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A possible proof of Cartan's magic formula $$L_X = i_X \circ d+d \circ i_X$$ is to follow the steps:

  • Show that two derivations on $\Omega^{\bullet}(M)$ commuting with $d$ are equal iff they agree on $\Omega^0(M)$.
  • Show that $L_X$ is a derivation on $\Omega^{\bullet}(M)$ commuting with $d$.
  • Show that $i_X \circ d + d \circ i_X$ is a derivation on $\Omega^{\bullet}(M)$ commuting with $d$.
  • Show that $L_X f = Xf = i_Xdf+ d i_Xf$ for all $f \in C^{\infty}(M)=\Omega^0(M)$.

I followed the sketch of proof, and the only point where I am stuck is when I have to prove that $L_X$ and $d$ commute.

More precisely, if I write $$L_Xd\omega =\frac{d}{dt}_{|t=0} \phi_t^* d\omega = \frac{d}{dt}_{|t=0} d\phi_t^* \omega = \lim\limits_{t \to 0} d \left( \frac{1}{t} \left( \phi_t^* \omega- \omega \right) \right),$$ is there an argument to permute $d$ and $\lim\limits_{t \to 0}$?

glS
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Seirios
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  • Is $\Omega(M)=\Omega^1(M)$ the space of 1-forms? – gofvonx Sep 28 '13 at 14:16
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    You can do it in coordinates, in which case I think it reduces to commuting $d/dt$ with $\partial/\partial x^i$, or you can consider the action of the form $d[(\phi_t^*\omega - \omega)/t]$ on any derivation $X$ as $t \rightarrow 0$ and show that it's equal to the action of $d\nu$, where $\nu$ is the Lie-differentiated form. – snar Sep 28 '13 at 14:33
  • @gofvonx: No, $\Omega^{\cdot}(M)$ is the graded algebra $\bigoplus\limits_{k \geq 0} \Omega^k(M)$. – Seirios Sep 28 '13 at 15:00
  • Is it really called "Cartan's magic formula"? – Pedro Sep 28 '13 at 15:45
  • @PedroTamaroff: Yes, but its origin stays mysterious http://math.stackexchange.com/questions/188630/what-is-magical-about-cartans-magic-formula – Seirios Sep 28 '13 at 16:28
  • "Show that two derivations on $\Omega^{\bullet}(M)$ commuting with $d$ are equal iff they agree on $\Omega^0(M)$" looks suspicious to me. Consider two derivations $D_1$ and $D_2$ commuting with $d.$ Now consider any non-exact 1-form $\alpha.$ I see no reason why $D_1\alpha=D_2\alpha.$ – Fallen Apart May 03 '16 at 14:30
  • @FallenApart: in case you're still wondering: locally $\alpha = \sum_i f \wedge dx_i$. – T'x Mar 09 '17 at 20:42
  • Ah typo, I meant $f_i$. – T'x Mar 09 '17 at 21:03
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    The reason $\mathcal{L}_V$ and $d$ commute is that $d$ commutes with (local) diffeomorphisms, so commutes with the flow generated by $V$, hence with its derivative $\mathcal{L}_V$. – Agol Apr 13 '17 at 06:00
  • I'm curious if these steps can be generalized to general differential graded algebras, see this question – Andrews Apr 20 '19 at 08:08

4 Answers4

17

Putting in the limit made it more mysterious. You're just using the fact that for smooth functions mixed partials are equal. That is, $$\frac{\partial}{\partial t}\frac{\partial}{\partial x^j}=\frac{\partial}{\partial x^j}\frac{\partial}{\partial t}.$$

EDIT: In particular, if you write it out in local coordinates, it's computed by taking partial derivatives with the respect to the variables $x^1,\dots,x^n$. Writing $\phi_t^*\omega=\sum f_I(x,t)dx^I$ (where $I=(i_1,\dots,i_k)$ ranges over length-$k$ multiindices), then $d\phi_t^*\omega=\sum \frac{\partial f_I(x,t)}{\partial x^j}\,dx^j\wedge dx^I$, etc.

Ted Shifrin
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Here is an attempt to answer my question following Ted Shifrin's comment:

First, we may write $\phi_t^* \omega(x)= \sum\limits_{I} f_I(x,t)dx^I$ where $I$ ranges over length-$k$ multiindices. Then

$$\begin{array}{ll} dL_X \omega & = d \frac{d}{d t}_{|t=0} \phi_t^* \omega = d \sum\limits_I \frac{\partial }{\partial t}_{|t=0} f_I dx^I \\ & = \sum\limits_I \sum\limits_{i=1}^n \frac{\partial}{\partial x_i} \frac{\partial}{\partial t}_{|t=0} f_I dx^i \wedge dx^I = \sum\limits_I \sum\limits_{i=1}^n \frac{\partial}{\partial t}_{|t=0} \frac{\partial}{\partial x_i} f_I dx^i \wedge dx^I \\ & = \sum\limits_I \frac{d}{d t}_{|t=0} df_I \wedge dx^I = \frac{d}{dt}_{|t=0} d \phi_t^* \omega = \frac{d}{dt}_{|t=0} \phi_t^* d \omega = L_Xd\omega \end{array}$$

Seirios
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2

The external differential does not depend on the choice of chart. Therefore $$d\phi_t^*\omega=\phi_t^*d \omega.$$ Taking derivative we get $$dL_X\omega=L_Xd\omega.$$

Anton Petrunin
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0

Let $\omega$ be a $k$ form.

Use the formula $L_X= \iota_X\circ d+ d\circ \iota_x$.

We have $L_Xd\omega=\iota_X\circ d\circ d \omega+ d\circ \iota_X \circ d\omega =d\circ \iota_X \circ d\omega $

as $d\circ d=d^2 = 0$

And

$dL_X\omega= d\circ(\iota_X\circ d\omega+ d\circ \iota_x\omega)=d\circ\iota_X\circ d\omega+d\circ d\circ \iota_x\omega=d\circ \iota_X \circ d\omega $

again using $d\circ d=d^2 = 0$

Alex C
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