PROOF OF DE RHAM’S THEOREM

PETER S. PARK

1. Introduction Let M be a smooth n-dimensional . Then, de Rham’s theorem states that the de Rham of M is naturally isomorphic to its singular cohomology with coefficients in R; in particular, 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 = ω for closed forms ω, and the latter, information about the “holes” in each given , which is made precise by singular . The natural will be given by a version of Stokes’ theorem, which describes a 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, 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- ∗ σ : ∆k → M and a k-form ω on a manifold M, we wish to integrate over ∆k the pullback form σ ω. 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 ∈ 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 ∈ 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 ∞ map. For the usual free 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 ∞ 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 ∞ ∞ ∞ Ker(∂ : Ck (M) → Ck−1(M)) Hk (M) = ∞ ∞ Im(∂ : Ck+1(M) → Ck (M)) ∞ is well-defined. Since the inclusion i : Ck (M),→Ck(M) satisfies ι ◦ ∂ = ∂ ◦ ι, we see that it induces ∞ 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 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 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, 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 ˆ 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 ∈ ∂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 ∈ ∂∆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 ∈ ∆k satisfying x = 0 and similarly fj coincides with f for x ∈ ∆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 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) ∈ ∆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 × {0}) ∪ (∂∆k × I) ⊂ ∆k × I, and define H0 : S → M by ( σ(x) if x ∈ ∆ , t = 0, · k H0(x, t) ·= −1 Hσ◦Fi,k (Fi,k (x), t) if x ∈ ∂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. For (x, 0) ∈ S such that x ∈ ∂i∆k, the inductive hypothesis implies that −1 −1 Hσ◦Fi,k (Fi,k (x), 0) = σ ◦ Fi,k(Fi,k (x)) = σ(x), −1 and indeed σ(x) and Hσ◦Fi,k (Fi,k (x), 0) agree. Next, suppose (x, t) ∈ S such that x ∈ ∂i∆k ∩ ∂j∆k, where without loss of generality we assume 0 ≤ i < j ≤ k. Note that we have the homeomorphism ∆k−2 → ∂i∆k ∩ ∂j∆K given by Fi,k ◦ Fj,k−1 = Fj,k ◦ Fi,k−1, and thus we can write x = Fi,k ◦ Fj,k−1(y) = Fj,k ◦ Fi,k−1(y) for some y ∈ ∆k−2. Then, by the inductive hypothesis, we have −1 Hσ◦Fi,k (Fi,k (x), t) = Hσ◦Fi,k (Fj,k−1(y), t) = Hσ◦Fi,k◦Fj,k−1 (y, t) and −1 Hσ◦Fj,k (Fj,k (x), t) = Hσ◦Fj,k (Fi,k−1(y), t) = Hσ◦Fj,k◦Fi,k−1 (y, t), −1 −1 showing that Hσ◦Fi,k (Fi,k (x), t) = Hσ◦Fj,k (Fj,k (x), t) as needed. Next, we wish to extend H0 to the larger domain ∆k × I. To do so, we compose it with a k continuous r : ∆k ×I → S; note that r can be defined by projecting from (x0, 2) ∈ R ×R for an interior point x0 ∈ ∆K . This gives us the continuous map H ·= H0 ◦ r : ∆k × I → M. However, while H is a homotopy from σ to some k-complex σ0 ·= H(·, 1) satisfying property (3), we are not guaranteed that σ0 is smooth. Thus, we need to modify H to be a homotopy to a smooth k-simplex. First, we check that H is smooth on ∂i∆k ×{1}. Indeed, note that ∂i∆k ×{1} is contained in S, so H coincides with H0 on each such set. Every boundary simplex σ ◦ Fi,k satisfies property (1) by the inductive hypothesis, so we have that H0 is smooth on all boundary simplices ∂i∆k. By Lemma 2.1, it follows that H0 is smooth on all of ∂∆k × {1}. Next, take a continuous 00 0 k extension σ of σ to an open set U ⊂ R containing ∆k. By Theorem 2.2, there exists a homotopy H : U × I → M from σ00 to a smooth map H(·, 1) : U → M, satisfying that H(x, t) = σ00(x) for 4 PETER S. PARK

0 all x ∈ ∂∆k. Restricting to ∆k × I, we obtain a homotopy H˜ : ∆k × I → M from σ to a smooth ˜ 0 k-simplexσ ˜ ·= H(·, 1), similarly satisfying that H(x, t) = σ (x) for all x ∈ ∂∆k. k ◦ Choose a u : R → M that has values 0 < u(x) < 1 for x ∈ ∆k and equals 1 on ∂∆k. We use the homotopies H and H˜ to define the homotopy   t  ◦ H x, u(x) if x ∈ ∂∆k or (x ∈ ∆k and 0 ≤ t ≤ u(x)), Hσ(x, t) ·=   ˜ t−u(x) ◦ H x, 1−u(x) if x ∈ ∆k and u(x) ≤ t ≤ 1. 0 ˜ ◦ Since H(x, 1) = σ (x) = H(x, 0), we have that Hσ is continuous on ∂∆k × I by the gluing lemma. t Moreover, Hσ(x, t) is given by the continuous function H(x, u(x) ) in a neighborhood of ∂∆k ×[0, 1). Finally, we show that Hσ is continuous on ∂∆k × {1}. For an arbitrary x0 ∈ ∂∆k, take an open neighborhood U ⊂ M of H(x0, 1). By continuity, there exists ε1 > 0 such that for all x ∈ ∆k and t ∈ [0, u(x)], we have that |(x, t) − (x0, 1)| < ε1 implies H(x, t/u(x)) ∈ U. Furthermore, since I is compact and H˜(x0, t) = H˜(x0, 0) = H(x0, 1) = Hσ(x0, 1) ∈ U for all t ∈ I, it follows that there exists ε2 > 0 such that for all x ∈ ∆k, we have that |x − x0| < ε2 implies H˜(x, t) ∈ U for all t ∈ I, ˜ t−u(x) and in particular, H(x, 1−u(x) ) ∈ U . Thus, |(x, t)−(x0, 1)| < min(ε1, ε2) implies that Hσ(x, t) ∈ U, which shows that Hσ is continuous at (x0, 1) as needed. We have that Hσ is a homotopy from Hσ(·, 0) = H(·, 0) = σ to the k-simplex defined by ˜ ◦ 0 ˜ Hσ(x, 1) = H(x, 1) =σ ˜(x) on x ∈ ∆k and Hσ(x, 1) = H(x, 1) = σ (x) = H(x, t) =σ ˜(x) on x ∈ ∂∆k (where t is arbitrary in the latter equality). This shows that Hσ is a homotopy from σ to the smooth k-simplexσ ˜, verifying property (1). Moreover, we have Hσ = H on ∂∆k × I by construction, verifying property (2).  Equipped with the above lemma, we can achieve our originally stated goal of proving that smooth singular homology is isomorphic to singular homology. ∞ Theorem 2.4. The map ι∗ : Hk (M) → Hk(M) is an isomorphism. Proof. Let i0 : ∆k,→∆k × I be defined by x 7→ (x, 0), and similarly, let i1 : ∆k,→∆k × I be defined by x 7→ (x, 1). Note that these are clearly smooth embeddings, and that for a k-simplex σ, the homotopy Hσ defined in Lemma 2.3 satisfies that Hσ ◦ i1 is a smooth k-simplex by property (1). ∞ This allows us to define a s : Ck(M) → Ck (M) by sending σ 7→ Hσ ◦ i1 and extending linearly. One observes that s is a chain map, since k ! k k X i X i X i s ◦ ∂(σ) = s (−1) σ ◦ Fi,k = (−1) Hσ◦Fi,k ◦ i1 = (−1) Hσ ◦ (Fi,k × IdI ) ◦ i1 i=0 i=0 i=0 k X i = (−1) Hσ ◦ i1 ◦ Fi,k = ∂(Hσ ◦ i1) = ∂ ◦ s(σ), i=0 ∞ where we have used property (2) of Hσ. Thus, s induces a homomorphism s∗ : Hk(M) → Hk (M), ∞ which we will show to be the inverse of ι∗ : Hk (M) → Hk(M). Note first that property (3) of Hσ shows that s ◦ ι = Id ∞ , so that in particular, s∗ ◦ ι∗ = Id ∞ . Ck (M) Hk (M)

For the converse result that ι∗ ◦ s∗ = IdHk(M), it suffices to show that a chain homotopy h :

Ck(M) → Ck+1(M) such that ∂ ◦ h + h ◦ ∂ = ι ◦ s − IdCk(M) exists. For a map from ∆` to some subset of a Euclidean space V , denote the affine transformation sending δi to vi ∈ V by A(v0, . . . , v`). Then, we define

Gi,k ·= A((δ0, 0),..., (δi, 0), (δi, 1),..., (δk, 1)), which is clearly a well-defined map Gi,k : ∆k+1 → ∆k × I. One can compute that the maps Gi,k ◦ Fi,k+1,Gi−1,k ◦ Fi,k+1 : ∆k → ∆k × I coincide as

(2.1) Gi,k ◦ Fi,k+1 = Gi−1,k ◦ Fi,k+1 = A((δ0, 0),..., (δi−1, 0), (δi, 1),..., (δk, 1)). PROOF OF DE RHAM’S THEOREM 5

Also, one can compute that the map (Fi,k × IdI ) ◦ Gj,k−1 : ∆k → ∆k × I is given by ( Gj+1,k ◦ Fi,k+1 if i ≤ j, (2.2) (Fi,k × IdI ) ◦ Gj,k−1 = Gj,k ◦ Fi+1,k+1 if i > j.

Define h : Ck(M) → Ck+1(M) by

k X i h(σ) ·= (−1) Hσ ◦ Gi,k. i=0 We see that k ! k k−1 X i X X i+j h ◦ ∂(σ) = h (−1) σ ◦ Fi,k = (−1) Hσ◦Fi,k ◦ Gj,k−1 i=0 i=0 j=0 k k−1 X X i+j = (−1) Hσ ◦ (Fi,k × IdI ) ◦ Gj,k−1 i=0 j=0     X i+j X i+j =  (−1) Hσ ◦ Gj+1,k ◦ Fi,k+1 +  (−1) Hσ ◦ Gj,k ◦ Fi+1,k+1 , 0≤i≤j≤k−1 0≤j

k ! k k+1 X i X X i+j ∂ ◦ h(σ) = ∂ (−1) Hσ ◦ Gi,k = (−1) Hσ ◦ Gi,k ◦ Fj,k+1 i=0 i=0 j=0  k+1  k+1  X X i+j X =  (−1) Hσ ◦ Gi,k ◦ Fj,k+1 −  Hσ ◦ Gj−1,k ◦ Fj,k+1 0≤i

k+1    X X i+j +  Hσ ◦ Gj,k ◦ Fj,k+1 +  (−1) Hσ ◦ Gi,k ◦ Fj,k+1 , j=1 0≤j j. Note however that the first and last subsums of the above together equal the negative of the expression for h ◦ ∂(σ). Also, by applying (2.1), one gets that the second and fourth subsums cancel out to

∂ ◦ h(σ) = Hσ ◦ G0,k ◦ F0,k+1 − Hσ ◦ Gk,k ◦ Fk+1,k+1

= Hσ ◦ A((δ0, 0),..., (δk, 0)) − Hσ ◦ A((δ0, 1),..., (δk, 1)) = Hσ ◦ i1 − Hσ ◦ i0. Thus, overall we have

h ◦ ∂(σ) + ∂ ◦ h(σ) = Hσ ◦ i1 − Hσ ◦ i0 = s(σ) − σ = ι ◦ s(σ) − σ, as needed. 

3. The Main Proof k Consider a closed k-form ω ∈ Ω (M) and a smooth k-simplex σ : ∆k → M. Define the of ω over σ by Z Z ω ·= σ∗ω, σ ∆k 6 PETER S. PARK

k P` considering ∆k ⊂ R as the domain of integration. Naturally, for a smooth chain c = i=1 ciσi ∈ ∞ Ck (M) we define the integral of ω over σ by ` Z X Z ω ·= ci ω. c i=1 σi One can prove an analogue of Stokes’ theorem for this notion of integral. ∞ k−1 Theorem 3.1 (Stokes). For c ∈ Ck (M) and ω ∈ Ω (M), we have Z Z ω = dω. ∂c c Proof. This is a straightforward derivation from the usual Stokes’ theorem. See [2, Theorem 18.12] for a detailed proof.  As mentioned in the introduction, the above version of Stokes’ theorem allows us to define a k k map I : HdR(M) → H (M; R). First, note that by the universal coefficient theorem, the natural k map H (M; R) → Hom(Hk(M), R) is an isomorphism, which allows us to define I as follows. Let k ∼ ∞ [ω] ∈ HdR(M) and [c] ∈ Hk(M) = Hk (M), where throughout the remainder of this exposition, we take c to be smooth by appealing to Theorem 2.4. Then, define I [ω] to be the cohomology class corresponding to the element of Hom(Hk(M), R) defined by Z [c] 7→ ω. c 0 Note that this map Hk(M) → R is well-defined, since for another smooth k-cycle c that represents [c], it follows from Theorem 2.4 that c − c0 = ∂c00 for a smooth (k + 1)-chain c00, so Z Z Z Z ω − ω = ω = dω = 0 c c0 ∂c00 c00 k by Theorem 3.1. Moreover, I is well-defined as a map from HdR(M), since for exact ω = dη, we have again by Theorem 3.1 that Z Z Z ω = dη = η = 0, c c ∂c and linearity is clear. We name I the de Rham map, which is natural in the following way. Proposition 3.2. The de Rham map I satisfies the following properties: (a) For a smooth map of smooth f : M → N, we have the commutative diagram

f ∗ k / k HdR(N) HdR(M)

I I  ∗  k f k H (N; R) / H (M; R). (b) For open sets U, V of M such that U ∪ V = M, we have the commutative diagram

k−1 δ / k HdR (U ∩ V ) HdR(M)

I I   k−1 ξ k H (U ∩ V ; R) / H (M; R), where δ (respectively, ξ) are the boundary maps in the Mayer–Vietoris sequences of de Rham cohomology (respectively, of singular cohomology). PROOF OF DE RHAM’S THEOREM 7

Proof. (a). For a smooth k-simplex σ and ω ∈ Ωk(N), we have Z Z Z Z ∗ ∗ ∗ ∗ ∗ ∗ I (f [ω])([σ]) = F ω = σ f ω = (f◦σ) ω = ω = I ([ω])◦f∗([σ]) = f (I ([ω]))([σ]). σ ∆k ∆k f◦σ Extend linearly to verify the desired property. k−1 (b). Let [ω] ∈ HdR (U ∩ V ). Appealing to the well-definedness of the Mayer–Vietoris sequence k k maps, choose ηU ∈ HdR(U) and ηV ∈ HdR(V ) such that ω = ηU |U∩V − ηV |U∩V , and in particular, δ([ω]) is equal to dηU on U and dηV on V . Likewise, let [b] ∈ Hk(M), where the representative chain b can be chosen to be smooth and satisfy b = bU + bV for smooth chains bU ∈ Ck(U) and bV ∈ Ck(V ), so that ∂∗([b]) = [∂bU ] = −[∂bV ], where ∂∗ denotes the map induced by the boundary map ∂ : Ck(M) → Ck−1(U ∩ V ) on chains. We have Z  Z  I (δ([ω]))([b]) = I (δ([ω]))([bU ]) + I (δ([ω]))([bV ]) = dηU + dηV bU bV and Z  Z  Z  Z  Z  Z  I ([ω])(∂∗([b])) = ω − ω = ηU − −ηV = ηU + ηV . ∂bU ∂bV ∂bU ∂bV ∂bU ∂bV But the left of both expressions are equal by Theorem 3.1, as are the right integrals. This holds for arbitrary b, so I (δ([ω]) = I ([ω]) ◦ ∂∗ = ξ ◦ I ([ω]), as needed.  We are now ready to prove our main theorem. Before we do so, we introduce the following k k terminology. A smooth manifold M is de Rham if I : HdR(M) → H (M; R) is an isomorphism for all k ≥ 0. An open {Ui}i∈I of M is a de Rham open cover if all finite intersections

Ui1 ∩ · · · ∩ Uir are de Rham. A de Rham basis is a de Rham open cover that is also a basis. With this terminology, de Rham’s theorem can be stated concisely. Theorem 3.3 (de Rham). Every smooth manifold M is de Rham.

Proof. First, we show that every smooth manifold M with a finite de Rham cover, say {U1,...,Ur} is de Rham. We prove this by induction on r. The base case r = 1 is trivial. Suppose that M has a de Rham cover {U, V }. We have the following commutative diagram of Mayer–Vietoris sequences:

k−1 k−1 / k−1 / k HdR (U) ⊕ HdR (V ) HdR (U ∩ V ) HdR(M)

I ⊕I I I    k−1 k−1 / k−1 / k HdR (U; R) ⊕ HdR (V ; R) H (U ∩ V ; R) H (M; R)

/ k k / k HdR(U) ⊕ HdR(V ) HdR(U ∩ V )

I ⊕I I   k k k / H (U; R) ⊕ H (V ; R) / H (U ∩ V ; R). The commutativity of the second-to-left square follows from Proposition 3.2(b), and commutativity for the other squares of the diagram follows from Proposition 3.2(a). The inductive hypothesis implies that the first, second, fourth, and fifth vertical maps are , so by the five lemma, the third vertical map is also an isomorphism, as needed. Second, we show that every smooth manifold M with a de Rham basis {Ui}i∈I is de Rham. Take −1 a smooth positive exhaustion f : M → R, i.e., a map such that f ((−∞, c]) is compact for all · −1 S c ∈ R. Then, for ` ∈ Z≥0, we can define A` ·= f ([`, ` + 1]) and note that M = `≥0 A`. The 0 · −1 1 3 open set A` ·= f ((` − 2 , ` + 2 )) contains A`, and can be written as a union of basis open sets Ui. 8 PETER S. PARK

Since these cover A`, finitely many of them (the union of which we denote as B`) cover A`. By our 0 work above, B` is de Rham. Since B` ⊆ A`, the open sets [ [ U ·= B` and V ·= B` odd `≥0 even `≥0 F are each a disjoint union of de Rham smooth manifolds B`. However, for a disjoint union j∈J Nj F of countably many de Rham manifolds Nj, the inclusion maps ιj : Nj → j∈J Nj induce an isomorphism between the direct product of de Rham (respectively, singular) cohomology groups of F Nj with the de Rham (respectively, singular) cohomology groups of j∈J Nj, and it follows from F Proposition 3.2 that these isomorphisms commute with the de Rham map, showing that j∈J Nj is de Rham. This verifies that U and V are de Rham. Furthermore, U ∩ V is de Rham, since it is the disjoint union of B` ∩ B`+1 for ` ≥ 0, and B` ∩ B`+1 has a finite de Rham open cover given by Ui ∩ Ui0 such that Ui is one of the finitely many basis open sets chosen in defining B` and Ui0 , one of the finitely many basis open sets chosen in defining B`+1. Thus, {U, V } is a finite de Rham cover of M, so our previous work shows that M is de Rham. n We are now ready to prove the theorem. First, note that every convex open set U ⊂ R is k de Rham. Indeed, by the Poincar´elemma, HdR(U) is 1-dimensional (generated by the constant k ∼ function 1 : M → R of value 1) for k = 0 and trivial for k > 0, and similarly, H (U; R) = Hom(Hk(U), R) is 1-dimensional (generated by the dual of any singular 0-simplex σ : ∆0 = {0} → M, which is automatically smooth and a choice for which we fix) for k = 0 and trivial for k > 0. Then, Z I ([1])([σ]) = σ∗1 = (1 ◦ σ)(0) = 1, ∆0 n showing that the map I is nontrivial and thus an isomorphism. Second, every open set U ⊂ R is de Rham. Indeed, U has a basis of Euclidean open balls, which are convex (and thus, de Rham) and consequently, whose finite intersections are convex (and thus, de Rham). Thus, this basis is de Rham, verifying that U is de Rham. Finally, every smooth manifold M has a basis of atlas n charts, and each finite intersection of such charts is diffeomorphic to an open set of R and is thus de Rham, so the basis is de Rham. This concludes our proof that M is de Rham.  4. Example: de Rham Cohomology of Sn We mention a brief application of de Rham’s theorem regarding differential forms on Sn. Since Sn is a connected orientable n-manifold, there exists a choice of (nonvanishing) volume form of Sn; this is a closed, but non-exact n-form, and thus, its cohomology class generates the top de n n ∼ Rham cohomology group HdR(S ) = R. Likewise, the cohomology class of the constant function 1 n 0 n ∼ : S → R of value 1 generates the 0th de Rham cohomology group HdR(S ) = R. One can then ask the question: for 0 < k < n, are there closed but non-exact k-forms on Sn? By de Rham’s k n k n ∼ theorem, we know that I : HdR(S ) → H (S ; R) = 0 is an isomorphism, showing that all closed k-forms on Sn are exact.

References [1] Glen E. Bredon, and geometry, Graduate Texts in Mathematics, vol. 139, Springer-Verlag, New York, 1993. MR1224675 [2] John M. Lee, Introduction to smooth manifolds, Second, Graduate Texts in Mathematics, vol. 218, Springer, New York, 2013. MR2954043 [3] Frank W. Warner, Foundations of differentiable manifolds and Lie groups, Graduate Texts in Mathematics, vol. 94, Springer-Verlag, New York-Berlin, 1983. Corrected reprint of the 1971 edition. MR722297

Department of Mathematics, Harvard University, Cambridge, MA 02138 E-mail address: [email protected]