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Given a Lie group.

Multiplication and inversion act infinitesimally at the identity by: $$\mathrm{d}\mu:\mathrm{T}_{(e,e)}(G\times G)\to\mathrm{T}_eG:(u,v)\mapsto u+v$$ $$\mathrm{d}\iota:\mathrm{T}_eG\to\mathrm{T}_eG:w\mapsto -w$$

How to prove these statements from scratch?

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  • This question was asked at least twice before: http://math.stackexchange.com/questions/904231/differential-of-the-inversion-of-lie-group?rq=1, http://math.stackexchange.com/questions/209682/pushforward-of-inverse-map-around-the-identity?rq=1 – Moishe Kohan Jan 31 '15 at 01:31
  • See also this http://math.stackexchange.com/questions/799211/lie-group-differential-of-multiplication-map. – PhoemueX Jan 31 '15 at 02:50
  • @studiosus: Aah so it needs some more advanced things like one-parameter subgroups and or the exponential map. – C-star-W-star Jan 31 '15 at 12:10
  • @PhoemueX: Nice answer of yours. It makes things more intuitively clear. :) – C-star-W-star Jan 31 '15 at 12:14
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    @Freeze_S: You should learn these in any case if you were to learn about Lie groups. – Moishe Kohan Jan 31 '15 at 12:14

1 Answers1

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Ok I got it meanwhile...

Inversion: Geometric Approach

Applying the multiplication infinitesimally gives: $$[\alpha]+\mathrm{d}\iota[\alpha]=[\alpha]+[\iota\circ\alpha]=\mathrm{d}\mu[(\alpha,\alpha^{-1})]=[\mu(\alpha,\alpha^{-1})]=[e]=0$$ (Note that all equality signs in this line are for real.)

Multiplication: Geometric Approach

One has the explicit identification: $$\mathrm{T}_eG\oplus\mathrm{T}_eG\cong\mathrm{T}_{(e,e)}(G\times G):\quad([\alpha],[\beta])\widehat{=}[(\alpha,\beta)]$$

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.)

Multiplication: 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 on sections: $$\mu\circ s_1=\mathrm{id}\quad\mu\circ s_2=\mathrm{id}$$

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.)

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