9
$\begingroup$

Let $\mathcal C$ be an accessible $\infty$-category, and let $ho(\mathcal C)$ be its homotopy category. I can think of two "trivial" reasons for $ho(\mathcal C)$ to be accessible:

  1. $ho(\mathcal C) = \mathcal C$;

  2. $ho(\mathcal C)$ is small with split idempotents.

Otherwise, I am aware of very few examples where $ho(\mathcal C)$ is accessible. Indeed, given that $ho(Spaces)$ is very far from accessible, I think I should expect that it is quite rare for $ho(\mathcal C)$ to be accessible.

However, I know of one interesting class of examples where $ho(\mathcal C)$ is "nontrivially" accessible. Let $k$ be a field.

  • Write $D(k)$ for "derived $\infty$-category of $k$", i.e. the category of chain complexes over $k$ localized ($\infty$-categorically) at the quasi-isomorphisms. This is a presentable $\infty$-category.

  • Then $ho(D(k))$ is the usual derived category of $k$, which is equivalent to the usual 1-category of graded $k$-vector spaces (though of course the triangulated structure is different), and so is obviously accessible (locally presentable, in fact).

This one class of examples has me second-guessing my expectation that taking homotopy categories rarely preserves accessibility.

Questions:

  1. What are some other examples of accessible $\infty$-categories $\mathcal C$ such that $ho(\mathcal C)$ is also accessible (which do not satisfy (i) or (ii) above)?

  2. Given such an example, is the functor $\mathcal C \to ho(\mathcal C)$ accessible (i.e. does it preserve $\kappa$-filtered colimits for some $\kappa$?)

  3. Can we (partially) characterize this condition, giving necessary and / or sufficient conditions for $ho(\mathcal C)$ and $\mathcal C \to ho(\mathcal C)$ to be accessible?

  4. Is there anything to be said in particular about the case where $\mathcal C$ is stable and presentable, or even more particularly the case where $\mathcal C$ is the derived $\infty$-category of a ring?

More broadly, at the moment it seems very mysterious to me that $ho(D(k))$ comes out to be accessible. I'd appreciate any perspective which makes this fact look less mysterious.

$\endgroup$
  • 4
    $\begingroup$ Give a look at Example 5.3 in Generalized Brown representability in Homotopy categories by Rosicky. The author shows that $\mathsf{Ho}(\text{SSet}_n)$ is accessible. $\endgroup$ – Ivan Di Liberti Feb 12 at 21:58
  • 1
    $\begingroup$ @IvanDiLiberti Thanks! Just to be clear, this is not the homotopy category of $n$-truncated spaces as one might naively guess (and the homotopy category of $n$-truncated spaces is not accessible for $n \geq 1$ because idempotents do not split -- the counterexample appearing in HTT 4.4.5.19, which I think is due to Freyd, uses only 1-truncated spaces). $\endgroup$ – Tim Campion Feb 12 at 22:29
  • 1
    $\begingroup$ Not directly related to the question, but a paper you might be interested, with a similar line of inquiry (for combinatoriality) is: arxiv.org/abs/1702.00240v2 $\endgroup$ – David White Feb 15 at 15:12
  • $\begingroup$ @DavidWhite Thanks! $\endgroup$ – Tim Campion Feb 16 at 1:24
  • $\begingroup$ @IvanDiLiberti Does your criterion here show, perhaps, that the derived category of a ring $R$ is not concrete (and hence not accessible) unless $R$ is a division ring? If so, that would be very interesting. $\endgroup$ – Tim Campion Feb 16 at 1:26
2
$\begingroup$

Here's something we can say which addresses a large class of examples.

Claim: Let $R$ be a commutative ring which is not zero-dimensional (i.e. there exists $r \in R$ which is not a unit or a zero-divisor). Then the derived category $ho(D(R))$ of left $R$-modules is not concrete (i.e. does not admit a faithful functor to $Set$) and in particular is not accessible.

The proof is a straightforward generalization of Freyd's argument that the homotopy category of spaces is not concrete, and in particular is robust enough that "$D(R)$" could be interpreted to have various "boundedness" conditions if one likes.

Beginning of Proof of Claim: Pick $r \in R$ which is neither a unit nor a zero-divisor. If $M$ is a left $R$-module, define the submodule $r^\alpha M$ for any ordinal $\alpha$ inductively by $r^0 M = M$, $r^{\alpha+1} M = rr^\alpha M$, and taking intersections at limit ordinals. Note that if $\phi: M \to N$ is any $R$-module map, then $\phi(r^\alpha M) \subseteq r^\alpha N$ for any $\alpha$.

Let $\alpha$ be an ordinal. For each $n \in \mathbb N$, let $W_\alpha^{(n)}$ be the set of strictly increasing sequences of ordinals $\alpha_0 < \alpha_1 < \dots < \alpha_m$ where $m \leq n$ and $\alpha_m \leq \alpha$. Let $F_\alpha^{(n)} = R\{W_\alpha^{(n)}\}$ be the free left $R$-module on generators given by $W_\alpha^{(n)}$, and define $M_\alpha^{(n)} = F_\alpha^{(n)} / K_\alpha^{(n)}$, where $K_\alpha^{(n)}$ is generated by those elements of the form $[\alpha_1,\dots,\alpha_n] - r[\alpha_0, \alpha_1, \dots, \alpha_n]$ (when $n = 0$ we interpret this to mean that $r[\alpha_0] \in K_\alpha^{(n)}$).

Lemma: For $n \in \mathbb N$, the map $M_\alpha^{(n)} \to M_\alpha^{(n+1)}$ is injective.

Proof: This is straightforward, using the fact that $r$ is not a right zero-divisor.

Define $M_\alpha = \cup_{n \in \mathbb N} M_\alpha^{(n)}$. We have a natural filtration $M_\alpha^{(0)} \subseteq \cdots \subseteq M_\alpha^{(n)} \subseteq \cdots M_\alpha$, and the associated graded is naturally identified with $\oplus_{n \in \mathbb N} \oplus^{ W_\alpha^{(n)} \setminus W_\alpha^{(n-1)}} (R/(Rr))$.

Lemma: Pick a set of "canonical" coset representatives in $R$ of the nonzero elements of $R/(Rr)$. Then every element of $M_\alpha$ may be written uniquely in the form $\sum_i r_i w_i$ where the $r_i$ are canonical coset representatives and the $w_i$ are distinct words of some $W_\alpha^{(n)}$.

Proof: The existence of such a representation is basically obvious. Suppose that two such representations denote the same element of $M_\alpha^{(n)}$, where $n$ is minimal: $\sum_i r_i w_i = \sum_j s_j v_j$. Then they have the same image in the associated graded, and so the $w_i$'s of length $n$ match up with the $v_j$'s of length $n$; since the choice of coset representatives has been normalized, their coefficients in fact coincide. Deleting these words, we get an identification of represntations in $M_\alpha^{(n-1)}$, contradicting the minimality of $n$.

Lemma: For all ordinals $\beta$, $r^\beta M_\alpha$ consists of those elements whose canonical representation as chosen above contains only words $[\alpha_0,\dots,\alpha_n]$ where $\alpha_0 \geq \beta$.

Proof: Using the canonical representation, this is now an easy transfinite induction. (Here we use the fact that $R$ is commutative, though.)

Conclusion of Proof of Claim: Note that $r^\alpha M_\alpha = R/(Rr) \neq 0$ canonically, because $r$ is not a right unit. Let $f_\alpha$ be the map $R/(Rr) = r^\alpha M_\alpha \subseteq M_\alpha$. Then for $\alpha < \beta$, there is no factorization of $f_\beta$ through $f_\alpha$. Pick $d \in \mathbb Z$ such that $(R/r)[d]$, each $M_\alpha[d]$, and the fiber $(M_\alpha/f_\alpha)[d-1]$ of $f_\alpha[d]$ are all in whichever flavor of $ho(D(R))$ we are working with. Then the maps $f_\alpha[d]$ are a proper class of pairwise-inequivalent weak cokernels of their fibers in $ho(D(R))$. But as Freyd shows, a proper class of pairwise-inequivalent weak cokernels out of a fixed object in a pointed category imply that the category is not concrete.


So the outline, as in Freyd's proof, is to use the fact that in $ho(D(R))$, basically every map is a weak cokernel (in fact, all maps are if we're working in the unbounded derived category).

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Here's a further generalization:

Claim: Let $\mathcal T$ be a triangulated category and with a $t$-structure such that $\mathcal T^\heartsuit$ is has coproducts, which are exact. Suppose there exists $M \in \mathcal T^\heartsuit$ and a monic endomorphism $M \overset r \rightarrowtail M$ in $\mathcal T^\heartsuit$ which is not an isomorphism, and such that $Hom(M,M/r)$ is an $r$-torsion $End(M)$-module. Then $\mathcal T$ is not concrete, and in particular not accessible.

Proof: For most of the proof, we work in $\mathcal T^\heartsuit$. Similar to before, for every ordinal $\alpha$, we define $W_\alpha$ to be the set of finite subsets of $\alpha+1$, and set $M_\alpha$ to be the cokernel of the map $\oplus^{W_\alpha} M \xrightarrow{1-s} \oplus^{W_\alpha} M$, where $s$ carries the $[\alpha_0<\dots< \alpha_n]$ -copy of $M$ to the $[\alpha_1<\dots<\alpha_n]$ copy of $M$ via the map $r$. A straightforward transfinite induction shows that the $[\alpha_0<\dots<\alpha_n]$'th structure map $M \to M_\alpha$ is in $r^{\alpha_0} Hom(M,M_\alpha)$, and in particular the $[\alpha]$th structure map is in $r^\alpha Hom(M,M_\alpha)$.

It is straightforward to see that the natural map $M_\alpha \to M_{\alpha+1}$ actually splits, and its cokernel $Q$ sits in an exact sequence $M/r \to Q \to M_\alpha \to 0$, where the map to $M_\alpha$ is obtained by deleting the last letter of each word (which is always $\alpha$ here), and the copy of $M/r$ corresponds to the generator $[\alpha]$ (moreover, the $\alpha$th structure map $M \to M_\alpha$ is nonzero and factors through $M/r$). We claim now that $r^{\alpha+1}Hom(M,M_\alpha) = 0$. This is proved by induction on $\alpha$ using the following

Lemma: Let $M$ be an object of an abelian category and $r: M \to M$ a map. If $0 \to A \to B \to C$ is exact, and if $Hom(M,C)$ is $r$-torsion, then any map $\phi \in r^{\alpha+1} Hom(M,B)$ factors (necessarily uniquely) through $A$, and the factored map lies in $r^\alpha Hom(M,A)$. Dually, if $A \to B \to C \to 0$ is exact and $Hom(M,A)$ is $r$-torsion, then if $r^\alpha Hom(M,C) = 0$ it follows that $r^\alpha Hom(M,B) = 0$, for $\alpha \geq 1$.

We apply the lemma to the exact sequences $0 \to M_\alpha \to M_{\alpha+1} \to Q \to 0$ and $M/r \to Q \to M_\alpha \to 0$ at the successor steps of a transfinite induction to conclude that indeed $r^{\alpha+1} Hom(M,M_\alpha) = 0$.

Now we conclude as before: the $\alpha$th structure map $M \to M_\alpha$ doesn't factor through the $\beta$th structure map $M \to M_\beta$ for $\beta > \alpha$ because then it would lie in $r^\beta Hom(M,M_\alpha) = 0$, and it is in fact nonzero. Since $M\to M_\alpha$ are weak cokernels in $\mathcal T$, they form a proper class of pairwise-inequivalent weak cokernels out of $M$, so that $\mathcal T$ is not concrete.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.