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.

July 26, 2011

The Interior Product

We have yet another operation on the algebra $\Omega(M)$ of differential forms: the “interior product”. Given a vector field $X\in\mathfrak{X}(M)$ and a $k$-form $\omega\in\Omega^k(M)$, the interior product $\iota_X(\omega)$ is the $k-1$-form defined by

$\displaystyle\left[\iota_X\omega\right](X_1,\dots,X_{k-1})=\omega(X,X_1,\dots,X_{k-1})$

That is, we just take the vector field $X$ and stick it into the first “slot” of a $k$-form. We extend this to functions by just defining $\iota_Xf=0$.

Two interior products anticommute: $\iota_X\iota_Y=-\iota_Y\iota_X$, which follows easily from the antisymmetry of differential forms. Each one is also clearly linear, and in fact is also a graded derivation of $\Omega(M)$ with degree -1:

$\displaystyle\iota_X(\alpha\wedge\beta)=(\iota_X\alpha)\wedge\beta+(-1)^{-p}\alpha\wedge(\iota_X\beta)$

where $p$ is the degree of $\alpha$. This can be shown by reducing to the case where $\alpha$ and $\beta$ are wedge products of $1$-forms, but rather than go through all that tedious calculation we can think about it like this: sticking $X$ into a slot of $\alpha\wedge\beta$ means either sticking it into a slot of $\alpha$ or into one of $\beta$. In the first case we just get $\iota_x\alpha$, while in the second we have to “move the slot” through all of $\alpha$, which incurs a sign of $(-1)^p=(-1)^{-p}$, as asserted.

July 26, 2011