1
$\begingroup$

Let $I$ be a small category and $\mathcal{D}=D^b_\infty(\mathbb{Z})$ the bounded derived $\infty$-category of abelian groups. Consider the $\infty$-category $\mathcal{C}:=\mathrm{Fun}(I,\mathcal{D})$. Define a bounded t-structure on $\mathcal{C}$ by lifting the one on $\mathcal{D}$, that is $\mathcal{C}^{\leq 0}=\mathrm{Fun}(I,\mathcal{D}^{\leq 0})$. This is well defined because mapping spaces in $\mathcal{D}$ are computed as an end : if $F\in \mathcal{D}^{\leq 0}$ and $G\in \mathcal{D}^{\geq 1}$ then we have $\mathrm{Map}(F(i),G(j))=0$ for all $i,j\in I$ hence the bifunctor $\mathrm{Map}(F(-),G(=))$ is trivial and its end must be too. The heart of this t-structure is equivalent to the nerve of the abelian category of functors $I\to \mathbb{Z}\mathrm{-Mod}$. I am interested in computing $$ \mathrm{Ext}^i_{\mathcal{D}}(F,G):=\pi_0 \mathrm{Map}_{\mathcal{D}}(F,G[i]) $$ for ordinary functors $F,G:I\to \mathbb{Z}\mathrm{-Mod}$. This seems similar to the situation of the computation of Ext groups between abelian groups seen as objects in the stable infinity category of spectra (which seems to be something quite standard ; note though that I know very little algebraic topology), so I was wondering if it has already been treated somewhere or if some methods would translate.

We can wonder wether $\mathcal{C}$ is the derived category of its heart; but showing it would anyway amount to doing the above computation I guess, by Lurie's recognition principle (Higher Algebra, 1.3.3.7).

If this can help, in my particular situation of interest, $I$ is the category of $\mathbb{Z}$-constructible sheaves on a smooth projective curve $X$ over a finite field and I am looking for instance at $F=\mathrm{Ext}_X^1(-,\mathbb{G}_m)^\dagger$ and $G=\mathrm{Ext}_X^2(-,\mathbb{G}_m)^D/H^1_{ét}(X,-)$ where $(-)^\dagger=\mathrm{Hom}(-,\mathbb{Q})$ and $(-)^D=\mathrm{Hom}(-,\mathbb{Q}/\mathbb{Z})$.

$\endgroup$
1
  • 2
    $\begingroup$ I think this is more appropriate for MathOverflow $\endgroup$
    – Matt
    Commented Oct 21, 2020 at 16:02

1 Answer 1

7
$\begingroup$

$\newcommand{\Z}{\mathbb Z} \newcommand{\Ch}{\mathrm{Ch}} \newcommand{\Fun}{\mathrm{Fun}}$ Let $\Ch(\Z)$ be the projective model category of chain complexes. It is well known that it presents $D_\infty(\Z)$.

Moreover, $\Fun(I,\Ch(\Z))$ with its projective model structure presents $\Fun(I,D_\infty(\Z))$, and of course $\Fun(I,D^b_\infty(\Z))$ is a full stable subcategory of $\Fun(I,D_\infty(\Z))$, hence its mapping spaces (in fact mapping spectra as well) are the same as the ones in the latter.

But now $\Fun(I,\Ch(\Z)) \cong \Ch(\Fun(I,\Z-\mathrm{Mod}))$, with the projective model structure everywhere, so the mapping spaces can be computed as usual : with projective resolutions (in $\Fun(I,\Z-\mathrm{Mod})$).

So your question is precisely a question of functor cohomology, which is well studied subject, with various techniques.

(note : this is for small $I$. I'm not sure how to make sense of the question when $I$ isn't small)

$\endgroup$
8
  • $\begingroup$ Thank you ! I feel like I should have seen this but I was getting tunnel vision because at the start I was considering the case where $I=D^b_{constr}(X,\mathbb{Z})$ (instead of $\mathrm{Sh}_{constr}(X,\mathbb{Z}))$. Would you know how to tackle that case ? $\endgroup$ Commented Oct 22, 2020 at 7:41
  • $\begingroup$ So the case where $I$ is a small $\infty$-category rather than a category ? $\endgroup$ Commented Oct 22, 2020 at 7:42
  • $\begingroup$ Yes exactly. In that case I guess there should be the projective model structure (following Mazel-Gee's $\infty$-model categories) on $\mathrm{Fun}(I,\mathrm{Ch}(\mathbb{Z}))=\mathrm{Fun}(ho(I),\mathrm{Ch}(\mathbb{Z}))$ so we are reduced to the previous case by taking $ho(I)$ instead of $I$ ? $\endgroup$ Commented Oct 22, 2020 at 7:54
  • $\begingroup$ I am not familiar at all with model $\infty$-categories, in particular I'm definitely not sure that in this setting, $\mathrm{Fun}(I,\mathrm{Ch}(\mathbb Z))$ presents the right thing. The fact that $W^{-1}Fun(I,Ch(\mathbb Z)) \simeq Fun(I,W^{-1}Ch(\mathbb Z))$ strongly uses the fact that $Ch(\mathbb Z)$ is a model category, so I couldn't say for sure. However, what is true is that $I\to ho(I)$ is cofinal and final so limits and colimits computed over $ho(I)$ coincide with the ones pulled back to $I$, so if your functors land in $\mathbb Z-Mod$ (and thus factor through $ho(I)$), I think their $\endgroup$ Commented Oct 22, 2020 at 8:04
  • 1
    $\begingroup$ No you are right, that much is true. But you're not interested in functors into $Ch(\mathbb Z)$, you're interested in functors into $D(\mathbb Z)$, right ? And the point is that it's not clear how to use $Fun(I,Ch(\mathbb Z))$ to compute things in $Fun(I,D(\mathbb Z))$ (for instance take $I= D^b(\mathbb Z)$ : the inclusion will be in the second object, but not in the first, not even up to weak equivalences) $\endgroup$ Commented Oct 22, 2020 at 8:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .