Understanding projection of vector field in homgeneous spaces I'm studying homogeneous spaces from the book of A.Arvanitoyeorgos, "An introduction to Lie groups and the geometry of homogeneous spaces".
Consider $G/H$ be a homogeneous space, and let $\pi:G\rightarrow G/H$ be the canonical projection.
We can consider the differential $d\pi_e:T_eG\rightarrow T_o(G/H)$ (and remember that $\mathfrak{g}\cong T_eG)$. Now we can compute what is the projection of a $X\in\mathfrak{g}$. The author says that, for any $g\in G$
$$X^{\star}_{gH}=\frac{d}{dt}(\exp{tX})gH\Bigr\rvert_{t=0}$$
where $\exp{tX}$ is the corresponing 1-parameter subgroup.
Now my doubt is: The author notice that $[X^{\star},Y^{\star}]=-[X,Y]^{\star}$;
Unfortunately I tried lots of computation for proving this identity but this result keeps arising as a magic trick; moreover my gut feeling is that this identity is not true, since if we reduce to the case $gH=o$ we should en up with an identity without the minus sign.
And this is a problem, because this identity is strongly used to obtain the Riemannian connection for a reductive homogeneous space.
So i'd like to understand whhy this result should be true, and in the other case why the computation of the Riemannian connection are correct (or, if they're not, what is the known formula).
Thanks in advance to anyone who will help me trying to understand this a little bit.
 A: I'm taking my treatment out of Mathematical Gauge Theory by Mark Hamilton.  The reason for this actually stems out of the general formalism of an action $\Phi:G\times M \to M$ of a Lie group $G$ into a manifold $M$.  This then induces the mapping $\Phi_*:\mathfrak{g} \to \mathcal{X}(M)$ of the Lie algebra into the space of smooth vector fields on $M$.  
If we fix a point $p\in M$ we can consider the orbit map $\phi_p: G\to M$ given by $\phi_p(g) = g^{-1}\cdot p$.  This then induces the action:
$$
X_p^* \;\; =\;\; d\left (\phi_p\right )_e(X)
$$
for an element $X \in \mathfrak{g}$.  Ultimately the minus sign is going to arise by demonstrating that $X \in \mathfrak{g}$ and $-X^*$ are $\phi_p$-related.  Notice for all $g\in G$ and noting that $X \in \mathfrak{g}$ is a left-invariant vector field:
\begin{eqnarray*}
d\left (\phi_p\right )_g(X_g) & = & d\left (\phi_p\right )_g \circ d\left (L_g\right )_e \left (\left . \frac{d}{dt} \right |_{t=0} \exp(tX) \right ) \\
& = & d\left (\phi_p\right )_g \left (\left . \frac{d}{dt} \right |_{t=0} g\exp(tX) \right ) \\
& = & \left .\frac{d}{dt} \exp(-tX) \left (g^{-1}\cdot p\right ) \right |_{t=0} \\
& = & -X_{g^{-1}\cdot p}^* \\
& = & -X_{\phi_p(g)}^*.
\end{eqnarray*}
Because $X$ and $-X^*$ are $\phi_p$-related we then have that 
$$
\phi_{p*}\left ([X,Y]\right ) \;\; =\;\; \left [ \phi_{p*}(X), \phi_{p*}(Y)\right ]
$$
which reduces to
$$
-[X,Y]^* \;\; =\;\; [-X^*, -Y^*] \;\; =\;\; [X^*, Y^*].
$$
Ultimately we see that the minus sign has to do with the fact that when we invert the element $g\exp(tX)$ we get $\exp(-tX)g^{-1}$. 
