the Lie bracket of vector fields
/ LEE /
If two vector fields are two winds, you might ask: does the order in which I follow them matter? Flow along X for a tick then along Y, versus Y then X — do I end up in the same place? The Lie bracket [X, Y] is the precise measure of this failure to commute. It is a third vector field that records the infinitesimal discrepancy between the two routes, and it vanishes exactly when the flows can be done in any order.
The cleanest definition treats vector fields as differential operators on smooth functions: X acts on f by directional differentiation Xf. Then the Lie bracket is the commutator [X, Y]f = X(Yf) - Y(Xf). It is a small miracle that this difference, which naively involves second derivatives, is actually a first-order operator — the second derivatives cancel by equality of mixed partials — so [X, Y] is again a genuine vector field. In coordinates, with X = sum X^i partial_i and Y = sum Y^j partial_j, the k-th component is [X, Y]^k = sum_i (X^i partial_i Y^k - Y^i partial_i X^k). Geometrically, [X, Y] at p is the leading-order displacement of the commutator of the two flows: flow time sqrt(t) along X, then Y, then back X, then back Y, and the net move is t[X, Y] plus higher order.
The bracket makes the vector fields on M into an (infinite-dimensional) Lie algebra: it is bilinear, antisymmetric [X, Y] = -[Y, X], and satisfies the Jacobi identity. It is the seed of Lie theory (on a Lie group the left-invariant fields close under bracket to give the finite-dimensional Lie algebra) and the operator behind the Frobenius theorem. Two flows commute, theta^X_s composed with theta^Y_t = theta^Y_t composed with theta^X_s, if and only if [X, Y] = 0; that equivalence is the whole reason the bracket matters. Note the bracket is not C^infinity-bilinear in the usual sense — [X, fY] = f[X, Y] + (Xf)Y picks up a derivative term, so it is not tensorial.
On R^2 take X = partial_x and Y = x partial_y. Then [X, Y] = partial_x(x) partial_y = partial_y, which is nonzero. So flowing right then shearing up does not equal shearing up then flowing right — the bracket partial_y is exactly their infinitesimal disagreement.
A nonzero bracket means the two flows do not commute.
The bracket is not tensorial: [X, fY] gains an (Xf)Y term, so its value at p depends on X and Y near p, not just at p. Coordinate fields partial_i always commute, [partial_i, partial_j] = 0.