Proof of Cartan's Magic formula

Question

I'm trying to prove Cartan's magic formula, $\mathcal{L}_{v}(X) = i_{v}(dX)+d(i_{v}X)$ for X a p-form.

My attempt so far:

We can choose a basis (t,$x^{i}$) such that $v$ = $d/dt$. In this basis

$\mathcal{L}_{v}(X) = \frac{\partial }{\partial t}(X_{\mu_{1}...\mu_{p}})$, $v^{\alpha} = \delta^{\alpha}_{0}$ and set $\frac{\partial }{\partial t} = \partial_{0}$

Then I evaluate:

$i_{v}(dX) = i_{v}((p+1)\partial_{[\nu_{1}}X_{\mu_{1}...\mu_{p}]}) = (p+1)v^{\alpha}\partial_{[\alpha}X_{\mu_{1}...\mu_{p}]}$

and:

$ d(i_{v}X) = d(v^{\alpha}X_{\alpha\mu_{2}...\mu_{p}}) = (p)\partial_{[\mu_{1}}v^{\alpha}X_{\alpha\mu_{2}...\mu_{p}]} = (p)v^{\alpha}\partial_{[\mu_{1}}X_{\alpha\mu_{2}...\mu_{p}]}$ (as $v^{\alpha}$ is constant)

Putting things together:

$i_{v}(dX)+d(i_{v}X) = (p+1)v^{\alpha}\partial_{[\alpha}X_{\mu_{1}...\mu_{p}]}+(p)v^{\alpha}\partial_{[\mu_{1}}X_{\alpha\mu_{2}...\mu_{p}]} = (p+1)v^{\alpha}\partial_{[\alpha}X_{\mu_{1}...\mu_{p}]}-(p)v^{\alpha}\partial_{[\alpha}X_{\mu_{1}\mu_{2}...\mu_{p}]} = v^{\alpha}\partial_{[\alpha}X_{\mu_{1}\mu_{2}...\mu_{p}]}$

Using $v^{\alpha} = \delta^{\alpha}_{0}$ we get $i_{v}(dX)+d(i_{v}X) =\partial_{[0}X_{\mu_{1}\mu_{2}...\mu_{p}]}$

I think this expression however is incorrect, as I don't see how $\partial_{[0}X_{\mu_{1}\mu_{2}...\mu_{p}]} = \partial_{0}X_{[\mu_{1}\mu_{2}...\mu_{p}]} = \partial_{0}X_{\mu_{1}\mu_{2}...\mu_{p}} =\mathcal{L}_{v}(X)$

Is this the correct approach to take or am I missing something obvious? Thank you very much!

Answer 1

I think taking a slightly more abstract route might be clearer in this context. Consider the following definition:

Let $M$ be a smooth manifold and $r\in\Bbb{Z}$. By a superderivation of degree $r$ acting on differential forms on $M$, we shall mean a collection of $\Bbb{R}$-linear maps $\theta^{(k)}:\Omega^k(M)\to\Omega^{k+r}(M)$ (though we usually don't put the index $k$, which is easily inferred from context) such that the "Leibniz rule" is satisfied:

• For every $\alpha\in \Omega^k(M), \beta\in\Omega^l(M)$, we have $\theta(\alpha \wedge \beta)= [\theta(\alpha)]\wedge \beta + (-1)^{kr} \alpha \wedge[\theta(\beta)]$.

Here, $\Omega^k(M)$ is defined to be the space of all differential forms on $M$ of degree $k$ (and by convention we put $\Omega^k(M)=\{0\}$ if $k<0$). There are two nice features about such superderivations:

• They are local: if two forms $\alpha$ and $\beta$ agree on an open set $U\subset M$, then $\theta(\alpha)$ and $\theta(\beta)$ also agree on $U$.
• They are completely determined by their action on smooth functions and $1$-forms; i.e if you know $\theta(f)$ and $\theta(dg)$ for every $f,g\in \Omega^0(M)=C^{\infty}(M,\Bbb{R})$ then you know $\theta(\omega)$ for every $k$-form $\omega$.

To prove the first bullet point, its a standard bump function argument. For the second assertion, write $\omega$ as a linear combination $\sum\omega_I dx^I$, where $I=i_1,\dots, i_k$ is an increasing multi-index; then the Leibniz rule and $\Bbb{R}$-linearity will completely determine the outcome.

Finally, one last thing to note:

If $\theta_1$ and $\theta_2$ are superderivations of degree $r_1$ and $r_2$, then we define their "supercommutator" $[\theta_1,\theta_2]:= \theta_1\circ \theta_2 -(-1)^{r_1r_2}\theta_2\circ \theta_1$. This is a super-derivation of degree $r_1+r_2$.

The proof is a direct calculation, and the funny sign makes everything work out in the end.

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For your actual question, note that the exterior derivative $d$ is a superderivation of degree $+1$, while the interior product $\iota_v$ is a superderivation of degree $-1$. Hence, we can consider their supercommutator \begin{align} \theta_v := [d,\iota_v] = d\circ\iota_v - (-1)^{(1)(-1)}\iota_v\circ d = d\circ \iota_v + \iota_v\circ d \end{align} This is now a super-derivation of degree $0$ (i.e it maps $k$-forms to $k$-forms, and satisfies the product rule with no minus signs).

It is also standard to check (starting with the flow definition) that the Lie derivative $L_v$ is also a super-derivation of degree $0$. So, in order to check $L_v = \theta_v$, all you have to do is verify that for all $f,g\in C^{\infty}(M)$, we have $L_v(f)=\theta_v(f)$ and $L_v(dg)= \theta_v(dg)$. But now this should be quite simple.

So, really the point of this answer was to show you that once you set up appropriate definitions, there are actually very few things you have to check: just check equality on functions and exact $1$-forms; because of the Leibniz rule this will then imply equality on all $k$-forms. So in a sense this is like a nice "uniqueness in extension" type of argument.

Answer 2

The point is that the indices of components are contracted with that of basis, i.e. \begin{equation} \partial_{[0}X_{\mu_1\dots\mu_p]}dx^0\wedge dx^{\mu_1}\dots dx^{\mu_p}=\partial_{0}X_{\mu_1\dots\mu_p}dx^0\wedge dx^{\mu_1}\dots dx^{\mu_p} \end{equation} as anti-symmetry can be 'absorbed' into the indices of basis.