Notation: Let $\mathcal T$ be a triangulated category, and let $\mathcal E$ be a full subcategory of $\mathcal T$. I write $\langle \mathcal E \rangle$ to indicate the smallest strictly full triangulated subcategory of $\mathcal T$ which contains $\mathcal E$.

Now, assume we are given triangulated categories $\mathcal T= \langle \mathcal E \rangle$ and $\mathcal T' = \langle \mathcal E' \rangle$, such that the "generating subcategories" $\mathcal E$ and $\mathcal E'$ are equivalent (or even isomorphic). Is it true that $\mathcal T$ is equivalent to $\mathcal T'$? Or, if not in general, is it true under some reasonable assumptions?


2 Answers 2


I don't think that there is any reasonable context in which one would expect this to be true. For example, you can take $\mathcal{T}$ to be the category of spectra and $\mathcal{T}'$ to be the derived category of $\mathbb{Z}$, then put $\mathcal{E}=\{S^0\}$ and $\mathcal{E}'=\{\mathbb{Z}\}$. Then $\mathcal{E}\simeq\mathcal{E}'$ but $\mathcal{T}\not\simeq\mathcal{T}'$.

If you want to insist that $\mathcal{E}$ is closed under (de)suspension, or regard $\mathcal{T}$ as a category enriched over graded abelian groups, then the above example breaks down, but there are other counterexamples. I think you can get some by considering the stable module categories of nonisomorphic $p$-groups with isomorphic mod $p$ Tate cohomology, such as $C_{p^2}$ and $C_{p^3}$.

  • $\begingroup$ Is it true that $\mathrm{Aut}_{\mathcal{T}}S^0 = \mathrm{Aut}_{\mathcal{T}'}\mathbb{Z}$? $\endgroup$
    – Leo Alonso
    May 7, 2014 at 10:54
  • $\begingroup$ Yes, provided you regard $\mathcal{T}$ and $\mathcal{T}'$ as ordinary categories and so consider only maps of degree zero. Then both endomorphism rings are just $\mathbb{Z}$ and so both automorphism groups are just $C_2$. In fact, even if we consider graded maps then the automorphism groups are the same, although the endomorphism rings are not. $\endgroup$ May 7, 2014 at 10:59

Maybe someone who knows more will come around and answer this in more detail, but I believe the short answer is:

No. This is why triangulated categories belong in the ashbin of history, and we should have been using dg-categories all along.

The example you should have in mind is essentially the minimal one: consider the triangulated category of compact dg-modules over two dg-algebras $A$ and $B$ which are not quasi-isomorphic but have isomorphic cohomology. These at least shouldn't be the same as triangulated categories (maybe someone can helpfully supply a proof), but $A$ and $B$ themselves are generating subcategories which are equivalent from the triangulated perspective, since the cohomology of $A$ and of $B$ are the relevant Ext-spaces.

For dg-categories on the other hand, this sort of property is essentially built-in from the start.

  • 5
    $\begingroup$ "the ashbin of history" is, perhaps, a little bit of an exaggeration... $\endgroup$
    – Leo Alonso
    May 7, 2014 at 10:41
  • 4
    $\begingroup$ I disagree completely with Ben's characterization of the role of triangulated categories. Of course, this may be a question of taste, but for me, triangulated categories are objects of study, while dg-categories are only instruments. $\endgroup$
    – Sasha
    May 7, 2014 at 11:14
  • $\begingroup$ @Sasha Obviously, the "ashbin" bit was something of a joke, but that seems like an oddly dismissive attitude toward dg-categories (and thus by extension $A_\infty$ categories). What makes them a fundamentally more interesting object to compensate for the technical problems like the one exemplified by this question (see, for example, section 2.2 of ens.math.univ-montp2.fr/~toen/swisk.pdf)? $\endgroup$
    – Ben Webster
    May 7, 2014 at 21:22
  • $\begingroup$ Hopefully a longer lasting link for the previous comment perso.math.univ-toulouse.fr/btoen/files/2012/04/swisk.pdf $\endgroup$ Dec 2, 2019 at 22:03

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