If we define a fundamental vector field, i.e.,
$$ X^\ast =\frac{d}{dt}|_0 \exp(tX)\cdot p $$ where $p\in M=G/K$,
Question 1 : then for $X,\ Y\in (T_eK)^\perp$, we have $$ X^\ast + Y^\ast = (X+Y)^\ast$$
How can we prove ?
From $$ X^\ast_m = \frac{d}{dt} \exp\ (tX)m=\frac{d}{dt} ( I + tX + t^2X^2/2 + ... )m=Xm,$$ is it followd ?
Question 2 : For $X\in T_eG$ we have $X=X^\top + X^\perp,\ X^\top\in T_eK,\ X^\perp \in (T_eK)^\perp$. Then we have $$ X_p^\ast = (X^\perp)_p^\ast$$
How can we derive ?
Thank you in advance.
For Question 1, I don't know of a direct proof, but here's an indirect one. First some notation. Let $\pi:G\rightarrow G/K$ denote the canonical projection, and let $R_g:G\rightarrow G$ denote right multiplication: $R_g(h) = hg$.
Proposition The map $\phi:T_e G\rightarrow T_[p] G/K$ is nothing but the map $ \pi_\ast \circ {R_p}_\ast:T_e G\rightarrow T_p G\rightarrow T_[p]G/K$.
Assuming the proposition for a moment, since both $\pi_\ast$ and ${R_p}_\ast$ are linear, $\phi$ must be as well.
Proof: The key observation is that $\pi$ is equivariant, so \begin{align*} (\pi\circ R_p) (\exp(tX)) &= \pi(\exp(tX)p)\\ &= \exp(tX)\cdot \pi(p)\\ &= \exp(tX)\cdot [p].\end{align*}
In short, $(\pi\circ R_p)(\exp(tX)) = \exp(tX)\cdot [p]$. Taking the derivative of both sides (via the chain rule) gives the proposition $\square$.
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As for question 2, either I'm missing something or its false. For example, when $G = SU(2)$ and $K = S^1$, so $G/K = S^2$, then $K$ acts on $S^2$ by rotations, keeping the north and south pole fixed. The action fields, then, associated to $X = i\in T_e S^3\cong Im \mathbb{H}$, is nonzero at all points other than the north and south pole. (For question $2$, if $p = e$, then it's true, but for other $p$, all bets are off.)