The fundamental theorem of Riemannian geometry implies that there is a unique symmetric (i.e., $\Gamma^{a}_{bc}=\Gamma^{a}_{cb}$, using a coordinate basis) connection on the three-sphere, $S^3$ which is compatible with its metric, $g$ (i.e. $\nabla g=0$). This is called the Levi-Civita connection.
The three-sphere, $S^3$ is the group manifold for the Lie group $SU(2)$. In terms of Lie algebra elements (i.e. left-invariant vector fields), the Killing form $Tr (T^a T^b)$ is a natural metric on $SU(2)$. Page 237 of 'Manifolds, Tensors and Forms' discusses this, saying that the Levi-Connection for this metric, $\nabla$, satisfies $\nabla_X Y=\frac{1}{2}[X,Y]$. However the Christoffel symbols in this case are not symmetric, rather they are antisymmetric ($\Gamma^{a}_{bc}=\frac{1}{2}f^{a}_{bc}$, where $f^{a}_{bc}$ are the structure constants of the Lie algebra).
This is different from the case where one has a holonomic or coordinate basis, where torsion-free implies $\Gamma^{a}_{bc}=\Gamma^{a}_{cb}$, The reason for this is the Lie algebra provides a non-holonomic basis of vector fields, since $[T^a,T^b]\neq0$. The general definition of a Levi-Civita connection is one for which the torsion tensor $T(X,Y)=\nabla_X Y+\nabla_Y X-[X,Y]$ vanishes, and this is true for the Killing form.
My main question is, given a metric, is it always possible to find a holonomic basis on the manifold, whereby the fundamental theorem would imply a symmetric connection? To be more precise, can one use the Killing form as a metric on $S^3$, but choose a basis which does not consist of left-invariant vector fields, but rather a holonomic basis, which would then give us a symmetric connection?