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Étale cohomology of schemes $X$ is constructed as follows: one associates to $X$ the so-called étale topos of $X$, and then one just takes the sheaf cohomology of that topos.

Is it possible to associate to each smooth manifold $M$ a "de Rham topos of $M$", whose sheaf cohomology yields the de Rham cohomology of $M$?

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    $\begingroup$ In algebraic geometry, yes: this is the crystalline topos. I would guess that the construction extends in a straightforward way to smooth manifolds. $\endgroup$
    – abx
    Commented Feb 7, 2022 at 13:27
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    $\begingroup$ Given any topological space, you can consider the topos of sheaves on that. This works fine for defining de Rham cohomology of manifolds by de Rham's theorem! $\endgroup$
    – Donu Arapura
    Commented Feb 7, 2022 at 15:05
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    $\begingroup$ As a reference for Donu Arapura's comment above, see math.stackexchange.com/q/673518. Čech cohomology and sheaf cohomology are the same thing over a manifold (ncatlab.org/nlab/show/…, Theorem 3.7). $\endgroup$
    – Jens Hemelaer
    Commented Feb 7, 2022 at 15:40
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    $\begingroup$ Just to point out a major difference: in the etale setting, you take cohomology of the constant sheaf, but you need to be smarter in the crystalline/de Rham setting (namely, you should take the structure sheaf). $\endgroup$
    – sdr
    Commented Feb 7, 2022 at 20:10

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One can define an analogue of the crystalline topos for smooth manifolds.

This is known as the de Rham stack of $M$.

One of the easiest constructions of the de Rham stack embeds smooth manifolds fully faithfully (using the Yoneda embedding) into the category of ∞-sheaves on affine smooth loci, the latter being defined as the opposite category of $\def\Ci{{\rm C}^∞} \Ci$-rings satisfying certain properties.

In this language, the de Rham stack of an ∞-sheaf $F$ is the ∞-sheaf $\def\dR{{\rm dR}} \dR(F)$ defined by $\def\Spec{\mathop{\rm Spec}} \def\red{{\rm red}} \dR(F)(\Spec A)=\dR(F)(\Spec(\red(A))$, where $\Spec A$ denotes the spectrum of a $\Ci$-ring (defined purely formally in this context) and $\red(A)$ denotes the quotient of $A$ by its ideal of nilpotent elements.

One can then prove that the commutative differential graded algebra of smooth functions on $\dR(M)$ is precisely the de Rham algebra of $M$.

The de Rham stack has other exciting properties: vector bundles (and, more generally, sheaves) on $\dR(M)$ can be identified with D-modules, etc.

The cited nLab article has the relevant pointers to the literature.

The de Rham stack is also closely related to the definition of the differential graded algebra of differential forms in synthetic differential geometry as the differential graded algebra of infinitesimal smooth singular cochains equipped with the cup product. See the nLab article differential forms in synthetic differential geometry for further pointers to the literature.

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  • $\begingroup$ Is there any explanation that quotient by the nilradical, an operation in algebra (instead of, say, some version of topological nilpotence), is still the correct object to consider in differential geometry? Does this construction extend to more general objects (I guess, say, orbifolds or diffeological spaces)? $\endgroup$
    – Z. M
    Commented Feb 7, 2022 at 22:13
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    $\begingroup$ @Z.M: Concerning the first question: normally, one uses finitely generated germ-determined C^∞-rings, which turns out to be a good analogue of finitely presented rings in algebraic geometry. Such rings are quotients C^∞(M)/I, where M is an ordinary smooth manifold and I is a germ-determined ideal, which can be defined as an ideal closed under locally finite (with respect to M) sums. From this description, it is easy to see that various possible variants of nilpotence coincide, essentially because nilpotence is determined on individual germs. $\endgroup$
    – Dmitri Pavlov
    Commented Feb 8, 2022 at 3:15
  • $\begingroup$ @Z.M: Concerning the second question, yes, certainly, since both orbifolds and diffeological spaces are special cases of ∞-sheaves on smooth loci, and embed into them fully faithfully. $\endgroup$
    – Dmitri Pavlov
    Commented Feb 8, 2022 at 3:16
  • $\begingroup$ For the nilradical, I thought of the following case: let $R$ be, say, the ($ C^\infty$-)ring of germs of smooth functions on $\mathbb R$ at $0$, and $I\subset R$ the ideal generated by germs $f$ such that all derivatives of $f$ at $0$ vanishes. I am not sure whether this kind of ideals should be taken into account. I asked because I am aware that Kähler differential does not coincide with ordinary differentials, and I am not sure in which part this discrepancy is somehow "solved". $\endgroup$
    – Z. M
    Commented Feb 8, 2022 at 17:26
  • $\begingroup$ @Z.M: Indeed, we do know an excellent way to solve the problem with Kähler differentials: the abstract notion of Beck module and its associated abstract notion of a Kähler differential produce correct answers in all cases. Specifically, for ordinary commutative rings they recover traditional Kähler differentials. For C^∞-rings they recover a very similar definition, except that now the Leibniz rule is formulated as a chain rule for arbitrary smooth functions, not just polynomials. $\endgroup$
    – Dmitri Pavlov
    Commented Feb 8, 2022 at 17:31

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