Lie brackets as infinitesimal commutator I've seen a couple statements justifying the idea that the Lie bracket represents the infinitesimal commutator of two vector field flows. See for example Lie bracket and flows on manifold or How to prove the Lie bracket is infinitesimal commutator. When the Lie bracket is computed on a Lie group, it can be seen as the Lie bracket of two left-invariant vector fields that extend given vectors at the identity, which is related to the commutator of the flows of those left-invariant vector fields. I'm wondering if there is a sense in which the Lie bracket is a commutator of two fixed vectors at the identity in a way that does not depend on their extension to the entire Lie group in a left-invariant way. Specifically, I am wondering about the truth of the following claim, which I expect is probably false.

Suppose $\gamma_1(t)$ and $\gamma_2(t)$ are smooth paths on a Lie group such that $\gamma_1(0)=\gamma_2(0)=e.$ Then
$$
\frac{d}{dt}\Big|_{t=0} [\gamma_1(\sqrt{t}),\gamma_2(\sqrt{t})]=[\gamma_1'(0),\gamma_2'(0)]
$$

where the first pair of square brackets $[-,-]$ represent the Lie group commutator. Is this true?
Edit: I came up with a little computation that doesn't resolve this question but seems to say something about it. By the Baker-Campbell-Hausdorff formula, $[e^X,e^Y]=e^{[X,Y]+\dots}$ where the dot represents "lower order terms". Also, because the differential of the exponential map is the identity, $\frac{d}{dt}\Big|_{t=0}e^{X(t)}=X'(0),$ where $X(t)\in T_e G$ is some vector at the identity that varies in time. Pick $X(t)$ and $Y(t)$ so that $e^{X(t)}=\gamma_1(t)$ and $e^{Y(t)}=\gamma_2(t).$ Then
$$
\begin{aligned}
\frac{d}{dt}\Big|_{t=0}[\gamma_1(\sqrt{t}),\gamma_2(\sqrt{t})]
&=\frac{d}{dt}\Big|_{t=0}\left[e^X(\sqrt{t}),e^Y(\sqrt{t})\right]\\
&\approx \frac{d}{dt}\Big|_{t=0}e^{[X(\sqrt{t}),Y(\sqrt{t})]}\\
&=\frac{d}{dt}\Big|_{t=0}[X(\sqrt{t}),Y(\sqrt{t})].
\end{aligned}
$$
On the other hand,
$$
\begin{aligned}
\left[\gamma_1'(0),\gamma_2'(0)\right]
&=\left[\frac{d}{dt}\Big|_{t=0}e^{X(t)},
\frac{d}{dt}\Big|_{t=0}e^{Y(t)}\right]\\
&=[X'(0),Y'(0)].
\end{aligned}
$$
This seems to suggest that my original claim might be equivalent to
$$
\frac{d}{dt}\Big|_{t=0}[X(\sqrt{t}),Y(\sqrt{t})]=[X'(0),Y'(0)].
$$
This definitely  holds for $X(t)=tX_0$ and $Y(t)=tY_0$, but might not hold generally. In this special case, the main claim seems to reduce to the statement on flows proven in the "Lie bracket and flows on manifold" link.
 A: One can't simply talk about the Lie algebra as the tangent space at the identity; you need to define a Lie bracket operation on the space as well. It is typically defined from the Lie bracket of vector fields, and in this definition one must appeal to left-invariant vector fields at some point. If you're comitted to avoiding them, you can alternatively define the Lie bracket using the adjoint representation.
It's instructive to look at the case of the group of invertible $n\times n$ matrices $GL(n,\mathbb{R})$. I'll assume assume we have established the standard identification of the Lie algebra $\mathfrak{gl}(n,\mathbb{R})$ with the space of $n\times n$ matrices with the matrix commutator. To prove the result here, we can use the following fact from calculus, which can be proven using Taylor's theorem:

Lemma: Let $f:[0,a)\to\mathbb{R}^n$ be a $C^\infty$ function with $f'(0)=0$. The function $g(t):=f(\sqrt{t})$ is $C^1$, and $g'(0)=\frac{1}{2}f''(0)$.

Let $A,B:[0,\epsilon)\to GL(n,\mathbb{R})$ be smooth matrix-valued paths with $A(0)=B(0)=I$, $\dot{A}(0)=X$, and $\dot{B}(0)=Y$. Using the product rule and the inverse rule $\frac{d}{dt}(A^{-1})=-A^{-1}\dot AA^{-1}$, you can expand out the second derivative of the group commutator at zero and all but the appropriate terms cancel:
$$
\frac{d^2}{dt^2}(ABA^{-1}B^{-1})_{t=0}=2(XY-YX)
$$
These two facts are enough to prove the result in $GL(n,\mathbb{R})$. This also implies the general case, since every Lie group is locally isomorphic to a Lie subgroup of $GL(n,\mathbb{R})$ for some $n$ (as a consequence of Ado's theorem).
