How to prove $ L_X(\omega(Y)) = (L_X\omega)Y + \omega(L_XY)$ starting from the fundamental definition of Lie derivative?

Question

The Lie derivative of a smooth real valued function $f$ along a vector field $X$, on a point $p$ in some smooth manifold is given as $$ L_X f(p) := \lim_{h\to 0} \frac{1}{h}\left[ f(\phi(p)) - f(p) \right]\label{Lief}\tag{1} $$ The Lie derivative of a vector field $Y$ along another vector field $X$, on a point $p$ in some smooth manifold is given as $$ L_X Y(p) =\frac{d}{dt}\left[\phi_{-t*}Y(p)\right] := \lim_{h\to 0} \frac{1}{h}\left[ (\phi_{-h*}Y)_p - Y_p \right]\label{Liev}\tag{2} $$ $\phi_t$ is the integral curve of the vector field $X$, with the push-forward map defined by

$$(\phi_{-h*}Y)_p = \phi_{-h*}Y_{\phi_h(p)}$$

Similarly, the Lie derivative of a one-form $\omega$ along a vector field $X$, is given by $$ L_X \omega(p) =\frac{d}{dt}\left[\phi_{t}^*\omega\right](p) := \lim_{h\to 0} \frac{1}{h}\left[ (\phi_{h}^*\omega)_p - \omega_p \right]\label{Lieo}\tag{3} $$

$$ (\phi_h^* \omega)(p)(X_p) = \omega(\phi_h(p)) (\phi_{h*}X_p) $$

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Now, I want to prove that the Lie derivative $$ L_X(\omega(Y)) = (L_X\omega)Y + \omega(L_XY) $$

The function $\omega(Y)(p) = \omega_p (Y_p)$, as $\omega(Y)(p)$ is a function it's transformation rule should be something like

\begin{align} L_X(\omega(Y)) &= \lim_{h\to 0} \frac{1}{h}\left[ (\phi_{h}^*(\omega(Y)))_p - (\omega(Y))_p \right] \label{LiewY}\tag{4} \end{align}

• Is the equation \eqref{LiewY} the right way to begin?, or the expression be more like \eqref{Lief} as $\omega(Y)$ is a real valued function over the manifold,
• I am not sure how $(\phi_{h}^*(\omega(Y)))_p$ in \eqref{LiewY} will look when simplified.

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This question has an answer here but in terms of Cartan's formula. I'd like to know how to start from the very basic definition of Lie derivative

Answer 1

A naive manner to prove basic formulae in Differential Geometry consists in writing everything in local frame and coframe. Let $\partial_1, \ldots, \partial_n$ denote a local frame around $p$ and $e^1, \ldots, e^n$ its dual coframe. Write our vector fields and forms in these bases: $$X = x^i \partial_i,\ Y = y^i \partial_i,\ \omega = \omega_i e^i.$$ But $\omega(Y) = \omega_i y^i$ is a function and the Lie derivative for function is just the usual directional derivative. So by the usual Leibniz rule, we have $$\mathcal{L}_X(\omega(Y)) = \mathcal{L}_X(\omega_i) y^i + \omega_i \mathcal{L}_X (y^i) = x^j \partial_j \omega_i y^i + \omega_i x^j\partial_j y^i.$$ It is well known that $\mathcal{L}_X(Y) = [X, Y]$. Thus $$ \mathcal{L}_X(Y) = (x^j \partial_j y^i - y^j \partial_j x^i)\partial_i\mbox{ and } \omega(\mathcal{L}_X(Y)) = \omega_i(x^j \partial_j y^i - y^j \partial_j x^i).$$ Finally, $\phi_t^{\star}\omega = \omega_i(\phi_t) e^i \circ d\phi_t$. Thus by the Leibniz rule again $$ \left.\frac{d}{dt}\right|_{t=0} \phi_t^{\star}\omega = \left.\frac{d}{dt}\right|_{t=0} \omega_i(\phi_t) e^i + \omega_i \left.\frac{d}{dt}\right|_{t=0} e^i \circ d\phi_t = \{x^j \partial_j \omega_i + \omega_j \partial_i x^j \} e^i.$$ Hence, $\mathcal{L}_X(\omega)(Y) = \{x^j \partial_j \omega_i + \partial_i x^j \omega_j\} y^i.$ Simplify terms to conclude.

A more sophisticate manner to prove the equality consists in using tensorial algebra. All your definitions of Lie derivatives are particular cases of the Lie derivative of tensor fields. For any tensor field $T$ on your manifold, the Lie derivative is given by $$ \mathcal{L}_X(T) = \left.\frac{d}{dt}\right|_{t=0} (\phi_{-t})_{\star} T_{\phi_t}$$ Now, it is easy to prove that the Lie derivative commutes with contractions and satisfies the following Leibniz rule for the tensor product: $$ \mathcal{L}_X(T \otimes S) = \mathcal{L}_X(T) \otimes S + T \otimes \mathcal{L}_X(S).$$ But $\omega(Y)$ is a contraction of $\omega \otimes Y$ and the terms of the right-hand side of your equality are given by the same contraction.