The diffential of commutator map in a Lie group

Question

Leb $G$ be a Lie group and $f:G\times G\rightarrow G$ be the commutator map $:(x,y)\mapsto xyx^{-1}y^{-1}$.

How to obtain the Lie bracket in the associated Lie algebra of $G$ from the derivatives of $f$?

(We know that the Lie bracket is defined via the adjoint representation.)

(I saw $df_{(e,e)}(X,Y)=[X,Y]$ somewhere. Thanks to @John for remarking this is false. But I think there is indeed a relation between the Lie bracket and derivatives of $f$. )

Answer 1

I think the differential at $(e,e)$ is actually zero:

First, if $m : G\times G \to G, (g,h)\mapsto gh$ is the multiplication map then $dm : T_e G \oplus T_e G \to T_e G$ is given by $(X,Y)\mapsto X+Y$.

Also if $i: G \to G, g \mapsto g^{-1}$ denotes the inversion map then we have $di : T_e G \to T_e G, X\mapsto -X$.

Your map $f$ is given by the composition $$ G\times G \to G\times G\times G \times G \to G\times G \to G \\ (g,h) \mapsto (g,h, g^{-1}, h^{-1}) \mapsto (gh, g^{-1} h^{-1}) \mapsto ghg^{-1} h^{-1} $$ so the differential is the composition $$ (X,Y) \mapsto (X,Y, -X,-Y) \mapsto (X+Y, -X-Y) \mapsto0. $$

Answer 2

Let $G$ be a Lie group with the neutral element $e$. Then, given two elements in its Lie algebra, $\xi,\eta\in\operatorname{Lie} G:=T_eG$, represented by (smooth) paths $x,y\colon (-\varepsilon,\varepsilon)\to G$ with $x(0)=y(0)=e$, one can define the Lie bracket as the second derivative of the commutator map: $$[\xi,\eta]=\frac{d^2}{dt^2}(x(t)y(t)x(t)^{-1}y(t)^{-1})\biggr|_{t=0}.$$ Note that the first derivative of the commutator vanishes, which makes the second derivative well-defined as an element of the tangent space $T_eG$.