Let G be a lie group . we know a Lie group is a group with a smooth manifold structure s.t both the multiplication map $m$ and group inversion map $i$ are smooth .
Now by identifying $T_{(e,e)}(G\times G)\simeq T_eG\oplus T_eG $ , I can show that the tangent map $T_{(e,e)}m:T_eG\oplus T_eG\mapsto T_eG$ is given by $T_{(e,e)}m.(X,Y)=X+Y$.
Now , I wanna use this result to show that the tangent map $T_ei:T_e(G)\mapsto T_e(G)$ is given by $T_ei.X=-X$ since it seems hard to directly derive this result.
My attempt is to define a map $m':G\times G \mapsto G$ by $ g.g^{-1}=e$.Then $T_{(e,e)}m'.(X,Y)=X+Y$= 0 .
Now the question is I don't know how to relate $T_{(e,e)}m'.(X,Y)=X+Y$= 0 to $T_ei.X=-X$.I am also not sure about the correct way to define $m'$. Could someone help me with it ? Thanks a lot.
Best Answer
Consider the composition $G\rightarrow G\times G\rightarrow G$ where the first map sends $g$ to $(g,g^{-1})$, and the second map is $m$: it sends $(g,h)$ to $gh$.
This composition is constant, so its differential must be $0$. On the other hand, if you compute the differential via the chain rule, you'll get the result you want.