Lie bracket of vector fields

Remarks

In local coordinates, with $X = \sum_i X^i \partial_i$ and $Y = \sum_j Y^j \partial_j$, $$[X, Y] = \sum_{j} \sum_{i} \left( X^i \frac{\partial Y^j}{\partial x^i} - Y^i \frac{\partial X^j}{\partial x^i} \right) \partial_j.$$ Example on $\R^2$: for $X = \partial_x$ and $Y = x \, \partial_y$, one gets $[X, Y] f = \partial_x(x f_y) - x f_{yx} = f_y$, so $[X, Y] = \partial_y$.

The product $f \mapsto X(Y f)$ alone is not a vector field. For $X = Y = \partial_x$ on $\R$, it is $f \mapsto f''$, and $(fg)'' = f'' g + 2 f' g' + f g''$, so the product rule fails. In the bracket the second-order terms cancel, and the product rule holds.

The bracket is bilinear and antisymmetric, and it satisfies the Jacobi identity $[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0$. So the smooth vector fields on $M$ form a Lie algebra. The bracket $[X, Y]$ equals the Lie derivative $\mathcal{L}_X Y$.