Lie bracket and flows on manifold

Suppose that $X$ and $Y$ are smooth vector fields with flows $\phi^X$ and $\phi^Y$ starting at some $p \in M$ ($M$ is a smooth manifold). Suppose we flow with $X$ for some time $\sqrt{t}$ and then flow with $Y$ for this same time. Then we flow backwards along $X$ for the same time, and then flow backwards along $Y$. All in all, we can define a curve dependent on $t$ as follows $$\alpha(t):= \phi_{-\sqrt(t)}^Y \circ \phi_{-\sqrt(t)}^X \circ \phi_{\sqrt(t)}^Y \circ \phi_{\sqrt (t)}^X$$

It is an exercise to show that $\frac{d}{dt}|_{t=0} \alpha(t) = [X,Y](p)$. In theory, this should be workable with just the chain rule, (assuming I know how to do these derivatives properly, which is something I’d like some clarification on) but this process is going to be excruciatingly long and painful, and it’s really just something I want to avoid if I can help it. Is there another way to do this computation that will be less painful and more illustrative of why exactly this works out?

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I’m not sure if this is the answer you’re looking for, since this is by computation. But it does not involve the chain rule, at least.

So… both sides of the equation are elements of the tangent space $T_pM$. To see that they are equal, we compute their action on a function $f:M \to \mathbb R$.

Now, by definition
$$
[X,Y](f)(p) = \lim_{h \to 0} \frac 1h \left[ (Yf) \circ \phi_h^X(p)-Y(f)\right] – \lim_{h \to 0} \frac 1h \left[ (Xf) \circ \phi_h^Y(p)-X(f)\right]
$$
The first term is equal to:
$$
\lim_{h \to 0} \frac 1h \left[ (\lim_{k \to 0}\frac 1k [f \circ \phi_k^Y(p)-f(p)]) \circ \phi_h^X(p)-[\lim_{k \to 0}\frac 1k f \circ \phi_k^Y(p)-f(p) ]\right]
$$
Setting $k=h$ (the functions are differentiable, so this shouldn’t change the answer), we get
$$
= \lim_{h \to 0} \frac{1}{h^2} \left[ f \circ \phi_h^Y \circ \phi_h^X (p) – f \circ \phi_h^X(p) -f \circ \phi_h^Y(p) + f(p) \right]
$$
Doing the same for the other term, we get:
$$
\lim_{h \to 0} \frac{1}{h^2} \left[ f \circ \phi_h^X \circ \phi_h^Y (p) – f \circ \phi_h^Y(p) -f \circ \phi_h^X(p) + f(p) \right]
$$
Subtracting them, most terms cancel and we get
$$
\lim_{h \to 0} \frac {1}{h^2}\left[ f \circ \phi_h^Y \circ \phi_h^X(p) – f \circ \phi_h^X \circ \phi_h^Y(p) \right]
$$
Now we use that $(\phi_h^X)^{-1}=\phi_{-h}^X$ to get
$$
\lim_{h \to 0} \frac {1}{h^2}\left[ f- f \circ \phi_h^X \circ \phi_h^Y \circ \phi_{-h}^X \circ \phi_{-h}^Y(p) \right]
$$
But this is just the derivative of $\alpha$ (we traverse in the opposite direction, but that’s okay)! Putting $t=h^2$ we get the result.

So all in all, this computation was not painfree, but it is clear why we need the square root signs.