Good morning to everyone, I am writing here because I need to understand better some topics about t-structures on triangulated categories.

Consider this statement: take a, b in $\mathbb{Z}$, $(\mathscr{C}^{\leq 0}, \mathscr{C}^{\geq 0})$ a t-structure on a triangulated category $(\mathscr{C}, [1], \partial)$ and $C$ an object of $\mathscr{C}$.

For any $n$ in $\mathbb{Z}$, $X$ in $Ob(\mathscr{C})$, I call

$$\Delta_{X}^{0} : \tau^{\leq 0}(X) \rightarrow^{T^{\leq 0}(X)} X \rightarrow^{T^{\geq 1}(X)} \tau^{\geq 1}(X) \rightarrow^{h_{X}^{0}} \tau^{\leq 0}(X)[1]$$

the distinguished triangle coming from the axioms of the t-structure.

Is the equality

$$T^{\geq a}(C) \circ T^{\leq b}(C) = 0$$

true if $b \lneq a$?

I tried to prove it, but I did not manage. Thank you in advance.

  • $\begingroup$ Couldn't you reduce to the case $a = 1$, $b = 0$ in which case this is one of the axioms for a t-structure? $\endgroup$
    – JHF
    Jan 14 at 20:43
  • $\begingroup$ Did you check it using induction on a - b? By the definition of the Tn's built by the T0 and T1. (Sorry for my bad English) Anyway I thank @JHF , for the quick hint. $\endgroup$
    – Federico
    Jan 15 at 13:06
  • $\begingroup$ My thought was to use the shift equivalences to shift $b$ to $0$, and then observe that the truncation factors through $T^{\geq 1}$. But I suppose the idea is not that different from yours. $\endgroup$
    – JHF
    2 days ago

I thought this answer.

By the axioms

$\bullet$ $\mathscr{C}^{\leq -1} \subseteq \mathscr{C}^{\leq 0}$

$\bullet$ $\mathscr{C}^{\geq 1} \subseteq \mathscr{C}^{\geq 0}$

$\bullet$ $Hom_{\mathscr{C}}(\mathscr{C}^{\leq 0}, \mathscr{C}^{\geq 1}) \cong 0$

I can prove that

$$\hspace{-130pt} Hom_{\mathscr{C}}(\mathscr{C}^{\leq n}, \mathscr{C}^{\geq m}) \cong 0$$

for any $n$ and $m$ integers such that $n \lneq m$, using induction on $m - n \in \mathbb{N}_{\geq 1}$.

By the relation $$\hspace{-90pt} T^{\geq a}(C) \circ T^{\leq b}(C) \in Hom_{\mathscr{C}}(\mathscr{C}^{\leq b}, \mathscr{C}^{\geq a}) \cong 0$$ I am done.

If there are some mistakes, please let me know.

Anyway, thanks for everyone who read this thread.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.