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?

  • 1
    $\begingroup$ I asked this a while ago here. If the accepted answer works for you, this can be closed as a duplicate. $\endgroup$ – Pedro Tamaroff Jan 11 at 11:23
  • $\begingroup$ @PedroTamaroff your link seem to take me to my post :) $\endgroup$ – ChikChak Jan 11 at 11:48
  • $\begingroup$ Link should work now. $\endgroup$ – Pedro Tamaroff Jan 11 at 12:20
  • $\begingroup$ @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? $\endgroup$ – ChikChak Jan 11 at 13:17
  • $\begingroup$ 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 $\mathbb{R}^2$. $\endgroup$ – 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 $f(x) \d x_I = \omega$.


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