# Introduction

In proposition 58 Chapter 1 in the excellent book O’Neill (1983), the author demonstrates that the Lie derivative of one vector field with respect to another is the same as the Lie bracket (of the two vector fields) although he calls the Lie bracket just bracket and does not define the Lie derivative preferring just to use its definition with giving it a name. The proof relies on a prior result where he shows a co-ordinate system at a point $p$ can be given to a vector field $X$ for which $X_p \neq 0$ so that $X = \frac{\partial}{\partial x_1}$.

Here’s a proof seems clearer (to me at any rate) and avoids having to distinguish the case wehere the vector field is zero or non-zero. These notes give a similar proof but, strangely for undergraduate level, elide some of the details.

# A Few Definitions

Let $\phi: M \longrightarrow N$ be a smooth mapping and let $A$ be a $0,s$ tensor with $s \geq 0$ then define the pullback of $A$ by $\phi$ to be

$\displaystyle \phi^*A(v_1,\ldots,v_s) = A(\mathrm{d}\phi v_1,\ldots,\mathrm{d}\phi v_s)$

For a $(0,0)$ tensor $f \in {\mathscr{F}}(N)$ the pullback is defined to be $\phi^*(f) = f \circ \phi \in {\mathscr{F}}(M)$.

Standard manipulations show that $\phi^*A$ is a smooth (covariant) tensor field and that $\phi^*$ is $\mathbb{R}$-linear and that $\phi^*(A\otimes B) = \phi^*A \otimes \phi^*B$.

Let $F : M \longrightarrow N$ be a diffeomorphism and $Y$ a vector field on $N$ we define the pullback of this field to be

$\displaystyle (F^*{Y})_x = D(F^{-1})_{F(x)}(Y_{F(x)})$

Note that the pullback of a vector field only exists in the case where $F$ is a diffeomorphism; in contradistinction, in the case of pullbacks of purely covariant tensors, the pullback always exists.

For the proof below, we only need the pullback of functions and vector fields; the pullback for $(0,s)$ tensors with $s \geq 1$ is purely to give a bit of context.

From O’Neill (1983) Chapter 1 Definition 20, let $\phi: M \rightarrow N$ be a smooth mapping. Vector fields $X$ on $M$ and $Y$ on $N$ are $F$related written $X \underset{F}{\sim} Y$ if and only if $dF({X}_p) = Y_{Fp}$.

# The Alternative Proof

By Lemma 21 Chapter 1 of O’Neill (1983), $X$ and $Y$ are $F$-related if and only if $X(f \circ F) = Yf \circ F$.

Recalling that $dF(X_p)(f) = X_p(F \circ f)$ and since

$\displaystyle dF_x d(F^{-1})_{Fx}(X_{Fx}) = X_{Fx}$

we see that the fields $F^*{Y}$ and $Y$ are $F$-related: $F^*{Y}_x \underset{F}{\sim} Y_{Fx}$. Thus we can apply the Lemma.

$\displaystyle (F^*{Y})(f \circ F) = (F^*{Y})(F^*{f}) = Yf \circ F = F^*(Yf)$

Although we don’t need this, we can express the immediately above equivalence in a way similar to the rule for covariant tensors

$\displaystyle (F^*{Y})(f \otimes F) = (F^*{Y})\otimes(F^*{f})$

First let’s calculate the Lie derivative of a function $f$ with respect to a vector field $X$ where $\phi_t$ is its flow

\displaystyle \begin{aligned} L_X f &\triangleq \lim_{t \rightarrow 0} \frac{\phi_t^*(f) - f}{t} \\ &= \lim_{t \rightarrow 0} \frac{f \circ \phi_t - f \circ \phi_0}{t} \\ &= \lim_{t \rightarrow 0} \frac{f \circ \phi (t,x) - f \circ \phi (0, x)}{t} \\ &= (\phi_x)'_0(f) \\ &= X_x(f) \\ &= (Xf)_x \end{aligned}

Analogously defining the Lie derivative of $Y$ with respect to $X$

$\displaystyle (L_X Y) \triangleq \frac{(\phi_t^*{Y}) - Y}{t}f$

we have

\displaystyle \begin{aligned} L_X(Yf) &= \lim_{t \rightarrow 0} \frac{\phi_t^*(Yf) - Yf}{t} \\ &= \lim_{t \rightarrow 0} \frac{(\phi_t^*{Y})(\phi_t^*{f}) - Yf}{t} \\ &= \lim_{t \rightarrow 0} \frac{(\phi_t^*{Y})(\phi_t^*{f}) - (\phi_t^*{Y})f + (\phi_t^*{Y})f - Yf}{t} \\ &= \lim_{t \rightarrow 0} \Bigg( (\phi_t^*{Y})\frac{\phi_t^*{f} - f}{t} + \frac{(\phi_t^*{Y}) - Y}{t}f \Bigg) \\ &= Y(L_X f) + (L_X Y)f \end{aligned}

Since $L_X f = Xf$ we have

$\displaystyle X(Yf) = Y(Xf) + (L_X Y)f$

Thus

$\displaystyle (L_X Y)f = Y(Xf) - X(Yf) = [X,Y]f$

as required.

# Bibliography

O’Neill, B. 1983. Semi-Riemannian Geometry with Applications to Relativity, 103. Pure and Applied Mathematics. Elsevier Science. https://books.google.com.au/books?id=CGk1eRSjFIIC.