Lie group, differential of multiplication map

Question

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.

Answer 1

One of the problems here is the Lee uses the projections $\pi_1 : G \times G \rightarrow G, (x,y) \mapsto x$ and $\pi_2$ defined analogously to identify $T_{(e,e)}(G\times G)$ and $T_e G \times T_e G$ using the isomorphism $v \mapsto (d (\pi_1)_e v, d (\pi_2)_e v)$.

Because of this identification, the action of $(X,Y)$ on $f$ is not immediately clear.

An alternative approach runs as follows:

As the hint to the problem suggests, it is easiest to calculate $dm_{(e,e)}(X,0)$ and $dm_{(e,e)}(0,Y)$ and then use the linearity of $dm_{(e,e)}$.

Furthermore, it is often easiest to calculate differentials by using curves, e.g. as follows:

Let $\varphi : (-\varepsilon, \varepsilon) \rightarrow G$ be a smooth curve with $\varphi(0) = e$ and with $\varphi'(0) = X$. It is easy to see (using the above isomorphism) that $\psi : (-\varepsilon , \varepsilon) \rightarrow G \times G, t \mapsto (\varphi(t), e)$ satisfies $\psi'(0) = (X,0)$.

Furthermore,
$dm_{(e,e)} (X,0) = dm_{(e,e)} \psi'(0) = (m \circ \psi)'(0) = \varphi'(0) = X$.

Analogously, you can calculate the other case. Using linearity, the result follows.

EDIT: By the way, your calculation should have been $dm_{(e,e)}(X,Y)f = (X,Y)(f \circ m)$ for $X,Y \in T_e G$. As noted above, the action of $(X,Y)$ on $f \circ m$ is not immediately clear, as you only know $(X,Y) = (d(\pi_1)_e v, d(\pi_2)_e v)$ for some $v \in T_{(e,e)}(G\times G)$ and what you really want to calculate is $v(f\circ m)$.

Answer 2

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.)