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Let $\mathcal{T}$ be a triangulated category having all infinite coproducts(such triangulated category is sometimes said to be cocomplete or satisfying the TR5 axiom). We call an object $G$ tilting if

  1. $G$ is compact, that is, $\mathrm{Hom}_{\mathcal{T}}(G, -)$ preserves all infinite coproducts;
  2. $G$ is a generator, that is, if $\mathrm{Hom}_{\mathcal{T}}(G, X[n])=0, \forall n\in\mathbb{Z}$, then $X=0$;
  3. $G$ is both rigid and corigid, that is, $\mathrm{Hom}_{\mathcal{T}}(G, G[n])=0, \forall n\neq 0$.

Now suppose our triangulated category $\mathcal{T}$ is cocomplete and has a tilting object $G$. I want to ask the following two questions.

Question 1: Is there a triangle equivalence between $\mathcal{T}$ and $\mathsf{D}(\mathsf{Mod}\text{-}\mathrm{End}_{\mathcal{T}}(G))$?

Question 2: If the answer to the first question is false, then I wonder is $\mathcal{T}$ algebriac? By algebraic I mean $\mathcal{T}$ is triangle equivalent to the stable category of some Frobenius exact category.

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  • $\begingroup$ Your question title says "algebraic" - I think a more common use for that term is to mean that a triangulated category is the stable category of an exact category, which is not what you're asking. I mention it because it might be that if there's a mismatch of terminology, that might not be helping your search. $\endgroup$
    – Jan Grabowski
    Commented 2 days ago
  • $\begingroup$ @JanGrabowski Thanks a lot for your such speedy reply! And yes, the algebraic in my original title Is a triangulated category admitting a tilting object algebraic? was a mis use. $\endgroup$
    – Chen Yifan
    Commented yesterday
  • $\begingroup$ It's pretty obviously false: consider a tilting object R in the bounded below/above derived category of R. $\endgroup$
    – Denis T
    Commented yesterday
  • $\begingroup$ @DenisT Thanks. These are indeed good counterexamples. But the problem is, neither the bounded above derived category nor the bounded below one has all infinite coproducts(they have some of infinite coproducts, nut not all of them). So what if I require that our triangulated category $\mathcal{T}$ has infinite coproducts, or even compactly generated with $\{ T[i] | i\in\mathbb{Z} \}$ a generating set of compact objects? $\endgroup$
    – Chen Yifan
    Commented yesterday
  • 2
    $\begingroup$ @ChenYifan The problem is that none of those additional assumptions are stated in your question. If you edit your question accordingly and ask exactly what you want to ask, it'll be easier to give a reasonable answer. $\endgroup$
    – Denis T
    Commented 17 hours ago

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