# Question about definition of symmetric affine connection from do Carmo

I have been self studying Riemannian Geometry using do Carmo, and am a bit stuck in the section regarding the introduction of a symmetric affine connection.

According to do Carmo, and affine connection $$\nabla$$ on a smooth manifold is symmetric when $$\nabla_X Y -\nabla_YX = [X,Y] \text{ for all }X,Y \in \chi(M)$$.

Although I think I understand this bit, the next part where he speaks about the result in coordinates confuses me. He uses the fact that the Lie Bracket is always 0 to conclude the symmetry of the lower indices of the Christoffel symbols, but how does he know the Lie bracket is always 0? Wouldn't that just imply that the Lie bracket of every pair of vector fields is zero? I realize that this is probably a result of the symmetric condition, but because the Lie bracket is used in its definition I dont understand how we can use it to conclude the Lie brackets are always zero.

Any help understanding this would be greatly appreciated!

Having $$[\partial_i,\partial_j]=0$$ (which holds simply because second order partial derivatives in Euclidean space commute) for all $$i$$ and $$j$$ does not imply that $$[X,Y]=0$$ for all fields $$X$$ and $$Y$$, because the Lie bracket is not bilinear over smooth functions, only over real scalars (that is, it is not a tensor). The derivation character of vector fields implies that $$[fX,gY]=fg[X,Y]+fX(g)Y-gY(f)X$$holds for all fields $$X$$ and $$Y$$, and smooth functions $$f$$ and $$g$$. So $$\left[\sum_iX^i\partial_i,\sum_jY^j\partial_j \right]= \sum_jX(Y^j)\partial_j -\sum_i Y(X^i)\partial_i.$$