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    • CommentRowNumber1.
    • CommentAuthorDmitri Pavlov
    • CommentTimeSep 16th 2024

    Created:

    Background

    See the article Kähler C^∞-differentials of smooth functions are differential 1-forms for the necessary background for this article, including the notions of C^∞-ring, C^∞-derivation, and Kähler C^∞-differential.

    Idea

    In algebraic geometry, (algebraic) differential forms on the Zariski spectrum of a [commutative ring RR (or a commutative kk-algebra RR) can be defined as the free commutative differential graded algebra on RR.

    This definition does not quite work for smooth manifolds: as already explained in the article Kähler C^∞-differentials of smooth functions are differential 1-forms, the notion of a Kähler differential must be refined in order to extract smooth differential 1-forms from the C^∞-ring of smooth functions on a smooth manifold MM.

    Thus, in order to get the algebra of smooth differential forms, the notion of a commutative differential graded algebra must likewise be adjusted.

    \begin{definition} A commutative differential graded C^∞-ring is a real commutative differential graded algebra AA whose degree 0 component A 0A_0 is equipped with a structure of a C^∞-ring in such a way that the degree 0 differential A 0A 1A_0\to A_1 is a C^∞-derivation. \end{definition}

    With this definition, we can recover smooth differential forms in a manner similar to algebraic geometry, deducing the following consequence of the Dubuc–Kock theorem for Kähler C^∞-differentials.

    \begin{theorem} The free commutative differential graded C^∞-ring on the C^∞-ring of smooth functions on a smooth manifold MM is canonically isomorphic to the differential graded algebra of smooth differential forms on MM. \end{theorem}

    Application: the Poincaré lemma

    The Poincaré lemma becomes a trivial consequence of the above theorem.

    \begin{proposition} For every n0n\ge0, the canonical map

    R[0]Ω(R n)\mathbf{R}[0]\to \Omega(\mathbf{R}^n)

    is a quasi-isomorphism of differential graded algebras. \end{proposition}

    \begin{proof} (Copied from the MathOverflow answer.) The de Rham complex of a finite-dimensional smooth manifold MM is the free C^∞-dg-ring on the C^∞-ring C (M)C^\infty(M). If MM is the underlying smooth manifold of a finite-dimensional real vector space VV, then C (M)C^\infty(M) is the free C^∞-ring on the vector space V *V^* (the real dual of VV). Thus, the de Rham complex of a finite-dimensional real vector space VV is the free C^∞-dg-ring on the vector space V *V^*. This free C^∞-dg-ring is the free C^∞-dg-ring on the free cochain complex on the vector space V *V^*. The latter cochain complex is simply V *V *V^*\to V^* with the identity differential. It is cochain homotopy equivalent to the zero cochain complex, and the free functor from cochain complexes to C^∞-dg-rings preserves cochain homotopy equivalences. Thus, the de Rham complex of the smooth manifold VV is cochain homotopy equivalent to the free C^∞-dg-ring on the zero cochain complex, i.e., R\mathbf{R} in degree 0. \end{proof}

    References

    v1, current