$d(I\omega) + I(d\omega) = \omega$ for differential forms

Let $$U$$ be a convex connected and open set in $$\mathbb{R}^n$$, such that $$0\in U$$. For every $$k$$-differential form $$\omega$$, $$\omega =\sum_{i_1\lt\dots\lt i_k} c_{i_1,\dots,i_k}(x)dx{_{i_1}}\wedge\dots\wedge dx{_{i_k}},$$ we define a new $$k-1$$ differential form, $$I\omega$$ as follows -

$$\begin{split} (I\omega) = \sum_{i_1\lt\dots\lt i_k} \sum_{j=1}^kx_{i_j} &\left( \int_0^1s^{k-1}c_{i_1,\dots,i_k}(sx)ds\right)\\ & dx_{i_1}\wedge\dots dx_{i_{j-1}}\wedge dx_{i_{j+1}}\wedge\dots\wedge dx_{i_k} \end{split}$$

Prove that $$d(I\omega) + I(d\omega) = \omega$$.

So I know that -

$$d\omega = \sum_{i_1\lt\dots\lt i_k} \sum_{j=1}^n\frac{\partial c_{i_1,\dots,i_k}(x)}{\partial x_j}dx_j\wedge dx_{i_1}\wedge\dots\wedge dx_{i_k}$$

and therefore -

$$I(d\omega) = \sum_{i_1\lt\dots\lt i_k} \sum_{j=1}^k x_{i_j}(\int_0^1\frac{\partial c_{i_1,\dots,i_k}(sx)}{\partial x_j}ds)dx_{i_1}\wedge\dots dx_{i_{j-1}}\wedge dx_{i_{j+1}}\wedge\dots\wedge dx_{i_k}$$

All I need now is to calculate $$d(I\omega)$$, and put it into $$d(I\omega) + I(d\omega)$$ and show that it is equal to $$\omega$$.

However, I'm not sure - How can I calculate $$d(I\omega)$$?

What is $$d(I\omega)=\sum_{i_1\lt\dots\lt i_k}x_{i_j}\sum_{j=1}^n\frac{\partial(\int_0^1s^{k-1}c_{i_1,\dots,i_k}(sx)ds)}{\partial d_{x_j}}dx_{i_1}\wedge\dots dx_{i_{j-1}}\wedge dx_{i_{j+1}}\wedge\dots\wedge dx_{i_k}$$?

Or perhaps, is there any other way(some tricks) so I won't need to calculate it explicitly?

• I asked this a while ago here. If the accepted answer works for you, this can be closed as a duplicate. – Pedro Tamaroff Jan 11 at 11:23
• @PedroTamaroff your link seem to take me to my post :) – ChikChak Jan 11 at 11:48
• Link should work now. – Pedro Tamaroff Jan 11 at 12:20
• @PedroTamaroff In the answer to your question, he assumes that the differential forms are on $[0,1]\times \mathbb{R}^n$. Why can we assume that? – ChikChak Jan 11 at 13:17
• I think sure your formula for $I(\omega)$ is missing a factor of $(-1)^{j-1}$ inside the sum. Try setting $\omega = dx \wedge dy$ in $\mathbb{R}^2$. – 0x539 Jan 15 at 17:33

Why I think you formula contains an error $$\newcommand{\d}{\mathop{}\!d}$$

Let $$n = 2$$ and $$\omega = \d x \wedge \d y$$. Then according to your formula

$$I \omega = x \left(\int_0^1 s \d s \right) \d y + y \left( \int_0^1 s \d s\right) \d x = \frac12 (x \d y + y \d x)$$

so $$\d (I \omega) = \frac12(\d x \wedge \d y + \d y \wedge \d x) = 0$$ and since $$\d\omega = 0$$ the equation you want to prove does not hold. However if you add a factor of $$(-1)^{j-1}$$ to the inner sum in the definition of $$I \omega$$ then $$I \omega$$ would be $$\frac12(x \d y - y \d x)$$ in this example and everything works out. The formula for $$I$$ then is $$(I\omega) = \sum_{i_1\lt\dots\lt i_k} \sum_{j=1}^kx_{i_j} \left( \int_0^1s^{k-1}c_{i_1,\dots,i_k}(sx) \d s\right) (-1)^{j-1} \d x_{i_1}\wedge\dots \wedge \d x_{i_{j-1}}\wedge \d x_{i_{j+1}}\wedge\dots\wedge \d x_{i_k}$$

Proving $$\d(I \omega) + I(\d \omega) = \omega$$

Since $$\d$$ and $$I$$ are bot additive, it is sufficient to verify this for $$\omega = f(x) \d x_{i_1} \wedge \dots \wedge \d x_{i_k}$$. By further reordering of the coordinates we can assume $$\omega = f \d x_1 \wedge \dots \wedge d x_k$$.
Let's denote $$\d x_1 \wedge \dots \wedge d x_k$$ by $$\d x_I$$ and $$\d x_1 \wedge \dots \wedge \d x_{j-1} \wedge \d x_{j+1} \wedge \dots \wedge d x_k$$ by $$\d x_{I - j}$$. We have

$$I \omega = \sum_{j=1}^k x_j \left( \int_0^1 s^{k-1} f(s x) \d s \right) (-1)^{j-1} \d x_{I - j}$$ and \begin{align} \d (I \omega) &= \sum_{i = 1}^n \sum_{j=1}^k \frac{\partial}{\partial x_i} \left[x_j \int_0^1 s^{k-1} f(s x) \d s \right] \d x_i \wedge (-1)^{j-1} d x_{I - j} \\ &= \sum_{i=1}^k \frac{\partial}{\partial x_i} \left[x_i \int_0^1 s^{k-1} f(s x) \d s \right] \d x_I \\ &\phantom{=}+ \sum_{i = k + 1}^n \sum_{j=1}^k x_j \left( \int_0^1 s^k \frac{\partial f}{\partial x_i}(s x) \d s \right) (-1)^{j-1} (-1)^{k-1} d x_{I - j} \wedge d x_i \tag{1} \end{align} We can simplify the first sum: \begin{align} \sum_{i=1}^k \frac{\partial}{\partial x_i} \left[x_i \int_0^1 s^{k-1} f(s x) \d s \right] &= \int_0^1 k s^{k-1} f(s x) \d s + \int_0^1 s^k \sum_{i=1}^k x_i \frac{\partial f}{\partial x_i} (s x) \end{align}

Now $$\d \omega = \sum_{i = k + 1}^n \frac{\partial f}{\partial x_i} \d x_i \wedge \d x_I = (-1)^k \sum_{i = k + 1}^n \frac{\partial f}{\partial x_i} \d x_I \wedge \d x_i$$ and \begin{align*} I(\d \omega) = (-1)^k \sum_{i = k+1}^n &\left[ \sum_{j=1}^k x_j \left( \int_0^1 s^k \frac{\partial f}{\partial x_i}(s x) \right) (-1)^{j-1} \d x_{I - j} \wedge \d x_i \right.\\ &\phantom{[}\left. + x_i \left( \int_0^1 s^k \frac{\partial f}{\partial x_i}(s x) \right) (-1)^k \d x_I\right] \tag{2} \end{align*}

The first sum in $$(2)$$ is exactly the negative of the second sum in $$(1)$$ - $$(-1)^k$$ vs $$(-1)^{k-1}$$ - so they cancel when adding $$I(\d \omega)$$ and $$\d (I \omega)$$ What remains is $$\left( \int_0^1 k s^{k-1} f(s x) \d s + \int_0^1 s^k \sum_{i=1}^n x_i \frac{\partial f}{\partial x_i} (s x) \right) \d x_I$$ Since the sum on the right hand side is the total derivative of $$f(s x)$$ with respect to $$s$$, partial integration shows that this is equal to $$f(x) \d x_I = \omega$$.