Let $G$ be a Lie group.
(a) Let $m : G \times G \to G$ denote the multiplication map. Using Proposition
3.14 of john lee smooth manifolds (The Tangent Space to a Product Manifold) to identify $T_{(e,e)}(G \times G)$ with $T_e(G) \bigoplus T_e(G)$, Show that:
(a) the differential $dm_(e,e)(X,Y)=X + Y$
(b) Let $i: G \to G$ denote the invertion map. show that $di_e: T_eG \to T_eG$ is given by $di_e(X)= -X$
This is problem 7-2 of John Lee Smooth Manifolds.
I already solved this wanna see if my solution is correct or not. thanks.
Here is my solution:
We have :
$dm_{(e,e)}(X,Y) = dm_{(e,e)}(X,0)+dm_{(e,e)}(0,Y)=d(m^{1})_e(X)+d(m^{2})_e(Y)$
Which $m^1:G \to G $ is defined by $x \mapsto m(x,e)$ , $m^2:G \to G $ defined by $y \mapsto m(e,y)$, but $m^1 = m^2 = Id_G$ ,so $dm_{(e,e)}(X,Y)=X+Y$ this proves (a). Put $ n= m \circ p \circ s$ which $s :G \to G \times G$ is defined by $x \mapsto (x,x)$ and $p:G \times G \to G \times G$ is defined by $(x,y) \mapsto (x,i(y))$ then $n$ is a constant map ,so:
\begin{align}
0 = dn_e(X)=dm_{(e,e)}(dp_{(e,e)}(ds_e(X)))=dm_{(e,e)}(dp_{(e,e)}(X,X))&=dm_{(e,e)}(X,di_e(X))\\&=X+di_e(X)
\end{align}
so $di_e(X)=-X$
Best Answer
I think your proof is very neat! It works :)
Maybe you could argue a bit better why $dm_{(e, e)} (x, 0) = (d m_1) _e(X) $, which follows in your compositional style by writing $m(x, e) = m \circ i_1$. Here $i_1 : G \to G \times G$ is defined by $i_1(x) = (x, e) $.