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 $\lim\limits_{t \to 0}$?

  • $\begingroup$ Is $\Omega(M)=\Omega^1(M)$ the space of 1-forms? $\endgroup$
    – gofvonx
    Commented Sep 28, 2013 at 14:16
  • 1
    $\begingroup$ 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. $\endgroup$
    – snar
    Commented Sep 28, 2013 at 14:33
  • $\begingroup$ @gofvonx: No, $\Omega^{\cdot}(M)$ is the graded algebra $\bigoplus\limits_{k \geq 0} \Omega^k(M)$. $\endgroup$
    – Seirios
    Commented Sep 28, 2013 at 15:00
  • $\begingroup$ Is it really called "Cartan's magic formula"? $\endgroup$
    – Pedro
    Commented Sep 28, 2013 at 15:45
  • 1
    $\begingroup$ 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 $\mathcal{L}_V$. $\endgroup$
    – Agol
    Commented Apr 13, 2017 at 6:00

4 Answers 4


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.

  • $\begingroup$ But isn't the exterior derivative more than a partial derivative of smooth functions since it also changes the degree of the differential form? $\endgroup$
    – gofvonx
    Commented Sep 28, 2013 at 14:54
  • $\begingroup$ I followed your comment in the answer below, it seems to work! $\endgroup$
    – Seirios
    Commented Sep 28, 2013 at 16:42
  • $\begingroup$ Maybe you should include the hint in your answer. $\endgroup$
    – Seirios
    Commented Sep 28, 2013 at 16:44

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

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


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


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


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


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