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I'm fairly confident that the following assertion is true (but I will confess that I did not verify the octahedral axiom yet):

Let $T$ be a triangulated category and $C$ any category (let's say small to avoid alarming my set theorist friends). Then, the category of functors $C \to T$ inherits a natural triangulated structure from T.

By "natural" and "inherits" I mean that the shift map $[1]$ on our functor category sends each $F:C \to T$ to the functor $F[1]$ satisfying $F[1](c) = F(c)[1]$ on each object $c$ of $C$; and similarly, distinguished triangles of functors $$F \to G \to H \to F[1]$$ are precisely the ones for which over each object $c$ of $C$ we have a distinguished triangle in $T$ of the form $$F(c) \to G(c) \to H(c) \to F[1](c).$$

The main question is whether this has been written up in some standard book or paper (I couldn't find it in Gelfand-Manin for instance). Perhaps it is considered too obvious and relegated to an elementary exercise. Mostly, I am interested in inheriting t-structures and hearts from $T$ to functor categories $C \to T$, and would appreciate any available reference which deals with such matters.

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    $\begingroup$ I don't think that this will work, essentially since the mapping cone does not define a functor from arrows in the homotopy category to the homotopy category. Compare also the discussion here: mathoverflow.net/questions/57904/… You might be interested in the notion of a stable derivator, which essentially assigns triangulated "functor categories" to all diagram shapes in a consistent way: arxiv.org/abs/1112.3840 $\endgroup$ – Bertram Arnold Dec 16 '20 at 22:49
  • $\begingroup$ Thanks @BertramArnold; I had initially hoped that we wouldn't need to say something for all possible diagram shapes, only the C-indexed ones, but it seems that triangulated categories are much stranger than I'd initially thought. $\endgroup$ – Vidit Nanda Dec 16 '20 at 23:18
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The statement is false.

For example, take $C=[1]\times [1]$ to be a square and $\mathcal{T} = h\mathsf{Sp}$ to be the homotopy category of spectra. Now consider the square $X$ with $X(0,0) = S^2$, $X(1,0) = S^1$, and the other values zero, and the other square $Y$ with $Y(1,0) = S^1$ and $Y(1,1) = S^0$. Take the maps $S^2 \to S^1$ and $S^1 \to S^0$ to be $\eta$, and consider the natural transformation $X \to Y$ which is given by multiplication by 2 on $X(1,0)=S^1 \to S^1 = Y(1,0)$.

If this map had a cofiber, then, from the initial to final vertex we would get a map $S^3 \to S^0$. Following the square one direction, we see that we would have some representative for the Toda bracket $\langle \eta, 2, \eta\rangle$. Following the other direction, we factor through zero. But this Toda bracket consists of the classes $2\nu$ and $-2\nu$; in particular, it does not contain zero.

[Of course, this example can be generalized to any nontrivial Toda bracket/Massey product in any triangulated category you're more familiar with.]

Indeed, the Toda bracket is exactly the obstruction to 'filling in the cube' for the natural transformation $X \to Y$.

Anyway- this is one of many reasons to drop triangulated categories in favor of one of the many modern alternatives (e.g. stable $\infty$-categories, derivators, etc.).


As for t-structures and so on, in the land of stable $\infty$-categories these are easy to come by. (See, e.g., Higher Algebra section 1.2.1 and Proposition 1.4.4.11 for various tricks for building these.)

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Dylan Wilson's example is excellent. Let me offer another one, with a more algebraic and "finitistic" flavor.

In my opinion, the simplest triangulated category $\mathcal{T}$ is the category of finite-dimensional vector spaces over a field $k$, with identity suspension (a.k.a. translation) functor and $3$-periodic long exact sequences as exact triangles. (This is actually the only triangulated structure carried by $\mathcal{T}$ up to equivalence.)

Let $C_2$ be the cyclic group of order $2$ (regarded as a category with just one object). Then the functor category $\mathcal{T}^{C_2}$ is the category of finitely generated modules over the group algebra $k[C_2]$. This is the same as the category of finitely generated projective modules over the so-called Auslander algebra $B$ of $k[C_2]$. By a result Freyd, if $\mathcal{T}^{C_2}$ were triangulated then $B$ would be self-injective.

If $k$ has characteristic $2$, $k[C_2]\cong k[\epsilon]/(\epsilon^2)$ is the algebra of dual numbers and $B$ is the endomorphism algebra of the $k[\epsilon]/(\epsilon^2)$-module $k\oplus k[\epsilon]/(\epsilon^2)$. This $B$ is not self-injective. Indeed, since $k$ has characteristic $2$, $k[\epsilon]/(\epsilon^2)$ is not semi-simple, so $B$ has global dimension $2$. If $B$ were self-injective it would have global dimension either $0$ or $\infty$.

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I believe I have a simpler counterexample, which I learned from Paul Balmer's course on tensor-triangular geometry last spring:

Claim The arrow category $\mathcal{T}^{\bullet \to \bullet}$ of a triangulated category $\mathcal{T}$ never has any triangulated structure unless $\mathcal{T} = 0$. Actually, we don't even need $\mathcal{T}$ to be triangulated here: if $\mathcal{T}$ is any additive category such that $\mathcal{T}^{\bullet \to \bullet}$ is triangulated, then $\mathcal{T} = 0$.

Proof: Suppose $\mathcal{T}$ is an additive category such that $\mathcal{T}^{\bullet \to \bullet}$ is triangulated. Let $a$ be an arbitrary object in $\mathcal{T}$, with identity morphism $1_a : a \to a$. Let $t$ denote the unique morphism $a \to 0$. Then $\require{AMScd}$ \begin{CD} a @>1_a>> a\\ @V 1_a V V @VV t V\\ a @>>t> 0 \end{CD} defines a morphism $\alpha : 1_a \to t$ in $\mathcal{T}^{\bullet \to \bullet}$. Note that $\alpha$ is an epimorphism. In any triangulated category, all epimorphisms are split, so let $\beta : t \to 1_a$ be a splitting of $\alpha$ (that is, $\alpha \circ \beta$ is the identity morphism of $t$). Then $\beta$ is a commutative diagram \begin{CD} a @>t>> 0\\ @V f V V @VVs V\\ a @>>1_a> a \end{CD} such that $1_a \circ f = 1_a$ (and $t \circ s = 1_0$). From this and the commutativity of the diagram, we see that $1_a = 1_a \circ f = s \circ t$ factors through $0$. Thus, $a = 0$. Since $a$ was arbitrary, $\mathcal{T} = 0$.

Edit: Of course we could make the statement even weaker: we only really needed that $\mathcal{T}$ has a zero object. But if $\mathcal{T}^{\bullet \to \bullet}$ is triangulated, then $\mathcal{T}$ must be additive, because it embeds as an additive subcategory of $\mathcal{T}^{\bullet \to \bullet}$ via $a \mapsto 1_a$.

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    $\begingroup$ This example is amazing! $\endgroup$ – Fernando Muro Dec 19 '20 at 16:18

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