I am familiar with the notion of generators in a locally small category as "separators" of morphisms. More precisely, if $\mathcal{C}$ is a locally small category, we say that a set of objects, $$ \{ G_{i} \in \mathcal{ob}(\mathcal{C}) : i \in I \}, $$ is a family of generators in the sense of Grothendieck if, for any pair of morphisms $f, g: X \rightarrow Y$ in $\mathcal{C}$ with $f \neq g$, there is some $i \in I$ and a morphism $h: G_{i} \rightarrow X$ such that $$ f \circ h \neq g \circ h. $$ In the case that the family of generators only has one element, this is equivalent to the claim that the set valued functor, $$ \text{Hom}(G, -) $$ is faithful. I am familiar with this idea in the context of, for example, Grothendieck categories.

I have recently started studying triangulated categories, and in particular compactly generated triangulated categories (see for example Bondal & Van den Bergh 2003). In that setting, there is a notion of an object being compact: If $\mathfrak{T}$ is a triangulated category (and hence additive) admitting all colimits, we say that an object $A \in \mathcal{ob}(\mathfrak{T})$ is compact if the functor $$ \text{Hom}(X, -) $$ preserves all filtered colimits. This clearly implies that the functor preserves all coproducts (since they are just filtered colimits of finite coproducts).

If $\mathfrak{T}$ be a triangulated category admitting all colimits. Let $\{ G_{i} \in \mathcal{ob}(\mathfrak{T}) : i \in I \}$ be a family of compact objects in the above sense. We say that this is a family of generators in the sense of Verdier if the following condition holds: If $X$ is an object of $\mathfrak{T}$, and if $\text{Hom}(G_{i}, X[n]) \simeq 0$ for all $i \in I$ and all $n \in \mathbb{Z}$, then $X \simeq 0$. Here the notation $[n]$ denotes an application of the translation functor given by the triangulated structure.

Let $\mathfrak{T}$ be a triangulated category admitting all colimits. Let $\{ G_{i} \in \mathcal{ob}(\mathfrak{T}) : i \in I \}$ be a family of compact objects generating $\mathfrak{T}$ in the sense of Verdier. Is this somehow equivalent to the family generating $\mathfrak{T}$ in the sense of Grothendieck?

I am quite familiar with the Grothendieck version of generating, but I have never seen this new sense outside the context of triangulated categories, which makes me think that if they are equivalent in this situation, it will depend heavily on the triangulated structure. Are these seemingly different notions of generators equivalent for compact objects in a triangulated category?

  • $\begingroup$ You should at least add shifts. So the question is : Do we have the equivalence $\{G_i, i\in I\}$ is a family of generators in the sense of Verdier iff $\{ G_i[n], i\in I, n\in\mathbb{N}\}$ is a family of generators in the sense of Grothendieck ? Otherwise, this is clearly false. $\endgroup$ – Roland Jun 9 '18 at 16:28
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    $\begingroup$ There's no such thing as a triangulated category admitting all colimits, unless all triangles are split. You can't ask for more than all small coproducts. The cones and suspensions are only weak colimits-they induce nonunique maps. $\endgroup$ – Kevin Carlson Jun 9 '18 at 21:15

As Kevin Carlson pointed out in his comment, there are virtually no triangulated categories with all colimits. The condition usually considered is the existence of all coproducts. Also, in the context of triangulated categories, an object $X$ is usually called compact if $\text{Hom}(X,-)$ preserves coproducts. In particular, this is the definition used in the paper of Bondal and Van den Bergh referred to in the question.

But in any case, even in simple standard examples of triangulated categories, the two notions of "generator" do not coincide.

In the derived category of $R$-modules, for $R$ some ring, take $G=R$, the free module. Then $G$ is a compact generator "in the sense of Verdier".

However, suppose $0\neq\alpha\in\text{Ext}^n(M,N)$ for some $R$-modules $M$ and $N$ and some $n>0$. Then $\alpha$ can be regarded as a nonzero map $M\to N[n]$ in the derived category. But there is no map $\beta:G\to M$ such that $\alpha\circ\beta\neq 0=0\circ\beta$. So $G$ is not a generator "in the sense of Grothendieck". (Since there are no nonzero maps $G[t]\to M$ for any $t\neq0$, this example still works if you insist that your generating set is closed under shifts as suggested by Roland in his comment.)

The reverse implication does hold. In fact, in any additive category, if $\{G_i\}$ is a set of generators "in the sense of Grothendieck" then for any nonzero object $X$ there must be a nonzero map $G_i\to X$ for some $G_i$ in order to separate the identity and zero maps $X\to X$.


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