TRIANGULATED CATEGORIES WITH SEVERAL TRIANGULATIONS

Paul Balmer

Abstract. We give a simple algebraic example of a fixed additive category K, with a fixed additive selfequivalence Σ : K → K having arbitrarily many structures of triangulated categories with Σ as suspension.

Introduction. Triangulated categories were introduced in Verdier’s PhD thesis [3] and by Puppe (without Verdier’s key Octahedron Axiom about composition). A triangulated category is a suspended category (i.e. an additive category K with an additive auto-equivalence Σ : K →^{∼} K called the suspension) plus a collection T of triangles which satisfies four well-known axioms, denoted (TR I)-(TR IV) in [3, Def. II.1.1.1, p. 93-94]. It is natural to wonder whether these axioms are intrinsical. Or : Can a given suspended category carry two different triangulations ? Of course, given a triangulated category (K, Σ, T ) with triangulation T , we can define the negative triangulation T^{−} as the class of those triangles (u, v, w)

(Δ) [figure: A; u; B; v; C; w; Σ(A)]

such that (−u, −v, −w) ∈ T . Those two triangulations T and T^{−} are different in general, already for K = K^{b}(ℤ) the category of bounded complexes of abelian groups, up to homotopy. Then, we could ask :

Can a suspended category (K, Σ) admit more than two triangulations ?

(That is : one triangulation and its negative.) Strangely enough, this question seems to remain unclear, even for a few experts of the subject, see for instance [2, Problem 3.4 and Def. 3.2]. The answer to this question is indeed “yes”, and first in a trivial way : Let (K, Σ, T ) be a triangulated category such that T and T^{−} are different. Choose an integer n ≥ 1. Consider the additive category K^{n} = K × ⋯ × K with the obvious suspension. Then K^{n} has at least 2^{n} different triangulations compatible with its suspension. Well, this certainly sounds like cheating, because we basically only used T and T^{−}. So, we would like to build examples, say, with an indecomposable category K. In fact, it is possible to deduce from the results of Sections 16 and 17 of Heller [1] that such an example is given by K the usual topological stable homotopy category, although a picky reader might object that Heller does not consider the Octahedron Axiom in loc. cit. In this short note, we give a simple algebraic example (see Theorem 7).

Acknowledgment. I sincerely thank Bernhard Keller for indicating the reference to Heller’s article and Jeroen Maes for correcting a mistake.

Definition 1. Let (K, Σ) be a suspended category. A global endomorphism α of (K, Σ) will be an endomorphism of the identity functor Id : K → K which commutes with Σ. In other words, it consists of a collection of endomorphisms α_{A} : A → A, for all objects A ∈ K, such that for any morphism f : A → B in K one has α_{B} f = f α_{A}, and such that α_{Σ(A)} = Σ(α_{A}) for any A ∈ K. A global automorphism will be an invertible global endomorphism. A global endomorphism α is pointwise nilpotent if for any A ∈ K, there is an n ∈ ℕ such that (α_{A})^{n} = 0.

Example 2. Let R be a commutative ring and let b ∈ R. Then multiplication by b gives a global endomorphism λ_{b} of K^{b}(R). It is a global automorphism when b is a unit.

Definition 3. Let (K, Σ, T ) be a triangulated category and let α be a global automorphism of (K, Σ). Define the class T_{α} as the collection of those triangles (u, v, w), like in (Δ) above, such that the twisted triangle (u·α_{A} , v , w) belongs to T . This condition is equivalent to any of the following : (α_{B} ·u , v , w) ∈ T , (u, v · α_{B} , w) ∈ T , (u , α_{C} · v , w) ∈ T , and so on : permuting α with the morphisms, and moving α around; this flexibility follows from Axiom (TR I) and Def. 1.

1991 Mathematics Subject Classification. 18E30.

Key words and phrases. Triangulated categories, triangulations.

Proposition 4. Let (K, Σ, T ) and α be as in Definition 3. Then (K, Σ, T_{α}) is a triangulated category.

Proof. [figure: D; v′; w′; B; α v; α u′; E; ·; v′′; ·; α t; A; u; α w; F; ·; w′′; ∃g; C; ·; u′′; ∃f] The proof is straightforward. The Composition Axiom (TR IV), for instance, can be checked by contemplating the diagram on the left (or the reader’s favorite picture instead). The arrows with a small dot are of degree 1. Start with a composition w = v ∘ u. Then choose triangles (u, u′, u′′), (v, v′, v′′), (w, w′, w′′) in T_{α}. The morphism t : E → Σ(C) is as always defined to be t := Σ(u′)v′′. The displayed octahedron is obtained for T from α w = α v ∘ u. It induces the wanted octahedron for T_{α} by “removing α”. For readability, we have dropped the indices of α, forced by the objects. □

Lemma 5. Let R be a commutative ring and b ∈ R^{×} a unit. Assume the existence of r ∈ R such that :

(1) the element r is not a zero divisor and (2) the element r does not divide 1 − b. Consider the category K = K^{b}(R) with its usual triangulation T . Consider the global automorphism λ_{b} of K (see Example 2). Then the triangulations T and T_{λ_{b}} are different.

Proof. Ab absurdo, assume that T_{λ_{b}} = T . In the category K, consider the morphism R → R given by multiplication by r as a morphism of complexes concentrated in degree 0. Let C(r) be its cone with the usual morphisms i : R → C(r) and p : C(r) → Σ(R). The triangle (b · r, i, p) is then exact. By (TR III), there must exist a morphism h : C(r) → C(r) which makes the following diagram commute :

[figure: R; r; R; i; b; C(r); p; ∃; h; Σ(R); R; b·r; R; i; C(r); p; Σ(R).]

The morphism h is characterized by two elements x, y ∈ R such that r · x = y · r which forces x = y by hypothesis (1). The commutativity (up to homotopy !) of the above diagram implies the existence of e, f ∈ R such that b = x + r · e and 1 = x + f · r. This gives 1 − b ∈ rR which contradicts (2). □

Example 6. Of course T_{−Id} = T^{−}. The Lemma shows that T^{−} ≠ T for K^{b}(ℤ) as claimed above.

Theorem 7. There exists a suspended category (K, Σ) which carries infinitely many different triangulations. Moreover, there exists such a (K, Σ) which cannot be decomposed as (K_{1}, Σ_{1}) × (K_{2}, Σ_{2}) with K_{1} and K_{2} non-zero.

Proof. Let S be a commutative domain with infinitely many units. Let R = S[X] be the polynomial ring with coefficients in S. Consider r = X ∈ R. It certainly satisfies conditions (1) and (2) of the above Lemma for any unit b ∈ S^{×} except for b = 1. Let us write T_{b} for T_{λ_{b}}. It is clear that (T_{b})_{c} = T_{b·c} for any b, c ∈ S^{×}. Therefore T_{b} = T_{c} forces T = T_{b^{−1}c} and thus b^{−1}c = 1 by the Lemma and the above comment. That is : all the triangulations T_{b} for b ∈ S^{×} are distinct.

For the “moreover part”, assume (K, Σ) = (K_{1}, Σ_{1})×(K_{2}, Σ_{2}) then the projection on the K_{1}-summand yields a global endomorphism β of K, see Definition 1. At the object R ∈ K, we have necessarily β_{R} = 0 or 1 − β_{R} = 0, since End_{K}(R) ≃ R and since R is a domain. Let us say β_{R} = 0 for instance. The object R generates K as a triangulated category, this forces β to be pointwise nilpotent (easy induction). But β = β^{2} is an idempotent, so we have β = 0 and thus K_{1} = 0. Similarly if 1 − β_{R} = 0, then K_{2} = 0. □

Problem 8. Is there a suspended category (K, Σ) admitting two triangulations T and T′ such that T′ ≠ T_{α} for any global automorphism α of (K, Σ)?

References.

1. Alex Heller, Stable homotopy categories, Bull. AMS 74 (1968) pp. 28-63.

2. J.P. May, The additivity of traces in triangulated categories, Adv. Math. 163, no. 1 (2001), pp. 34-73.

3. Jean-Louis Verdier, Catégories dérivées des catégories abéliennes, Astérisque 239, (1996).

Paul Balmer, D-Math, ETH Zentrum, 8092 Zürich, Switzerland

   E-mail address: balmer@math.ethz.ch