Lie algebra: Proof that the commutator of infinitesimal motions is an infinitesimal motion I am following Classical and Quantum Mechanics via Lie Algebras by Neumaier and Westra.
Setup
I am stuck at part of Thm 2.3.1. Consider the matrix group $\mathbb{G}$. The set of $\mathbb{G}$-motions is the set of maps $U:[0,1]\rightarrow \mathbb{G}$ with $U(0) = I$. if $U$ is a $\mathbb{G}$-motion then
\begin{align}
f = \frac{d}{dt}U(t)\Big\rvert_{t=0}
\end{align}
is called an infinitesimal motion. For example, if $U = e^{ft}$ then $\frac{d}{dt}U(t)\Big\rvert_{t=0} = f e^{ft}\Big\rvert_{t=0} = f$ so $f$ is an infinitesimal motion. The authors denote the set of all infinitesimal motions by $\mathbb{L} = \log \mathbb{G}$. The theorem in question state that

*

*$\mathbb{L}$ is a vector space and that if $f,g \in \mathbb{L}$ then $[f,g]=fg-gf \in \mathbb{L}$,


*for $f\in\mathbb{L}$ the adjoint mapping of $V\in\mathbb{G}$ on $f$: $\text{Ad}_V(f) = VfV^{-1} \in \mathbb{L}$


*$\text{Ad}_V [f,g] = \left[\text{Ad}_V f, \text{Ad}_V g \right]$
I understand points 2) and 3) and that $\mathbb{L}$ is a vector space. What I want help with is the proof that $[f,g] \in \mathbb{L}$. I am trying to work through this proof without relying on the Landau notation $O(t^2)$ at any point.
Example of the type of proof I'm looking for
For example, it can be proven that $\mathbb{L}$ is a vector space as follows: $f,g \in \mathbb{L}$, $\alpha, \beta \in \mathbb{K}$, the field. Since $f,g \in \mathbb{L}$ there exists $V_f(t)$ and $V_g(t)$ with $\frac{d}{dt}V_f(t)\Big\rvert_{t=0} = f$ and $\frac{d}{dt}V_g(t)\Big\rvert_{t=0} = g$. Consider then
\begin{align}
V_{\alpha,\beta}(t) = V_f(\alpha t)V_g(\beta t)
\end{align}
Then
\begin{align}
\frac{d}{dt}V_{\alpha,\beta}(t) &= \left(\frac{d}{dt}V_f(\alpha t)\right)V_g(\beta t)\Big\rvert_{t=0} + V_f(\alpha t)\left(\frac{d}{dt}V_g(\beta t)\right)\Big\rvert_{t=0}\\
&= \alpha f I + I \beta g = \alpha f + \beta g
\end{align}
so $\alpha f + \beta g \in \mathbb{L}$ so $\mathbb{L}$ is closed under addition and scalar multiplication. This proof didn't rely on the big-O notation at all.
The proof I'm having trouble with and my questions
The proof in the document that $[f,g] \in \mathbb{L}$ is as follows. From part 2) and the first part of part 1) we have that $V_g(t)fV_g(t)^{-1} - f \in \mathbb{L}$. It can be seen that $\frac{d}{dt}V_g(t)^{-1}\Big\rvert_{t=0} = -g$  So then
\begin{align}
V_g(t)fV_g(t)^{-1} - f = (1+gt+O(t^2))f(1-gt+O(t^2))-f = t[f,g] + O(t^2)
\end{align}
Then divide by $t$ and take $t\rightarrow 0$ to see that $[f,g] \in \mathbb{L}$.
I have two problems with this proof.
A) it relies on the big $O(t^2)$ notation. I would prefer a proof that didn't rely on that.
B) I can see that $\frac{V_g(t)fV_g(t)^{-1}-f}{t} \in \mathbb{L}$ but it doesn't seem obvious to me that the limit of this operator as $t\rightarrow 0$ should also be an element of $\mathbb{L}$.
Could anyone please help me find a proof which doesn't rely on big-O notation and also help me with this limit question?
 A: I adapted my answer from Theorem 3.20 in "Introduction to Lie Algebras" by Karin Erdmann and Mark J. Wildon. After writing the answer it does feel a bit too mathy instead of physicsy.  I'm not sure what the appropriate action is to do with the post then.
Consider $f,g \in \mathbb{L}$. Then there are $\mathbb{G}$-motions $V_f(t), V_g(t)$ with $\frac{d}{dt}V_f(t)\Big\rvert_{t=0} = f$ and $\frac{d}{dt}V_g(t)\Big\rvert_{t=0} = g$.
Consider the $\mathbb{G}$-motion:
\begin{align}
V(s,t) = V_g(s)V_f(t)V_g(s)^{-1}
\end{align}
Then take 
\begin{align}
\frac{d}{dt}V(s,t)\Big\rvert_{t=0} = V_g(s)fV_g(s)^{-1} \in \mathbb{L}
\end{align}
So $V_g(s)fV_g(s)^{-1} \in \mathbb{L}$ for all $s$. Note that it can be shown without too much trouble that $\frac{d}{ds}V_{g}(s)^{-1}\Big\rvert_{s=0} = -g$ by taking the derivative of $V_g(s)V_g(s)^{-1} = 0$ using the product rule.
Now notice that
\begin{align}
\frac{d}{ds}\left(V_g(s)fV_g(s)^{-1}\right)\Big\rvert_{s=0} = gf - fg = [g,f]
\end{align}
This however does not complete the proof that $[g,f] \in \mathbb{L}$. This is because we haven't proven that derivatives of sets of elements of the Lie Algebra parametrized by an $s$ are contained in the Lie Algebra.. That requires writing out the derivative above:
\begin{align}
\frac{d}{ds}\left(V_g(s)fV_g(s)^{-1} \right)\Big\rvert_{s=0} &= \lim_{h\rightarrow 0} \frac{V_g(s+h)fV_g(s+h)^{-1} - V_g(s)fV_g(s)^{-1}}{h}\Bigg\rvert_{s=0}\\
&= \lim_{h\rightarrow 0}\frac{V_g(h)fV_g(h)^{-1} - f}{h}
\end{align}
$V_g(h)fV_g(h)^{-1} \in \mathbb{L}$ by the above argument. Then subtracting $f$ gives us something again in $\mathbb{L}$ because the Lie Algebra is closed under addition, finally dividing by $h$ again gives us an element of the Lie Algebra because it is closed under multiplication by real numbers. This means we are taking a limit of elements of the Lie Algebra.
The question then is if limits of elements of the Lie Algebras are contained in the Lie Algebra. The answer is that they are. The textbook I mentioned indicates that this is so because $\mathbb{L}$ is a real vector subspace of $M_n(\mathbb{C})$, the space of finite dimensional complex matrices meaning that $\mathbb{L}$ is topologically closed, which implies that $\mathbb{L}$ contains its limit points. Thus it follows that $\mathbb{L}$ contains the derivative above and thus $[g,f] \in \mathbb{L}$.
The closedness of $\mathbb{L}$ basically follows from the fact that it is a finite dimensional vector space over $\mathbb{R}$ which is complete.
