On page 198 of Arnold's Mathematical Methods of Classical Mechanics, he asks the reader to prove Cartan's formula $$\tag{1}L_X=\mathrm{d}i_X+i_X\mathrm{d}$$ where $L_X$ is the Lie derivative wrt. $X$, $\mathrm{d}$ is the exterior derivative, and $i_X$ is the interior derivative (interior product).

I am aware of the "usual" proof, i.e. to show that the action of $L_X$ on functions and differentials agrees with that of the action of the rhs. of (1). From that one may show this equality holds on all $p$-forms.

However, Arnold offers a hint that completely changes the nature of the problem:

We denote by $H$ the "homotopy operator" associating to a $k$-chain $\gamma: \sigma\to M$ the $(k + 1)$-chain $H\gamma: (I \times \sigma) \to M$ according to the formula $(H\gamma)(t, x) = g^t\gamma(x)$ (where $I = [0, 1]$). Then $g^1\gamma - \gamma = \partial(H\gamma) + H(\partial\gamma).$

($g^t$ is the flow of $X$. I understand how to obtain the equation for the boundary of the homotopy of $\gamma$.)

Since we are given the equation for the boundary of $H\gamma$, I thought it would be a good idea to integrate the differential of a $k$-form $\omega$ over $H\gamma$ and use Stokes' theorem, followed by the equation for the boundary: $$\int_{H\gamma}\mathrm{d}\omega=\int_{\partial(H\gamma)}\omega=\int_{g^1\gamma-\gamma-H(\partial\gamma)}\omega$$ I don't see how this helps. Alternatively, on the right hand side, we can use the equation $\int_{f(D)}\omega=\int_Df^*\omega$ ($f^*$ is the pullback of $f$) to write $$\int_{H\gamma}\mathrm{d}\omega=\int_{I\times\sigma}(H\gamma)^*\mathrm{d}\omega$$ Unfortunately, I don't know how to calculate $(H\gamma)^*\mathrm{d}\omega$. How does one continue here?


1 Answer 1


First of all, I think the formula for the homotopy should be $$ g^1 \gamma -\gamma = \partial(H\gamma) - H(\partial \gamma).$$ As the left hand side is the top minus the bottom, the right hand side then should be the boundary of $I\times \sigma$ minus the cylindrical part of the boundary ($I\times \partial \gamma$).

Using that, we proceed to the proof. By fundamental theorem of calculus and $g^{s+t} = g^s g^t$, for any $k$-form $\omega$ we have

\begin{align} (g^1)^*\omega - \omega &= \int_0^1\frac{d}{dt} (g^t)^*\omega|_{t=s} \ ds \\ &=\int_0^1 (g^s)^* \frac{d}{dt} (g^t)^* \omega \bigg|_{t=0} ds \\ &= \int_0^1 (g^s)^* \mathcal L_X \omega \ ds. \end{align}

If we integrate this on $\gamma$, by Fubini's theorem, $$\int_\gamma \int_0^1 (g^s)^* \mathcal L_X \omega \ ds = (-1)^k\int_0^1 \int_\gamma H_s^* \mathcal L_X \omega \ ds,$$ where $H_s : \sigma \to M$ is given by $H_s(x) = H(s, x) = g^s \gamma(x)$. Note that $H_0 = \gamma$.

On the other hand, $$\begin{split} \int_\gamma (g^1)^* \omega - \omega &= \int_{g^1 \gamma - \gamma} \omega \\ &= \int_{\partial (H\gamma) - H(\partial \gamma)} \omega \\ &= \int_{H\gamma} d\omega - \int_{H(\partial \gamma)} \omega. \end{split}$$

Now we use that (prove later) for any $r$-form $\alpha$ and for $r-1$ chain $\gamma$ we have $$\tag{1} \int_{H\gamma}\alpha = (-1)^{r-1}\int_0^1 \int_\gamma H_s^* i_X \alpha\ ds,$$

then $$\begin{split} \int_\gamma (g^1)^* \omega - \omega &= (-1)^k\int_0^1 \int_\gamma H_s^* i_X d\omega \ ds - (-1)^{k-1} \int_0^1 \int_{\partial\gamma}H_s^* i_X \omega \ ds\\ &= (-1)^k \int_0^1 \int_\gamma H_s^* (i_X d\omega + di_X \omega) \ ds . \end{split}$$ Hence $$\int_0^1 \int_\gamma H_s^* \mathcal L_X \omega \ ds = \int_0^1 \int_\gamma H_s^* (i_X d\omega + di_X \omega) \ ds .$$ Note that if we consider the homotopy $G_t$, which is the restriction of $H\gamma$ to $[0, t]\times \sigma$, then similarly we have $$\int_0^t \int_\gamma H_s^* \mathcal L_X \omega \ ds = \int_0^t \int_\gamma H_s^* (i_X d\omega + di_X \omega) \ ds $$ for all $t >0$. Thus (using $H_0 = \gamma$) $$ \int_\gamma \mathcal L_X \omega= \int_\gamma (i_X d\omega + di_X \omega).$$ As the above holds for all $\gamma$, we have $$ \mathcal L_X \omega = i_X d\omega + di_X \omega.$$

It remains to prove $(1)$. Let $\alpha$ be an $r$ form. By definition, (call $H\gamma = F$)

$$\int_{H\gamma} \alpha = \int_{I \times \sigma} F^*\alpha$$

if we write in local coordinates $(t = x^0, x_1, \cdots, x_{r-1})$ in $I\times \sigma$ and $(y^1, \cdots, y^m)$ of $M$.

$$\begin{split} F^* \alpha (t, x)&= F^* \left(\sum_{|I| = r} \alpha_{i_1\cdots i_{r}} dy^{i_1} \wedge \cdots \wedge dy^{i_{r}}\right) \\ &= \sum_{|I| =r} \alpha_{i_1\cdots i_{r}} (F(t, x)) dF^{i_1} \wedge \cdots \wedge dF^{i_{r}} \\ \end{split}$$

Note that $$dF^{i_j} = \frac{\partial F^{i_j}}{\partial t} (t, x) dt + \sum_{p=1}^{r-1} \frac{\partial F^{i_j}}{\partial x^p} (t, x) dx^p = X^{i_j}(F(t,x)) dt + H_t^* dy^{i_j} $$

as $X = \frac{\partial F}{\partial t}$ by definition of $F = H\gamma$. Thus

$$\begin{split} F^* \alpha&= \sum_{|I| =r} \alpha_{i_1\cdots i_{r}} \left( X^{i_1} dt + H_t^* dy^{i_1}\right) \wedge \cdots \wedge \left( X^{i_{k+1}} dt + H_t^* dy^{i_{r}}\right) \\ &= \sum_{|I| =r} \alpha_{i_1\cdots i_{r}} \ dt \wedge H_t^* \left( \sum_{j=1}^{r}(-1)^{j-1} X^{i_j} dy^{i_1} \wedge \cdots \wedge \hat{dy^{i_j}} \wedge \cdots \wedge dy^{i_{r}}\right) \\ &= \sum_{|I| =k+1} \alpha_{i_1\cdots i_{r}} \ dt \wedge H_t^*\left( i_X (dy^{i_1} \wedge \cdots \wedge dy^{i_{r}})\right) \\ &= dt \wedge H_t^*(i_X \alpha) = (-1)^{r-1} H_t^* (i_X\alpha) \wedge dt. \end{split}$$ Thus $$ \int_{H\gamma} \alpha = (-1)^{r-1} \int_{I\times \sigma} H_t^*(i_X \alpha) \wedge dt = (-1)^{r-1} \int_0^1 \int_\gamma H_t^*(i_X \alpha) dt.$$

  • $\begingroup$ Thanks for your answer. I have some questions. Why does Fubini's theorem produce a factor $(-1)^k$? Also, how did you obtain $\mathrm{d}F^{i_j}$? I don't see how you got either term in the result. $\endgroup$
    – Ryan Unger
    Commented Oct 15, 2015 at 14:46
  • 1
    $\begingroup$ In the "usual" Fubini theorem, you basically ignore the order of intergration, $dx dy$ is the same as $dydx$. But if you treat that as a differential form, they are different. In our situation I am using $dt\wedge dx^1\wedge \cdots \wedge dx^k = (-1)^k dx^1 \wedge \cdots \wedge dx^k \wedge dt$. $\endgroup$
    – user99914
    Commented Oct 15, 2015 at 19:36
  • $\begingroup$ For your second question, note that the pullback is calculated by (if $F = (F^1, \cdots, F^m)$) $F^* dy^{i_j} = dF^{i_j}$. @0celo7 $\endgroup$
    – user99914
    Commented Oct 15, 2015 at 19:37
  • 1
    $\begingroup$ The first one comes from the definition as $F (t, x)= g^t \gamma(x)$, so $\frac{\partial F}{\partial t} (t, x) = \frac {d}{dt} g^t \gamma(x) = X(F(t, x))$. $\endgroup$
    – user99914
    Commented Oct 15, 2015 at 20:09
  • 1
    $\begingroup$ For the second one, the definition of $H_t$ is $H_t(x) = F(t, x)$. So $H_t^* dy^{i_j} = dH_t^{i_j} = \frac{\partial H_t^{i_j}}{\partial x^p} dx^p$. But $\frac{\partial H_t^{i_j}}{\partial x^p} = \frac{\partial F^{i_j}}{\partial x^p} $ @0celo7 $\endgroup$
    – user99914
    Commented Oct 15, 2015 at 20:12

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