= Solution
A <principal bundle> $\pi:P\to B$ with structure group $G$ has a smooth free right $G$-action, each fibre is a single orbit, and it is locally equivariantly isomorphic to $U\times G\to U$. A representation $\rho:G\to GL(V)$ defines the <associated vector bundle>
$$
E=P\times_\rho V=(P\times V)/\sim,\qquad
(p,v)\sim(pg,\rho(g)^{-1}v).
$$
Local sections of $P$ identify $E$ with $U\times V$, and changes of section act by $\rho$ on the fibre coordinates.
A <principal bundle> is \b[trivial if and only if it has a global smooth section]. Given $s:B\to P$, the map $(b,g)\mapsto s(b)g$ is an equivariant isomorphism $B\times G\to P$: freeness and transitivity give its unique inverse on each fibre, and local trivializations make that inverse smooth. Conversely a product bundle has section $b\mapsto(b,e)$. This proves <principal bundle trivialization by a global section>. A contractible paracompact base is a sufficient condition for triviality of such a bundle; local sections alone are insufficient on a general base.
A manifold is <parallelizable> when its <tangent bundle> is trivial, equivalently when it possesses a global smooth frame. The circle $S^1$ is an example, with its nowhere-vanishing unit tangent field. The sphere $S^3$ is another: regard it as the <unit quaternions>. At a <unit quaternion> $q$, the three tangent vectors $qi,qj,qk$ are independent and smoothly depend on $q$. <Quaternion> multiplication preserves the <norm>, so these form an orthonormal frame. This gives the <quaternionic left-invariant frame on the three-sphere>.
For any <Lie group> $G$ of dimension $n$, choose a <basis> $E_1,\ldots,E_n$ of $T_eG$. <Left translations> define
$$
X_i(g)=(dL_g)_eE_i.
$$
These fields are smooth; at every $g$, the differential of the <diffeomorphism> $L_g$ is invertible, so they are a <basis> of $T_gG$. Hence \b[every <Lie group> manifold is <parallelizable>], as in <parallelization of a Lie group by left translations>.
For a real <semisimple Lie group>, let $B(X,Y)=\operatorname{tr}(\operatorname{ad}X\operatorname{ad}Y)$ be the <Killing form>. It is nondegenerate and invariant under the adjoint action. Left translating $B$ gives a <bi-invariant pseudo-Riemannian metric>. For left-invariant fields, the <Koszul formula> reduces to
$$
2B(\nabla_XY,Z)
=B([X,Y],Z)-B([Y,Z],X)+B([Z,X],Y)
=B([X,Y],Z).
$$
Thus $\nabla_XY=[X,Y]/2$. The curvature convention of Question 1 gives
$$
\begin{aligned}
R(X,Y)Z
&=\frac14[X,[Y,Z]]-\frac14[Y,[X,Z]]-\frac12[[X,Y],Z]\\
&=-\frac14[[X,Y],Z].
\end{aligned}
$$
To obtain the <Ricci tensor>, trace the <linear map> $X\mapsto R(X,Y)Z$. Since $[[X,Y],Z]=\operatorname{ad}Z\,\operatorname{ad}Y\,X$,
$$
\boxed{\operatorname{Ric}(Y,Z)=-\frac14B(Y,Z)=-\frac14g(Y,Z).}
$$
This proves the <Killing-form Einstein metric> construction for every real <semisimple group>. If $G$ is compact semisimple, $-B$ is positive definite and its <Ricci tensor> is $-\tfrac14B=\tfrac14g$, giving a Riemannian <Einstein metric>. For a noncompact <semisimple group> the general Killing-form construction is indefinite. Thus the unrestricted assertion is understood in the pseudo-Riemannian sense appropriate here; positive definiteness is not obtained by simply taking $-B$ in the noncompact case.
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