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 $T_eG$?


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.

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    $\begingroup$ @Daco, Thanks for your comment. I still don't fully understand how to extract only $\left[X,\,Y\right]$ by taking the log because there will be an infinite series of nested brackets. Can you then give an explicit formula for $\left(\left[X,\,Y\right]\right)\left(f\right)$? $\endgroup$ – PPR Oct 20 '16 at 8:03
  • $\begingroup$ I might have missen something... once you have X and Y obtained from the logarithm you just define the commutator the usual way $\endgroup$ – Dac0 Oct 21 '16 at 7:12
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    $\begingroup$ Anyway, whatever you are looking for, for these kind of questions the answer is in some kind of variation of the BHC forrmula $\endgroup$ – Dac0 Oct 21 '16 at 7:15
  • $\begingroup$ First of all, as I said, I don't understand how to obtain $X$ and $Y$ from the Logarithm since there is also the infinite series of nested commutators. Secondly, $X$ and $Y$ are actually given, and what is unknown is $\left[X,\,Y\right]$, and an explicit formula for how it acts on a scalar $f:G\to\mathbb{F}$. $\endgroup$ – PPR Oct 21 '16 at 13:19
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    $\begingroup$ Ok, now I got what you were asking $\endgroup$ – Dac0 Oct 21 '16 at 14:25

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