Equivalence of two definitions of Lie bracket of vector fields $[X,Y]$ in terms of derivatives and differential of flows

Question

Lie bracket of vector fields is defined in two ways:

Let $\Phi^X_t$ be the flow associated with the vector field $X$, and let $d$ denote the tangent map derivative operator. Then the Lie bracket of $X$ and $Y$ at the point $x \in M$ can be defined as $[X, Y]_x := \lim_{t \to 0}\frac{(\mathrm{d}\Phi^X_{-t}) Y_{\Phi^X_t(x)} - Y_x}t = \left.\frac{\mathrm{d}}{\mathrm{d} t}\right|_{t=0} (\mathrm{d}\Phi^X_{-t}) Y_{\Phi^X_t(x)}$

or equivalently $[X, Y]_x := \left.\frac12\frac{\mathrm{d}^2}{\mathrm{dt}^2}\right|_{t=0} (\Phi^Y_{-t} \circ \Phi^X_{-t} \circ \Phi^Y_{t} \circ \Phi^X_{t})(x) = \left.\frac{\mathrm{d}}{\mathrm{d} t}\right|_{t=0} (\Phi^Y_{-\sqrt{t}} \circ \Phi^X_{-\sqrt{t}} \circ \Phi^Y_{\sqrt{t}} \circ \Phi^X_{\sqrt{t}})(x)$

How does one show by math that these two definitions are equivalent? And for the second definition, how does one show that $\left.\frac12\frac{\mathrm{d}^2}{\mathrm{dt}^2}\right|_{t=0} (\Phi^Y_{-t} \circ \Phi^X_{-t} \circ \Phi^Y_{t} \circ \Phi^X_{t})(x) = \left.\frac{\mathrm{d}}{\mathrm{d} t}\right|_{t=0} (\Phi^Y_{-\sqrt{t}} \circ \Phi^X_{-\sqrt{t}} \circ \Phi^Y_{\sqrt{t}} \circ \Phi^X_{\sqrt{t}})(x)$?

Answer 1

Let $c(t)= \psi_{-t} \circ \phi_{-t} \circ \psi_t \circ \phi_t(p)$. In his book A comprehensive introduction to differential geometry, Spivak shows that $c'(0)=0$ and $c''(0)=2[X,Y]_p$ (Prop. 15 and Thm. 16, p. 160-163).

To simplify the notations, he defines $$\begin{array}{l} \alpha_1(t,h)= \psi_t \circ \phi_h(p) \\ \alpha_2(t,h)=\phi_{-t} \circ \psi_h \circ \phi_h(p) \\ \alpha_3(t,h) = \psi_{-t} \circ \phi_{-h} \circ \psi_h \circ \phi_h(p) \end{array}$$ so that $c(t)= \alpha_3(t,t)$. Noticing that $$\begin{array}{l} \alpha_2(0,t)= \alpha_1(t,t) \\ \alpha_3(0,t)= \alpha_2(t,t) \end{array}$$ and for any $f \in C^{\infty}(M)$ $$\begin{array}{ll} \partial_t(f \circ \alpha_1) = Yf \circ \alpha_1 & \partial_t(f \circ \alpha_2)=-Xf \circ \alpha_2 \\ \partial_t(f \circ \alpha_3)=-Yf \circ \alpha_3 & \partial_h(f \circ \alpha_1)(0,h)=Xf(\alpha_1(0,h)) \end{array}$$ the identities $c'(0)=0$ and $c''(0)=2[X,Y]_p$ are shown using differential calculus.

Then, the identity $[X,Y]_p = \frac{d}{dt}_{|t=0} \psi_{-\sqrt{t}} \circ \phi_{-\sqrt{t}} \circ \psi_{\sqrt{t}} \circ \phi_{\sqrt{t}}(p)$ can be deduced from the previous one using the following lemma (Ex. 16, p. 176):

Lemma: Let $c : \mathbb{R} \to M$ be a smooth curve satisfying $c'(0)=0$. If $\gamma(t)=c(\sqrt{t})$, then $$\lim\limits_{h \to 0^+} \frac{\gamma(h)- \gamma(0)}{h}= \frac{1}{2}c''(0)$$

In fact, because we only consider $f \circ c$ where $f \in C^{\infty}(M)$, it is sufficient to show the lemma for $c : \mathbb{R} \to \mathbb{R}$. But $$c(h)=c(0)+ \underset{=0}{\underbrace{c'(0)}} \cdot h+ \frac{1}{2}c''(0) \cdot h^2+o(h^2)$$ implies $$\frac{c(\sqrt{h})-c(0)}{h}= \frac{1}{2}c''(0)+o(1).$$