In the introduction to his paper "Towards a non-abelian $p$-adic Hodge theory", Olsson says that for any ringed topos $(\mathcal{T},\mathcal{O})$ with $\mathcal{O}$ a sheaf of $\mathbb{Q}$-algebras, the category $\mathrm{dga}_{\mathcal{O}}$ of $\mathcal{O}$-dga's has a model category structure, where the weak equivalences are quasi-isomorphisms, and the fibrations are surjections with level wise injective kernel (injective as objects in the category of $\mathcal{O}$-modules).

Now, it seems to me the proof of this fact goes something along the following lines. Since the category of $\mathcal{O}$-modules has enough injectives, then the category of positively graded chain complexes has a similarly defined model category structure. One takes generating sets cofibrations and acyclic cofibrations in this model category of chain complexes, and one applies the 'free algebra functor' from complexes to dga's and the small object argument to get generating sets of cofibrations and acyclic cofibrations in the category of dga's.

My question is the following.

Do we really need to restrict to $\mathbb{Q}$-algebras here? Or will this argument work for any ringed topos $(\mathcal{T},\mathcal{O})$? For example, will the above definitions of weak equivalences and fibrations define a model category structure on the category of sheaves of $\mathbb{Z}/\ell^n$-modules in the étale topos of some scheme?

I can't see where the argument breaks down, but I may not have understood it well enough.

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    $\begingroup$ I know almost nothing about topoi, so I'm writing a comment rather than an answer. I'm not at all surprised that dga$_O$ is a model category in the rational situation, e.g. by analogy to Hinich's model structure on algebras over an operad in DGA. In the operad setting, you run into trouble moving away from characteristic zero basically because you lose Maschke's Theorem from representation theory. See Tyler Lawson's answer to this MO question for more on this: mathoverflow.net/questions/23269/… $\endgroup$ Jan 25, 2013 at 18:46
  • $\begingroup$ Random question: would it make sense to use simplicial algebras instead of dgas? Or are you dealing with unbounded dgas? In derived geometry this is what you do when not over Q. $\endgroup$
    – Jacob Bell
    Feb 24, 2013 at 11:29
  • $\begingroup$ Okay, so it seems that the answer is quite an emphatic no. However, I like the idea of using simplicial dga's instead, thanks for suggesting it! $\endgroup$
    – ChrisLazda
    Mar 6, 2013 at 13:38

1 Answer 1


What you suggest should work. We're going to transport the model structure on the category of chain complexes via the adjunction using the criteria given in this answer: Transporting model structures via adjunctions

(A proof of this transport theorem for our case, where adjunction functor is monadic, can be found, for example, as Lemma 2.3 in this paper of Schwede and Shipley, http://homepages.math.uic.edu/~bshipley/monoidal.pdf)

We have an adjunction between the functors

$$\text{Free}: \mathbf{Ch}_{\mathcal{O}} \rightarrow \mathcal{O}\text{-}\mathbf{dga}$$ and

$$\text{Forget}: \mathcal{O}\text{-}\mathbf{dga} \rightarrow \mathbf{Ch}_{\mathcal{O}}$$

where the first is left adjoint to the second. Since $\mathcal{O}$-modules form a Grothendieck abelian category, there is a combinatorial model structure on $\mathbf{Ch}_{\mathcal{O}}$ with the fibrations and weak equivalences you described. Every object in both categories is small, since they are presentable.

Now I need to show that everything that can be obtained by sequential limits from cobase changing the arrows $\text{Free}(g)$,where $g$ is a (generating) acyclic cofibration in $\mathbf{Ch}_{\mathcal{O}}$, is a quasi-isomorphism. I think this is true, but I'm not sure so I'll include my argument in case there's something wrong with it.

First I claim that the relative tensor product $\text{Free}(D) \otimes_{\text{Free}(C)} (-)$ is exact on chain complexes. Indeed, given an exact sequence of chain complexes $0 \rightarrow X \rightarrow Y \rightarrow Z \rightarrow 0$ I can form the diagram

alt text

All three columns and the top two rows are exact, since free algebras are free as graded modules and the exactness diagrams of chain complexes is determined by exactness as graded objects. Therefore the bottom row is exact (diagram chase or spectral sequence argument.) (The unadorned tensors are over $\mathcal{O}$).

Now I claim that cobase changing a morphism $\text{Free}(C) \rightarrow \text{Free}(D)$ where $C \rightarrow D$ is a monomorphism that's a quasi-isomorphism, gives a quasi-isomorphism. Indeed, we have a convergent spectral sequence $$ \text{Tor}_{p,q}^{H^*F(C)} (H^*F(D), H^*A) \Rightarrow H(F(D) \otimes^{\mathbb{L}}A) $$ But since $F(D)$ is a flat $F(C)$ module this converges to the cohomology of $F(D) \otimes_{F(C)} A$. On the other hand, $$ H^*F(C) \rightarrow H^*F(D) $$ is an isomorphism so the $E_2$-term collapses to an edge, and moreover the edge is just $H^*A$. The edge homomorphism is then an isomorphism, but the edge homomorphism is precisely the map induced by $A {\rightarrow} F(D) \otimes_{F(C)} A$, whence this map is a quasi-isomorphism, which was to be shown.

Since sequential colimits in the category of algebras are the same as those in the category of chain complexes, we already know that sequential colimits of quasi-isomorphisms are quasi-isomorphisms. The result follows.

(I didn't use that the arrow $F(C) \rightarrow F(D)$ was monic... so that worries me.)

  • $\begingroup$ (I used almost nothing about the category of $\mathcal{O}$-modules except that it's a Grothendieck abelian category with a well-behaved tensor product and I guess at one point I used that there are enough flat objects in the unbounded derived category; I could probably avoid the last part by using a more explicit spectral sequence argument involving the bar resolution of $A$). $\endgroup$ Feb 10, 2013 at 1:03
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    $\begingroup$ Another comment is that the proof given in the cited paper doesn't seem to be a proof at all... it basically just says there is an adjunction and so we can lift, which isn't always true. What am I missing? $\endgroup$ Feb 10, 2013 at 1:04
  • $\begingroup$ (cited paper as in the one in the OP) $\endgroup$ Feb 10, 2013 at 1:05
  • $\begingroup$ Thanks for the detailed answer, but I'm not sure I'm convinced. The link in David's comment suggests that in char p, dga's just don't form a model category. I can't see where the argument goes wrong though.. $\endgroup$
    – ChrisLazda
    Mar 6, 2013 at 13:40
  • $\begingroup$ The link shows only that the model structure where the fibrations are levelwise surjections and the weak equivalences are quasi-isomorphisms doesn't work. Here we specify cofibrations and quasi-isomorphisms levelwise. So there's no contradiction. $\endgroup$ Mar 6, 2013 at 14:27

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