In this video, Arnold says that the Cartan formula $$ \mathscr L_{\mathrm X} = d i_{\mathrm X} + i_{\mathrm X} d$$ is just an avatar of $(fg)' = fg' + f'g$.

Why ?

  • $\begingroup$ At the very least, if $f$, $g \in C^\infty(\mathbb{R})$, then $$ (fg)^\prime = \mathscr{L}_{\tfrac{d}{dx}}(fg) = d\left(i_{\tfrac{d}{dx}}(fg)\right) + i_{\tfrac{d}{dx}}d(fg) = d(fg)(\tfrac{d}{dx}) = (gdf+fdg)(\tfrac{d}{dx}) = (gf^\prime+fg^\prime)dx(\tfrac{d}{dx}) = gf^\prime+fg^\prime, $$ but that feels like little more than a notational shell game, i.e., if you know that $\tfrac{d}{dx}$ is a vector field, which is true precisely because it satisfies the Leibniz rule, then you know it satisfies the Leibniz rule because the the exterior derivative satisfies its own Leibniz rule? $\endgroup$ Mar 29, 2013 at 8:46
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    $\begingroup$ Apart from a bit of algebra and spelling out the definitions, the main analytic ingredients in the proof of Cartan's formula are the Leibniz rule for functions and the chain rule. Indeed, the Leibniz rule for the exterior derivative is nothing but the Leibniz rule for functions married with the exterior algebra, etc. I think the proof on pages 33ff in these notes shows this quite clearly. $\endgroup$
    – Martin
    Mar 29, 2013 at 12:38
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    $\begingroup$ Thanks for the link to the video! Also, what Martin said is the answer. $\endgroup$
    – Sam Lisi
    Mar 29, 2013 at 20:07
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    $\begingroup$ @Martin, I agree, there is nothing more. But in the video, Arnold gives a geometric proof of the Leibnitz formula. Is there an intuitive/geometric proof for the Cartan formula ? $\endgroup$
    – Damien L
    Mar 29, 2013 at 22:37
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    $\begingroup$ Related: math.stackexchange.com/q/1480545 (the answer there is actually better than here, I think). $\endgroup$ Jan 9, 2019 at 18:52

1 Answer 1


Wow. This video reminds me of how sad I am not to have heard Arnol'd speak while he was alive.

The following is an attempt to work out what he might have meant by the observation.

Suppose that $V$ is a smooth vector field on a manifold $M$. Let $C$ be a smooth $k$-chain in $M$ and let $\omega$ be a smooth $k$-form on $M$. Denote the flow of $V$ by $\phi_t$. Let $\Phi$ be the embedding of $I \times C \to M$ by $(t, x) \mapsto \phi_t(x)$, where $I = [0, T]$ is an interval and $t$ is the coordinate on it.

For any $(k+1)$-form $\Omega$ on $M$, we have $\Phi^*\Omega = \phi_t^*\Omega + dt \wedge i_V \Omega$ on $I \times C$.

Now, the boundary of $I \times C$ is $$-(\{0\} \times C) \cup I \times \partial C \cup \{T\} \times C.$$ If the signs are done correctly, this is the formula Arnol'd also calls the Leibniz formula for a Cartesian product of chains, that $\partial (A\times B) =\partial A \times B \cup A \times \partial B$.

Now, by Stokes' theorem $$ \int_{I \times C} \Phi^*d\omega = -\int_{0 \times C} \Phi^*\omega + \int_{I \times \partial C} \Phi^*\omega + \int_{T \times C} \Phi^*\omega. $$

If we plug in what we computed for $\Phi^*\Omega$ (on the LHS, $\Omega = d\omega$ and on the RHS, $\Omega = \omega$), we get $$ \int_{I \times C} \phi_t^*d\omega + dt \wedge i_V d\omega = - \int_{0 \times C} \omega + \int_{I \times \partial C} dt \wedge i_V \omega + \int_{T \times C} \phi_T^* \omega. $$

Let me first explain why, from this step, you get Cartan's magic formula. Differentiate both sides with respect to $T$. This becomes $$ \int_{C} d\omega + i_V d\omega = -0 + 0 + \int_C L_V \omega. $$ This is true for an arbitrary smooth chain $C$ so the equality holds at the level of $k$-forms pointwise.

Now, the reason this is the same proof is that we have drawn the same picture. Indeed, in both cases, the key step is to use the picture he drew on the board of $I \times C$ and determine its boundary, and differentiate this with respect to the $t$ parameter.

EDIT: here is a proof that $\Phi^*\Omega = \phi_t^*\Omega + dt \wedge i_V \Omega$ on $I \times C$.

Note first that by definition $\Phi^*\Omega|_{t,x} = \Omega( d\Phi(t,x) \cdot , d\Phi(t,x) \cdot, \dots, d\Phi(t,x) \cdot)$. Now, $d\Phi(t,x) = d\phi_t(x) + V(\phi_t(x)) \otimes dt$. In other words, $d\Phi(t,x) \, v = d\phi_t(x)\,v$ for each $v \in TC$ and $d\Phi(t,x)\,\partial_t = V(\phi_t(x))$.

Unenlightening proof 1: combine this expression for $d\Phi$ with $\Omega$ and you get the formula I claim.

Proof 2: use local coordinates. Choose a basis for $TC$ at the point $x$. We have that $\Phi^*\Omega$ is a form on $C \times I$, so we can write it as $\mu + dt \wedge \nu$ for $t$-dependent forms $\mu$ and $\nu$ on $I \times C$. To compute $\mu$, we evaluate $\Phi^*\Omega$ on the basis of $TC$, and get $\mu = \phi_t^*\Omega$. To compute $\nu$, we evaluate $\Phi^*\Omega$ on $\partial_t$ followed by a collection of vectors from our basis of $T_xC$. This gives us the $i_V \Omega$ term.

It would be great if someone could teach me a nice proof of this fact, actually.

OK, as I fill in more details, the link between the two proofs becomes more and more tenuous. I suspect that Arnol'd is saying something deeper than the similarity I am noticing. I think I tend to fall back into the more "algebraic" way of thinking of the exterior derivative, and he has a more geometric picture in mind.

  • $\begingroup$ Very nice answer ! Please, can you write the details of $\Phi^{\ast} \Omega = \phi_t^{\ast} \Omega + dt \wedge i_v \Omega$ ? $\endgroup$
    – Damien L
    Apr 5, 2013 at 15:52
  • $\begingroup$ @DamienL: I've written some details. I hope this helps explain that identity. $\endgroup$
    – Sam Lisi
    Apr 7, 2013 at 20:15
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    $\begingroup$ Thank you for the bounty. I am grateful to you for the video. $\endgroup$
    – Sam Lisi
    Apr 8, 2013 at 17:13

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