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PROOF of DE RHAM's THEOREM 1. Introduction Let M Be a Smooth N

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PROOF of DE RHAM's THEOREM 1. Introduction Let M Be a Smooth N PROOF OF DE RHAM'S THEOREM PETER S. PARK 1. Introduction Let M be a smooth n-dimensional manifold. Then, de Rham's theorem states that the de Rham cohomology of M is naturally isomorphic to its singular cohomology with coefficients in R; in particular, de Rham cohomology is a purely topological invariant. This fact is a manifestation of a fundamental relationship between the analytical and topological characteristics of a smooth manifold, where the former include information about solutions to differential equations of the formulation dη = ! for closed forms !, and the latter, information about the \holes" in each given dimension, which is made precise by singular homology. The natural isomorphism will be given by a version of Stokes' theorem, which describes a duality between de Rham cohomology and singular homology. Specifically, the pairing of differential forms and singular chains, which can taken to be smooth, yields a map from the kth de Rham cohomology group to the kth singular cohomology group with coefficients in R for each k ≥ 0. And in 1931, Georges de Rham himself proved that this map is in fact an isomorphism. In this exposition, we will give a proof1 of this beautiful and fundamental result. 2. Smooth Singular Homology Our desired relationship between singular cohomology and de Rham cohomology will be given by the integration of differential forms over singular chains. Specifically, given a singular k-simplex ∗ σ : ∆k ! M and a k-form ! on a manifold M, we wish to integrate over ∆k the pullback form σ !. However, pulling back differential forms requires the map in question to be sufficiently smooth, yet as maps, singular k-simplices are only continuous in general. To address this issue, we will show that singular homology can be computed using only smooth simplices instead of all continuous ones. m For a subset C ⊂ R , we say that a continuous map f : C ! M is smooth at a point p 2 C if there exists an open neighborhood of p in which f has a smooth extension. Furthermore, we say that f is smooth if it is smooth at every point p 2 C. Then, specializing to the case that C is m m equal to the standard k-simplex of R , defined by ∆k ·= [δ0; : : : ; δk] ⊂ R for the standard basis vectors δi, we define a smooth k-simplex of M to be a k-simplex σ : ∆k ! M that is smooth as a 1 map. For the usual free abelian group Ck(M) on all k-chains of M, we define Ck (M) to be the subgroup of Ck(M) generated by the smooth k-simplices of M. Define the smooth k-chains of M 1 to be the elements of Ck (M), i.e., formal sums of finitely many smooth k-simplices. It is clear from the above definitions that the boundary of a smooth (k + 1)-simplex is a smooth k-chain, so the kth smooth singular homology group 1 1 1 Ker(@ : Ck (M) ! Ck−1(M)) Hk (M) = 1 1 Im(@ : Ck+1(M) ! Ck (M)) 1 is well-defined. Since the inclusion i : Ck (M),!Ck(M) satisfies ι ◦ @ = @ ◦ ι, we see that it induces 1 a map ι∗ : Hk (M) ! Hk(M) defined by ι∗[c] = [ι(c)]. 1This exposition is based on the proof provided in [2, Chapter 18], which in turn is based on that given in [1, Section V.9]. For another proof of de Rham's theorem based on sheaf theory, see [3, Chapter 5]. 1 2 PETER S. PARK We would like to show that this map ι∗ is an isomorphism. We will first need to construct for an arbitrary simplex a homotopy to a smooth simplex such that when the homotopy for any simplex is restricted to a boundary face, this yields the homotopy of the simplex given by the boundary face. To do so, we will need the following lemma showing that for a map whose domain is the standard k-simplex, smoothness on each boundary face implies smoothness on the entire boundary simultaneously. m Lemma 2.1. For the standard simplex ∆k ⊂ R , let f : @∆k ! M be a continuous map whose restriction to each boundary face is smooth. Then, f is smooth as a map from the entire boundary @∆k to M. Proof. Recall the notation ∆k = [δ0; : : : ; δk], where the standard basis vectors δ0; : : : ; δk are the ^ vertices of the simplex. For 0 ≤ i ≤ k, let @i∆k = [δ0;:::; δi; : : : ; δk] denote the boundary face opposite of δi. Our hypothesis then is precisely that for every i and x 2 @i∆k, there exists an open ~ ~ neighborhood Ux 3 x and a smooth extension f : Ux ! M of f, i.e., f restricted to Ux \ @i∆k ~ gives f. We then need to show that for every x 2 @∆k, a smooth extension f can be chosen for all ~ boundary faces, i.e., such that f restricted to Ux \ @∆k coincides with f. Since x is contained in at least one boundary face of ∆k but cannot be contained in all of them, we can without loss of generality suppose that the boundary faces containing x are @1∆k;:::;@m∆k for some 1 ≤ j ≤ k, and in particular, that x is not contained in @0∆k. By our hypothesis, there ~ exists an open neighborhood Ui 3 x and a smooth map fi : Ui ! M whose restriction to Ui \ @i∆k Tj coincides with f. Let U = i=1 Ui. We now prove by induction on j that there exists a smooth ~ Sj map f : U ! M whose restriction to U \ i=1 @i∆k coincides with f. Since our claim is local, we can, by replacing U with a subneighborhood if necessary, suppose that f maps U to a coordinate n chart of f(x) that can be diffeomorphically identified with R . The base case j = 1 is trivial. Now, suppose that j > 1 and that there exists a smooth map f~0 : U ! M whose restriction to Sj U \ i=1 @i∆k coincides with f. Note that the boundary face @i∆k is precisely the intersection ~ of the hyperplane xi = 0 with @∆k. Consistent with this, define f : U ! M by 1 m 1 m 1 j−1 j+1 m 1 j−1 j+1 m f~(x ; : : : ; x ) = f~0(x ; : : : ; x )−f~0(x ; : : : ; x ; 0; x ; : : : ; x )+f~j(x ; : : : ; x ; 0; x ; : : : ; x ): ~ For every 1 ≤ i ≤ j − 1, the restriction of f to U \ @i∆k equals f~(x1; : : : ; xi−1; 0; xi+1; : : : ; xn) 1 i−1 i+1 n 1 i−1 i+1 j−1 j+1 n = f~0(x ; : : : ; x ; 0; x ; : : : ; x ) − f~0(x ; : : : ; x ; 0; x ; : : : ; x ; 0; x ; : : : ; x ) 1 i−1 i+1 j−1 j+1 n + f~j(x ; : : : ; x ; 0; x ; : : : ; x ; 0; x ; : : : ; x ) = f(x1; : : : ; xi−1; 0; xi+1; : : : ; xn); ~ i ~ since f0 coincides with f for x 2 ∆k satisfying x = 0 and similarly fj coincides with f for x 2 ∆k j ~ satisfying x = 0. An analogous argument shows that the restriction of f to U \ @j∆k is f~(x1; : : : ; xj−1; 0; xj+1; : : : ; xn) 1 j−1 j+1 n 1 j−1 j+1 n = f~0(x ; : : : ; x ; 0; x ; : : : ; x ) − f~0(x ; : : : ; x ; 0; x ; : : : ; x ) 1 j−1 j+1 n + f~j(x ; : : : ; x ; 0; x ; : : : ; x ) = f(x1; : : : ; xj−1; 0; xj+1; : : : ; xn); which proves the inductive step as needed. We will also need the following, often called Whitney's approximation theorem. PROOF OF DE RHAM'S THEOREM 3 Theorem 2.2 (Whitney). Let X be a smooth manifold with or without boundary, Y a smooth manifold without boundary, and f : X ! Y a continuous map smooth on a closed subset C ⊂ X. We have that F is homotopic relative to C to a smooth map X ! Y . Proof. [2, Theorem 6.26]. Now, we are ready to prove our desired lemma that constructs, for any simplex, a homotopy from it to a smooth simplex such that the homotopies respect restriction to boundary faces. k Lemma 2.3. For every k-simplex σ : ∆k ! M from the standard k-simplex of R to M, there exists a continuous map Hσ : ∆k × I ! M satisfying the following: (i) Hσ is a homotopy from σ(x) = Hσ(x; 0) to a smooth k-simplex σ~(x) = Hσ(x; 1). ∼ (ii) For each boundary face inclusion Fi;k : ∆k−1 = @i∆k,!∆k, Hσ◦Fi;k (x; t) = Hσ(Fi;k(x); t) for (x; t) 2 ∆k−1 × I. (iii) For σ a smooth k-simplex, Hσ is the constant homotopy, i.e., Hσ(x; t) = σ(x). Proof. For σ smooth, Hσ can be taken to be the constant homotopy, which is easily seen to satisfy properties (1) and (2). We prove the problem statement for not necessarily smooth σ by induction on the dimension of σ. An arbitrary 0-simplex σ : ∆0 ! M is smooth, so the base case is trivial. Suppose now that for all ` < j, we have constructed for each `-simplex σ0 a corresponding homotopy Hσ0 satisfying the lemma's properties. Consider an arbitrary non-smooth j-simplex σ : ∆k ! M. Define the subset S ·= (∆k × f0g) [ (@∆k × I) ⊂ ∆k × I, and define H0 : S ! M by ( σ(x) if x 2 ∆ ; t = 0; · k H0(x; t) ·= −1 Hσ◦Fi;k (Fi;k (x); t) if x 2 @i∆k: To show that H0 is continuous, we need to show that the functions of the above cases coincide on overlaps so that we can apply the gluing lemma.
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