I would like to know if there is a "moral" reason why in the definition of triangulated categories the "rotation axiom" TR2 requires that we have to add a negative sign to an arrow when we rotate the triangles? What was (presumably Verdier's) initial motivation to pose this negative sign in the arrow $−u[1]$ as stated in Wikipedia or standard literature? Does it come only from a convention or is there a deeper reason for it?
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2$\begingroup$ Perhaps this can help? math.stackexchange.com/questions/4154849/… $\endgroup$– GabrielCommented May 10, 2023 at 14:21
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$\begingroup$ If you don't put it you don't find examples (very very few indeed). $\endgroup$– Fernando MuroCommented May 10, 2023 at 17:50
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2 Answers
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One deeper reason is that this is what you get (no convention) when you have a triangulated category that comes from a stable $\infty$-category.
Ultimately, this boils down to the fact that when you take a loop in some space and reverse its direction, you get minus that loop in $\pi_1$. The connection to this becomes clear when you then examine what the rotation of triangles corresponds to in stable $\infty$-categories. It is worked out in detail in lemma 1.1.2.13 of Lurie's Higher algebra.
More historically, this comes from the example of the triangulated category of chain complexes, which was the motivating example. In this case, if I understand correctly, if you are careful about sign conventions, it is also forced on you. This is worked out in Lawson's note on sign conventions (https://www-users.cse.umn.edu/~tlawson/papers/signs.pdf), in paragraph 11.
answered May 11, 2023 at 8:44
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$\begingroup$ I'm confused a bit. lemma 1.1.2.13 in Lurie's HA follows from 2.10. so essentially if we regard in context of this philosophy the triangulated categories as shadows of stable infty cats, then the sign comes from flipping of explicit choices of two zero objects, right? That looks strange at first glance since as far as I know a zero object is unique. Or does here exactly matter that the uniqueness of zero objects is only given up to iso, so the sign "detects" these as different considered as "bare" objects, even though isomorphic objects? (or do I missing the issue?) $\endgroup$– JackYoCommented May 30, 2023 at 18:59
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1$\begingroup$ I think it's not about the zero object, but about the orientation of the square ! implicit in saying "a square X -> 0,0 -> Y gives a map X -> Omega Y" is the choice of an actual square, and thus of an orientation thereof. If you flip the square, you have to account for this in the morphism X -> Omega Y. Another way of phrasing it: Omega Y is the pullback of 0 -> Y <- 0. If you flip the defining pullback square, this gives you a certain map Omega Y -> Omega Y : this is not the identity (think in terms of spaces), it's the map that flips loops, i.e. -1 $\endgroup$ Commented May 31, 2023 at 8:10
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I cannot speak to Verdier's original motivations, but here is my (rather limited) understanding from back when I was trying to work my way through the basic theory as to why that particular minus sign is there.
Arguably, triangulated categories try to extract and axiomatize a minimal categorical setting that captures homological algebra, but I will focus on the homotopy category $K(\mathcal{A})$ (of some additive category $\mathcal{A}$) as a prototype for triangulated categories rather than $D(\mathcal{A})$.
The minus sign is necessary if you want to express compatibility between triangle rotation and triangle "distinction", i.e. it is reasonable that a rotated distinguished triangle in your to-be-constructed $\mathcal{T}$ should remain distinguished. Indeed it turns out that in $K(\mathcal{A})$ we do in fact have an equivalence: a triangle $$ X \xrightarrow{f} Y \xrightarrow{g} Z \xrightarrow{h} X[1] $$ is distinguished (i.e. isomorphic to some cone triangle) if and only if its rotated triangle is distinguished, but where "rotated" is slightly modified to mean $$ Y \xrightarrow{g} Z \xrightarrow{h} X[1] \xrightarrow{-f[1]} Y[1] $$ This equivalence is worked out in some detail in https://stacks.math.columbia.edu/tag/014S. And so the axioms of triangulated categories copy this property from $K(\mathcal{A})$.
I guess one could then also ask why cone triangles are the model for distinguished triangles. In a sense cone triangles are sort of a prototype for splitting because the associated triangle of every term-wise splitting sequence is isomorphic to some cone triangle, and vice versa (see https://stacks.math.columbia.edu/tag/014L). Moreover, every term-wise splitting sequence in turn gives pre-homologically rise to a connecting homomorphism, which after applying the homology functors becomes the usual connecting homomorphism in the corresponding LES. To sum up, cone triangles seem to capture some of the defining features of homological algebra.
But you should probably wait for a specialist to give a better/deeper answer.
