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    • CommentRowNumber1.
    • CommentAuthorDmitri Pavlov
    • CommentTimeSep 16th 2024

    Created:

    Idea

    In algebraic geometry, the module of Kähler differentials of a commutative ring R corresponds under the Serre–Swan duality to the cotangent bundle of the Zariski spectrum of R.

    In contrast, the module of Kähler differentials of the commutative real algebra of smooth functions on a smooth manifold M receives a canonical map from the module of smooth sections of the cotangent bundle of M that is quite far from being an isomorphism.

    An example illustrating this point is M=R, since in the module of (traditionally defined) Kähler differentials of C(M) we have d(exp(x))expdx, where exp:RR is the exponential function. That is to say, the traditional algebraic notion of a Kähler differential is unable to deduce that exp=exp using the Leibniz rule.

    However, this is not a defect in the conceptual idea itself, but merely a failure to use the correct formalism. The appropriate notion of a ring in the context of differential geometry is not merely a commutative real algebra, but a more refined structure, namely, a C^∞-ring.

    This notion comes with its own variant of commutative algebra. Some of the resulting concepts turn out to be exactly the same as in the traditional case. For example, ideals of C^∞-rings and modules over C^∞-rings happen to coincide with ideals and modules in the traditional sense. Others, like derivations, must be defined carefully, and definitions that used to be equivalent in the traditional algebraic context need not remain so in the context of C^∞-rings.

    Observe that a map of sets d:AM (where M is an A-module) is a derivation if and only if for any real polynomial f(x1,,xn) the chain rule holds:

    d(f(a1,,an))=ifxi(x1,,xn)dxi.

    Indeed, taking f(x1,x2)=x1+x2 and f(x1,x2)=x1x2 recovers the additivity and Leibniz property of derivations, respectively.

    Observe also that f is an element of the free commutative real algebra on n elements, i.e., R[x1,,xn].

    If we now substitute C^∞-rings for commutative real algebras, we arrive at the correct notion of a derivation for C^∞-rings:

    A __C^∞-derivation__ of a [[C^∞-ring]] $A$ is a map of sets $A\to M$ (where $M$ is a [[module]] over $A$) such that the following chain rule holds for every smooth function $f\in\mathrm{C}^\infty(\mathbf{R}^n)$:
    $$d(f(a_1,\ldots,a_n))=\sum_i {\partial f\over\partial x_i}(x_1,\ldots,x_n) dx_i,$$
    where both sides use the structure of a [[C^∞-ring]] to evaluate a smooth real function on a collection of elements in $A$.
    

    The module of Kähler C^∞-differentials can now be defined in the same manner as ordinary Kähler differentials, using C^∞-derivations instead of ordinary derivations.

    \begin{theorem} (Dubuc, Kock, 1984.) The module of Kähler C^∞-differentials of the C^∞-ring of smooth functions on a smooth manifold M is canonically isomorphic to the module of sections of the cotangent bundle of M. \end{theorem}

    Related concepts

    References

    v1, current

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTime1 day ago
    • (edited 1 day ago)

    added pointer to

    for proof (using choice) of the claim that d(exp(x))exp(x)dx

    Also added a cross-link to the entry Kähler differential, section Over smooth rings regarded as ordinary rings, which is largely duplicated here.

    diff, v4, current

    • CommentRowNumber3.
    • CommentAuthorDmitri Pavlov
    • CommentTime1 day ago

    Re #2: The approach presented in Kähler differential (taking Ω^1 = Der^*, i.e., the dual of the module of derivations) is different from the approach presented in this article, which takes Ω^1 to be the codomain of the universal C^∞-derivation. So neither entry duplicates the other, although perhaps some content could be moved between articles.