Octahedral Axiom

The octahedral axiom (TR4) is the most subtle axiom of triangulated categories. It relates the cones of a composition $g \circ f$ to the cones of $f$ and $g$ individually. The name comes from the octahedral diagram that appears when all the relevant distinguished triangles are drawn.

Statement

Theorem5.4Octahedral Axiom (TR4)

Let $\mathcal{T}$ be a triangulated category. Given morphisms $f : X \to Y$ and $g : Y \to Z$ , suppose we have distinguished triangles:

$X \xrightarrow{f} Y \to Q \to X[1]$ $Y \xrightarrow{g} Z \to R \to Y[1]$ $X \xrightarrow{g \circ f} Z \to S \to X[1]$

Then there exists a distinguished triangle $Q \to S \to R \to Q[1]$ making all the relevant diagrams commute. Geometrically, the four triangles form the faces of an octahedron.

RemarkIntuition

The octahedral axiom says: if $X \xrightarrow{f} Y \xrightarrow{g} Z$ , then the "quotient" $Z/X$ (i.e., the cone of $g \circ f$ ) fits into a triangle with $Y/X$ (cone of $f$ ) and $Z/Y$ (cone of $g$ ). This is analogous to the third isomorphism theorem in abelian categories: $(Z/X) / (Y/X) \cong Z/Y$ .

Examples

ExampleThird isomorphism theorem

In $D(\mathcal{A})$ , for submodules $X \subseteq Y \subseteq Z$ , the octahedral axiom applied to $X \hookrightarrow Y \hookrightarrow Z$ gives the triangle $Y/X \to Z/X \to Z/Y \to (Y/X)[1]$ , which is the derived version of $0 \to Y/X \to Z/X \to Z/Y \to 0$ .

ExampleOctahedral for mapping cones

In $K(\mathcal{A})$ , for chain maps $f : A \to B$ and $g : B \to C$ , the octahedral axiom gives a distinguished triangle $\mathrm{Cone}(f) \to \mathrm{Cone}(g \circ f) \to \mathrm{Cone}(g) \to \mathrm{Cone}(f)[1]$ . This can be verified directly using the explicit cone construction.

ExampleFiltrations and the octahedral axiom

The octahedral axiom ensures that filtered objects have consistent associated graded pieces. If $X$ has a filtration $0 = F_0 \to F_1 \to F_2 \to F_3 = X$ , the octahedral axiom relates the cones $F_1/F_0$ , $F_2/F_1$ , $F_3/F_2$ to $F_2/F_0$ and $F_3/F_1$ .

ExampleComposition of exact functors

The octahedral axiom ensures that the composition of two exact functors behaves correctly with respect to cones: the cone of a composition is related to the cones of the factors by a distinguished triangle.

ExampleStable homotopy theory

In the stable homotopy category, the octahedral axiom follows from the fact that cofiber sequences satisfy a "puppe sequence" property. For spaces $X \to Y \to Z$ , the cofibers are related by a cofiber sequence.

ExampleVerification in K(A)

The octahedral axiom can be verified explicitly in $K(\mathcal{A})$ by constructing the required triangle using mapping cones. The four distinguished triangles form an "octahedron" where opposite faces commute.

ExampleDG enhancement makes TR4 automatic

In a pretriangulated DG category, the octahedral axiom is a theorem, not an axiom: it follows from the functoriality of the cone construction at the DG level. This is one advantage of DG enhancements over the bare triangulated structure.

ExampleSemi-orthogonal decompositions

The octahedral axiom is used to prove that semi-orthogonal decompositions are well-behaved under refinement: if $\mathcal{T} = \langle \mathcal{A}, \mathcal{B} \rangle$ and $\mathcal{B} = \langle \mathcal{C}, \mathcal{D} \rangle$ , then $\mathcal{T} = \langle \mathcal{A}, \mathcal{C}, \mathcal{D} \rangle$ .

ExampleWeight structures

The octahedral axiom plays a role in the theory of weight structures (the "dual" of t-structures), ensuring that weight filtrations are compatible with composition.

ExampleLocalization sequences

The octahedral axiom is needed to show that localization sequences $\mathcal{S} \to \mathcal{T} \to \mathcal{T}/\mathcal{S}$ behave well: the localization of a composition of localizations is the expected localization.

ExampleK-theory of triangulated categories

The octahedral axiom ensures that the Grothendieck group $K_0(\mathcal{T})$ is well-defined: the relation $[Y] = [X] + [Z]$ from a triangle $X \to Y \to Z \to X[1]$ is consistent with composition.

ExampleVerdier duality

The octahedral axiom is used in proving properties of Verdier duality on $D^b_c(X)$ : the duality functor $\mathbb{D} = R\mathcal{H}om(-, \omega_X)$ preserves the octahedral relations.

RemarkIs TR4 redundant?

It is an open question (related to a conjecture of Amnon Neeman) whether the octahedral axiom follows from the other axioms in certain settings. In practice, every known triangulated category satisfies TR4, and it is automatically satisfied by categories arising from DG or stable $\infty$ -categorical enhancements.