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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?

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  • $\begingroup$ t-structures are as good for periodic triangulated categories as for non-periodic ones. In my opinion not very good. $\endgroup$ May 20, 2013 at 13:19
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    $\begingroup$ 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? $\endgroup$ May 20, 2013 at 16:43
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    $\begingroup$ "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 $X=0$. $\endgroup$ May 20, 2013 at 17:32
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    $\begingroup$ @Fernando: If the heart is trivial then the t-structure is degenerate and does not give you an abelian subcategory in $T$, so it does not answer the question. $\endgroup$
    – Sasha
    May 23, 2013 at 10:03
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    $\begingroup$ @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. $\endgroup$ May 23, 2013 at 15:54

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