Wrote this a week or two ago but didn’t post, it’s mostly to try to get my own head straight about previous correspondences on the issues of asking whether the functor of MR1 sheaves is “really” a derived stack, and asking whether it is represented by an algebraic stack. The main point is A) these are orthogonal issues, the first is a question about an extension of the definition of the MR1 functor to something that is probed by cdga’s, while the latter is a question about a property of the MR1 functor. And B) answering these two questions are independent tasks: even with an intelligent definition of a derived MR1 functor in hand, there is still a derived representability question to be settled. There is an implicit “I think this is how things work, but could be somewhat off” throughout. It sounds pedantic because it was useful for me to make myself write it that way.

MR1 as a Derived Stack TLDR: A statement like “$M_1(\Lambda)$ is defined for nonsense reasons as a derived stack, and in general will not be a better object than that” doesn’t quite compile, so here’s an attempt at a precise version of that sentiment.

First off, to get on the same page:

1) A derived 1-stack is an $\infty$-functor $k-cdga \to$ 1-groupoids (satisfying descent). It is a derived algebraic 1-stack if it is suitably covered by 1-groupoid quotients of derived affine schemes.
2) An ordinary 1-stack is a functor $k-alg \to$ 1-groupoids (satisfying descent). It is an ordinary algebraic 1-stack if it is suitably covered by 1-groupoid quotients of affine schemes.

A derived 1-stack restricts along $k-alg \subset k-cdga$ to an ordinary 1- stack, which behaves like its “reduced subscheme” (the yoga being that “derived directions” are like “nilpotent directions”). Conversely, an ordinary 1-stack extends via $C^\cdot \mapsto H^0(C^\cdot)$ to a derived 1-stack, which is like “thinking of a variety as a scheme”. (From now on I suppress the 1 in 1-stack)

The MR1 functor defined in STZ is a functor $k-alg \to$ 1-groupoids, so it is meaningful to ask whether it is an algebraic stack or merely a stack. It is not meaningful, however, to ask whether it is a derived stack or merely a stack.

Instead the meaningful question at the derived level is whether, rather than “thinking of the MR1 stack as a derived stack” in the trivial way, we can extend it to a derived stack in some nontrivial/interesting/meaningful way. That is, can we soup up the MR1 functor to a functor $k-cdga \to$ 1-groupoids that really uses the cdga structure in a nontrivial way. In that case, there is then a further issue of whether this functor is a derived algebraic stack or merely a stack. If it is a derived algebraic stack I believe it follows that the ordinary MR1 functor is an ordinary algebraic stack.

There is also a further orthogonal question of whether the MR1 functor or its derived version extend in a nontrivial/interesting/meaningful way to a (derived) n-stack, that is a functor $k-alg/k-cdga \to$ n-groupoids. This is the issue of removing the word “quasi-isomorphism” from the definition in STZ in an intelligent way, c.f paragraph 3 of the subsection “Perfect Complexes” in section 3.3 of Toen’s “Derived Algebraic Geometry” EMS survey.

Quotients and Orthogonal Complements TLDR: A statement of the precise questions to ask to show that the correctly-defined MR1 stack is algebraic, the observation that it’s not the case in general that rank-fixed things are the orthogonal complement of local systems but that something good enough is probably still true.

First, notation-fixing discussion of orthogonal complements in the triangulated and DG settings, following Drinfeld. Below if $A$ is a DG category $A^{tr}$ will denote the associated triangulated category.

If $T$ is triangulated and $Q \subset T$ a full triangulated subcategory, then its right orthogonal complement $Q^\perp$ is triangle-equivalent to the triangulated quotient $Q/T$ if $Q$ is right admissible. (Right-admissible is the condition that every $t \in T$ sits in a triangle $q \to t \to q’$ with $q \in Q$, $q’ \in Q^\perp$. This is equivalent to the inclusion $Q \to T$ having a right adjoint).

If $B$ is DG and $A \subset B$ a full DG subcategory, its naive right orthogonal complement is not necessarily equivalent to the DG quotient $A/B$ (here orthogonal means that Hom is acyclic). Instead let $(B’)^{\perp}$ be the right orthogonal complement with respect to larger DG categories $B’ \subset A’$ of semifree objects ($A’$ is $A$ with a rightward arrow underneath in Drinfeld’s notation, who tells us to think of these as categories of ind-objects.) The category $(B’)^{\perp}$ is equivalent to $A’/B’$ and $((B’)^\perp)^{tr}$ is equivalent to $(A’)^{tr}/(B’)^{tr}$. Moreover, $A^{tr}/B^{tr}$ embeds as a full subcategory of $(A’)^{tr}/(B’)^{tr}$, hence as a full subcategory of $((B’)^\perp)^{tr}$, which Drinfeld describes explicitly. It is not discussed under what circumstances this subcategory is in fact simply $(B^\perp)^{tr} \subset ((B’)^\perp)^{tr}$, i.e. when it is not necessary to introduce ind-objects. I think it’s reasonable to wonder if $B^{tr}$ being right-admissible guarantees this, so that the above construction is “merely” a way of forcing $B^{tr}$ to always be right-admissible (this may even be “obvious”). (Note: in the DG setting Drinfeld simply writes $B^\perp$ for $(B’)^{\perp}$, so his $B^\perp$ really lives in $A’$.)

Now fix a Legendrian \Lambda, let $Sh$ be the sheaf category with SS on $\Lambda$ localized at acyclics, $Loc \subset Sh$ the local system subcategory.

Note that the definition of the MR1 stack in STZ is really an invariant of triangulated categories rather than DG categories per se, i.e. the only morphisms it knows about are ones seen by the triangulated envelopes of the DG categories involved. The points of the MR1 stack are the objects of $Sh^{tr}/Loc^{tr} \cong (Sh/Loc)^{tr}$ satisfying the MR1 condition. In particular, for purposes of establishing whether the MR1 stack is an algebraic stack, the questions we want to ask can be asked at the level of triangulated categories.

Assume now that we are in a position (STZ, $\Lambda$ is an ASD,…) where we can embed Sh as a subcategory of DG representations of a quiver $Q$ whose sources are all “at the same level”, i.e. all reverse-oriented paths from a fixed vertex to any source are the same length. We then say a sheaf/Q-representation is source-acyclic if it is acyclic at the sources of $Q$. Then here are questions we want to answer:

1) Is $Loc^{tr}$ right-admissible in $Sh^{tr}$?
2) Is the subcategory of $(Loc^{tr})^\perp$ consisting of MR1 sheaves exactly the subcategory of source-acyclic sheaves that are MR1?

If the answers to 1 and 2 are yes they form an argument that the correct “localization-defined” MR1 stack is equivalent to the naïve “rank-fixing-defined” MR1 functor, which the STZ argument shows is algebraic. Note that if $Q$ has more than one source it is not in general true that $(Loc^{tr})^\perp$ only consists of source-acyclic sheaves: if $Q$ is the source-sink-source $A_3$ quiver, $0 \to k \leftarrow k \in $(Loc^{tr})^\perp$ if the right map is an iso. But of course this representation is not MR1.

In principle, we can answer these questions at the level of DG categories, but this A) is not required, since the MR1 stack only sees the triangulated categories, and B) does not obviously make life easier. It could make life easier in that passing from $Loc$ to $Loc$’ is in particular a way of “forcing” right-admissibility. The cost is that we then have to ask question 2 in the “ind-object” category $(Loc’)^\perp$. Maybe that’s not bad, in fact probably the MR1 conditions exclude sheaves that are really in Sh’ and not Sh, but at this point enough definitions are involved that I find it hard to keep my bearings.