3
$\begingroup$

Let $C$ be the category with objects being maps $h:M\to A$ where $A$ is a commutative graded $\mathbb{Q}$-algebra (cdga), $M$ is a differential graded (dg) $A$-module and $h$ is an $A$-dg-module map, and morphisms being all commutative squares

$$\begin{matrix} M & \to & A \\ f\downarrow & & \downarrow g\\ M' & \to & A' \end{matrix}$$

where $g$ is a map of cdga's and $f$ is a map of $\mathbb{Q}$-vector spaces such that $f(am)=g(a)f(m)$ for all $a\in A, m\in M$.

I would like to ask whether $C$ admits a model structure with weak equivalences being squares of the above form with $f,g$ quasi-isomorphisms. If it does, then what are the cofibrant objects of this structure? If it turns out there are several such structures, then I'd be mostly interested in the one where cofibrant objects are easier to describe.

$\endgroup$
5
  • $\begingroup$ I know this has been discussed in meta somewhere, but I can never seem to make the commutative diagrams I want to. Could you just copy your code from above and put it in a comment so I could steal it for my answer? $\endgroup$ Oct 4, 2011 at 2:44
  • 2
    $\begingroup$ David -- here you are: begin{matrix} M & \to & A \\ f\downarrow & & \downarrow g\\ M' & \to & A' \end{matrix} The result does not look too nice (the arrows are off and the letters $f,g$ are too large) but it's the best I could do. $\endgroup$
    – algori
    Oct 4, 2011 at 2:50
  • $\begingroup$ Thanks. I had to turn the \\'s into \\\'s, but then it worked. How did you center it? $\endgroup$ Oct 4, 2011 at 3:08
  • $\begingroup$ Ah, never mind. You just have to use the double dollar signs $\endgroup$ Oct 4, 2011 at 3:30
  • $\begingroup$ David -- yes, indeed. Sorry I didn't mention that. $\endgroup$
    – algori
    Oct 4, 2011 at 3:35

2 Answers 2

2
$\begingroup$

I'm not sure about your category of arrows, but it's certainly true the following result: giving a model category $\cal{C}$, you always have a model structure on the category of its arrows $\cal{C}^2$ with the following distinguished morphisms:

  1. Weak equivalences: pairs $(f, g)$ where both $f, g$ are weak equivalences.
  2. Fibrations: pairs $(f, g)$ where both $f, g$ are fibrations.
  3. Cofibrations: pairs $(f, g)$ where $f$ and the induced map $M'\sqcup_M A \longrightarrow A'$ are cofibrations

Besides being probably folklore, this result can also be found in A. Roig, "Model category structures in bifibred categories", JPAA 95 (1994), corollary 7.3.

So, in your case I guess you would need a substitute for the push-out $M'\sqcup_M A $, which perhaps it's not difficult to find.

The above result for the category of arrows $\cal{C}^2$ is a particular case of a more general one about building model category structures for bifibred categories (op.cit., theorem 5.1). This theorem comes together with its dual (theorem 5.1$^o$), that you obtain exchanging fibrations and cofibrations. So, though it's not written explicitely in the paper, you have another "natural" model structure on $\cal{C}^2$, exchanging (2) and (3) above, and putting a pull-back in the place of the push-out.

$\endgroup$
2
  • $\begingroup$ Agusti -- thanks. I may be missing something simple here, but: how does this apply to the question? The category I'm interested in is not the arrow category of any category: the maps in the top and bottom rows of the diagram in the question are maps of modules over different algebras. $\endgroup$
    – algori
    Oct 4, 2011 at 18:10
  • $\begingroup$ Well, I thought it could give you inspiration. I conjecture that an analogous model structure could work for your category and the only problem should be to define an analogous of the push-out which is needed in order to define cofibrations. But maybe I'm too daring. $\endgroup$ Oct 4, 2011 at 18:43
1
$\begingroup$

I'm not 100% sure, but I think this is very likely. Mark Hovey has recently studied model category structures on $Arr(\mathcal{C})$ where $\mathcal{C}$ is a bicomplete, closed, symmetric monoidal model category (e.g. the category of differential graded algebras). Some things can also be said with weaker conditions on $\mathcal{C}$. This is unpublished but he's been giving some talks on it and I imagine it will appear in the not-too-distant future. I'm not sure how much I should say without his permission, but you could always contact him directly. One of the points of his paper is that you can get lots of nice properties on $Arr(\mathcal{C})$. It's also possible others have studied this. Anyway, it turns out that the arrow category has a model category structure where the weak equivalences are squares

$\begin{matrix}M & \to & A \\\ f\downarrow & & \downarrow g\\\ M' & \to & A' \end{matrix}$

with $f,g$ weak equivalences in $\mathcal{C}$, and the cofibrations are squares as in your question with $f,g$ cofibrations in $\mathcal{C}$. This is the so-called injective model structure. There is also one where the weak equivalences and fibrations are squares with $f,g$ weak equivalences and fibrations respectively. This is the so-called projective model structure.

If you look into the paper Algebras and Modules in Monoidal Model Categories by Stephan Schwede and Brooke Shipley you'll see that if $\mathcal{C}$ satisfies something called the Monoid Axiom and is cofibrantly generated then you can get a model structure on $Mon(\mathcal{C})$ the monoids of $\mathcal{C}$. You can also get a model category structure on modules over a commutative monoid, and this seems to be the situation you're in. One thing which Hovey proves in his manuscript is that if $\mathcal{C}$ satisfies the monoid axiom (and the other hypotheses listed in the second sentence of this answer) then $Arr(\mathcal{C})$ under the injective model structure satisfies the monoid axiom.

Finally, the maps you're interested in--where the source is a dg module over the target (which is a cdga)--should also form a model category as the monoids in $Arr(\mathcal{C})$ under an appropriate product on $\mathcal{C}$. It turns out the correct product to work with is the pushout product, which can be found in Hovey's book Model Categories, but which I'll reproduce here:

For $f:X_0\rightarrow X_1$ and $g:Y_0\rightarrow Y_1$

$f \Box g : (X_0 \otimes Y_1) \coprod_{X_0\otimes Y_0}(X_1\otimes Y_0) \rightarrow X_1\otimes Y_1$

I'm really not comfortable saying more here, but please feel free to email me if you'd like some more details.

$\endgroup$
4
  • $\begingroup$ David -- thanks, this is interesting. However, which category $\mathcal{C}$ do you have in mind regarding this example? $\endgroup$
    – algori
    Oct 4, 2011 at 2:53
  • $\begingroup$ ... to be more precise, $f$ in the above diagram is not a map in any category I can think of (apart from complexes of vector spaces). $\endgroup$
    – algori
    Oct 4, 2011 at 3:04
  • $\begingroup$ I added some more details, and I can see how to answer your question about $f$. However, I'd rather do this over email. I've edited my profile to include my email address $\endgroup$ Oct 4, 2011 at 3:17
  • $\begingroup$ @algori: Hovey's paper is now available online: mhovey.web.wesleyan.edu/smithideals.pdf $\endgroup$ Oct 2, 2013 at 23:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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