MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $T$ is a triangulated category, then the formalism of $t$-structures gives a way to find abelian subcategories inside. You're supposed to find two strictly full subcategories, $T^{\le 0}, T^{\ge 0}$ so that

1) $T^{\ge 0}[-1] \subset T^{\ge 0}$ and $T^{\le 0}[1] \subset T^{\le 0}$

2) There are no nonzero maps from an object in $T^{\le 0}[1]$ to an object in $T^{\ge 0}$

3) For any object $K$ there's a triangle $A \to K \to B$ with $A \in T^{\le 0}[1]$ and $B \in T^{\ge 0}$.

But if $T$ is periodic, say $[n]$ is the identity, this does not seem like a recipe for success.

How do you find abelian subcategories of a periodic triangulated category?

share|cite|improve this question
t-structures are as good for periodic triangulated categories as for non-periodic ones. In my opinion not very good. – Fernando Muro May 20 '13 at 13:19
for $M$ some module category, I can recover $M$ from $D^b(M)$ by a t-structure. How would I recover $M$ from just the periodic complexes? – Vivek Shende May 20 '13 at 16:43
"t-structures are as good for periodic triangulated categories as for non-periodic ones" -- are they? I thought for an object $X$ belonging to the heart of a t-structure one was supposed to have $Hom(X,X[n])=0$ for all $n<0$. If the t-structure is periodic, this would imply $Hom(X,X)=0$, hence $X=0$. – Leonid Positselski May 20 '13 at 17:32
@Leonid, I still think they are. As you show, periodict triangulated categories only have t-structures with trivial heart, but non-periodic ones need not be better in general, that's why I say 'as good as'. – Fernando Muro May 23 '13 at 6:29
@Vivek, I don't quite understand the question in your comment above. Nevertheless, as you probably know, derived categories of abelian categories are very special among triangulated categories and $t$-structures are more relevant for them than for general triangulated categories. My comment on top is just a reflect of my perception that, unless you're working with a very special kind of triangulated categories, $t$-structures may not be so helpful. – Fernando Muro May 23 '13 at 15:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.