# The Unapologetic Mathematician

## 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.

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

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