MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider two (distinct) octahedral diagrams i.e. diagrams mentioned in the octahedron axioms of triangulated categories (with four 'commutative triangular faces' and four 'distinguished triangular faces'). Is it true than one can extend to a morphism of such diagrams: 1. any morphism of one of the 'commutative faces' of the octahedron 2 any morphism of the pair of morphisms whose target is the upper vertex of the octahedron (i.e. a morphism of commutative triangles not lying on the faces of the octahedrons)?

Is there any text where I could look for various facts of this sort?

P.S. It seems that the answer is 'no' in general. Having a morphism of 'commutative faces', one can extend it to a morphism of three neighbouring 'triangulated faces'. Thus one obtains morphisms of each of six vertices. Yet (all possible) compositions of edges of the 'first' commutative triangles and the neigbouring distinguished faces do not yield all edges of the octahedron; two of the edges (in the 'lower hat') are missing.

Yet it would be very interesting to know which additional conditions are needed in order for the morphism of the octahedrons desired to exist. I would be deeply grateful for any comments!!

share|cite|improve this question
up vote 7 down vote accepted

It is true in a Heller triangulated category aka $\infty$-triangulated category (although strictly speaking one only needs a 3-triangulation for octahedra) that any morphism between the bases of octahedra (by which I mean the 3 objects and two composable morphisms from which the octahedron is built) extends to a morphism between the octahedra (it is part of the axiomatics).

It turns out that the triangulated categories which turn up in "nature" all satisfy these stronger axioms. For instance the homotopy category of a stable model category is $\infty$-triangulated (a more general statement holds which is Theorem 2 in Maltsiniotis' preprint).

References are M. Künzer. Heller triangulated categories and G. Maltsiniotis. Catégories Triangulées Supérieures

share|cite|improve this answer
Greg, I think this is not exactly as you say. What you say is true for 'exact' octahedra in $\infty$-triangulated categories, which are in practice the octahedra arising from algebraic or topological models, but there are non-exact octahedra too, as Matthias Künzer points in a comment below. – Fernando Muro Feb 5 '11 at 21:31

The identity of commutative triangles does not lift to a morphism of Verdier octahedra in general, for it may happen that there exist two mutually nonisomorphic Verdier octahedra on the same commutative triangle. Cf. . This has been observed already by Amnon Neeman in the 90s (unpublished).

(The situation is better if one uses 3-triangles, i.e. "distinguished octahedra", in the sense of Heller/oo-triangulated categories. Cf. , Lem. 3.2. So there exists a choice of octahedra such that morphisms of commutative triangles lift to morphisms of these particular octahedra.)

share|cite|improve this answer
Yes, that's a nice point; thanks! – Mikhail Bondarko Sep 17 '10 at 23:13

(Note: This does not answer your question. Edit: It still doesn't answer your question, but I am attempting to make my notation less horribly broken, so future readers don't get too confused.)

I'm not entirely sure which parts of the diagram you are considering, so I'll try to fix some notation. We start with maps $X \to Y \to Z$, and my understanding of the octahedral axiom is that by taking cones, we get maps $Y/X \to Z/X \to Z/Y \to Y/X[1]$ that form a distinguished triangle. The objects can be arranged into an octahedral diagram, with alternating triangles that commute or are distinguished, but I don't know which object is the upper vertex.

Since the objects $Y/X$, $Z/X$, and $Z/Y$ together with the maps between them are only defined up to nonunique isomorphism, I don't see why one should get a map of octahedra from the given data. I have heard that one needs an extra "octahedron-gluing" axiom (or some underlying stable infinity or model category) to pass from a commuting diagram $(X \to Y \to Z) \to (X' \to Y' \to Z')$ to a map of octahedra. I think the problem is that the "commuting map of triangles" axiom produces two maps $Z/X \to Z'/X'$ from the commuting squares $(X \to Z) \to (X' \to Z')$ and $(Z/Y[-1] \to Y/X) \to (Z'/Y'[-1] \to Y'/X')$, and there is no reason for those two maps to coincide.

You might be able to construct a counterexample using information on page 6 in Toën's paper, where there is a list of problems with localization.

share|cite|improve this answer
The 'upper vertex' is $Y$. – Mikhail Bondarko May 13 '10 at 19:18
Thank you for the reference; it's interesting! Yet I doubt that 'gluing' and 'localization' has much to do with my question. These matters usually arise when one wants to relate distinct triangulated categories. Yet in my question I have a single triangulated category. The proof of my 'conjecture' seems to be quite easy. Having a morphism of 'commutative faces', one can extend it to a morphism of three neighbouring 'triangulated faces'. Thus one obtains morphisms of each of six vertices! And it seems that one really gets a morphism of octahedra this way. – Mikhail Bondarko May 13 '10 at 19:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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