The Unapologetic Mathematician

Mathematics for the interested outsider

The Lie Derivative on Forms

We’ve defined the Lie derivative L_XY of one vector field Y by another, X. This worked by using the flow of X to compare nearby points, and used the derivative of the flow to translate vectors.

Well now we know how to translate k-forms by pulling back, and thus we can define another Lie derivative:

\displaystyle L_X\omega=\lim\limits_{t\to0}\frac{1}{t}\left((\Phi_t)^*(\omega)-\omega\right)

What happens if \omega is a 0-form — a function f? We check

\displaystyle\begin{aligned}\left[L_Xf\right](p)&=\lim\limits_{t\to0}\frac{1}{t}\left(\left[(\Phi_t)^*(f)\right](p)-f(p)\right)\\&=\lim\limits_{t\to0}\frac{1}{t}\left(f\left(\Phi_t(p)\right)-f(p)\right)\\&=X_p(f)=Xf(p)\end{aligned}

That is, the Lie derivative by X acts on \Omega^0(M) exactly the same as the vector field X does itself.

I also say that the Lie derivative by X is a degree-zero derivation of the algebra \Omega(M). That is, it’s a real-linear transformation, and it satisfies the Leibniz rule:

\displaystyle L_X(\alpha\wedge\beta)=\left(L_X\alpha\right)\wedge\beta+\alpha\wedge\left(L_X\beta\right)

for any k-form \alpha and l-form \beta. Linearity is straightforward, and given linearity the Leibniz rule follows if we can show

\displaystyle L_X(\alpha_1\wedge\dots\wedge\alpha_k)=\sum\limits_{i=1}^k\alpha_1\wedge\dots\wedge\left(L_X\alpha_i\right)\wedge\dots\wedge\alpha_k

for 1-forms \alpha_i. Indeed, we can write \alpha and \beta as linear combinations of such k- and l-fold wedges, and then the Leibniz rule is obvious.

So, let us calculate:

\displaystyle\begin{aligned}L_X\left(\alpha_1\wedge\dots\wedge\alpha_k\right)=&\lim\limits_{t\to0}\frac{1}{t}\left((\Phi_t)^*\left(\alpha_1\wedge\dots\wedge\alpha_k\right)-\alpha_1\wedge\dots\wedge\alpha_k\right)\\=&\lim\limits_{t\to0}\frac{1}{t}\left(\left((\Phi_t)^*\alpha_1\right)\wedge\dots\wedge\left((\Phi_t)^*\alpha_k\right)-\alpha_1\wedge\dots\wedge\alpha_k\right)\\=&\lim\limits_{t\to0}\frac{1}{t}\left(\left((\Phi_t)^*\alpha_1\right)\wedge\dots\wedge\left((\Phi_t)^*\alpha_k\right)-\alpha_1\wedge\dots\wedge\alpha_{k-1}\wedge\left((\Phi_t)^*\alpha_k\right)\right)\\&+\lim\limits_{t\to0}\frac{1}{t}\left(\alpha_1\wedge\dots\wedge\alpha_{k-1}\wedge\left((\Phi_t)^*\alpha_k\right)-\alpha_1\wedge\dots\wedge\alpha_k\right)\\=&\lim\limits_{t\to0}\left(\frac{1}{t}\left(\left((\Phi_t)^*\alpha_1\right)\wedge\dots\wedge\left((\Phi_t)^*\alpha_{k-1}\right)-\alpha_1\wedge\dots\wedge\alpha_{k-1}\right)\wedge\left((\Phi_t)^*\alpha_k\right)\right)\\&+\alpha_1\wedge\dots\wedge\alpha_{k-1}\wedge\lim\limits_{t\to0}\frac{1}{t}\left(\left((\Phi_t)^*\alpha_k\right)-\alpha_k\right)\\=&\left(\lim\limits_{t\to0}\frac{1}{t}\left(\left((\Phi_t)^*\alpha_1\right)\wedge\dots\wedge\left((\Phi_t)^*\alpha_{k-1}\right)-\alpha_1\wedge\dots\wedge\alpha_{k-1}\right)\right)\wedge\alpha_k\\&+\alpha_1\wedge\dots\wedge\alpha_{k-1}\wedge L_X\alpha_k\\&=L_X\left(\alpha_1\wedge\dots\wedge\alpha_{k-1}\right)\wedge\alpha_k+\alpha_1\wedge\dots\wedge\alpha_{k-1}\wedge L_X\alpha_k\end{aligned}

So we see how we can peel off one of the 1-forms. A simple induction gives us the general case.

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July 13, 2011 - Posted by | Differential Topology, Topology

5 Comments »

  1. […] Lie derivative looks sort of familiar as a derivative, but we have another sort of derivative on the algebra of […]

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  2. […] . But since it takes -forms and sends them to -forms, it has degree one instead of zero like the Lie derivative. As a consequence, the Leibniz rule looks a little different. If is a -form and is an -form, I […]

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