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Let $X$ be a smooth variety.

  1. One can define the notion of a sheaf of twisted differential operators (TDO) on $X$. They "quantise" functions on $T^*X$. Examples include the usual sheaf of differential operators $\mathcal{D}$. TDOs are classified by the complex of sheaves $$K_1\ =\ \Omega_X^1\ \longrightarrow\ \Omega^{2,cl}_X$$ meaning that TDOs are parametrised by the first hypercohomology $H^1$ of this complex, and for each TDO its automorphisms are $H^0$. You should maybe think of the obstruction $ob_X$ to the existence of TDOs as living in $H^2$, but it's just that $ob_X=0$ for all $X$ (because of the existence of $\mathcal{D}$).
  2. One can also define chiral differential operators (CDO). They "quantise" functions on $\text{Arc}(T^*X)=\text{Map}(\text{disk},T^*X)$, and are even vertex algebras. They are classified by (see 5.3 of linked article) $$K_2\ =\ \Omega_X^2\ \longrightarrow\ \Omega^{3,cl}_X$$ in the same way as in 1, except that now the obstruction $ob_X=ch_2(T_X)$ does not vanish, so CDOs do not necessarily exist.
  3. ?

My question is whether this pattern continues, i.e. are there interesting geometric objects classified as above by complexes of sheaves $K_n$, and whether e.g. they are related to iterated loop spaces in the same way that CDOs are related to loop spaces. The obvious guess is of course $K_n=\Omega^n_X\to \Omega_{X}^{n+1,cl}$.

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    $\begingroup$ I alluded to this below but maybe it's worth mentioning here that a fill in the sequence problem that is sort of logically prior is obtained by asking what one is quantising. So step 0 is "poisson algebra", step 1 is "poisson vertex algebra" and step 2 is presumably "poisson algebra 'with more than one $\lambda$' ". If this is reasonable it shouldn't be too hard to write down the axioms and then the first thing to check would be that iterated arc spaces are examples of such objects. $\endgroup$
    – user108998
    Commented Sep 1, 2021 at 19:46
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    $\begingroup$ @EBz what quantizes Coisson algebras ("Poisson algebra with more $\lambda$") is incredibly hard to define $\endgroup$ Commented Sep 1, 2021 at 20:42
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    $\begingroup$ @Reimundo thanks, I certainly hadn't intended to suggest it was easy or that I knew how to do it. At any rate I do believe that whatever the object the OP wants to define is, it is not unreasonable to expect that they should quantize the relevant coisson algebra, in an appropriate sense. If this means that they are objects with a mysterious structure, then all the more reason to believe (as I do) that the problem is probably quite difficult $\endgroup$
    – user108998
    Commented Sep 1, 2021 at 21:01
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    $\begingroup$ @ReimundoHeluani aren't we just look for higher dimensional holomorphic factorization algebras (or chiral algebras)? these are not hard to define, just all examples are necessarily derived (but by now many interesting examples exist, see eg work of Brian Williams) $\endgroup$ Commented Sep 3, 2021 at 3:04
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    $\begingroup$ @ David Ben-Zvi, do you know how well the local structure of such things has been worked out (so the higher dim version of vertex algebras)? Certainly (as you mentioned) they must be quite derived, as even the corresponding punctured disc must be by Hartogs' type theorem. The kac-moody case is meant to be done in Faonte-Hennion-Kapranov but I couldn't find a (dg-) linear algebraic definition in there in general. $\endgroup$
    – user108998
    Commented Sep 3, 2021 at 11:10

1 Answer 1

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$\newcommand{\co}{\mathcal{O}} \newcommand{\H}{\mathrm{H}} \newcommand{\dR}{\mathrm{dR}} \newcommand{\GG}{\mathbf{G}}$ Let $k = \mathbf{C}$, let $X$ be a (smooth) variety over $k$, and let $X_\dR$ be its de Rham space (so that the cohomology of $\Gamma(X_\dR; \co_{X_\dR})$ is $\H^\ast_\dR(X/k)$. There is a canonical map $f: X \to X_\dR$, and the global sections of the fiber of the map $\co_{X_\dR} \to f_\ast \co_X$ is $\H^\ast(X; \Omega^{\bullet\geq 1}_{X/k})$. In other words, the map $\co_{X_\dR} \to f_\ast \co_X$ corresponds to the projection $\Omega^{\bullet}_{X/k} \to \co_X$ (whose fiber is $\Omega^{\bullet\geq 1}_{X/k}$, with $\Omega^i_{X/k}$ placed in homological degree $-i$). From this point of view (some of the indices may be off, let me know if you find mistakes):

  • The group $\H^2(X; \Omega^{\bullet\geq 1}_{X/k})$ may be understood as $\pi_0$ of the space of $\GG_a$-gerbes on $X_\dR$ which pull back to the trivial $\GG_a$-gerbe along $f$. Note that $\H^2(X; \Omega^{\bullet\geq 1}_{X/k}) = \H^1(X; K_1)$ in your notation. (The relation to TDOs comes from an identification of $\GG_a$-gerbes on $X_\dR$ which pull back to the trivial $\GG_a$-gerbe along $f$ with $\GG_m$-gerbes on $X_\dR$ which pull back to the trivial $\GG_m$-gerbe along $f$, using the exponential on the formal completion of $\GG_a$. Because $\mathrm{QCoh}(X_\dR)$ describes D-modules on $X$ (using the finite-type assumption on $X$), $\GG_m$-gerbes on $X_\dR$ correspond to twistings of $\mathrm{DMod}(X)$. See Gaitsgory-Rozenblyum's https://arxiv.org/abs/1111.2087 for more on this.)

  • Similarly, $\H^3(X; \Omega^{\bullet\geq 2}_{X/k}) = \H^1(X; K_2)$ in your notation, and this can be understood as $\pi_0$ of the space of $\GG_a$-$2$-gerbes on $X_\dR$ which pull back to the trivial $\GG_a$-$2$-gerbe along $f$.

  • In general (if I haven't snuck in a $\pm 1$ mistake), if we define the space of $\GG_a$-$n$-gerbes on $Y$ to be the space of maps $Y \to B^{n+1} \GG_a$, then $\H^{n+1}(X; \Omega^{\bullet\geq n}_{X/k}) = \H^1(X; K_n)$ in your notation, and this can be understood as $\pi_0$ of the space of $\GG_a$-$n$-gerbes on $X_\dR$ which pull back to the trivial $\GG_a$-$n$-gerbe along $f$.

For this to be a satisfying answer, we should relate $\GG_a$-$2$-gerbes on $X_\dR$ to chiral differential operators . This should be given by an algebraic version of transgression (which, in topology, would be a map $\H^\ast(Y) \to \H^{\ast-1}(LY)$ where $Y$ is an oriented manifold). I don't know a reference for this in the algebraic setting (but I would like to know of one if it exists!), but if you are interested I can try to sketch the idea. I think $K_n$ wouldn't be related to iterated free loop spaces, but rather to some algebraic analogue of $\mathrm{Map}(S^n, Y)$.

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    $\begingroup$ Ben Hennion in his work suggests using punctured formal n-disc (nb one here has higher cohomology of structure sheaf in degree n-1 so actually they take the corresponding dg object) and taking maps out of this. Resulting thing has Tate type structure etc, it's all in his thesis and papers derived from it. I don't think his work constructs versions of higher chiral diff ops though. Maybe a first question should be "what sort of object do these things quantize"? Should be n-dim version of poisson vertex algebra on maps from n-disc (no punctures) I'd guess $\endgroup$
    – user108998
    Commented Sep 1, 2021 at 18:11
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    $\begingroup$ Thanks! I was not aware of Hennion's work, it looks quite useful. [FWIW, a related fact is that if $X$ is a scheme, then $\mathrm{Map}(S^n, X)$ (with $S^n$ being the Betti stack corresponding to the $n$-sphere) is closely tied (via a shifted version of HKR) to the shifted tangent space $T[-n]X = \mathrm{Spec}\ \mathrm{Sym}_X(L_{X/k}[n])$.] $\endgroup$
    – skd
    Commented Sep 1, 2021 at 19:17
  • $\begingroup$ You're welcome! Btw for transgression I think one can construct (w/o orientation on X) transgression map $H^{j}(X)\rightarrow H^{j}(LX)$ (note lack of shift) quite naturally. Eg it takes function $z^{2}$ on line to function $\Sigma z_{i}z_{-i}$. So in this case PTVV version of AKSZ construction implies if X is d-symp then LX is also d-symp so formal punctured disc behaving as if $0$ dimensional (implicitly $G_{m}$-equivariant plus oriented as such by $dlog(z)$)- space. $\endgroup$
    – user108998
    Commented Sep 1, 2021 at 19:41
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    $\begingroup$ @EBz This picture with punctured discs etc has actually been described in talks by Nick Rozenblyum going back to at least 2013 (with titles like "Higher Differential Operators") though I don't think there's anything made public - it's the holomorphic version of the E_n quantization of shifted cotangents as skd writes, and I believe Brian Williams and collaborators are working on such objects (he's developed a beautiful theory of holomorphic field theories in the BV formalism) $\endgroup$ Commented Sep 3, 2021 at 3:08
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    $\begingroup$ @Pulcinella instead of classifying gerbes whose pullback is trivial, $H^2(\Omega_X^{\geq 1})$ classifies $\mathbb G_a$-gerbes on $X_{dR}$ along with a trivialization of its pullback to $X$. The trivialization is extra data reflected in $H^1(\mathcal O_X)$. $\endgroup$ Commented Aug 18, 2023 at 4:05

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