(Ex. 27.4 page 252 Loring Tu) (The differential of an action). Let $\mu: P \times G \rightarrow P$. For $g \in G$, the tangent space $T_gG$ may be identified with $l_{g*} \mathfrak{g} $, where $l_g:G \rightarrow G$ is left multiplication by $g \in G$ and $\mathfrak g = T_eG$ is the Lie algebra of $G$. An element of the tangent space $T_{(p,g)}P \times G$ is of the form
$$(X_P, l_{g*} A)$$
for $X_p \in T_pP$ and $A \in \mathfrak g$. The differential is given by
$$ \mu_*= \mu_{*,(p,g)} :T_{(p,g)}(P \times G) \rightarrow T_{pg} P$$
is given by
$$ \mu_*(X_p, l_{g*} A) = r_{g*} (X_p) + \underline{A}_{pg} $$
Definition: $\underline{A}$ is the fundamental vector field on $P$ associated to $A \in \mathfrak{g}$,
$$ \underline{A}_p = \frac{d}{dt}\Big|_{t=0} p \cdot e^{tA} \in T_pP$$
I am struggling in writing out the proof rigorously and neatly. I would be greatful if someone may spell this out.
Best Answer
So with my edits above I think I have came up with the answer: $$ \cdots \cdots \cdots $$ Let us identify $$T_{(p,g)}(P \times G) \cong T_pP \oplus T_gG$$ hence by linearity of differential, it suffices to compute each component. $$ \cdots \cdots \cdots $$ Let $(X_p, 0) \in T_pP \oplus T_gG$. $\varphi(t):(-\varepsilon, \varepsilon) \rightarrow P$ be a smooth curve, $\varphi(0) = p, \varphi'(0)=X_p$. Define $$\gamma(t):(-\varepsilon, \varepsilon) \rightarrow P \times G, \quad t \mapsto (\varphi(t), g)$$ Then \begin{align*} \mu_*(X_p,0) &= \mu_* \gamma_* (\frac{d}{dt} \Big|_{t=0}) \\ & =(r_g \circ \varphi)_* (\frac{d}{dt} \Big|_{t=0}) \\ &= (r_g)_* X_p \end{align*} $$ \cdots \cdots \cdots $$ To compute $(0,(l_g)_*A)$ we consider the curve, $$ \gamma(t): (-\varepsilon, \varepsilon) \rightarrow P \times G, (p, g \cdot e^{tA}) $$ Then we have \begin{align*} \mu_* \gamma_* (\frac{d}{dt} \Big|_{t=0}) &= (c_{pg})_*(\frac{d}{dt} \Big|_{t=0}) \\ &=: \underline{A}_{pg} \end{align*} where $c_{pg}(t):\Bbb R \rightarrow P$ is the map given by $pg \cdot e^{tA}$.