The Unapologetic Mathematician

Mathematics for the interested outsider

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.

July 26, 2011 - Posted by | 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 The Lie Derivative on Cohomology « The Unapologetic Mathematician | July 28, 2011 | Reply


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