty$-algebras ["Bimodule complexes via strong homotopy actions", Bernhard Keller, Algebr. Represent. Theory 3 (2000), no. 4, 357-376]. $\endgroup$ – Jeremy Rickard Jun 10 '16 at 11:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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