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Let $T$ be a triangulated category and

$$ X \xrightarrow{u} Y \xrightarrow{v} Z \xrightarrow{w} X[1]$$

an exact triangle (or distinguished triangle). TR 2 implies that then the two rotated triangles

$$ Y \xrightarrow{v} Z \xrightarrow{w} X[1] \xrightarrow{-u[1]} Y[1] $$

and

$$ Z[-1] \xrightarrow{-w[-1]} X \xrightarrow{u} Y \xrightarrow{v} Z $$

are also exact.
Is there any reason/motivation for the appearance of the minus signs for $u[1]$ and $w[-1]$? What is the reason to introduce them? Grows it out of a geometric analogy?

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    $\begingroup$ This axiom just codifies what happens in explicit examples. Consider the case of the derived category of a ring, with the explicit mapping cone construction, and compute what happens. Once you check it on your own, you won't forget it. $\endgroup$ Oct 6, 2021 at 22:04
  • $\begingroup$ @FernandoMuro: maybe I understand what you mean. In derived category of ring (ie the category we obtain after passing from category of chain complexes of modules to to it's homotopy category, we can wlog asume that $Z=Cone(u)$ and that $X[1]$ is a some "kind of suspension" of $X$. And what we want is that in every partial sequence consisting of four consecutive members in the distinguished sequence $$ ... \to X \xrightarrow{u} Y \xrightarrow{v} Z \xrightarrow{w} X[1] \to ...$$ $\endgroup$
    – user267839
    Oct 6, 2021 at 22:55
  • $\begingroup$ is homotopic to $A \xrightarrow{f} B \to Cone(f) \to Suspension(A)$. And if we consider the part $Y \xrightarrow{v} Z \xrightarrow{w} X[1] \xrightarrow{?} Y[1]$ then the explicite calculation must show that it can be only homotopic to $Y \xrightarrow{v} Z \to Cone(v) \to Suspension(Y)$ if functor $?$ in right arrow is $-u[1]$; in other words there is no such homotopy if $?$ was $u[1]$. That's your point, right? $\endgroup$
    – user267839
    Oct 6, 2021 at 22:55
  • $\begingroup$ Well $X[1]$ is already the suspension, and we shouldn't mistake homotopies for isomorphisms in the homotopy category, but apart from that, the idea in your second comment is OK, that's what you can prove. $\endgroup$ Oct 7, 2021 at 6:05

1 Answer 1

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Let $(\mathcal{D},T)$ be an additive category with an autoequivalence $T:\mathcal{D}\to\mathcal{D}$. One could theoretically modify the (TR2) axiom to instead require that $X\xrightarrow{u}Y\xrightarrow{v}Z\xrightarrow{w}TX$ is distinguished if and only if $ Y\xrightarrow{v}Z\xrightarrow{w}TX\xrightarrow{Tu}TZ$ is distinguished, thereby deleting the minus sign. As far as I can tell, up to deletion of certain minus signs in some formulas, nothing would change in the theory and existing results of triangulated categories. Nonetheless, if later we would like to equip $K(\mathcal{A})$—the homotopy category of cochain complexes with terms in an abelian category $\mathcal{A}$—with the structure of a triangulated category, we would then need to define $T=-[1]$, i.e., $T^n=(-1)^n[n]$, because of the way distinguished triangles work in $K(\mathcal{A})$. The (TR2) axiom written in the standard way (with the minus sign) allows to spell $T^n=[n]$ for the rotation functor in $K(\mathcal{A})$.

In one sentence: the minus sign in TR2 is because we want our notation to be comfortable with the primary examples of triangulated categories, $K(\mathcal{A})$ and $D(\mathcal{A})=K(\mathcal{A})[\text{qis}^{-1}]$. One could ask now: well, then why do we define $d_{X^\bullet[n]}=(-1)^nd_{X^{\bullet+n}}$? Without the $(-1)^n$ we could have (TR2) without the minus sign. I answer this question here. Briefly put, we need the minus sign in the formula for $d_{X^\bullet[n]}$ because this is what allows one to define the d.t.'s in $K(\mathcal{A})$ (hence in $D(\mathcal{A})$). Note that, so far, we did set up two definitions: (i) $d_{X^\bullet[n]}$ with some minus sign and (ii) the d.t.'s as specified in the linked answer. It turns out that as a consequence of these choices, a triangle $X\xrightarrow{u}Y\xrightarrow{v}Z\xrightarrow{w}X[1]$ in $K(\mathcal{A})$ is distinguished if and only if $ Y\xrightarrow{v}Z\xrightarrow{w}X[1]\xrightarrow{-u[1]}Z[1]$ is distinguished (as it is shown in the second paragraph of the proof of 014S).

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