Lie bracket on the algebraic Lie algebra of a matrix Lie group? Let $G$ be a Lie subgroup of ${\rm GL}_n(\mathbb C)$. Also let $\mathfrak g \subseteq M_n(\mathbb C)$ be the Lie algebra of $G$.
I believe that we may, if we wish, get a handle on $\mathfrak g$ as follows: for a small enough open neighborhood $U$ of the identity in $G$, the map $\log$ (defined by a power series) is a homeomorphism of $U$ onto a an open neighborhood $V$ of $0$ in $\mathfrak g$, with inverse $\exp$.
In particular, if I wish, I may DEFINE $\mathfrak g$ as the linear span inside $M_n(\mathbb C)$ of $\log U$ for $U$ small enough. Let's call this the algebraic definition of $\mathfrak g$.
How can I understand the Lie bracket on $\mathfrak g$ given this definition? (Of course I secretly know that it's given by $[X, Y] = XY - YX$, let's pretend I don't know this.) Campbell-Hausdoff gives me the multiplication structure on $G$ given the bracket on $\mathfrak g$; here I am looking for something like the reverse -- a Lie bracket structure on $\mathfrak g$ given the multiplication structure on $G$.
Here's a start: Campbell-Hausdorff tells me that
$$[\log(1 + x), \log(1 + y)] = 2\log[(1 + x)(1 + y)] - 2\log(1 + x) - 2\log(1 + y) + \mbox{higher-order terms}.$$
(This makes sense: if $x$ and $y$ commute, then so do $\log(1 + x)$ and $\log(1 + y)$, so that on one hand, the Lie bracket on the left is zero. On the other hand, if $1 + x$ and $1 + y$ commute, then the logarithm of their product is the sum of their logarithms, so that the right-hand side is zero on the nose.)
Is it obvious how to take this further? Is there a well-known formula for this kind of expansion?
 A: Well if you want something with $[X, Y]$ with $X,Y$ in the group $G$, indeed it does not exist since $G$ doesn't have a Lie algebra structure.
Instead, if you want to obtain the brackets of the Lie algebra $g$ from the multiplication rule over $G$ you just need to differentiate. In this differentation process you might consider elements of the group of the form $X = $exp$(tA)$ and $Y = $exp$(tB)$ and write something of the form
$$[ log( exp(tA)), log( exp(tB))]=[ log(X), log(Y)]=[A,B]$$ with $A,B$ in the Lie algebra and $X,Y$ are in the Lie Group. But you have to workout carefully al the process and arrive to something that -at best- is tautological
Other option is to work directly on the logarythm of the CHB formula as it is done here in Richtmayer but can be found also in Brian Hall I think.
The main reason why you pass to study Lie algebras instead of Lie Groups is that Lie algebras are linear spaces and algebras with a bilinear bracket that is easy to study intead of studying a full non linear space as the Lie Group normally is. So in general Lie groups are not vector spaces and are not algebras (no brackets), only their tangent space at the identity is.
Hope this clarify :)
