Show that $\mu_{∗,(e,e)}(X, Y ) = X + Y$

Question

This is an assignment: (a)Show that $\mu_{∗,(e,e)}(X, Y ) = X + Y$ for any $ (X, Y ) ∈ T_eG ⊕ T_eG$ where $G$ is a Lie group?

(b)Show that $ι_{∗,e}(X) = −X$ for any $X ∈ T_eG$

Can anyone help me with (a) I will then try to solve (b)

Answer 1

Since $\mu_{*}$ is linear,

$$\mu_{*,(e,e)}(X,Y) = \mu_{*,(e,e)}(X,0) + \mu_{*,(e,e)}(0,Y).$$

Let $\alpha(t)$, $\beta(t)$ be integral curves of $X$ and $Y$, respectively, starting at $e$. Then $c_1(t) := (\alpha(t), e)$ and $c_2(t) := (e, \beta(t))$ are integral curves through $(X,0)$ and $(0,Y)$, respectively, starting at $(e,e)$. Thus

$$\mu_{*,(e,e)}(X,0) = \frac{d}{dt}\bigg|_{t = 0} \mu(c_1(t)) = \frac{d}{dt}\bigg|_{t = 0} \alpha(t)e = \frac{d}{dt}\bigg|_{t = 0} \alpha(t) = X$$

and

$$\mu_{*,(e,e)}(0,Y) = \frac{d}{dt}\bigg|_{t = 0} \mu(c_2(t)) = \frac{d}{dt}\bigg|_{t = 0} e\beta(t) = \frac{d}{dt}\bigg|_{t = 0} \beta(t) = Y.$$

Therefore

$$\mu_{*,(e,e)}(X,Y) = X + Y.$$