My personal motivation: Let $\omega$ be a $C^1$ closed differential $k$-form on a smooth manifold $M$. Then there exists a $C^\infty$ closed differential $k$-form $\alpha$ and a $C^1$ exact form $d\beta$ such that

$$\omega = \alpha + d\beta.$$

This allows application of results from de Rham cohomology to the study of differential forms like $\omega$, which have less than $C^\infty$ regularity.

In his answer to my previous question, Ted Shifrin showed me that the above is indeed true, and that it follows from the result from de Rham's book described below. I'm trying to understand the details of that proof, but I'm stuck on a few things. Any help would be much appreciated.


Let $\Omega^k_r(M)$ be the space of $C^r$ differential $k$-forms on a smooth manifold $M$. In Theorem 12 (p. 68) of deRham's book, among other things he proves that there are a family of "smoothing operators" $R: \Omega^k_1(M) \to \Omega^k_\infty (M)$ and a cochain homotopy from $R$ to the identity:

$$R \omega - \omega = d A \omega + A d \omega,$$

where $A$ represents a family of operators $A_k:\Omega_1^k(M)\to\Omega_1^{k-1}(M)$. Furthermore, $R$ commutes with the exterior derivative $d$. (de Rham actually states this result more generally for currents, but I'm just interested in differential forms for now.)

Here is my sketch of de Rham's proof. First, show that such operators $\tilde{R}$ and $\tilde{A}$ can be defined for differential forms on Euclidean space $\mathbb{R}^n$. Next, let $U_i$ be a locally finite cover of $M$ by regular coordinate balls, with each $U_i$ diffeomorphic to the open unit ball $B_1(0) \subset \mathbb{R}^n$ via a diffeomorphism that extends to a diffeomorphism of a larger neighborhood $V_i$ to $B_2(0)$. Letting $f_i:M\to \mathbb{R}$ be a smooth bump function supported$\color{red}{^1}$ on a neighborhood of $\bar{U}_i$ and such that $f_i \equiv 1$ on a smaller neighborhood of $\bar{U}_i$, we define operators $R_i:\Omega^k_1(M)\to \Omega^k_\infty(M)$, $A_i:\Omega^k_1(M)\to \Omega^{k-1}_1(M)$ via

$$R_i \omega := \tilde{R}_i (f_i\omega) + (1-f_i)\omega, \qquad A_i \omega := \tilde{A}_i (f_i\omega),$$

where $\tilde{R}_i:\Omega^k_1(V_i)\to \Omega^k_\infty(V_i)$ and $\tilde{A}_i:\Omega^k_1(V_i) \to \Omega^{k-1}_1(V_i)$ are the operators constructed earlier on $V_i \approx \mathbb{R}^n$. Now define the operators $R^{(h)}$ and $A^{(h)}$ via

$$ R^{(h)}:= R_hR_{h-1}\cdots R_2 R_1, \qquad A^{(h)}:= A_h R^{(h-1)}$$

and finally define the smoothing operator $R:\Omega^k_1(M)\to \Omega^k_\infty(M)$ and operator $A:\Omega^k_1(M)\to \Omega^{k-1}_1(M)$ via

$$R:= \lim_{h\to \infty} R^{(h)}, \qquad A:= \sum_{h=1}^\infty A^{(h)},$$

and argue that these operators are well-defined using local finiteness of the cover $(U_i)$. Now assuming$\color{red}{^2}$ that

$$\forall i: R_i \omega - \omega = d A_i \omega + A_i d \omega,$$

it follows that

$$\forall h: R^{(h)}\omega - R^{(h-1)}\omega = d A_h R^{(h-1)}\omega + A_h d R^{(h-1)} = d A^{(h)} \omega + A^{(h)}d \omega,$$

where in the last equality we have assumed that $d$ commutes$\color{red}{^3}$ with $R_{(h-1)}$. Summing now from $h=1$ to $\infty$ and using the definition of $A$ and the fact that the left hand sum is telescoping, we find

$$R\omega - \omega = dA\omega + Ad\omega$$

as desired.

Questions (see superscripts above):

$\color{red}{1.}$ Why do we need $f_i$ to be supported on a neighborhood of $\bar{U}_i$, and equal to 1 on a smaller neighborhood of $\bar{U}_i$? It might only be relevant for the additional results in de Rham's book on $R\omega$ approximating $\omega$, but could it possibly also be relevant for the two questions below?

$\color{red}{2.}$ Why is this true? Calculating, we find

$$R_i \omega - \omega = \tilde{R}_i (f_i \omega) - f_i \omega = d \tilde{A_i}(f_i \omega) + \tilde{A_i}d (f_i\omega) = d A_i \omega + \tilde{A_i}d (f_i\omega) .$$

But $d (f_i\omega) = f_i d\omega+ df_i \wedge \omega$, so

$$R_i \omega - \omega = d A_i \omega + A_i d \omega + \tilde{A_i}(df_i \wedge \omega),$$

and hence the term $\tilde{A_i}(df_i \wedge \omega)$ seems to spoil the desired equality.

$\color{red}{3.}$ Why is this true? Similarly to question 2 above, for any $i$ we have

$$d R_i \omega = d \tilde{R}_i (f_i \omega) = \tilde{R}_i d (f_i \omega) = R_i d\omega + \tilde{R}_i (df_i\wedge \omega), $$

and so the term $\tilde{R}_i (df_i\wedge \omega)$ appears to spoil the desired equality.

Edit: construction of the local operators. Since this might be relevant, $\tilde{R}_i$ and $\tilde{A}_i$ can be constructed as follows, for any $\epsilon > 0$ (hereafter identifying $U_i$ with $\mathbb{R}^n$). Let $\tau_\epsilon:\mathbb{R}^n \to \mathbb{R}_{\geq 0}$ be a smooth "mollifier" function such that $\text{supp}(\tau)\subset B_\epsilon(0)$ and such that $\int_{\mathbb{R}^n}\tau_\epsilon(h) dh = 1.$ Now, identifying $h\in \mathbb{R}^n$ with a constant vector field, define $\tilde{R}_i:\Omega^k_1(\mathbb{R}^n)\to \Omega^k_\infty(\mathbb{R}^n)$ and $\tilde{A}_i:\Omega^k_1(\mathbb{R}^n)\to \Omega^{k-1}_1(\mathbb{R}^n)$ via

$$\tilde{R}_i:= \int_{\mathbb{R}^n}\omega_{x+h} \tau_\epsilon(h) dh$$

$$\tilde{A}_i:= \int_{\mathbb{R}^n}\int_0^1 (h\lrcorner \omega_{x+th})\tau_\epsilon(h) dt dh,$$

with these integrals to be understood as $\Omega(\mathbb{R}^n)$-valued integrals over $\mathbb{R}^n$, and with $\lrcorner$ denoting the interior product/contraction. To see that this works, use the Lie derivative $L_h \omega$ to write

$$\omega_{x+h}-\omega_x = \int_0^1 (L_h \omega)_{x+th} dt = \int_0^1 h \lrcorner (d\omega)_{x+th}dt + d\int_0^1 h\lrcorner \omega_{x+th}dt, $$

where all exterior derivatives are with respect to $x$ only.


$$\tilde{R}_i \omega - \omega = \int (\omega_{x+h}-\omega_x)\tau_\epsilon(h) dh$$ $$ = \int_{\mathbb{R}^n}\int_0^1 (h \lrcorner (d\omega)_{x+th})\tau_\epsilon(h)dt dh + d\int_{\mathbb{R}^n}\int_0^1 (h\lrcorner \omega_{x+th})\tau_\epsilon(h) dt dh $$ $$ = d \tilde{A}_i \omega + \tilde{A}_i d \omega,$$

as desired. (Note also that $d$ commutes with $\tilde{R}_i$ as claimed before. To see that $\tilde{R}_i\omega$ is indeed $C^\infty$, make the change of variables $\rho = x + h$ and differentiate with respect to $\rho$ under the integral defining $\tilde{R}_i$.)

  • $\begingroup$ I'm not going to answer much, but note you do have typos in your definitions of $A_i$ and $\tilde A_i$, as they should map to $\Omega^{k-1}_1$. The main point somehow should be that because $f_i$ is identically $1$ on a big closed subset of $V_i$, it follows that $df_i\wedge\omega$ vanishes identically on that set. Aren't there further properties of $\tilde A_i$? $\endgroup$ – Ted Shifrin Jun 23 '18 at 23:49
  • $\begingroup$ @TedShifrin thanks for the typo catches. I also edited the question to add explicit constructions of $\tilde{R}_i, \tilde{A}_i$. I understand that if $f_i\equiv 1$ on a big closed subset of $V_i$ then $df_i\wedge \omega \equiv 0$ on that subset, but what about the other forms $df_j \wedge \omega$ which are outside of their corresponding big subsets of $V_j$? Why doesn't that spoil things? $\endgroup$ – Matthew Kvalheim Jun 24 '18 at 1:24

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