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Let $G$ be a Lie group with multiplication map $\mu :G \times G \to G$, the inverse map $\iota : G \to G$, and identity element $e$. Show that the differential $d\mu_{(e,e)}:T_eG \times T_e G \to T_eG$ is addition e.g $d\mu_{(e,e)}(u,v)=u + v$.

Let $c : (-\varepsilon, \varepsilon) \to G$ be a curve such that $c(0)=e$ and $c'(0)=v$. Define $\alpha:(- \varepsilon, \varepsilon)\to G \times G$ as $\alpha(t)=(c(t), e)$. Now computing $$d\mu_{(e,e)}(u, 0) = \frac{d}{dt} \bigg|_0 \mu \circ \alpha(t) = \mu(\alpha(0))' = \mu'(\alpha(0))\cdot \alpha'(0)=\mu'(e,e) \cdot (u,0)$$

I'm trying to show that this is just $u$, but I cannot do anything with this $\mu'(e,e)$ term. How can I get rid of it?

Thorgott
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Bernard Lees
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    $\mu^{\prime}(e,e)$ is the same thing as $d\mu_{(e,e)}$, so you're only writing tautology. You have to incorporate the definition of $\alpha$ to give a direct description of $\mu\circ\alpha$. – Thorgott Feb 10 '22 at 15:28
  • The last part of snulty's answer at https://math.stackexchange.com/q/1740179/169085 exactly answers this. – Alp Uzman Feb 10 '22 at 17:14

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