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1. • CommentRowNumber1.
   • CommentAuthorDmitri Pavlov
   • CommentTime4 days ago
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   Author: Dmitri Pavlov
   Format: MarkdownItexCreated: ## 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]] $R$ (or a commutative $k$-algebra $R$) can be defined as the free [[commutative differential graded algebra]] on $R$. 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]] $M$. 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]] $A$ whose degree 0 component $A_0$ is equipped with a structure of a [[C^∞-ring]] in such a way that the degree 0 differential $A_0\to A_1$ 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 $M$ is canonically isomorphic to the differential graded algebra of smooth [[differential forms]] on $M$. \end{theorem} ## Application: the Poincaré lemma The [[Poincaré lemma]] becomes a trivial consequence of the above theorem. \begin{proposition} For every $n\ge0$, the canonical map $$\mathbf{R}[0]\to \Omega(\mathbf{R}^n)$$ is a [[quasi-isomorphism]] of [[differential graded algebras]]. \end{proposition} \begin{proof} (Copied from the [[MathOverflow]] [answer](https://mathoverflow.net/questions/38439/integrals-from-a-non-analytic-point-of-view/38479#38479).) The de Rham complex of a finite-dimensional smooth manifold $M$ is the free C^∞-dg-ring on the C^∞-ring $C^\infty(M)$. If $M$ is the underlying smooth manifold of a finite-dimensional real vector space $V$, then $C^\infty(M)$ is the free C^∞-ring on the vector space $V^*$ (the real dual of $V$). Thus, the de Rham complex of a finite-dimensional real vector space $V$ is the free C^∞-dg-ring on the vector space $V^*$. This free C^∞-dg-ring is the free C^∞-dg-ring on the free cochain complex on the vector space $V^*$. The latter cochain complex is simply $V^*\to V^*$ 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 $V$ is cochain homotopy equivalent to the free C^∞-dg-ring on the zero cochain complex, i.e., $\mathbf{R}$ in degree 0. \end{proof} ## References * [[E. J. Dubuc]], [[A. Kock]], _On 1-form classifiers_, Communications in Algebra 12:12 (1984), 1471–1531. [doi](https://doi.org/10.1080/00927878408823064). <a href="https://ncatlab.org/nlab/revision/smooth+differential+forms+form+the+free+C%5E%E2%88%9E-DGA+on+smooth+functions/1">v1</a>, <a href="https://ncatlab.org/nlab/show/smooth+differential+forms+form+the+free+C%5E%E2%88%9E-DGA+on+smooth+functions">current</a>

   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 $R$ (or a commutative $k$-algebra $R$) can be defined as the free commutative differential graded algebra on $R$.

   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 $M$.

   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 $A$ whose degree 0 component $A_0$ is equipped with a structure of a C^∞-ring in such a way that the degree 0 differential $A_0\to A_1$ 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 $M$ is canonically isomorphic to the differential graded algebra of smooth differential forms on $M$. \end{theorem}

   Application: the Poincaré lemma

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

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

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

   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 $M$ is the free C^∞-dg-ring on the C^∞-ring $C^\infty(M)$. If $M$ is the underlying smooth manifold of a finite-dimensional real vector space $V$, then $C^\infty(M)$ is the free C^∞-ring on the vector space $V^*$ (the real dual of $V$). Thus, the de Rham complex of a finite-dimensional real vector space $V$ is the free C^∞-dg-ring on the vector space $V^*$. This free C^∞-dg-ring is the free C^∞-dg-ring on the free cochain complex on the vector space $V^*$. The latter cochain complex is simply $V^*\to V^*$ 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 $V$ is cochain homotopy equivalent to the free C^∞-dg-ring on the zero cochain complex, i.e., $\mathbf{R}$ in degree 0. \end{proof}

   References

   • E. J. Dubuc, A. Kock, On 1-form classifiers, Communications in Algebra 12:12 (1984), 1471–1531. doi.

   v1, current