Let $G$ be a Lie group. Its Lie algebra, when considered as a vector space, is given by $T_e G$, the tangent space of $G$ at the identity element $e$.
However, in order to fully describe the Lie algebra associated with $G$ we must also describe the commutator $\left[\cdot,\,\cdot\right]:T_eG^2\to T_eG$ and show that it obeys the three axions: bilinearity, alternativity and the Jacobi identity.
Following Wikipedia, one defines first for any $g\in G$ left multiplication on $G$ by $g$ as $$ l_g:G\to G$$ given by $$ x\mapsto gx$$and a left-multiplication invariant vector field $X\in TG$ is one which obeys for every $\left(g,\,h\right)\in G^2$ $$ \left(d l_g\right)_h X_h = X_{gh} $$ where $\left(d l_g\right)_h:T_hG\to T_{gh}G$ is the differential of $l_g$ at $h$. We define $Lie\left(G\right)$ as the vector space of left-multiplication invariant vector fields.
Apparently $G\times Lie\left(G\right)$ is isomorphic to $TG$ via $$\left(g,X\right)\mapsto\left(g,X_g\right)$$ so that we get a $\natural$ isomorphism $Lie\left(G\right)$ with $T_eG$ given by $$Lie\left(G\right)\ni X \mapsto X_e\in T_eG$$This last isomorphism we call $\phi$.
Now apparently there is a Lie bracket structure on $TG$ which induces a Lie bracket structure on $Lie\left(G\right)$: for any $X$ and $Y$ in $TG$ we have $$ \left[X,\,Y\right]_{TG}= X\left(Y\left(\cdot\right)\right)-Y\left(X\left(\cdot\right)\right)$$ This in turn can be pushed forward to a define our desired Lie bracket on $T_e G \equiv \frak{g} $ via: If $X$ and $Y$ are any tangent vectors in $T_eG$ then $$\left[X,\, Y\right]_{T_eG}=\phi\left[\phi^{-1}\left(X\right),\,\phi^{-1}\left(Y\right)\right]_{TG}$$
My question is: is there now any easier way to describe $\left[\cdot,\,\cdot\right]:T_eG^2\to T_eG$? I tried for example to write it in coordinates and it was a mess. How do you "easily" see how $\left[X,\, Y\right]_{T_eG}$ acts on any scalar function $f$ given some $X$ and $Y$ in $T_eG$?
Best Answer
Indeed there is at least locally at the identity. The heart of the questio is given by the Baker Hausdorff Campbell formula, i.e. $$\exp\left(X\right)\exp\left(Y\right)=\exp\left(X+Y+\frac{1}{2}\left[X,Y\right]+\frac{1}{12}\left(\left[X,\left[X,Y\right]\right]+\left[Y,\left[Y,X\right]\right]\right)...\right).$$ Usually you read it in the opposite way deducing locally the multiplication law starting from Lie's parenthesis (you might want to remember that $\exp\left(X\right)$ can be thought as an element of the Lie Group and so $\exp\left(X\right)\exp\left(Y\right)$ it just the multiplication of two elements $g$ and $h$ of the Lie Group). The converse, i.e. recovering the Lie parenthesis from the multiplication law, is also true but involves the logarithm function and therefore it's really just local, but anyway can be done.