O'Neill Formula : Consider a fibration $\pi : (M ,g)\rightarrow M/G$ where $G$ has only one orbit type. Then we have $$ K_{M/G} (d\pi V, d\pi W) = K_M(V,W) + \frac{3}{4} | [V,W]^V |^2_g$$ where $V,\ W$ are horizontals and any vector field $Z$ on $M$ has a decomposition $$ Z = Z^V + Z^H$$ with a horizontal direction $Z^H$ and a vertical direction $Z^V$.
Setting : Now we apply this to the following example : $$G\rightarrow (G\times M, \frac{1}{t}Q+g) \rightarrow (M,g_t)$$ where $g\cdot (k,m)=(gk,gm)$ and $M$ admits $G$-action. We want to restate O'Neill Theorem at $(e,m)$ where $G\cdot m = G/H$ is principal. Define a fundamental vector field $$ X_m^\ast = \frac{d}{dt}|_0 \exp\ (tX)\cdot m$$
After some computation, we have that vertical field has the form $(X,X^\ast_m)$ and horizontal field has the form $(-tPX,X^\ast_m)$ where $X\in (T_eH)^\perp$.
Question : We wanto calsulate $\frac{3}{4} | [V,W]^V |^2_g$ : So $${\rm max}_{Z\in T_eG} \frac{|g([X^\ast,Y^\ast],Z^\ast) + tQ([PX,PY],Z)|^2}{ g(Z^\ast,Z^\ast) + \frac{1}{t} Q(Z,Z)}$$
Here $g(X^\ast , Y^\ast) = Q(PX,Y)$ where $P : T_eG\rightarrow T_eG$ is $Q$-symmetric. In fact $h(X,Y)=Q(PX,Y)$ is a left invariant and a right $H$-invariant.
Now we rewrite $g([X^\ast,Y^\ast],Z^\ast) $ in terms of Killing form : Let $w_Z(X^\ast) = g(Z^\ast,X^\ast)$ so that $$ dw_Z(X^\ast, Y^\ast ) = \frac{1}{2} [X^\ast g(Z^\ast,Y^\ast) - Y^\ast g(X^\ast, Z^\ast) - g(Z^\ast,[X^\ast,Y^\ast])]$$
We must prove $$\frac{1}{2} g([X^\ast,Y^\ast],Z^\ast)=dw_Z(X^\ast, Y^\ast )$$ which refers in Mueter's Ph. D. Thesis on Cheeger deformation (Proposition 1.3 is in its english-summary)
In fact, in case of $M=G,\ g=Q,\ t=1,\ P=id$, the equality does not hold. Am I wrong in some point ? Please help me.