Torsion free is equivalent to symmetric christoffel symbols in coordinate frames

Evaluating the torsion expression on ${\mathrm{\partial }}_i,{\mathrm{\partial }}_j$ gives $T({\mathrm{\partial }}_i,{\mathrm{\partial }}_j)={\sum }_k({\mathrm{\Gamma }}^k{}_{ij}-{\mathrm{\Gamma }}^k{}_{ji}){\mathrm{\partial }}_k$, because their bracket is zero. If $T=0$, basis independence forces each difference to vanish in every chart.

Conversely, symmetry of the coefficients on a chart makes every displayed basis value zero. Tensor bilinearity then gives $T(X,Y)=0$ there for arbitrary fields. A chart cover proves this globally. With no indices in dimension zero the condition is vacuous; in dimension one the only lower-index pair is already symmetric. A noncoordinate frame may have nonzero brackets, so dropping that hypothesis would invalidate the computation.