# The Unapologetic Mathematician

## Cartan’s Formula

It turns out that there is a fantastic relationship between the interior product, the exterior derivative, and the Lie derivative.

It starts with the observation that for a function $f$ and a vector field $X$, the Lie derivative is $L_Xf=Xf$ and the exterior derivative evaluated at $X$ is $\iota_X(df)=df(X)=Xf$. That is, $L_X=\iota_X\circ d$ on functions.

Next we consider the differential $df$ of a function. If we apply $\iota_X\circ d$ to it, the nilpotency of the exterior derivative tells us that we automatically get zero. On the other hand, if we apply $d\circ\iota_X$, we get $d(\iota_X(df))=d(Xf)$, which it turns out is $L_X(df)$. To see this, we calculate

$\displaystyle\begin{aligned}\left[L_X(df)\right](Y)&=\lim\limits_{t\to0}\frac{1}{t}\left(\left[(\Phi_t)^*(df)\right](Y)-df(Y)\right)\\&=\lim\limits_{t\to0}\frac{1}{t}\left(df((\Phi_t)_*Y)-df(Y)\right)\\&=\frac{\partial}{\partial t}\left[(\Phi_t)_*Y\right](f)\bigg\vert_{t=0}\\&=\frac{\partial}{\partial t}Y(f\circ\Phi_t)\bigg\vert_{t=0}\\&=YXf\end{aligned}$

just as if we took $d(Xf)$ and applied it to $Y$.

So on exact $1$-forms, $\iota_X\circ d$ gives zero while $d\circ\iota_X$ gives $L_X$. And on functions $\iota_X\circ d$ gives $L_X$, while $d\circ\iota_X$ gives zero. In both cases we find that

$\displaystyle L_X=d\circ\iota_X+\iota_X\circ d$

and in fact this holds for all differential forms, which follows from these two base cases by a straightforward induction. This is Cartan’s formula, and it’s the natural extension to all differential forms of the basic identity $L_X(f)=\iota_X(df)$ on functions.

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Posted by John Armstrong | Differential Topology, Topology

## 1 Comment	»

1. [...] Cartan’s formula in hand we can show that the Lie derivative is a chain map . That is, it commutes with the exterior [...]

Pingback by | July 28, 2011 | Reply

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