Flows and Lie brackets, $\beta$ not a priori smooth at $t = 0$ Let $X$ and $Y$ be smooth vector fields on $M$ generating flows $\phi_t$ and $\psi_t$ respectively. For $p \in M$ define$$\beta(t) := \psi_{-\sqrt{t}} \phi_{-\sqrt{t}} \psi_{\sqrt{t}} \phi_{\sqrt{t}}(p)$$for $t \in (-\epsilon , \epsilon)$ where $\epsilon$ is sufficiently small. Does it follow that$$[X, Y](f)(p) = \lim_{t \to 0} {{f(\beta(t)) - f(\beta(0))}\over t}?$$Thoughts. I know that $\beta$ is not a priori smooth at $t = 0$ because of the $\sqrt{}$ terms. But I do not know what do from here. Could anybody help?
 A: Let me please put $\alpha(t):=\beta(t^2)$. We are asked to prove
$$\left.\frac{d}{dt}\right|_{t=0}\alpha(\sqrt{t})=[X.Y]_p.$$
Fact: $$\left.\frac{d}{dt}\right|_{t=0}\alpha(t)=
\left.\frac{d}{dt}\right|_{t=0}\psi_{-t}\phi_{-t}\psi_t\phi_t(p)
=-\psi'_0\phi_0\psi_0\phi_0(p)-\psi_0\phi'_0\psi_0\phi_0(p)
+\psi_0\phi_0\psi'_0\phi_0(p)+\psi_0\phi_0\psi_0\phi'_0(p)=0.$$
Fact: $\alpha(t)$ has a Taylor expansion near $0$ which I notate as
$$\alpha(t)=\alpha(0)+\alpha'(0)(t)+\frac{\alpha''(0)}2t^2+O(t^3).$$
Note that I can drop the $t$ term since the coefficient is $0$.
Then $$\left.\frac{d}{dt}\right|_{t=0}\alpha(\sqrt{t})=\lim_{t\rightarrow0}
\frac{\alpha(\sqrt t)-\alpha(0)}t=\lim_{t\rightarrow0}
\frac{\left(\alpha(0)+\alpha'(0)(\sqrt t)+\frac{\alpha''(0)}2(\sqrt t)^2+O(t^{3/2})\right)-\alpha(0)}t=\frac{\alpha''(0)}2.$$
So it is necessary and sufficient to show $$\left.\frac{d^2}{dt^2}\right|_{t=0}\alpha(t)=2[X.Y]_p.$$
Now, for two vector fields $X$ and $Y$ with local coords $X^i\partial_i$, $Y^j\partial_j$, where here and henceforth I omit the summation symbol over indices, there is a convenient formula for the bracket
$[X,Y]=\gamma^i\partial_i$, namely
$$\gamma^i=X(Y(x_i))-Y(X(x_i))=X(Y^i)-Y(X^i)=(X^j\partial_jY^i-Y^j\partial_jX^i).$$
In a similar way (following this answer), we may write a Taylor expansion of the flow at $p=p^i$ (implicit summation):
$$\phi_t(p)=p+tX(p)+\frac{t^2}2X(X(x_i))(p)+O(t^3),$$
finding (summing implicitly)
$$\phi_t(p)=p^i + t \; X^i(p) + \frac{t^2}{2} X^j(p) \; \partial_j X^i(p) + O(t^3).$$
This is true, since
$$
\frac{d}{dt} \phi_t(p) = X^i(p) + t X^j(p) \partial_j X^i(p) + O(t^2) \\
= X^i(p + t \; X^i(p)) + O(t^2) \\
= X^i(\phi_t(p)) + O(t^2).
$$
The second (middle) equality here follows from Taylor expanding $X^i$. The third (i.e. last) follows from substitution.
With this understanding, we can actually compute
$$
 \psi_t \phi_t(p) = p^i + t X^i + \frac{t^2}{2} X^j \partial_j X^i + t Y^i + \frac{t^2}{2} Y^j \partial_j Y^i + t^2 X^j \partial_j Y^i +O(t^3),
$$
$$
 \psi_{-t} \phi_{-t}(p) = p^i - t X^i + \frac{t^2}{2} X^j \partial_j X^i - t Y^i + \frac{t^2}{2} Y^j \partial_j Y^i + t^2 X^j \partial_j Y^i +O(t^3).
$$
Now, we don't write out exactly the composition of these two fields but we write only the $t^2$ term. From pairing $\frac{t^2}{2} X^j \partial_j X^i$ with $p$ and likewise for $Y$, we get a contribution of
$$\frac{t^2}{2} X^j \partial_j X^i +\frac{t^2}{2} X^j \partial_j X^i + \frac{t^2}{2} Y^j \partial_j Y^i + \frac{t^2}{2} Y^j \partial_j Y^i.$$
From pairing $t X^i$ with $-t X^i$ and $t Y^i$ with $-t Y^i$ we cancel this term. From pairing $-t X^j$ with $t Y^i$ we cancel one of the two $t^2 X^j \partial_j Y^i$, and are left with
$[X,Y]_pt^2$ from $-t Y^i$ with $t X^j$ and the remaining $t^2 X^j \partial_j Y^i$. It is now also easy to see explicitly that the linear term vanishes and $$\alpha(t)=p+[X,Y]_pt^2+O(t^3).$$
Hence $\alpha''(0)=([X,Y]_p t^2)''(0)=2[X,Y]_p$; that's it.
