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Suppose I have an $\ell$-form in $\Bbb R^n$ $$\omega=\sum_{i_1<\cdots<i_\ell}\omega_{i_1\cdots i_\ell} dx_{i_1}\wedge\cdots\wedge dx_{i_\ell}$$

I will say this is written in the canonical form. Having written it canonically, define $I\omega$ to be the $\ell-1$ form that sends $x\in\Bbb R^n$ to

$$I\omega(x)=\sum_{i_1<\cdots<i_\ell}\sum_{\alpha=1}^\ell (-1)^{\alpha-1}\int_0^1t^{\ell-1} \omega_{i_1\cdots i_\ell}(tx)dt \; x_{i_\alpha}\,\cdot dx_{i_1}\wedge\cdots \wedge \widehat{dx_{i_\alpha}}\wedge \cdots\wedge dx_{i_\ell}$$

where the hat means the term is ommited. We can denote this by $dx_{I,\alpha}$ for brevity. By linearity, we can focus on the basic forms $$\omega=f \;\cdot dx_{i_1}\wedge\cdots\wedge dx_{i_\ell}\;\;;\;\;i_1<\cdots <i_\ell$$

Again, in this case we will say this is written canonically. For brevity, will denote by $dx_I$ the full, ordered wedge product $dx_{i_1}\wedge\cdots\wedge dx_{i_\ell}$. In this case $$I\omega(x)=\sum_{\alpha=1}^\ell (-1)^{\alpha-1}\int_0^1 t^{\ell-1} f(tx) dt\; x_{i_\alpha} \cdot dx_{I,\alpha}$$

Now, when we take the derivative of this, we get two parts by the product rule. First, $$\tag 1 \sum\limits_{\alpha = 1}^\ell (-1)^{\alpha-1} {\int_0^1 {{t^{\ell - 1}}} f\left( {tx} \right)dt\;\cdot d{x_{{i_\alpha }}} \wedge dx_{I,\alpha}}\\ = \int_0^1 {\ell {t^{\ell - 1}}} f\left( {tx} \right)dt \cdot dx_I$$ since we move the form $dx_{i_\alpha}$, $\alpha-1$ places, filling the gap. While on the other hand we get $$\sum\limits_{\alpha = 1}^\ell {{{\left( { - 1} \right)}^{\alpha - 1}}\sum\limits_{j = 1}^n {\int_0^1 {{t^\ell }{D_j}} f\left( {tx} \right)dt}\, {x_{{i_\alpha }}}\;\cdot dx_j\wedge dx_{I,\alpha}} $$

Now, consider the $\ell+1$ form $$d\omega = \sum\limits_{j = 1}^n {{D_j}f} \;\cdot d{x_j} \wedge d{x_{{i_1}}} \wedge \cdots \wedge d{x_{{i_\ell }}}$$

This is not written canonically, but nevertheless we would like to find $I(d\omega)$. I should be getting that this is $$I(d\omega )\left( x \right) = \sum\limits_{j = 1}^n {\int_0^1 {{t^\ell }{D_j}} f\left( {tx} \right)dt} {x_j}\cdot d{x_I} - \sum\limits_{\alpha = 1}^\ell {{{\left( { - 1} \right)}^{\alpha - 1}}\sum\limits_{j = 1}^n {\int_0^1 {{t^\ell }{D_j}} f\left( {tx} \right)dt} x_{i_\alpha }\;\cdot d{x_j} \wedge d{x_{I,\alpha }}} $$

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up vote 9 down vote accepted

Here is a way around signs and combinatorics. Consider differential forms on $[0,1]\times \mathbb R^n$. Each such a form decomposes to ($t$ is the coordinate on $[0,1]$) $$\alpha=\alpha_0(t)+dt\wedge\alpha_1(t),$$ where $\alpha_0$ and $\alpha_1$ are forms on $\mathbb R^n$ with $t$-dependent coefficients. Let $$J\alpha=\int_0^1(\alpha_1(t))dt\in\Omega(\mathbb R^n)$$ (i.e. we integrate the coefficients of $\alpha_1$ w.r.t. $t$). Clearly $$dJ\alpha=\int_0^1(d\alpha_1)dt.$$ Notice that $$d\alpha=d\alpha_0+dt\wedge(\frac{\partial\alpha_0}{\partial t}-d\alpha_1).$$ This gives us $$Jd\alpha=\int_0^1(\frac{\partial\alpha_0}{\partial t}-d\alpha_1)dt=\alpha_0(1)-\alpha_0(0)-dJ\alpha.$$ This is the formula you want. To see it, lef $F:[0,1]\times\mathbb R^n\to\mathbb R^n$ be given by $$F(t,x^1,\dots,x^n)=(tx^1,\dots,tx^n),$$ so that $$I\omega=J(F^*\omega).$$

We thus have $$dI\omega+Id\omega=dJF^*\omega+JF^*d\omega=(dJ+Jd)F^*\omega=(F^*\omega)_0(1)-(F^*\omega)_0(0),$$ and here $$(F^*\omega)_0(1)=\omega$$ and $$(F^*\omega)_0(0)=0$$ unless $\omega$ is a function, when it's rather $$(F^*\omega)_0(0)=\omega(0).$$

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What does $\partial/\partial t$ mean when applied to a form? – Pedro Tamaroff Aug 26 '13 at 7:29
@PeterTamaroff: $\partial/\partial t$ applied to each coefficient (I should have written $d/dt$, but that $d$ would get confused with de Rham's $d$) – user8268 Aug 26 '13 at 8:04
OK. And what do you call $\Omega(\Bbb R^n)$? – Pedro Tamaroff Aug 26 '13 at 8:10
$\Omega(\mathbb R^n)$=differential forms on $\mathbb R^n$ – user8268 Aug 26 '13 at 8:16
Hmm... and what do you mean by the subscript $0$? For example, $(F ^\ast\omega)_0$? I am used to the notation in Spivak's book (Calculus on Manifolds). – Pedro Tamaroff Aug 26 '13 at 8:20

I think I got it. For each $j$, consider the $\ell+1$ form $$\eta=D_j f\cdot\;dx_j\wedge dx_{i_1}\wedge\cdots\wedge dx_{i_\ell}$$

Write this as $$\eta =D_j f\cdot\;dx_{j_1}\wedge dx_{j_2}\wedge\cdots\wedge dx_{j_{\ell+1}}$$ where $j_1=j$, $j_{k+1}=i_k$. Then $$I\eta=\sum_{\alpha=1}^{\ell+1}(-1)^{\alpha-1}\int_0^1 t^\ell D_jf(tx)dt\; dx_{j_\alpha} dx_{J,\alpha}$$ becomes $$\begin{align} I\eta &= \sum\limits_{\alpha = 1}^{\ell + 1} {{{( - 1)}^{\alpha - 1}}} \int_0^1 {{t^\ell }} {D_j}f(tx)dt\;d{x_{{j_\alpha }}}d{x_{J,\alpha }} \cr &= \int_0^1 {{t^\ell }} {D_j}f(tx)dt\;{x_{{j_1}}}d{x_{J,1}} - \sum\limits_{\alpha = 2}^{\ell + 1} {{{( - 1)}^{\alpha - 2}}} \int_0^1 {{t^\ell }} {D_j}f(tx)dt\;{x_{{j_\alpha }}}d{x_{J,\alpha }} \cr &= \int_0^1 {{t^\ell }} {D_j}f(tx)dt\;{x_j}d{x_I} - \sum\limits_{\alpha = 1}^\ell {{{( - 1)}^{\alpha -1}}} \int_0^1 {{t^\ell }} {D_j}f(tx)dt\;{x_{{i_\alpha }}}d{x_{J,\alpha+1 }} \cr &= \int_0^1 {{t^\ell }} {D_j}f(tx)dt\;{x_j}d{x_I} - \sum\limits_{\alpha = 1}^\ell {{{( - 1)}^{\alpha - 1}}} \int_0^1 {{t^\ell }} {D_j}f(tx)dt\;{x_{{i_\alpha }}}d{x_j} \wedge d{x_{I,\alpha }} \end{align} $$

as desired.

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+1 I feel how difficult it was typing this answer. – Babak S. Aug 27 '13 at 11:02
One could define $I[\omega]_x(\xi_1,\dotsb,\xi_{k-1})=\int_0^1\omega_{tx}(x,t\xi_1,\dotsb,t\xi_{k‌​-1})dt$. It's coordinate-free. From V.I.Arnold's Mathematical Methods of Classical Mechanics. – Frank Science Jan 13 '14 at 3:50

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