Suppose $\theta\colon G\times M\to M$ is a transitive smooth *left* action of a compact Lie group $G$ on a manifold $M$ and $\pi\colon G\to M\cong G/K$ the corresponding smooth submersion for some closed subgroup $K$. Then $\pi$ is equivariant with respect to $\theta$ and *left* multiplication on $G$, i.e. for all $g,h\in G$

$$\theta(h,\pi(g)) = \pi \circ L_h(g)\qquad \Big(=\pi \circ R_g(h)\Big).$$

If $\langle\cdot\ , \cdot\rangle$ is a bi-invariant Riemannian metric on $G$ then $\pi$ induces a $G$-invariant metric $\gamma$ on $M$, turning $\pi$ into a Riemannian submersion. The vertical bundle $V$ (where each fibre is the nullspace of $d\pi_g$) and the metric $\langle\cdot\ , \cdot\rangle$ define a horizontal bundle $H$ by letting $H_g = V_g^\perp$ with respect to $\langle\cdot\ , \cdot\rangle_g$.

Given $X\in \mathrm{Lie}(G)$ (the set of *left*-invariant vector fields), consider the induced fundamental vector field $\mathcal{K}(X)\in\mathfrak{X}(M)$ defined for $p=\pi(g)\in M$ by

$$\mathcal{K}(X)|_p= d_t|_{t=0} \theta(\exp^G(tX),p)=d\pi_g\big(d(R_g)_e(X_e)\big).$$

**Question**: Is it correct that the horizontal lift of $\mathcal{K}(X)$ is the vector field $\tilde{Z}\in\mathfrak{X}(G)$ given by $$\tilde{Z}_g = \big(d(R_g)_e(X_e)\big)^H$$
i.e., the horizontal part of the *right* (!) translation of the tangent vector $X_e\in T_e G$?
Initially, I thought it should be the horizontal part of $X$ itself which is only true at $g=e$ it seems.

Even though taking the horizontal part commutes with left translation (by construction of the induced metric on $M$), the vector field $\tilde{Z}$ is **not** left-invariant (i.e. *not* in $\mathrm{Lie}(G)$) because in general $L_h\circ R_g\neq R_{hg}$. **Neither** is $\tilde{Z}$ right-invariant because taking the horizontal part does *not* commute with right translations.
This somehow feels strange to me.