In the paper "Silting mutation in triangulated categories" by Aihara and Iyama, I stumbled upon this nice definition( Definition 2.1) of a tilting/silting subcategory of a triangulated category $\mathcal{T}$. Let $\mathcal{M}$ be a subcategory of a triangulated category $\mathcal{T}$:

(1) We say that $\mathcal{M}$ is a silting subcategory of $\mathcal{T}$ if $Hom_{\mathcal{T}}(\mathcal{M}, \mathcal{M}[>0]) = 0$ and thick$(\mathcal{M}) = \mathcal{T}$.

(2) We say that $\mathcal{M}$ is a tilting subcategory of $\mathcal{T}$ if it is silting and $Hom_{\mathcal{T}}(\mathcal{M}, \mathcal{M}[< 0]) = 0$.

It is clear to me as it is stated on the paper that thick$(\mathcal{M})$ is the smallest thick subcategory of $\mathcal{T}$ containing $\mathcal{M}$. Where a thick subcategory is triangulated subcategory of $\mathcal{T}$ closed under direct summands. What is not clear to me is the notion of $Hom_{\mathcal{T}}(\mathcal{M}, \mathcal{M}[>0]) = 0$ (respectively $Hom_{\mathcal{T}}(\mathcal{M}, \mathcal{M}[\neq 0]) = 0$).

If I'm right $Hom_{\mathcal{T}}(\mathcal{M}, \mathcal{M}[> 0]) = 0$ means that for every $Hom_{T}(X,Y) =0$ for every $X \in \mathcal{M}$ and $Y \in \mathcal{M}[>0]$ where $\mathcal{M}[>0]$ stands for $\Sigma^{i}(\mathcal{M})$ for $i \in \mathbb{Z}^{+}, i \neq 0$ and $\Sigma$ is the translation functor associated to the triangulated category $T$. Is my understanding of this definition right? Also as a first example of tilting subcategory the stalk of complexes of a ring $A$ in the triangulated category $K^{b}(A)$ is considered. Can someone recall me the proper definition of a stalk complex of a ring in a bounded homotopy category of complexes? Thanks for the support!


1 Answer 1


$\operatorname{Hom}_{\mathcal{T}}(\mathcal{M}, \mathcal{M}[>0]) = 0$ means that $\operatorname{Hom}_{\mathcal{T}}\left(X, \Sigma^i(Y)\right) = 0$ for all objects $X,Y$ of $\mathcal{M}$ and all integers $i>0$.

A stalk complex is a complex with only one nonzero term. So "$A$ considered as a stalk complex" means the complex with $A$ in degree zero and with $0$ in all other degrees.

  • $\begingroup$ @Jeremy_Rickard Thank you so much for your answer! But the condition $\operatorname{Hom}_{\mathcal{T}}\left(X, \Sigma^i(Y)\right) = 0$ isnt supposed to be for all objects $X,Y \in \mathcal{M}$ instead of all objects $X,Y$ of $\mathcal{T}$? Seems more intuitive. $\endgroup$
    – Køb
    Sep 13, 2021 at 16:08
  • 1
    $\begingroup$ Yes, sorry, that’s what I meant. I’ll correct it. $\endgroup$ Sep 13, 2021 at 16:10

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