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

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 Posted by | Differential Topology, Topology | 5 Comments

   

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