TITLE: Lie group, differential of multiplication map
QUESTION [12 upvotes]: Let $G$ be a Lie group
Let $m:G\times G\rightarrow G$ denote the multiplication map. Show that the differential $dm_{(e,e)}:T_eG\oplus T_eG\rightarrow T_eG$ is given by
$$dm_{(e,e)}(X,Y)=X+Y$$
This question comes from Lee, introduction to smooth manifolds second edition, problem 7-2.
Here is my attempt:
$$dm_{(e,e)}(X,Y)f=(X,Y)f(m(e,e))=(X+Y)f(e)$$
I don't think I understand fully what I am doing. The notation makes me dizzy sometimes. My guess is that X+Y is a tangent vector in $T_eG$. Now this vector takes $f\in C^{\infty}(G)$ so I feel like maybe this is ok.

REPLY [5 votes]: Algebraic Approach
Consider the sections:
$$s_1:G\to G\times G:p\mapsto (p,e)\quad s_2:G\to G\times G:q\mapsto(e,q)$$
This makes the identification explicit:
$$\Psi:\mathrm{T}_eG\oplus\mathrm{T}_eG\to\mathrm{T}_{(e,e)}(G\times G):(\delta,\varepsilon)\mapsto \mathrm{d}_es_1\delta+\mathrm{d}_es_2\varepsilon$$
Do the key observation:
$$\mu\circ s_1=\mathrm{id}\quad\mu\circ s_2=\mathrm{id}$$
Applying this on test functions gives:
$$\left[\mathrm{d}\mu(\Psi(\delta,\varepsilon))\right]\varphi=\left[\mathrm{d}\mu\mathrm{d}s_1\delta+\mathrm{d}\mu\mathrm{d}s_2\varepsilon\right]\varphi\\
=\delta(\varphi\circ\mu\circ s_1)+\varepsilon(\varphi\circ\mu\circ s_2)=\delta(\varphi)+\varepsilon(\varphi)$$
(Note that making the identification concrete really paid off.)
Geometric Approach
This time the implicit identification suffices:
$$\mathrm{T}_eG\oplus\mathrm{T}_eG\cong\mathrm{T}_{(e,e)}(G\times G):\quad([\alpha],[\beta])\widehat{=}[(\alpha,\beta)]$$
So that tangent curves split accordingly into:
$$[(\alpha,\beta)]\widehat{=}([\alpha],[\beta])=([\alpha],0)+(0,[\beta])\widehat{=}[(\alpha,e)]+[(e,\beta)]$$
Exploiting linearity one obtains:
$$\mathrm{d}\mu[(\alpha,\beta)]=\mathrm{d}\mu[(\alpha,e)]+\mathrm{d}\mu[(e,\beta)]=[\mu(\alpha,e)]+[\mu(e,\beta)]=[\alpha]+[\beta]$$
(Note that the above identification is rather hard to check.)