I begin by saying that while I understand what a triangulated / derived category is pretty well, I know nothing about Higher Algebra stuff and not even $\infty$-categories.

I've heard some people say that stable $\infty$-categories are a "more natural" point of view to derived categories. Can anyone explain in simple terms (given my background) to me why this is so?

I study algebraic and arithmetic geometry but all possible motivations are welcome.


1 Answer 1


I already answered some version of this question in this answer, but let me try to expand a bit on the concrete advantages in mathematical practice. For understanding the following you need to take on faith that ∞-categories exist and have roughly the same properties as ordinary categories (which in fact are just a special example of ∞-categories).

The derived $\infty$-category of a scheme

One of the biggest advantages of stable ∞-categories compared to triangulated categories is that they work well in families. This already should be appealing to an algebraic geometer. But it turns out that this makes constructing stable ∞-categories easier than triangulated categories (contrary to what you might be expecting). For example let me sketch a construction of the symmetric monoidal ∞-category of a scheme (you can easily adapt it to algebraic spaces/stacks etc..)

First you need to do the affine case: the derived category of a ring. There are many ways of doing it. The most concrete is probably taking the 1-category of chain complexes and inverting quasi-isomorphisms (in the ∞-categorical sense) $$\mathscr{D}(R):=\operatorname{Ch}(R)[q.iso^{-1}]$$ Then you need to prove that it is functorial, symmetric monoidal etc. I personally favour another (equivalent) definition, which uses more technology but it makes all properties follow formally. I am taking as the definition of the derived ∞-category the stabilization of the animation of the category of finitely generated projective $R$-modules $$\mathscr{D}(R):=\operatorname{Sp}(\mathcal{P}_\Sigma(\operatorname{Proj}_R))\,.$$ Then $\mathscr{D}(R)$ has all the properties we want because $\operatorname{Proj}_R$ does. Moreover you can still show easily that $\mathscr{D}(R)$ is localization of the category of chain complexes at quasi-isomorphisms by using a clever trick, so you lost nothing in concreteness.

Our next step is to show that $\mathscr{D}(-)$ as a functor on affine schemes satisfies Zariski descent. Concretely it boils down to showing that the following square of stable ∞-categories is cartesian $$\require{AMScd} \begin{CD} \mathscr{D}(R) @>>> \mathscr{D}(R[1/f])\\ @VVV @VVV\\ \mathscr{D}(R[1/g]) @>>> \mathscr{D}(R[1/fg])\end{CD}$$ and you can prove this is true again because it is true for $\operatorname{Proj}_R$ (where it is elementary).


This is one of the amazing advantages of stable ∞-categories compared to triangulated categories: you can glue their objects. Using this now it is clear how to define the derived $\infty$-category of a scheme $X$ $$\mathscr{D}(X)=\lim_{\operatorname{Spec}R\subseteq X} \mathscr{D}(R)$$ where the limit is taken over the poset of open affine subsets of $X$. That is, we're saying that an object of $\mathscr{D}(X)$ is just the collection of an object of $\mathscr{D}(R)$ for each open affine subscheme plus suitable gluing data. Now by formal properties of limits this is automatically a symmetric monoidal $\infty$-category: we can take tensor products of elements in $\mathscr{D}(X)$ by taking them in any affine open and gluing back the resulting objects. If all you have are triangulated categories, constructing the symmetric monoidal structure on $h\mathscr{D}(X)$ is highly non-trivial.

Stability is a property

One great advantage of stable $\infty$-categories is that stability is a property, not a structure. To construct a triangulated category it's not enough to construct the category: you also have to come up with a shift functor, and a family of exact triangles etc. In stable ∞-categories you don't have to worry about that: you just construct a certain ∞-category and then check that, say the (canonically defined) functor $\Omega$ is an equivalence. This has advantages because for example it's clear what should be a stable symmetric monoidal ∞-category: it's just a symmetric monoidal ∞-category which is stable and such that the tensor product is exact in each variable. Try instead to come up with the notion of symmetric monoidal triangulated category. I strongly suspect you would not come up with all the required axioms (there is a compatibility between the shift and the tensor product which has some tricky signs).

Homotopy (co)limits

Triangulated categories rarely have colimits. They have direct sums and usually little more. Stable ∞-categories instead have all finite limits and colimits (and most of those you'll encounter in practice have all small limits and colimits). This gives you a huge flexibility in working in them. For example you can say that you can reconstruct the sheaves (say on a space $X$) from a sheaf on a neighborhood of a closed subset of $Z$ and a sheaf on the complement, plus some gluing data. This is classical for sheaves of abelian groups, but it gets tricky (although not impossible) to state for derived categories of sheaves. For derived ∞-categories, however, the same statement as for sheaves of abelian groups work, with the same proof.

Algebraic K-theory

A powerful motivation for me is that you cannot define the higher algebraic K-theory of a triangulated category. There are stable ∞-categories that have the same underlying triangulated category but not the same higher algebraic K-theory. As a person that finds algebraic K-theory very interesting, this would be enough for me to not want to work with triangulated categories altogether


You can actually have sheaves with values in an ∞-category. It makes perfect sense, say, to say that algebraic K-theory gives you a Zariski (or Nisnevich) sheaf of spectra on the category of schemes. This ended up allowing wonderful computations that would have been hard to do without stable ∞-categories. Perhaps not impossible, but most of the older work in the same vein used secretly something like dg-categories which has many of the same features (although not the same versatility), and in any case not triangulated categories.

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    $\begingroup$ Thanks. OK, I'm convinced! Although, you touch on this briefly, could you say a bit more about the relation to dg-categories? I had the impression that they addressed some of the same issues. $\endgroup$ Commented May 29, 2021 at 14:37
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    $\begingroup$ @DonuArapura I'm loathe to lengthen this already long answer, so I'll add a couple of things as a comment. From a modern perspective pretriangulated dg-categories are just $\mathbb{Z}$-linear stable ∞-categories (that is to say, stable ∞-categories $\mathcal{C}$ together with a "multiplication" $\operatorname{Perf}_{\mathbb{Z}}\times \mathcal{C}\to \mathcal{C}$). Therefore they are strictly less general than stable ∞-categories, although of course sometimes they are the right tool for the job. But I would argue that the proper home of both concepts (cont.) $\endgroup$ Commented May 29, 2021 at 15:31
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    $\begingroup$ (cont.) is the setting of ∞-categories, where you can really do all these homotopical constructions "as if you were doing ordinary category theory" and everything "just works™". Therefore I find more convenient to work with stable ∞-categories and using $\mathbb{Z}$-linear (or $\mathbb{Q}$-linear, etc.) as needed. Especially since some important maps are not $\mathbb{Z}$-linear (among which there's the "Tate-valued Frobenius map" $R\to R^{tC_p}$, which has been important recently in work by Scholze et al) $\endgroup$ Commented May 29, 2021 at 15:33
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    $\begingroup$ @DonuArapura Denis's response above is very good, but let me add one reason why you might choose to work with $\infty$-categories over dg-categories even if you cared only about characteristic zero (derived) objects. Consider the problem of defining the derived category of quasi-coherent sheaves on a derived scheme $S$. One way to do so is to take the limit of $\mathscr{D}(R)$ over all affine (derived) schemes $\operatorname{Spec} R$ mapping to $S$. This index category is an (unstable) $\infty$-category; in particular, it is not a dg category. (cont.) $\endgroup$
    – dhy
    Commented May 29, 2021 at 17:23
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    $\begingroup$ So if you want to define $\mathscr{D}(S)$ as a dg-category, you need to first define what it means to take a limit of dg-categories over an index $\infty$-category. It's pretty awkward to mix dg and $\infty$ in this way, and it's easier to just work $\infty$-categorically from the beginning. $\endgroup$
    – dhy
    Commented May 29, 2021 at 17:27

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